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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: Mach-Zehnder-- interferometers. And we have a beam splitter. And the beam coming in, it splits into 2. A mirror-- another mirror. The beams are recombined into another beam splitter. And then, 2 beams come out. One to a detector d0-- and a detector d1. We could put here any kind of devices in between. We could put a little piece of glass, which is a phase shifter. We'll discuss it later. But our story is a story of a photon coming in and somehow leaving through the interferometer. And we want to describe this photon in quantum mechanics. And we know that the way to describe it is through a wave function. But this photon can live in either of 2 beams. If a photon was in 1 beam, I could have a number that tells me the probability to be in that beam. But now, it can be in either of 2 beams. Therefore, I will use two numbers. And it seems reasonable to put them in a column vector. Two complex numbers that give me the probability amplitudes-- for this photon to be somewhere. So you could say, oh, look here. What is the probability that we'll find this photon over here? Well, it may depend on the time. I mean, when the photon is gone, it's gone. But when it's crossing here, what is the probability? And I have 2 numbers. What is the probability here, here, here, here? And in fact, you could even have 1 photon that is coming in through 2 different channels, as well. So I have 2 numbers. And I want, now, to do things in a normalized way. So this will be the probability amplitude to be here. This is the probability amplitude to be down. And therefore, the probability to be in the upper one-- you do norm squared. The probability to be in the bottom one, you do norm squared. And you get 1. Must get 1. So if you write 2 numbers, they better satisfy that thing. Otherwise, you are not describing probabilities. On the other hand, I may have a state that is like this. Alpha-- oh, I'll mention other states. State 1-0 is a photon in the upper beam. No probability to be in the lower beam. And state 0-1 is a photon in the lower beam. So these are states. And indeed-- think of superposition. And we have that the state, alpha beta-- you know how to manipulate vectors-- can be written as alpha 1-0. Because the number goes in and becomes alpha 0. Plus beta 0-1. So the state, alpha beta, is a superposition with coefficient alpha of the state in the upper beam plus the superposition with coefficient beta of the state in the lower beam. We also had this little device, which is called the beam shifter of face delta. If the probability amplitude completing is alpha to the left of it, it's alpha e to the i delta to the right of it, with delta a real number. So this is a pure phase. And notice that alpha is equal to e to the i delta. The norm of a complex number doesn't change when you multiply it by a phase. The norm of a complex number times a phase is the norm of the complex number times the norm of the phase. And the norm of any phase is 1. So actually, this doesn't absorb the photon, doesn't generate more photons. It preserves the probability of having a photon there, but it changes the phase. How does the beam splitter work, however? This is the first thing we have to model here. So here is the beam splitter. And you could have a beam coming-- A 1-0 beam hitting it. So nothing coming from below. And something coming from above. And then, it would reflect some and transmit some. And here is a 1-- is the 1 of the 1-0. And here's an s and a t. Which is to mean that this beam splitter takes the 1-0 photon and makes it into an st photon. Because it produces a beam with s up and t down. On the other hand, that same beam splitter-- now, we don't know what those numbers s and t are. That's part of designing a beam splitter. You can ask the engineer what are s and t for the beam splitter. But we are going to figure out what are the constraints. Because no engineer would be able to make a beam splitter with arbitrary s and t. In particular, you already see that if 1-0-- if a photon comes in, probability conservation, there must still be a photon. You need that s squared plus t squared is equal to 1 because that's a photon state. Now, you may also have a photon coming from below and giving you uv. So this would be a 0-1 photon, giving you uv. And therefore, we would say that 0-1-- gives you uv. And you would have u plus v norm squared is equal to 1. So we need, apparently, 4 numbers to characterize the beam splitter. And let's see how we can do that. Well, why do we need, really, 4 numbers? Because of linearity. So let's explore that a little more clearly. And suppose that I ask you, what happens to an alpha beta state-- alpha beta state if it enters a beam splitter? What comes out? Well, the alpha beta state, as you know, is alpha 1-0 plus beta 0-1. And now, we can use our rules. Well, this state, the beam splitter is a linear device. So it will give you alpha times what it makes out of the 1-0. But out of the 1-0 gives you st. And the beta times 0-1 will give you beta uv. So this is alpha s plus beta u times alpha t plus beta v. And I can write this, actually, as alpha beta times the matrix, s u t v. And you get a very nice thing, that the effect of the beam splitter on any photon state, alpha beta, is to multiply it by this matrix, s u t v. So this is the beam splitter. The beam splitter acts on any photon state. And out comes the matrix times the photon state. This is matrix action, something that is going to be pretty important for us. How do we get those numbers? After all, the beam splitter is now determined by these 4 numbers and we don't have enough information. So the manufacturer can tell you that maybe you've got-- you bought a balanced beam splitter. Which means that if you have a beam, half of the intensity goes through, half of the intensity gets reflected. That's a balanced beam splitter. That simplifies things because the intensity here, the probability, [INAUDIBLE] must be the same as that. So each norm squared must be equal to 1/2, if you have a balanced-- beam splitter. And you have s squared equal t squared equal u squared equal v squared equal 1/2. But that's still far from enough to determine s, t, u, and v. So rather than determining, them at this moment, might as well do a guess. So can it be that the beam splitter matrix-- Could it be that the beam splitter matrix is 1 over square root of 2, 1 over square root of 2, 1 over the square root of 2, and 1 over square root of 2. That certainly satisfies all of the properties we've written before. Now, why could it be wrong? Because it could be pluses or minuses or it could be i's or anything there. But maybe this is right. Well, if it is right, the condition that it be right is that, if you take a photon state, 1 photon-- after the beam splitter, you still have 1 photon. So conservation of probability. So if you act on a normalized photon state that satisfies this alpha squared plus beta squared equal 1, it should still give you a normalized photon state. And it should do it for any state. And presumably, if you get any numbers that satisfy that, some engineer will be able to build that beam splitter for you because it doesn't contradict any physical principle. So let's try acting on this with on this state-- 1 over square root of 2, 1 over square root of 2. Let's see. This is normalized-- 1/2 plus 1/2 is 1. So I multiply. I get 1/2 plus 1/2 is 1, and 1. Sorry, this is not normalized. 1 squared plus 1 squared is 2, not 1. So this can't be a beam splitter. No way. We try minus 1 over square root of 2. Actually, if you try this for a few examples, it will work. So how about if we tried in general. So if I try it in general, acting on alpha beta, I would get 1 over square root of 2 alpha plus beta and alpha minus beta. Then, I would check the normalization. So I must do norm of this 1 squared. So it's 1/2 alpha plus beta squared plus 1/2 alpha minus beta norm squared. Well, what is this? Let me go a little slow for a second. [INAUDIBLE] plus beta star. Plus alpha minus beta. Alpha star minus beta star. Well, the cross terms vanish. And alpha alpha star, alpha alpha star, beta beta star, beta beta star add. So you do get alpha squared plus beta squared. And that's 1 by assumption because you started with a photon. So this works. This is a good beam splitter matrix. It does the job. So actually-- Consider this beam splitters. Actually, it's not the unique solution by all means. But we can have 2 beam splitter that differ a little bit. So I'll call beam splitter 1 and beam splitter 2. Beam splitter or 1 will have this matrix. And beam splitter 2 will have the matrix were found here, which is a 1 1 1 minus 1. So both of them work, actually. And both of them are good beam splitters. I call this-- beam splitter 1. And this, beam splitter 2. And we'll keep that. And so we're ready, now, to think about our experiments with the beam splitter.
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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'll do the relativistic corrections. And all the corrections that I'll do today, I'll skip the easy but sometimes a little tedious algebra. It's not very tedious. Nothing that is pages and pages of algebra. It's lines of algebra. But why would I do it in lecture? No point for that. So let's see what we can do. This is the relativistic correction, the minus p squared. So could we write this for the relativistic correction? We're going to do first order correction, relativistic, of the levels n l ml. Let's put a question mark. Minus 1 over 8 m cubed c squared psi n l ml p to the fourth. Now recall that p to the fourth, the way it was given, is really p squared times p squared. You have four things that have to be multiplied. So it's not px to the fourth plus py to the fourth plus pz to the fourth is px squared plus py squared plus pz squared, all squared, just in case there's an ambiguity. That seems reasonable. The first order corrections should be found by taking the states and finding this. But there is a big question mark. And this kind of question is going to come up every time you think about these things. This formula, where I said the shift of the energy of this state is that state evaluated here, applies for nondegenerate perturbation theory. And if the hydrogen atom is anything, it's a system with a lot of degeneracies. So why can I use that, or can I use that? We have the hydrogen atom. I just deleted it here. So here, if you have n, for degeneracies you fix n. For degeneracies, you fix some value of n. And now you have the degeneracies between the various l's, for each l between the various m's. A gigantic amount of degeneracy. Who allows me to do that? I'm supposed to take that level three has nine states, remember? n square states. Well, we should do a 9 by 9 matrix here and calculate this. Nine sounds awful. We don't want to do that so we better think. So this is the situation you find yourself. Technically speaking, this is a problem in the degenerate perturbation theory. We should do that. And you better think about this every time you face this problem because sometimes you can get away without doing the degenerate analysis, but sometimes you can't. Yes? AUDIENCE: It's like rotationally symmetric. So you can mix terms with different [INAUDIBLE].. PROFESSOR: OK. So you're saying, basically, that this thing in this basis-- so we have nine states here. n equal 3. In these nine states, it doesn't mix them. So this is diagonal here. And what one is claiming by doing that is that this is a good basis, that delta H is already diagonal there. And don't worry, we can do it. In fact, that is true. And the argument goes like that. We know that p to the fourth, the perturbation, commutes with l squared. We'll discuss it a little more. And p to the fourth commutes with lz as well. So these are two claims. Very important claims. Remember, we had a remark that I told you few times few lectures ago. Very important. If you have a Hermitian operator that commutes with your perturbation for which the states of your bases are eigenstates with different eigenvalues, then the basis is good. So here it is. l squared commutes with p to the fourth. Why? Because, in fact, p to the fourth commutes with any angular momentum because p to the fourth is p squared times p squared. And p squared is rotational invariant. p squared commutes with any l. If that's not obvious intuitively, which it should become something you trust-- this is rotational invariant. p squared dot product doesn't depend on rotation. If you have a p and you square it or you have a rotated p and you square it, it's the same. So p to the fourth commutes with any component of angular momentum. So these two are written like great facts, but the basic fact is that p squared with any li is 0. And all this follows from here. But this is a Hermitian operator. This is a Hermitian operator. And the various states, when you have fixed n, you can have different l's. But when you have different l's, there are different eigenvalues of l squared. So in those cases, the matrix element will vanish. When you have the same l's but different m's, these are different eigenvalues of lz. So the matrix element should also vanish. So this establishes rigorously that that perturbation, p to the fourth, is diagonal in that subspace. So the subspace relevant here is this whole thing. And in this subspace, it's completely diagonal. Good. So generally, this kind of point is not emphasized too much. But it's, in fact, the most important and more interesting and more difficult point in this calculations. We'll have one more thing to say about this. But let's continue with this. I'll say the following. We use the Hermiticity of p squared to move one p squared to the other side. So Enl ml 1 is equal to minus 1 over 8m cubed c squared p squared psi nlm p squared psi nml-- nlm. OK. We move this p to the fourth. It was p squared times p squared. One p squared is Hermitian. We move it here. And then, instead of calculating a billion derivatives here, you use the fact that p squared over 2m plus v of r on the wave function is equal to the energy of that wave function that depends on n times the wave function. These are eigenstates. So p squared-- we don't want to take derivatives, and those expectation values can be replaced by a simpler thing. P squared on psi is just 2m En minus v of r psi. So Enlm 1 is equal to minus 1 over 8m cubed c squared. Here we have, well, the m's. Two m's are out, so we'll put a 2 and an mc squared. Yep. En minus v of r psi En minus v of r psi nlm nlm. OK. We got it to the point where I think you can all agree this is doable. Why? Because, again, this term is Hermitian, so you can put it to the other side. And you'll have terms in which you compute the expectation value on this state of E squared. E squared is a number, so it goes out, times 1, easy. En cross terms with vr is the expectation of v of r in this state, is the expectation of 1 over r in a state. That's easy. It comes from the Virial theorem. Then you'll have the expectation of v squared in a state, and that's the expectation of 1 over r squared in a state. You've also done it. So yes, getting all together, getting the factors right would take you 15 minutes or 20 minutes or whatever. But the answer is already clear. So let's write the answer. And the answer is that Enl ml 1 relativistic is minus 1/8 alpha to the fourth. That's our very recognizable factor. mc squared 4n over l plus 1/2 minus 3. Now, fine structure is something all of us must do at least once in our life. So I do encourage you to read the notes carefully and just do it. Just become familiar with it. It's a very nice subject, and it's something you should understand. So here, again, I have to do a comment about basis, and those comments keep coming because it's an important subject. And I want to emphasize it. So what is the reason? The reason I wanted to comment is because in a second, I'm going to do the spin orbit term. And in that case, I would like to work with a coupled basis. Here, I'm working with the uncoupled basis. And really, this thing is the expectation value of Hl relativistic in nl ml ms nl ml ms. This is really that. I wrote psi nl ml, so you should trust the first three labels. And ms goes for the ride. It's this spin. The operator you're putting here, delta H relativistic, has nothing to do with spin. Could not change the spin of the states. This has to be diagonal in spin. So this number you've computed is nothing else than this overlap in the uncoupled basis. So this calculation was uncoupled basis matrix element. And we saw that it's diagonal. In fact, this whole thing is nothing but the function of n and l. n and l. And independent of ml and ms. OK. That's what we've calculated. So here is the question. We could consider this in the coupled bases nlj mj. nlj mj. And the question is, do I have to recalculate this in the coupled basis or not? And here is an argument that I don't have to recalculate it. So I'm going to claim that this is really equal to that. Just the same. It's the kind of thing that makes you a little uneasy, but bear with me. Why should it be the same? Think of this as fixed n and l because this depends on n and l. If we have the hydrogen atom here, you'd take one of these elements, one of these states-- this is a fixed n, fixed l. And we're looking at fixed n, fixed l. Yes. There are lots of states here that have different ml and ms. But the answer doesn't depend on ml and ms. In this basis, we are also looking at that subspace, that multiplet, nl fixed. And they have reorganized the states with j and mj. In fact, with two values of j and several values of mj. But at the end of the day, the coupled basis is another way to describe these states coming from tensoring the l multiplet with a spin 1/2. So it gives you two multiplets, but they are the same states. So the fact that every state here is some linear combination of states in the uncoupled bases with different values of ml and ms that add up to mj. But this answer doesn't depend on ml and ms. So whatever linear combination you need, it doesn't change because the answer doesn't depend on ml and ms. So this must be the same as that. I'll give another argument. Maybe a little more abstract, but clearer perhaps. Think of this. So this can be-- in the notes, I explain that by changing basis and explaining why exactly everything works out. But there is no need for that argument if you think a little more abstractly. Think of this subspace. Because with fixed n and l, we have this subspace. In this subspace, the uncoupled basis makes the perturbation diagonal. But more than diagonal, it makes the perturbation proportional to the unit matrix, because every eigenvalue is the same. Because in this subspace, n and l is fixed. And yes, m and ms change, but the answer doesn't depend on that. So this matrix, delta H, in this subspace is proportional to the unit matrix. And when a matrix is proportional to the unit matrix, it is proportional to the unit matrix in any orthogonal basis. A unit matrix doesn't get rotated. So it should be a unit matrix here as well, and it should be the same matrix. So this is the same pair.
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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 NELSON: Well, today we're going to continue with the lecture that was started last time that had the scintillating and descriptive title "some thermodynamic processes". And we're going to continue along those lines, and really there a couple of things that I want to do to follow up what you saw last time. One is that you've got introduced and led through a couple of elementary thermodynamic cycles. And I want to do a little bit more of that today. And one of the -- there are really two reasons. One is that as you saw last time doing a thermodynamic cycle can be extremely useful, because if you have one path along which some change occurs, but along which it's not necessarily easy to calculate all of the thermodynamic changes along such a path, it may be irreversible, lots of variables may change. It can be easier instead of looking at the complicated path to devise a cycle that involves several easier steps, each one of which is facile for the calculation of what happens to the thermodynamic properties. So in that case, it's much easier to go through several elementary step, and then the one complicated path you know because going around the entire cycle state functions, you know that they won't have any change. So in that sense, going through a thermodynamic cycle can be useful because it helps you calculate thermodynamic qualities. But the other reason to go through the thermodynamic cycles and really to develop great facility with them is because there are just an awful lot of things in nature and things that we build that run in cycles, where we want to calculate the thermodynamics, right. If we build an engine for a car or anything else, it almost always is going to have some key element that's operating in a cycle, otherwise it won't keep going, right. But it's not just engines that we might build. It's biological engines, right. Things that either produce or consume biological fuel to do different sorts of processes. Also, of course of key importance to understand what the, in that case, biological thermodynamics are. And again, there'll be key elements that run in cycles and understanding those can be extremely important in understanding how cellular function works. OK, so what I'd like to do is go through one cycle, and then I'll suggest another for you to work through on your own. So consistent with the title of the lecture, we've got a section from your last lecture notes entitled "some thermodynamic cycles". And what we're going to do is go through a cycle, please note these office hours and locations, which I think wasn't specified before. So we're going to go through a thermodynamic cycle, and here's what I want to calculate when we do this. Delta u, delta H, familiar state functions, changes in their values, q, w, heat and work. And I also want to introduce one new function. I won't give it a name yet, but it's going to have the unusual form dq over T. OK, we'll call it special function. It's so special that we'll call this quantity dS. OK, we'll just go through this on a whim, and this is going to be foreshadowing something that'll take on tremendous importance in subsequent lectures, but I'm going to use the opportunity now to just introduce the behavior of it in going around a cycle. If that's for me, tell them I'm busy please. OK, so let's look at the cycle we'll go through. We're going to look at pressure volume changes, similar to what you saw before. This time, we're going to have initial and final volumes, V1 and V2 and different pressures where we might end up. And we're going to look at two ways to go to a final volume V2. One of them is going to end up at pressure p2 and the other is going to end up at pressure p3. One of them is going to be a reversible isothermal path. This is going to be our path we'll label A, reversible isothermal which is to say constant T, right? And we're going to start at temperature T1, so since it's constant temperature, this is going to end up at temperature T1. And then we're also going to consider a reversible adiabatic path. This'll be our path B, reversible adiabatic. And then we'll close this circle, this cycle. This is going to end up at a different temperature by the way. You saw this last time in a slightly different way. Last time what you saw is we compared isothermal and adiabatic paths that ended up at the same final pressure, and what you saw is that therefore, they ended up in different final volumes. So this is just a little bit different from that. So this is going to end up at a different temperature, we'll call it T2. And then we're going to close the cycle. And so this is a constant volume path then. This is path C, constant volume. OK? So now let's go around the cycle and just compare notes on what happens to the thermodynamic quantities as we do that. So here's path A, it's isothermal. And I didn't specify, but let's make sure to do so, we've got a, it's going to be an ideal gas. OK, so A is constant temperature. What does that mean for delta u in path A? Zero right? Ideal gas, no temperature change, right has to be zero because da is Cv dT for an ideal gas. dT is zero, there's no temperature change. OK, how about delta H, zero Cp dT for an ideal gas. dT is zero. All right, now, it's reversible. So that means we can immediately write down an expression for the work. Since it's reversible, it's negative p dV right? And since it's an ideal gas then we can replace p, right, with RT over V, right? We want to do that because we have too many variables here. We've already got dV we'll get rid of p as an additional variable and replace it with V which is already in here. So it's minus R T1 dV over V, right? I can put in T1 because we're a constant temperature. so now we can just integrate straight away and find that the work in path A it's just the integral of that quantity minus integral of R T1 going from V1 to V2 dV over V. so it's minus R T1 log V2 over V1. All right, so let's just do a reality check here. The way I've got this drawn, the volume is going up in the process. It's an expansion. So is this system doing work on the surroundings, or are the surroundings doing work on the system? Somebody say it real loud. Yes, the system is working on the surroundings, right? It's pushing out against them. Work is defined as the work that the surroundings do on the system. So this is the negative number, right? V2 is bigger than V1. This is positive. This whole thing is negative. All right, q has to be just the opposite of this because we've already figured out that there's no change in u. This is just q plus w. There's w, q has to be R T1 log of V2 over V1. OK, so that means heat is being imparted to the system, right, from the surroundings in a manner that compensates exactly the amount of work done. OK, All right, so these are the thermodynamic quantities that you're familiar with already. Let's just quickly look at our special function. It's not going to be hard to calculate because it's a constant temperature path, so that T in the bottom is just T1, right. You've already looked at q, right. So dq A over T1 it's just R log V2 over V1 right. So there's our special function for this particular path. OK, so that's path A. Now let's compare to path B. So if path B is an adibat, what's zero? q, it's adiabatic, that means there's no heat exchanged between the system and the surroundings. So what happens, q B is zero. There is a change in temperature, right? We've got, if we say it's a mole of gas, it's going from p1, V1, and T1 to one mole of our gas at p3, V2, and T2. So unlike before now temperature is going to change. So that means that delta u isn't zero this time. du, it's an ideal gas. So this is Cv dT and of course we can just integrate this straight away. So delta u B is Cv times T2 minus T1, right. and similarly dH is Cp dT right. So delta H in pathway B is just Cp T2 minus T1 okay? All right, what we've already got that q is zero. Delta u is q plus w. Delta u isn't zero. So this must be equal to work, right? And now we can look at our special function, but since this is zero right dq B is zero, along path B, so our special function, dS, so we're going to go, dq B over T is equal to zero. OK? OK, finally let's look at our third path, this constant volume path, that's going to connect T1 and T2, right? Constant volume, what's zero? STUDENT: [INAUDIBLE] PROFESSOR NELSON: A little more noise here. What's zero? STUDENT: Work. PROFESSOR NELSON: Work, great. So, path C constant volume means our work in past C is zero, right? Now, our temperature is going to change from T2 to T1, and we just saw what happens to u and H when that happens going from T1 to T2, and it's no different. It's an ideal gas going along these path. So we can immediately write delta u C is Cv times T1 minus T2. Delta Hc C is Cp times T1 minus T2, right? Work is zero. Delta u is w plus q, work plus heat. This is zero, this isn't. This must be heat q B, right. So again, there are our familiar thermodynamic quantities, but now let's also look at our special function. It's not going to be zero this time because we have non zero heat exchange between the system and the environment, right. So are integral of dq C over T, right, which is our heat, but in this case we can do this calculation easily because since work is zero, we can equate dq with du, right. So we can write this as our integral from T1 to T2, du over T. du is just Cv dT, right. So this is just Cv log T2 over T1. OK? So now we've completed the cycle. So let's just compare. Let's compare what happened in path A to what happened in paths B and C. Yes? STUDENT: [INAUDIBLE] PROFESSOR NELSON: Ah sorry, yes of course. Thank you. Yes, oh yes, so. STUDENT: [INAUDIBLE] PROFESSOR NELSON: And it's right because, thank you again, it's because we're going from T1 to T2. So I'm getting confused by the oh sorry, we're going from T2 to T1. So it's in reverse of the order of the subscripts and I let this confuse me. There it is. There it is. There. Thank you, any other details which should be pointed out? All right, then let's proceed. So what we're going to do is our comparison of what happened on pathway A to what happened in combined pathways B and C, right? So here are results for pathway A, right, for delta u zero delta H zero. I didn't actually explicitly write it or did I? Let's just write it again. Here's our work in path A, and here's our heat exchange in path A. Now let's look at the sum of B and C. So B plus C, here is delta uB Cv times T2 minus T1. Delta uC Cv times T1 minus T2, the sum is zero right. That's the some of delta u in these two paths, and if we look at delta u in just path A, it's zero. And it's going to be the same without delta H. The only difference is it'll be Cp instead of Cv, but there it is for pathway B. There it is for a pathway C. So the state functions that we're familiar with are doing what we expect they ought to be doing, right? If you go around in a cycle, starting and ending at the same place, the state functions have to stay the same. They only depend on the state the system is in. They don't depend on the path, and that's what we've shown, right. Similarly if we go from this point to this point through path A or if we do it through the combination of these two paths, and the change in u and H in any state function, those have to be the same. It turns out they're zero in both cases, and that's what we've seen, right? Now let's compare what happens to work and heat. So here are expressions for work and heat. They're opposites for the pathway A. Here's pathway B and C. In C the work is zero. In B it isn't, it's Cv times T2 minus T1, right. The work isn't equal to the work in pathway A, right, because you know work is not a state function it depends on the path right, and there are different amounts of work done on the system, or done by the system on the surroundings in these two different processes. They're both expansions. They'll both have net work done on the surroundings, but not the same amount, right. And of course it's going to be the same with heat. We've already seen that delta u is zero. So we know that in each case the heat is going to be the opposite of the work, but the work isn't the same in these two different ways of getting from here to here, right. So let's just see it explicitly. Here's our qA. Here's heat exchanged in pathway A and in pathway B heat is zero, and in pathway C, here is qC it's Cv T1 minus T2. So again, for both heat and work we don't get the same result. Now let's look at our special function, right. So here's path A. We found that it's R log V2 over V1. Pathway B is zero. Pathway C it's Cv log T1 over T2. All right, let's just look at that a little more carefully. R log V2 over V1. B plus C, Cv log of T1 over T2, right. Just some weird combination of functions, right? But, of course, it's not quite like that. Why? Because this path is an adiabatic path. And you already saw last time there was this relationship between the temperature and volume changes along an adiabatic path. So let me just write that down. Adiabatic reversible path. What you saw is that T1 over T2 was V2 over V1 to the power R over Cv. Isn't that something? So what does that mean? We take this Cv and put in the exponent here, right. And we put this R up in the exponent here. What are we going to discover? Those two things are equal right, T1 over T2 to the power Cv is V2 over V1 to the power R, so dS over path A equals dS over paths B plus C, okay? So what this suggests is that whatever this funny special function is, at least for this one cycle that we've, tried it's behaving like a state function, right. It seems like its change is independent of path. Start here, end up here, do it through two different pathways, end up in the same place, right. OK, now that's all the foreshadowing I'm going to give it for right now. We will certainly come back to this very special function shortly. Before I move on, I'm just going to put on the board another cycle, and I'm going to urge you to work through that on your own. It's worked through already, let's see, I believe it's worked through in the notes, yes. So if you need the help of the notes it's in there, but I would urge you to work through it on your own, and it's the following: let's start with the same path A, all right? So we've got our reversible adiabatic path right. And now, try working through what happens if we close the cycle in a different way, right, like this. So here's a path that's constant pressure, and here is a path that's constant volume, similar to the constant volume path that we did before, but not between the same pressures, right. So we can label this one D and this one E, and I urge you to just try going through that and verifying for yourself that again the state functions will behave the way state functions should, and you can see whether this special function once again behaves like a state function, right. And of course you should expect to see that the work and the heat again won't behave like state functions which they are, then you'll see you have different results for those, depending on the path, OK? Any questions about going through these cycles and using expressions like this and so forth? Expressions like this you know turn up in various places, like in the equations sheet that appears at the end of exams, right? Which means that you may not need to be committing it to memory. On the other hand it means that you should, it's the sort of thing that you should be familiar with so that the need to use it is something that you'll conjure when you're working on a problem and there's a reversible adiabat, and you have temperatures and volumes that change that you might like to relate to each other. Cycles that are suggested that you go through and aren't gone through in class also sometimes turn up on exams too, by the way. So that said, let's move on to the next topic, which is thermochemistry. So, so far what you've seen in most of the examples in the class are essentially mechanical kinds of changes, pressure volume sorts of changes, where the system is doing essentially a kind of mechanical work, pressure volume work on the surroundings or vice versa and heat is being exchanged, and you're calculating the basic thermodynamics just like we went through, of processes of that sort. Now I'd like to start introducing something that's really central to chemical thermodynamics, mainly chemistry, right. Let's start looking at chemical reactions, right, and understanding what the changes in thermodynamic properties are that occur when chemical reactions occur, and the immediate application is to calculate the change in enthalpy, which, as you've seen it, at constant pressure which is the condition under which an awful lot of chemistry is done, that's just the heat. Which means we're going to calculate heats of reaction, right. You're running some reaction. It's in the atmosphere. It's at constant pressure. You know, you mix acid and base together and the thing heats up like crazy, right. Or other things might react spontaneously. You can feel something cooling off right. I mean simple examples of these happen when you, you know, if you buy cold or hot packs. You break the seal between them and feel the thing get cold, for example, right. What's happening there? Well, the selection of reactants has been done judiciously to provide either heat or to provide something that's cool. And all of that is coming out of the heat of reaction, whether it's positive or negative. Of course the biggest practical application is burning of fuel. You might use the heat to, and convert it into pressure volume work, right? So if it's an internal combustion engine, you'll do a reaction. There will be the heat of reaction. There'll be a volume change, depending on the conditions under which is occurs, and you can extract work through the reaction and so forth. So let's just to see how you run through some of those calculations. Now let me just ask who here took 5.111? And who here took 5.112? OK, now you've seen some thermal chemistry in 5.111 and 5.112. So I I'm going to do some review of that, but I'm also going to call on the thermochemistry that I'm pretty sure you're familiar with. So I won't go painstakingly over every element of the notes here. Subsequently we'll go on into additional topics that you haven't seen, and I'll treat them in full detail. OK, so let's do some thermochemistry. What we want to do is we'd like to be able to predict for any reaction what's the heat of reaction. And so what we're going to calculator is delta H. So for any kind of reaction we should be able to do that, right? And we're going to treat constant pressure situations certainly at first and most of the time, right? Constant pressure, right, reversible delta H is going to give us our heat of reaction, OK. And then if we can also determine delta u, then we know this, we know delta u is q plus w, then we can determine work as well, right? And typically, we'll be treating at least some cases where we're dealing with ideal gases in which case we can easily get delta u. And then we'll be able to in a very straightforward way get w. So it's a really powerful and simple formalism, once we set up what is needed the go through it, right? So let's just consider any reaction. The one that I've got written down for you is it's the reverse of the rusting of iron, right? So iron left to own devices in the atmosphere in the presence of a little bit of water, and the atmosphere will start combining with the water form iron oxide. There's an equilibrium between the two. So we'll have an iron oxide species. It's a solid. It's combining with hydrogen which is a gas to give iron, also a solid and water, right? So the way I've written this we're imagining the iron, the solid iron immersed in the water as a liquid, could be calculated otherwise as well, but in this case were imaging the water as liquid and going in this direction then it's forming iron oxide and evolving hydrogen gas, okay? And our heat of reaction or enthalpy of reaction is defined as the enthalpy at constant pressure. Isothermal conditions, temperature won't change and reversible work. For Isothermal reaction constant pressure reversible, right? Of course, they're all sorts of conditions under which a reaction could be wrong in the lab or outdoors or however, right. But this is the way we're going to define delta H of reaction. We want to have that definition clear because in fact we're going to, we might want tabulate heats of reaction, right, and of course want to know what the conditions are for the tabulated values apply. And we're going to want to calculate them from other quantities, and again, we're going to need to know each case what are the relevant conditions? OK, so what happens? Well, we should be able to get this. We should be able to calculate delta H. It's a state function. If we know the enthalpy of the products minus the enthalpy of the reactants, right. It's a state function. And we can we can do this in principle except for one important detail, which is that enthalpy, just like energy u isn't measured on an absolute scale but on a relative scale, right? You know, if you want to measure the potential energy of something in a gravitational field, you have to define the zero somewhere, right, because it's arbitrary. You can set it anywhere you want. It's the same with enthalpy. Enthalpy is just u plus p V. So there is an arbitrary set point that needs to be defined, right? Because what you actually measure in the lab are changes in enthalpy, just like what you measure when you look at energy change of some sort, you measure the change in energy, right. The absolute number that you assign to it is something that's arbitrary. You have to set what the zero is. And so there's a well-understood convention for what the zero is. What we define as zero is the enthalpy of every element in its natural state at room temperature and ambient pressure. In other words, we choose a convention for the zero of entropy, so that then we can write entropies of products and reactants always referring to the same standard state. And then we calculate changes, the convention is understood with respect to what is the zero, right. And so our tabulated values, they'll all work. They'll all refer to the same standard state. And we'll be able to use the formalism that we set up. So our reference point for H, it's 298.15 Kelvin, one bar and in that standard state our molar enthalpy is defined as zero for every element in its stable form. OK, very important. OK, now, given that reference point, all we need to do to get the value of enthalpy that we're going to use for each reactant and each product is calculate how much does the enthalpy change to form it from the elements in their stable form at room temperature and pressure, right. So we're going to replace the H or we're going to put in for the H what we'll call our heat of formation or delta H of formation starting from the elements in their stable states at room temperature and pressure, for each of these things. And we can do that because now we're going to have instead of just sort of H, it's going to be delta H of formation for each of these things, is going to appear in our calculation of H, but that's OK. Delta H of formation means the enthalpy of this compound minus the enthalpy of its constituent elements in their most stable state at room temperature and pressure. But we've defined the enthalpy of those elements in their stable state at room temperature and pressure as zero, right? So we're just subtracting, in effect, zero, right, from the enthalpy of the product, but of course it's important have that established because the heat of formation is something you could measure, right? You could run the reaction, take solid iron, gaseous oxygen, form iron oxide, measure the heat of formation of it, tabulate it. We know it. We know it forever, right. And for any of the other compounds. So in other words, by defining that reference state, we can then figure out or measure heats of formation of a vast number of compounds. We can tabulate them. We can know them, and then when we have reactions that inter-convert different compounds, we can calculate the heat of reaction is just the difference between the heat of formation of the reactants, and the heat of formation of the products, right. So let's just write that out. So first of all, for example, right, molar enthalpy of hydrogen gas at 298.15 K is zero. Hydrogen gas it's in its most stable state, right at room temperature and pressure. That little zero, that little superscript means one bar always. OK. A molar enthalpy of or whatever, iron, as a solid at 298.15 Kelvin is zero. Iron as an element is a solid. That's it's most stable state at room temperature and pressure, right, and so on. And then we can figure out heats of formation. Now in this particular reaction, I've got hydrogen gas, iron solid. Those already are elements in their most stable forms at room temperature and pressure. But this is a compound, right, it has some non-zero heat of formation from the elements. So is water, right? So I need to find out the heats of formation of the iron oxide and the water. And if I do that then I can find out the change in enthalpy of this reaction. It's just going to be the heat of formation of these three moles of water, minus the heat of formation of the iron oxide. OK. So how am I going to do that? Well, I need to write the reaction that forms that compound from it's elements, right? And I want to tabulate that for an enormous number of materials, right? So for example, if I want to look at HBr, there's a simple case, right, hydrogen bromine. What's my heat of formation? Delta Hf. It's going to have our little zero, right? How do I calculate it? Well, I need to write the reaction that forms this from its constituent elements. What is it? Well, 1/2 H2 as a gas, temperature, at one bar, plus 1/2 bromine, no actually bromine is a liquid at room temperature and one atmosphere. They form HBr which is a gas at room temperature and one bar, right. I can measure that. And the heat of reaction for this, delta H of reaction is equal to delta H of formation, of HBr as a gas at our temperature and one bar, okay? So that's how we determine our heats of formation. We measure them for all these compounds. And then we go back to reactions like this, and we can just very simply determine the heat of reaction because all we're doing is the following cycle. We go from reactants to products and there's some heat of reaction, and here's the cycle that we have. We go to the elements in their standard states, in both cases. So we're really just doing our delta H is the negative sum of delta H of formation for the reactants, all right? Because here what we're doing is we're going to take apart our reactants and form the elements from them, right? So written this way, for example, I'm going to pull these things apart, and I'm going to have solid iron, and I'm going to have gaseous oxygen, and I know the enthalpy of that reaction. That's just given by the heat of formation, the enthalpy of formation of iron oxide. Here, now I'm going to take those same elements, and I'm going to make the products, right. I know it's going to work because I already had this reaction, so all the, you know, I'm going to conserve atoms in the right way. So here, I'm going to have delta H, is just the sum for all the products of delta heat of formation, right? Here I'm going to put together all the products. So this is a positive heat of formation. This is the negative heat of formation, right? In other words, I've got reactants, and I've got products. What's delta H of reaction? It's delta H of formation of the products minus delta H of formation of the reactants. That's the important message, delta H of formation of the products, minus delta H of formation of the reactants. Any questions? All right, tomorrow I'll go through the remaining details. The truth is after this, it's all arithmetic, right? We're just going to go through it and execute it a little bit. And then we'll move on and talk about how we'd actually make some of these measurements of delta H and compare them to what we calculate. See you tomorrow.
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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 osw.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/8-01sc-classical-mechanics-fall-2016/8.01sc-fall-2016.zip
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So we're going to return to our one-dimensional elastic collision with no external forces. So we have object 1 moving with velocity V1 initial and object 2 maybe it's moving this way with V2 initial i hat, again, on a frictionless surface. And we'll call that our initial state. And here you can imagine we're going to use a ground reference frame. So both objects are moving. And our final state has object 1-- well, we don't know, again, which way it's going. We can just say it bounced back. And object 2 also bounced back. But the goal of our problem, of course, is to determine these vectors. And by knowing the vectors, we know which way they go. Now because energy and momentum are constant, let's write down our two equations. And I'm going to write them down, again, in terms of components. So we have 1/2 m1 V1x initial squared plus 1/2 m2 V2x initial squared equals 1/2 m1 V1x final squared plus 1/2 m2 V2x final squared. Now we're going to do some algebraic manipulations here. So the first thing I'm going to do is just eliminate these halves because it's not necessary. And I don't want to rewrite this equation. And this is our fact that our energy is constant. And now our condition that momentum is constant, we'll write this-- now, I'm going to leave a little space here intentionally. And our condition that momentum is constant is m1 Vx initial plus m2 V2x initial equals m1 V1x final plus m2 V2x final. Now this energy equation can be factored in by bringing all the m1 terms to one side and the M2 terms to the other side. So when I write that, I'll need a little room. I have m1 V1x initial squared minus V1x final squared. And that's equal to m2 V2x final squared minus V2x initial squared. So I've just brought those terms over to the other side. Now likewise I'll do the same thing down here. I have m1 V1x initial minus V1x final. And that's equal to m2 V2x final minus V2x initial. Now here comes the algebraic trick in which I'm going to linearize these systems. This is a squared minus b squared, which factors into a plus b times a minus b. So let's give ourselves a little room. m1 V1x initial plus-- let's put the minus sign first. minus V1x final times V1x initial plus V1x final. Factored that term. We have the same factoring on the other side. So it's just identical, V2x final minus V2x initial times V2x final plus V2x initial. Now let's call this equation 1a and our momentum factored as 2a. Now if you notice, the momentum piece is appearing exactly there and exactly here. So when I divide 1a by 2a-- and I'll just symbolically represent that-- then these two pieces cancel. And that leads to just this term equal to that term. And the significance, as you'll see when I write it out, 1 of x 1x final equals V2x final plus V2x initial. I've solved the quad-- I've eliminated the squared terms, linearized the system. Now I still want to write this equation in another way. Another important point to notice is that this equation is independent of mass. Now what I want to do is write this in terms of those concepts of relative velocity we had. Remember just to motivate this, V relative by definition was V1 minus V2. So let's write this in terms of the initial and the final. So in order to do that, we have to bring this initial term over to here and this final term over to there. And so this equation, which will give it a number 3, and now we'll modify that by calling it 3a. We have V1x initial minus V2x initial. Now notice the sign. I'm going to want to keep the order of 1 and 2. So I have to put a minus sign, V1x final minus V2x final. And when written this way, this is the initial component of the relative velocity. And in there is the final component of relative velocity. So by combining these two equations, I have this remarkable result that V relative initial is minus V relative final. And this condition is a very powerful tool for simply analyzing one-dimensional elastic and inelastic collisions. I'd like to even give this a name. I'd like to call it the energy momentum equation. Now there's a lot of significant things about this. So let's just think about it for a moment. We have that V relative initial in magnitude is equal to V relative final. And so right away, this gives us some insight into any collision. We can see whether a collision, if we know what the relative initial velocity is, we know that the final relative velocity has the same magnitude but simply switches direction. And that's a powerful tool in which to analyze collisions without doing a lot of algebra.
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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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JOANNE STUBBE: We spend a lot of time looking at-- all proteins are made from amino acid building blocks. And so one question I get, and students hate this, is, why do I need to know? I make them memorize the side chains of the amino acids. Why do I do that? Because the side chains of the amino acids are the key to the way all the catalysts in your body function. A catalyst simply enhances the rate of conversion of some small molecule into some other small molecule, and it can enhance the rate of the interconversion by 10 to the 15th fold. So if you didn't have that catalyst, you couldn't do anything. So amino acid side chains play a key role in catalysis. So thinking about the chemical properties of the amino acid side chains is key to understanding all the transformations in your body. So, what is the pKa of imidazole? You remember that? Huh? GUEST SPEAKER: I don't. I don't. JOANNE STUBBE: How can you not remember that? [LAUGHTER] Anyhow, the pKa of imidazole is close to seven, OK. So those physiological-- everything in the body is controlled. The pH has to be controlled. And so that means that you can protonate or deprotonate it. So knowing that is key to thinking about how the chemical reaction is going to work. The amino acid side chains now-- everybody thought there were 22 amino acids. But now we know almost all these amino acids can be modified once they get into the protein, so that's called post-transational modification. So now we have, probably, another 250 modifications. And one of the modifications that is essential, not for the catalysis part of proteins, but for the structural part of proteins, is hydroxylation of the amino acid proline. So if you don't hydroxylate proline, then you can't make this molecule called collagen. And collagen is a structural protein. It's 25% of all humans' protein, OK. And it's found extracellularly. Gram per gram it has the strength of steel. It has very complicated biosynthetic pathway. It has amazing tensile strength. It's found in cartilage and teeth and bone. And a key component of collagen is this hydroxylated proline. Because without it, you can't form the actual collagen structure. So collagen is this long-- most proteins, if you look at the structures, they look like little balls. They're globular. But collagen is a fibrillar protein, so it's very long. it's probably the longest protein, too. It's 3,000 angstroms long. And it has three chains initially, and they they're left-handed sort of helixes but not real helixes, and they have to wind around each other to form a right-handed helix. This all happens inside the cell, then somehow has to get to the outside of the cell. People are studying that now. And then it forms additional fibrils, and they become insoluble. And that provides the strength-- the extracellular structures provide the strength that maintain the cell's shape and viability of the cell. And a key component of all that is hydroxylation of proline. And how did they find that? This goes back to, again, misregulation. So, in the 1600s, or whenever they used to sail the ocean blue with no food, they didn't have enough vitamin C. So they didn't have any citrus fruit. So vitamin C has the vitamin ascorbate, and ascorbate plays a key role in the chemistry of allowing the proline to become hydroxylated. So it turns out, to get the proline to be hydroyxlated, you use, again, metal catalyzed reactions. So here's iron-2. And if that iron-2 reacts with oxygen, if it gets oxidized with iron-3, it can't react. So the function of this vitamin is to keep the iron-2 in the reduced state. So there's an example. People study that. Took them many, many, many years, like 200 years later, when they really understand the details of how this post-translational modification actually occurs. And we understand a lot about that chemistry now. And this was the first one of these modifications discovered, and now they're finding this modification everywhere. So lots of amino acids turn out to be hydroxylated, not just proline. So these modified amino acids now, because the technology we have is so mind boggling, we can find a needle in a haystack. Whereas in the beginning we were just trying to figure out what the amino acids were, now we have very, very sensitive analytical methods which allow us to see all of these modifications. Then the key question is, what is the function of the modification? And that's not so easy in terms of thinking about regulation, anyhow. And that's the focus of a lot of people's efforts right now.
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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 for several lectures, we've been talking about the XY model, which is a collection of two component spins, for example defined on a square lattice, but it could be any type of lattice. And today we are going to compare and contrast that with the case of a solid. More specifically for simplicity, an isotropic solid. And a typical example of an isotropic solid in two dimension is the triangular lattice. So something like this. What both of these systems have in common is that there is an underlying continuous symmetry that is broken. In the case of the XY model, at low temperatures all of the spins would be pointing more or less in the same direction. Potentially with fluctuations, let's say, go around the direction where they're supposed to be pointing-- being, let's say, here in the vertical-- characterized by some kind of an angle. Let's call it theta. And essentially the uniform state at 0 temperature can be characterized by any theta. And because of the deformations, it's captured through these low energy Goldstone modes. And the probability of some particular deformation will be proportional at low temperature to the integral, let's say, in d dimensions of gradient, the change in theta within different spins squared. Of course, with some considerations as we discussed about topological defects, et cetera, being buried in this description. Now, similarly, away from 0 temperature the atoms of a solid will fluctuate around the position that corresponds to the minimum. And so the actual picture that you would see at any finite temperature would be distorted with respect to the perfect configuration and there would be some kind of a vector [INAUDIBLE] u at each location that would describe this fluctuation. Again, there is no cost in making a uniform distortion theta that is the same across space. And similarly here, there is no cost in uniformly translating things. So these would both imply that this row can [? continue ?] [INAUDIBLE] leads to Goldstone modes. And the corresponding energy cost in the case of the isotropically distorted system is typically returned as an integral involving elastic moduli, as we've seen. And the traditional way to write that is in this form where the same reason here that it wasn't theta that was appearing but gradient of theta, because a uniform theta does not make any cross. These strain fields are related derivatives of the distortion. But now we remember that this u is a vector and this strain field uij it was one half of the derivative of uj in the i direction derivative of ui in the j direction symmetrized. Actually, probably better to write this symmetrically as beta h equals to this quantity. And then the typical type of calculation that we did for the XY model was to go in terms of the independent spin wave modes. So we go to Fourier space. This becomes an integral ddq 2 pi to the d. The derivative here gave q squared. And then I had something like the Fourier transformed theta squared. Now I can also more clearly represent this in the Fourier description. I will get something like 1/2 integral vdq 2 pi to the d. And a little bit of work shows that this form Fourier transforms to something that is proportional to mu. Again, it has to be proportionality to q squared because the energy should go to 0 as q goes to 0 for very long wavelength modes. And so we'll have something like q squared mu tilde of q squared. This is a vector squared. u is a vector. q is a vector. And there is another component that gives you mu plus lambda. And consistent with rotational symmetry, we can have a term that is q dotted with u tilde of q, the whole thing squared. And again, you can reason that you can describe the isotropic system in terms of just two elastic moduli is that the only rotational invariant quantities are q squared, u squared, and q.u, that those are the only terms that you can [? find. ?] Now the next thing. Once you know this form, you can immediately say something about the fluctuations. So you say that expectation value of theta tilde q theta tilde q prime, if I assume that this is a Gaussian theory and these are the only fluctuations that I'm considering, the answer is going to be 2 pi to the d delta function q plus q prime. And then I will get a factor of 1 over kq squared. The corresponding thing here is a bit more complicated. Because as we said, this quantity u tilde is a vector. And I can look at the correlation between, say, the i-th component of that vector along mode q, the j-th component of that vector along mu 2 prime and ask what this is. And again, only the same values of q are coupled together, so we get the usual formula over here. If this term was the only term in the story, that is if we were dealing with vectors q and u that are orthogonal to each other, then we would just get the 1 over q squared. And so there is actually a term that is like that. But because of this other term, there is the possibility that q and u are in the same direction, in which case, these two costs will add up and I will get something that is proportional to 2 mu plus lambda. And so then I will get 2 mu plus lambda q squared. And here I will have qi qj. This becomes q to the 4th. Let me make sure that I did not write this incorrectly. Yes, I did write it incorrectly. It turns out that there is a mu plus lambda out here too. But essentially there's just a small complication that we get because of the fact that you have a vector u. But overall the scaling is something that goes like 1 over q squared, which is characteristics of Goldstone modes here with additional vectorial things to worry about. Let me make sure I separate these two. Now one of the things that we have been concerned with is whether these fluctuations destroy long range order, so for long range order the information that we would like to have is that the entirety of the system is roughly pointing in the same direction, meaning that if I know that this beam here is pointed in the vertical direction, if I go very far away, it is still more or less pointed out-- pointing in the same direction. So for that, we could look, for example, as the degree which [? with ?] these two spins? Yes, question? AUDIENCE: [INAUDIBLE] q squared [INAUDIBLE]? PROFESSOR: This is 4. Is that what you're worried about? AUDIENCE: Yeah. It should be 2, I guess. PROFESSOR: No, because I have two powers of q out front. AUDIENCE: Right. PROFESSOR: So I think in the notes what I have is that I have actually written things this way. I put the factor of mu q squared out front. And then I had this as qi qj over q. And I think-- AUDIENCE: [INAUDIBLE] q squared [INAUDIBLE]? PROFESSOR: q squared. Let me one more time check to make sure that that's not-- actually there's a minus sign. All right. AUDIENCE: Wouldn't there be another factor of mu to compensate for the factoring out? PROFESSOR: Yeah. So let's check this in the following fashion. So I can look at modes where u and q are perpendicular to each other. When u and q are perpendicular to each other, I don't have this there. And so then what I should have is just a cost that is 1 over mu q squared. OK. So let's see if that comes about. What I can do is let's say imagine that I'm looking at a q that is oriented along the x direction. And I look at u's that are in the y direction. And I correlate two u's in the y direction. So I will definitely get a delta ij here. I would have a 1. Here I would get a factor of nothing, because this term is absent. The q's that I'm looking at don't have any component along the direction of u that I'm having. So this is absent and I will get a 1 over mu q squared, which is consistent with that. Now let's look at the case where I look at u's that are aligned with the vector q. Let's say again in the x direction. Then, as far as the energy is concerned, both terms are present and the cost should be 2 mu plus lambda. So the answer that I should ultimately get should be 1 over 2 mu plus lambda. So let's check. So here q squared-- and cancels with this q squared. All of the q's are in the direction of the indices that I'm looking at. So that cancels. This is 1. So I will get 1 minus mu plus lambda over 2 mu plus lambda. So then I will get 2 mu plus lambda minus mu minus lambda. So I will get a mu in front that cancels this. And I will be left with just the 2 mu plus lambda. So we've checked that this is correct. All right. So this is a proof, if you like, by induction. You knew that the only forms that are consistent are delta ij and qi qj over q squared. So we can give them two coefficients, a and b, and then adjust a and b so that these two limits that I considered are reproduced. So-- OK. I didn't have to resort to my notes if I was willing to spend the corresponding three minutes here. So going back here, in order to see the degree of alignment of the two spins that we have over here, what we need to do is to do is to look something like cosine of, say, theta at location x minus theta at location x prime, which is non other than the real part of the expectation value of e to the theta at location x minus theta at location x prime. And assuming that everything is governed by this Gaussian rate, the answer is going to obtain by computing this Gaussian as exponential of minus 1/2 the average of theta x minus theta x prime squared. So then my task is to replace theta x and theta x prime in terms of these theta tilde in Fourier space. And then this will give me some expectation that involves this average. And when I complete that, eventually I need to Fourier transform 1 over q squared. And we've seen that the Fourier transform of the 1 over q squared is the Coulomb interaction. And so ultimately this becomes exponential of minus of 1 over k. The Coulomb interaction as a function of x minus x prime with the appropriate cutoff included in this expression. And then the statement that we always make is that this entity at large distances, the Coulomb interaction saying three dimension, falls off as 1 over separation. So in three dimensions, this at large distances goes to a constant 1 over k is something that is proportional to temperature. So after doing enough temperature, this entity at large distances goes to a constant. So there is some knowledge about the correlation between these things that is left. Whereas in two dimensions and below, when I go to sufficiently large distances the Coulomb interaction diverges. This whole thing goes to 0 and information is lost. So this is the in content that Goldstone modes destroy through long range order in two dimensions and below. Except that we saw that in two dimensions, because the Coulomb interaction grows logarithmically, this correlation fall off according to this description only as a power law with an exponent connected to k. And then the discussion that we had for the XY model was that at high temperatures you have exponential decay in two dimensions. At low temperature you have power law decay. Although there is no true long range order, there has to be a phase transition. And we saw that this transition was described by the unbinding of these topological defects. So we would like to state that something similar to that is going on over here. The only reason I went through this was to figure out what the analog of this entity should be in the case of the solid. So the analog of our theta is this vector u, the distortion. What should be the analog of e to the i theta? And the things that you should keep in mind is that theta, I can't really tell it's difference between theta and theta plus 2 pi, theta of 4 pi, et cetera. So this e to the i theta captures that. If theta goes to theta plus a multiple of 2 pi, e to the i theta remains invariant. So it's a good measure of ordering. It's better than theta itself. So similarly, there's a better measure of deviation from perfect order here for the case of the lattice. You say, well, why should I worry about that since I can measure what u is from the perfect position? Well, my answer is this point here. Did it come from here or from there? Should you use for u this or should you use that? You don't know. So the u that you have the to use similar to the theta that you're using is somewhat arbitrary. And the arbitrariness of it is given by unit vectors along the original lattice. That is, u and any u that is increased or decreased by some lattice vector of the original lattice is the same thing. You can't take them apart in the same sense that theta and multiple of 2 pi you cannot. So then that suggest that I can construct some kind of an order parameter, I call it row G, which is like this. This e to i something. It's e to the iu but u is a vector. And what I do I multiple by a G, which is an inverse lattice vector. So G is an inverse lattice vector. And essentially the inverse lattice vectors are defined to have the property that when you multiply them with any of the original lattice vectors-- which are, let's say, parametrized by two integers, m and n-- the answer is going to be some multiple of 2 pi. So you can see that irrespective of whether I choose u as distortion from one lattice position or any other lattice position that happens to be nearby, this measure is going to be giving you the same face. So in the same way as before for the spins where I asked whether spins at large distances are correlated long range, I can do the same thing here and define a long range correlations for a solid by looking at e to the iG dotted with, let's say, u at position x minus u at position x prime. And x and x prime are two points on the lattice. And again, the u's that I have are Gaussian distributed. So this I will complete the square. I will have exponential minus 1/2. Well, OK, let me be a little bit careful with this. I have G alpha G beta over 2 because I have two factors of G dot u. I will write them using Einstein's summation convention as G alpha times u alpha component times the u beta component. So once more what we can do is to then write these u's in terms of their Fourier vectors. So what I will have is exponential minus G alpha G beta over 2. I will need to integrate over q's. Because of these factors, I will get something like 2 minus 2 cosine of qx minus x prime. So whenever you go to Fourier space, you will get e to the iq x minus e to the iq x prime. Different one for here with different q primes, but when you take the average, q prime is set to minus q so then you get this form as usual. And then I will have the average Fourier transform use, which is what I wrote over there. I will get a 1 over mu q square delta i delta alpha beta minus q alpha q beta q squared mu plus lambda over mu plus lambda. There is a little bit of subtlety here. So I will do something this is not quite kosher but gives me the right answer. Certainly I can take this G alpha a beta inside here. When it multiplies delta alpha beta, I will simply get G squared. I really G alpha G alpha. Sum over alpha. It will give me G squared. Here what I will get is a G dot q squared because I will have two Gq's divided by q squared. Ultimately I have to integrate over all values of q, including all angle values. So there will be, as far as the angle is concerned, something like a cosine squared here. And the rather unconventional thing to do is to angular averages that first and write this as G squared over 2. Once you do that, then essentially you can see that this structure just gives you a constant. And what you will need to do is the usual Fourier transformation of 1 over q squared. So eventually you can see that the answer is going to be proportional to minus G squared. I will get a factor of 2 mu u from here. Once I have written this as G squared over 2, I will get 1 minus 1/2 of this quantity. So that becomes 2 mu plus lambda-- or 4 mu plus lambda minus mu minus lambda. So I will get a 3 mu plus lambda divided by 2 2 mu plus lambda. And so there's a factor of 4 here. So that's just the numerics that goes out front. The key to the whole thing will be the usual integration of 1 over q squared. Just as we had before, I will get again the same Coulomb interaction as a function of x minus x prime with appropriate lattice cut-off introduced so that the short distance behavior is controlled. This whole thing has a name. It's called a Debye-Waller factor. Well, almost. In 3D. And the point is that if I'm looking at this whole thing in three dimensions, at large enough separation this will go to a constant, just as we had before. And I will conclude that this entity, as I look at two points that are very far apart, at least that's sufficiently low temperature and all of my mus and lambdas are scaled inversely with temperature, so this whole thing is proportional to temperature. Essentially, you can see that this information at large distance goes to a finite constant, telling me that there's long range orders preserved. Again, when I go to two dimensions, you would say that this, at large distances, grows logarithmically. And so the correlations will fall off, but fall of as a power law. Now for the case of a solid, usually what you do in order to probe for the order is you shine x-rays and you see what comes out. So essentially, you will see that there will be some scattering. And the way that the wave vector of the light has changed is characterized by a vector q. And we discussed this already, that the structure factor s of q is going to be related to expectation value of the phase that you get from the different points. So I will get a factor of e to the i q. the location of the different atoms, let's call them r, of x. I have to do a sum over x. I have to scatter from all of the atoms. There's some overall form factor that if I'm not really interested, but this whole thing has to be first squished. So once I square this, I will be getting averages between pairs of points. Because of translational symmetry I can relate to that sum over pairs of points to a center of mass that typically would pick up the number of points that I have. And then the expectation value of e to the something like sum over all points, e to the i q.r r of x minus r of 0 average. This average I can certainly take out here. Yes? STUDENT: Sorry. Is r what you're calling u before? PROFESSOR: No. So r is the actual location of where the particle is location. So r, let's say for a particle that at zero temperature was sitting at lattice point-- that is, let's say, in two dimensions labeled by m and n-- it is shifted by an amount that is u that I would assign to that point. So the r's that I put here are the actual physical position of the atom. Which is composed of where it was sitting at 0 temperature plus a little bit more. So when I do this, indeed I will get a sum over all positions. This could be a label mn, whatever, of e to the iq. The difference between the two r's would be, first of all, the difference between these two at 0 temperature, which would be some lattice vector of some variety. Let's call r 0 of x. And that quantity has [INAUDIBLE] fluctuations. The fluctuations go into e to the iq.u of x minus mu of 0. Now if I'm doing this at 0 temperature or very close to 0 temperature where this is either 1 at 0 temperature or not too much fluctuating at high temperatures, I will essentially be adding a lot of e to iqr's. These are essentially phases. They can be positive, negative, all over the complex plane. And typically you would expect that for a randomly chosen q, the answer from adding all of these things will be 0. So the only time we're close to 0 temperature, you expect this to be something that is significant. Scattering is when q is one of these inverse lattice vectors. So if I was really doing this at 0 temperature, this would be 1. And the answer that I would get is essentially I would get these delta function back peaks at the locations that correspond to the inverse lattice spacing. And that's presuming something that you have seen in crystallography or whatever. You take a solid and you scatter light from it and you will back points. So what I expect is that when I have a q that is roughly close to G but not necessarily sitting exactly at that point, what I would get is a scattering that these proportional to an integral over the entire lattice of something like how far I am away from the right position. And I should really write this as a sum over lattice points, but I can approximate this by an integral. The integral again has the property that it will give me 0 unless q is exactly sitting at G. And then for here, I evaluate this quantity when q is close to G according to what I have calculated up there. So I will get exponential of minus G squared-- What do I have? Over 4 mu, three mu plus lambda, 2 mu plus lambda. The Coulomb interaction as a function of x. Yes? STUDENT: Are you missing an overall factor of m? PROFESSOR: That's why I wrote the proportionality. Because what I'm really interested is what you see. And what you see is how things are varying as a function of q. So I'll tell you what the answer is. So you do the Fourier transform from this. And in d equals to 3, what you find at 0 temperature, as we said, you will get Bragg spots at the locations of G's that correspond to the inverse lattice vectors, which actually themselves form a lattice. So this is essentially the space of q. And these are the different values of G. Now what happens is that the strength of each one of these delta functions is then modulated by this factor evaluated at the corresponding G. And since it is proportional to G squared, you find that the spots that are close to the origin are well defined. As you go further and further away, the become very weaker. And at some point you cease to see them. So that's the-- the Debye-Waller factor describes how these things vanish as you go further and further away. And essentially, you find that they go and diminish as e to the minus G squared. But clearly, this kind of scattering from a crystal is very different from the scattering that you expect from an liquid, which is essentially you will have a bright spot at q close to 0 and things that would symmetrically fall off, maybe with a couple of oscillations due to the size of the particles, at cetera. Now what happens in d equals to 2 is kind of interesting. Because in d equals to 2, I'm doing a two dimensional integral, and this thing is growing as a power law. So basically what I end up having to deal with is something like e to the i q minus G x. And something that falls off as a over x to some power that I will simply cause eta G. So related to that combination. And when you integrate this, just dimensionally you can see that it is an integral that scales as x to the power of 2 minus eta. x is inverse to q minus G. So this is going to scale as 1 over q minus g to the power of 2 minus eta. So as you go along some particular direction and ask what you see, you find that will see peaks close to the locations of the lattice vectors. So let's say they are equally spaced along the particular axis that I'm picking. In three dimension, you would have been seeing delta function at each one of these positions whose strength is governed by this Debye-Waller factor. In two dimensions, you will essentially see power laws close to each one of them. But as you go further, the exponent of the power law changes. So at some point you even cease to have a divergence. But you can essentially ignore anything like that [? happening. ?] So you can see that because we don't have true long range order into two dimensions because of these logarithmic divergences, et cetera, the traditional picture that we have of a solid that is characterized by Bragg scattering and delta function peaks is modified, but still it is something that looks very different from the liquid. So we expect that as we raise temperature there is a phase transition between scattering of this form and something that looks like liquid. And you can roughly guess that essentially it is going to be related with these etas becoming larger and larger so that these divergences disappear. But what is the mechanism? So let's go back and see what the mechanism was that we discovered for the XY model. So we are going to look at d equals to 2 from now on. That discussion so far was general. In d equals to 2, we said that we have topological defect. And the story here was that, as I have been discussing, the angle is undefined up to 2 pi. So it would configuration of angles that kind of radiate away from the origin. And the idea was that if I take a circuit and integrate around this circuit-- let's call this the s gradient of theta-- the answer does not need to come back to 0. It will be some multiple of 2 pi . Some integer multiple. And similarly here, we said that the value of u is undefined up to a lattice spacing. That leads to topological defects, which I find it very difficult to draw for the triangular lattice, so I'll draw it for the square lattice. So at each one of these positions there's a particle sitting. And you look out here. You have a perfect square lattice. You look out here. You have a perfect square lattice. But clearly it is not a perfect square lattice. And the analog of the circuit that I drew over here that encloses a singlularty-- there's clearly some kind of a defect sitting over here-- is that I can start with some point on the lattice and perform what is called a Burgers Circuit. What Burgers Circuit is is some walk that you would take on a lattice, which in a perfect lattice would bring you back to your starting point. So for example-- that was not a good starting point. Let me start from here. Let's say I take four steps up-- one, two, three, four. I take five steps to the right. One, two, three, four, five. I reverse my four steps. So I go four steps down. One, two, three, four. I reverse my five steps and I go five steps left. One, two, three, four, five. And you see that I did not end up where I was. I ended up here. And a failure to close my circuit has to be a lattice vector. And this is called d and it's called a Burgers Vector that characterize this defect, which is called a dislocation. So the uncertainty that I have in assigning a value for u because of being on a lattice or deforming a lattice gives rise to these kinds of ambiguity that are similar to the topological defects that you had over here. Whereas the defects here were characterized by an integer, the defects here are characterized by a lattice vector. And here the simplest things were, say, plus or minus. Clearly here on the square lattice the simplest things are plus b, minus b along the x direction and plus b minus b along the y direction. For example, on the triangular lattice it would be plus minus at any of the three directions that would define the lattice. So calculating the distortion field here was actually an easy matter because we said that the gradient of theta was something like 1 over r. And so the gradient of theta, which was 1 over r, I could in fact write-- we saw as vector that is orthogonal to the vector that is going out of the plane. So we call that a z hat. It was also orthogonal to the direction of motion away from the defect, which we could characterize by looking at the gradient. The gradient vector for this distortion is clearly in the radial direction. And the answer that we had here for one defect was ni log of r minus ri. And since I'm taking the gradient it doesn't matter. I can't put or not put the cut off. And if I had multiple ones, I would simply sum over it. So this was the contribution to the distortion that came from a collection of defects such as this. Now you can see that essentially in some continuum sense, all I'm doing here is similar. I can be taking a big circuit. And as I go along that big circuit, I can take a gradient-- So let's-- yeah. The s. gradient of the distortion field. Distortion field is, of course, a vector. So it has components that I can write as the x component or the y component. So this alpha could be the x component of the distortion or the y component of the distortion. And once I complete the circuit, here the answer was 2 pi n. Here the answer is a lattice vector, which is this Burgers Vector that could be in any direction that you want. Now as far as mathematics is concerned, this line and this line for each component are exactly the same. So the solution that I can write for gradient of u I can just copy from here for each component. So I can write that gradient of the u that is due to a collection of dislocations is something like z hat crossed with curl of a sum over all of the potential locations off my defect. Rather than n, I have b over 2. And it's a vector log of minus ri over [INAUDIBLE]. So if I have a collection of these dislocations at different places on my lattice-- locations ri, strengths, b alpha which is a vector-- then the distortion field that they would generate is given by this. It's very just following the answers that the [INAUDIBLE]. So then what we did for the case of the overall system-- in order to find how the different topological defects interact with each was to calculate beta h, which was an integral of gradient of theta squared. Yes? STUDENT: So the b alpha, you have an index i [? also? ?] PROFESSOR: b alpha have index i also. Like like the ni have index i. So maybe I write it in this fashion. Yeah. So this is the cost that we have to evaluate, except that we have to be somewhat careful with the meaning of this gradient of theta in that the gradient of theta, as we said, has a contribution that is from regular spin waves, the Goldstone modes, that can be deformed back to everybody pointing in the same direction without topological defect. Plus a contribution due to this topological defect that are categorized in that case by ni. Now similarly here for the case of the solid, we have the beta h. Slightly more complicated integral d 2x. We have mu uij uij plus lambda over 2 uii ujj. And what we can do is to say that our strain field uij has a component that is like the Goldstone modes that we have been calculating so far, essentially treating everything as Gaussians. Except that I will write it as phi ij. And then a part that would come from the distortion field u bar that I calculated. So I take that distortion field and then take derivatives of it to symmetrize appropriately to construct the strain that sums from them. So we substituted this form over here. And we found that beta h that we got had a part that was simply relate to the Goldstone modes. This was the part that could be treating as a Gaussian. And then we had a part that corresponded to the interactions among these defects. And that part we saw had the character of charges ni and nj plus minus 1 characterizing these topological defects having a Coulomb interaction [? between them ?]. And then, of course, this i less than j, we had to worry about what was happening when i was j. And that we considered to be the contributions of the core energy that, once exponentiated, we described as y. So it's beta h which is log of that would be log of y. And actually this k bar, by the way, was simply 2 pi k. And it was 2 pi k because each one of these charges is 2 pi n. So there's 2 pi 2 pi. But the Coulomb interaction really should be log divided by 2 pi. And so k bar becomes 2 pi k times this. So I can do the same thing over here. And what I find happens is that my beta h gets decomposed as follows. There will be a part that is simply the original expression now for the field that is well behaved and has no dislocation piece. And then you wouldn't be surprised because this structure is no different from the other structure. There's essentially a gradient of this distortion field squared. And you can see that if I take a second derivative of this distortion field or one derivative of this, effectively I will have two derivatives of a log, which will give me a delta function. So essentially these things do behave like what you would get from Coulomb type of potential. [INAUDIBLE] plus [INAUDIBLE] if the potential is 0. So not surprisingly, I we get a minus k bar sum over pairs of where these dislocations are located. And then I will have b at location i, b at location j. And then I would have a log of ri minus rj with some cut off. Except that these b's are actually vector. So this term is a dot product of the two b's. And it turns out that when you go through the algebra, there is another term, which is bi. ri minus rj, bj dotted with ri minus rj divided by ri minus rj squared. So the charges that we had in our original theory were scalar quantities. These defects were characterized by a scalar value of n. And they were interacting with something that was like the ordinary Coulomb potential. Whereas now we are looking at a system that is characterized by charges that are vectors. And it turns out that what we have here is the vectorial analog of the Coulomb potential. And again, if you think about an isotropic system and vectors, this is a vector that you can form, but b.r is another vector that you can form. And so both of them do appear. And once you go through the whole algebra and through the inverse Fourier transform, et cetera, you get this additional contribution to this vector Coulomb interaction. That's really is the only difference. And of course, you will again get a sum over all locations of the core energies for creating these defects. And the value of k bar here is related to these parameters by mu mu plus lambda divided by pi 2u plus lambda. So this combination controls what you have. So the next thing that we did was to construct a perturbation. We essentially said that if I have these pairs of charges that can spontaneously appear at an energy cost but an entropy gain, they will effectively weaken the overall Coulomb potential. For example, if I have to test charges then there will be some polarization of the medium, some reorientation of these charges that appear, that weakens the effective charge here. And we could perturbatively calculate what the correction was, except that we found that that correction, which involves an integration over the separation of these things. There was an integration that was potentially divergent for us no matter how small we made the core energies here. And so what we did was eventually to construct an RG. And the RG was that the value of this parameter k bar-- actually more usefully its inverse that is related to temperature-- changed as a function of integrating out short distance degrees of freedom. It was something like 4 pi-- I think it was 4 pi cubed, it doesn't matter-- y squared. And that the scaling of to core energy itself was determined by 2 minus pi k, which in terms of this k bar becomes 2 minus k bar over 2 times y. And we can do exactly the same thing over here, except for the complication of having to deal with the charges that are vectorial. And you get that the strength of the Coulomb interaction, it's inverse that is related temperature gets re-normalized by exactly the same process. That is, there will be charges that will appear. All the mathematics you can see is clearly similar. And there will be a reduction, which then you can think of as an increase in temperature, which will be proportional to y squared. I don't know what the constant of proportionality is. And that dy by dl-- so essentially I focus on the simplest type of dislocations that I would have of unit spacing-- has exactly the same [INAUDIBLE], 2 minus k bar over 2 y. It turns out that there is actually a difference. And the difference is as follows. That while we did not calculate the next correction in this series, I said that this correction came from essentially what would a single dipole do, which would be some contribution that is order of y squared. And I expect that there will be correction if I look at a configuration that has pairs of dipoles. And this will change this by order of y to the fourth and this by order of y cubed. Whereas if I look at something like a triangular lattice, you can see that if I have two dislocations out there as test dislocations pointing out in opposite directions, which would have some kind of a interaction such as this. And I ask how that interaction is modified by the presence of dipoles of dislocation in the medium. I could put a pair of dislocations integrate over the separation between them. I will get, essentially, the same type of mathematical structure that would ultimately give me this. But then the next order term in this series is not for dislocations because I can have a neutral configuration of three dislocations, such as this. And so once you do that, you will find that the next correction here is order of y cubed and order of y squared here. But generically, both of them have a phase diagram for RG flows that involves the inverse of the interaction of these charges. That is a temperature-like quantity. And there is a critical value of that inverse, which is 1/4. And what happens is that-- let's see, this other axis is y. Anything smaller than 1/4 y tends to go to 0. Anything larger, y tends to get larger. And there is going to be some kind of a separatrix that he describes the transition between flows that go down here and the flows that go away. And presumably, as I heat up my original system-- my lattice, different types of lattices-- at low temperature, I can figure out what this combination of mu an lambda is. That gives me some value of k. And there is some kind of the core energy for the vertices that I can calculate. So there will be some point in this phase diagram, which at low enough temperature I will go over here, which corresponds to a system that has an effective logarithmic interaction between dislocations. As I change the temperature, I will proceed around some trajectory such as this because everything will change. And presumably, at some point I will intersect this and I will go to the other phase. Now the characteristic of the other phase was that this parameter k was going to 0. It was essentially the logarithmic interaction was disappeared. And you can see that that parameter k going to 0 means that mu, the shear modulus, has to go to 0. So basically this phase out here is definitely characterized by 0 shear modulus. So again, the two parameters mu and lambda, mu is the one that gives you the cost of trying to shear in a system. And essentially the presence of the resistance to shear is what defines a solid for you. So this is a solid to something that does not have rigidity transition. It turns out that there is one subtlety here. I don't know whether it's good to mention or not. But so in the case of alpha this superfluidity, for example, we said that this coupling strength as you approach the transition rose to a universal value. In these units it will be 4. But that it approaches that universal value with a logarithmic singularity. Sorry, with a square root singularly. Sorry. And we saw that this was experimentally observed in films of superfluid film. I also said that on the other side you have a finite correlations. The correlations will be decaying exponentially. And this correlation length has this very unusual signature, which is that it diverged as a square root of t minus tc in the exponent. It was giving rise to these essentially singularities. While everything kind of looks identical between this RG and the other RG, it turns out that once you keep track of these corrections, these corrections actually are important. And for the vectorial version of the Coulomb gas, you find that they shear modulus that is finite in this phase reaches its final value, which is not universal because the thing that the universal is that combination that involves both mu and lambda. But rather than stc minus t to the 1/2, stc minus t to some exponent that is called mu bar, which is 0.36963 [INAUDIBLE]. So this is a kind of subtle thing that is buried in these recursion relationships once you go to a higher order. And accompanying that is a correlation length that also diverges with some behavior dominated by the same exponent. So once people understood the original [? Koster ?] [INAUDIBLE] transition, it was kind of a very natural next step for them to think about these locations in a two dimensional solid and say that in the same manner that the unbinding of these defects got rid of the residual long range order that was left in the XY model. That this kind of ordering that we said exists for the two dimensional solid, this appears as a result of unbinding of dislocations and you get a liquid. Turns out that that's not correct. And it was pointed out by Bert Halperin I think that if you look at a solid, there's two things that you. You have the translational order, but there's also orientational order. That is, you can look, for example, at a collection of spins here and the bonds are pointing along the x and y directions and you can look down here and the bonds are pointing in the x and y directions. And from the picture that I drew over here, you can kind of see that once I insert this dislocation which corresponds to an additional line, the positions here are distorted. Once I have many of these inserted, it is kind of obvious that I will lose any idea of how the position of this lattice point and something out there is related. It's not so obvious that inserting lots and lots of these lines will remove the knowledge that I have about the orientation of the bonds. And the insight that we will follow up-- and maybe it's probably a good idea to do it next time-- is that actually once the dislocations unbind, we will lose this kind of ordering. If I do this in two dimensions, these are power law decays at low temperatures. Once the dislocations unbind, this becomes an exponential decay, as I would expect. But I can also define locally some kind of an orientation. Let's again call it theta. But theta is, let's say, the orientation of the bond that connects two neighboring spins. Now what I can do is to look at the correlation of orientations at different positions. And it turns out that if I'm doing something like a triangular lattice, I can't just pick the angle and correlate the angle from one point to another point because let's say on the square lattice the angle itself, I don't know whether it was coming from a bond that was originally in the x direction or in the y direction. So this is unknown up to a factor of 90 degrees. In the triangular lattice it is an unknown up to a factor of 60 degrees. So what I should do is I should define e to the 6 i theta at each location as a measure of the orientational order. You can see that, again, if I'm at 0 temperature, I can reorient the lattice any way that I like. This phase would be the same across the entire system. So what I'm interested is to find out what is the expectation value of this object when I look at points that are further apart. And again, clearly if I'm at 0 temperature these thetas are the same. This will go to 1 at finite temperature. Because of the fluctuations it will start to move a bit. And what we can show-- and you will do it next time-- is that as long as you are in the 0 temperature phase you will find that this goes to a constant, even in two dimensions. But once the dislocations unbind, what we will show is that it doesn't decay exponentially, but then it starts to decay as 1 over x to some power, let's call it eta theta. And then at some higher temperatures, it eventually starts to decay exponentially. So there is the original solid in two dimension that we started that has true long range order in the orientations. And that is reflected in these Fourier transforms that we make. So we said that when we do the Fourier transform and look at the x-ray scattering, let's say from a two dimensional crystal, you were seeing these kinds of pictures. And then these were becoming weaker as we went further and further out. But now you can see that this picture clearly has a very, very defined orientational aspect to it. Now once we go to this intermediate phase, all of these peaks disappear. But what you find is that when you look at this, rather than getting a ring that is uniform, you will a see a ring that has very well- defined variation. One, two, three, four, five, six. Six-fold symmetry to it. So this phase has no knowledge of where the particles are located, but knows the orientations. It's a kind of a liquid crystal. It actually has a name. It's called hexatic. But it's one of the family of different types of materials that have no translational order but some kind of orientational order that are hexatics. So the transition between this hexatic phase to this fully disordered phase, it turns out in two dimension to be another one of these dislocation-- well, topology defect unbinding transition. So maybe we will finish with that next time around.
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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: Uncertainty. When you talk about random variables, random variable Q, we've said that it has values Q1 up to, say, Qn, and probabilities P1 up to Pn, we speak of a standard deviation, delta Q, as the uncertainty, the standard deviation. And how is that standard deviation defined? Well you begin by making sure you know what is the expectation value of the-- or the average value of this random variable, which was defined, last time, I think I put braces, but bar is kind of nice sometimes too, at least for random variables, and it's the sum of the Pi times the Qi. The uncertainty is also some expectation value. And expectation value of deviation. So the uncertainty squared is the expectation value, sum over i, of deviations of the random variable from the mean. So you calculate the expected value of the difference of your random variable and the mean squared, and that is the square of the standard deviation. Now this is the definition. And it's a very nice definition because it makes a few things clear. For example, the left hand side is delta Q squared, which means it's a positive number. And the right hand side is also a positive number, because you have probabilities times differences of quantities squared. So this is all greater and equal to zero. And moreover, you can actually say the following. If the uncertainty, or the standard deviation, is zero, the random variable is not that random. Because if this whole thing is 0, this delta squared, delta Q squared must be 0 and this must be 0. But each term here is positive. So each term must be 0, because of any one of them was not equal to zero, you would get a non-zero contribution. So any possible Qi that must have a Pi different from 0 must be equal to Qbar. So if delta cubed is equal to 0, Qi is equal to Q as not random anymore. OK, now we can simplify this expression. Do the following. By simplifying, I mean expand the right-hand side. So sum over i, Pi Qi squared, minus 2 sum over i, Pi Qi Q bar plus sum over i, Pi Q bar squared. This kind of thing shows up all the time, shows up in quantum mechanic as well, as we'll see in a second. And you need to be able to see what's happenening. Here, you're having the expectation value of Qi squared. That's the definition of a bar of some variable, you'd multiply with variable by the exponent of [INAUDIBLE]. What is this? This a little more funny. First, you should know that Q bar is a number, so it can go out. So it's minus 2 Q bar. And then all that is left is this, but that's another Q bar. So it's another Q bar. And here, you take this one out because it's a number, and the sum of the probabilities is 1, so it's Q bar squared as well. And it always comes out that way, this minus 2 Q bar squared plus Q bar squared. So at the end, Delta Q, it's another famous property, is the mean of the square minus the square of the mean. And from this, since this is greater or equal than 0, you always conclude that the mean of the square is always bigger than the-- maybe I shouldn't have the i here, I think it's a random variable Q squared. So the mean, the square of this is greater or equal than Q bar squared. OK. Well, what happens in quantum mechanics, let give you the definition and a couple of ways of writing it. So here comes the definition. It's inspired by this thing. So in quantum mechanics, permission operator Q will define the uncertainty of Q in the state, Psi O squared as the expectation value of Q squared minus the expectation value of Q squared. Those are things that you know in quantum mechanics, how you're supposed to compute. Because you know what an expectation value is in any state Psi. You so Psi star, the operator, Psi. And here you do this thing, so it's all clear. So it's a perfectly good definition. Maybe it doesn't give you too much insight yet, but let me say two things, and we'll leave them to complete for next time. Which is claim one, one, that Delta Q squared Psi can be written as the expectation value of Q minus absolute expectation value of Q squared. Like that. Look. It looks funny, and we'll elaborate this, but the first claim is that this is a possible re-writing. You can write this uncertainty as a single expectation value. This is the analog of this equation in quantum mechanics. Claim two is another re-writing. Delta Q squared on Psi can be re-written as this. That's an integral. Q minus Q and Psi. Look at that. You act on Psi with the operator, Q, and multiplication by the expectation value of Q. This is an operator, this is a number multiplied by Psi. You can add to this on the [? wave ?] function, you can square it, and then integrate. And that is also the uncertainty. We'll show these two things next time and show one more thing that the uncertainty vanishes if and only if the state is an ideal state of Q. So If the state that you are looking for is an ideal state of Q, you have no uncertainty. And if you have no uncertainty, the state must be an ideal state of Q. So those all things will come from this planes, that we'll elaborate on next time.
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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 the lectures I'm going to record today, we're going to conclude cosmology. And then we're going to begin talking about a another system in which we saw the Einstein field equations using asymmetry. So this is one where we will again consider systems that are spherically symmetric but we're going to consider them to be compact. In other words, things that are sort of, rather than filling the whole universe, the spacetime that arises from a source that's localized in some particular region in space. So let me just do a quick recap of what we did in the previous lecture that I recorded. So by arguing solely from the basis of how to make a spacetime that is as symmetric as possible in space but has a time asymmetry, so the past and the future look different, we came up with the Robertson-Walker metric, which I will call the RW metric, which has the form I've written there on the top one line. There were actually several forms for this. I've actually written down two variants of it on the board here. Key thing which I want to highlight is that there is, in the way I've written it on the first line, there's a hidden scale in there, r zero, which you can think of as, essentially, setting an overall length scale to everything that could be measured. There is a parameter, either k or kappa. k is either minus 1, 0, or 1. Kappa is just k with a factor of that length scale squared thrown in there. We define the overall scale factor to be-- it's a number. And we define it so that its value is 1 right now. And then everything is scaled to the way distances look at the present time. A second form of that-- excuse me-- which is essentially the change of variables is you change your radial coordinates to this parameter chi. And depending on the value of k, what you find then is that the relationship between the radius r and the coordinate chi, if k equals 1, you have what we call a closed universe. And the radius is equal to the sign of the parameter chi. If you have an open universe, k equals minus 1, then the radius is the sinh of that parameter chi. And for a flat universe, k equals 0, they are simply 1 is just the other modulo, a choice of units with the R of 0 there. All this is just geometry. OK. So when you write this down, you are agnostic about the form of a and you have no information about the value of k. So to get more insight into what's going on, you need to couple this to your source. And so we take these things. And using the Einstein field equations, you equate these to a perfect fluid stress energy tensor. What pops out of that are a pair of equations that arise in the Einstein field equations. I call these F1 and F2. These are the two Friedmann equations. F1 tells me about the velocity associated with that expansion premise. There's an a dot divided by a squared. We call that the Hubble parameter, h of a squared. And it's related to the density of the source of your universe as well as this kappa term. OK. And as we saw last time, we can get some information about kappa from this equation. The second Friedmann equation relates the acceleration of this expansion term, a double dot-- dot, by the way, d by dy. A double dot divided by a is simply related to a quantity that is the density plus 3 times the pressure of this perfect fluid that makes up our universe. We also find by requiring that local energy conservation be held, in other words that your stress energy tensor be divergence free, we have a constraint that relates the amount of energy-- the rate of change of energy in a fiducial volume-- to the negative pressure times the rate of change of that fiducial volume. And this, as I discussed in the last lecture, is essentially nothing more than the first law of thermodynamics. It's written up in fancy language appropriate to a cosmological spacetime. As we move forward, we find it useful to make a couple of definitions. So if you divide the Hubble parameter squared by Newton's gravitational constant, that's got the dimensions of density. And so we're going to define a critical density to be 3h squared over 8 pi g. And we're going to define density parameters, omega, as the actual physical density is normalized to that critical density. And when you do this, you find that the critical-- the first Friedmann equation can be written as omega-- oh, that's a typo-- omega plus omega curvature equals 1 where omega curvature-- pardon the typo here. Omega curvature is not actually related to identity but it sort of plays one in this equation. It is just a parameter that has the proper dimensions to do what is necessary to fit this equation. And it only depends on what the curvature parameter is. So remember that this kappa is essentially minus 1, 0, or 1 module of factor of my overall scale. So this is either a positive number, 0, or a negative number. All right. So let's carry things forward from here. Mute my computer so I'm not distracted by things coming in. We now have everything we need using this framework to build a universe. Let's write down the recipe to build a universe, or I should say to build a model of the universe. So first thing you do is pick your spatial curvature. So pick the parameter k to be minus 1, 0, or 1. Pick a mixture of species that contribute to the energy density budget of your model universe. So what you would say is that the total density of things in your universe is a sum over whatever mixture of stuff is in your universe. You will find it helpful to specify an equation of state for each species. Cosmologists typically choose equation of state that has the following form. OK. So you require your species-- if you follow this model that most cosmologists use, each species will have a pressure that is linear in the density. If you do choose that form, then when you enforce local conservation of energy, what you will then find is that for every one of your species, there is a simple relationship. There's a simple differential equation that governs how that species evolves as the scale factor changes. This can be immediately integrated up to find that the amount of, at some particular moment, the density of species i depends on how it looks. So 0 again denotes now. It is simply proportional to some power of the scale factor where the power that enters here can be simply calculated given that equation of state parameter. Once you have these things together, you're ready to roll. You've got your Friedmann equations. You've got all the constraints and information you need to dig into these equations. Sit down, make your models, have a party. Now what we really want to do-- OK. We are physicists. And our goal in doing these models is to come up with some kind of a description of the universe and compare it with our data so that we can see what is the nature of the universe that we actually live in. OK. So what is of interest to us is how does varying all these different terms change-- well, we're going to talk about various observables. But really, if you think about this model, how does it change the evolution of the scale factor? Everything is bound up in that. That is the key thing. OK. So if I can make a mathematical model that describes a universe with all these various different kinds of ingredients, yeah, I can sit down and make-- if I make a really complicated thing, I probably can't do this analytically so that's fine. I will make a differential equation integrator that solves all these coupled differential equations. And I will then make predictions for how AFT evolves depending upon what the different mixtures of species are, what the curvature term is equal to, all those together, and make my model. But my goal as a physicist is then to compare this to data. And so what I need to do is to come up with some kind of a way this will then be useful. We then need some kind of an observational surrogate for a of t. What I would like to be able to do is say, great, model A predicts the following evolution of the scale factor. Model B predicts this evolution of the scale factor. Can I look at the universe and deduce whether we are closer to model A or closer to Model B? And in order to do that, I need to know how do I measure a of t. And as we're going to see, this really boils down to two things. I need to be able to deduce if I look at some event in the universe, if I look at something, I want to know what scale factor to associate with that. I need to measure a. And I need to know what t to label that a with. So this kind of sounds like I'm reading from the journal of duh. But if I want to do this, what it basically boils down to is I need to know how to associate an a with things that I measure and how to associate the t with the things that I measure. That's what we're going to talk about today. What are the actual observational surrogates, the ways in which we can go out, point telescopes and other instruments at things in the sky, and deduce what a of t is for the kind of events, the kind of things that we are going to measure? Let's talk about how I can measure the scale factor first. OK. So let's ignore the fact there's a t on here. How can I measure a? We saw a hint of this in the previous lecture that I recorded. So recall-- pardon me. I've got a bit of extra junk here in my notes. Recall that in my previous lecture, we looked at the way different kinds of densities behaved. But I've got the results right here. For matter which had an equation state parameter of 0, what we found was that the density associated with that matter, it just fell as the scale factor to the inverse third power. That's essentially saying that the number of particles of stuff is constant. And so as the universe expands, it just goes with the volume of the universe. If it was radiation, we found it went as a scale factor to the inverse fourth power. And that's consistent with diluting the density as the volume gets larger provided we also decrease the energy per particle of radiation per photon, per graviton, per whateveron. If I require that the energy per quantum of radiation is redshifted with this thing, that explains the density flaw that we found for radiation. And so that sort of suggests that what we're going to find is that the scale factor is directly tied to a redshift measure. OK. I just realized what this page of notes was. My apologies. I'm getting myself organized here. Let's make that a little bit more rigorous now. OK. So that argument on the basis of how the density of radiation behaves. It's not a bad one as a first pass. It is quite indicative. But let's come at it from another point of view. And this allows me to introduce in a brief aside a topic that is quite useful here. So we talked about Killing vectors a couple of weeks ago. Let's now talk about a generalization of this known as Killing tensors. So recall that a Killing vector was defined as a particular vector in my space time manifold such that if I Lie transport the metric along that Killing vector, I get 0. This then leads to the statement that if I put together the symmetrized covariant gradient of the Killing vector, I get 0. Another way to write this is to use this notation. Whoops. OK. So these are equations that tell us about the way that Killing vectors behave. A Killing tensor is just a generalization of this idea to an object that has more than one index line, to a higher rank tensorial object. So we consider this to be a rank one Killing tensor. A rank n Killing tensor satisfies-- so let's say k is my Killing tensor. Imagine I have n indices here if I take the covariant gradient of that Killing tensor and I symmetrize over all n indices. That gives me 0. This defines a Killing tensor. Starting with this definition, it's not at all hard to show that if I define a parameter, k, which is what I get when I contract the Killing tensor, every one of its indices with the four-velocity of a geodesic. If my u satisfies the geodesic equation-- or this could be-- let's write this as a momentum. Which you say is the tangent to a world line. Could be either a velocity or a momentum. So if I define the scalar k by contracting my Killing tensor with n copies of tangent to the world line, and that thing satisfies the geodesic equation, then the following is true. You guys did this on a homework exercise for when we thought about a spacetime-- you did something similar to this, I should say, for a spacetime containing an electromagnetic field. We talked about how this works for the case of a Killing vector. Hopefully you can kind of see the way you would do this calculation at this point. Now the reason I'm doing this aside is that if you have a Friedmann-Robertson-Walker spacetime, search spacetimes actually have a very useful Killing tensor. So let's define k with two indices, mu nu. And this is just given by the scale factor. Multiplying the metric u mu, u mu, u nu. Where this u comes from the four-velocity of a co-moving fluid element. So this is the four-velocity that we use to construct the stress energy tensor that is the source of our Friedmann equations. So here's how we're going to use this. Let's look at what we get for this Killing vector. Excuse me, this Killing tensor when I consider it's a long a null geodesic. We're going to want to think about null geodesics a lot, because the way that we are going to probe our universe is with radiation. We're going to look at it with things like telescopes. These days people are starting to probe it with things like gravitational wave detectors. All things that involve radiation that moves on null geodesics. So let's examine the associated conserved quantity that is associated with a null geodesic. So let's say v-- let's make it a p, actually. So it's going to be a null geodesic, so we're going to imagine it's radiation that is following. It has a four-momentum, pu. And let's define k, case of ng, from my null geodesic. That is going to be k mu nu, p mu, p nu. Let's plug-in the definition of my Killing tensor. So this is a square root of t, g mu nu, p nu, p nu. This is zero. It's a null geodesic. Then I get u mu p nu, u nu p mu. Now remind you of something. Go back to a nice little Easter egg, an exercise you guys did a long time ago. If I look at the dot product of a four-momentum and a four-velocity, what I get is the energy associated with that four-momentum as measured by the observer whose four-velocity is u. So what we get here is two copies of the energy of that null geodesic measured by the observer who is co-moving. So what this null geodesic-- what this quantity associated with this null geodesic is two powers of the scale factor times the energy that would be measured by someone who is co-moving with the fluid that fills my universe. Energy of p mu, as measured by u mu. And remember, this is a constant. So as this radiation travels across the universe-- as this radiation travels across the universe, the product of the scale factor and the energy associated with that radiation as measured by co-moving observers is a constant. So this is telling us that the energy, as measured by a co-moving observer, let's say it is emitted at some time with a scale factor is a. When it propagates to us, we define our scale factor as 1, the energy will have fallen down by a factor of 1 over a. So this makes it a little bit more rigorous, this intuitive argument that we saw from considering how the density of radiation fell off. What we see is the energy is indeed redshifting with the scale factor. So if I use the fact that the energy that I observe-- if I'm measuring light, light has a frequency of omega-- what I see is the omega that I observe at my scale factor, which I define to be 1, normalized to that when it was emitted, it looks like the scale factor when it was admitted. Divided by a now, a observed. I call this 1. I can flip this over, another way of saying this is that, if I write it in terms of wavelengths of the radiation, the wavelength of the radiation and when it was emitted versus the wavelength that we observe it tells me about the scale factor when the radiation was emitted. Astronomers like to work with redshift. They like to work with wavelength when they study things like the spectra of distant astronomical objects. And they use it to define a notion of redshift. So we define the redshift z to be the wavelength that we observe, minus the wavelength at the radiation that when it is emitted divided by the wavelength when it was emitted. Put all of these definitions together, and what this tells me is that the scale factor at which the radiation was emitted is simply related to the redshift that we observe. So this at last gives us a direct and not terribly difficult to use observational proxy that directly encodes the scale factor of our universe. Suppose we measure the spectrum of radiation from some source, and we see the distinct fingerprint associated with emission from a particular set of atomic transitions. What we generally find is some well-known fingerprints of well-characterized transitions, but in general they are stretched by some factor that we call the redshift z. Actually, you usually stretch by-- when you go through this, you'll find that what you measure is actually stretched by 1 plus z. You measure that, you have measured the scale factor at which this radiation was measured-- was emitted. So this is beautiful. This is a way in which the universe hands us the tool by which we can directly characterize some of the geometry of the universe at which light has been emitted. This is actually one of the reasons why a lot of people who do observational cosmology also happen to be expert atomic spectroscopists. Because you want to know to very high precision what is the characteristics of the hydrogen Balmer lines. Some of the most important sources for doing these tend to be galaxies in which there's a lot of matter falling onto black holes, some of the topics we'll be talking about in an upcoming video. As that material falls in, it gets hot, it generates a lot of radiation, and you'll see things like transition lines associated with carbon and iron. But often all reddened by a factor of several. You sort of go, oh, look at that, carbon falling onto a black hole at redshift 4.8. That is happening at a time when the scale factor of the universe was 1 over 4.8-- or 1 over 5.8, forgot my factor of 1 plus there. So you measure the redshift, and you have measured the scale factor. But you don't know when that light was emitted. We need to connect the scale factor that we can measure so directly and so beautifully to the time at which it was emitted. We now have a way of determining a, but we need a as a function of t. And in truth, we do this kind of via a surrogate. Because we are using radiation as our tool for probing the scale factor, we really don't measure t directly. When we look at light and it's coming to us, it doesn't say I was emitted on March 27th of the year negative 6.8 billion BC, or something like that. We do know, though, that it traveled towards us at the speed of light on a null geodesic. And because it's a null geodesic, there's a very simple-- simple's a little bit of an overstatement, but there is at least a calculable connection. Because it's moving at the speed of light, we can simply-- I should stop using that word-- we can connect time to space. And so rather than directly determining the time at which the radiation was emitted, we want to calculate the distance of the source from us that emitted it. So rather than directly building up a of t, we're going to build up an a of d, where d is the distance of the source. And if you're used to working in Euclidean geometry, you sort of go, ah, OK, great. I know that light travels at the speed of light, so all I need to do is divide the distance by c, and I've got the time, and I build a of t. Conceptually, that is roughly right, and that gives at least a cartoon of the idea that's going on here. But we have to be a little bit careful. Because it turns out when you are making measurements in a curved space time, the notion of distance that you use depends on how you make the distance measurement. So this leads us now to our discussion of distance measures in cosmological spacetime. So just to give a little bit of intuition as to what's the kind of calculation we're going to need to do, let me describe one distance measure that is observation, which is about useless, but not a bad thing to at least begin to get a handle on, the way different parameters of the spacetime come in and influence what the distance measure is. So let's just think about the proper distance from us to a source. So let's imagine that-- well, let's just begin by first, let's write down my line element. And here's the form that I'm going to use. OK, so here's my line element. This is my differential connection between two events spaced between one another by dt, d chi, d theta, d phi, all hidden in that angular element, the omega. Let's imagine that we want to consider two sources that are separated purely in the radial direction. So my angular displacement between the two events is going to be 0. So the only thing I need to care about is d chi, and let's imagine that I determine the distance between these two at some instant-- So then you just get ds squared equals a squared or zero squared d chi squared, and you can integrate this up and you get our first distance measure, d sub p equals scale factor. Your overall distance scale are zero and chi. So Carroll's textbook calls this the instantaneous physical distance. Let's think about what this means if you do this. This is basically the distance you would get if you took a yardstick, you put one end at yourself, you're going to call yourself chi equals 0, you put the other end of the yardstick at the object in your universe at some distance chi in these coordinates, and that's the distance. d sub p is the distance that you measure. It is done, you're sort of imagining that both of the events at the end of this yardstick. You're sort of ascertaining their position at exactly the same instant, hence the term instantaneous, and you get something out of it that encodes some important aspects of how we think about distances in cosmology. So notice everything scales with the overall length scale that we associated with our spatial slices with our spatial sector of this metric, the r0. Notice that whatever is going on your scale factor, your a, your distance is going to track that. As a consequence of this, two objects that are sitting in what we call the Hubble flow, in other words, two objects that are co-moving with the fluid that makes up the source of our universe. They have an apparent motion with respect to each other. If I take the time derivative of this, the apparent is just a dot, or 0 chi, which is equal to the Hubble parameter times dp. Recall that the Hubble parameter is a dot over a, and if I'm doing this right now, that's the value of the Hubble parameter now. So this is the Hubble Expansion Law, the very famous Hubble Expansion Law. So we can see it hidden in this-- not even really hidden, it's quite apparent in this notion of an instantaneous physical distance. Let me just finally emphasize, though, that the instantanaeity that is part of this object's name, instantaneous measurements, are not done. As I said-- I mean, this is it sounds like I'm being slightly facetious, but it's not. The meaning of this distance measure is, like I said, it's a yardstick where I have an event at me, the other end of my yardstick is at my cosmological event. Those are typically separated by millions, billions of light years. Even if you could-- OK, the facetious bit was imagining it as a yardstick. But the non-facetious point I want to make is we do not make instantaneous measurements with that. When I measure an event that is billions of light years away, I am of course measuring it using light and I'm seeing light that was emitted billions of years ago. So we need to think a little bit more carefully about how to define distance in terms of quantities that really correspond to measurements we can make. And to get a little intuition, here are three ways where, if you were living in Euclidean space and you were looking at light from distant objects, here are three ways that you could define distance. So if spacetime were that of special relativity-- well, let's just say if space were purely Euclidean. Let's just leave it like that. Here are three notions that we could use. One, imagine there was some source of radiation in the universe that you understood so well that you knew its intrinsic luminosity. What you could do is compare the intrinsic luminosity of a source to its apparent brightness. So let's let f be the flux we measure from the source. This will be related to l, the luminosity, which-- suspend disbelief for a moment, we want to imagine that we know it for some reason. And if we imagine this is an isotropic emitter, this will be related by a factor of 4 pi, and the distance between us and that source. Let's call this d sub l. This is a luminosity distance. It is a distance that we measure by inferring the behavior of luminosity of a distant object. Now it turns out, and this is a subject for a different class, but nature actually gives us some objects whose luminosity is known or at least can be calibrated. In the case of much of what is done in cosmology today, we can take advantage of the behavior of certain stars whose luminosity is strongly correlated to the way that-- these are stars whose luminosity is variable, and we can use the fact that their variability is correlated to their luminosity to infer what their absolute luminosity actually is. There are other supernova events whose luminosity likewise appears to follow a universal law. It's related to the fact that the properties of those explosions are actually set by the microphysics of the stars that set them. More recently, we've been able to exploit the fact that gravitational wave sources have an intrinsic luminosity in gravitational waves, the dedt associated with the gravitational waves that they emit, which depends on the source gravitational physics in a very simple and predictable way that doesn't depend on very many parameters. I actually did a little bit of work on that over the course of my career, and it's a very exciting development that we can now use these as a way of setting the intrinsic luminosity of certain sources. At any rate, if you can take advantage of these objects that have a known luminosity and you can then measure the flux of radiation in your detector from these things, you have learned the distance. At least you have learned this particular measure of the distance. That's measure one. Measure two is imagine you have some object in the sky that has a particular intrinsic size associated with it. You can sort of think of the objects whose luminosity you know about as standard candles. Imagine if nature builds standard yardsticks, there's some object whose size you always know. Well, let's compare that physical size to the angular size that you measure. The angle that the object sub tends in the sky is going to be that intrinsic size, delta l divided by the distance. We'll call this distance d sub a, for the angular diameter distance. Believe it or not, nature, in fact, provides standard yardsticks type so that we can actually do this. Finally, at least as a matter of principle, imagine you had some object that's moving across the sky with a speed that you know. You could compare that transverse speed to an apparent angular speed. So the theta dot, the angular speed that you measure, that would be the velocity perpendicular to your line of sight divided by distance. We'll call this d sub m, the proper motion distance. So if our universe were Euclidean, not only would it be easy for us to use these three measures of distance, all three of them would give the same result. Because this is all just geometry. Turns out when you study these notions of distance, in an FRW spacetime, there's some variation that enters. Let me just emphasize here that there is an excellent summary on this stuff. I can't remember if I linked this to the course website or not. I should and I shall. An excellent summary of all this, really emphasizing observationally significant aspects of these things come from the article that is on the archive called Distance Measures in Cosmology by David Hogg, a colleague at New York University. You can find this on the Astro PH archive, 9905116. It's hard for me to believe this is almost 21 years old now. This is a gem of a paper. Hogg never submitted it to any journal, just posted it on the archive so that the community could take advantage of it. So the textbook by Carroll goes through the calculation of d sub l. On a problem set you will do d sub m. We're going to go through d sub a, just so you can see some of the way that this works. Let me emphasize one thing, all of these measures use the first Friedmann equation. So writing your Friedmann equation like so. i is a sum of all the different species of things that can contribute including curvature. So recall that even though curvature isn't really a density, you can combine enough factors to make it act as though it were a density. You assume a power law. So this n sub i is related to the equation of state parameter for each one of these species. And let's now take advantage of the fact that we know the scale factor directly ties to redshift. I can rewrite this as how the density evolves as a function of redshift. So when you put all of this together, this allows us to write h of a as h of z. This is given by the Hubble parameter now times some function e of z. And that is simply-- you divide everything out to normalize to the critical density, and that is a sum over all these densities with the redshift waiting like so. Let's really do an example, like I said. So if you read Carroll, you will see the calculation of the luminosity distance. If you do the cosmology problem set that will be-- I believe its p set 8, you will explore the proper motion distance. So let's do the angular diameter distance. Seeing someone work through this, I think, is probably useful for helping to solidify the way in which these distance measures work and how it is that one can tie together important observables. So let's consider some source that we observe that, according to us, subtends an angle, delta phi. Every spacial sector is spherically symmetric, and so we can orient our coordinate system so that this thing, it would be sort of a standard ruler-- what we're going to do is orient our coordinate system so that object lies in the theta equals pi over 2 plane. The proper size of that source-- so the thing is just sitting in the sky there-- the proper size of the source, you can get this in the line element. The delta l of the source will be the scale factor at the time at which it is emitted, r0. So this is using one of the forms of the FRW line elements I wrote down at the beginning of this lecture. And so the angular diameter distance, that's a quantity that I've defined over here, it's just the ratio of this length to that angle. Let's rewrite this using the redshift. Redshift is something that I actually directly observe, so there we go. This is not wrong, but it's flawed. So this is true. This is absolutely true. Here's the problem, I don't know the overall scale of my universe and this coordinate chi doesn't really have an observable meaning to it. It's how I label events, but I look at some quasar in the sky and I'm like, what's your chi? So what we need to do is reformulate the numerator of this expression in such a way as to get rid of that chi and then see what happens, see if we have a way to get rid of that r0. Let's worry about chi first. We're going to eliminate it by taking advantage of the fact that the radiation I am measuring comes to me on a null path. Not just any null path. I'm going imagine it's a radial one. We are allowed to be somewhat self-centered in defining FRW cosmology, we put ourselves at the origin. So any light that reaches us moves on a purely radial trajectory from its source to us. So looking at how the time and the radial coordinate chi are related for a radial null path we go into our FRW metric. I get this. So I can integrate this up to figure out what chi is. So this is right. Let's massage it a little bit to put it in a form that's a little bit more useful to us. Let's change our variable-- change our variable of integration from time to a. So this will be an integral from the scale factor at which the radiation is emitted to the scale factor which we observe it, i.e. now. And when you do that change of variables your integral changes like so. Let's rewrite this once more to insert my Hubble parameter. Now let's change variables once more. We're going to use the fact that our direct measurable is redshift. And so if we use a equals 1 over 1 plus z, I can further write this as an integral over redshift like so. And that h0 can come out of my integral. So this is in a form that is now finally formulated in terms of an observable redshift and my model dependent parameters. The various omegas that, when I construct my universe model, I am free to set. Or if I am a phenomenologist, that are going to be knobs that I turn to try to design a model universe that matches the data that I am measuring. r0, though, is still kind of annoying. We don't know what this guy is, so what we do is eliminate r0 in favor of a curvature density parameter. So using the fact that omega curvature-- go back to how this was originally defined-- it was negative kappa over h0 squared. That's negative k over r0 squared h0 squared. That tells me that r0 is the Hubble constant now, divided by the square root of the absolute value of the curvature, at least if k equals plus or minus 1. What happens when it's not plus or minus 1, if it's equal to 0? Well, hold that thought. So let's put all these pieces together. So assembling all the ingredients I have here, what we find is the angular diameter distance. There's a factor of 1 over 1 plus z, 1 over the Hubble constant. Remember, Hubble has units of 1 over length-- excuse me, 1 over time, and with the factor of the speed of light that is a 1 over length, so one over the Hubble parameter now is essentially a kind of fiducial overall distance scale. And then our solution breaks up into three branches, depending upon whether k equals minus 1, 0, or 1. So you get one term where it involves minus square root the absolute value of the curvature parameter times sine. That same absolute value of the curvature parameters square root. So here's your k equals plus 1 branch. For your k equals 0 branch, basically what you'll find when you plug-in your s of k is that r0 cancels out. So that ends up being a parameter you do not need to worry about, and I suggest you just work through the algebra and you'll find that for k equals 0, it simply looks like this. And then finally, if you are in an open universe-- that is supposed to be curvature-- what we get is this. So this is a distance measure that tells me how angular diameter distance depends on observable parameters. Hubble is something that we can measure. Redshift is something we can measure. And it depends on model parameters, the different densities that go into e of z, and-- which I have on the board right here-- the different densities that go into e of z. My apologies, I left out that h0 there-- and your choice of the curvature. When you analyze these three distances here is what you find. You find that the luminosity distance is related to the proper motion distance by a factor of 1 plus z, and that's related to the angular diameter distance by a factor of 1 plus z squared. So when you read Carroll, you will find that 1 plus z factor there-- excuse me, 1 plus z to the minus 1 power turns into a 1 plus z-- is the camera not looking at me? Hello? There we go. So that 1 over 1 plus z turns into a 1 plus z. When you do it on the p set, you do the proper motion distance, so it will just be no 1 plus z factor in front of everything. So the name of the game when one is doing cosmology as a physicist is to find quantities that you can measure that allow you to determine luminosity distances, angular diameter distances, proper motion distances. Now it turns out that the proper motion distance is not a very practical one for basically any cosmologically interesting source. They are simply so far away that even for a source moving essentially at the speed of light, the amount of angular motion that can be seen over essentially a human lifetime is negligible. So this turns into something that-- hi. MIT POLICE: [INAUDIBLE] SCOTT HUGHES: That's OK. Yeah, I'm doing some pre-recording of lectures. [LAUGHS] I was warned you guys might come by. I have my ID with me and things like that, so. MIT POLICE: That's fine. Take care. MIT POLICE: You look official. SCOTT HUGHES: [LAUGHS] I appreciate it. So those of you watching the video at home, as you can see, the MIT police is keeping us safe. Scared the crap out of me for a second there, but it's all good. All right, so let's go back to this for a second. So the proper motion distance is something that is not particularly practical, because as I said, even if you have an object that is moving close to the speed of light this is not something that even over the course of a human lifetime you are likely to see significant angular motion. So this is generally not used. But luminosity distances and angular diameter distances, that is, in fact, extremely important, and a lot of cosmology is based on looking for well understood objects where we can calibrate the physical size and infer the angular diameter distance, or we know the intrinsic brightness and we can determine the luminosity distance. So let me just give a quick snapshot of where the measurements come from in modern cosmology that are driving our cosmological model. One of the most important is the cosmic microwave background. So, vastly oversimplifying, when we look at the cosmic microwave background after removing things like the flow of our solar system with respect to the rest frame of-- excuse me, the co-moving reference frame of the cause of the fluid that makes up our universe, the size of hot and cold spots is a standard ruler. By looking at the distribution of sizes that we see from these things, we can determine the angular diameter distance to the cosmic microwave background with very high precision. This ends up being one of the most important constraints on determining what the curvature parameter actually is. And it is largely thanks to the cosmic microwave background that current prejudice, I would say, the current best wisdom-- choose your descriptor as you wish-- is that k equals 0 and our universe is in fact spatially flat. Second one is what are called type 1a supernova. These are essentially the thermonuclear detonations of white dwarfs. Not even really thermonuclear, it's just-- my apologies, I'm confusing a different event that involves white dwarfs. The type 1a's are not the thermonuclear explosions of these things. This is actually the core collapse of a white dwarf star. So this is what happens when a white dwarf it creates or in some way accumulates enough mass such that electron degeneracy pressure is no longer sufficient to hold it against gravitational collapse, and the whole thing basically collapses into a neutron star. That happens at a defined mass, the Chandrasekhar mass, named after one of my scientific heroes, Subramanian Chandrasekhar. And because it has a defined mass associated with it, every event basically has the same amount of matter participating. This is a standard candle. So there's a couple others that I'm not going to talk about in too much detail here. While I'm erasing the board I will just mention them. So by looking at things like the clustering of galaxies we can measure the distribution of mass in the universe that allows us to determine the omega m parameter. That's one of the bits of information that tells us that much of the universe is made of matter that apparently does not participate in standard model processes as we know them today-- the so-called dark matter problem. We can look at chemical abundances, which tells us about the behavior of nuclear processes in the very early universe. And the last one which I will mention here is we can look at nearby standard candles. And nearby standard candles allow us to probe the local Hubble law and determine h0. "Probble," that's not a word. If you combine "probe" and "Hubble" you get "probble." And when I say nearby, that usually means events that are merely a few tens of millions of light years away. It's worth noting that all of these various techniques, all of these different things, you can kind of even see it when you think about the mathematical form of everything that went into our distance measures, they're all highly entangled with each other. And so to do this kind of thing properly, you need to take just a crap load of data, combine all of your data sets, and do a joint analysis of everything, looking at the way varying the parameters and all the different models affects the outcome of your observables. You also have to carefully take into account the fact that when you measure something, you measure it with errors. And so many of these things are not known. Turns out you can usually measure redshift quite precisely, but these distances always come with some error bar associated with them. And so that means that the distance you associate with a particular redshift, which is equivalent to associating a time with a redshift, there's some error bar on that. And that can lead to significant skew in what you determine from things. There's a lot more we could say, but time is finite and we need to change topic, so I'm going to conclude this lecture by talking about two mysteries in the cosmological model that have been the focus of a lot of research attention over the past several decades. Two mysteries. One, why is it that our universe appears to be flat, spatially flat? So to frame why this is a bit of a mystery, you did just sort of go, eh, come on. You've got three choices for the parameter, you got 0. Let's begin by thinking about the first Friedmann equation. I can write this like so, or I can use this form, where I say omega plus omega curvature-- I'm going to call that omega c for now-- that equals 1. The expectation had long been that our universe would basically be dominated by various species of matter and radiation for much of its history, especially in the early universe. If it was radiation dominated, you'd expect the density to go as a to the minus 4. If it's matter dominated, you expect it to go as a to the minus 3. Now, your curvature density goes as a to the minus 2. And so what this means is that if you look at the ratio of omega curvature to omega, this will be proportional to a, for matter, a squared for radiation. If your universe-- in some sense, looking at these parameter k of the minus 1, 0, 1, that's a little bit misleading. It's probably a little bit more useful to think about things in terms of the kappa parameter. And when you look at that, your flat universe is a set of measure 0 in the set of all possible curvature parameters that you could have. And physicists tend to get suspicious when something that could take on any range of possible random values between minus infinity and infinity picks out zero. That tends to tell us that there may be some principle at play that actually derives things to being 0. Looking at it this way, imagine you have a universe that at early times is very close to being flat, but not quite. Any slight deviation from flatness grows as the universe expands. That's mystery one. Mystery two, why is the cosmic microwave background so homogeneous? So when we look at the cosmic microwave background, we see that it has the same properties. The light has the same brightness, it has the same temperature associated with it, to within in a part in 100,000. Now the standard model of our universe tells us that at very early times the universe was essentially a dense hot plasma. This thing cooled as the universe expanded, much the same way that if you have a bag of gas, you squeeze it very rapidly, it will get hot, you stretch it very rapidly, it will cool. There's a few more details of this in my notes, but when we look at this one the things that we see is that in a universe that is only driven by matter or by radiation-- so the matter dominated and radiation dominated picture-- it shows us that points on opposite sides of the sky were actually out of causal contact with each other in the earliest moments of the universe. In other words, I look at the sky, and the patch of sky over here was out of causal contact with the patch of sky over here in the earliest days of the universe. And yet they had the same temperature, which suggests that they were in thermal equilibrium. How can two disparate, unconnected patch of the sky have the same temperature if they cannot exchange information? You could imagine it being a coincidence if one little bit of the sky has the same temperature as a bit of another piece, but in fact, when you do this calculation, you find huge patch of the sky could not communicate with any other. And so how then is it that the entire sky that we can observe has the same temperature at the earliest times within a part in 100,000. You guys will explore this on-- I believe it's problem set eight. The solution to both of these problems that has been proposed is cosmic inflation. So what you do is imagine that at some earlier moment in the universe, our universe was filled with some strange field, and I'll describe the properties of that field in just a moment, such that it acted like it had a cosmological constant. In such an epic, the scale factor of the universe goes as exponentially with the square root of the size of the cosmological constant. What you find when you look at this is that this-- it still, of course, goes as a to the minus 2-- but a to the minus-- well, sorry. Let me back up for a second. So my scale factor in this case you'll find goes are a to the minus 2, and that's because the density associated with the cosmological constant remains constant. So even if you start in the early universe with some random value for the curvature, if you are in this epic of exponential inflation, just for-- you know, you have to worry about timescales a little bit, but if you do it long enough you can drive this very, very close to 0. So much so that when you then move forward, let's say you come out of this period of cosmic inflation and you enter a universe that is radiation dominated or matter dominated, it will then begin to grow, but if you drive it sufficiently close to 0 it doesn't matter. You're never going to catch up with what inflation did to you. On the P set you will also show that if you have a period of inflation like this, then that also cures the problem of piece of the sky being out of causal contact. So when you do that, what you find is that essentially everything is in causal contact early on. It may sort of come out of causal contact after inflation has ended, more on that in just a moment, and then things sort of change as the universe continues to evolve. OK so it looks like recording is back on. My apologies, everyone. So as I was in the middle of talk, I talked a little bit too long in this particular lecture so we're going to spill over into a little bit of an addendum, just a five-ish minute piece that goes a bit beyond this. Doing this by myself is a little bit weird, I'm tired, and I will confess, I got a little bit rattled when the police came in to check in on me. Let's back up for a second. So I was talking about two mysteries of the modern cosmological model. One of them is this question of why the universe is so apparently flat. The spatial sector of the universe appears to be flat. And we had this expectation that the universe is either radiation dominated or matter dominated, which would give us the density associated with radiation. If it was radiation dominated, then the density of stuff in our universe would fall off as the scale factor to the fourth power. If it's matter dominant, it's scale factor to the third power. When you define a density associated with the curvature, it falls off as scale factor to the second power. And so if we look at the ratio of the curvature density to any other kind of density, it grows as the universe expands. So any slight deviation from flatness we would expect to grow. And that's just confusing. Why is it when we make the various measurements that we have been making for the past several decades, all the evidence is pointing to a universe that has a flatness of 0? If you sort of imagine that the parameter kappa can be any number between minus infinity to infinity, why is nature picking out 0? Another mystery is why is the cosmic microwave background so homogeneous? We believe that the universe was a very hot dense plasma at very early times. It cooled as the universe expanded, and when we measured the radiation from that cooling expanding ball of plasma, what we find is that it has the same temperature at every point in the sky to within a part in 100,000. But when one looks at the behavior of how light moves around in the early universe, if the universe is matter dominated or radiation dominated, what you find is that a piece of the sky over here cannot communicate with a piece of sky over here. Or over here, or over here. You actually find that the size of the sky that, if I look at a piece of sky over here, how much of the universe could it talk to, it's surprisingly small. So how is it that the entire universe has the same temperature? How is that they are apparently in thermal equilibrium, even if they cannot exchange information? So I spent some while talking to myself after the cameras went out. So I'll just sketch what I wrote down. A proposed solution to this, to both of these mysteries, is what is known as cosmic inflation, something that our own Alan Guth shares a lot of the credit for helping to develop. So recall, if we have a cosmological constant, then the scale factor grows exponentially and the density of stuff associated with that cosmological constant is constant. As the universe expands, the energy density associated with that cosmological constant does not change. If our universe is dominated by such a constant, then the ratio of density is associated with curvature, the density associated with cosmological constant, actually falls off inversely with the curvature scale squared, and the curvature scale is growing exponentially, it means that omega c is being driven to zero relative to the density in cosmological constant, as e to a factor like this. It's being exponentially driven close to 0. You'd do do a little bit of work to figure out what the timescales associated with this are, but this suggests that if you can put the universe in a state where it looks like a cosmological constant, you can drive the density associated with curvature as close to 0 as you want. Recall that a cosmological constant is actually equivalent to there being a vacuum energy. If we think about the universe being filled with some kind of a scalar field at early times, it can play the role of such a vacuum energy. Without going into the details, one finds that in an expanding universe there is an equation of motion for that scalar field. How the field itself behaves is driven by a differential equation, looks like this. Here's your Hubble parameters, so this has to do with a scale factor in here. v is a potential for this scalar field, which I'm not going to say too much about. Take this guy, couple it to your Friedmann equations, and the one that's most important is the first Friedmann equation. What you see is that v of phi is playing the role of a cosmological constant. It's playing the role of a density. So if we can put the universe into a state where it is in fact being dominated by this scalar field, by the potential associated with the scalar field, it will inflate. A lot of the research in early universe physics that has gone on over the past couple decades has gone into understanding what are the consequences of such a such a potential. Can we make such a potential? Do the laws of physics permit something like this to exist? If the universe is in this state early on, what changed? How is it that this thing evolves? Is there is an equation of motion here so that scalar field is presumably evolving in some kind of a way? Is there a potential, does nature permit us to have a field of the sort with a potential that sort of goes away after some time? Once it goes away, what happens to that field? Is there any smoking gun associated with this? If we look at the universe and this is a plausible explanation for why the universe appears to be spatially flat and why the cosmic microwave background is so homogeneous. Is there anything else that we can look at that would basically say yes, the universe did in fact have this kind of an expansion? Without getting into the weeds too much, it turns out that if the universe expanded like this, we would expect there to be a primordial background of gravitational waves, very low frequency gravitational waves. Sort of a moaning background filling the universe. And so there's a lot of experiments looking for the imprints of such gravitational waves on our universe. If it is measured, it would allow us to directly probe what this inflationary potential actually is. I'm going to conclude our discussion of cosmology here. So some of the quantitative details, the way in which inflation can cure the flatness problem and cure the homogeneity problem you will explore on problem set seven. Going into the weeds of how one makes it a scalar field, designs a potential and does things like this beyond the scope of 8.962-- there are courses like this, and I would not be surprised if some of the people in this class spent a lot more time studying this in their futures than I have in my life. All right, so that is where we will conclude our discussion of cosmology.
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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. BOGDAN FEDELES: Hi, and welcome to 5.07 Biochemistry Online, the biochemistry course on MIT OpenCourseWare. I'm Dr. Bogdan Fedeles. Let's metabolize some problems. Now today we're going to do a really nice problem. This is Problem 1 in Problem Set 2. Now, this is a problem about elucidating the primary structure of a protein. Problems like this one that we're about to discuss are a lot of fun because they're really biochemistry puzzles. We're given a number of pieces of data or clues, if you want, and then we'd have to use, not only our biochemical sense, but also deduction and elimination process in order to come up with a final answer. Well, in practice, elucidating the primary structure of a protein is now a largely automated process and utilizes high resolution mass spectrometry. Some of the traditional chemical methods that we're discussing here are still occasionally useful. But most importantly, the logical step-wise process through which we analyze and use each piece of data to construct a big picture result is really representative of the process by which we make discoveries in biochemistry. Before we begin, let me just say that this problem assumes familiarity with the structures and abbreviations of the 20 natural amino acids. Feel free to pause this video and review the relevant chapters in the book and the lecture notes before continuing. One important tool that we have for elucidating the primary structure of proteins is proteases. Proteases are enzymes that can hydrolyze the peptide bonds of a polypeptide chain. Now, proteases that can cleave in the middle of a polypeptide chain are also called endopeptidases. Now you notice a lot of these names that end in "ase" denote enzymes, and peptidase means an enzyme that acts on a peptide. It's the enzyme that hydrolyzes the peptide bond. Endo, in this case, refers to the fact that it acts in the middle of a polypeptide chain. Now, we're going to be learning in this problem about trypsin and chymotrypsin. These are proteases that cleave in the middle of a polypeptide chain. They are endopeptidases. One important feature of proteases is that they are specific. They don't just cut any which one peptide bond, but rather they recognize a specific sequence of amino acids. In the case of trypsin, for example, we are told that it cleaves adjacent to positively charged amino acid. As you know, positively charged amino acids would be lysine or arginine. So trypsin will always be cutting after arginine or lysine. Let's take a look. Now, here is a polypeptide chain with amino acid residues, R1, R2, R3, R4. Now, let's look at this peptide bond right here in the middle of the chain. Now let's say in order for trypsin to cut, to hydrolyze this peptide bond, it means that R2, the amino acid residue adjacent to it, should be a positively charged one. So if R2 is lysine or arginine, then this bond here becomes a good substrate for trypsin. And what it's going to do, it's going to use a water molecule-- it's going to put here trypsin-- and it's going to hydrolyze forming a carboxyl end and an amine end of this peptide bond. So we're getting this is a carboxyl end and this is the amine end of that original peptide bond. So what I want you to remember is that if we're having a reaction with trypsin on a polypeptide chain, then the carboxy end of the peptide that results, which is this particular amino acid, is going to be one of either lysine or arginine. So in other words in a trypsin digest, all the smaller peptides that we obtain are going to end in arginine or lysine except, of course, the very end of the chain, which might have a different amino acid at its carboxyl end. Now, this consideration we've just made about the trypsin digest, in fact, answers part 1 of the problem, which is asking what is common about all these peptides generated by a trypsin digest. And as we've just explained, all these peptides should be ending with a positively charged amino acid such as lysine or arginine. This problem also mentions another protease chymotrypsin, which is a protease that has a different specificity from trypsin. It actually cleaves after amino acids that are either very hydrophobic and large or aromatic. So let's write that down and remember. So if we were to do a digest with chymotrypsin, then our R2, the residue that's recognized by the protease, is going to have to be either aromatic, so phenylalanine, tyrosine, or tryptophan, or something that's large and hydrophobic such as leucine, isoleucine, or even valine sometimes. Put this in parentheses. So if any of these residues are at this position, R2, then this protease, chymotrypsin, is going to generate two peptides. And once again, the resulting peptide, R2 at the carboxy end is going to be one of these amino acids. That's going to be the signature of a chymotrypsin digest. Now, keeping these things in mind, we're ready to tackle the rest of the problem. Question 2 asks about the use of DTT, or dithiothreitol. Now, DTT is a commonly used reducing agent, which can reduce disulfide bridges in proteins. There are many reasons why we want to use DTT. For example, when proteins form disulfide bridges, you may shield certain amino acids from being accessible by proteases, and therefore, they're not going to be cleaved and we'll get a mixture of peptides. So it makes our analysis and our results very difficult. Now, disulfide bridges can also hold two peptides together that have no other covalent attachment between them. So in that case, we get one fragment instead of two fragments, and once again, it complicates our analysis of the protein. But more importantly, because we're typically purifying proteins in the air, in an oxygen atmosphere, proteins can acquire disulfide bonds, which weren't there in the beginning. So in that case, we can get very unusual results, unreproducible and artefactual. That's why using DTT can prevent formation of spurious disulfide bridges. And finally, DTT is also useful to tell if there were any disulfide bridges in the protein to begin with. Because if we're looking at the analysis before and after using DTT, we can tell if the results change, and that will tell us whether disulfide bridges were there to begin with. Now, let's take a look at the DTT chemistry. Here we have a disulfide bond or bridge in a protein, and we're going to treat this with DTT, which looks like this. This is DTT or dithiothreitol. Now, if we substitute the SH groups with OH's, you notice we're going to have four OH's and four carbons. That's just an alcohol derived from a sugar, which is called threose That's why the threitol part of the name. All right. So when we do this chemical reaction, this disulfide bridge is going to transfer between the two sulfur atoms of DTT. They're going to form a intramolecular sulfide bond. So our cysteines are going to get reduced, and from DTT, we're going to get this intramolecular sulfide. And because entropical considerations, make the formation of the six-member ring very, very easy, then this equilibrium will typically shift to the right. Of course, we're also using vast quantities of DTT to make sure that the entire protein becomes reduced, and then our agent is going to pick up this [INAUDIBLE] off the bond. This sums up the question 2 of the problem. Next, we're going to look at distinguishing between a couple of different peptides that we generate during our analysis of our mystery protein. So we're told that we're isolating by HPLC peptides of the following composition. One has tryptophan, phenylalanine, valine, aspartate, lysine, cysteine. That's peptide one. Another one has phenylalanine, serine, cysteine, and an unknown amino acid. Finally, the third one has alanine and lysine. All right. Now, obviously, these peptides have different compositions, so if you could just put them through mass spectrometry, we're going to get different masses, so we can tell very quickly which one is which. But if we don't have mass spec available, we can also tell them apart only using UV-Vis spectroscopy. So all you need to remember is that an amino acid like tryptophan that we have here, W, has a very strong absorption in the 280 to 300 nanometers. Whereas, amino acids like phenylalanine, present here or here, they absorb only around 260 nanometers. Most of the other amino acids don't absorb in this range at all. So if we were to plot the UV-Vis spectra of these peptides, this is going to be our absorption, and this is going to be lambda in nanometers, the wavelength. So let's look in the range from 200 to 400. 300 is about here. 250 is about here. And let's label these. Let's say this peptide is red, this peptide is blue, and this one is green. All right. So now the red peptide, as I told you, contains both tryptophan and phenylalanine, so it's going to absorb both around 260 and both around 280 to 300. So it's going to have a pretty big hump in this area from 250 to 300. Now, the blue peptide only has phenylalanine, so above 280 or so, nanometer is going to drop off. So it's going to look more like this, whereas, the green one, well, it has neither phenylalanine or tryptophan, so below 240 or so, it's going to have no UV absorption at all. So it's going to drop off right here. So basically, if we're just looking, say, around 260 nanometers, we should see a stronger peak from the red peptide and a weaker peak from the blue one. But if we're going to look around 300, we should only see the red peptide. So by just using the UV-Vis absorption, we can tell these three peptides apart. Now we're ready to figure out the structure of our mystery protein. This is exactly what the last part of the problem is asking. So we're going to take each one of the clues provided-- A, B, C, and D-- and analyze and see what information we can derive from each one of them. The first piece of information that we're given is the result of a total hydrolysis of our peptide with six molar HCl. So we're told we get the following amino acid composition. We get two phenylalanines, one methionine, alanine, valine, two lysines, one serine, two cysteine, and one aspartate. Now, recall this is not a comprehensive list because the hydrolysis and six smaller HCl may destroy some of the amino acids. And specifically, we know for sure something like tryptophan, threonine, and tyrosine. They're going to be destroyed, and we're not going to see them here. So these amino acids may still be present in our protein, but we're not going to be able to see them in this situation. Now, let's analyze these amino acids and see what we can derive from this. So first of all, we have cysteines. So two cysteines, it means our protein can form disulfide bridges. So from the get go, two cysteine, it means we can form an internal disulfide bridge or our protein contains a disulfide bridge between two otherwise unconnected peptides. So here are some possibilities. For example, our peptide is one chain, and our cysteines are not actually connected by a disulfide bridge. Another possibility is we do have a disulfide bridge between them, like that. Or yet another possibility is that we have two pieces, two polypeptide chains, and the disulfide bridge is the only thing that's holding them together. Now, in each one of these cases, obviously, these polypeptide chains are oriented. So in one end, they're going to have the amine group. Let me draw it here, NH3. Here we're going to have two, and the other end is going to be the carboxyl group, COO minus. And once again, in this case, we're going to have two of them. All right. Now, how can we narrow down from these couple of possibilities? Well, the fact that we're getting two lysines here is a very important clue. Now remember, we're told that our peptide, this mystery protein, it actually comes as a result of a trypsin digest. And if you recall our discussion earlier, we said that upon trypsin digest, all the smaller peptides that were obtained will end in lysine or an arginine. The fact that we have two lysines here, it means both of them they need to be at the carboxy ends of peptides. Because if they were in the middle of a chain, the trypsin would have cut that chain in half. So two lysines means we've got to have two carboxy ends. So therefore, this seems to be the only possibility in which we can accommodate two lysines. Basically, the two carboxy ends that we see here, each one has to be a lysine. This possibility only has one carboxy end, so the other lysine that we have to place will not be at the end, so would not be compatible with a trypsin digest. Same applies for this case. So these possibilities are not consistent with our data. So we know already that our protein must look like this, two polypeptide chains, each one ends with lysine, and there must be a disulfide bridge. The second clue we're given is that the Edman degradation of the protein yields valine. Now, as you know, Edman degradation is a chemical reaction by which we can digest the protein from the N terminus, from the amino terminus of a polypeptide chain. Now as you recall, we have established in the first part that we have two amino termini in our protein. So the fact that we only get one amino acid and that is valine, it says that the other amino terminus might be blocked or somehow unavailable for the Edman degradation. So let's update the structure of our protein to take into account the second clue. So we said we have two polypeptide chains. There's a disulfide bond in between them. Now, each one of them ends with a lysine. This is a carboxy end, and now we know from the Edman degradation that amino end of one of them has to be valine. The other one-- we're going to put a box like that-- is a blocked end. So the amino terminus is not available. So this is as much as we can tell from these first two clues. We're given the products of the chymotrypsin digest of our protein after it was previously treated with DTT. So we know first DTT is going to cleave the disulfide bond, and then chymotrypsin, as we talked previously, is going to cut after large, hydrophobic, or aromatic amino acids. Now, we're told we're getting five smaller peptides. Let's take a look. Here are the five fragments. One is tryptophan, valine; one is cysteine, phenylalanine, another one is aspartate lysine, another one is methionine, alanine, cysteine, and lysine; and finally, the last one has serine, phenylalanine, and something that's not an amino acid. It's going to make it like a small x. All right. Now, from what we know about chymotrypsin digest, we should be able to orient these peptides basically to tell which amino acid is at the N terminus and which amino acid is at the C terminus of each one of them. Now the first one. Well, we know from the second clue that valine is at the N terminus, so that makes it pretty easy. Then the sequence has to be V-W. So valine is the N terminus, W is the C terminus, and as we said, tryptophan is one of the amino acids recognized by chymotrypsin, so it's going to end up at the carboxy terminus. C-F, that has to go C and F, phenylalanine, another chymotrypsin amino acid that's left of this carboxy terminus. D-K. Now, neither of these amino acids is recognized by chymotrypsin, but we remember from clue one that D-K must be the carboxy terminus. So the sequence can only be D-K. Now, here M, A, C, and K, none of these is actually on our recognition list for chymotrypsin, but once again, we know that K must be in the carboxy end. So for now we're going to have M, A, and C in a particular order, which we cannot establish just yet and K at the carboxy end. And finally, we do an S, F and x. Well, x is not even an amino acid, and F is an amino acid that's left of the carboxy terminus by chymotrypsin, so it's probably x, S, and F. Now, what about x? We're told that x is not an amino acid and is hydrophobic and it has a molecular weight of 256 Daltons. All right. So in order to figure out what x is, we have to read back the beginning of the problem, which tells us that we're looking at a protein that's associated with a plasma membrane. And one way for proteins to associate with a plasma membrane is to be modified, to incorporate a fatty acid, such as palmitate. So perhaps x, which is blocking the N terminus of this peptide is a fatty acid. Now, the general formula for fatty acids is something like this, CH3 CH2 repeated, say, n times, and then COOH. So let's see what would be the fatty acid that has the molecular weight 256 Daltons? Well, the mass of a methyl group is 15. The mass of this carboxylate is 45, so we have about 60 plus 14n equals 256 or 14n equals 196. Therefore, n is 14. So the fatty acid that would fit these criteria-- it's hydrophobic, it has a mass of 256-- will be CH3, CH2 14 times COOH, or the fatty acid was 16 carbons. This is palmitic acid. Now, the answer that we got, palmitic acid, was anticipated in the text of the problem because it gave us an example one way to associate proteins to the plasma membrane is to form a covalent linkage with a fatty acid such as palmitate. But how does a protein associate with a plasma membrane when it has a palmitic acid residue as part of it? Well, let's take a look at a diagram. Here we have a representation of the plasma membrane, where we have the phospholipids like the hydrophilic head pointing inside and outside the cell and the hydrophobic tails of the fatty acids lined up to each other. Now, imagine we have a protein that's modified to contain one of these fatty acids. Then this fatty acid can just insert right next to the phospholipids of the plasma membrane. And that way it tethers the protein right at the plasma membrane. The final clue that we're given in order to figure out the structure of our mystery protein is the digest with an inorganic agent this time. It's called cyanogen bromide. Cyanogen bromide reacts quite specifically with methionines, and it cleaves the peptide bond after methionine leaving behind an unnatural amino acid called a homoserine lactone. Now, let's take a look at what happens when we treat our protein with cyanogen bromide. So we're told we're getting the following peptides, A, K and W, F, V, D, K, C and F, S, C, and an unnatural amino acid. As I just explained to you, the unnatural amino acid probably is this homoserine lactone. So it's most likely, instead of this amino acid in the actual sequence, we had a methionine. So we know these four residues-- F, S, C, and M-- go together. All right. Now, we can start putting together the clues from part 3 and 4 and try to figure out the final structure of our protein. So the peptide A, K, we know has to have K at the C terminus. Now, from part 3, we found out that there was a peptide that contained M, A, C, and K, after the chymotrypsin digest. So now we know K has to be the carboxy end and A has to be right before K, so in order to get an A-K peptide, then A has to be right after methionine because that's where cyanogen bromide is going to cleave the peptide bond. So the only sequence that we can have here is going to be C followed by M followed by A followed by K. So when we treat the cyanogen bromide, we're going to be cleaving this bond between M and A, generating M as a homoserine lactone. Now, we also know that M has to be in this small peptide, and we know it has to be C, M as a sequence. So therefore, S and F have to be on the amino terminus of this peptide. And we also have a clue from part 3, which said that S, F, and this non amino acid moiety were in the same peptide. So from here we said that, well, the sequence there, it's probably x modifying the amino S and then F with a carboxy end. So putting these three things together, we can come up with a sequence for this strand, which is going to be x, S, F, then C, M and then A and K. So this is probably one of the peptide chains in our mystery protein. Then, of course, the other is going to be composed of these amino acids. Now, we already know something about the sequence of these. For example, we know V goes before W. We also know D goes before K, and also know C goes before F. Now, we know this has to be the carboxy end of the peptide chain. Let's write it here, COO minus. And V has to be the amino end. So then C-F, there's no other way, has to be in the middle. So the only possible sequence for our second chain is going to be V, W, C, F, D, and K. Now that we've established the exact sequence of each one of these peptide chains, then we can put together the final structure of our mystery protein. So I'm going to transcribe these here, the first chain V, W, C, F, D, and K. And we know the cysteine is going to have our disulfide bridge to the other cysteine, which goes C, M, A, K, and then F, and then S. And now we know the N terminus of this peptide, we're going to have our fatty acid residue CH2 14 times CH3. So let's just mark, once again, the carboxy ends here and the amino end here. So this is the final answer for our problem and the structure of our mystery peptide. Well, that's it for this problem. I hope you enjoyed this little protein mystery hunt. Now, remember that the strategy that we used here in which we logically string together pieces of data to build a big picture, it's really the same strategy that has been used and is being used right now to advance our knowledge about living systems and their underlying biochemistry.
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https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/8.01sc-fall-2016.zip
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Here, we will talk about in the calculation of angular momentum. People often forget that angular momentum can be calculated for any object, even for an object that's traveling in a straight line and not rotating at all. For example, here I have an object with mass m moving at a speed v along a street line. Another important thing to keep in mind is that angular momentum does not have a definite value. It depends on the choice of origin, which is arbitrary, although some choices are easier to calculate or more useful than others. Here, I will choose this completely random point to be my origin. Notice that it can be any point in space. Angular momentum is a lot like torque. They both involve a cross product of a distance vector with another vector. For angular momentum, it's the distance vector r, the vector from the origin to the object, cross p, the momentum of the object, or mv. Once again, we can write the magnitude of this cross-product as r times mv times the size of the theta between the two. But it's often more helpful to think about this as r times sine theta times mv. In other words, the r sine theta is the component of the position vector that's perpendicular to the direction of the momentum. Let's practice with a few other examples. If I have a ball moving up and I have my origin at the side, then this is the perpendicular distance. So the angular momentum is r perpendicular times mv. A reference point that's the same horizontal distance away from the object will see the same angular momentum. Notice that in this case, the angular momentum is not changing as the ball moves, because the perpendicular distance is not changing with time. In this case, if the ball is moving at an angle, we again, have to take the perpendicular component of the position vector to find the angular momentum. This second reference point is now a different distance away. So the angular momentum is larger. Now, let's talk about the sign of angular momentum. You can calculate the sine of a cross product with the right hand rule. Make sure your vectors are tail to tail when you compare the directions. So in this case, we have that r cross v points into the page. In the case of circular motion, r and v are always perpendicular. So we can just multiply the magnitudes together. And the sine tells us which way we're going around the circle. If we consider out of the page to be positive and the angular momentum is positive, then the object is circulating counter-clockwise. If the angular momentum is negative, the object is circulating clockwise.
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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: OK. We're going to take 10 more seconds. OK. Does someone want to explain how they got the right answer? We have a Faculty of 1,000 research bag for them. Do you want to hand that up and the bag, too? We're just-- AUDIENCE: So the theme of light gives enough energy for the electrons to be ejected. And the amount of energy for that is 4.3 eV. And then it also has kinetic energy of 7.9 eV. So you just add the two. So 4.3 plus 7.9 is 12.2 eV. CATHERINE DRENNAN: Thanks. Let's bring it back. OK. Thank you. All right. We'll have lots more practice with this today, and we'll get the hang of doing these problems. So let's just jump in and get started. We're still continuing to think about the photoelectric effect, and think about light as a particle. So we're going to finish up with the photoelectric effect, and we're going to have a little demo on that in a few minutes. Then we're going to go on. If light is, in fact, quantized, and you have these photons, then photons should have momentum. And so we'll talk about that. Then we've talked about white as a particle. And most of you are probably pretty OK with matter being a particle. But what about matter being a wave? So we're going to talk about matter being a wave. And if we have time at the end, we're going to start on the Schrodinger equation, which we're going to continue with on Friday. So I'll just say that sometimes, I am a little overly ambitious, and I put things on the handout that I'm not really sure I'm going to get to. Just because I've never gotten into it before doesn't mean that I won't get into it this time. So if I don't finish everything on a handout, bring your handout to the next class, and we'll just continue from there. And there'll be a new handout then as well, so just a heads up on that. All right. So let's continue with the photoelectric effect and get good at doing these kinds of problems. So let's look at these particular examples. We have three different examples here. We have the energy of an incoming photon must be equal or greater to that threshold energy or that work function, in order for an electron to be ejected. So in this case, the energy is greater than the work function. So tell me whether an electron will be ejected or will not be ejected. And you can just yell it out. What do you think? AUDIENCE: Will. CATHERINE DRENNAN: Yes. So an electron is ejected. It will be ejected. What about this scenario over here, where the energy is less than that threshold energy? Is or is not ejected? AUDIENCE: Is not. CATHERINE DRENNAN: Yes. OK now we have another scenario. We have three photons, each of which have half of the energy needed, half of that threshold energy. But you have three of them. So will an electron is or is not ejected? AUDIENCE: Is not. CATHERINE DRENNAN: Is not. OK. So three photons each that have half the energy does not add up. You cannot add it. It will not eject an electron. So let's just think about it for a minute. Suppose the threshold knowledge for passing an exam is answering three specific questions correctly. Suppose over here, we have the answer to one of the questions, but not to the other two. Over here, we have an answer to the middle one, but not the first or the second. And over here, we have the answer to the third, but not the first or the second. So everyone knows the answer to a different question. Will there be the threshold energy, a threshold knowledge, to pass this test? No. Everyone needs to have the threshold knowledge themselves to be able to pass. Everyone has to overcome that critical amount of knowledge to be able to pass the test. So that's the same thing here. You can't add it up. Now, with the test here at MIT, if everyone has that threshold knowledge, and a really high level of the threshold knowledge, everyone can get an A. So the more people with the threshold knowledge, the more tests that are passed, and the more the course is passed by people. So the more photons coming in with that threshold energy, the more electrons being ejected. But you can't add up if you have photons that don't have enough, if they're not greater than, the threshold energy. You won't eject an electron. So everyone needs to meet that threshold criteria. You can't add things up. OK. So here's just some useful terminology for solving problems on this problem set. And there will also be problems on problems set two related to this topic. So photons-- also called light, also called electric magnetic radiation-- may be described by their energy, by their wavelength, or by their frequency. Whereas electrons, which are sometimes also called photoelectrons, may be described by their kinetic energy, their velocity, and, as you'll see later, by their wavelength. So you'll be given problems where, given different pieces of information, you have to think about how you're going to convert it. You've got to think about am I talking about a photon? Am I talking about an electron? And you also want to think about units. You'll sometimes be told about energy in eVs and sometimes be told about energies in joules. So this is a conversion factor. All conversion factors are given to you on the exam. You do not need to memorize any kind of conversion factor. But you need to be aware, when someone said joules, what's that a unit for, or eV, what's that a unit for. All right. So now we're going to do and in-class demonstration of the photoelectric effect. But before we actually do the experiment, we're going to predict what the experiment will show. Always dangerous to do that, so we'll hope it works after we do the prediction. All right. So we're going to be looking at whether we're going to get an injection of an electron from a zinc surface. And we're given the threshold energy, or the work function, of zinc. Every metal-- this is a property of metals. They're different, as we saw last time. So this is 6.9 times 10 to the minus 19th joules. And we're going to use two different light sources that are going to have different wavelengths. And we'll predict whether they have enough energy to meet this threshold to go over the threshold and inject an electron. So the two different sources, we have a UV lamp with a wavelength of 254 nanometers and a red laser pointer with a wavelength of about 700 nanometers. OK. So before we do the experiment, let's do some calculations to see what we expect. So first, we want to see what the energy, or calculate what the energy, of the photon will be that's a emitted by the UV lamp. And I will write this down. So what do we know? We know a bunch of things already. We know that energy is equal to Planck's constant times the frequency. We also know that the frequency is related to wavelength by c, the speed of light. And then we can put those two things together to say the energy, then, is also the Planck's constant times the speed of light divided by the wavelength. So we can use that last equation to do a calculation, and figure out the energy that's associated with that particular wavelength of light. So here we have energy. We're going to write in Planck's constant, 6.626 times 10 to the minus 34. And the units are joules times seconds and the speed of light, 2.998 times 10 to the 8 meters per second. And we want to divide this, then, by the wavelength. So we have the wavelength here that we're using first is a 254 times 10 to the minus 19 meters. Oh sorry-- 9 meters. Thank you. I wrote down 19. I'm like, wait a minute. That's not right. OK. OK. So then we can do the calculation out. And here is where I got excited about 19. We have 7.82 times 10 to the minus 19 joules. And if we look at the equation, we'll see that the meters are going to cancel. The seconds cancel, and we're left with joules, which is good, because we want an energy. So joules is a good thing to have. So there, we can do a simple calculation. And we can look and say, OK, if the energy, then, associated with that wavelength is 7.82 times 10 to the minus 19th joules, then we ask, is this greater or less than the threshold energy? And it's greater than that. So it does have enough energy. It should eject an electron. So we can try that out and see. Now we can look at what happens with the red laser pointer and see whether that should have the energy that's needed. And so I will just write these things down here instead of writing it again. So that was our UV. So now our red light, we have 700 times 10 to the minus 9 meters, or 700 nanometers. And so here is our answer for the UV. And our answer for the red light should be 2.84 times 10 to the minus 19 joules. And I'll move this up a little so people can see that. So does that have enough energy to eject an electron? AUDIENCE: [INAUDIBLE] CATHERINE DRENNAN: No, that should not work, because that's less than the threshold energy that's needed. All right. So we'll do one more calculation just for fun. And then we'll do the experiment. So the last calculation we'll do is we'll think about the number of photons that are emitted by a laser in 60 seconds if you have an intensity of one milliwatt. And a milliwatt is equal to 10 to the minus 3 joules per second. So we can just do that calculation over here. So we have 1.00 times 10 to the minus 3 joules per second, 1 photon, and here, this is for the red laser. So we'll use the number that we just calculated over here. So we have 2.84 times 10 to the minus 19th joules for the red laser and times 60 seconds. And we should get 2.1 times 10 to the 17 photons. So that's how much photons, if we hold it for 60 seconds, that were going to be shooting at our metal's surface. So these are the kind of calculations that you'll be doing on these kind of problems. And now let's see how well the experiment works. So we're going to bring out our demo TAs, who are going to tell you about this demo. And we're going to try to do some fancy stuff with this document camera to project it on the screen. So this is all very exciting. Oh, I guess I should put that down, the number, in case you couldn't see it-- 2.1 times 10 to the 17. All right. So let's bring-- you've got the mic. GUEST SPEAKER 1: OK. So we've got our metal plate here that Eric's got in his hand. And what he's doing right now is he's rubbing it with a little bit of-- what is that, actually? ERIC: It's just steel wool. CATHERINE DRENNAN: Steel wool. GUEST SPEAKER 1: OK. So that's just going to get the aluminum oxide, because sometimes-- you guys will get to it. But sometimes you can get a reaction of aluminum with the moisture in the air, and that's going to cause aluminum oxide. So he's getting get rid of that. And now we've put this on a-- what is this? ERIC: [INAUDIBLE]. GUEST SPEAKER 1: What do you call it? A detector of some kind. So basically, when he charges this, what's going to happen is that you have this plate, and you have this joint. And they're both going to be electrically negative, because you've introduced some electrons. And they're going to repel each other, because they're both negative. Two negative charges repel each other. So you're going to see some space develop as Eric's done. Now, what he's doing is he's got a plastic rod here that he's charging with the fur. And he's introducing those electrons onto the plate. So now we've got a negatively charged plate, and you can see that by the fact that you see some repulsion between that rod and the rest of the detector, which is actually working out pretty nicely. So once-- CATHERINE DRENNAN: So say this experiment is very weather-dependent. If it's really humid or too dry, it doesn't work nearly as well. But today, today's good weather. Today's good weather for this experiment, not so much good for sunbathing outside, but good weather for this experiment. GUEST SPEAKER 1: Although we have UV lamps, so maybe. CATHERINE DRENNAN: That's true. GUEST SPEAKER 1: OK. So now we've got a charge. CATHERINE DRENNAN: That's the green laser pointer. Let's get the red. GUEST SPEAKER 1: It's underneath here, I think. CATHERINE DRENNAN: Oh, yeah. GUEST SPEAKER 1: OK. So now-- CATHERINE DRENNAN: We could do the calculation for the green. If you want to do the calculation for the green, we can try it later. GUEST SPEAKER 1: Eric's got a red laser pointer in his hand. He's going to shine it. And we're going to see that nothing happens, because as we calculated, the energy of these photons is not enough to get over the threshold of this particular metal. CATHERINE DRENNAN: So if electrons were being ejected, you should see it move. GUEST SPEAKER 1: And we'll do that one more time. Maybe the green one will work. It doesn't. CATHERINE DRENNAN: All right. Well, now we have to see if the UV-- we built it up. The UV should-- GUEST SPEAKER 1: So hopefully this works. CATHERINE DRENNAN: --work. Let's see. GUEST SPEAKER 2: [INAUDIBLE] GUEST SPEAKER 1: OK. So oh-- maybe-- AUDIENCE: [INAUDIBLE] CATHERINE DRENNAN: Oh. GUEST SPEAKER 1: Oh, well, I guess it worked. CATHERINE DRENNAN: It did work. You could sort of see that. GUEST SPEAKER 1: So maybe we can charge it up again while I talk about it. CATHERINE DRENNAN: Yeah, sometimes the charge [INAUDIBLE]. GUEST SPEAKER 1: The UV lamp, obviously, has enough energy in each of these photons. So when you shine that light at the metal, you have the electrons on the surface, which are being ejected. And if those electrons get ejected, then the whole system becomes neutral. If the systems become neutral, then that rod can go back and is no longer feels a repulsion, because the two parts are no longer negative. So once we charge this up again, maybe we can go to the other side and-- I think it's good. It's good. CATHERINE DRENNAN: Yeah, that's good. Oh-- GUEST SPEAKER 1: It will be fine. GUEST SPEAKER 2: Wavering. CATHERINE DRENNAN: OK. GUEST SPEAKER 1: OK. Now we're just going to try it again. And yay. CATHERINE DRENNAN: Yay. GUEST SPEAKER 1: We got it. [APPLAUSE] CATHERINE DRENNAN: OK. Great. We can just leave this here. All right. And I think he held it for 60 seconds, so you know how many photons were coming off, too, if you want to do that calculation. So again, the photoelectric effect was really important at this time in understanding the properties that were being observed, to help us understand about this quantized energy of particles, that light had this particle-like property. It had this quantized energy. And you needed a certain amount of it to eject an electron from a metal surface. So we all know that light is a wave. But now there's this evidence that, even though it's pretty much this massless particle, that it still has particle-like properties. So light is a really amazing thing. This doesn't really show up very well. It's a view of the Stata Center. Stata Center always has some really spectacular sunlight coming around it sometimes. All right. So now, if this is true, that means that photons that have this quantized energy should have momentum as well. And so Einstein was thinking about that. And so he reasoned that this had to be true. There had to be some kind of momentum associated with them. And so momentum, or p, here is equal to Planck's constant times the frequency divided by the speed of light, c. And since the speed of light is equal to the frequency times the wavelength of the light, then the momentum should be equal to Planck's constant divided by the wavelength. So this is really-- we're talking about momentum in terms of wavelength, this inverse relationship here. This was just a kind of a crazy idea to be thinking about momentum, when you're talking about light. And this really came out of the photoelectric effect. And also, there were some experiments done by Arthur Compton that also showed that you could sort of transfer this momentum. And so that's again the particle-like property. So it's a really exciting time. OK. So we're going to now move to matter. So we've been talking about light and how light has this dual, particle, wavelike properties. But what about matter? So we accept that matter has particle-like properties. But what about as a wave? So enter de Broglie into this area. And so he was following what Einstein was thinking about. And he said, OK, so that's pretty cool. If you have momentum is equal to Planck's constant divided by wavelength, if you could think of things that have wavelengths as having momentum. And he said, or I can rewrite this equation, that wavelength equals Planck's constant divided by momentum. And we know something about momentum. We know that momentum is often associated with something's mass times its velocity. So therefore, I should be able to rewrite this equation again in terms of wavelength being equal to Planck's constant divided by a mass and a velocity. And here, we are expressing wavelengths in terms of masses. So this was really something. And this was basically his PhD thesis. I think it maybe had more pages than that, but this would have probably been enough, this sort of cover page. This is my PhD thesis. And Einstein said that he had lifted the corner of a great veil with really just manipulating what was known at the time and rearranging these equations and presenting relationships that people hadn't really put together before. So he ended up winning a Nobel Prize, basically, for his PhD thesis, which is a fairly rare thing to have happen. But this was really an incredible time. OK. So if this is true, if you have equations that relate wavelengths to mass, and particles have wavelike properties, how come we don't see this? How come this isn't part-- how come no one noticed the particle going by and this wavelength associated with it? So why don't we observe this wavelike behavior if, in fact, it is associated with particles? So let's think about this a minute. And we can consider why, when you go to Fenway Park-- and you should, because it's fun-- and you watch someone throw a fastball, why you don't see a wave associated with that fastball. So we can consider a fastball and that the mass of a baseball is about 5 ounces, or 0.142 kilograms. And the velocity of a fastball is around 94 miles per hour, or 42 meters per second. And so we can do a little calculation and figure out what the wavelength associated with that ball should be. So wavelength should be Planck's constant over the mass times the velocity of the ball. And we can plug in these values. And here's Planck's constant again. And now you'll note I did something with the units. So instead of joule seconds, I substituted joules with kilograms meters squared seconds to the minus 2. And that's what's equal to a joule. And I'm going to do that so I can cancel out my units. And again, all of this will be provided on an equation sheet. You do not need to remember all of these conversions. And so over the mass of the baseball and the velocity of the baseball-- and we're going to put the velocity in meters per second so our units can cancel out. And so I'll just cancel units out. So we're canceling our kilograms. We're canceling one of the meters, and canceling all of the seconds. And we have one meter left, which is good, because we're talking about wavelengths. So that's the unit we should have. And the wavelength is 1.1 times 10 to the minus 34 meters. That is a really small number times 10 to the 34. And it is, in fact, undetectably small. OK. So now why don't you try your hand at this, and we'll try a clicker question. Yeah, it's very tiny. All right. Let's take just 10 more seconds. Oh, or five seconds. OK. Awesome. It went away. That's OK. So they're in-- 97%. I like 97%. That's a good number. So again, you want to think about this and just realize the relationship, the equation, involved. And so thinking about-- oops, I switched pointers. I like the green better. So think about the relationship between the velocity of the ball and the wavelength. And so Wakefield, who was an knuckleballer, is the winner here, with the longest wavelength. But still, the number for this is 1.4 times 10 to the minus 34. And so this is still undetectably small. So of course, no one had noticed this property before. But it still, it still exists. So when you're talking about a baseball, the wavelength is really not very, relevant to you, because it is this incredibly small, undetectable number. But if you're talking about an electron, it's entirely different. So now, if we think about a gaseous electron traveling at 4 times 10 to the 6 meters per second, and so that's associated with an eV of about 54. So we have this electron traveling with this velocity. And now, if we do this calculation, so if we use Planck's constant divided by the mass of the electron-- and that's known, in another great experiment-- and its velocity, now we can calculate out the wavelength. And it's 2 times 10 to the minus 10, or about two angstroms. Now, 2 angstroms is a relevant number, when you're talking about an electron, because an electron is in an atom. And atoms tend to be-- you have diameters 0.5 to 4 angstroms. So now the wavelength is on the same scale as the size of the object you're talking about. And so when that's true, all of a sudden, the wavelength-- the wavelike property becomes super important to thinking about this. So for an electron that is a particle, it's really important to think about its wavelike properties. And so people were saying, OK, if electrons are waves, then maybe we should see other wavelike properties, such as diffraction. Diffraction, we talked about last time, is an important wavelike property of constructive interference, destructive interference. So people looked to see whether there were diffraction-like properties, and in fact, there are. So we had observed, then, the first was observing diffraction of electrons from a nickel crystal. And then JP Thomson showed that electrons that pass through gold foil again produced a diffraction pattern. So again, this was a wavelike property. So you might think Thomson, that sounds a little familiar to me. Didn't she just talk about that last week? And yes, here there are two important Thomsons in this story. And this is a father and son team. And so JJ Thomson won a Nobel Prize in 1906 for showing that an electron is a particle. He discovered an electron. And then in 1937, his son wins a Nobel Prize for showing-- son just had to be like, Dad, I'm going to show you're wrong. An electron is, in fact, a wave. But I think they were both happy. I think they both got along, no father-son rivalry. I think this is one of the cooler stories in science, how this father, son both had kind of opposite discoveries, which both ended up being true, and really changed the way we thought about matter. All right. So we have light as a particle and as a wave. We have matter, particularly electrons, as particles and waves. And now we are ready for a way to think about how to put this together. So before we move on and talk about the Schrodinger equation, I just want to take a break from history for a minute, because some of you are like, OK, well, this is really cool for the father and son team, but what about today? What's happening today? So let's take a break from history for a second and talk about why you should care about small particles. Small particles of special properties, if they're on the subatomic scale, their properties are different. If you have very, very few atoms, versus many atoms, the things with very few atoms have special properties. So why should you care about that? Why should you care about the energies that we can get out of the Schrodinger equation? So why should we care about the Schrodinger equation or quantum mechanics? So there are many reasons, but I will share one with you. And this is a segment in their own words. So you're going to hear from Darcy, who was actually a former TA for 5.111. So she is associated with this class. She actually just got her PhD in the spring from MIT, and she now works at Google. But in this short, she's going to tell you about research in Moungi Bawendi's Lab, and why you should care about quantum dots, which are small collections of atoms. So I'm going to try to switch over now and hope that our demo before didn't screw up the sound. But we'll see what we can do. And I think it should be good. [VIDEO PLAYBACK] - My name is Darcy Wanger, and I work as a graduate student in the Bawendi Lab at MIT. I work with quantum dots in my research. Quantum dots are really, really tiny particles of a semiconductor. So we're talking like 4 nanometers in diameter. In a particle that small, there are only 10,000 or so atoms, which seems like a lot of atoms if you're comparing to something like water, which only has 3 atoms in it. But if you compare it to something you can actually hold in your hand, which has a lot of atoms in it, 10,000 is actually a pretty small number. So a particle this small has really strange properties. Different things start to matter when you get really small. And just like an atom, a quantum dot, or semiconductor nanocrystal, has discrete energy levels. So if an electron is sitting at this energy level, and it absorbs light, an electron can get excited to a higher energy level. And then, when that electron relaxes back down to the ground state, it emits light. And the energy of that light is exactly the difference between these two energy levels. The difference between the energy levels is related to the size of the dot. So in a really small quantum dot, the energy levels are far apart. So the light it emits is higher energy, because there's a large energy difference between the energy levels. If we use a larger quantum dot, the distance between the energy levels is smaller, so the light it emits is lower energy, or redder. People in our lab are working to make quantum dots bind to a tumor. So when a doctor goes in to remove a tumor, they can see the shining of the UV light on it, and see whether it's all gone when they've taken out the tumor. They can also use quantum dots to label other things other than tumors, like pH or oxygen level or antibodies or the other drugs that are treating the cancer tumor. Each of those can be different colors. So if you shine a light on that whole area, you can see, oh, that orange spot, that's some cancer cells. Oh, and that green tells me that the pH is above 7.4. So it's pretty cool that we can use the idea of energy levels in something so applicable like surgery, where it can actually be used to track things and make it easy for doctors to see what's going on while they're doing a surgery. [END PLAYBACK] CATHERINE DRENNAN: OK. So that's an example for course 5 research. [APPLAUSE] And you can see all these credits online. I will mention that some of those nice animations were done by a former graduate student in the chemistry department. So these videos, even the art was done by chemists, which is a lot of fun. OK. So let's introduce the Schrodinger equation. And we'll spend some more time on this as we go along, on Friday. So we needed now-- we had learned a lot about wave particle duality and about these subatomic particles. And we needed a way to think about it. We needed a theory to describe their behavior. And classical mechanics had some flaws in with respect. So we needed a new kind of mechanism. We needed quantum mechanics. So here, if we're thinking about particles that are really small like electrons, we need to consider the wavelike properties. It's really important when you have a wavelength that is so similar to the size of the object that you're thinking about. So the Schrodinger equation really became to quantum mechanics like Newton's equations were to classical mechanics. So what is the Schrodinger equation? So here's a picture of Schrodinger. And he looks so happy. I would be happy, too, if I had come up with this equation, I think. So here's the simplest form of the equation that you will probably ever see. And so we'll just define some of these terms. So we have wave function, psi. And over here is the binding energy, and that's the energy of binding an electron to a nucleus. And then an H with a hat, we have our Hamiltonian operator. And in this course, you will not be solving this equation. We're just going to be talking about what sorts of things came out of this equation. So I'm going to give you a little bit longer version of the equation now. And so again, you're thinking about the electron. It has these wavelike properties. And it's somewhere in the atom, not crashing into the nucleus. And it needs to be defined in three dimensions. And it has momentum, so it's moving. So we need to think about this as an equation of motion in a three-dimensional space. And the equation is going to change. The math will change, depending on where the electron is located, which you won't know exactly. So this is a very hard problem. But it's not totally without anything to do with classical mechanics. And if we write the longest version you'll see, at least in this course, for the hydrogen atom, I just want to show this to point out that there are some terms from classical mechanics in here. This is Coulomb's energy, also sometimes called potential energy. So we saw Coulomb's force before. Here is Coulomb's energy. So some of the classical mechanics is contained within this, but it expands from classical mechanics to consider the wavelike properties of the electrons. So whenever I talk about this, I always feel like I want to have something better to say about really what this is doing and where it came from. In terms of what it's doing, how is solving this helping you? What are you learning from solving this? So one thing you're learning from solving this is you're finding E. And that's really important, the binding energy of the nucleus and the electron. And we saw before that, if you just used simple classical mechanics, you have a positive and negative charge that are close to each other. Why don't they come and crash into each other? We want to know how they are bonded to each other, what's the real energy of that association. We also saw, with the photoelectric effect, that it's not that easy to get an electron to eject from a metal surface. So it's bound in there. And what is that actual binding energy? So that comes out of the Schrodinger equation. This E here is the binding energy. And also, solving it will tell you about the wave function or, as chemists like to talk about, orbitals, where the electrons are, in what orbitals. So this is the information you get out. And importantly, it works. It matches experiment. So chemists are experimentalists. We love experiments, and we see this data, and we want to understand it. And the Schrodinger equation helps us understand it. It correctly predicts binding energies and wave functions, and it explains why the hydrogen atom is, in fact, stable, where you don't have crashing or exploding of the hydrogen atom. So where did this equation come from that works so well? How did Schrodinger come up with this? And this is always sort of the puzzle when I teach this. I feel like I should have something profound to say about where this came from. And so I've done a little reading and looked, and I thought the best explanation for this that I ever saw came from Richard Feynman. And when he was asked how Schrodinger came up with this equation, he said, "it is not possible to derive it from anything you know. It came out of the mind of Schrodinger." And I thought that pretty much summed it up. So sometimes-- after class last week, on Wednesday, someone came down and said, you know, the Thomson experiment, discovering the electron, why didn't someone else do that experiment? It seemed like it's not a cathode ray. And you have to have a little phosphorous screen. Why didn't someone else discover the electron? And some of these other-- de Broglie rearranged some equations, did it in a way that no one else was thinking, but still. Or plot solving the equation of a straight line. No one else was thinking about it some way, using other people's data. They just sort of saw things in data that other people didn't. But you think why didn't someone else see that, too? When it comes to the Schrodinger equation, the question is why didn't someone else or lots of people come up with it? I think the question really is, how did Schrodinger come up with it? At least, that's the question to me. And I have never really-- that's the best explanation I have. It just came out of his mind. OK. So we're many years later. We've had the Schrodinger equation for a while. So this is an old story, right? Well, maybe for the hydrogen atom, but this is still actually a very active area of research. Oh, my startup disk is full. Thank you. Let's go back to that. All right. So I just thought-- I always like to give you examples of current research on these areas. And so I know a number of you were interested in potential of being chemical engineering majors, undergrads. And I'll tell you about a new professor who started about a year ago, Heather Kulik. And her research group is really interested in using a quantum mechanical approach to study materials and to study proteins. But when you get to things like proteins, there's thousands and thousands of atoms around. Forget multiple electrons, we're talking about multiple atoms with multiple electrons, huge complexes. How can you give a quantum mechanical analysis of things that are so large? And this is really important. I mean, I think that one of the big problems moving forward is solving the energy problem and doing it in a way that doesn't destroy our environment, so new batteries, new electrodes, new materials. We need to understand the properties of different metals to understand what will make those good electrodes. And to really understand them, we need a quantum mechanical approach. But these are big areas. There's a lot of things to consider here. So Heather is interested in coming up with improving algorithms, improving the computation, to really give a quantum mechanical analysis to systems that have a lot of atoms in them. So if you're interested in this area, you're not too late. You don't have to go back to the early 1900s. There's still a lot to do in this area. OK. So very briefly now, let's just look at the Schrodinger equation we saw from the hydrogen atom. So we'll go back to understanding quantum mechanical analysis of photosynthesis-- amazing, don't understand how it works. That would be great if we did. That would really solve a lot of energy problems. But we'll just go to hydrogen atom, one electron back. So if you solve the Schrodinger equation for this-- and I think I did this in college, not in freshman, chemistry, but somewhere along the line-- you'll come up with this term. So again, this is the binding energy. We just want to know about how the electron is being held by the nucleus. And there are some terms in here. We have the electrons mass-- that's known, the electrons charge. We have a permittivity constant and Planck's constant. And if you look at this, you go, wait a minute. That's a constant. That's a constant. That's a constant. That's a constant. We can simplify that. And we will. And that is the Rydberg's constant, 2.18 times 10 to the minus 18th joules. So now it doesn't look quite as scary. We can just substitute this RH. That makes us feel a lot better. It's one number that will be given on the equation sheet, so we don't even have to remember it. And now we can rewrite this in terms of the binding energy. So again, the binding energy, this is a constant. So now this turns into minus RH over n squared. And n, what is n? n is a positive integer 1, 2, 3, up to infinity. And what's its name? What is n? You can you yell it out. Some of you know. AUDIENCE: [INAUDIBLE] CATHERINE DRENNAN: Yeah. The principle quantum number, that's right. So the principle quantum number comes out of the Schrodinger equation. And that's how we can think about it. And again, here are these ideas. The binding energies are quantized. This is a constant over here. So the principle quantum number comes out of the Schrodinger equation. All right. So now, next time, we're going to think more about the Rydberg constant. And we're going to do a demonstration next Friday of the hydrogen atom spectrum to show that the Schrodinger equation, in fact, can explain binding energies. So that's on Friday, and that's our first clicker competition. So come. Be ready with your clickers. You can sit in recitations. You can share answers before clicking in. It's not cheating. It's teamwork. OK. See you 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: E less than v0. So you have an incident wave, e to the ikx, incident, and a reflected wave that you have, e to the minus ikx-- remember minus the other face-- and e to the 2i delta of E. So this is the incident wave, and this is the reflected wave. They correspond to your Ae to the ikx plus Be to the minus ikx. Remember, when the energy was less than v0, the range of B over A was minus e to the 2i delta. And since I take A equals 1, you get this thing. So suppose I construct a psi incident of x less than 0 and t as the sum 0 to infinity dk f of k e to the ikx e to the minus iEt over h bar. Whew. So I superimpose the incident thing here. Then the reflected one should be superimposed too and would be 0 to infinity, a minus in front because there's a minus, dk f of k e to the minus ikx e to the 2i delta of E e to the minus iEt over h bar. Whew. That's the reflected wave superimposed. So now you've constructed everything. Here the reflected wave is more interesting than the transmitted wave, because there's no real big transmitted wave. It just whistles out. But the reflective thing is interesting. If you're doing the experiment, you send in a particle, you want to see what you get back. That's going to tell you what kind of potential you can expect it encountered. So let's do the stationary phase for this one, for the reflected. Let's see how it moves. We know how the incident moves. The incident moves with x equals-- we've done it there-- hk0 over m t. But how about this one? Well, this one you would have to do d dk of minus kx plus 2 delta of E minus Et over h bar, all that at k0 equals 0. And you'll probably remember that this thing was in your midterm and your first test. You had this wave, and you had to analyze, what did stationary phase do? And it does that. So what do you get? Well, when you take the derivative, you have to take the derivative of delta with respect to energy, that's delta prime, and then derivative of energy with respect to t. Let me save you a little time. The answer is minus h bar k0 m t minus 2h bar delta prime of E. OK. That's what you get. That's how this packet moves. And what does it do really? Remember, this is defined for x less than 0. So this is valid-- forget about this little term here-- this is valid for t positive. For t positive, you're going to get this to satisfy. So this is a big wave packet for t positive. It's the reflected wave. That's what you would expect. This is the reflected. Now, if this factor was not here, it is as if, well, the incoming packet hit the origin at t equals 0. And this will be perfect bouncing in which the packet gets reflected. And at t equals 0, it starts to move to the left. And as t increases, it's moved more and more to the left. You see it there, because x must be negative. But if there is this term, it really doesn't start to move to the left until t is bigger than that so that x is negative. So only at t equal to this amount of time the packet reflects. So there's a delay, and the delay is 2h bar delta prime of E. So this is a technology people use in scattering theory to figure out what kind of potential you have, figure out how much things get delayed from the bouncing. Now, this derivative-- we've plotted it there-- delta prime of E. You get a big delay for low energy, for energies near v0, and in the middle it's not so big.
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https://ocw.mit.edu/courses/7-014-introductory-biology-spring-2005/7.014-spring-2005.zip
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OK, so I'd like to go now to the next segment of the course. Think you can probably appreciate little bit better this triangle I had on before about how what biochemists did was they tended to break cells open, look at the component parts, through other things in there, and then proteins. But an awful lot of stuff having to do with function is proteins, and what geneticists, the discipline of genetics would do, which made mutants of living organisms, and that looked at how function was affected by mutating individual genes, how those were both very powerful approaches. Genetics told you what was really important, and biochemistry told you how it worked at a molecular level, but the real problem is knowing whether this thing you had doing something in the test tube was actually the one that did it in life. And I think it'll do when Arthur Kornberg isolated the very first DNA polymerase, he was able to copy DNA. And he got a Nobel prize, and it was the first enzyme that could copy DNA, and then John Karens made a mutant that was lacking the enzyme. And the organism was still alive. So therefore, it couldn't have been the DNA polymerase that was copying the chromosome. It actually turned out to be a DNA repair enzyme. So, if you can actually unite genetics and biochemistry, if you can take a mutant that's broken in a function and you found your protein was missing, or vice versa, then you had a very, very powerful insight because you connect your knowledge of what was physiologically important to the biochemistry that you're doing. But it was really, really hard for years, and only in very rare occasions did some geneticists have a mutant that suggests so strongly that some biochemists would look, or some biochemists would have such a powerful result that they talked to a geneticist and seen if anybody had found the result. And the power of recombinant DNA, although it started a biotech industry, and it made possible the sequencing of the genome, there's another level up one higher in conceptual understanding. And what it did was, it let you go back and forth between here and here. You wouldn't have any problem now if I gave you the sequencing of the gene, you can order it. You could stick it in a cell and make massive amounts of the protein. You purified it twenty-fold and it would be pure, whereas before you might have had to purified it fifteen thousand-fold out of 1,000 g of cells, and you would have had to been a very good biochemists with 15 steps in order to purify it. So, you can go from the sequence of the gene to the protein, or if we got a protein and we wanted to know which gene we'd just sequence a bit of the protein, use that genetic code, work ourselves backwards to some possible sequences, then go looking for the sequences, and then go find the gene. So, what recombinant DNA allows us to do is close that loop. You can go from genetics to biochemistry at back and forth. Now everybody does everything instead of it being isolated disciplines, which it was when I entered the field. So, all of the stuff depended on the development to clone particular pieces of DNA. And I want to make clear right at the beginning, there's a couple of uses of cloning that are in popular usage right now. What we're talking about in this lecture is cloning a piece of DNA. What that means, is I'm going to take a particular segment of DNA, say, cut it here, and cut it there. And I'm going to take that piece, and I'm going to do something to it that lets me amplify it and make many, many copies of that piece of DNA. And cloning of anything else you make a whole lot of copies. So that's one use of the word cloning. The other use, which you see in the popular press all the time, is cloning an individual, but not being there but you would take the nucleus from the cell of the individual, he put that nucleus into an egg that didn't have its own nucleus moved, and now you hope what you get out of that is an organism that has all the same genetic content as the starting individual. And in fact, although it sounds very good in paper, is you're probably beginning to see it's not the panacea that people thought it was, or that we'd have to worry that in 10 years, all of my MIT students would be clones of the brightest person in the class or anything like that, because other stuff happens, because unless you go on to advanced biology courses, but there are modifications of DNA. There's all sorts of stuff that happens to it, so it's not identical. And so many of these cloned organisms, like Dolly the sheep, that was famous, died early with, I forget, arthritis and things. So, there are a lot of problems on that score. But that's the other use of the word cloning and that's not what these next three lectures that we are talking about cloning a piece of DNA. And that was the big problem that faced the field, certainly when I was an undergrad and even when I was a grad student I was interested in synthesizing pieces of DNA, and it was one of those things that people said, why are you doing it? Well, because you could try to do it. What if you got a piece of DNA, like Gobind Khorana, who's my colleague, who got the Nobel prize for synthesizing the first gene. He synthesized it. It was a tRNA gene that was 120 nucleotide base pairs long or something. He synthesized it. He'd shown you could do it. But you couldn't do anything with it. And there were sort of two big problems. One was the fact that this DNA, although it's not its a monotonous tetranucleotide. It's pretty hard to tell. Each one of these things is a base pair, and human DNA has 3 billion of those. And a bit down here, doesn't look very different than the bit out there. And it certainly wouldn't looks very different than the bit of DNA that's 2 billion base pairs over on the other side of campus or something. So, there is no way to take DNA and cut it reproducibly, so you to get fragments. What you could see from first principles was what you would need was magic scissors. And what would the scissors look like? Well, it would have to be scissors that could be sequenced because there's nothing else different. You know it's a regular backbone, and it's only four nucleotides. So, if you wanted to cut DNA in particular places, you had to have scissors that could see a sequence. And furthermore, you can see they couldn't just, there are the hydrogen bonding parts of ANT or GNC because those are stuck together there in the middle of the DNA. So, you'd need scissors that could somehow find a sequence and make a cut. And those were found. I'm going to till you about those. They're called restriction enzymes. So that was part of the thing. The other thing was, imagine I could cut out this fragment. And they gave it to you, and I said, great. Now I've got it. Would you make me a lot of copies of this DNA? Could you do that? Let's say, you now know how to transform the principle that we saw back with Avery. We could take naked DNA and put this fragment into the cell. Would it replicate? What do you think? Anybody remember? No? OK, we talked about some other languages, right? But one of the things that's in the DNA is the genetic code with all the genes. And we can find the reading frame. Remember when we talked about an origin of replication. I said that was sort of, at least for E. coli there's one origin. In eukaryotes the origins are spaced out along the DNA. And every time you have a round of replication, it starts with one of those origins and then goes. So the chance of this piece of DNA by chance is going to have an origin is pretty small. So if I put it into an organism, it's going to sit there, if you're lucky, make it degraded because it's got [blend ins? , or even if I made it into circle it probably wouldn't replicate because it probably doesn't have the word in the DNA that says start a round of DNA replication. So, the other overarching principle of DNA replication is you somehow have to take the fragment of DNA that you're looking at, and you have to attach it to an origin of replication. Now, if you have an origin of replication and you have a fragment of DNA, and you put it in the cell, now you'll get a lot of copies of that piece of DNA. So that's what recombinant DNA is all about in a really, really simple form. I'm just going to take you now into increasing sort of levels of detail. So, let me just sort of give you just sort of a really broad view of this cloning, and then we'll sort of start to dive in to some of the fancier techniques that have come out of this. We'll talk about DNA sequencing, and PCR, and stuff in the next lecture. So the first principle here is to cut the DNA, and I know you may think this is sort of baby talk, but this is how I think. If you really think about this stuff, this is what it really is. With sequence specific molecular scissors, these have the rather odd name. They're called restriction enzymes. I don't know if any of you know why they're called restriction enzymes, and although I'm sure that some of you have used them in your op to cut up pieces of DNA. But what that does, these are enzymes, as I'll tell you, that recognize a particular sequence. And they always cut at that sequence. In the value of that is you can reproducibly cut DNA exactly the same spot, and the spots are specified by whatever sequence that particular pair of scissors knows how to read. Then the next thing we have to do is we need to join the piece of DNA to an origin of replication. So the thing that carries the origin of replication is called a vector. And usually, not always, these are circles. We'll consider the ones that grow in E. coli are most of the time, or in bacteria, mostly circles. They are the ones that are broadly used for most cloning. So, we'll talk about those. What makes a vector? When it has to have is an origin of DNA replication. They usually have something else. We could call it a selectable marker, but something like a drug resistance. If any of you have done cloning in a UROP usually it's something like a gene for making the cell ampicillin resistant, or tetracycline resistant, some antibiotic that would normally kill the cell. So you can tell, does that cell have that vector or not because the cell starts at either ampicillin sensitive. It acquires the vector that's replicating it, also acquires the gene that gives it the drug resistance. But if we're going to cut that, if we're going to join a piece of DNA to that, we can't join into a circle without breaking it. So, we need to cut the vector at a unique site. And we would use a restriction enzyme for that. And you can also see in designing a vector, you'd want something that only has one site. So, what we would have achieved from this conceptually as we've now got this, this is the vector, its origin of replication here, and let's say ampicillin resistance, for example, as a selectable marker, the gene for that could be encoded here. And we have the fragment of, if you want to that down, just put it down on the floor, I think. We've made the point at this stage. Thanks. What one has to do is to join this piece of DNA to that. And we'll go to the molecular details of this. But, we'll join the fragment to the vector, and actually this was something that was already in molecular biologists' toolkits, have been studying DNA replication. That's DNA ligase. When we finish an Okazaki fragment, we had to seal [UNINTELLIGIBLE] and the enzyme that did that was an enzyme called DNA ligase. So, molecular biologists basically had the scotch tape or the glue to join stuff back together. What they were missing for many, many years where the sequence specific molecular scissors. So at this stage, if we were doing the recombinant DNA, we now have a vector. We now have a piece of DNA joined to it. In fact, we probably have a whole other mess of things that happens along the way. But at the moment, they're in a test tube. So, if we want to have this thing grow, what do we have to do next? We are going to have to get the DNA from outside the cell inside the cell. That's the word; we need to transform the DNA into a cell. Again, the word transform, that goes back to those transformation experiments with the Streptococcus pneumonia going from smooth to rough, and you are taking stuff from the cell that transformed them from rough to smooth, whatever, that's where the word came from, but we now know it's getting naked DNA inside the cell where it can be replicated. And then the next thing we need to know is, what cells have acquired this vector that at least as the vector. We'll settle for that in the beginning, and to do that, you need to select for the marker on the vector. In the case of this one, we would start with a strain that killed my ampicillin, and then we just play it out and ask for guys that are ampicillin resistant. And, you can see that there is another class of problem because if we had uncut vector, and there would probably be, for sure, some of that in our mix, that would make the cells ampicillin. And if we had an insert, it would also be ampicillin. So, if we really wanted to get into this, we'd have to do some more work to sort out what's on there. But that's the basic stuff. I suspect most of you know this practically since kindergarten. But that's the overall framework into which, now, I'm going to start layering different pieces of detail. And the next part, again, some of you may know. I don't think it will be a totally foreign concept. You are probably familiar with this, that what are these restriction enzymes? The actual word is restriction endonuclease. They are often usually called restriction enzymes in a lab [parlons?]. Nuclease is something that cuts the nucleic acid, and endonuclease is one that doesn't need a free end. So, it can cut in the middle of the sequence instead of nibbling at the end. That would be an exonuclease. So, these things have names that tend to be something like ECO-R1, which has something to do with where they are derived from. And a typical one, one of the very first ones that is still in really wide use, is ECO-R1. And this recognizes the sequence G A A T T C. Now, you'll notice that if you read the sequence in this way it's the same sequence when you read it on the other strand. It's called a palindrome but be careful because palindromes in English, those are words that you read from the front to the back; they're the same. In an English letter, it doesn't matter whether it's in A here or an A there. But you guys know something about DNA structure. There's a five to three prime polarity. So, reading this way doesn't look at all the same. It's totally different. But reading in this strand, we say that's five to three. The thing that's identical is the reciprocal sequence on the other side. So there is this, you see G A A and G A A but it's not like the English word palindrome, so get yourself mixed up about that. Anyway, what this will then, what this enzyme then does, is it cuts to the side here. It cuts symmetrically. And what it generates them is a G three prime hydroxyl. Remember the ribose? If we have, say, an A there, this is the three prime position, and that's the five prime position in the sugar. So, it leaves a three prime hydroxyl, and is also then leaves a five prime phosphate. So, we'd have A A T T C here, and then on the other side, we would get the reciprocal thing. So, we'd have G with a three prime hydroxyl, and then over A A T T C like that. So now we've got a break here. We can pull those apart. But one of the nice things you can see right from this is that we're generating five prime single stranded ends, and this one is the sequence A A T T C. This is A A T T C here, and these guys, if they could get together and line up as they would here, they'd be able to form hydrogen bonds. So, if you take an enzyme like ECO-R1, and we took, let's say, a circle that had a single ECO-R1 site, G A A T T C, if we cut it with the restriction enzyme we would make [nicks?]. And if we kept a cold, all that we'd have is DNA nicks. And if we warm things up a little bit, there's only four hydrogen bonds that are holding that together. So, the thing would linearize and just flop around in the breeze. If we cooled it slowly, the thermodynamically most favorable state, the lowest energy state, would be with those ends coming back together. So, we could then add DNA ligase. If we added these up and added DNA ligase, we could reverse the process and go back and forth, ECO-R1 to cut it, DNA to ligate it. And then, the beauty of recombinant DNA is this rejoining part doesn't see what's out here or what's out there. All it sees is the little ends that are generated by an ECO-R1 site. So they take some of my DNA, and I'll cut it up. I'll get a zillion ECO-R1 fragments, but they'll all have the same little overhanging bit that's complementary to the vector. So if I take a vector cut with ECO-R1, and I take some of my own DNA and I mix them, I can get a little fragment, get in between the vector, and it does exactly that joining that I was diagramming right here. So again, it was the discovery of these restriction enzymes that made possible almost all the stuff that's happened in biology since 1975. The development of restriction enzymes was essentially, I was a postdoc at Berkeley at that point and the labs, Stan Cohen at Stanford, Herb Boyer at UCSF, and a two others around the country were working on this. They were almost all labs that had worked on bacterial plasmids. Plasmids are little circles of DNA, so the labs that started were ones who have been busy studying little circles of DNA that usually carry drug-resistances between cells. And so that was happening while I was a postdoc. And when I got to MIT and '76 the technology was just beginning. I was one of the first labs trying to cut pieces of DNA and join them back together. So, it's a pretty recent development. At that point, DNA sequencing hadn't been invented. The idea that you could pull out a piece of DNA and do something with it or produce a protein was just a thought. It didn't exist. So it's hard to overemphasize how critical the discovery of these restriction enzymes were. Now, I just want to tell you where they came from, or how people found them. And they'll try and do this quickly because I know some of you get impatient with history. But this is really important because it's very easy to make fun of basic research. You can ridicule anything pretty easily, and you might just ask because I'm telling you the story. Somebody proposed doing this. I'm going to tell you the experiment that basically is the basis of the biotech industry, and would you have been smart enough to recognize that it was the discovery of a phenomenon called restriction, restriction in bacteriophage growth on bacteria? And it was, here are actually a couple of EM's and these little plasmids. This is an electron micrograph one. In these little circles it's been shadowed. And this is actually artificially colored, but that was the kind of plasmids that people were cutting up. So, as I said, trying to get through this DNA, and the stuff, what's made possible the sequencing at the Whitehead Genome Center and stuff that I'll tell you about is going on. I didn't really set this one up. But that's Eric Lander who teaches 701 in the fall. I told you a picture from that DNA 50th. Well, they had a banquet at the end of it, and I was there. This is [Savandi Pabo? from Europe who is sequencing the chimp genome. And that's Francis Collins who is head of the entire Human Genome Project. This is Evelyn Witkin, who was a big discoverer of early DNA repair events. And I put that one in because it was sort of interesting. There was Eric, and Savandi, and Francis were talking about what would happen when they knew the sequence of the chimp genome, which wasn't done. And there was an advertisement, a poster advertising Jim Watson's latest book. And they ripped that in half, and were writing notes on the back all the way through dinner. So if you want to see what scientists on the cutting edge, including someone who teaches 701 the fall looks like when they are not teaching 701, there is a picture. So anyway, the discovery of restriction enzymes was Salvador Luria, who I've mentioned. He was a member of the biology department, and one of our Nobel Prize winners. He started the cancer center. He also trained at Jim Watson, when I showed you that picture. This is Salvador standing over here. Another thing that Salvador did, he was a Nobel Prize winner but he thought introductory biology. So I am basically following in the footsteps of Salvador. He wrote a book called, even though he was a Nobel Prize winner, a book called 36 lectures in introductory biology. And some universities, the intro to biology is taught by whoever is at the bottom of the food chain. The most junior professor gets stuck with intro to biology. And here it's the other way. I mean you're getting Eric and Bob, for example, Weinberg to teach in the fall, tells you that. And really where that comes from is the fact that Salvador Luria had such an interest in replication. So he trained at Jim Watson, started the cancer center here, and he also carried out this phenomenon of restriction. And to get this working with bacteriophage. And I know a [couple he wrote? , he didn't like to see old guys on porches. So I got freaked out, and I took this next picture out for this morning. Oh, I've got to show it anyway for two reasons. This is Salvador sitting on a porch at Cold Spring Harbor with Max Delbrook who started, really, much of the work on bacteriophage that gave us the underpinnings of microbiology. And then put it partly on A because it shows the informality of the molecular biology culture which persists to this day, and also because Salvador had such an impish sense of humor. You would have really enjoyed had he been teaching this course. Anyway, Salvador was studying this bacteriophage. Remember we talked about it? And they basically [with syringe? they injected their DNA into the cell. There's an electron micrograph. The DNA is up top there. It goes in, and then the DNA takes over the cell, reprograms it, and makes babyphage. And I showed you how we make plaques. So that was what Salvador was studying. And it's a little like what we are talking about with Mendel. He didn't have very many techniques available to him at the time. He couldn't sequence DNA. He couldn't do a lot of things. But he could [plate phage? and count, and things like that. And what Salvador was looking at, he had a bacteriophage, and he had two strains of bacteria. I'll call them A and B. OK, here comes the experiment that founded the biotech industry. You ready? You going to fund me? All right, so what I propose doing is I'm going to grow the phage on strain A, and now I'm going to plate on strain A and strain B. I laid awake all last week thinking of this experiment. So what did I get? I got a lot of plaques on strain A, probably something like 109 or 1010 per mil because that's what this phage lysate usually looks like once you've grown them up. And if I plate them over on strain B, no phage, maybe an occasional phage. So, I bristly the bacteriophage can grow mine on strain A. It can't grow on strain B most of the time, but some variant has managed to figure out how to grow on strain B. Give our foray into genetics, I think many of you would think, probably as I suspect Salvador did at the time, the things mutated. It's learned. It's made some change in its genomes that's allowed it to grow on strain B. So, I wonder if it learned to grow on strain B, couldn't grow on strain A? So, basically what he took was the phage from that experiment, and then he plated them on strain A and strain B. And, well, as you might guess, since it was growing on strain B, lots of plaques, and there were lots of plaques over here. OK, so it didn't forget how to grow on strain A. So, better check over here, too, need a control experiment, so take this guy, plate it out, strain A, strain B, some plaques over here. That's not a surprise going in strain A. We are back to where we started from. It doesn't sound like a mutant, does it? And if it was a mutant, everybody should have been the same. Instead, when you grew a phage on strain A, it didn't have the ability to grow on strain B. But if you gave it a chance to grow on strain B, most of them wouldn't make it. But if it ever did, it had now acquired the ability to grow on strain B. So it could still grow on both. But if you take somebody who had been growing, something had been growing at strain A, it lost it. And so, the idea there was then that it wasn't a mutation. Something was happening in the strain B that enabled it to grow on strain B. And if it ever got away from that environment, [it? lost it. So, the phenomenon was called restriction. It wasn't a mutation. It was something else. And it turned out, then, that restriction was due to an enzyme. An example of this kind of thing, then, would be this ECO-R1 activity that's able to cut at a very specific sequence. Now, if you are going to have a set of molecular scissors inside of you that could cut a G A T C sequences, you'd have a problem unless you did something else because every one of your G A T C sequences would be cut by the restriction enzymes. So what the cells that have a restriction enzyme have, as a modification enzyme that recognizes the same sequence, and then modifies it in some way that makes it resistant to the restriction enzyme. And in the case of the ECO-R1, it puts on a methyl group on this A. You might have thought that that would interfere with base pairing, but it doesn't because adenine looks like this. These are the guys that do the base pair refining, if you look back, and you'll see that you could put a methyl group in there. It wouldn't interfere with the base pairing, but would allow this to go. So, it was the discovery of this phenomenon of restriction of bacteriophage grew on one strain, not on another. It could learn to grow on the other strain. It could lose that acquisition. It was that phenomenon. People, didn't have any other reason other than it was an interesting problem in biology to understand it. Once they understood the basis of it, another whole world opened up because you could see from basic principles, now I could cut any piece of DNA. I could generate these little overhanging sticky ends. I could take a plasmid. People knew about those. I could find one that only has one restriction site. I could stick things in. Now I've attached an origin of replication to each of those pieces. And I'm in business. I can now, for the first time ever, take a particular piece of DNA and make as many copies as I want. And that was an absolute transformation to the way people were able to think about biology. So I'm going to just kind of give you an idea of how people would start. So the way people began and still began most things is they'd call it, the usual term is you call constructing a recombinant DNA library. And there are a variety of different ways of doing this. But this principle is the same. We'd take the DNA from whatever organism you're interested in, [and studying?]. And we cut with some restriction enzyme. And this restriction enzyme will cut wherever there happened to be sites. They might be close together. They might be far apart. But whatever, still generate some characteristic set of fragments. And we'll now have, in this case, fragment number one, fragment number two, fragment number three, fragment number four, and so on. And if it's my DNA, there's a lot of fragments. And of course, they're all mixed up. I can't tell where any of them are. They're just all mixed together in the test tube. Then we'll take that vector that we've opened. And now, we'll mix all of these fragments together with this vector. And then we join it just the way [that it's cut? . And now what we'll get, it's a collection of plasmids that have different inserts. So, one of the plasmids will have that fragment number one. Another one of them will have fragment number two, number three, and so on. Then this whole thing is what's known as a library. You can see if it's DNA from me, there were 3 billion base pairs to start. Given the human DNA, G A T C sequences are pretty common. You can calculate the frequency yourself for how many sites on average there would be for a restriction enzyme within a piece of DNA and figure out roughly how the fragments there would be in the library of human DNA. So, we are partway there. We can now make a library. We can make it from bacterial DNA. We can make it from human DNA. But the next thing that people had to learn how to do was to figure out how to find a particular fragment that had the gene that you are interested in. And there's a whole variety of things. I mean, ultimately today since the human genome is sequenced, you go on a computer and a type and you find it because the sequence is all known. But the only reason we can do that is because of all the work that was done in between. So, I'll give you several ways of doing this. But one of the ways I think you can see very easily, and it's actually going back to the term complementation. Remember complementation? We had something that was mutant, and then we'd put in a wild type gene, and fixed it up again. So, for example, suppose I was studying histidine biosynthesis in E. coli, and I wanted to find the gene that encoded the enzyme that I had just disabled in my histidine minus mutant. So, if I have a [hisoxotroph?], I'll call it a [his G? gene, for example, that's one of the genes involved in making histidine. So, since it's a histidine auxotroph, if I have it on just minimal glucose plates, and I streak it out, it's not going to grow. But if I grow it on minimal glucose plus histidine, then it will be able to grow, right? So, I've got a variant of this organism. It's got a single mutation in it that's affecting one gene. And because I don't have that gene, I can't grow on minimal. If I made a library of E. coli DNA, which is going to have a lot of fragments as well, and I took that library and I put it into this mutant, I'm going to get a big mess of things, all the different plasmids with all the different fragments that go into that mutant. How am I going to find the one that I want? Anybody see that? It's not that hard. I'm the mutant. I can't grow on minimal because I can't make this enzyme. Therefore, I can't make histidine. What do I need? How could you fix me up if you were a doctor? What's the gene we want out of here? The one that makes that particular enzyme that makes histidine. Yeah, take the whole library, stick it in this mutant. If the gene coming in encodes DNA polymerase, I'm not going to help this guy. It still won't grow in minimal, take a gene involved in making part of the cell wall would help. But if I put in the, I get a fragment of DNA that includes the his G plus gene, and I put it in here, it's going to grow up. If it has the plasmid that has, or let's say the vector that has the his G plus gene. So what you've done is a really, sort of, you've used that principle of complementation that some of you were sort of wondering about when we were doing genetics. So, you'd break a copy of a gene. In the things we talked about, we bring in a whole chromosome that included in it just a wild type copy of the gene. With recombinant DNA, we can really narrow it down in the extreme. We can bring in a piece of DNA that is only the gene that's broken. And we can take the gene back to the wild type. One thing, just to close, you'll see, if you remember back when I talked about language that are not universal, although the genetic code is universal, promoters and things are not. So I couldn't ever do this with a human DNA, could I, because it wouldn't get expressed. So, we need some other ways of finding those. We'll talk about those on the next lecture, OK?
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https://ocw.mit.edu/courses/10-34-numerical-methods-applied-to-chemical-engineering-fall-2015/10.34-fall-2015.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. WILLIAM GREEN JR: All right today, I would like to do a Kinetic Monte Carlo example with you in class and, also, talk a little bit about the connection between the master equations, master equation solution, the normal kinetic equations we use, and the-- how you handle these trajectories, you'll get how to Kinetic Monte Carlo. All right, so let's just pick a simple example. As we mentioned, in the real world, it's pretty easy for these examples to get completely out of hand, because the number of possible states can be so gigantic. Let's all do a really super, super simple one here. To try to keep that underhand, not worry about that. When you do the real problems, that will be a major issue-- is if he number of states get out of hand. So the simple example I'm gonna do is a real one from inside your body. So inside your body, you have a thing called low density lipoprotein. And there's little vesicles of fat, lipids, inside your body, actually, in your blood vessels. And these are bad. If you go to the doctor, they measure your LDL and your HDL. And if your LDL number is too high then they yell at you for eating too much fat food. And they make you exercise and stuff like that. This has happened to me, so I know. And there's a reason for this, is because the LDL is susceptible to oxidation. And if too many of those lipid molecules peroxidize, it causes inflammation in your blood vessels. And that can cause your blood vessels to constrict and stuff that eventually can lead to heart attacks. And so that's why your doctor is alarmed if you have too much LDL, because you have a lot more material available that could be oxidized, that could kill you. There's actually a really complicated, interesting story about what vitamin E does in the LDL, and how it interacts with vitamin C. And if you take my kinetics class, I'll tell you about that. But we'll do the simplest case now. There's no vitamin E, there's nobody Vitamin C. You just have some lipid, and it's oxidizing. And let's model what's happening. So what you have is a little bubble of fat. So out here is water. And here's my fat which I'll politely call lipid. That sounds like not quite as fat, if you call it lipid, slimy. And the lipid is susceptible to attack by this reaction. So if I have some peroxy radical, and I have the lipid molecules, which are some hydrocarbon, they can react and make peroxide plus a radical. And in your blood vessels, if you're not under the water or being killed, you have a lot of oxygen. And so the carbon set of radicals you form immediately react with oxygen to make the peroxy radical back. And so the net process-- we can kind of combine these steps, because this is super fast. So it's like a microsecond times scale. For that to happen in your body, because the oxygen concentration is so high in the blood. And so this reaction is really roo, plus RAH, plus O2, to rooh, plus roo. So the radical is just catalyzing the oxidation of your lipid into the peroxide. When this level gets too high, then your body will detect it. There's an inflation, and you'll have all kinds of trouble. OK, so we like to understand this. Now, these vesicles are very small, so there's a probability you might have only a very, very, small number of radicals in there. Also, your body has a lot of antioxidants in it, like vitamin C and vitamin E and some other ones too. And the antioxidants are trying to get rid of radicals. So the number of radicals you'd expect the concentration-- background concentration is very tiny. And because of that, this is tiny and the concentration is tiny. So the concentration times the volume might also be tiny. So the total number of radicals might be very, very small. And so therefore, you might need to worry about the stochastic kinetics being different than the continuum kinetics. OK, so that's the idea. Now, on top of that, you also have a lot of these vesicles in your body. So you might be able to somehow work up some continuum model, treat them all. And in fact, I think that's what people did originally, is they just said well, let's take the total amount of LDL you got in your body and do a continuum model. But as you will see, it's different, because the fact that the vesicles are small, each one might only have, say, one radical in it. And that will actually behave quite differently than if you had a lot of radicals. All right so, this is the kinetics, and the oxygen is almost always high concentration, so I'm not going to worry about that. And I'm going to assume that we're not going let your lipid oxidize so much that the amount of lipid changes significantly. So let's just, for now, we can just ignore this. OK, maybe later we might go back and figure out how you'd account for the fact that these cells are being consumed, which would change the rates a little bit. So we'll just assume that concentration is constant. So you have a tiny amount of radicals. And if you had the radicals in your vesicle, they're creating roh. And you see, the ro is catalytic. And so what we're gonna see is basically D. If you did it in classical kinetics, you'd write DR dt this is basically equals some constant times roo. And then, I lumped these concentrations into this constant. [INAUDIBLE] That all right? All right, so that's the import process. We also have another important process, is that if you get two radicals in there, and they bump into each other, they can kill each other off. So two peroxy radicals can combine and make stable molecules. And we don't care what stable molecules they really are. What's important is that they get rid of the radicals, because the radicals are catalyzing the unfavorable oxidation. In this program I started writing, I called this k3. This one is k4, reaction four. There's two other things happening. Out here, I had some peroxy radicals. They're floating around the water. Some of them might go into my lipid. I'll call that process K1. There's some time constant, which new radicals arrive. And then, I also have the possibility these guys could diffuse out. So I call that k2. And k1, I'll just treat as a constant. So as soon as a background concentration radicals that's constant, and so this is a sum-- every 10 seconds or something, a radical arrives in a vesicle. And this will depend on the concentration, the number of ro's. So the more you have inside here, the faster the rate which they go out. That all right? And this just mass transfer. But from the point of view of the equations, it doesn't care if it's mass transfer or chemical reaction. They arrive in the same way. All right, so that's the system. And so let's look at the-- I should try to write the MATLAB code for this. And maybe, we can try to finish this together in class. So I have four processes-- arrival of new radical, diffusion of the radical and the vesicle, reaction to make the peroxide, and self-destruction of the radicals. And I wrote down what each of these does. So the first one increases the number of radicals by one. The second one gets rid of a radical because it's diffused away. The third one increases the amount of peroxide. The last one gets rid of two radicals. So we're going to try to compute a trajectory by the Kinetic Monte Carlo method. And trajectory is going to be some time points, the number of radicals at that time point, the number of peroxides I have at that time point. And as I run, I'm going to get many, many, different time points. And each one will have a different number of radicals or a different number of peroxides. And that's what I want to compute. And then, at the end, after I have trajectories like that, big tables of these things, I want to somehow put it together to figure out on average what happened or something, try to understand it. And so let's see, so my inputs are the initial numbers of radicals and peroxides in the vesicle, the vector of rate coefficients, the max amount of time I want to run the trajectory for-- not clock time, but time in the simulation. So it's like how many microseconds or milliseconds or seconds I look at my lipid protein and see what it's doing. And the maximum number of steps, because I don't want to get a trajectory list that's 10 to the ninth long, because over a while, my memory, my computer would cause me trouble. So here they are, all the setups. Everything is fine so far. So here's the loop. So the outermost loop is just to keep track of how many steps we have, which is going to be the number of entries we have in the trajectory table. And the second one is really-- what we care about-- is well, the time is less than the maximum time of the simulation. First, we want to figure out the time [INAUDIBLE] until something happens. So following Joe Scott's notes, we compute this quantity called A, which is the sum of all the rates. And that's equal to the rate of the first process. And then, they're in the second process, which depends on the number of radicals. And the rate of the third process all depends on the radicals. And the rate of the fourth process, which is a number radicals times the number of radicals minus one. Now k4 has to have units of per second. But bi-molecular rates would normally have units of volume per second, per mole even-- volume is per molecule per second. And so that k4 has to really be a normal kind of k divided by the volume. And we're going to have to figure out what to do about that in a minute. And then, we do the formula to get the-- Gillespie's formula for how to sample from an exponential decay, as good as that is. So Gillespie thinks that the probability that something is going to happen, or the probability, the time, until the next thing happens is going as e to the negative a times t. So we're trying to sample from e to the negative, a, t. And that's what that crazy log of the random number is doing-- it's sampling from that distribution. And so we get a [INAUDIBLE] that how long we've waited from the time the last thing happened until the next thing happened. And now, we have to figure what happened. So there's four different processes that could have happened. A radical could have arrived. A radical could have left. A radical could have reacted with one of my hydrocarbon molecules, with one of my lipids, or the two radicals could have killed each other. So we list these guys out. A is the sum of all those processes. P1 is the probability that the first process happened, that a new radical arrives. So it's going to be the k for the first step, divided by a. That one was zero water, because it's subterfuging it from outside. So it's just a k of 1. Probably the second step is k2 times n rad. That's the rate of the second step divided by the total rate of all the processes. And I'm going to add that onto p1 to make a vector of possible things that could happen. So I'm picking a random number from zero to one. And I want to make a little bar. Here's the probability the first process happened. And here's the probability that the second process happened. And here's the probability the third process happened. And here's the probability the fourth process happened. I know that these guys have to add up to one, because I computed that something happened. And so I'm going to try to compare my random number from zero to one to these breaks and sort it out into these four possibilities, and then figure out which thing happened. OK, so this is the second step of Gillespie's algorithm. All right, so maybe you guys can help me finish the code here. If you want to go to Broadway in Chicago, you can go there. So I need to write the next one. So p3 is equal p2 plus what? So step three is-- sorry? Say again. AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: All right, that's enough, right? So now, I get to choose a random number, so, say, x is equal to rand. And then, I have to say if x is less than p1, than something. What's it gonna be? Then step number one happened. So that means that n rad is equal to n rad plus one, so one radical transported in. If x is less than p2-- [INAUDIBLE] Is that good? AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: [INAUDIBLE] space. Thank you. What happens here? So this is b, n rad is equal to n rad minus one? AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: I think they also take care of that, right? Is this right? X is less than p3, than nroh. Else and rand. Does that look right? Do I need an end? AUDIENCE: Semi-colon. WILLIAM GREEN JR: Semi-colon. Semi-colon, thank you. That would be really bad. All right, yes, Ziggy? AUDIENCE: So in the first problem, you have step size longer than that [INAUDIBLE] size. WILLIAM GREEN JR: So that's not good. It should be less than. Thank you. Any more bugs? AUDIENCE: I think you need spaces after the brackets. WILLIAM GREEN JR: Spaces after the brackets-- where? Is that here? AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: [INAUDIBLE] separate line. AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: I can do that. It's easier to read anyway. So that's good. I definitely agree with that. [INTERPOSING VOICES] WILLIAM GREEN JR: Is that right? AUDIENCE: Yeah. WILLIAM GREEN JR: You like this? With all these [INAUDIBLE] sets, I don't need any n's, is that right? [INAUDIBLE] after the other like that, no problem. Just one end at the end. [INTERPOSING VOICES] WILLIAM GREEN JR: That's good. All right, so we're OK with this? We think this is going to work? AUDIENCE: [INAUDIBLE] [INTERPOSING VOICES] WILLIAM GREEN JR: All right, so now, we've computed what's going to happen. Now, we need to store results somehow. So I'm gonna say trajectory is equal to-- or trajectory of steps-- so step was zeros. Now, step is one. [INAUDIBLE] step is one, to start with. [INTERPOSING VOICES] WILLIAM GREEN JR: Does this look good? They had only commas, is that right? No commas? AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: All right, one step is zero. I already have the first line filled in here. [INTERPOSING VOICES] WILLIAM GREEN JR: Is that good? We'll see. I don't know if it's OK or not, but I'll believe you. All right, so right now, we have a program that maybe or maybe not might actually compute something. Is that right? OK, now, we to figure out-- are we going to run this program? So we need an initial. So any suggestions? [INTERPOSING VOICES] WILLIAM GREEN JR: All right, so let's do an initial-- how about n rad is equal to 1 and nroh is equal to 0. Is that OK? We're cool with that? And how about k? So we need a vector of k's. Yup? AUDIENCE: For your line 18, I think that's just the [INAUDIBLE] WILLIAM GREEN JR: Yes, thank you. Yes? All right, so let's think about what's reasonable for the k's. So I think we can make the k for this stuff diffusing in very, very small. So I don't know-- [INAUDIBLE] make a small, 1e minus 5 something. Isn't that small-- only once an hour. You think it might be a little too small? And then, the next one is how fast do things diffuse out? Now, what's that going to physically depend on? It should be something like the diffusion. Diffusion [INAUDIBLE] get back to the wall, back to the outside. So it should be-- by one radical in here, what's the average time for it to diffuse back out? Any ideas? So if it does, say, just a random walk, maybe-- then you say that delta x squared is like d times tt. OK, so delta x squared, maybe make this like the radius of our lipid vesicle, so that delta t is like r squared over d. Seems like a reasonable scaling. And we actually want to rate, so we want the other way around. So k2 is going to be something like d over r squared. All right, so we need to pick a size of our LDL vesicles. I don't actually know what they are. I used to know this, but I don't remember. Anybody want to make a guess, how big is a vesicle inside your blood vessel? [INTERPOSING VOICES] AUDIENCE: Micron. WILLIAM GREEN JR: Micron, OK. So let's pick a mircron. So let's say r is equal to 10 to the minus 6 meters. What's a diffusion constant in liquid phase? [INAUDIBLE] AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: [INAUDIBLE] meter squared per second. OK, so feasible. [INTERPOSING VOICES] WILLIAM GREEN JR: Anybody know if it's meter squared or centimeter squared? What? [INTERPOSING VOICES] WILLIAM GREEN JR: You guys did the one of the drug. What was the drugs one? Was it 10 to the minus 6, meter squared or centimeter squared? [INTERPOSING VOICES] WILLIAM GREEN JR: Meters squared. OK, so for a light, smaller molecule, it's reasonable. So this number should be something like 10 to the minus 5, divided by 10 to the minus 12, so something like 10 to the 7. 1, 8, and 7. All right, now for the reaction, roo plus RH, we can go look it up on the [INAUDIBLE].. They'll have a list of reactions like that. For now, let's just guess. And so let's say that k3-- k of roo plus rh-- these guys typically have 8 factors of 10 to the eighth leaders per mole second. And they have ea's, about 15-- 15 k cals per mole over rt. OK so now, we need to figure out how to turn this into the k3 we want. Now, the k3 we want is really-- K3 is like this-- times the concentration of the rh, because we took the rh out of the problem, because we want this to have units of per second. Is that right? And this is in units of volume per meters cubed per mole second. And this would be moles per meter cubed. And so it's per second, which looks right. OK so now, we need to have-- I understand we have a calculator? We can try to calculate this number for room temperature. And this does not mean room temperature is r time t-- actually for body temperature. And the concentration of hydrocarbon-- it's actually the number of h's, because every h can be attacked in a hydrocarbon. So you typically have densities of 0.8 grams per centimeter cubed. It's for organic. And you typically have two h atoms per 14 grams of ch2 groups, because you only see h2 groups in there. S two h atoms. And then, if we're going to try to keep moles here, we need 6 times 10 to the 23rd atoms per mole. All right, and this is centimeters cubed, but this one has leaders, so I need 10 to the 3 centimeters cubed [INAUDIBLE].. So again, verifying what your high school teacher taught you, that the first thing is to learn units. [INAUDIBLE] So we take all these numbers. This is the RH, and this is the k. And we can multiply them all together, and we should get something reasonable, maybe. And since we have MATLAB we can make it do it for us. So let's just do that. So I think it's-- you have to help me, I can't read this very well. 1ea star exponential, [INAUDIBLE] 15,000, slash 1.987 times-- what's body temperature? 40ec you told me? So it's 310 Kelvin maybe-- times all these numbers, 0.8 times 2. Divide by 10 to the 23rd. More factors in there. OK, so 5 times 10 to the minus 25. We think we got this right? AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: But I want to get moles, because I have this in moles-- it's [INAUDIBLE]. Three times mole [INAUDIBLE] This is really mole of [INAUDIBLE] atoms. OK, so we'll try and see what happens. So 5 times 7 minus 25. It does seem pretty small, so we have a problem. And then the last reaction-- those reactions are typically 10 to the fifth, liters per mole second. This is for a recombination of peroxy radicals. So now, we have to get rid of the volume unit. This is for roo-- sorry, [INAUDIBLE] Yeah? Yes, question? AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: One mole per [INAUDIBLE].. Very good-- that's right, because [INAUDIBLE] that's just two atoms. It's actually two moles. Yes? That would change the [INAUDIBLE] Only a factor of 10 to the 23rd. So we'll get back. Thank you. Area 0.3, somewhere like it. OK, so now let's go do the same thing with this one, ro plus roo. So typically these reactions, the normal way they're written are numbers that are about 10 to 5th meters per mole second, for two radicals recombining. So now, I need to figure out how can I change the volume here. And the thing that is key is that I know the volume of this lipid particle. So what I really want is this. My k4 is going to be equal to normal k, divided by the volume of the reactor, because the rate we would write normally, which would be droodt-- it would be like negative 2k roo plus roo, times the moles, the concentration of roo squared. This how we would normally write it. And so we have to have the-- this unit is all right. So this is the volume of the reactor. So we were having nroo over v, so we might need Avogadro's number over here, because the rate here is in moles. But now, we are going to do single molecules. Yeah? AUDIENCE: Do you ever actually use k4 in calculations? WILLIAM GREEN JR: Yeah, for a. I think an a is there. Good question. All right, so we think this should be like a volume per molecule or something like that, for one molecule. So this should be k4, should be 10 to the 5th, meters per mole second, divided by 4/3 pi r cubed. And then, we need a mole-- [INAUDIBLE] 23rd molecules. And we're gonna have to make sure that the leaders and these guys match up, which is not going to be so easy. You might use 10 to the minus 6 meters. This is going to be meters cubed, so I need 10 to the 3 liters per liter cubed. And I think that will all work out to be per second. So let's try it. So this would be 10 to the 5th over 4/3 pi is about 4-- 10 to the negative 18, 6 times 10 to the 23rd. So all the big ones cancel each other out, is a 1 over 24. Oh, I lost it. That's bad. OK, so one over 24,000. It's a lot easier to do with a lot of people checking your work. All right, so four times n minus 5. So this is what we think k is. Anything else we need. We need a t-max, you want to guess the time? You want to wait? The time for the main reaction, the time constant is like seconds, right? So if we made 1,000 seconds, a lot of stuff is going to happen, right? So maybe 1,000 seconds will be OK for t-max. So let's try this all. Where's the main function. The main function-- [INAUDIBLE] my n initial was simple 1, 0. And k, which was decided was 1e minus 5, [INAUDIBLE],, 2.3, 4e, 25. Now, if you just look at these rates for a second, I think you can see a problem we're going to have. How many steps are we going to do? I don't know. 10,000. The time scales here-- this 27 is pretty fast. So the stuff is going to diffuse out of there, because it's going to disappear, and then nothing is gonna happen. Yeah? AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: Centimeters squared, I thought so, yeah. Thank you, so we think this is 10 to the minus 9. So this is actually [INAUDIBLE] lower, is that right? We're still gonna have a problem. But it's not quite bad [INAUDIBLE] problem. So let's fix that. Um, so I think our problem is that the 1e minus 5 is actually too slow. So radicals are diffusing in from outside at some rate, and it's got to be a rate so that if there was no reaction, basically, the concentrations of the stuff here and here might be about the same, same order of magnitude, of the stuff that dissolved in the water, dissolved in the lipid. Maybe a couple of [INAUDIBLE] are different, but not a million times different. And so we need to have the rate of this stuff coming in to be something reasonable, that's going to be consistent with having a chance of having some radical in there. So now we to guess what we think the real outside world radical concentration is. So maybe-- I don't know, what? 1e minus 10 moles per liter? 1e minus 6 moles per liter? I don't know. Any biologists here? Do you know how many free radicals you have in your body? AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: They can measure it, right? So it can't be zero. People talk about reactive oxygen species in your body, ROS. Yes, so I don't know what it is, 10 to the minus 6, maybe-- 10 to the minus 8. So we had the number for-- 1e3 e3 is for going out from the-- we have that rate leaving is 1e3 per second times the number in there divided by the volume, really. Well, it's times the number, that's the rate that's leaving. But we would think of this as a k. I don't know how to think of this, actually. But for sure, the volume matters. So in order for us to compute this, we use the r squared. Now, another way you look at this is say well, is diffusion times the gradient in the concentration? AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: [INAUDIBLE] Well, it's OK. They would live in their little tiny thing for a millisecond, and then they diffuse out. I think that actually sounds very reasonable. It's only a micron. It's a very tiny, little thing. So the question is how do we complete the reverse? So if we have stuff outside, and suppose we didn't have anything inside, we'd have some rate of diffusion of the radical from the outside coming in. How do we think about it? It's really got to be the inverse of this. So if we think the time constant is 10 to the minus 3-- a millisecond to come out, it's probably about a millisecond to come in. I don't know. You think it's got to maintain whatever initial concentration we put there, one per volume, then it's got to be about the same. So this can't be as tiny as 10 to the minus 5. So if we take the average concentration as one, then this should 10 to the 3, I think. If we think the average concentration is 10, then it should be 10 to the 4. If we think the average concentration this 100, then it should be 10 to the five. So let's try 10 to the 3. So every millisecond, something comes in or goes out, some radical comes in and out. You guys buy this? Would you defend this to your boss? You can blame it on Professor Green. All right, here goes nothing. Do you think it's going to work? [LAUGHTER] Aww, aww, line 29. What's wrong? What's wrong with that? Parentheses off? AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: Maybe I didn't save the latest version, let's try again. Ah, I didn't say the latest version. That's what it is. Now, here goes nothing. That [INAUDIBLE] help. Leave me alone. All right, trajectory, the first column is the times, so it uses the x-axis. And the last column is the oxidation, the products. So let's see if anything peroxidized or not. [LAUGHTER] [INTERPOSING VOICES] WILLIAM GREEN JR: Oh, it should have all the zeros, right? So how can we do it when we [INAUDIBLE] that way. That's really weird. Try it again. [INTERPOSING VOICES] WILLIAM GREEN JR: The time is 2 plus tau. AUDIENCE: [INAUDIBLE]. WILLIAM GREEN JR: I guess if you look at the [INAUDIBLE] So trajectory-- so it only [INAUDIBLE] 2 points. That's why it looks like that. I don't know why it doesn't show the zeros. Maybe they'll show on top of the origin there. So why did it only give us two points? AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: Yeah, also-- seems like a 307 oxidations, but why 307 all of a sudden? That makes no sense, right-- because it should be one at a time. I should be seeing one at a time coming in. AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: Yes, it's very worrying. I agree, so we have a bug. So where is our bug. This is where I need your help. OK guys, let's figure out why didn't this work. So I'm storing-- by stepping the steps, they're going up. Ah, I know what's wrong. I need this step thing inside the loop here. So all of this-- so let's see if I can explain this. I'm storing them according to the step, but right now, the wild loop is just zipping around, and then, it ends at the step and just gets one. So I'd rather start again. It's gonna work this time? You guys don't have any faith. If people say that scientists and engineers don't have any faith, I think we have more faith than anybody else. We believe stuff like this is going to work. It's doing something now. Who knows what? [INTERPOSING VOICES] WILLIAM GREEN JR: Yup? AUDIENCE: [INAUDIBLE] time until your arrival time, 7:15 [INAUDIBLE] WILLIAM GREEN JR: [INAUDIBLE] is, not good, right? AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: Yeah, it's a very good point. I have a nested loop that should not be, right? AUDIENCE: [INAUDIBLE] You don't necessarily want to sample every arrival time. You might want to sample [INAUDIBLE] So you might want to have two. You don't necessarily want to [INAUDIBLE] every single-- WILLIAM GREEN JR: Yes, yes, I agree. AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: OK, so let's see how we can fix this. So we really want-- is it really a jump out? If the number of steps gets too large, we just want to jump out, so we want to go to? What do you think? [INTERPOSING VOICES] WILLIAM GREEN JR: Break, that's what we want. So while is not the correct thing to do here. AUDIENCE: Why would t [INAUDIBLE].. WILLIAM GREEN JR: So we're stepping time, every time we'll do the steps. And we just want a break. [INAUDIBLE] step greater than max steps. AUDIENCE: Why don't you do it with a while loop [INAUDIBLE] WILLIAM GREEN JR: What happens if you do a wild loop like that? Do you guys know? Does it work? [INTERPOSING VOICES] WILLIAM GREEN JR: It'll work. OK, that's easy. Double and-- I tired it with and this morning. It didn't work, so that's why I stopped doing it. Stuff like that will kill you, doesn't it? All right, we gotta get rid of one of the ends at the end. AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: I think it's likely to cause a problem, why it's taking 10 million years. So really, is there a reason it should be nested loop, right? Yes. AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: All right, try again. [INTERPOSING VOICES] WILLIAM GREEN JR: That's too fast. Try that again. That's not good. See we have a [INAUDIBLE] here. AUDIENCE: Oh no. WILLIAM GREEN JR: Oh, jumps to one. That doesn't look very good. Try that again. Boom. AUDIENCE: Oh! WILLIAM GREEN JR: Oh, that was pretty interesting, so much more like what you expect, so that at some time steps, I have one, two, three, or four, or five radicals in this vesicle. I only have one time separate because of the nine. So I think this actually looks like what I expect. I should see it jumping up and down, stuff goes in and out, sometimes it reacts, sometimes it goes away. So this one is correct. I don't know what's happened with the other one. So I'll have to figure out what's wrong with that equation. All right, we're almost done time. Well, I want to talk about for just one minute is-- what are we gonna do after we have this working? So we get this working, we have the trajectory, we really want to run 10,000 trajectories, because all we're doing is sampling from the probability, just the distribution of time. So we need to run zillions of them. So this whole thing we just wrote would be inside a loop that runs a lot of different cases, because every time you run this, because the random number, you're going to get a different trajectory. And what you care about is some kind of average behavior. Or you might want to know what's the percentage chance that I'm going to have [INAUDIBLE] oxidized and die? So you might say I want to count what fraction of the trajectories end up with number of peroxide greater than some number, that means I'm dead. And if that probability is too high, then I know I'm in big trouble. So that'd be one possible thing. You might have whatever objective function you want, but you need to run a lot to get good statistics for anything. So you're gonna have to embed this whole calculation inside a loop, and then, you're going to get a zillion of these trajectories out. And you figure out how are you going to analyze them in order to figure out what you want? So one way is if you could add the trajectories up-- so I have a trajectory-- suppose I have the number roh's versus time. And I do it once, and I have none. Then I have one, and then I have two, and then, I wait longer, and I get three. And then, I wait longer, I get four. Whatever, something like that. That's what it really looks like for one trajectory. And then, the next time I run it, it starts in a slightly different time. And this time it runs longer before something else happens. And then, it gets here, and then maybe, it goes over here. And I have a lot of them like that. I have 10,000 trajectories, all look like that. So if I could add them, then I can do an average, to get an average trajectory, so that would be one possible thing. Or I might want to histogram, so I might pick one time point. So like after 20 minutes, I want a histogram of what this looks like here or here. One of them has four peroxides, and one of them has five peroxides. And then the other 9,999-- some of the three, some of the whatever. Yes, question? AUDIENCE: [INAUDIBLE] you're not gonna get the same time points-- WILLIAM GREEN JR: You're not going to get the same time points, that's right. So see the first time it stopped, the first time point was here, the second time, the first time point was here. And they'll all be different. So one is I was asking a few of the students before the class started, there's got to be a program in MATLAB that will let you generate these kind of plots with the flat lines. Or even a linear interpolation would be OK too. But you need to have some way to add them up as continuous functions, because they all have different time points. So that's one practical issue about it, because you want to pick some special time you care about, and you want to know what does the trajectory say, what does this trajectory say? But you actually only have numbers here, just really you have that number and this number and this number. That's all you got. Yeah? AUDIENCE: You can plot like a-- an out of time to get to n amount of roh's and then, that would be [INAUDIBLE] WILLIAM GREEN JR: OK, so then you plot [INAUDIBLE] versus time instead, that's right. So that you just think of what you want. You're going to have all this trajectory information. And then you need to figure out what do you want, and then, what are you going to plot? And what are you trying to compute? And you might want to compute not only the number but the standard deviation of that number, because you don't know how many trajectories you have to run before the center of deviation is narrow enough that you'll be confident with it. So this is a general problem with this whole approach-- is that what you're getting out are just samples of things that happen. It's just like if you were doing a Monte Carlo, and you just got some of the energy values from the hydrogen peroxide calculation you guys did, you have 47 of those energy values. What are you going to do with that? So you have to figure out, how are you going to take the [INAUDIBLE] of this calculation. It's sampling from the real solution, so it should be OK. But what is it? How are you going to handle that and use it as a means-- it's sort of like as if you ran a zillion experiments, and then, what would you do with all that data? So that's like issue number one. Now alternatively, you can rewrite this as the master equation and solve it as an ODE, in which case, you'll get explicitly p of each of these ends. So n rad, nroh time-- it would be continuous in principle, but actually, the ODE solver gives you out random time points also. So you get a similar kind of thing out from the ODE solver, except it'll give you probability for every possible range of these guys. So if you have one radical, two radicals, three radicals, four radical, fie radicals, and number of peroxides from one to 1,000, or however many peroxides you get, you're going to get numbers. So you have all these numbers, all these probabilities, at different times. Again, what are you going to do with that? So one thing people do a lot is they would plot, say, the number of roh's versus time. So at different time points, compute the average over all your trajectories. Were with, if you had this solution, you can compute that as nrooh, p, nrooh. So like this, you're gonna have some kind of average if you do it that way. And either one are fine, but you have to think of what you want, and then in both cases, you'll probably be interested in the dispersion. So you [INAUDIBLE] want to worry about the n squared too. We'll talk more about this on Monday. All right and I posted a homework problem that was given last year about this, for Kinetic Monte Carlo for catalysis. It's used a lot in heterogeneous catalysis. And if you have extra time, feel free to do the problem. It's a really good problem. Those of you who are feeling like you don't have any extra time, and you're about to kill yourself, please don't kill yourself. And instead, just send an email to me or Professor Swan asking for an extension. And we can give you some more time to finish up the homeworks that's due tonight. All right, talk to you later.
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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 my comments. First one-- if the psi n of t can be chosen to be real, the geometric face vanishes. So why is that? The geometric face is, remember, the integral of this new factor, which was i psi n of t psi n of t dot. You had to integrate that thing to get the geometric face. We explained that this quantity in general is-- well, in general, it's always imaginary, this quantity. And then with an i, this is a real quantity which we've been using. But if this is imaginary-- look at this. This is imaginary. How can it be imaginary if psi is real? It can't be. If psi is real, imagine doing that integral. You can kind of imagine. So it has to be 0 if psi is real. Yes? AUDIENCE: [INAUDIBLE]. Why did we say that was imaginary? PROFESSOR: We proved this was imaginary. We did calculate the derivative with respect to time of psi with psi, and we proved that this was imaginary. And now I'm saying if this was imaginary but if psi is real, the only possibility is that this is 0. But you can prove, in fact, that this is 0, if this is real. I can do it in a second. It's i integral psi n of p-- because it's real-- d dt of psi n of t. Output an x here, and this is dx. That's what it is if psi is real. Otherwise there would be a star here. But if it's real, there's no star. But this is just the integral of 1/2 of d dt of psi of x and t squared dx. The d dt is that. And then the d dt goes out of the integral-- so this is i over 2 d dt of the integral of psi squared of x and t dx. But that integral is 1, so the derivative is 0. So there's no geometric face. So if you have real instantaneous eigenstates, don't even think of Berry's phase. There's another case where you don't get a Berry's phase. So when I'm speaking of Berry's phase at this moment, I mean the Berry's phase from a closed pathing configuration space. So if the configuration space is one-dimensional, the Berry phase vanishes. Barry's phase vanishes for 1v configuration space. Why is that? Well, what do we have to do? The Berry's phase will be the integral over the closed path in the configuration space psi of r d, d r-- because there's just one side of r-- d r. You have to integrate that thing. That is the Berry phase-- is the integral of the Berry connection over-- this is capital R-- over the space. But now, if this configuration space is one-dimensional-- this is r-- a closed path is a path that goes like this and black. It retraces itself. So it integrates this quantity with increasing r, and then it integrates the same quantity with decreasing r, and the two cancel, and this phase is equal to 0. So you cannot get a Berry's phase if you have one dimension. You cannot get the Berry's phase if you have real instantaneous eigenstates. But you can get a Berry's phase in two dimensions, in three dimensions, and there are several examples. I will mention one example, and and then leave Barry's phase for some exercises. So in 3D, for example-- 3D's nice. A 3D configuration space is the perfect place to confuse yourself, because you have three dimensions of configuration and three dimensions of space. So nice interplay between them-- r1, r2, r3. And then you have an integral-- the Berry's phase is an integral-- over a closed path here. So let's call gamma and let's call this surface, s, whose boundary is gamma. So what is the Berry phase? The Berry phase is the integral over gamma of the Berry connection d r. But what is that? That is, by Stokes' theorem, the integral over the surface of the curl of the Berry connection times the area. And so the area on that surface. Stokes' theorem-- remember in E&M? So here was a vector potential gives you a magnetic field-- so the integral of a along a loop was equal to the flux of the magnetic field through the surface. And now the berry phase along the loop is equal to the integral of the-- we should call it Berry magnetic field? No. When people think berries-- curvature. As in the sense that the magnetic field is the curvature of that connection. So this is called the Berry's curvature, but you think about magnetic field-- Berry's curvature. So the Berry's curvature-- people go with d is the curl sub r of the Berry connection a of r. It's the magnetic field, so it's the integral of d over that surface. So they're nice analogies. So one example you will do in recitation-- I hope-- you have Max over there-- is the classic example of a spin in a magnetic field. So I will just say a couple of words as an introduction, but it's a very nice computation. It's a great exercise to figure out if you understand all these things. It's done in [INAUDIBLE] explicitly in some ways, and it's a great thing to practice. So if you want the little challenge-- I know you're busy with papers and other things. but the idea is that you have a magnetic field-- the cogs of a magnetic field, B0n-- along some direction n, and you have a spin 1/2 particle pointing in that direction n. But then you start letting it vary in time-- that unit vector. So this unit vector varies in time and traces a path. And if it's idiomatic process, the spin will remain in the instantaneous energy eigenstate, which is spin up in this direction, and it will track the magnetic field, basically. There will be some small probability it flips, but it's small. And if this is an idiomatic process, that error can be small. So as you move around here, there will be a geometric face, and the geometric face is quite nice. So for the spin up state-- m+ state-- if it follows this thing, the geometric phase for the spin up state in this closed path in configuration space-- this is the configuration space of the magnetic field-- will just be given by minus 1/2. This comes from spin 1/2 times an invariant of this loop, which is the solid angle traced by this loop. So the geometric phase that's acquired by this motion is proportional just to the solid angle of this loop. It's a very nice result. Shows how geometric everything is.
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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: This recitation on mass spec is supposedly associated with reactive oxygen species. So [INAUDIBLE],, which happens all the time in this course, because we can't describe all the techniques as we go along. So what I'm going to do is just give you a two second overview of what you need to think about to put the paper you read into the big picture. I don't think the paper-- the paper also explains it. And this week we're going to focus on the mass spec paper, which is mostly sort of trying to figure out the technology, and then the next week is focused on the biology. And so the major unsolved problem-- so everybody and his brother is using mass spectrometry as a tool nowadays. There has been a revolution in mass spectrometry. [INAUDIBLE] The instrumentation is cheaper. The mass spectrometric methods have just really taken off. And people didn't even know who mass spectrometrists were, but they're starting to win major prizes, because it's revolutionized what we can do. I was talking to somebody yesterday, and they just got a mass on a protein that's 3.3 million. How do you get a protein into the gas phase that's 3.3 million? Right. Doesn't that sort of blow your mind? Anyhow, it's been a revolution. And we're going to be looking at-- this module seven, which is on reactive oxygen species, and we've been talking about the question of homeostasis. And so one of the things with these reactive oxygen species is they are used by us to kill bacteria, viruses, or parasites. But now, in the last five years or so, everybody's focusing on the fact that here are these reactive oxygen species that play a key role in signaling, which is everywhere, and the signaling process we're going to be looking at next time and is alluded to in this particular paper is epidermal growth factor receptor and epidermal growth factor. There are hundreds of these proteins that have receptors that are involved in growth and epidermal growth factor receptor [INAUDIBLE] of successful cancer therapeutics. So it's interesting what happens up here, what happens down here, how do you control all of that, and people are studying this. So we've already seen cystine is unique. And if you have a reactive oxygen species, and we'll see that the reactive oxygen species we'll be looking at, when we're going to be looking at a number is actually superoxide. So that's one electron reduced oxygen, which in the presence of protons can rapidly disproportionally give[?] oxygen gas to hydrogen peroxide. And hydrogen peroxide can react with cystines to form sulfenic acids, which is the subject of the paper you had to read. And so the question is how prevalent is this, and the question is, is this important and interesting in terms of regulation inside the cell? And so the key issue is-- even cystines aren't all that stable, you know, if you have proteins with cystines, and you let it sit around for a long time, you could form-- and the protein's concentrated, you could form disulfides. It's not a straightforward reaction, but you can form disulfides. The question is if you had hydrogen peroxide inside the cell, which you do, can you form sulfenic acids, and do they have a consequence biologically? OK, and that's the question we're going to address next time. And so the issue is this is unstable. So if you want to develop a method to look for this species, and you start cracking open cells, and you start working it up, what happens is this falls apart and reacts and gets destroyed. And an example of this is the area of DNA therapeutics and DNA drug interactions, therapeutics that interact with DNA. For decades, you see lesions on your DNA. How do you determine what the lesions are? Mass spec has been a major method to look at that. Almost all the lesions in the early days were complete artifacts of the analytical chemistry to work them up. They had to get them into some form that you could stabilize the lesion and then analyze it. And what was happening because they weren't careful enough and quantitative enough, they changed it to something else. And so the data in the early years was all completely misinterpreted. So the issue in this paper is that other people had developed this, and Kate Caroll has taken this on. Can we have a way of derivativizing this [? mentally ?] inside the cell, because if you disrupt-- if you disrupt this by cracking open the cells and trying to purify things, it undergoes further reaction, and what this can undergo further reaction to is SO2 minus, sulfonic acids or SO3 minus. Sulfinic acids and sulfonic acids. OK. And it turns out this reaction is also reversible with hydrogen peroxide a lot of people are looking at that at this stage is irreversible. Anyhow. So the question is can you develop methods to look at all of these things. And in fact Tannenbaum, who was in the chemistry department, but also [INAUDIBLE],, he is looking at nitrosation of SHs, again forming a reactive species, and he's developed new methods sort of like Carroll has to try to specifically look at these modifications. And in the end, what you want to do, and this is the key, you might be able to detect this-- the question is, is this interesting? So you have to have a way to connect this back to the biology inside the cell. And that's what the second paper is focused on. So what we're doing today is simply looking at the technology that's been developed to try to get a handle, how do you look at sulfenylation, you're not really focusing on the biology of the consequences. And so what we're using is mass spec. And we're using a method of mass-- how many of you have done mass spec? So if you know something and I say something wrong, you should speak up, because I'm not a mass spec expert. And actually, I've got a whole bunch of information from, say, the Broad, and I thought it was not very good. So we need a way of trying to figure out that you're going to see-- there's hundreds of variations on the theme. I'm going to give you a very simplified overview of what things you need to think about. And so if I say something that you don't agree with, tell me. OK, so when looking at mass spec-- this didn't exist when I was your age-- using soft ionization methods, and what does that mean? It means that you don't want your molecules to crack. So the issue is that what mass spec is about-- so really looking at mass spec, and the key issue of what you wind up looking at is mass to charge. OK. So m over z. OK, so the problem is how do we get something charged enough so that the mass is small enough so that you can see it, taking a look at the mass analyzer, which is going to be part of all mass spectrometers. OK. So there are two different ways you could change the mass to charge. You could dump an electron in. And if you dump an electron in, that produces radical species, which can then fragment. We want to avoid that. That's not soft ionization methods. But how can we control this? The way we can control this is dumping in protons. So what we do is we can control it by adding protons or by subtracting protons. And we'll see that the different methods we're going to be looking at, we'll see there are two main methods that most of you have probably heard about your classes. One is electrode spray ionization, so ESI. And I think, if you're in Brad's lab, they have a lot of these. Yesterday's class had people that had used these, but really didn't know much about what's inside the machine. So this is the kind of thing I think your generation, if you're going to use this as a tool, need to roll up your sleeves and understand a lot more about what's going on, and MALDI, maser MALDI. Matrix Assisted Laser Desorption-- it will become clear why it's called that in a minute. So these are the two methods. And what we do is we can protonate, so that we can move this into the analyzer range, where we can actually read it. So what we'll see is the analyzer-- I'm going to show you sort of what the three parts of a mass spectrometer are-- can only read 1,000 to 2,000 daltons. OK, so if you look at your protein, much, much bigger. So you're going to have to stick a lot of charges on there to be able to see anything. So that's the whole thing, and the question is, how do you do it by one method or by using the other method? OK, so all mass specs have sort of the same components. And you can go to websites. The Broad does have a website, and what the Broad will tell you is what all these spectrometers are, but I don't think they do a particularly good job telling you what's useful for what, and why it's useful, which is, I think, what you need to use if you're only going to use it fleetingly and then move out. So you have a source. So you have an inlet. How do you get your sample from the liquid phase or the solid phase into the gas phase? OK, so that's going to be that. And so what is the distinct ionization method? And we will see that there are many ways that you can ionize, and we're just going to briefly look at in a cartoon overview of how this happens. And then so once you ionize it, it needs to move from the source. So you need to have ion movement into the analyzer. So this is the mass analyzer. And this becomes important. And we will see in a second that there are many methods to do the mass analysis, mass to charge analysis, and then after you do this, you have a detector. And then, furthermore-- and I think this is a big part of it now, if you're doing wholesale anything, you have to have a really sophisticated method of data analysis. And so that's the other thing that I get frustrated about all the time. So you see people-- I mean, people do experiments where they've spent-- last year, somebody spent three months trying to get all the proteins out of a cell, 10,000 proteins out of the cell by mass spectrometry right. Now, because the technology is changing, they can do it in four days. But what do you do with all this information? And how do you use this information in a constructive way, and how do you know if it's correct or not? So those are the kinds of things. I think if you're going to use this-- I think everybody is going to be using this technology. You need to educate yourself about how to look at this. OK, so that's what the issue is. And so we have a source, an analyzer, and a detector. OK, so this is just a cartoon of that, which describes this in more detail. And I think he put this on the web. I think he put the PowerPoint on the web. I was doing this at the last minute yesterday. So it's different from the handout I gave you that's written out. This is a PowerPoint. OK, so you can go back and look at this, but one of the other things I wanted to say is that sometimes when you analyze your mass, you want to analyze it further, and that was true-- many of you might not have caught it, but that was true in the analysis that was carried out in this paper. Did anybody recognize that you had to analyze this using more than one mass spec? Did you look at the data carefully enough? So also you probably didn't read the supplementary information, which also is critical to think about. I mean, if you want to look at the methods, you need to get in there and roll up your sleeves and look at them. So we're going to see that the methods that people often use is they don't look at the whole protein, but they degrade it down into pieces. So then you can find here a whole bunch of pieces, OK. But that doesn't tell you anything. The mass does tell you something. It might tell you whether it's sulfenylated or hopefully, you can distinguish between any other modification, but it doesn't tell you the location of the sulfenylation. And so you can do a second method. So you could have some other gas. There are many ways to do this that you bring this in to now take a peptide. So you pick one mass charge. You throw in something that's going to degrade it by fragmentation, and then I'll show you in a minute we understand what kind of-- using certain methods, we understand the fragmentation patterns, which actually allow you to sequence the amino acids. And the reason I'm bringing that in is when I first got to MIT, Klaus Biemann was in the lab, and I did many experiments with him. And these are the first experiments that were done to sequence peptides by mass spec as opposed to doing Edman sequencing, which the mass spec was actually better, and there are pluses and minuses, but I noticed from looking at the literature, people were still using the same method that he developed. So this is just a cartoon. And it just shows you that there are many ionization methods. We're focusing on these two, FAB, fast atom bombardment. We didn't have any of these when I was your age. Fast atom bombardment was something I used a lot because I've worked on DNA drug interactions, and it allows you to look at nucleic acids. And a lot of these other methods don't. I mean, we're focused on proteomics in this particular paper, and then mass analyzer. So you have time of flight. I think Brad's lab has MALDI time of flight. So what does that mean? You've got a long tube in here, and what happens is you have mass to charge, and they're different sizes. And so the smaller ones fly faster. They don't want to keep away from the walls, but the smaller ones fly faster than the bigger ones. So that helps you differentiate between all the ions you're actually looking at. I guess somebody just told me you guys just got a new quadrupole ion trap. Anyhow, if you want to look at this, I have notes on all these things. But I think this is something you'd have to study in detail. And so while I have pictures of them all and how you can differentiate one from the other, I think it doesn't really mean that much to me, because I don't know enough about the physics of how they were designed. I mean, this really has revolutionized what you can do. OK. So that's the components of all mass spectrometry. What I want to do is very just briefly look at the ESI and then look at the MALDI and then show you what the issues are in general, and then we'll focus right in on the paper, and the recitation I did on Thursday, we didn't quite get through all of it. We got through most of it, but then we'll continue next week and also attach this to the biology, which is the second paper, the nature chemical biology paper also written by the Carroll group. ESI. So that's the one we want to look at next. And so that's up there. This is a cartoon of how this works. So what do you do, and how do you do this? So the first thing is you have your protein of interest, which I'll call the analyte, because we want to charge. Lots of times you put it under more acidic conditions, pH 6 or something, 6 1/2, depends on the protein. So you get more charge states. And if you're trying to look at something big, you need a lot of charges on there to get it into this mass range of 1 to 2,000 to be able to see it using this method, and apparently what you do here-- can you see this? Have you done this? Can you see this capillary? Can you look at what's going on? AUDIENCE: I don't think so. JOANNE STUBBE: OK. So I was just wondering, because I haven't ever. So it's all closed off. It's in a box, and so you can't-- there's not like a thing where you can watch what's going on? AUDIENCE: Not that I've seen. JOANNE STUBBE: OK, because I think it's sort of amazing. How do you get this huge protein and solution into the gas phase? Right. I mean, that, to me, is like mind boggling, OK? I mean, these guys were geniuses. And you know, there's been a number of Nobel Prizes for this, but I wouldn't have a clue how to do something like that. So what you do is apparently, you put it down a capillary and then you spray it out, and then you have to-- so what you get at the end of this, this plume of spray, apparently you've got a lot of the analytes and a lot of solvent molecules, and then the goal is during this process, to get into the analyzer is to get rid of-- to separate all the analytes mixed together into a single analyte and remove all the solvent. OK, so that's the goal. And apparently, according to the people that were here yesterday, this is taken from, I think, sort of one of the papers that was first out. This is the way they did it in the old days. I don't know if they still do it this way, but the goal is really, to get a single analyte with no solvent on it. OK, and so the question is, how do you do this, and the chamber they had was at atmospheric pressure, and then they had a potential and pressure gradient, which allowed it to get into the mass, before the mass analyzer. So you start here with the initial spray, and then as you go farther, you remove some water molecules. You finally get to the place where you've removed enough water molecules that all these positively charged species come together, they're incredibly unhappy. And then they fragment apart. I mean, that's the way they describe it. It sounds reasonable. So you get smaller and smaller till eventually, you get to a place where you have an analyte that you can look at specifically and the water has been removed, and that's what you look at. OK. So again, we need to be in the range of 1 to 2,000. So that's the way these things work. Although, I think, again, how you get to looking at the single ions I think in different mass spectrometers. And so what the issues are, I think, are shown here, and this is the beauty of this methodology. So if you have a protein of 10,000 molecular weight, you couldn't see it, because the mass analyzer is limited. So you have to go all the way down to eight charges on it to be able to see it. And then, you divide that by that, and you get-- what do you have? You have to do some corrections, but you get something that's this size. OK, but you can see it now because of all the charges on it, but the beauty is if you add more charges, you get another peak. And you get another peak. And it all has the same information, and it just differs by the number of charge. So you have all this information. You can use that-- all these informations together to give you a very accurate mass on this system. So this method by analyzing all the data, and now the computers do this, I guess, routinely can give you a very accurate mass. So if you look at this printout, it doesn't look like that. This is what it looks like. And what do you think's going on there? So we look at mass charge, and we're in the range of 1,000 to 2,000 daltons. And then what is this all-- what is all of these peaks associated with? Anybody got a clue? AUDIENCE: Isotopes. JOANNE STUBBE: Yeah, so isotopes. So where are we seeing isotopes before? So these are mostly stable isotopes. We spent recitation two and three looking at radio isotopes. OK. I would say, you know, radioactivity is pretty important. Stable isotopes are extremely important to mass spectrometry. So if you get into this, you're going to be able-- you'll see that being able to label things with different kinds of stable isotopes is key to really deconvoluting the complexity when you're looking at a whole proteome and thousands of peptides. We're getting down-- it becomes very complicated, and you have to be able to compute what you expect based on the normal natural abundance isotopic distribution. So that's the key thing. So we look at the normal isotopic distribution. And if you look at that, I think in the next one, I show you an example of that. So what are the isotopes-- you probably can't read this here, but if you pull out your computer, you'll see this. So we have C12, C13. OK, we have hundreds of amino acids with carbons. So you have C12 and C13. C12 is 99%. C13 is 1%. That's an actual abundance. OK, so every one of these has different natural abundance. We know what they are. In fact, if you're an organic chemist, you can measure isotope of x using a mass spectrometer, if you have something that's really accurate, which we do. I've measured a lot of C13 isotope effects, using a mass spectrometer, based on differences in natural abundance and changes. Yeah. AUDIENCE: [INAUDIBLE] JOANNE STUBBE: The what? AUDIENCE: The natural abundance of deuterium? JOANNE STUBBE: Yeah, I think it's up here. So it's up here. I think it's-- let's see, 3%. Yeah, protons deuterium 3%. AUDIENCE: Would you expect a huge distribution from that? JOANNE STUBBE: You see isotope effects on everything. You see-- if you do mass spec, I mean, this is something I think that's not appreciated, and you have a linker with deuteriums in it, and even if you chromatograph it, you change the chromatographic properties based on the deuterium, and so you might think it's migrating here, and it doesn't. It has an isotope effect on how it migrates. So yeah, you need to pay attention to all of this stuff. OK, and it seems like a small amount, but the beauty is that it is a small amount, but it's incredibly informative, and we have very powerful computers that can allow us to do the analysis. So we do have protons. You see deuterium used. You saw deuterium used in this paper you read today. They did CD3 and CH3's. OK, you can also see the tritium. OK, that's much smaller. I don't know what the ratio is, but you can look at it. But you also-- this one is also incredibly important and is widely used in proteomics-- N14 and 15, and people do isotopic labeling. So they might see N15 labeled lysine or arginine or deuterated lysine or arginine. And why do you think they would deuterate the lysine or the arginine or N15 label it? What do we know about lysine and arginine in terms of thinking about proteins and analysis of proteins? What do you think about lysine and arginine? You've seen it several times over the course of this semester, and you probably saw it in 5.07. AUDIENCE: [INAUDIBLE] JOANNE STUBBE: What? AUDIENCE: The protons will exchange? JOANNE STUBBE: Well, now as you put it-- no. So that that's true if it was on a hydrogen and a nitrogen, it would exchange, but they put the deuteriums in on carbon, so they're not exchanging. OK, so why that would happen in any amino acid, why lysine and arginine? And the reason is that almost all-- and this was also done in this paper, you don't work on the huge protein. You cleave it to pieces. And you cleave it into pieces, and where you cleave is with trypsin, which is the major-- you've seen this used now over and over again. That's a major thing you use because it cleaves next to basic amino acids. So these become really important in labeling experiments, if you read much mass spec data, or if you look at Alice Ting's work, everything is N15 and deuterium labeled, and lysine and arginine to try to make sure they have coverage of the whole proteome, which is what her lab actually looks at. OK. So we have isotopic labels, and we can take advantage of these, and we can calculate what the distribution should look like, OK, of the isotopes should be, depending on what the-- we know what the sequence is. We know what the abundance is. And so you can calculate the whole mass spec. So let's see. So there's going to be a number of things that we want to do, and what we're going to be describing today and the next time is a "workflow." These are the words that people use all the time, and "platform." And what we're trying to do in the case of the Carroll papers is simply look at whether the protein is modified or not. But as with most post-translational modifications, do you think this is going to be 100% modified? No. In fact, it's only partially modified. That adds to the complexity of understanding whether the biology is interesting or not, so what you have then is something that's modified and something that's more non-modified. So then the question is, how do you tell how much is modified and how much is non-modified? If this enhances the rate only a factor of two, and this is 99.8%, of this, are you ever going to be able to see an effect of this modification? That's the question that you have to focus on, and everybody and his brother is doing experiments like this. We will see in a second, hundreds of post translational modifications, and the question is what are they doing in terms of thinking about the biology of the system. OK, so what's the platform? What's the platform we're going to use? So there are two ways you can look at this. So we have a protein that has been modified. You're going to-- if you had a huge protein, and you only had a single OH on it, even if it was 100%, and the protein was, say, 300,000 molecular weight, you might not be able to see it. You need to do a calculation to see whether you could see it or not. If you have a small protein of molecular weight 30,000, or whatever-- I think the 22,000 or 23,000 like glutathione peroxidase, used in this paper, you could see it. So you could look at the protein directly. But how else could you do this? You would enrich. If you were doing this in the whole cell, you would want to separate this away from everything else. OK, so to do that, you want to be able to have a way to stabilize this, OK, and that's what this paper is all about, and then not only to stabilize it, but to separate the stabilized form out. So where does this happen? And in this particular cartoon, where do you see post translational modifications? Probably the most popular one is phosphorylation. So we have signaling cascades in kinases. And in fact, if you look at the epidermal growth factor receptor, it's a tyrosine kinase, and it gets phosphorylated and is regulated. And this sulfenylation is supposed to be on top of the phosphorylation. So you have multiple post-translational modifications that can affect activity. So Forest White, for example, in BE, works on kinase signaling cascades. And so he's developed a method, as have others, to be able to pull phosphorylated proteins out of a crude gemisch. OK. So, you know, if you look at this, here he's got iron bound to a phosphate and bound to some bead. So the iron's bound to some chelate around the bead, just like your nickel affinity column, which then binds to the protein. But this raises the issue that I was discussing in class, which I spent a lot of time on over and over again, but you need to think about, do you think these bonds are tight, how tight do you think those bonds are? What do you need to think about for this kind of analysis to work? It's the same thing with nickel affinity column that you talked about when you were looking at purification of proteins. AUDIENCE: It has to be stable enough. JOANNE STUBBE: It has to be stable enough. That's the key. So you have to undergo ligand exchange. It's got to-- if you didn't have-- when you start, you don't have phosphorylated form of your protein around. You have nothing. You have water there. OK, so the waters have to undergo exchange, so the phosphate can then bind, but it's an equilibrium, and so up and down the column is coming off and on. Yeah. AUDIENCE: [INAUDIBLE] JOANNE STUBBE: It could. I mean, so it's a question of what out competes what. It's a question of relative Kds. So what you have to do is study all of this to figure out how to optimize this, how did they arrive at this? Probably somebody did a lot of studies. OK. This is a new method. I don't know how new it is, but it's a method I don't know that much about, again, of pulling phosphates out. So that's one way. So you have-- so you usually have an affinity purification. And if we look at the Carroll paper, what she does in the next paper is she's going to figure out a way-- she's derivatized, she's made a dimedone derivative, which stabilizes the sulfenic acid, and then she attaches something to it that's going to allow us to affinity purify that. We'll come back and talk about that later. So what are they using over here? They're using-- this is-- if you look at histones that get acetylated or methylated, they have an antibody that's specific for the acetylated lysine, so they use antibodies to pull something out. So that's a method-- the second way of pulling things out are using antibodies. That's quite frequently used. And what did they use in this paper? Did it detect the modified sulfenic acid? Does anybody remember? Did you read the paper carefully enough? AUDIENCE: Like, a anti-dimedone antibody? JOANNE STUBBE: Yeah, so they use an antidimedone antibody. OK, so that becomes really critical that you know that your antibodies are actually working effectively. So we have antibodies, and then, another thing that people are interested in this department, the Imperiali lab, is sugars. We have sugars everywhere. OK, we don't really understand the function of these sugars. We understand some of them, but it's amazingly complex. And what we have are proteins called lectins, and any of you heard Laura Kiessling talk, maybe undergraduates wouldn't have done this, but she discovered a new lectin and discovered the basis, the structure the sugar that binds to this lectin. And so you can selectively move that type of sugar. Again, it's an equilibrium. So they're coming off and on, but it binds, hopefully, enough so that the other stuff washes through, and you enrich in the protein of interest. So these are sort of some of the tricks that are actually used. We're going to see, in the case of the Carroll paper, next time we use click chemistry to make something with a biotin on it, because biotin you all know can bind to streptavidin, which has pluses, and it has minuses, but it allows you to pull things out more easily, because the interaction is so tight. So you could do this-- the workflow could be on the intact protein, or it could be on peptides. OK. And so the bottom half of this graph shows what happens after you treat this with trypsin. So with trypsin, and you're always cleaving next to lysine or arginine. So the C terminus of your protein is always a lysine or an arginine. And you can find that more easily if you deuterate or N15 label it. That's what people routinely do in the [? Broad. ?] And then you have, I think this is the most amazing thing, so you have a protein. And then you have an HPLC column. Have any of you done HPLC? And so do you think-- you could have a protein of 300,000 molecular weight, and look at the separation of your peptides. But if you look at any one of these things, do you think it's pure? So it's not pure. So every one of these peaks, if it's 300,000 molecular weight, you can calculate-- the reason people use trypsin is-- does anybody know why use trypsin, besides that cleaves at lysines in its specifics? Why do people use trypsin as a thing to cleave a big thing down into a little thing? AUDIENCE: What's the rationale for cleaving it? [INAUDIBLE] JOANNE STUBBE: So the rationale for cleaving it is just to make it smaller and easier to analyze. That's the rationale for cleaving it. So a peptide, a small peptide. But the question is, how big is the small peptide that's easy to analyze? And so that's the rationale. It gives you a distribution of peptides that's pretty good, that are all accessible to mass spec methods. So I don't know what the distribution is, but you know, people have done that calculation. And so almost always the peptides fly, whereas if you use other things, and you have something much bigger, it might not get ionized in the appropriate way or in a quantitative way, and you completely miss it. So the trypsin has been most successful. But each one of these little peaks is not one peak. You'll see when you put it into the mass analyzer, and if you read this paper carefully, you will see they got multiple mass charge species, which then they associated with specific peptides, OK. They know the sequence of their protein. And then they always use tosyl phenyl chloro ketone. Why do they use that? Anybody have an any idea? So in the experiments where they're doing the trypsin cleavage, they put in tosyl phenyl chloro ketone. Anybody know why? OK. No good. This is something that-- so tosyl phenyl chloro ketone is an alpha halo ketone. So it's activated for nucleophilic attack, and what you do is you have an acylated N terminus and an aromatic, and that's specific for chymotrypsin, like proteases. And so what this does is that covalently modifies the active site of chymotrypsin, and kills chymotrypsin. If you choose the wrong time to cleave with trypsin, you don't start getting cleavage next to hydrophobics, which then makes the analysis of the peptides much more complex. So the analysis of the peptide, a lot of people have done a lot of peptide chemistry, and I was telling this story before. I always go off on tangents. But Stein and Moore won the Nobel Prize. Maybe this is what you do when you get old, but Stein and Moore won the Nobel Prize, you know, in the 1950s, the 1950s, for separating amino acids. Do you know that they had a three story column of Dowex that was composed of anion exchange Dowex and cations? It was all polystyrene backbones of anion and cation polystyrenes, to be able to separate the amino acids. OK. And when you do that, of course, it gets stuck on the resin. Your recovery's out of the bottom of this chromatography. You need tons of stuff to put on the column in the first place. And this is what's happened. I mean, you have a little tiny HPLC column that has huge number of theoretical place that allows you amazing separations. I mean, again, the technology is sort of mind boggling, what you can do now. OK, so what you're doing here is then you're just asking the question, if you have a post-translational modification, x, you can either look at the entire protein. And so you could probably tell it was modified, but telling the location of the modified location, you can't, or you can treat it with trypsin. And then you get, again, with trypsin, you have little pieces. And one of these little pieces will have an x on it. And then you can define it. And then if you want to do sophisticated analysis, you can hit it-- use a second mass spectrometer, and actually sequence this. OK. So I think the next one just briefly goes to MALDI. And MALDI-- so Matrix Assistant Laser Disorption-- have any of you ever done that? OK. So where do you do that? Do you do that in [INAUDIBLE] lab? AUDIENCE: No, in the undergrad lab. JOANNE STUBBE: Oh, OK, because this is Brad's new thing. OK. OK, so you're looking at peptides. OK, so what do you use as the matrix? AUDIENCE: We used some aromatic acid. I don't remember. JOANNE STUBBE: OK. So you probably used sinapinic acid. AUDIENCE: [INAUDIBLE] JOANNE STUBBE: OK. So this is so-- you're using a different one still from this one-- this is-- I don't know. I got this idea somewhere. I don't know. So when I've done this-- I did do this maybe 10 years ago-- I've looked at a lot of peptides. We went through five or six of them before we found one that really worked well. So I don't know how state of the art has become, you know what it is. But the other one in the book that I got this from was, again, an acid. And so what is the idea? So the first thing you have to do is you have to ionize. So the way you do that is you mix your matrix and solution with your protein of interest, your analyte, then you evaporate it. So you have a solid on a little plate. And then you use a laser beam at 337 nanometers. And the light is absorbed by whatever the matrix is and causes you to have a plume of material. This is, again, amazing to me that the protein goes into the gas phase. And then, you have to go through this, go into the analyzer. Did you do time of flight? OK, so you have time of flight. So you guys know what it is then. And in the end, you do detection. So, again, the protocol is the same, but the method is different, and this is widely used and easy really easy to use nowadays. So the issue then is this is what you face when you're looking at a whole proteome. So you just can't calculate the mass of all the proteins from the gene sequences. Why? Because almost every single amino acid in your proteins are modified. So that adds complexity to all of this. So de-convoluting the mass spec becomes more complicated. So this just shows you, you don't need to look at this, but if you look at cystine, you could form disulfides. You can attach a prenyl group, an isoprene group on it. You can attatch palmitic acid on it. You can sulfenylate it. You can nitrosate it. So you have many, many modifications of the amino acids that are chemically reactive and involved on catalysis, and then not only involving catalysis, they are involved in regulation. So that then adds to the complexity of trying to deconvolute what the mass spec, I think, is actually telling you. And then, sorry, it went backwards. And so then what that does is tells you-- whoops. I'm just completely discombobulated here. OK, so what that does is that, again, you're just adding different masses on to all of these amino acids. The problem is that you have modified, and you have unmodified. And the question is what's the distribution? OK, and so if you have a very non-abundant protein, and most of it's unmodified, it's going to be much harder to find. So these are just things you need to think about, and your technology to look needs to be extremely well worked out, so that when you look and you don't find something, you know what the lower limits of detection are. So here we are at our system. Now we're into the Carroll paper, and so what we're looking at is sulfenic acids, degenerated by hydrogen peroxide. We'll see-- do you think that's a fast reaction, hydrogen peroxide with a cystine? Anybody have any intuition? I think these reactive oxygen species you're going to find are not so intuitive about the chemical reactivity. I'll give you a table with what we think we know in general. But I think it's not so intuitive. If you look at the rate constants for reaction of a hydrogen peroxide with a cystine it's 1 per molar per second, really slow. OK. So then the question you have to ask yourself, so this was something that was debated in the literature for 15 years. Is this so slow that this could never happen inside the cell? Because I just gave you a second order rate concept. So we have two molecules interacting at the concentration, this could be high. This is really low. You can calculate the rate constant for the actual reaction. It's really, really slow. OK, so we'll see that there are some proteins, peroxiredoxins that are in humans, are there in quite high levels that can increase this rate to 10 to the fourth per molar per second. So there's a huge rate increase but you need to think about all this kinetic stuff to really understand if this modification can happen inside the cell. Otherwise, well, if it can't happen, why are you wasting your time looking for it? Which is what a lot of people are doing scientifically. OK, so let me see what the next-- OK. So now we're into making a reagent that can specifically modify this, or specifically modify this. OK. So the reagent that they chose-- she didn't invent this reagent-- was dimedone. And this reagent specifically interacts with sulfenic acids. It doesn't react with the free cystine. So you've got to study all of this. And if you're going to use this as a reagent inside the cell, you want it to be fast. You don't want to take 30 hours to do the reaction. You want it to be over fast, and you want it to happen at pH 7. So how do you think this reaction works? Where's the most reactive part of this molecule? AUDIENCE: Those two protons? JOANNE STUBBE: So two protons. So this these have low pKas, so you can easily form the enolate. Depends on the details, the experimental details. And now you have this, and what you end up with is this molecule. And so the question is, does this go in 5%? You need something that goes in quantitative yield at pH 7, rapidly. OK, we're going to come back and talk about what the issues are, because the issues are even harder if you want this region to work inside the cell. OK, we're doing this on glutathione peroxidase, which is what he's using as a model to see if all of this stuff works. OK. So what you really want to do if you're thinking about regulation in the end, is you want to know how much is in each form, and you know, if you read hundreds of papers published on methods trying to figure this all out, but what she did in this case, was she developed a second reagent with an iodo group. OK. And as you can see, what is the product of the reaction? The product of the reaction is the same as the product of this reagent. But this reagent does not react with sulfenic acid. OK, so you get no reaction. So how does this reaction work? What do you think? The what? AUDIENCE: SN2. JOANNE STUBBE: So it could work by an SN2, but the way probably works is it attacks the iodine. So you form-- this is probably the mechanism from what's been done in the literature. So you attack this, and you form this, which then gets attacked by the enolate. So it doesn't really matter what the mechanism is, but the key thing is for this to react-- if you're interested in a mechanism, which I am, it does matter what it is. So the key thing is now you have the same reagents. So how could you ever use it attached? How could you ever use it to distinguish sulfenylation from a cystine. So what did they do in this paper? AUDIENCE: [INAUDIBLE] JOANNE STUBBE: Yeah, so they put the deuterated form on this. So what they did then was in this paper, so you got to keep these straight, if they see deuteriums present, so they made this deuterium label, and this protonated so now you have a mass difference of 6. OK. And in the system, they're using glutathione peroxidase, which has three cystines in it. And one of the cystines is more reactive than the other two, but for proof of concept they mutated two of these cystines into serine initially, so you only had a single reactive cystine, but then they went back and studied the whole protein. OK, so let me just introduce you to this, and then we'll come back and talk about this next time. Let me just do one more thing. OK, so here is the difference in mass between these two species. So this is what you're looking at. If they start out with deuterium labeled dimedone, the peak that they observe is going to be associated with sulfenylation, and if they start out with the protonated material, the peak they observe is going to be associated with the [INAUDIBLE] group. OK, so that's the idea. And then what they did was they simply took their protein, and they have, in this case, 50 micromolar of their protein, and then they increase the concentration of hydrogen peroxide. They don't really talk very much about how they design the timing, but they use, you know, two equivalents. So they use variable amounts of hydrogen peroxide. And what you can see is the maximum amount. So now what you're using, we talked about this before, but we're using anti dimedone antibodies for the detection. And here, they're starting with no hydrogen peroxide. So you don't see any dimedone derivative, and then you increase the concentration. But you get to the highest concentration here that they looked at. So it's 100 micromolar versus 50 micromolar in the protein they used. But what did this immediately tell you? Did any of you look at this data very carefully? What is this? This guy here is associated with a [INAUDIBLE] group that is only reacted with iododimedone, so if you got 100% yield, what does that tell you? This tells you the maximum amount of material you're going to observe. So if you look at this peak, and you look at that peak, you can't do this by eyeball. You need to do this quantitatively. The phosphor images or methods that allow us to do this quantitatively. What do you see? AUDIENCE: It's not at the max. JOANNE STUBBE: Yeah, it's not at the max. And so what we'll do next time-- so these are sort of controls, and the question is how effective is this reagent, and if you start hanging stuff off of your dimedone over here, are you going to change the rate of modification? Can it get into the active site where this SOH actually is, these are the kinds of things we're going to talk about next time when we look a little bit more at the details of the reaction with this, and you should look at the reaction with gap dehydrogenase, which is another control enzyme they ended up looking at it, because what they do is address what the issues are that you're going to encounter when you get into something real that you care about. And that's much more complicated. OK so that's it.
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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: Last time, we looked at the case of a Bose gas, and we found that there was this condensation phenomena into the ground state. Let's go over it one more time. So the idea was that when you have a Bose system, the occupation of the different one-particle states, which you can specify for a non-interacting system in the grand canonical ensemble, these entities are independent for each one of the one-particle states and have a form that is related to the chemical potential and the energy of these states through this form, epsilon [? kd ?] in the one-particle state, and z standing for e to the beta mu, chemical potential divided by temperature. OK? Now, if we go and look at this in the limit where z is much less than 1, which is appropriate to high temperatures or when the case where the gas is mostly classical, you can expand this. You can see that in the lowest order, it is z e to the minus beta epsilon of k. So it looks like a standard Boltzmann weight that we would use in classical statistical mechanics. But then there are higher order terms. And in particular there would be a correction that is e to the minus beta epsilon of k squared and so forth that would modify this result. But I think this is something that I mentioned. If you look at a form such as this at high temperatures, it's like a Boltzmann weight. So the state that has this lowest energy, in this case if we say epsilon of k is h bar squared k squared over 2m corresponding to k equals to 0 for the lowest energy, has the highest mean occupation number of weight. OK? Now that is always the case. This is a function that, as epsilon of k becomes larger, no matter whether you expand it or not, becomes less and less. The other limit that is of interest to us is how big can z be? We said that the largest it can be is when it approaches 1. And when it is 1, this quantity, the occupation number, is e to the beta epsilon of k minus 1. Again, something that is largest as k becomes smaller, and you go to the lowest energy. But there is the difficulty that the one energy that corresponds to k equals to 0, if I look at this formula, it is 1 divided by 0. It just doesn't make too much sense. Now, of course, we said that typically we don't want to work in this variable. We want to figure out things as a function of the number of particles or the number of particles divided by volume density, which is a nice intensive quantity. OK? So typically, what should we do? We said that in the grand canonical ensemble, N is the sum over k these expectation values. I should really put an average here, but we said that in the thermodynamic sense, fluctuations being of the order of square root of N, we are going to ignore. So this is what you would normally call the number that we have in the system. And then you have to replace this sum in the limit where we have a large system with an integral. So we have the integral d cubed k 2 pi cubed V. And then we have this N of k, which is 1 over z inverse e to the beta epsilon of k minus 1. Now if I go like I did in the line above to the limit where z is much less than 1, then the integral that I have to do is simply the integral of z e to the minus beta epsilon k h bar squared k squared 2m. We know what this is. This is simply, up to a factor of z, these are the integrals that have given me these factors of 1 over lambda cubed. Actually, there's the factor of V here, so it is V over lambda cubed. So normally, you would then solve for z, and you would find that z is the density times lambda cubed. And this is a formula that we have seen a number of times, that the chemical potential is kt log of n lambda cubed. OK? Fine. But now we are interested in the other limit. Now in the other limit, we have this difficulty that this quantity, if I look at it as a function of z, this is the quantity that we would call f-- it is related to f 3/2 of z. So this is going to be 1 over lambda cubed f 3/2 plus of z. It's a function that increases with z up to the limiting value that we are allowed to have, which is at z equals to 1. It's a function that starts linearly, as we have discussed, but then up here, it comes to a finite value of zeta 3/2. OK? So basically, we know that when I evaluate this for finite z, I'm less than where I was here. So this is certainly less than 1 over lambda cubed zeta of 3/2, the limiting value that we have over here. So now I have a difficulty. This function over here, I can only make it so big. But I need to make it big enough so that I can put enough particles, the total number of particles that I have in the system. So what is happening over here? What is happening is that when I make this density larger and larger, ultimately when I hit this point at z equals to 1, as we have discussed there is an uncertainty as to what happens to this expression when z is equal to 1, and you are at this state that corresponds to k equals to 0. Right? I don't know how many things are over there. So I can have, in principle, the freedom to take advantage of the fact that I don't know what 1 over 0 is to call that the number that I need to make up the difference between here and there. OK? And so, in principle, when z goes to 1, then this becomes 1 over lambda cubed, or V over lambda cubed, zeta of 3/2, obtained by essentially doing the integration with z equals to 1, evaluating it at this possible limit. And then whatever is left, I will put in the k equals to 0 state. OK? And you will say, well, what does that explicitly mean? Well, what it means is that I'm really evaluating things at the value of z that is extremely close to 1 so that in the thermodynamic limit where N goes to infinity I wouldn't know the difference. But it is really something that vanishes only in that thermodynamic limit of N going to infinity. And we saw that if I kind of look at the occupation number of the k equals to 0 state, what it is is z inverse, which is e to the minus beta mu minus 1. And if I'm in the limit where z is very small, or beta mu is-- sorry, z is very much close to 1 or beta mu is very small, then I can expand this. And this becomes approximately, when mu goes to 0, minus 1 over beta mu. OK? And I want this to be the N0, what is left over when I subtract from the total number the amount that I can put in to the ground state. OK? And so you can see that this is a quantity that is extensive. It is proportional to the number of particles. And mu is the inverse of that, so it is infinitesimal. And if I choose that value of mu that is, essentially, pushing z infinitesimally close to 1, then I can ensure that I have a finite occupation over there. OK? So what does it mean? If I could actually look at these occupation numbers, then I am at temperatures less than Tc of n. And we saw that basically we could get what the temperature is when this happens by equating this combination g over lambda cubed, in principle, over-- I had said g equals to 1. I could repeat it with g not equal to 1. And n, which is g over lambda cubed zeta of 3/2, rather than f 3/2 plus, this defines for me this temperature Tc of n. OK? Now, if I could look at the occupation numbers when I was at temperature less than Tc, what would I see? I would see that the occupation number as a function of k is exactly the form that we have over here with z equals to 1, 1 over e to the beta epsilon of k minus 1. But then at k equals to 0, which I can indicate by some kind of delta function, I have everybody else, which is this N minus V over lambda cubed zeta of 3/2. OK? Actually, let's remember that our k and momentum are related to each other. So rather than asking what the density is as a function of this label k, I can maybe ask what the density is for things that have different momentum P. And according to this formula, it is going to be beta P squared over 2m, this kind of modified version of the Boltzmann weight, and a delta function at P equals to 0 there. The rest of the particles are up here. Now, when people saw the signature of Bose-Einstein condensation, it was precisely this. So this is the figure that shows this from the experiment of Eric Cornell and group, and you see two thermometers. They are cooling the temperature of this gas of atoms that is confined in a trap. The way that they are cooling it is that they are evaporating away the high energy particles. And then the rest of the particles are cooled down because they thermalize, and the highest energy is always removed. But in the process, the density in the trap is also being reduced. And if the density goes down, then Tc of n will also go down. So what is plotted over here on the right thermometer is the Tc of n that you would calculate from that formula. And on the right thermometer is actually the temperature of the particles in the gas. And you can see that right at some point, then the Tc of the particles in the gas goes below Tc of n. And what is being plotted over here is as follows. So you have the particles that are in the trap at some temperature. We remove the trap, and then the particles are going to escape. The particles that have zero momentum are going to stay where they are. The particles that have large momentum are going to go further away. So let's say after one second you take a picture. And what you see over here from the picture is over here there are the particles that had most momentum and ran further away. While at the center, you see the particles that had small momentum. You can see that always, even at high temperature, the zero momentum is the peak. But once you go below this condition of Tc of n, in addition to the normal peak that is over here, you get some additional peak that is appearing at P equals to 0, which is the condensation that you have into that one single state. OK? The other thing that I started to do last time but let's complete and finish today is I wanted to show you the heat capacity of this system. OK. So how do we do that? So the first formula that we had, the one that I used over here if I were to divide by V, is that the density, in general, is given by g over lambda cubed f 3/2 plus of z. OK? And we said that if z goes to 1, then I have to interpret that appropriately that this is really just excited states, and there's going to be a separation into the ground state and the excited states. So there is really two portions of n when we are below Tc. OK? But then there is also a formula for pressure. Beta P was g over lambda cubed f 5/2 of z. Again, these f's we had defined-- f plus m minus 1 of z was 1 over m minus 1 factorial integral 0 to infinity dx x to the m minus 1 z inverse e to the x plus 1-- minus 1. OK? Fine. But in order to calculate the heat capacity, I need to work with the energy. And we said that, quite generally, whether you have classical system, quantum gas, bosons, fermions, the energy and pressure are related simply by this formula, which is ultimately a consequence of the scaling of the energy being proportional to momentum squared that we have over there. OK? So let's look at this formula. What do I have? For T that is less than Tc of n, my z is equal to 1. Pressure comes entirely from particles that are moving around. So although the density breaks into two components, in principle I guess I would say that the pressure also breaks into two components. But the component that is at k equals to 0 is not moving around. It is not giving you any contribution to the pressure. So what do we? We have P is kt, taking the beta to the other side. I have g over lambda cubed, and then this f plus evaluated at z equals to 1, which is this zeta function of 5/2. OK? So fine. So what do I have for energy? Energy is going to be at 3/2 V times this formula, kd T g over lambda cubed zeta 5/2. And I noticed that lambda, being h square root of 2 pi mk T, scales like T to the minus 1/2. So 1 over lambda cubed scales like T to the 3/2 plus this is something that's scaling like T to the 5/2. And this is simply proportional to the volume and temperature to the [INAUDIBLE] fifth power. It knows nothing about the density. Why? Because if you were at the same temperature and volume, put more particle in the system, all of those additional particles would go over here, would make no contribution to pressure or to energy. You wouldn't see them. So once you have this, the heat capacity at constant volume, dE by dT, is simply 5 over 2T of whatever energy is. And there is a reason that I write it in this fashion. So what you would say is that if I were to plot the heat capacity at constant volume as a function of temperature, well, it's natural units are of the order of kB. You can see that this T and this T I can cancel out. So I have something that has units of kB. It is extensive proportional to volume. I can really certainly write it in this fashion. And what I have is that the curve behaves like T to the 3/2. So this is proportional to T to the 3/2. And presumably, this behavior continues all the way until the temperature at which this system ceases to be a Bose-Einstein condensate. So the low temperature form of the heat capacity's calculation is very simple. Now, at very high temperatures, we know that the heat capacity is going to be 3/2 N kT, right? So ultimately, I know that if I go to very high temperatures, my heat capacity is going to be 3/2 N, if I divide it by [? kB. ?] OK? Now, there is a reason that I drew this line over here less than the peak here. Well, in order to calculate the heat capacity on this side, we'll have to do a little bit more work. You can no longer assume that z equals to 1. z is something that is less than 1 that is obtained by solving N equals to g over lambda cubed f 3/2 plus of z. OK? But I can still presumably use these formulae. I have that the energy is 3/2 PV. So I have V kT times-- what is the thing that I have on the right-hand side? It is g over lambda cubed, just as I have over here except that I have to put f 5/2 plus of z. OK? And if I want to calculate the heat capacity, what I need to do is to take a derivative of the energy with respect to temperature. And then I notice, well, again, right like I had before, whatever appears here is proportional to T to the 5/2. So when I take a derivative with respect to temperature, I essentially get 5/2 this whole quantity divided by T. So just like I had over here, I will write the result as 3/2 V kT g over lambda cubed. And then I have the 5/2 1 over T of f 5/2 plus of z. But you say that's the wrong way of doing things because in addition to this combination out front be explicitly proportional to T to the 5/2 power, this factor z here is implicitly a function of temperature because it is related to density through that formula N equals to g over lambda cubed f 5/2 plus of z. So you should not forget to take the derivative of what is over here with respect to T. OK? So what do I get when I take a derivative of this with respect to T? Well, the argument is not T, it is z. So I will have to take a dz by dT, and then the derivative of f 5/2 with respect to z. And we saw that one of the characteristics of these functions was that when we took the derivative of each one of them, we generated one with a lower argument. So the derivative would be f 3/2 plus of z. But in order to have this [? latter ?] property, I had to do a z d by dz, and so I essentially need to restore this 1 over z here. So you can see that, really, the difference between what I had at low temperatures and high temperatures, at low temperatures this was just a number that I was using over here, zeta of 5/2, and I didn't have to bother about this other term. Now, we can figure out what the behavior of this term is certainly at small z because we have the expansion of these functions at small z. And if you do sufficient work, you will find-- well, first of all, at the largest values of z-- sorry, at the smallest value, z going to 0, you can confirm that the heat capacity in units of N kB is simply 3/2. So you will come to this line over here, and then the first correction that you calculate you'll find is going to take you in that direction. Actually, if you do the calculation for bosons and fermions at high temperatures, you can convince yourself that while bosons will go in this direction, fermions, the heat capacity, will start to go in the other direction. But OK, so you have this behavior. And then what we haven't established and the reason that I'm still doing all of this algebra is that, OK, this curve is going up, where is it going to end up? Is it going to end up here, here, or right where this curve hits? OK? And what I will show you is that this has to end up right at this point. And actually, in order to show that, all we need to do is to figure out what these terms do when I evaluate them at z equals to 1. And because these things are telling me what is happening as I am approaching to this point, the question that I ask is, what is the limiting behavior of this function as I get to z equals to 1? But the limiting value of this one I know. It is precisely what I want over here. It is zeta of 5/2. So I have to somehow show you that this next term is 0. Well, this next term, however, depends on dz by dT. How can I get that? Well, all of these things are done under the conditions where the volume is fixed, or the number of particles is fixed, if the density is fixed. So z and T are related through the formula that says N over V, which is the density, is g over lambda cubed f 3/2 plus of z. OK? So if I take the system and I change its temperature, while keeping the number of particles, volume, fixed so that the density is fixed, if I take d by dT on both sides, on the other side I'm assuming that I'm doing the calculations under conditions where the density does not change. And then what do I have over here? I have to take the temperature derivative of this. There's an implicit derivative here since this is T to the 3/2 proportionately. So when I take a derivative, I will get 3 over 2T, and then I will have f 3/2 plus of z. But then I have to take the derivative that is implicit in the dependence on z here. And that's going to be identical to this, except that when I take a derivative of f 3/2, I will generate f 1/2 of z. OK? So since the left-hand side is equal to 0, I can immediately solve for this, and I find that T dz by dT 1 over z is minus 3/2 f 3/2 plus of z divided by f 1/2 plus of z. OK? So essentially, what I can do is I can multiply through by T. So this T will get rid of this T. I can bring a T, therefore, here. And this combination over here I can replace from what I have there with minus 3/2 f 3/2 plus of z divided by f 1/2 half plus of z. Why is any of this useful? Well, it immediately gives the answer that I need. Why? Because f 1/2 is the derivative of this function. And you can see that I drew it so that it has infinite derivative when z hits 1. And we had indeed said that this function zeta of m, which is the limit of this when z goes to 1, only exists for m that is larger than 1, and it is divergent for m that is less than 1. So when z hits 1, this thing goes to infinity, and the fraction disappears. And so basically, the additional correction that you have beyond this becoming zeta of 5/2, which is what we have over here, vanishes. So I have shown that, indeed, whatever this curve is doing, it will hit that point exactly at Tc of n at the same height as the low temperature curve. And you can do a certain amount of work and algebra to show that it goes there with a finite slope. So that's the shape of the heat capacity curve for the Bose-Einstein condensate. AUDIENCE: Professor? PROFESSOR: Yes? AUDIENCE: So it seems that f 1/2 goes to infinity, but it's [? on ?] the half, so it is half comes from the [INAUDIBLE] 3. So-- PROFESSOR: Exactly. AUDIENCE: --you have higher dimensional system, then it seems that [INAUDIBLE] won't vanish there. [INAUDIBLE]. PROFESSOR: So if you have a higher dimensional system, indeed the situation would be different. And I guess the borderline dimension would need to be-- since this is a V over 2 minus 1-- AUDIENCE: [INAUDIBLE] PROFESSOR: V over 2 minus 1 being 1, which is the borderline, will give you a dc of 4. So something else happens above 4. Yes? AUDIENCE: So it seems that the nature of this phase transition, indeed it changes with the dimension. PROFESSOR: Exactly, yes. The nature of this phase transition does depend on [INAUDIBLE]. Good. OK? So we will come back to this heat capacity curve a little bit later because I want to now change directions and not talk about Bose-Einstein condensation anymore, but about a phenomena that is much more miraculous in my view, and that is superfluidity. So the first thing that is very interesting about helium is its phase diagram. So we are used to drawing phase diagrams as a function of pressure and temperature, which have a separation between a liquid and a gas that terminates at a critical point. So we have seen something like this many times, where you have on the low pressure/high temperature side a gas. On the higher pressure/lower temperature side, you have a liquid. And they terminate, this line of coexistence between the two of them, at a critical point. And for the case of helium, the critical point occurs roughly at 5.2 degrees Kelvin and something like 2.6 atmospheres. OK? So far, nothing different from, say, liquid water, for example, and steam, except that for all other substances, when we cool it at the low enough temperatures, we go past the triple point, and then we have a solid phase. For the case of helium, it stays liquid all the way down to 0 temperature. If you put sufficiently high pressure on it, ultimately it will become a solid. But at normal pressures-- so this maximal pressure is something of the order of 26 atmosphere. For pressures that are less than 26 atmosphere, you can cool the liquid all the way down to 0 temperature, and it stays happily a liquid. So let's think about why that is the case. Well, why do things typically become solids anyway? The reason is that if I think about the interaction that I have between two objects, two atoms and molecules, we have discussed this in connection with condensation before that what we have is that at very large separations there is a van der Waals' attraction. Typically, the dipole moments of these atoms would be fluctuating, and the fluctuations would somehow align them temporarily in the manner that the energy can be reduced. At very short distances, the electronic clouds will overlap, and so you would have something such as this. And then, typically, you will have the function that joins them and look something like this. OK? And this is no different for the case of helium. You have here essentially a hard core appearing at the distance that is of the order of 2.6 angstroms. There is a minimum that is of the order of 3 angstroms. Now, the depth of this potential is not very big for helium. It is only of the order of, say, 9 degrees Kelvin. And the reason is that the strength of this van der Waals' attraction is proportional to the polarizability that scales typically with the number of electrons that you have in your atom. And so it is, in fact, proportional to your atomic number squared. And with the exception of hydrogen, this is going to be the smallest value that you are going to get for the van der Waals' attraction. And hydrogen does not count because it forms a molecule. So helium does not form a molecule. You have these [? balls. ?] They have these attractions. Now, typically things become solid as you go to 0 temperature is because you can find the minimum energy in which you put everybody at the minimum energy separation. So you form a lattice. Everybody is around this separation, and they're all happy and that solid minimizes the energy. So why doesn't helium do the same thing? Maybe it should just do it at a lower temperature because the scale of this is lower. Again, the reason it doesn't do that is because of quantum mechanics. Because you can say that, really, I cannot specify exactly where the particle are because that would violate quantum mechanics. There has to be some uncertainty delta x. And associated with delta x there is going to be some kind of a typical momentum, which is h bar over delta x. And there will be some contribution to the kinetic energy of particles due to this uncertainty in delta x that is of this form. And if I make the localization more precise, I have to pay a higher cost in kinetic energy. Well, presumably the scale of the energies that I can tolerate and still keep these things localized is of the order of this binding energy, which is of the order of, as we said, 9 or 10 degrees Kelvin. Let's say 10 degrees Kelvin. OK? Now, the other property that distinguishes helium from krypton and, let's say, all the other noble gases is that its mass is very small. So because this mass is small, this uncertainty in energy becomes comparatively larger. OK? And so if you put the mass of the helium here and you use this value of 10 degrees K converted to the appropriate energy units, you ask what kind of a delta x would be compatible with this kind of binding energy, you find that the delta x that you get is of the order of 0.5 angstroms. OK? So basically, you cannot keep this in the bottom of the potential if you have uncertainty that is of this order. So the atoms of helium cannot form a nice lattice and stay in order because that registry would violate the quantum uncertainty principle. They are moving around. And so you have a system that has been [? melted ?] due to quantum fluctuations rather than thermal fluctuations. OK? So that's all nice, except that something does happen when you cool helium. And actually, it is interesting that helium provides us not only with one liquid that persists all the way to 0 temperature, but two liquids because helium has two isotopes. There is helium-4, which is the abundant one, and there is helium-3, which is a much less available isotope. But the number of protons being different in the nucleus of these means that helium-4 is a boson, while helium-3 is fermion. So we have the opportunity to observe both a Bose system that is liquid at low temperature and a Fermi liquid at [INAUDIBLE] temperatures. Now, the case of the Fermi liquid is not really that different in principle from the Fermi liquid of the electrons in copper that we already discussed. There are some subtleties. In the case of the bosons, [? well, ?] we saw that for bosons there is potentially this kind of phase transition. In reality, what happens is that helium has a transition between two forms of the liquid at a temperature that is of the order of 2.18 degrees Kelvin. So we can call this helium I and this helium II, OK, two different forms of the helium-4 liquid that you have. So how was this seen experimentally? I kind of told you how in these experiments of the Bose-Einstein condensation, part of the way that they cooled the system was to remove the high-energy particles, and the remaining particles cooled down. So that's the common method that actually people use for cooling various things. It's called evaporative-- evaporative cooling. And for the case of helium, it is actually reasonably simple to draw what it is. You have a bath that is full of helium that is reasonably nicely thermally isolated from the environment. So you want to cool it, remove the energy from it. And the thing that you do is you pump out the gas. OK? So this is a coexistence because this part up here on top of the liquid helium is not vacuum. What happens is that there is evaporation of helium, leaving the liquid and going into the gas. And once they are in the gas, they are pumped out. But every time you grab a helium from here to put it into the gas, you've essentially taken it from down here and moved it to infinity, so you lose this amount of binding energy. It's called latent heat. Essentially, this is the latent heat, and it is actually of the order of 7 degrees Kelvin because typically it has to be the depth of the typical positions of these particles. OK? So for each atom that you pump out, you have reduced the total energy that you have in the liquid, and the liquid cools down. And what is the trajectory? You always actually remain on this coexistence line. The pressure is changing. The temperature is lowering. But you're always remaining on the coexistence of the gas and liquid line. So you follow what is happening over here, and you find that as you are approaching here, suddenly this system starts to bubble up. It is just like having a kettle that is boiling. Indeed, it is, again, in like the case of kettle, there are these bubbles that are carrying the steam that is going to go over here and being pulled out, and so it is bubbling. And then suddenly, [INAUDIBLE] heat 2.18 degrees Kelvin, and the whole thing quiets down. And below that, it stays quiet. And so let's see if we can show that. We saw the phase diagram. So this is the liquid. Let's see if we can get the sound also. [VIDEO PLAYBACK] -[INAUDIBLE]. This rapid [INAUDIBLE] evaporates [INAUDIBLE] helium cools, until at 2 degrees above absolute 0, a dramatic transformation takes place. Suddenly, we see-- PROFESSOR: Everything quiets. --[INAUDIBLE] stops and that the surface of the liquid helium is completely still. The temperature is actually being lowered even further now, but nothing is happening. Well, this is really one of the great phenomenon in 20th-century physics. PROFESSOR: OK. I'll show you this movie at the next lecture. But this is just a segment of it that shows the important property of this fluid, helium II, which is the superfluid. So now you're above the temperature it was boiling, but it quieted down. You can see that the fluid started to go through the capillaries that were at the bottom of this. So at the bottom of this glass there were fine capillaries. As long as you were up here, the capillaries were sufficient to keep the fluid above. Once you went below, the fluid went through as if there is no resistance. And that's the superfluid property. -Superfluidity and superconductivity were baffling concepts for scien-- [END PLAYBACK] PROFESSOR: OK, we'll go back to that later. So let's list some of the properties of this superfluid state. OK. So what this last experiment showed is that it appears to have zero viscosity. So what's the way that you would calculate the viscosity of a fluid? And one way to sort of do that relies on like what we had over here, passing the fluid through some pore or capillary. In the case of this experiment, you have to really make this capillary extremely fine so that practically nothing will go through it. But this could be a quite general way of measuring viscosity of some fluid. So you have some fluid. You can measure its viscosity by trying to pass it through a capillary by exerting pressure on one side and measuring, in essence, the difference between the two pressures that you have on the two sides as a measure of the force that you are exerting on this. And once you do that, then these pistons enhance the velocity-- sorry, the fluid, are moving through with some kind of velocity. And some measure of viscosity, after you divide by various things that have dimensions of length and pressure, et cetera, would be related to the velocity divided by delta P-- is to achieve the same velocity, how much pressure difference you have to put. Now, for this experiment of superfluid helium, you'll find that essentially you can get a finite velocity by typically infinitesimal pressure difference. So when you try to use this method to calculate viscosity, you find that, to whatever precision you can measure, essentially this value of eta is not distinguishable from 0. OK? Yes? AUDIENCE: The way you defined it, it would be infinity? PROFESSOR: Then I probably defined it-- [? no? ?] Delta P goes to 0 for a finite V. Yes, so viscosity-- AUDIENCE: So what you're essentially writing is the Reynolds number, yeah? PROFESSOR: Exactly. AUDIENCE: OK. PROFESSOR: All right? But there is another experiment that appears to indicate that there is a finite viscosity, and that is a famous experiment due to Andronikashvili. And basically, this is the other experiment. You imagine that you put your fluid inside a container. And then you insert in this container a set of plates, could be copper plates or something else, that are stacked on top of each other. So essentially, you have a stack of these plates that have a common axis, and there's a rod that is going through them. And what you created here is a torsional oscillator in the sense that this rod that holds all of these things together has a particular preferred orientation. And if you start, let's say-- well, it may or may not have a particular orientation-- but if you start to rotate it, it will certainly have a moment of inertia that resists the rotation that is coming from all of the plates if it was in vacuum. So if I could measure the moment of inertia of these objects, either through torsion or some other mechanism, if it is a torsional oscillator I could figure out what the frequency is. If I knew the [? spinning ?] constant that wants to bring the angle to some particular value, I would divide that by the moment of inertia. The square root of that would give me the frequency. So the frequency is a measure of the moment of inertia. So if I had this thing in vacuum, I could figure out what the moment of inertia of all of the plates was. Now, if I add the fluid to this, then as I start to rotate this, some of the fluid will certainly move with the plates. And so the moment of inertia that you would measure in the presence of the fluid is larger than it is in the absence of fluid. And that additional moment of inertia tells you about how much of the fluid is, in some sense, stuck to the plates, and that's another different measure of the viscosity. OK? So if we sort of look at this experiment, we would imagine that if I measure the moment of inertia as a function of temperature, something should happen when I hit Tc. Above Tc, I'm getting some value that is measuring, essentially, the moment of inertia of the plates plus the fluid that is moving around, [? we do. ?] But what happens when you hit Tc? If the superfluid doesn't care about things around it, you would imagine that this would drop down to the value that it has in vacuum. In reality, what it does is something like this. So it does go to the value that you would have in vacuum but only at 0 temperature. So at any finite temperature we know Tc. There is indeed some fraction of the superfluid that is trapped by these plates. So what's happening? OK, so this is one [INAUDIBLE]. The other property-- maybe I will squeeze it somewhere here. There are a whole bunch of things that indicate that there is a coupling between temperature and mechanical properties, so-called thermo-mechanical couplings. What you find in this experiment, where the fluid spontaneously goes, let's say, from this side to this side, so the direction is like this, is that when you measure the temperatures you will find that the temperature of this one goes up, and the temperature of this one goes down. So somehow the motion led to some temperature differences. A dramatic one that, again, maybe we'll show in the movie next time is the fountain effect. What you do is you-- let's imagine that, again, you have a vat of helium, and inside of it you put a tube that is, let's say, shaped like this. And you have the fluid that is occupying everything here. Then inside this tube, you put some material that is a good absorber of heat, such as, let's say, a piece of carbon, and then you shine laser on it so that this thing becomes hot. So in this case, the temperature goes up, and what you find happens is that the fluid will start to shoot up here like a fountain. It's called the fountain effect because of that. And essentially, whereas here pressure was converted to temperature, here temperature, the heating, gets converted to pressure. So this is like-- I don't know if you've ever done this, where you put your hand under water, and if you squeeze your hand and the water shoots up. So it's kind of putting pressure here, causing a fountain that is shooting up like that. OK? So there is this coupling between temperature and-- actually, this issue of boiling-not boiling is also a manifestation of a coupling between temperature and mechanical properties, which is whether the fluid is boiling or not boiling. OK? So given everything that we have been talking about so far, it is tempting to imagine that there is a connection between all of this and the Bose-Einstein condensation. AUDIENCE: [INAUDIBLE]. PROFESSOR: Yes? AUDIENCE: Do superfluids have no vorticity or can they? PROFESSOR: Yes, they can have vorticity, but it is of a special kind. It is quantized vorticity. And so what you can certainly do is you can take a vat of helium and set it into rotation, and then you will create vortex lines inside this superfluid. AUDIENCE: OK. I was wondering if there was an analogy to superconductivity and magnetic fields. PROFESSOR: The analog of the magnetic field is the rotation. AUDIENCE: Right. PROFESSOR: Yes. AUDIENCE: OK. PROFESSOR: So in that sense, yes. OK? OK. So similarities between BEC, Bose-Einstein Condensation, and superfluidity. I would say that perhaps the most important one, although I guess at the beginning of the story they didn't know this one, is that helium-3, which is a fermion, does not become superfluid around this same temperature of 2 degrees Kelvin, which is where helium-4 becomes superfluid. What is special about this is, well, you would say that once you have mass and density, you can figure out what the Bose-Einstein temperature is. You had this formula that was n lambda cubed is equal to zeta of 3/2. So that gives you some value of kB Tc being h squared 2 pi mass of helium, and then you have a zeta of 3/2 and n somewhere. 3-- I guess this n-- oops, n-- yeah, it's correct-- n h cubed 2 pi m kB Tc to the 3/2 being zeta of 3/2. So 2 pi m kB Tc is proportional to n zeta of 3/2 to the 2/3 power. OK. So I had a picture here, which was the typical attraction potential. The typical separation of helium atoms is something like 3.6 angstroms. How do I know that is because the density I can figure out is typically 1 over the volume per particle, and this volume per particle is roughly 3.6 angstroms cubed. It's something like 46.2 angstrom cubed. So I can put that value of n over here. I know what the mass of helium is. And I can figure out what this Tc is, and I find that Tc is 3.13 degrees K. You say, well, it's not exactly the 2 pi and something that we had. But given that we completely ignored the interactions that we have between this, and this is the result that we have for [? point ?] particles of mass m with no interactions, this is actually very good. You would say, OK, this is roughly in the right range. And presumably if I were to include the effects of interactions, potentially I will get an interaction, temperature, that is roughly correct. So this is another reason. The third reason is that you have automatically, when you are talking about the Bose-Einstein condensation, a relationship between pressure and temperature. So remember the formula that we had for the Bose-Einstein condensation, beta P was something, where P is kT 1 over lambda cubed zeta of 5/2, right? So once you specify the temperature, you've also specified the pressure. They go hand in hand. So then you can start to explain a lot of these phenomena, such as here you increase the temperature, pressure had to go up, and it was the increased pressure than did that. It kind of tells you something about what is going on in this experiment over here, where the temperature goes up where you have higher pressure. It actually also explains why things stop bubbling when you go into the superfluid. Because let's back out and think of why things are bubbling when you put your kettle on top of the oven. And the reason is that there are variations in temperature. There could be a local hotspot over here, and in this local hotspot, eventually there will a bubble that forms, and the bubble goes away. Why is it that it forms here is because heat gets spread out through diffusion. If I strike a match or even if I have a blow torch here, you won't feel the heat for some time. But if I do this, you hear my hand clapping much more rapidly. So pressure gets transmitted very efficiently through sound, whereas heat goes very slowly through diffusion. And that's why when you boil something, essentially you will have the diffusion of heat, local hotspots, bubbles. But if temperature changes are connected to pressure changes, then for the superfluid, having a local hotspot is like doing this. It creates immediately a pressure change, and pressure changes are rapidly taken away. And that's why this could also explain what's going on over here. So I think this number two was first noted by Fritz London. Maybe we'll talk about that a bit more next time around. Based on all of these similarities, Laszlo Tisza, who spent many years at MIT after being in Europe, proposed the two-fluid model. So again, roughly inspired by what we saw for the case of the Bose-Einstein condensate, we saw that for the Bose-Einstein condensate, below Tc you had to make a separation for the number of things that were in the ground state, k equals to 0, and all of the others that were in the higher excited states. Pressure, heat capacity, everything came from the excited component, whereas the component that was at the k equals to 0 state was not making any contribution to these phenomena. So what he proposed was that somehow, again, similarly when you are in the superfluid, you really have two components, two fluids that are in coexistence. There is a superfluid component, and there is a normal component. And as you change temperature, these two components get converted to each other. So just like the case of the Bose-Einstein condensate, it is not like you have so many of this fluid, so many of that fluid. These two things are always interchanging. And then what is happening in this experiment, presumably, is that the superfluid component goes through, the normal component leaves behind. The superfluid component is the component that, in some sense, has less entropy, is at lower temperature. So when it goes here, it cools down this part. This part then goes up in temperature. Again, there is no way that you can push all of the superfluid here and have normal here because these things get converted. So some amount of superfluid goes through here. Some amount gets converted from the normal back to the superfluid. So there is this exchange that is constantly taking place in the system. And so on the basis of this model and having different velocities for the normal and superfluid component, Tisza could explain a number of these things. OK? So I will mention some important differences at the end of this lecture and leave their resolution for next. So while those were similarities between BEC and superfluidity, there are very important differences. One important observation immediately is that if I try to compress superfluid helium or normal helium, it doesn't matter. It is just like trying to compress any other fluid such as water. It does not like to be squeezed. So helium liquid is approximately, of course not entirely, incompressible, while BEC, we saw that the pressure is only a function of temperature. It doesn't know anything about the density. So you can change the density. All of the additional particles will go to go to the k equals to 0 state. It doesn't really care. So BEC is infinitely compressible. And of course, this is immediately a manifestation of the interactions. In the case of the BEC, we didn't put any interactions among the particles. We can put as many of them as we like into the single state k equals to 0. Of course, real heliums, they have the self-avoiding-- the interaction between them. They cannot go [? through ?] each other. So that's an important difference. Number 2 is that detailed shapes/ T dependences of heat capacity and density of the superfluid part are different. OK. So I have up there what the heat capacity of a BEC should look like. Now, people have measured the heat capacity of helium as a function of temperature, and something does indeed happen at Tc. And what you find the shape is is something like this. It goes to 0, and then it does something like this. And so while it doesn't look that different from what I have up there, except that when you look at the behavior down here, at the transition, over there it went to a finite value. Here it diverges as a log. So when you go to Tc, it actually goes to infinity. And because of the shape that this has, this is sometimes called the lambda transition. Now, there's lots of issues about interesting things happening at the transition, and we won't be able to resolve that. But one thing that we should be able to figure out, based on what we know already, is the behavior as things go to 0 temperature. And I said that for the Bose-Einstein condensate, the temperature dependence is T to the 3/2. What is important is that for the superfluid, it is actually proportional to T cubed. And T cubed, if you go back in the course, is a signature of what was happening for phonons. So Landau looked at this and said there must be some kind of a phonon-like excitation going on, which is different from this case [INAUDIBLE] excitations that we have. The other thing that we had was that if we look at the fraction that is in-- let's say the density that is in the k equals to 0 state as a function of temperature, what happens for the superfluid-- sorry, what happens for the Bose-Einstein condensate-- for the Bose-Einstein condensate, let's use the green-- essentially, the curve will come down to Tc linearly, and then it goes up, and it reaches its asymptotic value. From this result, the fraction that is excited, the difference to 1 is proportional to 1 over lambda cubed. So this goes like T to the 3/2. You can ask whether what was observed in the Andronikashvili experiment, eventually this measure also, the fraction that is in the superfluid component or excited states, does this curve match that curve? But if I take this curve and put it over here, first of all, down here it is different. It goes like Tc minus T to the 2/3 power rather than linear that you have over here. That, again, is less [? worry ?] than the fact that over here it goes to 0 proportional to T to the fourth. Again, something that needs explanation. And finally, I will write down one statement, and then maybe we'll explain next time, is that BEC with the spectrum of excitations, which is k squared over 2m, cannot be superfluid. There is an interesting reason for that that we will explain. And indeed, we'll see that having excitations that are [INAUDIBLE] like is compatible with superfluidity. This kind of excitation is not. So why is that, et cetera, we'll talk about next time.
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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, an announcement that I must make, after developing all these friendships and working together, we will, next Tuesday, once again, enter a brief adversarial relationship. And we will have quiz number two. What we will cover on that quiz will be about 50%, perhaps a little bit more, on symmetry. And I'll remind you in just a minute what we've covered since the last quiz. And the second half or 40% of the quiz will be on the formalism of tensors and tensor manipulation. And obviously, you've had a chance only to do a very few problems. So these will be pretty basic questions and nothing tricky or very difficult. So let me remind you very, very quickly just by topics what we've covered. We began the period following the first quiz discussing how we can take the plane groups, two-dimensional symmetries that are translationally periodic, and defining a third translation to stack up the plane groups to give us a three-dimensional periodic arrangement of symmetry elements. In doing so, we found that we had to pick that third translation such that the symmetry elements that were in the plane group, and those were the things that made the lattices of the plane groups special, made them square or hexagonal or rectangular, that was true because there was symmetry there. Once again, symmetry and the specialness of a lattice are two inseparable parts of the story. But if we did this, the symmetry elements had to coincide. Otherwise, we would reproduce new symmetry elements in the base of the cell and wreck everything. So when we did that, there were a very limited number of ways of doing it. In the course of doing that, we fell headlong over screw axis in the same way that we fell over glide planes as soon as we put mirror lines in a two-dimensional lattice. The interesting thing was that that exercise of stacking the two-dimensional plane groups gave us almost all of the lattice types that exist in three dimensions. And the reason for that is we showed very directly that inversion requires nothing. So therefore, any of the point groups that were used in deriving the plane groups, when augmented by inversion to give a three-dimensional point group, are going to require nothing more than the lattices that we derived by stacking the plane groups. And that took care of just about everything except for the cubic symmetries. And we disposed of those in very short order. So that led us to the 14 space lattices, so-called bravais lattices. And then we proceeded to work with those lattices and put in the point groups that could accommodate them. Having discovered glide planes, we then, if we had done the process systematically, would have replaced mirror planes by glide planes. And having discovered screw axes, we replace pure rotation by screw axis rotations. And then doing what we did but didn't have to do very often, mercifully, for the plane groups, we said one could also interweave symmetry elements. And that gave a few additional space groups. We did not derive very many of them. What is important at this stage is simply to realize that their properties and applications are very similar to what we discovered for the two-dimensional plane groups. The notation is a little bit different. Because there can be symmetry elements parallel to the plane of the paper so that in depicting the arrangement of symmetry elements you need a little [? chevron-type ?] device to indicate whether you've got a mirror plane or a glide plane in the base of the cell. And similarly with two-fold axes, and that's about the only thing that would be in the plane of the paper if the principal symmetry is normal to the paper, we need a device for indicating the elevation and whether the axis is a two-fold rotation axis or a screw. So there's a geometrical language for describing the arrangement of the space groups. We had a few problems in determining space group symbols and interpreting what these mean. But again, I don't think I would be sadistic enough to ask you to derive a complicated three-dimensional space group. I don't know. Maybe over the weekend, if I get to feeling mean, I might ask you something in those directions. But it will not be anything terribly formidable. Then finally, we looked at the way in which space groups can be used to describe three-dimensional periodic arrays of atoms. And that again was strictly analogous to what we did in two dimensions. The description of the three-dimensional arrangement of atoms, particularly if it is too complicated to be readily appreciated in a projection, would have to be given in terms of the space group, the type of position that is occupied, either special or the general position, and then depending on how many degrees of freedom are available in those positions to give you the coordinates of the element or just the x-coordinate or the x and y-coordinate. And then you run to the international tables. You don't have to work this out yourself. And you find the list of coordinates that are related by symmetry for that particular special position and then plug in x and y to get all the coordinates of the atoms in the unit cell. To go to the practical aspects of this, usually only simple structures can be projected with any clarity that lets them be interpreted. And even with a computer these days one can show an orthographic projection of the structure and then in real time manipulate it so you can see this beautiful construction with colored shiny spheres in it rotating around. But if it's a complicated structure, even that doesn't do you very much. It works for NaCO and zinc sulfide. But you know what those look like. And so you don't really need this very attractive, real time interaction. What I found works well is to take the structure and slice it apart in layers like a layer cake and look at a limited range of the one translation, generally, one of the shorter translations, and examine the coordination of atoms within that layer. And so by slicing the structure away and looking at it in several layers that are stacked on top of one another, the way I work between the ears, usually lets me gain an appreciation of what's going on in that structure. Something different might work for you. Matter of fact, you might be a gifted person who can see the things spinning around on a computer screen and understand exactly what it is. But everybody's mind works in a slightly different way. OK, that is roughly what we'll be covering on symmetry. Then we got into the connection between symmetry and physical properties. And we defined what we meant by a tensor. A tensor of the second rank is something that relates to vectors. We got into a notation involving the assumption that any repeated subscript, and there's subscripts all over the place in tensor properties. Any repeated subscript is automatically implied to be summed over from 1 to 3. We then derived the laws for transformation of a tensor. Namely that, if you specify the change of coordinate system by a direction cosine scheme, cij, where cij is the cosine of the angle between x of i prime and x sub j, that, since these direction cosines constitute unit vectors, there had to be relations between them. Namely the sum of the squares of the terms in any row or column had to be unity. The sum of corresponding terms in any row or column had to add to 0. Because these represented dot products of unit vectors in a Cartesian coordinate system. And then using this array of direction cosines, we found that the way in which a second ranked tensor transformed was that each tensor element was given by a linear combination of all of the elements of the original tensor. And out in front of the tensor was a product of two direction cosines, the subscripts of which were determined by the indices of that element. A vector could be regarded as a first ranked tensor. And everybody understands the law for transformation of a vector. Namely that each component of the vector is a linear combination of every component of the original vector with one direction cosine out in front. Tensor of second rank has a product of two direction cosines summed over all the elements of the original tensor. Third ranked tensors, which, believe it or not, we'll get into after the quiz in considerable detail, and even shudder fourth ranked tensors are going to involve summations that involve scads of terms and products of three and four direction cosines respectively. So we won't get very far into tensors. We will perhaps have a couple of questions on the nature of anisotropy that's implied by these relations. And I think, if you looked at the problem sets to this point, that's as far as we'll go on the quiz. And we will not throw something at you that you've not seen before. So it'll be merciful. We'll pull out all the stops in the last quiz. Because then you'll be leaving. And you can hate me. But I never have to deal with you again. So that will not make any difference. All right, I have handed back all of the problem sets except two, which are partly done. I never again will ask anybody to draw out patterns of rotoinversion and rotoreflection axes. Because they go on and on and on. And I am so tired of staring at little circles and trying to tell which ones are right and which one are wrong. But that's something else we have in common. You probably hated looking at them when you were drawing them out for yourself. So let me tell you what's going to go on for the next several days. Tomorrow is a holiday, not for instructors who are making up a quiz, however. So I will be in my office from early morning on for the rest of the day. So if you want to come by to ask questions, what I will be doing primarily is finishing up grading the problem sets. And if you want to pick up your problem sets and ask a question about those, I am there at your disposal hopefully with no ringing phone calls and no committee meetings and nothing of that sort. If I have finished the problem sets and you haven't gotten yours yet, what I will do is tape an envelope to my door and have them tucked in there. So you can come by any time over the weekend and pick them up if you want to get hold of them and look them over at your leisure. I have to tell you, looking ahead in the schedule, I, unfortunately, bad timing, I have to be away on Monday. So if you have a last minute question or you need to be calmed down from a last minute panic, you can catch me Tuesday morning. I'll be in all of the earlier part of the day. But if you do have questions, please come in on Friday. Because I'll have lots of time to spend with you. So your problem sets, anything turned in to this point and turned in today, should be on my door, if you haven't called for it no, later than the beginning of the week. OK, any questions? Any puzzles that you want to go over before we launch into new material? OK, what we have done with second ranked tensor properties is to derive a whole set of symmetry constraints on something aij that represents a physical property of a crystal. And one of the things that we've seen, let's look at conductivity once more since that's something we're all familiar with and can readily appreciate, that the components of the current density vector depend on all of the components of the applied electric field, voltage per unit length. And in as much as for anything other than a crystal of cubic symmetry, the numbers in this array, at least some of them, are different. And that gives two surprising results. One is that the direction of J relative to the applied electric field will change as you apply the field in different directions in the crystal. And moreover, the magnitude of the current is going to change as you change the orientation of field. That's true of every crystal in a crystal system other than cubic. So the thing that I would like to address today is the interesting matter of what is the nature of this anisotropy and if a consequence of a second ranked tensor relation is that, if we apply an electric field, the current doesn't go in the same direction as the applied field. It runs off in some other direction as non-intuitive as that might seem. First of all, there are two directions. If we talk about a property varying with direction, which direction are we talking about, the direction of the applied field, the direction of the resulting current flux, or something halfway in between, split the difference? Well, we can answer that question by doing a couple of thought experiments. Suppose we had decided that we want to measure the electrical conductivity along the 1, 0, 0 direction of an orthorhombic crystal. What would we do? Well, we get a hold of that crystal. And we would cut surfaces on it such that this was the direction of 1, 0, 0, if that's what we wanted to do. We would glom on electrodes and put a voltage across this crystal, this chunk of crystal that we had cut out. So when we say we're going to measure the property along 1, 0, 0 or 1, 1, 1, what we really mean is that we're going to cut the crystal normal to that direction and then apply some sort of electrodes or attach some sort of probe. So when we talk about the direction of the property, the direction of the property will be the direction of the applied vector. And we had a very fancy name for the applied vector. We said we could think of that as a generalized force. And what happens is a generalized displacement. Notice that, when we apply the field, there are all sorts of components to J. And so the direction of the property, although we would get a number out of this experiment, the property really is a tensor. And there are nine different quantities involved in it. OK, so the direction of the property is going to be the direction in which we apply the generalized force of the stimulus. When we do that, however, that is going to unequivocally, at least in this experiment, place the electric field in a direction along the crystallographic orientation of interest. Because the electric field in a pair of electrodes like this extends normal to those electrodes. The current flow, the charge per unit area per unit time, goes off in some direction like this. Well, conductivity relates current flow to applied electric field. So would we say that the conductivity, when we apply field in that direction, is simply J/E, magnitude of J over magnitude of E since these are both vectors question mark? The answer to that, if we do this in terms of a thought experiment, is no. That is not what we're measuring. What we can imagine is that between our electrodes is some window of unit area between those electrodes. And what we're going to measure as the current flow is the amount of charge per unit area per unit time that passes between the electrodes, namely the charge per unit area per unit time that gets through this window and reaches the other electrode. That's what we're going to measure. And what we're going to want to call the conductivity is the ratio of this part of the current flux, let me call that J parallel, over the magnitude of E. And that's what we'll mean by the conductivity in the direction in which we've applied the field. The actual flow of current goes off in some other direction. But what we measure, at least in this experiment, is the component of that charge flux that goes through a window, that pair of unit area, that's parallel to the electrode. And we don't really care if the current is going off in a different direction. We don't even care if it jumps around in majestic loop-de-loops. What we're going to measure is the amount of that charge motion that moves normal to our window and gets through one unit area from one electrode to another. So the conductivity and the direction of E then is going to be measured by the parallel part of the resulting generalized displacement divided by the magnitude of the applied generalized force. If you're not convinced and completely overwhelmed by that, let me give you another experiment. Another tensor relation, another tensor property, is the diffusivity of a material. And if you put a material in a concentration gradient dc dx, you produce a flux of matter. And the proportionality constant is the diffusion coefficient. And this has units of length per unit time if you put in the units of the concentration gradient and the flux. Well, dc dx is a gradient. That's a vector really. So a proper tensor relation, instead of a scalar relation like this, would be the change of concentration with the i-th coordinate. And out in front would be a diffusion coefficient Dij. And this tensor relation would give us the components of the mass flux J1 J2 J3. Actually, thermodynamically we shouldn't be talking about a concentration gradient. We should be talking about a gradient in free energy. The concentration works well enough. And the concentration is what one would measure. So what might a typical experiment be? If you wanted to know what the diffusion was in the 1, 1, 1 direction of some crystal, you would, again, cut a plate that is normal to the direction of interest. And then your boundary conditions can be different. But what you would typically do for a simple experiment is to put some sort of solute on the surface of the crystal. And what you would see after heating the sample up for a period of time is that the material would have migrated into the sample. And if you plotted concentration as a function of distance along the normal to the plate that you've cut, this would do something like that. What can you say about this process after you've done the experiment? All that you can say is that some front of concentration has advanced parallel to the surface. And you have no idea what the individual atoms have done. Then they could be doing loop-de-loops or little hip hops. And all that you measure is the rate at which material advances in the direction of the current gradient. So again, I think I've convinced you, hopefully, that what you're measuring is the flux of matter unit area per unit time that is in the direction of the applied concentration gradient. And you do not measure, and in fact, in this instance, I don't think I could come up with even a very clever experiment that would let you measure in a single experiment the magnitude and direction of the net flux. All right, so if we're measuring a property then like conductivity, the value of the conductivity in the direction of the applied field is going to be the part of the conductivity that is parallel to the field divided by the magnitude of the field. We all agree on that. Let's have a show of hands. Fine. I saw just a few tired hands saying, OK, I believe, I believe, let's get on with it. All right, let me now put this using our tensor relationship into a nice analytic form that we can not only do something with but can gain some insight. The components of the charge density vector are given by the conductivity tensor sigma 1, 1 times the component of the electric field E1 plus sigma 1, 2 times the component of field E2 plus sigma 1, 3 times E3. Or in general, I don't have to write out every equation. We can say that the i-th component of J, our old friend the second ranked tensor property, again, is given by sigma ij times each component of applied field. If we specify the direction relative to the same coordinate system in which the electric field is being applied, we can specify that direction by a set of three direction cosines, l1, l2, and l3. So we can write as the three components of the electric field E sub i as the magnitude of E times the direction cosines l1, l2, and l3. So here we now have the components of the electric field. If these are the components of the electric field, we can say that J sub i is going to be equal to sigma ij times E sub j. And I can write for E sub j the magnitude of E times the direction cosines l sub j where these are the same direction cosines of the applied field. OK, now we have two relations in reduced subscript form that we can do something with. The magnitude of the conductivity in the direction of E, that is in the direction that has direction cosines li, is going to be J parallel over the magnitude of E. We can write that as J.E, except that involves the magnitude of J, the magnitude of E times the cosine of the angle between them. And so if I just want J parallel, I'd want to divide this by the magnitude of E. And that will give me just the part of J that falls parallel to E. And then down here in the denominator, I'll have magnitude of E. So if I tidy this up a little bit to bring everything on the same level, we can write the dot product of J and E as Ji Ei, like J1 E1 plus J2 E2 plus J3 E3. And then I will have in the denominator magnitude of E squared. But we know what Ei is. It's going to be magnitude of E times li. I know how to find the i-th component of J. That's going to be sigma ij times E sub j over magnitude of E squared. And I can write E sub j as magnitude of E times l sub j times l sub i times magnitude of E divided by magnitude of E squared. Magnitude of E drops out. Again, the conductivity is linear the way we've defined it. It shouldn't depend on the magnitude of E. And what I'm left with then is a very simple and a rather profound relation. It says that in a direction specified by the set of direction cosines l sub i that the magnitude of conductivity in that direction is going to be li lj times sigma ij. So that is an expression that involves the three direction cosines in which you're applying the field and measuring the property. But this is a linear combination of every one of the elements in the original tensor. So if I write this out, because this is an important relation, it's going to be l1 squared sigma 1, 1 plus l2 squared sigma 2, 2 plus l3 squared sigma 3, 3. And then there will be cross-terms of the form 2 l1 l2 sigma 1, 2 plus 2 l2 l3 times sigma 2, 3 plus 2 l3 l1 times sigma 1, 3. That has made an assumption. And probably I should not have done it at this point. That's assuming that the term l1 l2 sigma 1, 2 and the term l2 l1 sigma 2, 1 can be lumped together. And maybe I shouldn't write it that way. Because that assumes that the tensor is symmetric. And it doesn't have to be for some properties. So let me back off and write it without any assumptions. So this would be sigma 1, 2 l1 l2 plus sigma 1, 3 l1 l3 plus sigma 2, 1 times l2 l1 and then corresponding terms-- let me put this one down here-- sigma 2, 1 times l2 l1 plus sigma 3, 1 l3 l1. And then the one that I'm missing is sigma 2, 3 times l2 l3 plus sigma 3, 2 times l2 l3. So there is an explicit statement of this more compact expression written in reduced subscript notation. OK, so this is something that is going to tell us in a very neat sort of way what the anisotropy of the property is. You pick the direction in which you're interested. And this will tell you the value of the property in that direction and, therefore, by inference how things will change with direction. Let me quickly, and then I'll pause to see if you have any questions, give you an interpretation of the meaning of some of these elements of the tensor. We've really not been able to do that. We've been able to use them to give the value of the property in a given direction but not their individual meaning. Suppose I ask what the value of the property would be along the x1 direction. So suppose we ask what value of the conductivity would we measure along x1 for a given tensor sigma ij. If that is the case, l1 is equal to 1. The angle between the x1 axis and x2 is 0. So l2 is 0. 90 degrees cosine of 90 degrees is zero. l3 would be equal to 0. So if I put this set of direction cosines into this complicated polynomial, the term l1 squared sigma 1, 1 stays in. But l2 is 0. l3 is 0. And I've got either an l2 or an l3 or both in all of these other terms. So along the x1 direction, the only term that stays in is sigma 1, 1 squared times l1 squared. But l1 is unity. So along the x1 direction the value of sigma is sigma 1, 1. And by inference the value of the property that we would measure along x2 would be sigma 2, 2. And the value of the property along x3, guess what, sigma 3, 3. So in this three by three array of numbers for sigma ij, there's a very direct meaning to the diagonal values of the tensor. These are the values that you would measure along x1, along x2, and along x3 directly. So the off diagonal terms are going to be saying something about how the extreme values of the property are aligned relative to our reference axes. OK, any comment or questions at this point? Yeah, [INAUDIBLE]? AUDIENCE: [INAUDIBLE]. PROFESSOR: Yeah, we've defined those as the cosines of the angles between the reference axes. And this is the direction in which we're applying our electric field or our generalized forced temperature gradient, concentration gradient, magnetic field, or what have you. OK, I've got some time left. So let me push this further into a different form by using the elements of the tensor to define a locus in space. Again, what I'll do is something that may seem silly. But we'll see that there are some very useful and profound conclusions to be drawn from it. Let me take the elements of the tensor and use them as coefficients to define a surface. I'm going to take sigma 1, 1 and multiply it by the product of the coordinate x1 times x1. What I'm viewing this now is a variable. I'm going to take sigma 2, 2 and multiply it by the coordinate x2 and x2 again. Sigma 3, 3 and I multiply that by x3 times x3. And then I'll have these cross-terms, sigma 1, 2, x1, x2, and so on. Or in short, I'm going to define a function sigma ij xi xj. And xi and xj are running variables. x1, x2, x3, they can extend from 0 to infinity. But now I'm going to take this sum of nine terms and I'm going to set it equal to a constant. And what constant could be neater and cleaner and more abstract than unity? But I could say that some function of x1, x2, x3 equal to a constant is going to define a locus of points in space, which satisfy that equation. And we do that all the time. We specify some functional relationship that defines in the space some surface f of xi xj equals a constant. And that's done all time in mathematics. I've yet to encounter yet a department of mathematics that did not have in its corridors some glass case that was filled with yellowing plaster figures, that some of them cracking or already cracked, that represented exotic surfaces, like elliptic paraboloids and things of that sort. I think the last time I walked down the corridor of building two there really was one of those in our mathematics department in the little door that opened out onto the great court about halfway down that corridor. But you've seen these things I'm sure, elliptic cylinders, elliptic cones, and all sorts of glorious things. This function, though, is a very special one. This is a quadratic equation in the second order of coordinates. And this is referred to as a quadratic form. And sometimes for short, since that's something of a mouthful, one refers to this as a quadric. And we are going to very shortly demonstrate that this particular function tells you everything you could possibly want to know about the variation of a tensor property with direction, how the magnitude of the property changes, how the direction of the resulting current flow or flux is oriented. And this is, consequently, called, when applied to a tensor, the representation quadric. Because it really does represent anything you would like to know about the physical property. This will be demonstrated in short order. Let's first notice, though, that there's only a limited number of surfaces that can be represented by this equation. One of them is one that we're all familiar with, an ellipsoid. And that's most easily represented when we've referred it to the principal axes of the ellipsoid. So an ellipsoid is a quasi-sausage like thing like this. These sections perpendicular to x3, x1, and x2 are all ellipsoids. And if this semi-axis is a and this semi-axis is b and this semi-axis is c, the equation of the surface in that special orientation is x1 squared over a squared plus x2 squared over b squared plus x3 squared over c squared equals 1. OK, the surface that we have defined involves tensor elements. And if the equation is ever going to get into this form, we've had to get rid of these cross-terms. And we are going to have principal axes that are some function of the tensor elements that would be out here in the form of some aggregate sigma ij prime. So we'll have to see how one could go from a tensor that describes an ellipsoid in a general orientation to something that has been diagonalized and has the coordinate system taken along the principal axes. But that's only one possible surface that we could encounter. The second one is one that we might encounter that, when we put it in diagonal form, has this form, x1 squared over a squared plus x2 squared over b squared minus z squared over c squared equals 1. This is something that has a peculiar shape. It's sort of an hourglass like figure. It has an ellipsoidal cross-section. This is x3. This is x1. This is x2. It has an ellipsoidal section perpendicular to x3. But it has, as a cross-section in planes that are parallel to x3, paraboloids. And the shape of the paraboloids depends on which particular section that you take through x3. This is a surface that's called a hyperboloid. And it's all one continuous surface. So this is called a hyperboloid of one sheet, sheet in the sense of surface. And then down in here when x3 is 0, it would again be an ellipsoidal cross-section but a little bit smaller than when we increase x3 above 0. OK, and the other surface, another surface, a third kind of surface is something that occurs when the terms in front of two coordinates are negative, x2 squared over b squared x3 squared over c squared equals 1. This is something that looks like two surfaces nose to nose. The cross-section of these two different sheets are ellipsis. The cross-sections passing through-- let's see. This would be the shape for x1. That would be the special direction. And then this could be x2 and this x3. OK, this is called an hyperboloid of two sheets. And it has the property that the distance to the surface in orientations in between the asymptotes of these hyperbolic sections, these radii, are imaginary. And between the asymptotes, if we take a section, we would have two hyperboloids in a cross-section. And again, a range of directions between the asymptotes in which the radius would be imaginary in directions along the coordinate which has the positive sign we'd have a minimum radius. And this would get progressively larger and go to infinity along the asymptote. Finally, there's a third surface in which, by extension, all three terms in x1, x2, and x3 have negative signs. And what does this look like? AUDIENCE: [INAUDIBLE] PROFESSOR: No, it does. It's something that's called an imaginary ellipsoid. So called because the radius in all direction is imaginary. And what does it look like? You just have to use your imagination. That's all. Because it's an imaginary ellipsoid. Is that something that we could ever get if we were using the elements of a property tensor? The answer, he says with a big wink, yes. It's unusual but yes. You can get physical properties whose magnitude as a function of direction do this. All right, I think that is a sufficiently formal and deadening component to our presentation that it would be appropriate to take a break and stretch a little bit. When we return, I will show you an amazing connection between the anisotropy of a physical property and this representation surface. So called because it tells you how the property is going to change.
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https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/8.01sc-fall-2016.zip
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For a rocket and an external force, we had the rocket equation, which we wrote as mass of the rocket-- now remember, this is a function of time-- times the derivative of the rocket dr dt. And we also had this second term that came from the ejecting fuel. Now from our mass conservation equation, we can rewrite this equation as a differential relationship that the change of mass of the fuel in time is equal to minus the change of the rocket. It doesn't matter which side we put the minus sign on. So now I want to interpret this equation in a different way, and it will come back to what we mean by our system. Let's first off bring this other term over to this side. So we have plus dMr dt u. And that's equal to Mr dVr dt. Now separately, let's make this substitution again and go back to our fuel term. So that's minus dMf dt u equals Mr dVr dt. Now notice that over here we have mass times the acceleration of the rocket. So if we just rethought our system as simply the rocket-- Mr-- then we have two forces acting on the rocket. The external force might be the gravitational field, but we have a new term here which we're going to refer to as thrust. And so our thrust can be thought of as an external force simply on the rocket as a system. And that's equal to minus the f dt u. Now again, if I chose j-hat up-- let's just look at this in components-- then our thrust as a vector, we'll write it as a y-component, is equal to minus dMfuel dt. Now what is that relative speed? Well, the fuel is being ejected backwards, so that's minus mu j-hat. The external force, by the way, would be minus mg j-hat if it's near the surface of the Earth. So we have another minus u j-hat. And we see that we have a positive thrust force in the vertical direction that is giving us an additional force other than the gravitational field. And in Cartesian-- in our unit vectors here, we have minus mg plus dMfuel dt times u is equal to Mr dV ydt of the rocket. And so this is the rocket equation in components.
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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: Good morning. Today's lecture will deal with Zeeman effect. And then we'll get started with a semi-classical approximation. So Zeeman in effect is the last topic we do with respect to the hydrogen atom and the corrections of perturbation theory and with WKB or the semi-classical approximation, we begin a new chapter in 806. Zeeman effect. So this is an effect having to do with an atom in a magnetic field. It was discovered by a Dutch physicist, Peter Zeeman, who lived from 1865 to 1943, who was a Dutch. The work was actually done in 1896 at a time where there was very little idea of quantum mechanics to be, and he got a Nobel Prize for in the year 1902. So what he discovered was the spectral lines seemed to split in the presence of a magnetic field. It's an old result, therefore, 100 years old. Its explanation and understanding took about a couple of decades, because you couldn't do it without quantum mechanics. So nobody could quite figure out what had happened, but was very, very important in its time. It still remains very important. People use the Zeeman effect all the time. In fact, it's used nowadays in studies of astrophysics, studies of the sun, the sunspots. You know, there are places in the sun, where the temperature is a little lower, and that's places where the magnetic field lines in the sun sort of breakout from the interior to the exterior. And it's interesting, because the sun produces this cycle of solar spots, and it happens, because the sun doesn't rotate uniformly, has a different rotation speed in the equator, faster than in the poles, so the magnetic field lines that go from north to south in the sun get tangled up, and they start breaking up and doing all these things. So people want to know what is the magnetic field in the solar sunspot. And in fact, they observe the weak magnetic fields away from the solar sunspots, and then they see the spectrum of an atom. And suddenly, as you move inside the spot, the field, the spectrum, the lines split. And they can measure the magnetic fields very accurately. They're of the order of 3,000 Gauss, 2,000, 3,000 Gauss. And it's pretty interesting work. So this Zeeman effect remains very important. So what is the Zeeman effect? We have Zeeman effect here. We have the magnetic field interacting with the electron, and the electron now has two magnetic moments, a magnetic moment associated with the orbital motion. This looks completely like the classical formula of the magnetic moment due to a particle that goes in circles. It produces a current, and that current is proportional to the angular momentum of the rotating particle. And of course, there is the magnetic moment due to the spin that has a factor of 2 here. This g factor, we've discussed before. S. So H Zeeman, you put an external magnetic field into the atom, a constant uniform external magnetic field, and you have a minus mu dot b, so you have a interaction mu l plus mu s, dot B. So this is e over 2mc. That's the typical factor here, and a recognizable l plus 2s. So it's not l plus s. It's l plus 2s times b. And many times, we think of B, align the axis so that it is in the z direction. So this turns out to be e over 2mc lz plus 2sz times B. So this is the Zeeman Hamiltonian. But this is part of a story of an atom. So if we want to think of the hydrogen atom properly, we must consider and reconsider what was the Hamiltonian there. And we had an H for the hydrogen atom. That was an H 0. That was the familiar one, p squared over 2m minus e squared over r. Then we had a fine structure Hamiltonian, delta H, fine structure. Let's put fs for fine structure. Those were the relativistic terms, the Darwin term, and the spin orbit coupling. The three of them constituted what we call defined structure Hamiltonian. And now we have a Zeeman effect. I probably should go delta H Zeeman, because it's an addition here to the term we had before. It refers to what I call just H Zeeman. But in the context of the hydrogen atom, we should call it delta H Zeeman. And now, we have to rethink. And the reason the Zeeman effect is non-trivial for us, and it's a very interesting and somewhat challenging example of what we have to do in perturbation theory, is that we cannot forget about the fine structure. So if it would be just this, it would be kind of simple. But we have the whole thing. So we have to make an approximation sometimes. And we're going to consider two interesting cases, the very weak Zeeman effect and the very strong Zeeman effect. I will not consider the intermediate Zeeman effect, not because it's not interesting, but because there's very little you can do to simplify it, and to think about it. You basically have to go ahead and diagonalize the large matrix. So while it's important, and if your life depended on this, and your research dependent on this, you would do it. For us, we have a lot to learn from the weak case and the strong case, lots of concepts, and we'll leave the intermediate one for later. So how can we decide how to treat these terms? Well, we should look physically at what's happening. We have a magnetic field, an external magnetic field. But the fine structure constant taught you that there is something like an internal magnetic field in the atom. It's that magnetic field responsible for spin orbit coupling is that magnetic field that the electron sees when it's going around the proton. There's a static electric field, but whenever you are in motion, a static electric field in the lab also has a magnetic field by relativity, and you do see a magnetic field. You could also imagine it as you are the electron, and the problem is going around you, and creates at the center of the loop, a magnetic field. In any case, you've looked at that magnetic field. And we can call it the internal magnetic field due to spin orbit. As one exercise in the homework, you've been asked to estimate the value of that internal magnetic field. Is it a Gauss, 1,000 Gauss, 10,000 Gauss? How much it is for a typical level? So there's a number here that this interest. And now we can decide whether we have a weak Zeeman effect or a strong Zeeman effect, by looking how your external magnetic field compares with this little magnetic field. So we'll have a weak Zeeman effect, A. Zeeman. When B is much smaller than B internal. And therefore, the effects of Zeeman is going to be smaller than fine structure. So really, when you look at this line, and you have the first two terms, you should think of these two terms as your new known Hamiltonian. And the Zeeman term, at its perturbation. Yes. You've solved for the fine structure coupling and the shift. So this is the known thing. These are going to be the known states with known energies. You only know them the first order, but you know them. And then the Zeeman effect will be a perturbation theory on this. This is the weak Zeeman effect. How about the strong Zeeman effect? So B will be strong Zeeman. This time, the magnetic field associated with the Zeeman effect, the external magnetic field, is much smaller than the magnetic field responsible for spin orbit coupling. And this time, what are we going to do? Well, we will take the Hamiltonian. It will be H 0. The strong Zeeman effect means this Zeeman Hamiltonian is a lot more important than the spin orbit coupling. So we'll add delta H Zeeman here. And we hope this will be our known Hamiltonian, because anyway, the Zeeman stuff is much stronger now than fine structure. So this should be our non-Hamiltonian. You should complain, no, this is not known. I haven't done Zeeman, but we'll look at it. And once we have our known Hamiltonian here, spin orbit has to be rethought. Fine structure has to be rethought as a perturbation.
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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: This is Louis-- L-O-U-I-S d-e Broglie. And this is not hyphenated nor together. They are separate. And the d is not capitalized apparently too. And it's 1924, the photon as a particle is clear, and the photon is also a wave. And Louis de Broglie basically had a great insight in which he said that if this is supposed to be a universal or a real basic physical property that photons are waves and particles, we knew them as waves and now we know they're particles. But if they are dualed with respect to each other, both descriptions are in different regimes and in a sense, a particle at the end of the day has wave attributes and particle attributes. Wave attributes because it interferes and is described by waves. And particle attributes is because it has a definite amount of energy, it comes in packets, they cannot be broken into other things-- this could be a more general property. And in a sense, you could say that de Broglie did a fundamental step almost as important as Schrodinger when he claimed that all matter particles behave as waves as well. Not just the photon, that's one example, but everybody does. So associated to every modern particle, there is a wave. But that is quite interesting because in quantum mechanics, you have the photon and it's a particle, but it's associated to a wave and if you are a little quick, you say, oh sure, the electromagnetic wave, but no, in quantum mechanics, it's the probability amplitude to be some work. Those are the numbers we tracked in the mass and the interferometer, the probability to be sampled. We didn't track the waves or a single photon, the wave was a wave of probability amplitude, something they didn't know at all about yet at that time. So de Broglie's says just like the photons have properties of particles and properties of waves, every particle has properties of waves as well and every wave has a property of particles. But what is left unsaid here is yes, you have a wave, but a wave of what? And we've already told you a little bit, the answer has to do with probability waves. So it's very strange that the fundamental equation for a wave that represents a particle is not an electric field or a sound wave or this, it's for all of them is a probability wave. Very, very surprising. But that's what de Broglie's ideas led to. So if you had a photon, you would say it's a particle, and when you think of it as a particle, you would say it's a bundle of some energy and some momentum. And if you think of it as a wave, you would say it has a frequency. And that's a particle wave duality or in some sense, a particle wave description of this object-- you have a particle and a wave at the same time. When we have this, we have a particle wave duality. And de Broglie said that this is universal for all particles. Universal. And it appeared the name of matter waves. These are the matter waves that we're going to try to discuss. These are the waves of something that are probability amplitudes we're going to try to discuss. So you could say wave of what? What. And that comes later, but the answer is probability amplitudes, those complex numbers whose squares are probabilities. So just like we had for a photon, de Broglie's idea was that we would associate to a particle a wave that depends on the momentum. So remember, the Compton wavelength was a universal-- for any particle, the Compton wavelength is just one number, but just for photons, the wavelength depends on the momentum, so in general, it should be dependent on the momentum. So we say that for a particle of momentum p, we associate a wave-- a plane wave, in fact-- a plane wave, so we're getting a little more technical, with of lambda equals h over p, which is the de Broglie wavelength-- de Broglie wavelength. So it's a pretty daring statement. It was his PhD thesis and there was no experimental evidence for it. It was a very natural conjecture-- we'll discuss it a lot more next lecture-- but there are very little evidence for it. So experiments can a few years later, and people saw that you could interfere or diffract electrons. They would behave, colliding into lattices like waves, and those are rather famous experiments of Davisson and Germer. So particles, just like you do as an interference effect-- a two slit interference effect in which you have a screen, a slit and a screen, and you shine photons and then you get an interference effect over here because of the wave nature of photons, or in quantum mechanics, you would say, because there are probability amplitudes, that are complex numbers that have to be interfered between the possibilities of the two paths, because every photon goes through both paths at the same time, these experiments of interference, or two slit interference, were done for electrons. And then, eventually, they've been done for bigger and bigger particles, so that it's not just something that you do with elementary particles now. There's experiments done about three years ago-- I will put on the web site or on the notes some of these things so that you can see them, but now you can throw in molecules here, molecules that have a weight of now 10,000 atomic mass units, like 10,000 protons, like hundreds of-- 430 atom molecules, and you can get an interference pattern, so it's pretty ridiculous. It's almost like, you one day so you throw a baseball and you're going to see an interference pattern, but, you know, we've got to things with about 10,000 hydrogen atoms and de Broglie wavelengths of 1 picometer, which are pretty unbelievable. So the experiments are done with those particles and in fact with electrons. People do those experiments and they're in very beautiful movies in which you see those electrons hitting on the screen and then-- I'll give you some links so you can find them as well-- and you see one electron falls here and it gets detected and two electrons, three electrons, four electrons, five electrons, six electrons-- by the time you get 10,000 electrons, you see lots of electrons here, very well here, lots of electrons here, and the whole interference pattern is created by sending one electron at that time in an experiment that takes several hours and it's reduced to a movie of about one minute. So particles, big particles interfere, not just photons interfere. So those particles have some waves, some matter waves discovered by de Broglie, and next lecture, we're going to track the story from de Broglie to the Schrodinger equation where the nature of the wave suddenly becomes clear.
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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: Last time we spoke about the Stern-Gerlach experiment, and how you could have a sequence of Stern-Gerlach boxes that allow you to understand the type of states and properties of the physics having to do with spin-1/2. So the key thing in the Stern-Gerlach machine was that a beam of silver atoms, each of which is really like an electron with a magnetic moment, was placed in an inhomogeneous strong magnetic field, and that would classically mean that you would get a deflection proportional to the z-component of the magnetic moment. What was a surprise was that by the time you put the screen on the right side, it really split into two different beams, as if the magnetic moments could either be all the way pointing in the z-direction or all the way down, pointing opposite to the z-direction, and nothing in between. A very surprising result. So after looking at a few of those boxes, we decided that we would try to model the spin-1/2 particle as a two-dimensional complex vector space. What is the two-dimensional complex vector space? It's the possible space of states of a spin-1/2 particle. So our task today to go into detail into that, and set up the whole machinery of spin-1/2. So we will do so, even though we haven't quite yet discussed all the important concepts of linear algebra that we're going to need. So today, I'm going to assume that at least you have some vague notions of linear algebra reasonably well understood. And if you don't, well, take them on faith today. We're going to go through them slowly in the next couple of lectures, and then as you will reread this material, it will make more sense. So what did we have? We said that the spin states, or the possible states of this silver atom, that really correspond to an election, could be described by states z comma plus and z colon minus So these are the two states. This state we say corresponds to an angular momentum Sz hat. Sz-- I can put it like that-- of h-bar over 2, and this corresponds to Sz equals minus h-bar over 2. And those are our two states. The z label indicates that we've passed, presumably, these atoms through a filter in the z-direction, so that we know for certain we're talking about the z-component of angular momentum of this state. It is positive, and the values here again have the label z to remind us that we're talking about states that have been organized using the z-component of angular momentum. You could ask whether this state has some angular momentum-- spin angular momentum-- in the x-direction or in the y-direction, and we will be able to answer that question in an hour from now. So mathematically, we say that this statement, that this state, has Sz equals h-bar over 2 means that there is an operator, Sz hat-- hat for operators. And this operator, we say, acts on this state to give h-bar over 2 times this state. So when we have a measurement in quantum mechanics, we end up talking about operators. So this case is no exception. We think of the operator, Sz, that acts in this state and gives h-bar over 2. And that same operator, Sz, acts on the other state and gives you minus h-bar over 2 times the state. You see, an operator on a state must give a state. So in this equation, we have a state on the right, and the nice thing is that the same state appears on the right. When that happens, you say that the state is an eigenstate of the operator. And, therefore, the states z plus, minus are eigenstates of the operator Sz with eigenvalues-- the number that appears here-- equal to plus, minus h over 2. So the relevant physical assumption here is the following, that these two states, in a sense, suffice. Now, what does that mean? We could do the experiment again with some Stern-Gerlach machine that is along the x-axis, and say, oh, now we've got states x plus and x minus and we should add them there. They are also part of the possible states of the system. Kind of. They are parts of the possible states of the system. They are possible states of the system, but we shouldn't add them to this one. These will be thought as basis states. Just like any vector is the superposition of a number times the x-unit vector plus a number times the y-unit vector and a number times the z-unit vector, we are going to postulate, or try to construct the theory of spin, based on the idea that all possible spin states of an electron are obtained by suitable linear superposition of these two vectors. So, , in fact, what we're going to say is that these two vectors are the basis of a two-dimensional vector space, such that every possible state is a linear superposition. So psi, being any possible spin state, can be written as some constant, C1 times z plus plus C2 times z minus where these constants, C1 and C2 belong to the complex numbers. And by this, we mean that if any possible state is a superposition like that, the set of all possible states are the general vectors in a two-dimensional complex vector space. Complex vector space, because the coefficients are complex, and two-dimensional, because there's two basis vectors. Now this doesn't quite look like a vector. It looks like those things called kets. But kets are really vectors, and we're going to make the correspondence very clear. So this can be called the first basis state and the second basis state. And I want you to realize that the fact that we're talking about the complex vector space really means these coefficients are complex. There's no claim that the vector is complex in any sense, or this one. They're just vectors. This is a vector, and it's not that we say, oh this vector is complex. No. A complex vector space, we think of as a set of vectors, and then we're allowed to multiply them by complex numbers. OK, so we have this, and this way of thinking of the vectors is quite all right. But we want to be more concrete. For that, we're going to use what is called a representation. So I will use the word representation to mean some way of exhibiting a vector or state in a more concrete way. As something that any one of us would call a vector. So as a matter of notation, this being the first basis state is sometimes written as a ket with a 1. Like that. And this being this second basis state is sometimes written this way. But here is the real issue of what we were calling a representation. If this is a two-dimensional vector space, you're accustomed to three-dimensional vector space. What are vectors? They're triplets of numbers. Three numbers. That's a vector. Column vectors, it's perhaps easier to think about them. So column vectors. So here's what we're going to say. We have this state z plus. It's also called 1. It's just a name, but we're going to represent it as a column vector. And as a column vector, I'm going to represent it as the column vector 1, 0. And this is why I put this double arrow. I'm not saying it's the same thing-- although the really it is-- it's just a way of thinking about it as some vector in what we would call canonically a vector space. Yes AUDIENCE: So do the components of the column vector there have any correspondence to the actual. Does it have any basis in the actual physical process going on? Or, what is their connection to the actual physical [INAUDIBLE] represented here? PROFESSOR: Well, we'll see it in a second. It will become a little clearer. But this is like saying, I have a two-dimensional two0 vector space, so I'm going to think of the first state as this vector. But how do I write this vector? Well, it's the vector ex. Well, if I would write them in components, I would say, for a vector, I can put two numbers here, a and b. And this is the a-component and b-component. So here it is, ex would be 1, 0. And ey would be 0, 1. If I have this notation then the point a, b is represented by a and b as a column vector. So at this moment, it's just a way of associating a vector in the two-dimensional canonical vector space. It's just the column here. So the other state, minus-- it's also called 2-- will be represented by 0 1 1. 1 And therefore, this state, psi, which is C1 z plus plus C2 z minus will be represented as C1 times the first vector plus C2 times the second vector. Or multiplying, in C1, C2. So this state can be written as a linear superposition of these two basis vectors in this way-- you can write it this way. You want to save some writing, then you can write them with 1 and 2. But as a vector, it's represented by a column vector with two components. That's our state. Now in doing this, I want to emphasize, we're introducing the physical assumption that this will be enough to describe all possible spin states, which is far from obvious at this stage. Nevertheless, let's use some of the ideas from the experiment, the Stern-Gerlach experiment. We did one example of a box that filtered the plus z states, and then put it against another z machine, and then all the states went through the up. Which is to say that plus states have no amplitude, no probability to be in the minus states. They all went through the plus. So when we're going to introduce now the physical translation of this fact, as saying that these states are orthogonal to each other. So, this will require the whole framework, in detail, of bras and kets to say really, precisely-- but we're going to do that now and explain the minimum necessary for you to understand it. But we'll come back to it later. So this physical statement will be stated as z minus with z plus. The overlap, the bra-ket of this, is 0. The fact that all particles went through and went out through the plus output will state to us, well, these states are well normalized. So z plus, z plus is 1 Similarly, you could have blocked the other input, and you would have concluded that the minus state is orthogonal to the plus. So we also say that these, too, are orthogonal, and the minus states are well normalized. Now here we had to write four equations. And the notation, one and two becomes handy, because we can summarize all these statements by the equation Ij equals delta Ij. Look, if this equation is 2 with 1 equals 0. The bra 2, the ket 1. This is 1 with 1 is equal to 1. Here is is 1 with 2 is equal to 0, and 2 with 2 is equal to 1. So this is exactly what we have here. Now, I didn't define for you these so-called bras. So by completeness, I will define them now. And the way I will define them is as follows. I will say that while for the one vector basis state you associate at 1, 0, you will associate to one bra, the row vector 1, 0. I sometimes tend to write equal, but-- equal is all right-- but it's a little clearer to say that there's arrows here. So we're going to associate to 1, 1, 0-- we did it before-- but now to the bra, we think of the rho vector. Like this. Similarly, I can do the same with 2. 2 was the vector 0, 1. It's a column vector, so 2 was a bra. We will think of it as the row vector 0, 1. We're going to do this now a little more generally. So, suppose you have state, alpha, which is alpha 1, 1 plus alpha 2, 2. Well, to this, you would associate the column vector alpha 1, alpha 2. Suppose you have a beta state, beta 1, 1 plus beta 2, 2. You would associate beta 1, beta 2 as their representations. Now here comes the definition for which this is just a special case. And it's a definition of the general bra. So the general bra here, alpha, is defined to be alpha 1*, bra of the first, plus alpha 2*, bra of the second. So this is alpha 1* times the first bra, which we think of it as 1, 0, plus alpha 2* times the second bra, which is 0, 1. So this whole thing is represented by alpha 1*, alpha 2*. So, here we've had a column vector representation of a state, and the bra is the row vector representation of the state in which this is constructed with complex conjugation. Now these kind of definitions will be discussed in more detail and more axiomatically early very soon, so that you see where you're going. But the intuition that you're going to get from this is quite valuable. So what is the bra-ket? Alpha-beta is the so-called bra-ket. And this is a number. And the reason for complex conjugation is, ultimately, that when these two things are the same, it should give a positive number. It's like the length squared. So that's the reason for complex conjugation, eventually. But, for now, you are supposed to get a number from here. And the a reasonable way to get a number, which is a definition, is that you get a number by a matrix multiplication of the representatives. So you take the representative of alpha, which is alpha 1*, alpha 2*. And do the matrix product with the representative of beta, which is beta 1, beta 2. And that's alpha 1*, beta 1 plus alpha 2*, beta 2. And that's the number called the inner product, or bra-ket product. And this is the true meaning of relations of this kind. If you're given an arbitrary states, you compute the inner product this way. And vectors that satisfy this are called orthonormal because they're orthogonal and normal with respect to each other in the sense of the bra and ket. So this definition, as you can see, is also consistent with what you have up there, and you can check it. If you take I with j, 1, say, with 2-- like this-- you do the inner product, and you get 0. And similarly for all the other states. So let's then complete the issue of representations. We had representations of the states as column vectors-- two by two column vectors or row vectors. Now let's talk about this operator we started with. If this is an operator, acting on states, now I want to think of its representation, which would be the way it acts on these two component vectors. So it must be a two by two matrix, because only a two by two matrix acts naturally on two component vectors. So here is the claim that we have. Claim, that Sz hat is represented-- but we'll just put equal-- by this matrix. You see, it was an operator. We never talked about matrices. But once we start talking about the basis vectors as column vectors, then you can ask if this is correct. So for example, I'm supposed to find that Sz hat acting on this state 1 is supposed to be h-bar over 2 times the state 1. You see? True. Then you say, oh let's put the representation, h-bar over 2, 1 minus 1, 0, 0. State one, what's its representation? 1, 0. OK, let's act on it. So, this gives me h-bar over 2. I do the first product, I get a 1. I do the second product, I get a 0. Oh, that seems right, because this is h over 2 times the representation of the state 1. And if I check this, and as well that Sz on 2 is equal minus h-bar over 2, 2-- which can also be checked-- I need to check no more. Because it suffices that this operator does what it's supposed to do of the basis vectors. And it will do what it's supposed to do on arbitrary vectors. So we're done. This is the operator Sx, and we seem to have put together a lot of the ideas of the experiment into a mathematical framework. But we're not through because we have this question, so what if you align and operate the machine along x? What are the possible spin states along the x-direction? How do you know that all that the spins state that points along x can be described in this vector space? How do I know there exists a number C1, C2 so that this linear combination is a spin state that points along x. Well, at this moment, you really have to invent something. And the process of invention is never a very linear one. You use analogies-- you use whatever you can-- to invent what you need. So, given that that's a possibility, we could follow what Feynman does in his Feynman lectures, of discussing how to begin rotating Stern-Gerlach machines, and doing all kinds of things. It's an interesting argument, and it's a little hard to follow, a little tedious at points. And we're going to follow a different route. I'm going to assume that you remember a little about angular momentum, and I think you do remember this much. I want to say, well, this is spin angular momentum. Well, let's compare it with orbital angular momentum, and see where we are. You see, another way of asking the question would be, well, what are the operators Sx and Sy. Where do I get them? Well, the reason I want to bring in the angular momentum is because there you have Lz, but you also have Lx and Ly. So angular momentum had Lz, just like we had here, but also Lx and Ly. Now these spin things look a lot more mysterious, a lot more basic, because, like Lz, it was xpy minus ypx. So you knew how this operator acts on wave functions. You know, it multiplies by y, takes an x derivative, or it's a dd phi. It has a nice thing, but Sz on the other hand, there's no x, there's no derivatives. It's a different space. It's working in a totally different space, in the space of a two-dimensional complex vector space of column vectors with two numbers. That's where it acts. I'm sorry there's no dd x, nothing familiar about it. But that's what we have been handed. So this thing acts on wave functions, and thus natural things. Well, the other one acts on column vectors. Two-by-two-- two component column vectors, and that's all right. But we also know that Lz is Hermitian. And that was good, because it actually meant that this is good observable. You can measure it. Is Sz Hermitian? Well, yes it is. Hermeticity of a matrix-- as we'll discuss it in a lot of detail, maybe more than you want-- means you can transpose it complex conjugated, and you get the same matrix. Well that matrix is Hermitian. So that's nice. That maybe is important. So what other operators do we have? Lx and Ly. And if we think of Lx as L1, Ly as L2, and Lz as L3, you had a basic computation relation. Li with Lj was equal to i epsilon ijk Lk-hat-- oops i-hbar. And this was called the algebra of angular momentum. These three operators satisfy these identities. i and j are here, k is supposed to be summed over-- repeated in this is our sum from 1, 2, and 3. And epsilon ijk is totally anti-symmetric with epsilon 1, 2, 3 equal to plus 1. You may or may not know this epsilon. You will get some practice on that very soon. Now for all intents and purposes, we might as well write the explicit formulas between Lx, Ly equal i-hbar Lz. Ly Lz equals i-hbar Lx. And Lz Lx-- there are hats all over-- equal i-hbar Ly. So we had this for orbital angular momentum, or for angular momentum in general. So what we're going to do now is we're going to try to figure out what are Sx and Sy by trying to find a complete analogy. We're going to declare that S is going to be angular momentum. So we're going to want that Sx with Sy will be i-hbar Sz. Sy with Sz will be i-hbar Sx. And finally, Sz with Sx is i-hbar Sy. And we're going to try that these things be Hermitian. Sx and Sy. So let me break for a second and ask if there are questions. We're aiming to complete the theory by taking S to be angular momentum, and see what we get. Can we invent operators Sx and Sy that will do the right thing? Yes. AUDIENCE: What's the name for the epsilon ijk? I know there's a special name for the [INAUDIBLE]. PROFESSOR: What's the name? AUDIENCE: There's a [INAUDIBLE] tensor. PROFESSOR: That's right. Let [INAUDIBLE] to be the tensor. It can be used for cross products. It's very useful for cross products. It's a really useful tensor. Other questions. More questions about what we're going to try to do, or this so far. Yes. AUDIENCE: When you use the term representation, is that like the technical mathematical term of representation, like in algebra? PROFESSOR: Yes. It's representation of operators in vector spaces. So we've used the canonical vector space with column vectors represented by entries one and numbers. And then the operators become matrices, so whenever an operator is viewed as a matrix, we think of it as a representation. Other questions. Yes. AUDIENCE: Will we talk about later why we can make an analogy between L and S? Or is it [INAUDIBLE]? PROFESSOR: Well you see, this is a very strong analogy, but there will be big differences from orbital angular momentum and spin angular momentum. And basically having to do with the fact that the eigenvalues of these operators are plus minus h-bar over 2. And in the orbital case they tend to be plus minus integer values of h-bar. So this is a very deep statement about the algebra of these operators that still allows the physics of them to be quite different. But this is probably the only algebra that makes sense. It's angular momentum. So we're going to try to develop that algebra like that, as well here. You could take it to be an assumption. And as I said, an experiment doesn't tell you the unique way to invent the mathematics. You try to invent the consistent mathematics and see if it coincides with the experiment. And this is a very natural thing to try to invent So what are we facing? We're facing a slightly nontrivial problem of figuring out these operators. And they should be Hermitian. So let's try to think of Hermitian two-by-two matrices. So here is a Hermitian two-by-two matrix. I can put an arbitrary constant here because it should be invariant on their transposition, which doesn't change this diagonal value in complex conjugation. So c should be real. d should be real. For the matrix to be Hermitian, two-by-two matrix, I could put an a here. And then this a would have to appear here as well. I can put minus ib, and then I would have plus ib here. So when I transpose a complex conjugate, I get this one. So this matrix with abc and d real is Hermitian. Hermiticity is some sort of reality condition. Now, for convenience, I would put a 2c and a 2d here. It doesn't change things too much. Now to look at what we're talking about. We're talking about this set of Hermitian matrices. Funnily, you can think of that again as a vector space. Why a vector space? Well, we'll think about it, and in a few seconds, it will become clear. But let me just try to do something here that might help us. We're trying to identify Sx and Sy from here so that this commutation relations hold. Well, if Sx and Sy have anything to do with the identity matrix, they would commute with everything and would do nothing for you. So, I will remove from this matrices then trying to understand something having to do with the identity. So I'll remove a Hermitian matrix, which is c plus d times the identity-- the two-by-two identity matrix. This is a Hermitian matrix, as well. And I can remove it, and then this matrix is still Hermitian, and this piece that I've removed doesn't change commutators as they appear on the left hand side. So if you have an Sx and an xy here, and you're trying to do a computation, it would not contribute, so you might as well just get rid of them. So if we remove this, we are left with-- you're subtracting c plus d from the diagonal. So here you'll have c minus d. Here you'll get b minus c, a minus ib, and a plus ib. And we should keep searching for Sx and Sy among these matrices. But then you say, look, I already got Sz, and that was Hermitian. And Sz was Hermitian, and it had a number, and the opposite number on the other diagonal entry. If Sx and Sy have a little bit of Sz, I don't care. I don't want these to be independent matrices. I don't want to confuse the situation. So if this thing has something along Sz, I want it out. So since precisely this number is opposite to this one, I can add to this matrix some multiple of Sz and kill these things in the diagonal. So add the multiple and Sz multiple, and we finally get this matrix. 0, a minus ib, a plus ib, and 0. So we've made quite some progress. Let's see now what we have. Well, that matrix could be written as a times 0, 1, 1, 0 plus b times 0, minus i, i, 0. Which is to say that it's this Hermitian matrix times a real number, and this Hermitian matrix times a real number. And that makes sense because if you take a Hermitian matrix and multiply by a real number, the matrix is still Hermitian. So this is still Hermitian because these are real. This is still Hermitian because a is real, and if you add Hermitian matrices, it's still Hermitian. So in some sense, the set of Hermitian matrices, two-by-two Hermitian matrices, is a real vector space with four basis vectors. One basis vector is this, another basis vector is this, the third basis vector is the Sz part, and the fourth basis vector is the identity that we subtracted. And I'm listing the other two that we got rid of because physically we're not that interested given that we want Sx and Sz. So, Sx and Sy. But here it is. These four two-by-two matrices are sort of the linearly independent Hermitian matrices. You can think of them as vectors, four basis vectors. You multiply by real numbers, and now you add them, and you got the most general Hermitian matrix. So this is part of the subtlety of this whole idea of vector spaces of matrices, which can be thought of as vectors sometimes, as well. So that's why these matrices are quite famous. But before we just discuss why they are so famous, let's think of this. Where we're looking for Sx and Sy, and we actually seem to have two matrices here that could do the job, as two independent Hermitian two-by-two matrices. But we must add a little extra information. We don't know what the scale is. Should I multiply this by 5 and call that Sx? Or this by 3? We're missing a little more physics. What is the physics? The eigenvalues of Sx should also be plus minus h over 2. And the eigenvalues of Sy should also be plus minus h over 2. Just like for Sz. you could have started the whole Stern-Gerlach things thinking of x, and you would have obtained plus minus h over 2. So that is the physical constraint. I have to figure out those numbers. Maybe Sx is this one, as y is this one. And you can say, oh, you never told us if you're going to get the unique answer here. And yes, I did tell you, and you're not going to get a unique answer. There are some sign notations and some other things, but any answer is perfectly good. So once you get an answer, it's perfectly good. Of course, we're going to get the answer that everybody likes. And the convention is that happily that everybody uses this same convention. Questions. AUDIENCE: So I have a related question, because at the beginning we could have chosen the top right entry to be a plus ib and the bottom left to be a minus ib and that would have yielded a different basis matrix. PROFESSOR: Right, I would have called this plus and minus. Yes. AUDIENCE: Are we going to show that this is the correct form? PROFESSOR: No, it's not the correct form. It is a correct form, and it's equivalent to any other form you could find. That's what we can show. In fact, I will show that there's an obvious ambiguity here. Well, in fact, maybe I can tell it do you, I think. If you let Sx go to minus Sy, and Sy goes to plus Sx, nothing changes in these equations. They become the same equations. You know, Sx would become minus Sy, and this Sx-- this is not changed. But, in fact, if you put minus Sy and Sx as the same commutator then this one will become actually this commutator, and this one will become that. So I could change whatever I get for Sx, change it from minus Sy, for example, and get the same thing. So there are many changes you can do. The only thing we need is one answer that works. And I'm going to write, of course, the one that everybody likes. But don't worry about that. So let's think of eigenvectors and eigenvalues now. I don't know how much you remember that, but we'll just take it at this moment that you do. So 0, 1, 1, 0 has two eigenvalues, and lambda equals 1, with eigenvector 1 over square root of 2, 1, 1. And a lambda equals minus 1 with eigenvector 1 over square root of 2, 1, minus 1. The other one, it's equally easy to do. We'll discuss eigenvectors and eigenvalues later. Minus i, i, 0, 0, plus a lambda equals one eigenvector, with components 1 over square root of 2, 1, and i. I'm pretty sure it's 1 and i. Yes, and a lambda equals minus 1, with components 1 over square root 2, 1, minus i. Now I put the 1 over square root of 2 because I wanted them to be normalized. Remember how you're supposed to normalize these things. You're supposed to take the row vector, complex conjugate, and multiply. Well, you would get 1 for the length of this, 1 for the length of this. You would get one for the length of this, but remember, you have to complex conjugate, otherwise you'll get 0. Also, you will get one for the length of this. So these are our eigenvalues. So actually, with eigenvalues lambda equals 1 and minus 1 for these two, we're in pretty good shape. We could try Sx to be h-bar over 2 times 0, 1, 1, 0. And Sy to be h-bar over 2, 0, minus i, i, 0. These would have the right eigenvalues because if you multiply a matrix by a number, the eigenvalue gets multiplied by this number, so the plus minus 1s become plus minus h-bar over 2. But what are we was supposed to check? If this is to work, we're supposed to check these commutators. So let's do one, at least. Sx commutator with Sy. So what do we get? h-bar over 2, h-bar over 2-- two of them-- then the first matrix, 0, 1, 1, 0 times 0, minus i, i, 0, minus 0, minus i, i, 0 times 0, 1, 1, 0. Which is h-bar over 2 times h-bar over 2, times i, 0, 0, minus i, minus minus i, 0, 0, i. And we're almost there. What do we have? Well, we have h-bar over 2, h-bar over 2. And we've got 2i and minus 2i. So this is h-bar over 2 times h-bar over 2, times 2i times 1, minus 1. And this whole thing is i h-bar, and the other part is h-bar over 2, 1, minus 1, which is i h-bar as z-hat. Good, it works. You know, the only thing that could have gone wrong-- you could have identified 1 with a minus, or something like that It would have been equally good. Once you have these operators, we're fine. So one has to check that the other ones work, and they do. I will leave them for you to check. And therefore, we've got the three matrices. It's a very important result-- the Sx Sy, and Sz. I will not rewrite them, but they should be boxed nicely, the three of them together, with that one there on top of the blackboard. And of course by construction, they're Hermitian. They're famous enough that people have defined the following object. Si is defined to be h-bar over 2, sigma i, the power of the matrix sigmas. And these are poly matrices, sigma one is 0, 1, 1, 0. Sigma two is 0, minus i, i, 0. And sigma three is equal to 1, minus 1, 0, 0. OK, so in principle-- yes, question. AUDIENCE: Is it at all significant that the poly matrices are all squared [INAUDIBLE]? PROFESSOR: Yes, it is significant. We'll use it, but at this moment, it's not urgent for us. We'll have no application of that property for a little while, but it will help us do a lot of the algebra of the poly matrices. AUDIENCE: [INAUDIBLE] eigenvalues, right? PROFESSOR: Sorry? AUDIENCE: Doesn't that follow from the eigenvalue properties that we've [INAUDIBLE] plus or minus one. Because those were both squared. [INAUDIBLE]. PROFESSOR: That's right. I think so. Our eigenvalues-- yes, it's true. That the fact that the eigenvalues are plus minus 1 will imply that these matrices squared themselves. So it's incorporated into our analysis. The thing that I will say is that you don't need it in the expression of the commutators. So in the commentators, it didn't play a role to begin with. Put it as an extra condition. Now what is the next thing we really want to understand? Is that in terms of plain states, we now have the answer for most of the experiments we could do. So in particular, remember that we said that we would have Sx, for example, having states x plus minus, which are h-bar over 2 plus minus, x comma plus minus. The states along the x-direction referred like that would be the eigenstates of the Sx operator. But we've calculated the states of the Sx operator-- they're here. The Sx operator is h-bar over 2 times this matrix. And we have those things. So the plus eigenvalue and the minus eigenvalue will just show up here. So let me write them, and explain, in plain language, what these states are. So the eigenstate with lambda equal 1-- that would correspond to h-bar over two-- so the x plus corresponds to this vector. So what is that state? It's that vector which, if you want more explicitly, it's the z plus, plus z minus. This is the state 1 over square root of 2, 1, 1. The x minus is z plus, minus z minus. As you see on that blackboard, it's 1 minus 1. So here it is. The states that you were looking for, that are aligned along x-- plus x or minus x-- are not new states that have you to add to the state space. They are linear combinations of the states you've got. We can invert this formula and write, for example, that z plus is 1 over square root of 2, x plus, plus 1 over square root of 2, x minus. And z minus is 1 over square root of 2, x plus, minus 1 over square root of-- minus the square root of 2 is already out, I'm sorry-- minus x minus. So actually, this answers the question that you had. For example, you put a z plus state, and you put an x filter-- what amplitude do you have to find a state in the x plus, given that you start with a state on the z plus? Well, you put an x plus from here. You get 1 from this and 0 from this one because the states are always orthogonal. The states are orthogonal-- you should check that. And therefore, this is 1 over square root of 2. If you ask for x minus with respect to z plus, that's also 1 over square root of 2. And these are the amplitudes for this state to be found in this, for this state to be found in them. They're equal. The probabilities are 1/2. And that's good. Our whole theory of angular momentum has given us something that is perfectly consistent with the Stern-Gerlach experiment, and it gives you these probabilities. You can construct in the same way the y states. So the y states are the eigenstates of that second matrix, Sy, that we wrote on the left. So this matrix is Sy, so its eigenstates-- I'm sorry, Sy is there. Sy is there. The eigenstates are those, so immediately you translate that to say that Sy has eigenstates y plus minus, whose eigenvalues are plus minus h-bar over 2y plus minus. And y plus is equal 1 over square root of 2, z plus-- and look at the first eigenvector-- plus iz minus. And, in fact, they can put one formula for both. Here they are. So, it's kind of neat that the x1s were found by linear combinations, and they're orthogonal. Now, if you didn't have complex numbers, you could not form another linear combination of this orthogonal. But thanks to these complex numbers, you can put an i there-- there's no i in the x ones-- and the states are orthogonal, something that you should check. So again, you can invert and find the z states in terms of y, and you would conclude that the amplitudes are really the same up to signs, or maybe complex numbers, but the probabilities are identical. So we've gotten a long way. We basically have a theory that seems to describe the whole result of the Stern-Gerlach experiment, but now your theory can do more for you. Now, in the last few minutes, we're going to calculate the states that are along arbitrary directions. So here I produced a state that is along the x-direction plus, and along the x-direction minus. What I would like to construct, to finish this story, is a state that is along some arbitrary direction. So the state that points along some unit vector n. So here is space, and here's a unit vector n with components nx, , ny, and nz. Or you can write the vector n as nx ex plus ny ey plus nz ez. And I would like to understand how I can construct, in general, a spin state that could be said to be in the n direction. We have the ones along the z, x, and y, but let's try to get something more general, the most general one. So for this, we think of the triplet of operators S, which would be Sx, Sy, and Sz. Now you can, if you wish, write this as Sx-hat ex vector, plus Sy-hat ey vector, plus Sz hat ez vector. But this object, if you write it like that, is really a strange object. Think of it. It's matrices, or operators, multiplied by unit vectors. These vectors have nothing to do with the space in which the matrices act. The matrices act in an abstract, two-dimensional vector space, while these vectors are sort of for accounting purposes. That's why we sometimes don't write them, and say we have a triplet. So this product means almost nothing. They're just sitting together. You could put the e to the left of the x or to the right. It's a vector. You're not supposed to put the vector inside the matrix, either. They don't talk to each. It's an accounting procedure. It is useful sometimes; we will use it to derive identities soon, but it's an accounting procedure. So here's what I want to define. So this is a crazy thing, some sort of vector valued operator, or something like that. But what we really need is what we'll call S-hat n, which will be defined as n dot S. Where we take naively what a dot product is supposed to mean. This component times this component, which happens to be an operator. This times this, this times that. nx Sx plus ny Sy, plus nz Sz. And this thing is something very intuitive. It is just an operator. It doesn't have anymore a vector with it. So it's a single operator. If your vector points in the z-direction, nx and ny z, and you have Sz because it's a unit vector. If the vector points in the x-direction, you get Sx. If the vector points in the y-direction, you get Sy. In general, this we call the spin operator in the direction of the vector n-- spin operator in the direction of n. OK, so what about that spin operator? Well, it had eigenvalues plus minus h-bar over 2 along z, x, and y-- probably does still have those eigenvalues-- but we have to make this a little clearer. So for that we'll take nx and ny and nz to be the polar coordinate things. So this vector is going to have a theta here on the azimuthal angle phi over here. So nz is cosine theta. nx and ny have sine theta. And nx cosine phi, and this one has sine phi. So what is the operator Sn vector hat? Well, it's nx times Sx. So, I'll put a h-bar over 2 in front, so we'll have nx sigma x, or sigma1, plus ny sigma2, , plus nz sigma3. Remember the spin operators are proportional h-bar over 2 times the sigmas-- so sigma1, sigma2, sigma3. And look what we get. h-bar over 2. Sigma1 has an nx here, nx. Sigma2 has minus iny plus iny. And sigma3, we have a nz minus nz. So this is h-bar over 2, nz is cosine theta, nx minus iny-- you'd say, oh it's a pretty awful thing, but it's very simple-- nx minus iny is sine theta times e to the minus i phi. Here it would be sine theta, e to the i phi, and here we'll have minus cosine theta. So this is the whole matrix, Sn-hat, like that. Well, in the last couple of minutes, let's calculate the eigenvectors and eigenvalues. So what do we get? Well, for the eigenvalues, remember what is the computation of an eigenvalue of a matrix. An eigenvalue for matrix a, you write that by solving the determinant of a minus lambda 1 equals 0. So for any matrix a, if we want to find the eigenvalues of this matrix, we would have to write eigenvalues of Sn-hat. We have to ride the determinant of this, minus lambda i, so the determinant of h-bar over 2 cosine theta, minus lambda, minus h-bar over 2 cosine theta, minus lambda. And here, it's sine theta, e to the minus i phi, sine theta e to the i phi, the determinant of this being 0. It's not as bad as it looks. It's actually pretty simple. These are a plus b, a minus b. Here the phases cancel out. The algebra you can read in the notes, but you do get lambda equals plus minus h-bar over 2. Now that is fine, and we now want the eigenvectors. Those are more non-trivial, so they need a little more work. So what are you supposed to do to find an eigenvector? You're supposed to take this a minus lambda i, acting on a vector, and put it equal to zero. And that's the eigenvector. So, for this case, we're going to try to find the eigenvector n plus. So this is the one that has Sn on this state-- well, I'll write it here, plus minus h over 2, n plus minus here. So let's try to find this one that corresponds to the eigenvalue equal to plus h-bar over 2. Now this state is C1 times z plus, plus C2 times z minus. These are our basis states, so it's a little combination. Or it's C1, C2. Think of it as a matrix. So we want the eigenvalues of that-- the eigenvector for that-- so what do we have? Well, we would have Sn-hat minus h-bar over 2 times 1, on this C1, C2 equals 0. The eigenvector equation is that this operator minus the eigenvalue must give you that. So the h-bars over 2, happily, go out, and you don't really need to worry about them anymore. And you get here cosine theta minus 1, sine theta e to the minus i phi, sine theta e to the i phi, and minus cosine theta minus 1, C1, C2 equals 0. All right, so you have two equations, and both relate C1 and C2. Happily, and the reason this works is because with this eigenvalue that we've used that appears here, these two equations are the same. So you can take either one, and they must imply the same relation between C1 and C2. Something you can check. So let me write one of them. C2 is equal to e to the i phi, 1 minus cosine theta over sine theta C1. It's from the first line. So you have to remember, in order to simplify these things, your half angle identities. Sorry. 1 minus cosine theta is 2 sine squared theta over 2, and sine theta is 2 sine theta over 2 cosine theta over 2. So this becomes e to the i phi sine theta over 2, over cosine theta over 2, C1. Now we want these things to be well normalized, so we want C1 squared plus C2 squared equal to 1. So, you know what C2 is, so this gives you C1 squared times 1 plus-- and C2 you use this, when you square the phase goes away-- sine squared theta over 2, cosine squared over 2 must be equal to 1. Well, the numerator is 1, so you learn that C1 squared is equal to cosine squared theta over 2. Now you have to take the square root, and you could put an i or a phase or something. But look, whatever phase you choose, you could choose C1 to be cosine theta over 2, and say, I'm done. I want this one. Somebody would say, no let's put the phase, e so to the i pi over 5. So that doesn't look good, but four or even worse, this phase will show up in C2 because C2 is proportional to C1. So I can get rid of it. I only should put it if I really need it, and I don't think I need it, so I won't put it. And you can always change your mind later-- nobody's going to take your word for this. So, in this case, C2 would be sine theta over 2, e to the i phi. It's nice, but it's [INAUDIBLE]. And therefore, we got this state n plus, which is supposed to be cosine theta over 2, z plus, and plus sine theta over 2, e to the i phi, z minus. This is a great result. It gives the arbitrarily located spin state that point in the n-direction. As a linear superposition of your two basis states, it answers conclusively the question that any spin state in your system can be represented in this two-dimensional vector space. Now moreover, if I take that theta equals 0, I have the z-axis, and it's independent of the angle phi. The phi angle becomes singular at the North Pole, but that's all right. When theta is equal to 0, this term is 0 anyway. And therefore, this goes, and when theta is equal to 0, you recover the plus state. Now you can calculate the minus state. And if you follow exactly the same economical procedure, you will get the following answer. And I think, unless you've done a lot of eigenvalue calculations, this is a calculation you should just redo. So the thing that you get, without thinking much, is that n minus is equal to sine theta over 2 plus, minus cosine theta over 2, e to the i phi minus. At least some way of solving this equation gives you that. You could say, this is natural, and this is fine. But that is not so nice, actually. Take theta equal to pi-- no, I'm sorry. Again, you take theta equal to 0. Theta equal to 0-- this is supposed to be the minus state along the direction. So this is supposed to give you the minus state. Because the vector n is along up, in the z-direction, but you're looking at the minus component. So theta equals 0. Sure, there's no plus, but theta equals 0, and you get the minus state. And this is 1, and phi is ill-defined-- it's not so nice, therefore-- so, at this moment, it's convenient to multiply this state by e to the minus i phi, times minus 1. Just multiply it by that, so that n minus is equal to minus sine theta over 2, e to the minus i phi, plus, plus cosine theta over 2, minus. And that's a nice definition of the state. When theta is equal to 0, you're fine, and it's more naturally equivalent to what you know. Theta equal to 0 gives you the minus state, or z minus. I didn't put the zs here, for laziness. And for theta equal to 0, the way the phase phi doesn't matter. So it's a little nicer. You could work with this one, but you might this well leave it like that. So we have our general states, we've done everything here that required some linear algebra without doing a review of linear algebra, but that's what we'll start to do next time.
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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: OK, why don't we you get started again? I can understand why there is an air of excitement in the room since tomorrow's a holiday. But we still got two hours before the day is over. Any questions on what we have done up to this point? No, that's good. Let me erase some of this art work then and ask why one would indulge in this rather bizarre exercise of taking the elements of something that represents a physical property and constructing a surface out of them. Well, the name suggests that maybe these surfaces, be they hyperboloids of one or two sheets or imaginary ellipsoids, have something to say about the property that the tensor that formed the coefficient of these functions is doing. So let me take the case of an ellipsoid. That would be the case where all the coefficients are positive. And let's let this be x1 and this x2. And let me ask a first question. If this is to be a well deserved function, does the surface transform in exactly the same way as the tensor elements? In other words, if we take the tensor in one coordinate system and then change the coordinate system to a different coordinate system, are the coefficients in front of x1, x2, and x3 still the elements of the tensor in that new coordinate system? So let me show you that this is, in fact, the case. Suppose our original relation, sticking to conductivity, is that x sigma ij xi xj equals 1. And then for some reason or another, we change axes. So the new equation will be some different coefficients. Because as we change axes, they're going to wink on and off, still the same surface, but now referred to different axes xi prime and xj prime still equal to 1. And let me now use the reverse transformation to put xi prime in the terms of xi. If we do that, we can say that xi prime is cli x sub l prime. That's the reverse transformation. So notice the inverted order of the direction cosine subscripts. And then xj prime is going to be equal to cmj times x sub ep prime. To find this, let's just rearrange these terms in a trivial fashion. Then I will have cx sigma ij cil cmj times xl prime xm prime equals 1. And what is this? This is a summation of direction cosines where the first subscript goes with the subscript on sigma prime. What did I do wrong? This is mj. Why is that not coming first? AUDIENCE: [INAUDIBLE] PROFESSOR: Well, what I would like to get this in the form of is a summation of sigma ij prime over two dummy indices l and m. What did I do wrong? Ah, yes, right, right, right, right, this is li. Thank you somebody said it. And my nose was right in it. And I didn't notice it. OK, this is a summation of all the original elements sigma ij prime times dummy indices l and m. And this actually is sigma ijn. So it's the same set of coefficients. And that is a technicality, which probably you wouldn't want to worry about. So anyway, the elements of the tensor transform in the same way that the equations for the surfaces transform. So if you change coordinate system, the coefficients that you should use should be the elements that are in the transform tensor. OK, now I'm going to ask a question. Suppose I define some direction by a set of direction cosines li and I ask the value of the radius of the surface in that direction. So this is some radius vector. The radius vector will have components, R1, R2, R3 or Ri. And each of those components will be equal to the magnitude of R times the direction cosines li. Yes, sir? AUDIENCE: Could you briefly rego over what you just stated before you erased it [INAUDIBLE]? PROFESSOR: Oh, OK. I'm sorry. I said aij xi xj equals 1 is the original equation for the quadric. If we change the coordinate system, we're going to have some new elements aij prime times xi prime xj prime. Because we've got a new coordinate system. And therefore, the coefficients in front of the equation for the quadric have to change. And now what I did was to express xi prime and xj prime in terms of xi and xj using the reverse transformation. And xi prime in terms of the original x's will be c sub l sub i x sub l prime. Usually, we write it xl equals cil x sub l. This is the reverse transformation where the order of the subscripts is reversed. So this will give me x of i prime. The expression for x sub j prime will be given by cmj x sub j where m is a dummy index. And that's equal to 1. And if I just rearrange this, then I will have cli cmj times aij prime equals 1 times xl xj. And this is going to be alx lxm. And this will be alm times xl xm equals 1, which is the equation for the quadric in the original coordinate system. I think that was better when I was standing in front of it so you couldn't see it. It's just that the surface transforms formally to the surface that we would create if we used the new tensor elements for the same change of coordinates system, which you would probably will be willing to take my word for. AUDIENCE: [INAUDIBLE] PROFESSOR: This is a direction cosine for the-- AUDIENCE: [INAUDIBLE] right below. PROFESSOR: Right below? AUDIENCE: [INAUDIBLE] PROFESSOR: cli cmj aij prime xl xm, and then I collected the cli cmj times alm. And that is the transfer. Yes? AUDIENCE: In the previous [INAUDIBLE], you defined the derivitave side as being xi prime equals cil xl. [INAUDIBLE]? PROFESSOR: You can say that x prime is cil times x sub l. And if we want to use the same direction cosine scheme but do it in reverse, we would say that x sub l is equal to c-- yeah, OK, x sub i is cli times x sub i, right, no, times x sub l-- is that right-- x sub l prime. AUDIENCE: On the second line to the third line, you say that xi prime equals cli [INAUDIBLE]. It's either xi equals xcli xl prime or xi prime equals cil. PROFESSOR: OK, you want to say this. xi prime and cil becomes xl, yes, yeah. And then mj times x-- that's x sub m. OK, and then this says that cil cjm times aij prime should be alm, right, OK, OK. So it's OK then. OK, that was supposed to be a small point that everybody would accept. But now it's done correctly. OK, let's get back to this, which is more hair raising and exciting. What is the radius of the quadric in a given direction that we specify by three direction cosines l1, l2, l3? So there's some radius from the center of the quadric out to the surface in that particular direction specified by three direction cosines. And we know what those three components are, sub i, are going to be in terms of the direction cosines, magnitude of R times li. And those points at the terminus of the radius vector go to one point on the surface of the quadric. So these values of R are coordinates that satisfy the surface of the quadric. So we can say that just substitute Ri for the different values xi in the equation for the quadric. And this says that sigma ij magnitude of R times li, that would correspond to x sub i times magnitude of R times l sub j. That's the magnitude of xj. That should be equal to 1. And if I rearranged this slightly, this says that the magnitude of R squared is going to be equal to 1 over sigma ij times li lj. So the radius of the quadric gives me not the value of the property in that direction. This is the value of the property that will occur in the direction specified by the direction cosines lj. The radius of the quadric is going to be equal to 1 over the square root of the value of the property in that direction. So this is why it's called the representation quadric. If you construct the surface from the tensor elements, you will have defined a quadratic form, which has the property that, as you go in different directions look at the distance out to the surface of the quadric in that direction, that that distance squared is going to be equal to 1 over the property in that direction or, alternatively, the radius is going to be 1 over the square root of the value of the property. There is an enormous implication here. This says that the value of the property, if the quadratic form is an ellipsoid, the value of the property as we go around in different directions in the crystal is going to be a smooth, uniformly varying function. They're going to be no lobes sticking out. They're going to be no dimples, no lumps. It's going to be an almost monotonous property, not very interesting. It's going to change in a uniform way. And in fact, we could put the two surfaces side by side. If this is the value of the quadric as a function of direction, if we make a polar plot of the property as a function of direction, it's going to look sort of like the reciprocal of this. The minimum value of the property will be in the direction of the maximum value of the radius of the quadric. And the maximum value of the property is going to go in the direction of the minimum value. So the value of the property as a function of direction is going to be a quasi-ellipsoidal sort of variation but not really an ellipsoid. It's going to go as this inverse square of the radii in ellipsoid. But the thing is it's going to be something that varies uniformly between extreme values of the maximum and minimum value of the property. We are, I assure you, for higher ranked tensor property going to look at some absolutely wild surfaces with surfaces that do have lobes and extreme values and very, very irregular variation of properties with directions but not for second ranked tensor properties. There will be a few variations on this theme, which are also interesting to touch upon. In principle, we can get other quadratic forms from the tensor if some of the elements of the tensor are negative. And in particular, what would we see-- and I'll draw this relative to the principal axes of the surface-- what would we see for an hyperboloid of one sheet? This is the sort of surface that might result if one principal value of the tensor had a negative value. Well, this is the quadric. And this is a radius in one particular direction. What would this say about the property? Well, let me, to make it clear, look at one of these sections through the quadric that our hyperbola. In directions like this, we have a radius that is a minimum out to the surface. In other directions, the radius gets progressively larger. And then if we plot the reciprocal of the square root of that, that says that, in this direction, we get the maximum value of the property. And as the direction approaches the asymptote, the reciprocal will go down to 0. But how did I get two y-axes here? OK, so the maximum will be in the direction of the minimum radius. And then it will go down to 0 for the asymptote. And that's going to be symmetrical on either side of this principle axis. So what then are the radii in directions outside of the asymptotes of this hyperboloid? Within this range, the radius is imaginary. But if you square it, you get a negative number. So within these two lobes, the value of the property is positive. As you go away from the asymptotes, you'll get another lobe like this and another lobe like this where the value of the property is negative. How in the world could you get anything that looked like that? Well, a good example of a property that has this behavior is thermal expansion. And let me give you two examples. The structure of selenium and tellurium is a hexagonal structure. And there are chain-like molecules that are pairs of bonds that rise up around a threefold screw axis. So this is two coordinated atoms in the structure just spirals up in this triangular spiral around the threefold screw axis. So this is a material in which the bonding is very, very anisotropic. The bonding within these covalently bonded spirals, which are like springs that you might put on your screen door in the summertime, the bonding is very strong. Between these individual molecular chains, the bonding is very weak. So what happens when you heat this stuff up? It expands like the dickens in the direction of these weak bonds. But the spirals are, in part, held in this extended form by repulsive interactions in like chains. So when these spirals move apart as a result of large thermal expansion in the plane normal to the chain, the chains relax a little bit. So you have a large positive thermal expansion here. But in one direction along the normal to the hexagonal in the structure, the thermal expansion is negative. And that gives you a variation of property with direction that looks exactly like this. Negative value of the property, that means the structure, contracts in that direction, positive values this way. The structure expands. So there's a good example of this. There's another example of a material, which, again, has a negative thermal expansion in one direction. And this is calcium carbonate, which is a complicated hexagonal structure. But it looks very, very much like the structure of rock salt in a distorted form. These are the calcium atoms. And calcite is CaCL3 And the calciums are in a face-centered cubic arrangement just like in rock salt. The carbonate groups are very tightly bonded little triangles with the carbon in the middle. And these things are arranged on the edges of the cell. And this is going to be very schematic. These triangles are normal to the body diagonal of the cells. So you have one family of triangles that are all parallel to one another. On the next edge, the triangles are anti-parallel to this first orientation. So think of rock salt. Tip it up on its body diagonal. Every place you have a chlorine anion, place a triangle in an orientation that is perpendicular to the body's diagonal of the rock salt structure. Again, these tightly bonded little triangles don't do much as you increase temperature. But the bonding between the calcium and the triangles is rather weak. So again, you find that the structure expands in one direction so strongly that it contracts to the other direction so calcium carbonate. The calcite form also has the distinction of having a negative thermal expansion coefficient. We'll talk quite a bit about expansion coefficients as one example of a tensor property. Interestingly, and we'll say more about this and I'll give you some references when we come to this point, can you get a material that has a negative thermal expansion coefficient in all directions? No, that seems as though it would violate in a flagrant way some vast law of thermodynamics that things have to increase their volume when you increase temperature. There were, however, a few odd ball materials that over a very, very limited temperature range would contract but only by a tiny amount in all directions. And then one of the very interesting discoveries of recent years done primarily by a crystal chemist named Art Slate, who's out at the University of Oregon, he discovered a family of materials that have large negative expansion coefficients in all directions over a considerable range of temperatures, like 100 degrees. So these are materials, when you heat them up, they contract amazing as that seems. And there's a structural reason for it. These are tetrahedral frameworks in which one corner of a tetrahedron is dangling. And as you heat it up, this tetrahedral corner can get closer to other tetrahedra, which takes energy to do. But in so doing, you actually are changing the net volume occupied by the solid so very, very anomalous set of compounds. More of these have been found. There are probably a dozen materials now that have negative thermal expansion coefficients in all directions. So is the imaginary ellipsoid a viable representation quadric for real materials? Yes, in the case of thermal expansion coefficients. OK, so the representation quadric then is a dandy device for seeing with one function how a property will vary with direction. For an ellipsodial quadric, the variation of the property itself with direction is not really ellipsodial but quasi-ellipsodial in that there's a small principal axis, a large principal axis. And the large value of the property goes with the short axis of the quadric. So we have a nice device for representing the value of the property as a function of direction. It looks as though we've lost all information about the direction of the resulting vector, the generalized displacement. It looks as though that's not in here at all. Well, there was an ad years ago for a bottle spaghetti sauce, which offends many cooks who are very fiercely proud of their own spaghetti sauce. So here's a wife who's using the canned stuff. And her husband comes home and looks at it very, very skeptically and says, where's the basil? And the housewife says, it's in there. It's in there. And he sniffs again and says, where is the basil? And she says finally, it's in there. It's in there. Well, this is a similar situation. Where is the direction of the generalized force? And I say, it's in there. It's in there. Not obvious, but it's in there. So let me now show you where it is lurking. I think we have time to carry this through. Let me describe how it's embodied in the quadric and then prove to you that this is indeed the case. And I'll use a general quadric in the form of an ellipsoid. That looks more like an egg than an ellipsoid. It has a pointed end. OK, this is something that's called the radius normal property. And what it says, in words, is that, if you pick a particular direction relative to the quadric, we know that its length is going to be inversely proportional to the square root of the value of the property. If you want to know what the direction of the resulting vector is, for example, in our relation for conductivity now getting very warm, it says the current flow is given by a linear combination of every component of the electric field. The radius normal properties is that, if you want to know where J is for this particular direction, look at the point where the radius vector intersects the surface of the quadric and, at that point, construct a perpendicular to the surface, which, in general, will not be parallel to R. And this will be the direction of J. It won't give you the magnitude. But it'll give you the direction of it. OK, let me now prove to you that the quadric does have this property. In our conductivity relation, the direction of J is going to be given by the tensor relation. Namely that J sub i is equal to sigma ij times E sub j, which can be written as sigma ij times the direction cosines of E sub j times the magnitude of E. I'm going to want to compare this expression, with which you're very familiar term by term, with the normal to the surface that we can compute from its derivative at that point. So this will say that J1 is equal to sigma 1, 1 l1 times the magnitude of E plus sigma 1, 2 times l2 times the magnitude of E plus sigma 1, 3 times l3 times the magnitude of E. And I don't need to write much more. J2 will be sigma 2, 1 times l1 magnitude of E plus sigma 2, 2 l2 times the magnitude of E and so on. What is going to be the normal to the surface as a function of direction? OK, to do this I'm going to say that the normal to the surface is going to be, if I have some function of xyz, the normal to that function is the gradient. Let me say that that's G of xyz. And we can say then that the x1 component of the normal is not going to be equal to but proportional to the gradient of the equation for the quadric with respect to x1, dx2, and dx3. So it's going to be proportional to the differential of the function that gives us the surface with respect to x1 times i plus the differential of the function with respect to x2 times j plus dF dx3 times k. So this is the normal entirely. So if we split this into components, N1 is going to be equal to dF dx1. And if I differentiate the equation for the quadric, that's going to be 2 sigma 1, 1 times x1. Remember the equation for the quadrant is sigma 1, 1 times x1 squared plus sigma 1, 2 times x1 x2. So if I differentiate-- let me write it down. If I take those terms and differentiate with respect to x1, I'm going to get 2 sigma 1, 1 times x1. Here I'll get sigma 1, 2 times x2. But then down in this line where I have a sigma 2, 1 x2 x1, if I differentiate with respect to x1, I'll have another term sigma 2, 1. And if I differentiate with respect to x3, I'll have sigma 1, 3 plus sigma 3, 1 times x3. And a similar thing for the x2 component of the normal, that's going to be proportional to the gradient with respect to x2. And that's going to be equal to sigma 1, 2 plus sigma 2, 1 times x1 plus 2 sigma 2, 2 times x2 plus sigma 2, 3 plus sigma 3, 2 times x3 and similarly for N3. OK, I've made my case if I can demonstrate that each component of J, J sub i, is proportional to each component of the gradient to the function that defines the quadric. If you look at them, they're close but no cigar. The first term is OK. I've got a 2 out in front of sigma 1, 1 for the x1 term. Here, though, I have the sum of two off-diagonal terms. And for this expression, I have just sigma 1, 2. And down here I've just the single term sigma 2, 1. So the conclusion we're forced to draw is that these two expressions would have components of a vector that are parallel to one another if sigma 1, 2 was equal to sigma 2, 1. Because if these were equal, I could write just twice sigma 1, 2. And down here I could write twice sigma 2, 1 and do that all the way through these two expressions. And then this factor of 2 would just be part of the proportionality constant. So the radius normal property will be true or valid only for symmetric tensors. That is it's going to be true only if sigma ij is equal to sigma ji. We mentioned when we first started talking about second ranked tensors that a tensor does not have to be symmetric. And showing that the tensor is symmetric has nothing to do with symmetry. Because it's symmetric in an algebraic sense not in the literal sense of geometrical symmetry. And there are some properties, the thermal electricity tensor is one notable example, where the tensor is a second ranked tensor but it decidedly is not symmetric. So this is OK for most properties but not all. OK, so here are properties of the representation surface that we can construct from the representation quadric. And what I would like to do following this and after the inevitable unpleasantness of a quiz is to look at some specific properties that illustrate second ranked tensor properties. And in particular, I would like to look at some other sorts of second ranked tensor properties that represent generalized forces, namely stress and strain. You've seen these before in other contexts perhaps not actually defined rigorously in terms of tensors. Because, obviously, stress and strain can be regarded as generalized forces. A stress is something that you can apply to a solid. And that will cause various things to happen. In addition to mechanical behavior, different properties and effects can result. So that is going to involve a generalized force, which is a second ranked tensor. And a thing that might happen could be a vector. It could be another second ranked tensor. And this is going to introduce us to the nasty world of higher ranked tensor properties and their representation surfaces. So this is a nice place to quit and pause for a quiz. When we resume, we'll look at stress and strain in terms of our tensor algebra.
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https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2013/8.04-spring-2013.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. ALLAN ADAMS: Hi everyone. Welcome to 804 for spring 2013. This is the fourth, and presumably final time that I will be teaching this class. So I'm pretty excited about it. So my name is Allan Adams. I'll be lecturing the course. I'm an assistant professor in Course 8. I study string theory and its applications to gravity, quantum gravity, and condensed matter physics. Quantum mechanics, this is a course in quantam mechanics. Quantam mechanics Is my daily language. Quantum mechanics is my old friend. I met quantum mechanics 20 years ago. I just realized that last night. It was kind of depressing. So, old friend. It's also my most powerful tool. So I'm pretty psyched about it. Our recitation instructors are Barton Zwiebach, yea! And Matt Evans-- yea! Matt's new to the department, so welcome him. Hi. So he just started his faculty position, which is pretty awesome. And our TA is Paolo Glorioso. Paolo, are you here? Yea! There you go. OK, so he's the person to send all complaints to. So just out of curiosity, how many of you all are Course 8? Awesome. How many of you all are, I don't know, 18? Solid. 6? Excellent. 9? No one? This is the first year we haven't had anyone Course 9. That's a shame. Last year one of the best students was a Course 9 student. So two practical things to know. The first thing is everything that we put out will be on the Stellar website. Lecture notes, homeworks, exams, everything is going to be done through Stellar, including your grades. The second thing is that as you may notice there are rather more lights than usual. I'm wearing a mic. And there are these signs up. We're going to be videotaping this course for the lectures for OCW. And if you're happy with that, cool. If not, just sit on the sides and you won't appear anywhere on video. Sadly, I can't do that. But you're welcome to if you like. But hopefully that should not play a meaningful role in any of the lectures. So the goal of 804 is for you to learn quantum mechanics. And by learn quantum mechanics, I don't mean to learn how to do calculations, although that's an important and critical thing. I mean learn some intuition. I want you to develop some intuition for quantum phenomena. Now, quantam mechanics is not hard. It has a reputation for being a hard topic. It is not a super hard topic. So in particular, everyone in this room, I'm totally positive, can learn quantum mechanics. It does require concerted effort. It's not a trivial topic. And in order to really develop a good intuition, the essential thing is to solve problems. So the way you develop a new intuition is by solving problems and by dealing with new situations, new context, new regimes, which is what we're going to do in 804. It's essential that you work hard on the problem sets. So your job is to devote yourself to the problem sets. My job is to convince you at the end of every lecture that the most interesting thing you could possibly do when you leave is the problem set. So you decide who has the harder job. So the workload is not so bad. So we have problem sets due, they're due in the physics box in the usual places, by lecture, by 11 AM sharp on Tuesdays every week. Late work, no, not so much. But we will drop one problem set to make up for unanticipated events. We'll return the graded problem sets a week later in recitation. Should be easy. I strongly, strongly encourage you to collaborate with other students on your problem sets. You will learn more, they will learn more, it will be more efficient. Work together. However, write your problem sets yourself. That's the best way for you to develop and test your understanding. There will be two midterms, dates to be announced, and one final. I guess we could have multiple, but that would be a little exciting. We're going to use clickers, and clickers will be required. We're not going to take attendance, but they will give a small contribution to your overall grade. And we'll use them most importantly for non-graded but just participation concept questions and the occasional in class quiz to probe your knowledge. This is mostly so that you have a real time measure of your own conceptual understanding of the material. This has been enormously valuable. And something I want to say just right off is that the way I've organized this class is not so much based on the classes I was taught. It's based to the degree possible on empirical lessons about what works in teaching, what actually makes you learn better. And clickers are an excellent example of that. So this is mostly a standard lecture course, but there will be clickers used. So by next week I need you all to have clickers, and I need you to register them on the TSG website. I haven't chosen a specific textbook. And this is discussed on the Stellar web page. There are a set of textbooks, four textbooks that I strongly recommend, and a set of others that are nice references. The reason for this is twofold. First off, there are two languages that are canonically used for quantum mechanics. One is called wave mechanics, and the language, the mathematical language is partial differential equations. The other is a matrix mechanics. They have big names. And the language there is linear algebra. And different books emphasize different aspects and use different languages. And they also try to aim at different problems. Some books are aimed towards people who are interested in materials science, some books that are aimed towards people interested in philosophy. And depending on what you want, get the book that's suited to you. And every week I'll be providing with your problem sets readings from each of the recommended texts. So what I really encourage you to do is find a group of people to work with every week, and make sure that you've got all the books covered between you. This'll give you as much access to the texts as possible without forcing you to buy four books, which I would discourage you from doing. So finally I guess the last thing to say is if this stuff were totally trivial, you wouldn't need to be here. So ask questions. If you're confused about something, lots of other people in the class are also going to be confused. And if I'm not answering your question without you asking, then no one's getting the point, right? So ask questions. Don't hesitate to interrupt. Just raise your hand, and I will do my best to call on you. And this is true for both in lecture, also go to office hours and recitations. Ask questions. I promise, there's no such thing as a terrible question. Someone else will also be confused. So it's a very valuable to me and everyone else. So before I get going on the actual physics content of the class, are there any other practical questions? Yeah. AUDIENCE: You said there was a lateness policy. ALLAN ADAMS: Lateness policy. No late work is accepted whatsoever. So the deal is given that every once in a while, you know, you'll be walking to school and your leg is going to fall off, or a dog's going to jump out and eat your person standing next to you, whatever. Things happen. So we will drop your lowest problem set score without any questions. At the end of the semester, we'll just dropped your lowest score. And if you turn them all in, great, whatever your lowest score was, fine. If you missed one, then gone. On the other hand, if you know next week, I'm going to be attacked by a rabid squirrel, it's going to be horrible, I don't want to have to worry about my problem set. Could we work this out? So if you know ahead of time, come to us. But you need to do that well ahead of time. The night before doesn't count. OK? Yeah. AUDIENCE: Will we be able to watch the videos? ALLAN ADAMS: You know, that's an excellent question. I don't know. I don't think so. I think it's going to happen at the end of the semester. Yeah. OK. So no, you'll be able to watch them later on the OCW website. Other questions. Yeah. AUDIENCE: Are there any other videos that you'd recommend, just like other courses on YouTube? ALLAN ADAMS: Oh. That's an interesting question. I don't off the top of my head, but if you send me an email, I'll pursue it. Because I do know several other lecture series that I like very much, but I don't know if they're available on YouTube or publicly. So send me an email and I'll check. Yeah. AUDIENCE: So how about the reading assignments? ALLAN ADAMS: Reading assignments on the problem set every week will be listed. There will be equivalent reading from every textbook. And if there is something missing, like if no textbook covers something, I'll post a separate reading. Every once in a while, I'll post auxiliary readings, and they'll be available on the Stellar website. So for example, in your problem set, first one was posted, will be available immediately after lecture on the Stellar website. There are three papers that it refers to, or two, and they are posted on the Stellar website and linked from the problem set. Others? OK. So the first lecture. The content of the physics of the first lecture is relatively standalone. It's going to be an introduction to a basic idea then is going to haunt, plague, and charm us through the rest of the semester. The logic of this lecture is based on a very beautiful discussion in the first few chapters of a book by David Albert called Quantum Mechanics and Experience. It's a book for philosophers. But the first few chapters, a really lovely introduction at a non-technical level. And I encourage you to take a look at them, because they're very lovely. But it's to be sure straight up physics. Ready? I love this stuff. today I want to describe to you a particular set of experiments. Now, to my mind, these are the most unsettling experiments ever done. These experiments involve electrons. They have been performed, and the results as I will describe them are true. I'm going to focus on two properties of electrons. I will call them color and hardness. And these are not the technical names. We'll learn the technical names for these properties later on in the semester. But to avoid distracting you by preconceived notions of what these things mean, I'm going to use ambiguous labels, color and hardness. And the empirical fact is that every electron, every electron that's ever been observed is either black or white and no other color. We've never seen a blue electron. There are no green electrons. No one has ever found a fluorescent electron. They're either black, or they are white. It is a binary property. Secondly, their hardness is either hard or soft. They're never squishy. No one's ever found one that dribbles. They are either hard, or they are soft. Binary properties. OK? Now, what I mean by this is that it is possible to build a device which measures the color and the hardness. In particular, it is possible to build a box, which I will call a color box, that measures the color. And the way it works is this. It has three apertures, an in port and two out ports, one which sends out black electrons and one which sends out white electrons. And the utility of this box is that the color can be inferred from the position. If you find the particle, the electron over here, it is a white electron. If you find the electron here, it is a black electron. Cool? Similarly, we can build a hardness box, which again has three apertures, an in port. And hard electrons come out this port, and soft electrons come out this port. Now, if you want, you're free to imagine that these boxes are built by putting a monkey inside. And you send in an electron, and the monkey, you know, with the ears, looks at the electron, and says it's a hard electron, it sends it out one way, or it's a soft electron, it sends it out the other. The workings inside do not matter. And in particular, later in the semester I will describe in considerable detail the workings inside this apparatus. And here's something I want to emphasize to you. It can be built in principle using monkeys, hyper intelligent monkeys that can see electrons. It could also be built using magnets and silver atoms. It could be done with neutrons. It could be done with all sorts of different technologies. And they all give precisely the same results as I'm about to describe. They all give precisely the same results. So it does not matter what's inside. But if you want a little idea, you could imagine putting a monkey inside, a hyper intelligent monkey. I know, it sounds good. So a key property of these hardness boxes and color boxes is that they are repeatable. And here's what I mean by that. If I send in an electron, and I find that it comes out of a color box black, and then I send it in again, then if I send it into another color box, it comes out black again. So in diagrams, if I send in some random electron to a color box, and I discover that it comes out, let's say, the white aperture. And so here's dot dot dot, and I take the ones that come out the white aperture, and I send them into a color box again. Then with 100% confidence, 100% of the time, the electron coming out of the white port incident on the color box will come out the white aperture again. And 0% of the time will it come out the black aperture. So this is a persistent property. You notice that it's white. You measure it again, it's still white. Do a little bit later, it's still white. OK? It's a persistent property. Ditto the hardness. If I send in a bunch of electrons in to a hardness box, here is an important thing. Well, send them into a hardness box, and I take out the ones that come out soft. And I send them again into a hardness box, and they come out soft. They will come out soft with 100% confidence, 100% of the time. Never do they come out the hard aperture. Any questions at this point? So here's a natural question. Might the color and the hardness of an electron be related? And more precisely, might they be correlated? Might knowing the color infer something about the hardness? So for example, so being male and being a bachelor are correlated properties, because if you're male, you don't know if you're a bachelor or not, but if you're a bachelor, you're male. That's the definition of the word. So is it possible that color and hardness are similarly correlated? So, I don't know, there are lots of good examples, like wearing a red shirt and beaming down to the surface and making it back to the Enterprise later after the away team returns. Correlated, right? Negatively, but correlated. So the question is, suppose, e.g., suppose we know that an electron is white. Does that determine the hardness? So we can answer this question by using our boxes. So here's what I'm going to do. I'm going to take some random set of electrons. That's not random. Random. And I'm going to send them in to a color box. And I'm going to take the electrons that come out the white aperture. And here's a useful fact. When I say random, here's operationally what I mean. I take some piece of material, I scrape it, I pull off some electrons, and they're totally randomly chosen from the material. And I send them in. If I send a random pile of electrons into a color box, useful thing to know, they come out about half and half. It's just some random assortment. Some of them are white, some of them come out black. Suppose I send some random collection of electrons into a color box. And I take those which come out the white aperture. And I want to know, does white determine hardness. So I can do that, check, by then sending these white electrons into a hardness box and seeing what comes out. Hard, soft. And what we find is that 50% of those electrons incident on the hardness box come out hard, and 50% come out soft. OK? And ditto if we reverse this. If we take hardness, and take, for example, a soft electron and send it into a color box, we again get 50-50. So if you take a white electron, you send it into a hardness box, you're at even odds, you're at chance as to whether it's going to come out hard or soft. And similarly, if you send a soft electron into a color box, even odds it's going to come out black or white. So knowing the hardness does not give you any information about the color, and knowing the color does not give you any information about the hardness. cool? These are independent facts, independent properties. They're not correlated in this sense, in precisely this operational sense. Cool? Questions? OK. So measuring the color give zero predictive power for the hardness, and measuring the hardness gives zero predictive power for the color. And from that, I will say that these properties are correlated. So H, hardness, and color are in this sense uncorrelated. So using these properties of the color and hardness boxes, I want to run a few more experiment's. I want to probe these properties of color and hardness a little more. And in particular, knowing these results allows us to make predictions, to predict the results for set a very simple experiments. Now, what we're going to do for the next bit is we're going to run some simple experiments. And we're going to make predictions. And then those simple experiments are going to lead us to more complicated experiments. But let's make sure we understand the simple ones first. So for example, let's take this last experiment, color and hardness, and let's add a color box. One more monkey. So color in, and we take those that come out the white aperture. And we send them into a hardness box. Hard, soft. And we take those electrons which come out the soft aperture. And now let's send these again into a color box. So it's easy to see what to predict. Black, white. So you can imagine a monkey inside this, going, aha. You look at it, you inspect, it comes out white. Here you look at it and inspect, it comes out soft. And you send it into the color box, and what do you expect to happen? Well, let's think about the logic here. Anything reaching the hardness box must have been measured to be white. And we just did the experiment that if you send a white electron into a hardness box, 50% of the time it comes out a hard aperture and 50% of the time it comes out the soft aperture. So now we take that 50% of electrons that comes out the soft aperture, which had previously been observed to be white and soft. And then we send them into a color box, and what happens? Well, since colors are repeatable, the natural expectation is that, of course, it comes out white. So our prediction, our natural prediction here is that of those electrons that are incident on this color box, 100% should come out white, and 0% should come out black. That seem like a reasonable-- let's just make sure that we're all agreeing. So let's vote. How many people think this is probably correct? OK, good. How many people think this probably wrong? OK, good. That's reassuring. Except you're all wrong. Right? In fact, what happens is half of these electrons exit white, 50%. And 50% percent exit black. So let's think about what's going on here. This is really kind of troubling. We've said already that knowing the color doesn't predict the hardness. And yet, this electron, which was previously measured to be white, now when subsequently measured sometimes it comes out white, sometimes it comes out black, 50-50% of the time. So that's surprising. What that tells you is you can't think of the electron as a little ball that has black and soft written on it, right? You can't, because apparently that black and soft isn't a persistent thing, although it's persistent in the sense that once it's black, it stays black. So what's going on here? Now, I should emphasize that the same thing happens if I had changed this to taking the black electrons and throwing in a hardness and picking soft and then measuring the color, or if I had used the hard electrons. Any of those combinations, any of these ports would have given the same results, 50-50. Is not persistent in this sense. Apparently the presence of the hardness box tampers with the color somehow. So it's not quite as trivial is that hyper intelligent monkey. Something else is going on here. So this is suspicious. So here's the first natural move. The first natural move is, oh, look, surely there's some additional property of the electron that we just haven't measured yet that determines whether it comes out the second color box black or white. There's got be some property that determines this. And so people have spent a tremendous amount of time and energy looking at these initial electrons and looking with great care to see whether there's any sort of feature of these incident electrons which determines which port they come out of. And the shocker is no one's ever found such a property. No one has ever found a property which determines which port it comes out of. As far as we can tell, it is completely random. Those that flip and those that don't are indistinguishable at beginning. And let me just emphasize, if anyone found such a-- it's not like we're not looking, right? If anyone found such a property, fame, notoriety, subverting quantum mechanics, Nobel Prize. People have looked. And there is none that anyone's been able to find. And as we'll see later on, using Bell's inequality, we can more or less nail that such things don't exist, such a fact doesn't exist. But this tells us something really disturbing. This tells us, and this is the first real shocker, that there is something intrinsically unpredictable, non-deterministic, and random about physical processes that we observe in a laboratory. There's no way to determine a priori whether it will come out black or white from the second box. Probability in this experiment, it's forced upon us by observations. OK, well, there's another way to come at this. You could say, look, you ran this experiment, that's fine. But look, I've met the guy who built these boxes, and look, he's just some guy, right? And he just didn't do a very good job. The boxes are just badly built. So here's the way to defeat that argument. No, we've built these things out of different materials, using different technologies, using electrons, using neutrons, using bucky-balls, C60, seriously, it's been done. We've done this experiment, and this property does not change. It is persistent. And the thing that's most upsetting to me is that not only do we get the same results independent of what objects we use to run the experiment, we cannot change the probability away from 50-50 at all. Within experimental tolerances, we cannot change, no matter how we build the boxes, we cannot change the probability by part in 100. 50-50. And to anyone who grew up with determinism from Newton, this should hurt. This should feel wrong. But it's a property of the real world. And our job is going to be to deal with it. Rather, your job is going to be to deal with it, because I went through this already. So here's a curious consequence-- oh, any questions before I cruise? OK. So here's a curious consequence of this series of experiments. Here's something you can't do. Are you guys old enough for you can't do this on television? This is so sad. OK, so here's something you can't do. We cannot build, it is impossible to build, a reliable color and hardness box. We've built a box that tells you what color it is. We've built a box that tells you what hardness it is. But you cannot build a meaningful box that tells you what color and hardness an electron is. So in particular, what would this magical box be? It would have four ports. And its ports would say, well, one is white and hard, and one is white and soft, one is black and hard, and one is black and soft. So you can imagine how you might try to build a color and hardness box. So for example, here's something you might imagine. Take your incident electrons, and first send them into a color box. And take those white electrons, and send them into a hardness box. And take those electrons, and this is going to be white and hard, and this is going to be white and soft. And similarly, send these black electrons into the hardness box, and here's hard and black, and here's soft and back. Everybody cool with that? So this seems to do the thing I wanted. It measures both the hardness and the color. What's the problem with it? AUDIENCE: [INAUDIBLE] ALLAN ADAMS: Yeah, exactly. So the color is not persistent. So you tell me this is a soft and black electron, right? That's what you told me. Here's the box. But if I put a color box here, that's the experiment we just ran. And what happens? Does this come out black? No, this is a crappy source of black electrons. It's 50/50 black and white. So this box can't be built. And the reason, and I want to emphasize this, the reason we cannot build this box is not because our experiments are crude. And it's not because I can't build things, although that's true. I was banned from a lab one day after joining it, actually. So I really can't build, but other people can. And that's not why. We can't because of something much more fundamental, something deeper, something in principle, which is encoded in this awesome experiment. This can be done. It does not mean anything, as a consequence. It does not mean anything to say this electron is white and hard, because if you tell me it's white and hard, and I measure the white, well, I know if it's hard, it's going to come out 50-50. It does not mean anything. So this is an important idea. This is an idea which is enshrined in physics with a term which comes with capital letters, the Uncertainty Principle. And the Uncertainty Principle says basically that, look, there's some observable, measurable properties of a system which are incompatible with each other in precisely this way, incompatible with each other in the sense not that you can't know, because you can't know whether it's hard and soft simultaneously, deeper. It is not hard and white simultaneously. It cannot be. It does not mean anything to say it is hard and white simultaneously. That is uncertainty. And again, uncertainty is an idea we're going to come back to over and over in the class. But every time you think about it, this should be the first place you start for the next few weeks. Yeah. Questions. No questions? OK. So at this point, it's really tempting to think yeah, OK, this is just about the hardness and the color of electrons. It's just a weird thing about electrons. It's not a weird thing about the rest of the world. The rest of the world's completely reasonable. And no, that's absolutely wrong. Every object in the world has the same properties. If you take bucky-balls, and you send them through the analogous experiment-- and I will show you the data, I think tomorrow, but soon, I will show you the data. When you take bucky-balls and run it through a similar experiment, you get the same effect. Now, bucky-balls are huge, right, 60 carbon atoms. But, OK, OK, at that point, you're saying, dude, come on, huge, 60 carbon atoms. So there is a pendulum, depending on how you define building, in this building, a pendulum which is used, in principle which is used to improve detectors to detect gravitational waves. There's a pendulum with a, I think it's 20 kilo mirror. And that pendulum exhibits the same sort of effects here. We can see these quantum mechanical effects in those mirrors. And this is in breathtakingly awesome experiments done by Nergis Malvalvala, whose name I can never pronounce, but who is totally awesome. She's an amazing physicist. And she can get these kind of quantum effects out of a 20 kilo mirror. So before you say something silly, like, oh, it's just electrons, it's 20 kilo mirrors. And if I could put you on a pendulum that accurate, it would be you. OK? These are properties of everything around you. The miracle is not that electrons behave oddly. The miracle is that when you take 10 to the 27 electrons, they behave like cheese. That's the miracle. This is the underlying correct thing. OK, so this is so far so good. But let's go deeper. Let's push it. And to push it, I want to design for you a slightly more elaborate apparatus, a slightly more elaborate experimental apparatus. And for this, I want you to consider the following device. I'm going to need to introduce a couple of new features for you. Here's a hardness box. And it has an in port. And the hardness box has a hard aperture, and it has a soft aperture. And now, in addition to this hardness box, I'm going to introduce two elements. First, mirrors. And what these mirrors do is they take the incident electrons and, nothing else, they change the direction of motion, change the direction of motion. And here's what I mean by doing nothing else. If I take one of these mirrors, and I take, for example, a color box. And I take the white electrons that come out, and I bounce it off the mirror, and then I send these into a color box, then they come out white 100% of the time. It does not change the observable color. Cool? All it does is change the direction. Similarly, with the hardness box, it doesn't change the hardness. It just changes the direction of motion. And every experiment we've ever done on these, guys, changes in no way whatsoever the color or the hardness by subsequent measurement. Cool? Just changes the direction of motion. And then I'm going to add another mirror. It's actually a slightly fancy set of mirrors. All they do is they join these beams together into a single beam. And again, this doesn't change the color. You send in a white electron, you get out, and you measure the color on the other side, you get a white electron. You send in a black electron from here, and you measure the color, you get a black electron again out. Cool? So here's my apparatus. And I'm going to put this inside a big box. And I want to run some experiments with this apparatus. Everyone cool with the basic design? Any questions before I cruise on? This part's fun. So what I want to do now is I want to run some simple experiments before we get to fancy stuff. And the simple experiments are just going to warm you up. They're going to prepare you to make some predictions and some calculations. And eventually we'd like to lead back to this guy. So the first experiment, I'm going to send in white electrons. Whoops. Im. I'm going to send in white electrons. And I'm going to measure at the end, and in particular at the output, the hardness. So I'm going to send in white electrons. And I'm going to measure the hardness. So this is my apparatus. I'm going to measure the hardness at the output. And what I mean by measure the hardness is I throw these electrons into a hardness box and see what comes out. So this is experiment 1. And let me draw this, let me biggen the diagram. So you send white into-- so the mechanism is a hardness box. Mirror, mirror, mirrors, and now we're measuring the hardness out. And the question I want to ask is how many electrons come out the hard aperture, and how many electrons come out the soft aperture of this final hardness box. So I'd like to know what fraction come out hard, and what fraction come out soft. I send an initial white electron, for example I took a color box and took the white output, send them into the hardness box, mirror, mirror, hard, hard, soft. And what fraction come out hard, and what fraction come out soft. So just think about it for a minute. And when you have a prediction in your head, raise your hand. All right, good. Walk me through your prediction. AUDIENCE: I think it should be 50-50. ALLAN ADAMS: 50-50. How come? AUDIENCE: [INAUDIBLE] color doesn't have any bearing on hardness. [INAUDIBLE] ALLAN ADAMS: Awesome. So let me say that again. So we've done the experiment, you send a white electron into the hardness box, and we know that it's non-predictive, 50-50. So if you take a white electron and you send it into the hardness box, 50% of the time it will come out the hard aperture, and 50% of the time it will come out the soft aperture. Now if you take the one that comes out the hard aperture, then you send it up here or send it up here, we know that these mirrors do nothing to the hardness of the electron except change the direction of motion. We've already done that experiment. So you measure the hardness at the output, what do you get? Hard, because it came out hard, mirror, mirror, hardness, hard. But it only came out hard 50% of the time because we sent in initially white electron. Yeah? What about the other 50%? Well, the other 50% of the time, it comes out the soft aperture and follows what I'll call the soft path to the mirror, mirror, hardness. And with soft, mirror, mirror, hardness, you know it comes out soft. 50% of the time it comes out this way, and then it will come out hard. 50% it follows the soft path, and then it will come out soft. Was this the logic? Good. How many people agree with this? Solid. How many people disagree? No abstention. OK. So here's a prediction. Oh, yep. AUDIENCE: Just a question. Could you justify that prediction without talking about oh, well, half the electrons were initially measured to be hard, and half were initially measured to be soft, by just saying, well, we have a hardness box, and then we joined these electrons together again, so we don't know anything about it. So it's just like sending white electrons into one hardness box instead of two. ALLAN ADAMS: Yeah, that's a really tempting argument, isn't it? So let's see. We're going to see in a few minutes whether that kind of an argument is reliable or not. But so far we've been given two different arguments that lead to the same prediction, 50-50. Yeah? Question. AUDIENCE: Are the electrons interacting between themselves? Like when you get them to where-- ALLAN ADAMS: Yeah. This is a very good question. So here's a question look you're sending a bunch of electrons into this apparatus. But if I take-- look, I took 802. You take two electrons and you put them close to each other, what do they do? Pyewww. Right? They interact with each other through a potential, right? So yeah, we're being a little bold here, throwing a bunch of electrons in and saying, oh, they're independent. So I'm going to do one better. I will send them in one at a time. One electron through the apparatus. And then I will wait for six weeks. [LAUGHTER] See, you guys laugh, you think that's funny. But there's a famous story about a guy who did a similar experiment with photons, French guy. And, I mean, the French, they know what they're doing. So he wanted to do the same experiment with photons. But the problem is if you take a laser and you shined it into your apparatus, there there are like, 10 to the 18 photons in there at any given moment. And the photons, who knows what they're doing with each other, right? So I want to send in one photon, but the problem is, it's very hard to get a single photon, very hard. So what he did, I kid you not, he took an opaque barrier, I don't remember what it was, it was some sort of film on top of glass, I think it was some sort of oil-tar film. Barton, do you remember what he used? So he takes a film, and it has this opaque property, such that the photons that are incident upon it get absorbed. Once in a blue moon a photon manages to make its way through. Literally, like once every couple of days, or a couple of hours, I think. So it's going to take a long time to get any sort of statistics. But he this advantage, that once every couple of hours or whatever a photon makes its way through. That means inside the apparatus, if it takes a pico-second to cross, triumph, right? That's the week I was talking about. So he does this experiment. But as you can tell, you start the experiment, you press go, and then you wait for six months. Side note on this guy, liked boats, really liked yachts. So he had six months to wait before doing a beautiful experiment and having the results. So what did he do? Went on a world tour in his yacht. Comes back, collects the data, and declares victory, because indeed, he saw the effect he wanted. So I was not kidding. We really do wait. So I will take your challenge. And single electron, throw it in, let it go through the apparatus, takes mere moments. Wait for a week, send in another electron. No electrons are interacting with each other. Just a single electron at a time going through this apparatus. Other complaints? AUDIENCE: More stories? ALLAN ADAMS: Sorry? AUDIENCE: More stories? ALLAN ADAMS: Oh, you'll get them. I have a hard time resisting. So here's a prediction, 50-50. We now have two arguments for this. So again, let's vote after the second argument. 50-50, how many people? You sure? Positive? How many people don't think so? Very small dust. OK. It's correct. Yea. So, good. I like messing with you guys. So remember, we're going to go through a few experiments first where it's going to be very easy to predict the results. We've got four experiments like this to do. And then we'll go on to the interesting examples. But we need to go through them so we know what happens, so we can make an empirical argument rather than an in principle argument. So there's the first experiment. Now, I want to run the second experiment. And the second experiment, same as the first, a little bit louder, a little bit worse. Sorry. The second experiment, we're going to send in hard electrons, and we're going to measure color at out. So again, let's look at the apparatus. We send in hard electrons. And our apparatus is hardness box with a hard and a soft aperture. And now we're going to measure the color at the output. Color, what have I been doing? And now I want to know what fraction come out black, and what fraction come out white. We're using lots of monkeys in this process. OK, so this is not rocket science. Rocket science isn't that complicated. Neuroscience is much harder. This is not neuroscience. So let's figure out what this is. Predictions. So again, think about your prediction your head, come to a conclusion, raise your hand when you have an idea. And just because you don't raise your hand doesn't mean I won't call on you. AUDIENCE: 50-50 black and white. ALLAN ADAMS: 50-50 black and white. I like it. Tell me why. AUDIENCE: It's gone through a hardness box, which scrambled the color, and therefore has to be [INAUDIBLE] ALLAN ADAMS: Great. So the statement, I'm going to say that slightly more slowly. That was an excellent argument. We have a hard electron. We know that hardness boxes are persistent. If you send a hard electron in, it comes out hard. So every electron incident upon our apparatus will transit across the hard trajectory. It will bounce, it will bounce, but it is still hard, because we've already done that experiment. The mirrors do nothing to the hardness. So we send a hard electron into the color box, and what comes out? Well, we've done that experiment, too. Hard into color, 50-50. So the prediction is 50-50. This is your prediction. Is that correct? Awesome. OK, let us vote. How many people think this is correct? Gusto, I like it. How many people think it's not? All right. Yay, this is correct. Third experiment, slightly more complicated. But we have to go through these to get to the good stuff, so humor me for a moment. Third, let's send in white electrons, and then measure the color at the output port. So now we send in white electrons, same beast. And our apparatus is a hardness box with a hard path and a soft path. Do-do-do, mirror, do-do-do, mirror, box, join together into our out. And now we send those out electrons into a color box. And our color box, black and white. And now the question is how many come out black, and how many come out white. Again, think through the logic, follow the electrons, come up with a prediction. Raise your hand when you have a prediction. AUDIENCE: Well, earlier we showed that [INAUDIBLE] so it'll take those paths equally-- ALLAN ADAMS: With equal probability. Good. AUDIENCE: Yeah. And then it'll go back into the color box. But earlier when we did the same thing without the weird path-changing, it came out 50-50 still. So I would say still 50-50. ALLAN ADAMS: Great. So let me say that again, out loud. And tell me if this is an accurate extension of what you said. I'm just going to use more words. But it's, I think, the same logic. We have a white electron, initially white electron. We send it into a hardness box. When we send a white electron into a hardness box, we know what happens. 50% of the time it comes out hard, the hard aperture, 50% of the time it comes out the soft aperture. Consider those electrons that came out the hard aperture. Those electrons that came out the hard aperture will then transit across the system, preserving their hardness by virtue of the fact that these mirrors preserve hardness, and end up at a color box. When they end at the color box, when that electron, the single electron in the system ends at this color box, then we know that a hard electron entering a color box comes out black or white 50% of the time. We've done that experiment, too. So for those 50% that came out hard, we get 50/50. Now consider the other 50%. The other half of the time, the single electron in the system will come out the soft aperture. It will then proceed along the soft trajectory, bounce, bounce, not changing its hardness, and is then a soft electron incident on the color box. But we've also done that experiment, and we get 50-50 out, black and white. So those electrons that came out hard come out 50-50, and those electrons that come out soft come out 50/50. And the logic then leads to 50-50, twice, 50-50. Was that an accurate statement? Good. It's a pretty reasonable extension. OK, let's vote. How many people agree with this one? OK, and how many people disagree? Yeah, OK. So vast majority agree. And the answer is no, this is wrong. In fact, all of these, 100% come out white and 0 come out black. Never ever does an electron come out the black aperture. I would like to quote what a student just said, because it's actually the next line in my notes, which is what the hell is going on? So let's the series of follow up experiments to tease out what's going on here. So something very strange, let's just all agree, something very strange just happened. We sent a single electron in. And that single electron comes out the hardness box, well, it either came out the hard aperture or the soft aperture. And if it came out the hard, we know what happens, if it came out the soft, we know what happens. And it's not 50-50. So we need to improve the situation. Hold on a sec. Hold on one sec. Well, OK, go ahead. AUDIENCE: Yeah, it's just a question about the setup. So with the second hardness box, are we collecting both the soft and hard outputs? ALLAN ADAMS: The second, you mean the first hardness box? AUDIENCE: The one-- are we getting-- no, the-- ALLAN ADAMS: Which one, sorry? This guy? Oh, that's a mirror, not a hardness box. Oh, thanks for asking. Yeah, sorry. I wish I had a better notation for this, but I don't. There's a classic-- well, I'm not going to go into it. Remember that thing where I can't stop myself from telling stories? So all this does, it's just a set of mirrors. It's a set of fancy mirrors. And all it does is it takes an electron coming this way or an electron coming this way, and both of them get sent out in the same direction. It's like a beam joiner, right? It's like a y junction. That's all it is. So if you will, imagine the box is a box, and you take, I don't know, Professor Zwiebach, and you put him inside. And every time an electron comes up this way, he throws it out that way, and every time it comes in this way, he throws it out that way. And he'd be really ticked at you for putting him in a box, but he'd do the job well. Yeah. AUDIENCE: And this also works if you go one electron at a time? ALLAN ADAMS: This works if you go one electron at a time, this works if you go 14 electrons at a time, it works. It works reliably. Yeah. AUDIENCE: Just, maybe [INAUDIBLE] but what's the difference between this experiment and that one? ALLAN ADAMS: Yeah, I know. Right? Right? So the question was, what's the difference between this experiment and the last one. Yeah, good question. So we're going to have to answer that. Yeah. AUDIENCE: Well, you're mixing again the hardness. So it's like as you weren't measuring it at all, right? ALLAN ADAMS: Apparently it's a lot we weren't measuring it, right? Because we send in the white electron, and at the end we get out that it's still white. So somehow this is like not doing anything. But how does that work? So that's an excellent observation. And I'm going to build you now a couple of experiments that tease out what's going on. And you're not going to like the answer. Yeah. AUDIENCE: How were the white electrons generated in this experiment? ALLAN ADAMS: The white electrons were generated in the following way. I take a random source of electrons, I rub a cat against a balloon and I charge up the balloon. And so I take those random electrons, and I send them into a color box. And we have previously observed that if you take random electrons and throw them into a color box and pull out the electrons that come out the white aperture, if you then send them into a color box again, they're still white. So that's how I've generated them. I could have done it by rubbing the cat against glass, or rubbing it against me, right, just stroke the cat. Any randomly selected set of electrons sent into a color box, and then from which you take the white electrons. AUDIENCE: So how is it different from the experiment up there? ALLAN ADAMS: Yeah. Uh-huh. Exactly. Yeah. AUDIENCE: Is the difference that you never actually know whether the electron's hard or soft? ALLAN ADAMS: That's a really good question. So here's something I'm going to be very careful not to say in this class to the degree possible. I'm not going to use the word to know. AUDIENCE: Well, to measure. [INAUDIBLE] ALLAN ADAMS: Good. Measure is a very slippery word, too. I've used it here because I couldn't really get away with not using it. But we'll talk about that in some detail later on in the course. For the moment, I want to emphasize that it's tempting but dangerous at this point to talk about whether you know or don't know, or whether someone knows or doesn't know, for example, the monkey inside knows or doesn't know. So let's try to avoid that, and focus on just operational questions of what are the things that go in, what are the things that come out, and with what probabilities. And the reason that's so useful is that it's something that you can just do. There's no ambiguity about whether you've caught a white electron in a particular spot. Now in particular, the reason these boxes are such a powerful tool is that you don't measure the electron, you measure the position of the electron. You get hit by the electron or you don't. And by using these boxes we can infer from their position the color or the hardness. And that's the reason these boxes are so useful. So we're inferring from the position, which is easy to measure, you get beaned or you don't, we're inferring the property that we're interested in. It's a really good question, though. Keep it in the back of your mind. And we'll talk about it on and off for the rest of the semester. Yeah. AUDIENCE: So what happens if you have this setup, and you just take away the bottom right mirror? ALLAN ADAMS: Perfect question. This leads me into the next experiment. So here's the modification. But thank you, that's a great question. Here's the modification of this experiment. So let's rig up a small-- hold on, I want to go through the next series of experiments, and then I'll come back to questions. And these are great questions. So I want to rig up a small movable wall, a small movable barrier. And here's what this movable barrier will do. If I put the barrier in, so this would be in the soft path, when I put the barrier in the soft path, it absorbs all electrons incident upon it and impedes them from proceeding. So you put a barrier in here, put a barrier in the soft path, no electrons continue through. An electron incident cannot continue through. When I say that the barrier is out, what I mean is it's not in the way. I've moved it out of the way. Cool? So I want to run the same experiment. And I want to run this experiment using the barriers to tease out how the electrons transit through our apparatus. So experiment four. Let's send in a white electron again. I want to do the same experiment we just did. And color at out, but now with the wall in the soft path. Wall in soft. So that's this experiment. So we send in white electrons, and at the output we measure the color as before. And the question is what fraction come out black, and what fraction come out white. So again, everyone think through it for a second. Just take a second. And this one's a little sneaky. So feel free to discuss it with the person sitting next to you. [CHATTER] ALLAN ADAMS: All right. All right, now that everyone has had a quick second to think through this one, let me just talk through what I'd expect from the point of these experiments. And then we'll talk about whether this is reasonable. So the first thing I expect is that, look, if I send in a white electron and I put it into a hardness pass, I know that 50% of the time it goes out hard, and 50% of the time it goes out soft. If it goes out the soft aperture, it's going to get eaten by the barrier, right? It's going to get eaten by the barrier. So first thing I predict is that the output should be down by 50%. However, here's an important bit of physics. And this comes to the idea of locality. I didn't tell you this, but these armlinks in the experiment I did, 3,000 kilometers long. 3,000 kilometers long. That's too minor. 10 million kilometers long. Really long. Very long. Now, imagine an electron that enters this, an initially white electron. If we had the barriers out, if the barrier was out, what do we get? 100% white, right? We just did this experiment, to our surprise. So if we did this, we get 100%. And that means an electron, any electron, going along the soft path comes out white. Any electron going along the hard path goes out white. They all come out white. So now, imagine I do this. Imagine we put a barrier in here 2 million miles away from this path. How does a hard electron along this path know that I put the barrier there? And I'm going to make it even more sneaky for you. I'm going to insert the barrier along the path after I launched the electron into the apparatus. And when I send in the electron, I will not know at that moment, nor will the electron know, because, you know, they're not very smart, whether the barrier is in place. And this is going to be millions of miles away from this guy. So an electron out here can't know. It hasn't been there. It just hasn't been there. It can't know. But we know that when we ran this apparatus without the barrier in there, they came out 100% white. But it can't possibly know whether the barrier's in there or not, right? It's over here. So what this tells us is that we should expect the output to be down by 50%. But all the electrons that do make it through must come out white, because they didn't know that there was a barrier there. They didn't go along that path. Yeah. AUDIENCE: Not trying to be wise, but why are you using the word know? ALLAN ADAMS: Oh, sorry, thank you. Thank you, thank you, thank you, that was a slip of the tongue. I was making fun of the electron. So in that particular case, I was not referring to my or your knowledge. I was referring to the electron's tragically impoverished knowledge. Yeah. AUDIENCE: But if they come out one at a time white, then wouldn't we know then with certainty that that electron is both hard and white, which is like a violation? ALLAN ADAMS: Well, here's the more troubling thing. Imagine it didn't come out 100% white. Then the electron would have demonstrably not go along the soft path. It would have demonstrably gone through the hard path, because that's the only path available to it. And yet, it would still have known that millions of miles away, there's a barrier on a path it didn't take. So which one's more upsetting to you? And personally, I find this one the less upsetting of the two. So the prediction is our output should down by 50%, because a half of them get eaten. But they should all come out white, because those that didn't get eaten can't possibly know that there was a barrier here, millions of miles away. So we run this experiment. And here's the experimental result. In fact, the experimental result is yes, the output is down by 50%. But no, not 100% white, 50% white. 50% white. The barrier, if we put the barrier in the hardness path. If we put the barrier in the hardness path, still down by 50%, and it's at odds, 50-50. How could the electron know? I'm making fun of it. Yeah. AUDIENCE: So I guess my question is before we ask how it knows that there's a block in one of the paths, how does it know, before, over there, that there were two paths, and combine again? ALLAN ADAMS: Excellent. Exactly. So actually, this problem was there already in the experiment we did. All we've done here is tease out something that was existing in the experiment, something that was disturbing. The presence of those mirrors, and the option of taking two paths, somehow changed the way the electron behaved. How is that possible? And here, we're seeing that very sharply. Thank you for that excellent observation. Yeah. AUDIENCE: What if you replaced the two mirrors with color boxes, so that both color boxes [INAUDIBLE] ALLAN ADAMS: Yeah. So the question is basically, let's take this experiment, and let's make it even more intricate by, for example, replacing these mirrors by color boxes. So here's the thing I want to emphasize. I strongly encourage you to think through that example. And in particular, think through that example, come to my office hours, and ask me about it. So that's going to be setting a different experiment. And different experiments are going to have different results. So we're going to have to deal with that on a case by case basis. It's an interesting example, but it's going to take us a bit afar from where we are right now. But after we get to the punchline from this, come to my office hours and ask me exactly that question. Yeah. AUDIENCE: So we had a color box, we put in white electrons and we got 50-50, like random. How do you know the boxes work? ALLAN ADAMS: How do I know the boxes work? These are the same boxes we used from the beginning. We tested them over and over. AUDIENCE: How did you first check that it was working? [INAUDIBLE] ALLAN ADAMS: How to say-- there's no other way to build a box that does the properties that we want, which is that you send in color and it comes out color again, and the mirrors behave this way. Any box that does those first set of things, which is what I will call a color box, does this, too. There's no other way to do it. I don't mean just because like, no one's tested-- AUDIENCE: Because you can't actually check it, you can't actually [INAUDIBLE] you know which one is white. ALLAN ADAMS: Oh, sure, you can. You take the electron that came out of the color box. That's what we mean by saying it's white. AUDIENCE: [INAUDIBLE] ALLAN ADAMS: But that's what it means to say the electron is white. It's like, how do you know that my name is Allan? You say, Allan, and I go, what? Right? But you're like, look that's not a test of whether I'm Allan. It's like, well, what is the test? That's how you test. What's your name? I'm Allan. Oh, great, that's your name. So that's what I mean by white. Now you might quibble that that's a stupid thing to call an electron. And I grant you that. But it is nonetheless a property that I can empirically engage. OK, so I've been told that I never ask questions from the people on the right. Yeah. AUDIENCE: Is it important whether the experimenter knows if the wall is there or not? ALLAN ADAMS: No. This experiment has been done again by some French guys. The French, look, dude. So there's this guy, Alain Aspect, ahh, great experimentalist, great physicist. And he's done lots of beautiful experiments on exactly this topic. And send me an email, and I'll post some example papers and reviews by him-- and he's a great writer-- on the web page. So just send me an email to remind me of that. OK, so we're lowish on time, so let me move on. So what I want to do now is I want to take the lesson of this experiment and the observation that was made a minute ago, that in fact the same problem was present when we ran this experiment and go 100%. We should have been freaked out already. And I want to think through what that's telling us about the electron, the single electron, as it transits the apparatus. The thing is, at this point we're in real trouble. And here's the reason. Consider a single electron inside the apparatus. And I want to think about the electron inside the apparatus while all walls are out. So it's this experiment. Consider the single electron. We know, with total confidence, with complete reliability, that every electron will exit this color box out the white aperture. We've done this experiment. We know it will come out white. Yes? Here's my question. Which route did it take? AUDIENCE: Spoiler. ALLAN ADAMS: Not a spoiler. Which route did it take? AUDIENCE: Why do we care what route? ALLAN ADAMS: I'm asking you the question. That's why you care. I'm the professor here. What is this? Come on. Which route did it take? OK, let's think through the possibilities. Grapple with this question in your belly. Let's think through the possibilities. First off, did it take the hardness path? So as it transits through, the single electron transiting through this apparatus, did it take the hard path or did it take the soft? These are millions of miles long, millions of miles apart. This is not a ridiculous question. Did it go millions of miles in that direction, or millions of miles in that direction? Did it take the hardness path? Ladies and gentlemen, did it take the hard path? AUDIENCE: Yes. ALLAN ADAMS: Well, we ran this experiment by putting a wall in the soft path. And if we put a wall in the soft path, then we know it took the hard path, because no other electrons come out except those that went through the hard path. Correct? On the other hand, if it went through the hard path, it would come out 50% of the time white and 50% of the time black. But in fact, in this apparatus it comes out always 100% white. It cannot have taken the hard path. No. Did it take the soft path? Same argument, different side, right? No. Well, this is not looking good. Well, look, this was suggested. Maybe it took both. Maybe electrons are sneaky little devils that split in two, and part of it goes one way and part of it goes the other. Maybe it took both paths. So this is easy. We can test this one. And here is how I'm going to test this one. Oh, sorry. Actually, I'm not going to do that yet. So we can test this one. So if it took both paths, here's what you should be able to do. You should be able to put a detector along each path, and you'd be able to follow, if you've got half an electron on one side and half an electron on the other, or maybe two electrons, one on each side and one on the other. So this is the thing that you'd predict if you said it went both. So here's what we'll do. We will take detectors. We will put one along the hard path and one along the soft path. We will run the experiment and then observe whether, and ask whether, we see two electrons, we see half and half, what do we see. The answer is you always, always see one electron on one of the paths. You never see half an electron. You never see a squishy electron. You see one electron on one path, period. It did not take both. You never see an electron split in two, divided, confused. No. Well, it didn't take the hard path, didn't take the soft path, it didn't take both. There's one option left. Neither. Well, I say neither. But what about neither? And that's easy. Let's put a barrier in both paths. And then what happens? Nothing comes out. So no. So now, to repeat an earlier prescient remark from one of the students, what the hell? So here's the world we're facing. I want you to think about this. Take this seriously. Here's the world we're facing. And when I say, here's the world we're facing, I don't mean just these experiments. I mean the world around you, 20 kilo mirrors, bucky-balls, here is what they do. When you send them through an apparatus like this, every single object that goes through this apparatus does not take the hard path, it does not take the soft path, it doesn't take both, and it does not take neither. And that pretty much exhausts the set of logical possibilities. So what are electrons doing when they're inside the apparatus? How do you describe that electron inside the apparatus? You can't say it's on one path, you can't say it's on the other, it's not on both, and it's not on neither. What is it doing halfway through this experiment? So if our experiments are accurate, and to the best of our ability to determine, they are, and if our arguments are correct, and that's on me, then they're doing something, these electrons are doing something we've just never thought of before, something we've never dreamt of before, something for which we don't really have good words in the English language. Apparently, empirically, electrons have a way of moving, electrons have a way of being which is unlike anything that we're used to thinking about. And so do molecules. And so do bacteria. So does chalk. It's just harder to detect in those objects. So physicists have a name for this new mode of being. And we call it superposition. Now, at the moment, superposition is code for I have no idea what's going on. Usage of the word superposition would go something like this. An initially white electron inside this apparatus with the walls out is neither hard, nor soft, nor both, nor neither. It is, in fact, in a superposition of being hard and of being soft. This is why we can't meaningfully say this electron is some color and some hardness. Not because our boxes are crude, and not because we're ignorant, though our boxes are crude and we are ignorant. It's deeper. Having a definite color means not having a definite hardness, but rather being in a superposition of being hard and being soft. Every electron exits a hardness box either hard or soft. But not every electron is hard or soft. It can also be a superposition of being hard or being soft. The probability that we subsequently measure it to be hard or soft depends on precisely what superposition it is. For example, we know that if an electron is in the superposition corresponding to being white then there are even odds of it being subsequently measured be hard or to be soft. So to build a better definition of superposition than I have no idea what's going on is going to require a new language. And that language is quantum mechanics. And the underpinnings of this language are the topic of the course. And developing a better understanding of this idea of superposition is what you have to do over the next three months. Now, if all of this troubles your intuition, well, that shouldn't be too surprising. Your intuition was developed by throwing spears, and running from tigers, and catching toast as it jumps out of the toaster, all of which involves things so big and with so much energy that quantum effects are negligible. As a friend of mine likes to say, you don't need to know quantum mechanics to make chicken soup. However, when we work in very different regimes, when we work with atoms, when we work with molecules, when we work in the regime of very low energies and very small objects, your intuition is just not a reasonable guide. It's not that the electrons-- and I cannot emphasize this strongly enough-- it is not that the electrons are weird. The electrons do what electrons do. This is what they do. And it violates your intuition, but it's true. The thing that's surprising is that lots of electrons behave like this. Lots of electrons behave like cheese and chalk. And that's the goal of 804, to step beyond your daily experience and your familiar intuition and to develop an intuition for this idea of superposition. And we'll start in the next lecture. I'll see you on Thursday.
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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 what I'm going to try to do now is set up again this equation and do the analog of what we're doing there and try to determine this function fk in some nice way. All right. So let's think of this equation. I want to do it in pieces. So psi of r is going to be equal to some formula. And then it has to be equal to this right hand side. Let's write the e to the ikz the way we've done it before. It's up there. So I'll write it here. Square root of 4 pi over k square, sum of l equals 0 to infinity, square root of 2l plus 1, i to the l yl 0 1 over 2 i e to the ikr minus l pi over 2 over r minus e the minus ikr minus l pi over 2. Wow. It's tiring this r. Plus f of k of theta e to the ikr over r. OK. So what do we have here? We've written the right hand side of this equation. I copied it. I have not done anything except taking r much greater than a. Because otherwise in the plane wave into the ikz, I could not have expanded the Bessel functions unless I took r greater than a. But that's good, because we now have our waves. OK. We have our waves there. Now, look at this right hand side. Where is the incoming wave in this right hand side? The incoming wave is here. That's the only term that is incoming, because this is an outgoing wave, and this is an outgoing wave. So if I want to write the left hand side, the incoming wave of the left hand side has to be equal to this wave. And of course, the outgoing wave of the left hand side will also have to be equal to whatever is outgoing here, but the incoming must be this. So I'm going to write this left hand side and already use this and put 4 pi over k square sum of l equals 0 to infinity 2l plus 1 i to the l y l 0 1 over 2 pi, big parentheses, and one outgoing wave and one incoming wave minus ikr minus l pi over 2 over r. And here, I don't know what to put, but I've put already there on the left hand side of this equation for psi of r for the full solution, a wave that matches the right hand side, because it has the same incoming wave. And now, I'm going to use some physical intuition to guess what we'll have to put on this part. This is the step that requires a little imagination, not too much, because we already did something similar here. So what's happening here and here is intuition I think you should keep after weeks of this course when it's all forgotten, there's some intuition that you should keep. And it's about this scattering happening for each partial wave independent. Yes. AUDIENCE: 2i. PROFESSOR: 2i, yes. No, 2 pi there. Yes. Thank you. So here it is. This is a solution, and we've got the intuition already. I will justify this later, of course, very precisely. But I think that this one you need to have a little bit of an intuition of what you should do. And first, we said each l works separately to create a solution of the Schrodinger equation. That's superposition, and it's [INAUDIBLE].. Each l is working separately. Each l is like a scattering problem. Each l has a wave that comes in and a wave that comes out, because these things, j and n, have waves, and they have an in and out. So these have some in wave and some out wave. And if each wave works separately, it has an in wave and then out wave, in a scattering problem, these waves must have the same amplitude, because otherwise they wouldn't have the same probability current, and probability would get stuck. So this must be an outgoing wave having this same amplitude as this wave. And by the argument we have here, it just differs by a phase. So we'll put here e to the ikr minus l pi over 2 plus 2i delta l over r. So that this wave, spherical wave, that it's outgoing, it has the same amplitude as this one, and cannot be the same. The only difference can be a phase shift, and that's the phase shift. So your picture is scattering in three dimensions. Looks like, OK, you threw in a plane wave and out came a spherical wave out. The other picture that is more consistent with the way you solve it is that you have an infinite set of partial waves for different l's, each one scattering, the l equals 0, the equal 1, the l equal 2, all of them scattering. So this corresponds to an [? ansatz ?] in terms of phase shifts, and now you can say you've parameterized your ignorance in a physical way. You've discovered that all that characterizes the scattering is, as it was in one dimension, a phase shift. In one dimension, there was a single phase shift, because you didn't have all these general solutions that you had in three dimensions. Your energy eigenstates were momentum eigenstates there were non-degenerate really. There was just a couple of momentum eigenstates, wave in and waves out. Here is infinitely degenerate. There's spherical stuff. So there's a phase shift for each value of l. So we've parameterized the physics of the scattering problem in terms of phase shift, and now, it's interesting to try to figure out what is this quantity after all in terms of the phase shifts. It's already here. We just have to solve it. So from this equation, I now can say that this term cancels with this term, and now, I can solve for this term f of theta e to the ikr by collecting these other two terms together. And therefore, fk of theta e to the ikr over r is equal, and it's exactly the same thing I did here, cancel here, pass, and the mathematics is going to be completely analogous, except that they have to carry all that sum there. No big deal. So what is it? Square root of 4 pi over k, sum from l equals 0 to infinity, 2l plus 1, i to the l y l0. 1 over 2i. OK, 1 over 2i. OK, I'll do it this way. e to the 2i delta l minus 1 e to the ikr e to the minus il pi over 2. I think I got everything there. I put the first term on the right hand side, I moved it to the left hand side. The coefficient was all the same. This was the coefficient they both had the same coefficient. Then I just have to subtract these two exponentials over r. So I did forget the r. So I just subtract the two exponentials. Both exponentials have the e to the ikr minus l pi over 2. And the difference is that the first exponential on the top has the extra e to the 2i delta l minus 1, and the other one doesn't. So what do we get here? This part is e to the i delta sine delta, and this part is e to the minus i pi over 2 to the power l, which is minus i to the l, and i to the l times minus i to the l is happily just 1. i times minus i is 1, and 1 to the l is 1. So this term and this term cancel. So finally, and I can cancel happily they r dependence is all the same. I can cancel this r dependence, this r dependence. And finally, we've got fk of theta equals square root of 4 pi over k sum from l equals 0 to infinity square root 2l plus 1 y L0 of theta e to the i delta l sine delta l. So that said, that's our formula for fk in terms of the phase shift. So what have we achieved? We want f of k, because that gives us the cross section. What we have figured out is that the calculation of fk really requires knowing the phase shift. And the phase shifts are defined by that formula over there, where we have estimated how one wave is connected to the other one, the incoming and the outgoing for a given fixed f, for a given partial wave, how they're offset by this phase shift.
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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: This is the second recitation on cholesterol, and it's really focused on this question of how do you sense cholesterol in a membrane? So that's really a tough problem. And they've developed new tools, and that's what we're going to be talking about-- what the tools are, and whether you would think they were adequate to be able to address this question about what kinds of changes in concentration of cholesterol. Number one, can you measure them? And number two, what effects do they have, in terms of whether you're going to turn on cholesterol biosynthesis and uptake, because you need more cholesterol, or you're going to turn the whole thing off? So we've been focusing, as we've described in the last few lectures, in the endoplasmic reticulum. And what would the cholesterol-- what kinds of changes in cholesterols did they see in the experiments they were doing in this paper? What were the range of changes that they saw? AUDIENCE: 3% to 10%? JOANNE STUBBE: Yeah, so see, something low. Say they were trying to do this same experiment in the plasma membrane-- how do we know it's the ER membrane that does this sensing? That's what the whole paper is focused on, that's what everything we've focused on in class. Say you wanted to do a similar kind of experiment in the plasma membrane, do you remember what I said about the levels of cholesterol? So they distributed throughout the cell, in all membranes. Where is the most cholesterol? So if you don't remember, it's the plasma membrane. So say, instead of having 7% or 8% of the lipids cholesterol, say you had 40%-- that's an over-exaggeration-- do you think this kind of an experiment would be hard to do, that they've talked about in this paper? So you would want to do this-- if you tried to do the same experiment with the plasma membrane? So the key issue that you need to think about, is go back and look at the changes-- they did a whole bunch of different experiments. The numbers are squishy, but they came up with numbers that reproduced themselves, I thought, in an amazing way. But now say you wanted to do this in the plasma membrane, where the levels of cholesterol are much higher. Do you think it would be easy to do? Using the same tech techniques that are described, that we're going to discuss, or not? And what would the issues be? Yeah? AUDIENCE: So they had to deplete the cholesterol from the membrane, and so that would probably be hard to deplete it to a level that's low enough, so that you don't get the activity. Right? JOANNE STUBBE: So, I don't know. So that's an interesting question. So you'd have to deplete-- so that's going to be it, we're going to have to control the cholesterol levels. But what change-- if you looked at the changes in levels of cholesterol in the ER, how much did they change? They change from what to what? From-- 2% to 7%. Say that you were in that same range of change that was going to turn on a switch in the plasma membrane. And say you could control the levels. Do you think it would be easy to see that? So you start with 40%, say, that's the norm. Say the change was very similar to what you see in the change in the ER-- do you think that would be easy to detect? No, because now you have two big numbers, and there's a huge amount of error in this method of analysis. So those are the kinds of things I'm trying to get you to think about. I don't know why it's the ER-- I mean, everybody's focused on the ER. Could cholesterol and other organelles have a different regulatory mechanism? Or somehow be connected, still, to what's going on in the ER? Could be-- I mean, you start out with the simplest model you can get and you test it, but then as you learn more, or we have more and more technology, we learn new things, you go back and you revisit and rethink about what's going on. So the key question is, it's really this switch of having cholesterol that keeps it in the membrane, or not having cholesterol. And the question is, what are the differences in the levels that allow turn on of cholesterol-- biosynthesis and LDL biosynthesis, which then allows uptake of cholesterol from the diet? OK, so that's the question. And what does this look like? And people hadn't measured this by any method, and this model I've gone through a number of times in class today, so I'm not going to go through it again. Hopefully you all know that in some form in your head, or you have the picture in front of you so you can remember it. So these are the questions I want to pose, and I want you guys to do the talking today. And what I'm going to do is, I have most of the figures on my PowerPoint, so we can bring them up and look at them. And you can tell me what you see. And then everybody might be seeing something different-- and so we're thinking about this differently, and maybe we come to some kind of consensus about whether these experiments were carried out well or not. So one of the first things-- so these will be the general things, and then we'll step through them. But they wanted to perturb the cellular cholesterol levels. And how did they end up doing that? Did that make sense? We talked a little bit about this already. I mean, what did they use as tools to do that? AUDIENCE: [INAUDIBLE] JOANNE STUBBE: So you need to speak louder, because I really am deaf. Sorry. AUDIENCE: So just, right here, they were careful of the amount of cholesterol present in this? JOANNE STUBBE: So that's one place, so they can deplete cholesterol for the media. But then what did they do? So the whole paper is about this-- how did they control the [INAUDIBLE]? Let's assume that they can do that, and they got good at that. I think a lot of people have used that method, and so they can deplete media. So how did they deplete cholesterol? There was some unusual ways to deplete cholesterol in this paper. Did any of you pick up on that? AUDIENCE: A chemical that could bind to cholesterol. JOANNE STUBBE: So did you think that was unusual? Did any of you look up what that was? AUDIENCE: It was a kind of carbohydrate that can bind to cholesterol. JOANNE STUBBE: Yeah, so but what was interesting about it, it was hydroxypropyl-- remember HP, cyclodextrin. We're going to look at this in a minute. But what do we know-- what was the other molecule they used to add cholesterol back? AUDIENCE: Another form of that molecule is-- JOANNE STUBBE: So methyl-cyclodextrin-- I'm going to show you the structure, but they aren't very different. So have any of you ever heard of cyclodextrin before? People won the Nobel Prize for that, Don Cram won it, Breslow spent his whole life studying host guest interactions. So you guys, I don't know what you teach you now anymore, but that used to be something that was taught a lot, host guest interactions, trying to understand weak non-covalent interactions as the basis for understanding catalysis. But to me, that was-- immediately when I saw this, what the heck's going on? So then I Googled it, and immediately-- and I don't know anything about hydroxypropyl-- you Google it, you look it up. And then you look at it, and if you were a chemist and you were really interested in the molecular interactions, you might make a model of it. And then see, what is the difference between that one little group, when you look at the structure, it's amazing. And that's the basis of most of the experiments. So you need to believe that they figured that out. And that's not in this paper, so if you really cared about it you would have to go back and read earlier papers, and see what are the experiments that led them to focus on these molecules? How else did they end up getting cholesterol levels back into the cell? Do you remember what the other method was? So we'll come back and we'll talk about this in a minute-- so that was one of the methods. AUDIENCE: They added two kind of sterols. JOANNE STUBBE: OK, so they did add two kind of sterols-- and they tried to figure out, this is another unknown, what was the difference between the sterols? Simply a hydroxyl group. OK, so if you looked at this, cholesterol is this guy. And then they had something like this guy-- 25, and remember where [INAUDIBLE] the side chain, hanging out of the little [? cheer ?] system you have. I don't think they learned very much from that. And in fact, in your problem set, you had all of these different cholesterol analogs. I mean, I think we still really don't get it. That's complicated-- we talked about this in class. You have these transmembrane helices-- what is it that's actually the signaling agent? So people are still asking that question, and we haven't quite gotten that far. But if you've read the reading, for HMG CoA reductase degradation, which is what we we're going to be talking about in class, the signaler is not the sterile, it's lanosterol. OK, and where have you seen lanosterol? The biosynthetic pathway has lanosterol sitting in the middle. It's not all that different, structurally, from cholesterol. You need to go back in, they all have four-membered rings, they have different extra methyl groups. So people are trying to sort that out. I don't think we really know. But how well? So you're right, they use sterols. They didn't use that, they didn't see very much difference with the sterols. What was the other way, which is sort of unusual, that they added cholesterol back into the system. So they could add it back with the methyl cyclodextrin-- they told you that that worked, and if you believe that-- and you look at the data-- it looked like that was happening. Nobody remembers? OK, well, we'll get to that in a little bit. OK, so the question we're focusing on is what are the changes in concentrations of cholesterol in the ER? So what method did they use to try to separate the ER membranes from all the other membranes? AUDIENCE: They first separated the [INAUDIBLE]---- JOANNE STUBBE: They separated the what? AUDIENCE: The sterols and the nucleus in the [INAUDIBLE].. JOANNE STUBBE: OK, so that's good. You can separate out the nucleus, and you could do that by ultracentrifugation-- we've seen that used in different kinds of ultracentrifugation. We've seen the different particles, the lipoproteins in the diet, how do we separate those? We talked about that in class briefly, you haven't had any papers to read. But what was the method of separation? If you look at all those particles-- remember we had a little cartoon of all the particles, and we focused on LDL, which is the particle that has the most cholesterol. So that's why everybody is focusing on that. What was the basis of the separation? AUDIENCE: Was it sucrose screening? JOANNE STUBBE: Was the what? AUDIENCE: Was it a sucrose screening-- the ultracentrifugation? JOANNE STUBBE: You need to-- AUDIENCE: Did they use a sucrose screening, like ultracentrifugation? JOANNE STUBBE: Yeah, ultracentrifugation. But how did the-- AUDIENCE: For the sucrose screening? JOANNE STUBBE: Yeah, OK, so they have different density gradients. , OK so that's going to be a key thing, and that's because if you look at the composition, they have different amounts of proteins, different amounts of fats. And they have different-- they float differently. So that's the method that they're going to use here. Is that a good method? Can you think of a better method? So in order to understand the switch for cholesterol, you've got to be able to measure the changes in cholesterol. Not an easy problem, because cholesterol is really insoluble in everything. And so how much is really in there, and how does it change under different sets of conditions? So is this a good method? What do you think? We'll look at the method in a little more detail, when I pull up the figures, but what did you think when you read the paper? AUDIENCE: Seems a pretty good method, other than that they're slightly different any other like properties different from the membrane than say, press on golgi bodies and ER. So it's like the only one I can think of. JOANNE STUBBE: Yeah, so the question is, you could you separate? Even separating the nucleus from the cytosol is not so trivial. But these methods are really gross methods, and during the centrifugation, things diffuse. So if you're having close separations, it's a equlibrating down this thing. And so you're getting your proteins, or your lipids are spreading out. Is there anything else any of you experience with insoluble-- this is what we're dealing with, is an insoluble mess, and how do you how do you separate things in a way that you have control over it so that you can address the key questions in this paper? Nobody thought about anything else? Did you like this method? Were you convinced by the data? AUDIENCE: I mean, like I couldn't necessarily think of something better. I don't know, I guess the thing that sketches me out the most about it just like how-- I'm not really familiar with the method. I haven't done this myself, so I don't know how that process affects the membrane integrity. JOANNE STUBBE: So that's an incredibly important question, because lipids confuse. They can mix. The question is, what are the rate constants for all of that? And we don't really teach very much in the introductory courses about lipids, and they're partitioning between other membranes and fusion, and all that stuff. But if you think about it, that's what the cell is, right? How do you get a plasma membrane, and all these membranes around all these little organelles-- that's an amazing observation. And we've seen in class already, what have we seen to get LDL receptor from here to the plasma membrane? How do we have to do that? We had to use these little vesicles. So you're generating something over here, it goes through the Golgi stack. Again, another set of membranes has got to come out the different levels of the Golgi stack. And then it's still got to get into the plasma membrane, and fuse, and dump its cargo. So I think it's an amazing process. And people interested in evolution, this is one of the major things people are focused on is, how can you make cells, little fake cells, artificial cells, that can replicate themselves. You can make it, and they're going to have to divide and fuse. And it's exactly the same problem here. And so this question of fluidity is an extremely important question. And a lot of people that focus on lipids-- which is not a popular thing to study, because it's so hard-- it's incredibly important. And people that look at membrane proteins, they almost always have lipids on them. And when you do them yourself, you have a detergent, which is not a real lipid-- does that change the property? So all of these questions, I think, are really central to what happens in the membranes, which is a lot of stuff inside the cell. So I think it's good to question what they did. I think their results turned out to be quite interesting. But we'll come back-- I think that was a hard problem. And so we'll come back and we'll look at this. And so then, let's say that we could end up separating things. Then the question is, what was the key type of measurement they made, where they could correlate the changes in cholesterol levels-- we talked about, you can control perhaps the cholesterol levels with the cyclodextrin. But then, how did they correlate the changes in the cholesterol levels in the membrane with this transcriptional regulation? Which, that is what happens with the steroid-responsive element-binding protein, the transcription factor. So what happens in that process? What are the changes in the SRE BP dependent on the concentrations of the cholesterol? And how did they take advantage of that in answering this question about what the cholesterol levels were that allowed you to turn on transcription of LDL receptor, and HMG CoA reductase. So what's the major assay? We'll look at that, as well. So if you go back and you look at the model, what happens in this model? All right, here we go-- what happens in this model? What's happening to SREBP? AUDIENCE: It has completely changed and exposed [INAUDIBLE]. JOANNE STUBBE: No, that's SCAP-- SCAP, that's this guy. OK? So SCAP, that's a key player. That's what we talked about. I know the names are all confusing. You're going to need to write these down to remember. The names are very confusing. Yeah? AUDIENCE: So the SCAP SREBP, whatever you call it, complex move signal g-apperatus then part of it's cleaved and moves to the nucleus? JOANNE STUBBE: Right, so how could you take advantage of that? This is the key observation that they're taking advantage of, to ask the question-- since this whole process is dependent on the concentration of cholesterol. If you have high cholesterol, there's no way you want this to happen-- you want to shut it off. If you have low cholesterol, you want to turn these guys on. So this movement is the key. And what do we see, if we look at what happens to this protein, SREBP, what happens to it during this process? It gets cleaved. And how could you monitor that cleavage? How do they do it in the paper? AUDIENCE: They used a-- was it a [? florifor-- ?] or is that the homework? JOANNE STUBBE: They could use a [? florifor, ?] they didn't do that. They did a what? AUDIENCE: They were able to separate the [INAUDIBLE] gel? JOANNE STUBBE: So it can be operated by a gel. So to me, this is quite an easy assay. Because if you look at this-- I don't remember what the molecular weight is, but it's a lot smaller over here. And so, that turns out to be a great assay. So that part of their analysis, I think, was a really smart part of the analysis. And so then the question becomes, can you quantitate all of this? So if you have a lot of cholesterol, this doesn't happen. And so everything is bigger, and resides in the membrane. You could even probably look at that. Whereas, when the cholesterol is really lower, things go there. And it's everything in between. The question is, what is the concept-- can you measure if you have X% cholesterol in the ER, how much do you have to decrease it to see a change or a switch in where this protein goes? So I think the experimental design is actually amazingly creative. But then you see the data of the other side. And what I want to do now is focus on what the issues are. So we're going to come back and look at, how did they look at SREBP? So you could look at this a number of ways-- you could look at this by protein gel directly. How else do people look at proteins using westerns? What's a western? Anybody know what a western analysis is? Didn't I ask you that at the beginning of class? How else do you detect proteins? You've seen this in the first half of the semester a lot. Yeah, antibodies. So if you have antibodies to this-- and we'll talk about this, because the western analysis, which people use all the time, and there are so many issues with it, that I think I want you to think about what the issues are. And then you correlate the two-- changing the levels of cholesterol. Which they measure by mass spec after separation and purification of lipids, and the cleavage. And they plot the data, and that's where they got the analysis from. So the first thing that you want to do-- the first thing, and the key to everything, is separation of the membranes. And so, this is a cartoon of when you put something, you load something on the top, and you have a gradient, and the gradient could be made of a number of things. Have any of you ever run these kinds of gradients? OK, so you can make them out of glycerol, you can make them out of sucrose-- did anybody look at how these gradients were made? Did you read the experimental carefully enough to look at that? Yeah, how do you make a sucrose gradient? You have no idea? But yeah, so layering. So what you really like to do is have a continuous gradient, or something. But sucrose is incredibly viscous. So if you were trying to make a linear gradient, which you could do by mixing two things of different concentrations-- if you could get them to stir really well, and then add it in, and you could generate a gradient. But it's so hard to do, that what happens is they end up layering it. So they make X%, Y%, Z%, they put it down. And then they try to layer something on top of it. And then they put whatever the interest in at the top, and then they centrifuge it. So what are the issues? Do you think this is what the gradient would look like? So what are the issues when you're doing this, when you layer it? And this is why the data-- which we'll talk about in a minute-- is the data, or part of the issue is this method. That's why you need to think about the method. And there are better ways to do this. And it really depends on what you're trying to separate. So if this band-- say these were two bands, you wouldn't really get very much separation at all. If there were two separate things that sedimented under these conditions very close together. So what would happen when you're sedimenting this? Does anybody have any idea how long it takes? Do you think you'd do this in a centrifuge, you spin it for three minutes, and then-- so sometimes you sediment these things for 16, 20 hours. So what happens during the sedimentation? That might make this more challenging, in terms of separating what you want to separate? AUDIENCE: I'm not sure, but it [INAUDIBLE] diffusion. JOANNE STUBBE: Yeah, so exactly, you have diffusion. And even when you've layered things on top of each other like that, you start to have diffusion. And if you shake up the tube a little bit, it's all over. So how do you prepare these things is not-- so people still use these methods, but I would like to see better methods. And so they tried one method with sucrose, and then that wasn't good enough. We'll look at the data. So they went to a second method. And where did they come up with this? I have no idea where they came up with this, but there was an MD PhD student in our class who had seen this and one of his classes, and they use it and some blood test. So I think that's probably where these guys got it from, because Brown and Goldstein are both MDs. But again, it's just another way to make a gradient. And I'm not sure why this gradient works as effectively as it does. But the first gradient didn't work so great, and we'll look at that data. So then they added on a few more steps, because they weren't happy with the level of separation. So looking at membranes, I think this is going to be more and more looking at membranes, because membranes, you have two leaflets-- the lipids and the leaflets are different. Do you think that affects the biology? I guarantee you it affects the biology in ways that we would really like to understand that I don't think we understand very well. When you isolate a membrane protein, have any of you ever isolated a membrane protein? So you have an insoluble-- it's in this lipid system. How do you think you get it out, so you can go through the steps, a protein purification that you've talked about, or you have probably done in an introductory lab course? What is the first thing you need to do? Yeah, solubalize it. And how do you solubalize it? AUDIENCE: With a detergent. JOANNE STUBBE: Yeah, with some kind of detergent. It's like what you saw with a kilo microns, or the bile acids that we talked about. So you can use different-- and people have their own favorite detergents. But again, that changes things. But otherwise, you can't purify anything unless you happen to have a membrane where the only protein in the membrane is the one you're interested in, which, of course, doesn't exist. So anyhow, they went through that. And then what did they end up seeing? So they went through different steps, and they separate them into different-- the supernate, or the light and the heavy membrane fractions. And then they have to analyze it. And so the question is, how do they analyze to tell how well these separations actually worked? What was the method that they did to determine whether they separated the ER from the plasma membrane, from the Golgi stacks, from the lisosomes, from the peroxisomes. So they have all we have all these little organelles in there. What did they do to test each one of these fractions? Let me ask you this question-- how do you think they got the-- how do you how did they get the material out of these gradients to do the experiments that I was just talking about. So they want to analyze what's in each of these bands. How did they get it out of this tube? AUDIENCE: Would they use a Pasteur filter? JOANNE STUBBE: So what do you think? You just stick it down in and suck it out? Well, I mean, yes, so what do you think? You could do that-- you open the top, you stick it in, you carefully stick it in. If you can see it. Lots of times you can see these lipids, because they're opaque, or something. So you can see. Or, if you still hope your sucrose layers, lots of times they layer in between the different concentrations of the sucrose, and you see white stuff precipitating. So you could conceivably stick a pipe head from the top and suck it out. AUDIENCE: But that would perturb all the other layers. JOANNE STUBBE: Absolutely it would perturb all the other layers. So here you're doing something-- it's already a very hard experiment, because they're all being perturbed anyhow, because of diffusion. So is there any other way you could think about separating these things? And so, the hint is that they use plastic tubes. So these things are not glass. Most centrifuges-- AUDIENCE: Freeze it? Cut it? JOANNE STUBBE: Well, so you don't do that, that could be-- OK, so you could. But you then have to, if you were cutting it, you still have to get it out of the tube. Unless you had a saw that didn't have any vibrations when you were cutting it, of course, which would not happen. But if you look here in this cartoon, so I gave you this, what are they doing here? They're sticking a syringe in through the side of the tube. And that's still what people use. So you can suck out-- if you can see something. So you have to be able to see in some way to know where to suck it out, so you might have a way, actually, in doing ultracentrifugations. I think with the lipids you can see them by eyeball, but you might look at absorption. If they have proteins, you could monitor absorption through the gradient, and that might tell you how to fractionate things. But anyhow, that's also an issue. Because before they can do the next step in the analysis, they've got to get the material out. So they've got the material out in each of these steps, and then, how do they look at this? They can pull it out. So what are they looking for? To tell them how effective this method is. AUDIENCE: Maybe some specific markers for each protein. JOANNE STUBBE: Exactly. So what are they-- to do that, what they're going to have to do is, before we look at the details of the method, I want to go through a western blot. So what do we know about a western blot? AUDIENCE: I have a quick question about the method here. JOANNE STUBBE: About the which method? AUDIENCE: The lysis method [INAUDIBLE] ball bearing homogenizer. So they're literally putting these cells in something like a bunch of ball bearings? JOANNE STUBBE: Yeah, you could do that. There's a lot of ways to crack open cells. I don't know which one's the best-- mammalian cells are really easy to open. Sometimes what I like to do is freeze and thaw them-- sometimes you have like a little mortar and pestle, or something like that. But that's-- I mean, yeast cells, you roll them. You have to have enough cells so you can do something. If you only have a tiny amount of cells, it makes it really challenging with beads, because it covers the beads. AUDIENCE: Do you have any issues with any of the different types of membranes that-- JOANNE STUBBE: Sticking to that? Absolutely. I'm sure you have to look at all of that kind of stuff. So how you choose, that's an important thing to look at, how you choose to crack open the cells. And it's the same with bacterial cells-- there are three or four ways to crack open the cells. And I can tell you only one of them really works efficiently. And a lot of people, when they use some of the others, they do something and they assume it works, but they never check to see whether the cell walls have been cracked open. A lot of times they haven't, and so what you get out is very, very low levels of protein, because you haven't cracked open the cell. So figuring out-- mammalian cells are apparently, I haven't worked with those myself, but they're apparently much easier to disrupt than bacteria. Or if you look at fungi-- fungi are really hard to crack open, yeast. So anyhow, that's an important thing to look at. So every one of these things, again, the devil is in the details. But when you're doing your own research, it doesn't matter what method you're looking at. The first time around, you need to look at it in detail, and convince yourself that this is a good way to chase this down. And you look at it in detail the first time around. And when you convince yourself it's working really well, and doing what you want to do, then you just use it. And that's the end of it. You don't have to go back and keep thinking about this over and over again. So the method we're going to use is a western blot. So we've got this stuff out, and have you all run SDS page shells? OK, so SDS page shells separate proteins how? AUDIENCE: Based on size... JOANNE STUBBE: By the what? AUDIENCE: It separates into a a charge gradient, and then-- not a charge gradient, but-- JOANNE STUBBE: Not charge. AUDIENCE: That's what drives the protein, but... JOANNE STUBBE: Right, but it's based on size, because it's coded-- every protein ratio is coded with this detergent, sodium dodecyl sulfate, which makes them migrate pretty much like the molecular weight. But if you've done these, it's not exactly like the molecular weight. You can do standards where you know the molecular weight, you can do a standard curve, and then you see where your protein migrates. And sometimes they migrate a little faster, sometimes a little slower, but it's OK. So you run this, and then what do you do? Does anybody know what you do next, to do a western? AUDIENCE: You need to use the membrane to... JOANNE STUBBE: Right, so the next thing they did was they used-- I'm going to put all of these up-- so they transferred it to a membrane. And why did they have to transfer it to a membrane to do this analysis? This is an extra step. And it turns out-- we're going to look at an antibody interacting with a protein. Why don't we just look at the antibody interacting with the protein to start with? AUDIENCE: It doesn't have access to the protein. JOANNE STUBBE: Right, it doesn't have very good access. It's really not very efficient. So people found, pretty much by trial and error, that you needed to transfer this to a membrane. I mean, we have hundreds of kinds of membranes. How did they choose nitrocellulose? If any of you have one run westerns, you remember what kind of a membrane you used? Did you use nitrocellulose? You do this in undergraduate class, don't you? You don't do a western? We used to do-- AUDIENCE: Did it once in undergrad class. JOANNE STUBBE: Yeah, in what kind of a membrane? Was it in biology? AUDIENCE: Yes, biology. JOANNE STUBBE: So what membrane? Do you remember what the membrane was? AUDIENCE: I think it was-- it was not nitrocellulose. JOANNE STUBBE: It's not nitrocellulose. So this PVDF, polyvinyl difluoride is the standard one that people use now. It works much better than nitrocellulose-- this paper is really old, and so they're looking at nitrocellulose. So then they do this. And then, what do they do next? They have an antibody-- we'll look at the details of this in a minute-- that can recognize the protein, that can find it on the membrane. And then what we're going to see is-- you still can't see anything really, because you don't have very much material there. And you can't observe-- you don't have enough to stain, oftentimes, by Coomassie, so you're going to have to amplify the signal. So then you're going to make an antibody to an antibody. And then you have to figure out how to, then, amplify the signal. And we'll look at that in a second. Is this what-- you ran a western, is this what westerns look like? AUDIENCE: I remember, we first [INAUDIBLE] non-specific proteins to occupy the sites. JOANNE STUBBE: Yeah, so that's good, you have to block everything, if you're using crude extract. So in this case, we would be using the crude mixture-- well, not a crude mixture, it's been fractured by the ultracentrifugation that's been fractionated. But you still have mixtures of proteins in there. Have any of you ever looked at westerns in a paper? Or even the papers you had to read? The paper on the PC-- go look at the PCK-- PCSK9 paper, that had westerns in it. What do you see? Do people show you something that looks like this? And if they did show you that, what would it look like? So you have an antibody that's specific for the protein of interest, whatever that is-- supposedly specific. What do you see? What do you think you see? Do you think antibodies are specific? I think I have an example of a typical western. AUDIENCE: I don't think they're as specific as [INAUDIBLE] JOANNE STUBBE: Yeah. Yeah. So when you look at a paper, you should pay attention to this when you read a paper, if you're doing anything in biology, what do you see? You never see a gel, ever. What you see is a slice of a gel where they cut off this-- the way they cut up all this stuff and all this stuff. The reason they do that is because it's a hell of a mess. So let me just show you a typical-- I don't care what kind of an antibody you're using, in crude extracts, it's a mess. Because you have non-specific interactions. We'll just look at that. So that would be something like you might see-- depending on how much antibody you have. So when you see this, the reason everybody reports data like that now. So it looks like it's really clean, but in reality-- I think if it is dirty as that, then in my opinion, I would make you publish the whole gel. But people don't do that. They just cut off the little band they're interested in-- they can see it change in concentration using this method. But you should be aware of the fact that antibodies in general aren't as specific as you think they're going to be. Yeah? AUDIENCE: Are they required to report the whole gel in supplementals? JOANNE STUBBE: I mean, I think, it probably depends on the journal, and it probably depends on the reviewer. But I would say, we're going away from data-- is something that is a pet peeve for me. And all the data, which I think is all right, is published in supplementary information, as opposed to the paper. I think if you have something really dirty, you should publish in the paper, in the main body of the paper. If you have something that's really clean, and it looks like that, it's fine with me. You don't even have to publish it, if you could believe what people were saying. Because people know what this looks like, a lot of people-- everybody uses westerns. But if it's a real mess, then you need to let your reader know that this is not such an easy experiment, and it's not so clear-cut. That's what your objective is, is to show people the data from which you drew your conclusions. And then they can draw their own conclusions, which may be different. So let's look at the apparatus to do this. So how do you get from here to here? So you have a gel, you run the gel, a polyacrylamide gel-- what do you do? AUDIENCE: Put the membrane on the gel. JOANNE STUBBE: So you put the membrane on the gel. And what do you do? AUDIENCE: [INAUDIBLE] applying charges to. JOANNE STUBBE: Yeah, so you're transferring it based on applying a voltage across this system. So here's your gel. And here's your membrane, nitrocellulose membrane. And then they have filter paper above the gel, and below the membrane. Why do you think they have the filter paper there? When you ran the gel, did you have filter paper? AUDIENCE: Yes. JOANNE STUBBE: Yeah. How do you think they decide how to do this transfer? Do you think is a straightforward? Do you run it for an hour, do you run it for five hours, do you run it for 15 minutes? What is the voltage you use to do the transfer? Do you think any of that is hard to figure out? So how do you figure that out? Somebody told you that this is a good way to do it? Yeah, so that might be a place you start. So you do it because somebody gave you a recipe. But then what do you need to do to make sure this recipe is correct? AUDIENCE: Find out what conditions that work for what you're working on. JOANNE STUBBE: Right, and then how do you do that? So that's true, every protein is going to be different. And if you have a protein-- if you have a clean protein, versus a mess of proteins, and you try to do this transfer, the transfer conditions will be different. So for example, if you really want to look at the concentration of something inside the cell, in the crude extracts, you never compare it to a standard with clean protein, because this transfer is different. So you need-- in the back of your mind, if you care about quantitating this, you need to understand the basis of the transfer. So why do you think they have these filter papers here? So this goes back to what controls you would do to see whether your transfer was working. So what would you look for? Did you do this? What did you do? What did you do with the filter papers in your-- AUDIENCE: You want to filter all to the SDS molecules... JOANNE STUBBE: You did what? AUDIENCE: You want to filter all-- JOANNE STUBBE: No, that's not what you do. I mean, you might want to do some of that, too, but in terms of thinking about whether your transfer is successful-- figuring out the conditions to blot from the gel to a piece of paper is not trivial. And there is a standard way that you do this, initially, to try. But then you have to make sure that that method is working. And lots of times it doesn't work. So it's something that's going to be experimentally determined. So the question is, what would you think would happen if you did this for six or seven hours? Whereas, a normal blot would take two hours? AUDIENCE: Would be transferred onto the filter paper? JOANNE STUBBE: Right, it would go right into the filter paper, or even off the filter paper. So what you do is you take the filter paper out, you look for protein being bound. What about the gel? What do you do with the gel after your experiment's over? AUDIENCE: Make sure a protein's not on it? JOANNE STUBBE: Right, make sure that the protein is not on it. So these are simple controls, but these are the controls you always do until you work out the conditions to make sure this works. And it's pretty critical to make sure you have good transfer. So then, so this is the antibody thing that they do. Has anybody thought about these kinds of assays? You've seen them, I think, already in class. But what's wrong with this picture? The target protein, what's wrong with this picture in the target? So here's your nitrocellulose filter paper. What's wrong with this cartoon? Should be unfolded, yeah. So you're doing SDS page, it's unfolded. So then we react it with an antibody. Presumably we have a good antibody, but you've already learned in the first half of this course that having really good antibodies is not so trivial-- you can get them, but most of the time they are not specific if you're looking at crude extracts. They have little epitopes they recognize, if you're using monoclonals that could be present in other proteins. And furthermore, how are you detecting something? An antibody as a protein, it has absorption of 280. Again, this is too low to see, so putting an antibody on it is still going to be too low to detect. So how do you detect your signal? So have you done this? I'm surprised they don't do this in your introductory class-- they don't do westerns, at all. So what you're looking at is an antibody to an antibody. So you put your antibody on, that's specific for your protein. And then you make an antibody in another organism that can specifically recognize antibodies in general. So if this is to a mouse, you make it to go and isolate that. And then what you do is derivatize the second antibody with what? A protein? That can function as a catalyst. AUDIENCE: Why can't you just derivatize the first antibody? JOANNE STUBBE: Well, what? What did you say? AUDIENCE: It's more expensive? JOANNE STUBBE: Well, no, I don't know whether it's more expensive or not. But-- AUDIENCE: Well, because you'd have to derivatize every primary antibody. JOANNE STUBBE: So you'd have the derivatize every primary antibody, and so this is a standard procedure. You could derivatize the primary antibody. So that's not a bad question. And so what you're doing now, you can buy these commercially, so they have rabbit, rabbit, mouse, whatever, antibodies. And the key is the amplification of the signal, and you use enzymes to amplify the signal. Does anybody know what the enzymes are, what the enzymes do to amplify the signal? AUDIENCE: You can covert the molecule to a blue molecule... JOANNE STUBBE: To something that's colored. So does anybody know what that horseradish peroxidase-- have you ever heard of horseradish peroxidase? So that's a heme iron-- we're going to be talking about heme irons pretty soon, and hydrogen peroxide. It makes a chemically very reactive iron oxide species, that can oxidize a dye that changes color. And it has extremely high extinction coefficients. So you can see it, and it does it catalytically and the lifetime of the dye is long enough. So it accumulates, and you can get really amplification of your signal. Or you can use a phosphatase that liberates something that's highly colored, again, and you can see it. So this is a standard method that everybody uses. And so, that's our gel. So now we're looking at sort of-- at the end already-- but we're looking at these gels, and what do you see through the different steps? So if we look through the first gradient, through the sucrose gradient, that gets us through DNE. And if you look, say, at lane E-- our goal is to separate proteins that are specifically localized in each one of these membranes. So you need to believe that's true, that people have selected the right group of proteins to look for. And you notice they do more than one. So they look at multiple proteins. Why do you think-- do you think it's easy to select the proteins to look for? And why or why not? So, they obviously have selected a group of proteins, and I think most people would agree that they've selected a good group of proteins. But what do we know now about proteins, do they stay in one place? No, they move around. But some might be present in very low amounts, sometimes in much higher amounts. And so you need to have more than one protein as a control to make sure you're looking in the right region. And what do you see in E? If you look over here, it tells you what the organelle is. And if you look at this protein, this is localized to the lisosomes-- we talked about that in class. If you looked at this protein, it's localized to the peroxisomes. So in addition to the ones we care about, the ER proteins, we're also getting proteins that are localized in other membranes. So that's when they went to the next method, and they added on another gradient to try to separate out, again, the lysosomal and the peroxisomal proteins. And you can see they were pretty successful at this. There's none of these proteins left in this gradient. So that's good. And they took it a step further. Do you remember what this is? What are they looking for down here, in this? AUDIENCE: Enzymatic activity. JOANNE STUBBE: Yeah, so enzymatic activity is localized in certain organelles. So they again did a second experiment to look at all of that. So they were very careful in this, they figured out how to separate. And that's the key thing for them to analyzing the concentration of cholesterol in these membranes. And what they looked at-- we're over time-- but is the concentration of cholesterol compared to the total amount of lipids. And how did they do that analysis? Gene Kennedy, who's at Harvard Medical School-- he's in his 90s, now-- really trained all the lipid chemists in the whole country. And they figured out many years ago how to separate lipid fractions with methanol, chloroform extract, something that you guys probably haven't though about at all. But we're really pretty good at separating things, and it's nothing more than an extraction like you do as organic chemist to purify and separate things. We've figured that out. And so then they use mass spec to allow them to quantitate the amount of glycerol. And then in the end, so they use mass spec, these western blots, and they can change the concentration of the cholesterol and do the experiments over and over again, to see what happens. And when they do that, this is the picture of cyclodextrin. So you can see the only difference is this group here versus that with a methyl. And one, so this is hydroxypropryl-- hydroxypropionyl cyclodextrin-- so it's like a cavity like this. And the other only other change here is a methyl group, removing that. And they have very different properties about binding and releasing cholesterol, which somebody had to do a lot of studying on to be able to ensure that they can use it to remove cholesterol, and then to add it back to the media. And so you have to think about the exchange kinetics, you have to think about a lot of things. This is not trivial to set this up, to figure out how to control the levels of cholesterol. And then what they do is, this is like a typical assay, and this is the end. What you can do is this, removes cholesterol, and you can see it change. This reports on low levels of cholesterol, which is happening over here, allows the protein to move to the nucleus where it's smaller. And that's how they do the correlation-- the correlation between the levels in the nucleus and the levels of cholesterol. So I thought this was a pretty cool paper. And these kinds of methods, I think, will be applicable to a wide range of things if people ever do biochemistry, looking at the function of membranes. So, OK, guys.
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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: Our subject today then is Fermi's golden rule. So that's what we're going to develop. Fermi's golden rule. Fermi's golden rule. And this has to do with the study of transitions. And typically, the interesting and sophisticated thing about this subject is that you have a transition from some initial state to a state that is part of a continuum of states. That's what makes it complicated. The transition from one discrete state to another discrete state with a perturbation is kind of a simple matter to do. But when you can go into a continuum, you have to integrate over the set of final states, and that makes it a lot more interesting. So we go from a discrete state into a continuum. And that makes it somewhat challenging. So we will consider this in two forms. It's worth considering the case of what we call constant perturbations. And you might say, well, aren't we doing time dependent perturbation theory? Yes, we are. But this kind of perturbation, you will think of it as saying that H is H0 plus a V that is time independent. But the way we think of it is that here is time, here is time 0, and here is V. V turns on at time equals 0. So there's a little bit of time dependence. There was no V before time equals 0, and there is a V after time equals 0. So it's almost like the Hamiltonian changes. And we want to see what transitions we get. Because always the subject is the subject of transitions. And we will see an application of this later today. The other case is what is called a harmonic perturbation, in which H is equal to H0 plus delta H. And delta H is going to be harmonic. 2H prime. That's conventions to put a 2 in there. It's convenient. Times cosine omega t. And this will be for t between 0 and t0. Sorry if some people can't see this. Too far to the right. 2H prime cosine of omega t when t is between 0 and t0. And it's 0 otherwise. And H prime, of course, is time independent as well. Time independent. So these are the two main cases when we will consider transitions. And we'll do the first one in a lot of detail today. The subject is trying to get the transitions understood for this case. And after that, next time, we will do the case of the harmonic perturbation that brings in a few new issues. But then, all of the rest is the same. So once you've understood how to do the constant transition, the harmonic perturbation is going to be easier. So constant perturbations will be useful to understand, for example, the phenomenon of auto-ionization. Some atoms sometimes ionize spontaneously, and this has to do with this subject. Harmonic perturbations has to do with even a more popular subject, which is atoms interacting with radiation. You send in an electromagnetic field. That is a harmonically varying perturbation, and that's going to allow you to calculate transitions. And by the time we're done with this, you will be experts in calculating atomic transitions. So one subject we need since we're going to be doing transitions between a discrete state or continuum is to describe the continuum. And for that, for the continuum, we use the concept of a density of states in the continuum. Now, we will be considering momentum eigenstates. And the momentum, you know very well the momentum of a free particle is a continuum. Takes absolutely continuous values. So there is no way you can count them or you can tell how many there are per momentum range. It's like saying how many numbers are there from 0 to 1. The way you have to do that always is by adding an extra parameter you wish you didn't have to add. And that's kind of things that we have to do in physics sometimes. We have to add parameters. You consider when we discussed last time the delta function perturbation, we broadened it, and we were able to calculate. And then we saw that the broadening size didn't matter for the result. Therefore, good. So here, we'll do the same. You know that the momentum eigenstates of a particle in an interval or in a circle are quantized. And then you can count quantized states. So we will put the whole world in a box. A big box. Size of the galaxy, size of the Earth, size of the laboratory maybe is big enough. And we'll put this parameter l there for the size of the box. And we will get the density of states because now the states can be counted. Afterwards, by the time we're all done with the transition rates of the Fermi's golden rule, we hope that length is going to disappear. So that's something we will see that it happens in calculations. And there will be good reasons why it happens. So we'll put the world in a big box. It's not that big the way I draw it, but it's supposed to be very big. Length l. And you think of the quantization of momentum by considering a wave function that would be normalized and has-- it's a momentum eigenstate. It has momentum in the x direction, y direction, and z direction. And that wave function, if all the sides of this world, this cubical world with length l, that wave function is properly normalized. You square it, and the norm is equal to 1 times this factor squared. You integrate it. It's all good. So this is a nice wave function in which the integral psi squared V cubed x over the box is 1. And then what else do we do? We decide that we have to consider periodic boundary conditions. Well, think of it as a torus, properly speaking, in which each direction is a circle. So the wave function repeats itself after x increases by l, after y increases by l, or after z increases by l. You could have chosen a box with a finite big wall at the end. It would make no difference for the counting at the end of the day. So the conditions are that kx multiplied by l should be a multiple of nx 2 pi times nx, ky times l, 2 pi times ny, kz times l, 2 pi nz. And therefore, the total number of states can be calculated by taking a little differential of this thing. We say, well, if we let kx vary by a little bit, dkx from this equation times l is 2 pi dnx dky times l, 2 pi times dny, and dkz times l is 2 pi times dnz. So if I were to consider a little interval of momentum defined by dkx, dky, dkz, a little cube in momentum space, the number of states in this cube would be dn, which is dnx, dny, dnz. And this is equal to l over 2 pi cubed d cubed k, which is dkx, dky, dkz. I say, if in this momentum range the quantum number nx contains this number of values, dnx, the quantum number ny can take a set of values, and the same for the quantum number in the z direction. The total set of quantum numbers is the product of them. So there is that many states in this little cube of momentum space-- that is, momentum space that is between some k and a little more, called dk. Well, that's a famous formula. And the n is equal to l over 2 pi cubed d cubed k. Some people know this by heart. But now we want to write this in the terms of a density of states as a function of energy. So if this is the number of states in some interval in momentum space, I can try to convert this into saying, well, all these states, because the momentum is varying within some little bounds, the energy is varying within some little bounds. So let's figure out how much the dE is. And then the total number of states will be given in this range by the number of states per unit energy multiplied by this thing. So this is the density of states, which is states per unit energy. And many times, this is the quantity we really want. We have here an energy E as well. So we think of this V cubed k as a little cube in momentum space. If momentum space has an origin here, you can imagine states with some momentum, and then a little cube in here saying how much the momentum varies. So all these states have some momentum and vary a little bit in the momentum. Therefore, they all basically have some energy up to some little variation. So how do we connect these two things? Well, you remember, E is equal to h squared k squared over 2m. And how about d cubed k? We'll try to think of d cubed k physically as all the states that have momentum k. And now, this is a space diagram now. This is x, y, and z. So here are the states with momentum k. They point in this direction. So this is the direction the states are pointing. And now we think of them as having some possible angle here. d omega. So this is the direction. So let me not say they're x, y, and z. Let me use angles here. Theta and phi, representing the azimuthal and polar angle of this direction. So we have a little range here and a little range of magnitude of momentum. So let me see if I can draw it kind of nicer. You can imagine a little cone as it grows, and the last part is a little thick piece of the cone here of thickness dk. So d cubed k is the volume of that little pillbox here. And it's k square d omega times dk. So that's d cubed k. And now to relate it to the energy we have here, this equation. So we take a differential. And it's d energy is h squared kdk over m. So d cubed k here is k dk, k d omega times k vk, and this is the same k dk here. So we can write it as k d omega. And for k dk we have m over h squared d energy. So we're almost done with our computation. We go back to this equation. And we have rho of E dE, or dn, is equal to l over 2 pi cubed d cubed k. But now we have what d cubed k is. It's m over h squared k d omega dE. So this is our formula. And this is rho of E dE. And here, for example, you have rho of E. Actually, if it were rho of E, I should have here k expressed in terms of E. Just makes the formula a little more messy. But if you think of it as a function of the energy, you have to say the words properly. What is rho of E? Rho of E is the density of states per unit energy, but when you're only counting states that have a momentum within an angle d omega. So this is the density of states per unit energy, because this gives you states. So this must be states per unit energy. But you're only looking at the states that are within an angle d omega. If you were to look at the density of all states that have energy E, you would have to integrate over omega. But many times, we want to make a transition. You have an atom and you're sending an electromagnetic wave. And you want to make a transition, and you want to see how many electrons, for example, are kicked out in some direction. So you need the density of states within some solid angle. So this is a relatively useful thing to have, not to integrate, so that you can keep control over your states and orient them at a given angle. So we've done a little bit of basic preparation. We've said what the Fermi golden rule aims to do, and it requires transition to the continuum where we'll have to use density of states. And here is an example of how you calculate the density of states.
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https://ocw.mit.edu/courses/10-34-numerical-methods-applied-to-chemical-engineering-fall-2015/10.34-fall-2015.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: OK, let's go ahead and get started again. So we had a nice review on Monday of some of the different concepts that we've talked about up to this point in the course. We've covered linear algebra, solutions of systems of nonlinear equations. There's been a lot of fundamentals involved in that. Things are going to get more applied as the sorts of problems we're trying to solve get more complicated. Complicated problems require very sophisticated methods to solve them. It's not going to be our expectation that you're necessarily able to implement all of these sophisticated methods. So for example, you implemented Newton's method on your last homework assignment. That'll be the last time we ask you to do that by hand. You'll utilize tools built into Matlab to solve other systems of nonlinear equations going forward. You understand how it works and what the limitations are. You understand how to build different tools that will be inputs to these functions like the function you're trying to find the root for, or an explicit expression for the Jacobian. You even understand how to set certain parameters associated with these methods, and what those particular parameter values mean. Like setting the function norm tolerance, or setting the step norm tolerance associated with these methods. Setting the number of iterations you're going to let the method go. You've implemented it, so you understand what these things mean, what the consequences of choosing them are, how they might increase the time associated with computations that you're going to conduct. So we've done a lot of fundamentals. Maybe that was difficult for some of you who haven't seen this before. Now things are becoming more about the practice of numerical methods. How do we actually solve things reliably, and that's what today's lecture will talk about. We discussed solving systems of nonlinear equations and needing good initial guesses for our methods. Bad initial guesses can lead to unpredictable results. Good initial guesses, we know the Newton-Raphson method is going to double the number of accurate digits each iteration, and we should converge very rapidly to a solution. So it seems like having good initial guesses is really ideal. And this topic, homotopy and bifurcation, will cover precisely those issues, and the practice of getting these good initial guesses. I think is one of the most useful things you'll learn in this class, actually. It's a really nice way of going about solving complicated problems. Before we begin, I want to address a question I got via email last night. That's always good if these questions go on [? Piaza ?] so everyone can read them, but I thought it was a very nice question about how do you find-- the question was how do you find multiple roots using the Newton-Raphson method. So if I have a polynomial, where the function doesn't cross through the origin, but it just touches the origin, and goes back up. That's a root with multiplicity in the polynomial. And can the Newton-Raphson method find that? Because the derivative of the function is 0 at the root. So who says the Newton-Raphson method will be able to find those roots? What do you think? Yes? And who says it won't be able to find those roots? Or who doesn't know? You don't know. That's OK. That's a good question, right. Do you know or don't you know? So we know the Newton-Raphson method relies on knowing the derivative of the function at different points in order to take our next Newton-Raphson step towards that solution. It may be the case, the derivative is 0 at the root, but nearby the root, the derivative may not be 0. And then the Newton-Raphson steps are going to be fine. They're are always going to keep stepping down towards the root. If you go through an analysis of how the algorithm converges, you'll find out that you only get linear convergence in that case, instead of quadratic convergence. So there will be some penalty to pay for the fact that the derivative is 0 at the root itself, but it's not going to be so bad. The method can still converge to those sorts of solutions. Not all is lost there, but the rate of convergence may be not as good as you expected it to be. And the same is true in the multi-dimensional case as well. It's a little hard to imagine what a multiple root is going to look like. It's something like tangent curves in the solution of plane. So if I have f1 of x1 and x2 equal 0, that's a curve in the x1, x2 plane. And I have f2 in that same plane equal to 0. The place where they touch might be one of these multiple roots. That's a place where the determinant of the Jacobian may be equal to 0. You'll have the same sort of circumstance. Nearby, the Jacobian can be non singular, and you can take steps towards the root, but there's a worst case scenario where you're coming in on a direction that's parallel that belongs to the null space of the Jacobian at the root, and then everything's going to slow down. You can still converge, but it may be very slow as a consequence. I thought that was a really nice question. It's a detailed question about how Newton-Raphson works. It's not necessarily going to fail under these weird circumstances. They don't come up too often. They're going to come up in our discussion today, actually. It's not going to fail, but it may converge very slowly to the solution. OK? Good. I'm going to remind you, the last thing we did with Newton-Raphson methods, we talked about one way of fixing them. In lieu of having good initial guesses, we may not want to take the full Newton-Raphson step. We may want to do these quasi Newton-Raphson methods instead. One of those was this damp Newton-Raphson where we introduced some damping parameter. We don't take the full step, we take some shorter step, and we try to make the absolute value over function become as small as possible. That's an optimization problem. It's as hard as finding the root of the system of equations that we're looking at. So we don't solve that optimization problem exactly. That's crazy. Instead we use this backtracking line search method where we change our damping parameter in some systematic way until we get to a point where we're satisfied. Where we say OK, in the process of changing this, we've actually found a smaller value of our function, and we accept that as our best guess for where the next iterate is supposed to go. And this sort of approach can give you global convergence for the algorithm, which is great, but it's not globally convergent to roots. It's globally convergent to minima, asymptotes, and roots as well. But it's globally convergent. So it'll terminate. The algorithm will stop, and when it stops, you can ask is this a root or not? Do I think this is sufficiently close to a root or not? If it's not sufficiently close to a root, then you know you terminated somewhere like a minima, or you rushed out towards an asymptote, and maybe you need to start with a new initial guess for your solution. These quasi Newton-Raphson methods are very useful. This is what Matlab uses in order to-- it uses a more sophisticated version of this called the Dogleg method, but it's the same principle. It's trying to make sure you're not taking steps that are too big. Trying to make sure you're taking steps in the right direction so that you're approaching the roots of your function. Or at least reducing the absolute value of your function as close to zero as you can get it. Are there any questions about this? Anything we need to follow up on in the previous lecture? No. OK, good initial guesses. This is the key, really, for solving systems of nonlinear equations. It's the key for doing optimization problems too. We need to have some idea of what our solution looks like, and we want guesses that are close to those solutions so that maybe we are in this region of local convergence for the Newton-Raphson method, and everything converges very quickly. So where do they come from? This is an important question, and I'll show you how you can get them. There's another issue too, which is related to this, which is the non-linear equations, they can have multiple roots. If we do optimization problems-- that's our next topic-- they can have multiple minima or maxima that we're trying to seek out. Depending on our initial guess, our numerical method may find one of these roots or another root. It may find one of these minima or another minima. Maybe we're looking for the global minimum, so we need to look at them all. Are there methods that we can use to find them all? So we'll talk about continuation in homotopy, and we'll talk about bifurcation. And it turns out there are methods one can use to find all the roots for a system of non-linear equations. In fact, the homework assignment you have this week will ask you to do ternary phase equilibrium problem. Vapor liquid phase equilibrium with three components. There are five parts to this problem. Four of them, I think, are relatively straightforward. They're about finding the pure component roots or randomly guessing some initial conditions for your nonlinear solver in order to find all of the different azeotropes in the system. There's a fifth part that asks you to actually implement this homotopy method, and try to find all the roots in one go with your algorithm. Like the other homeworks you've had, that part's a little more tricky. Usually these homeworks have one thing that's a little bit more tricky than everything else. That part is a little bit more tricky, but it's really instructive to try to do it. We'll learn how to do it today. You can find all of the coexisting compositions of this ternary phase diagram, simultaneously, with one go through this algorithm. So let's talk about continuation first. This is an interesting idea, and it relates to trying to find roots of these equations, having good initial guesses. Here's a simple example. We have a third degree polynomial, and it has three roots. We can look, graphically, because this is one dimension, and see that one of these roots is near minus 1 and 1/2, another root is near a 1/2, and another root is near 1. And so I can go to my non-linear equation solver, and give it initial guesses, minus 1/2, 1/2, and 1, and I'm probably going to find these roots pretty reliably. But in many dimensions, we can't do these sorts of graphical approaches. You can't look at this function. It may be difficult to actually get enough data to even look at it if it's a two-dimensional set of functions. It's hard to see where these roots are, and so continuation is a method of transforming the problem into an easier one to solve. So let's take this problem, this third order polynomial, and let's transform it. Let's replace the 2 with a 2 times the lambda, and let's try to find the roots of this polynomials as lambda grows from 0 to 1. This is sort of a weird idea, but we actually know when lambda is equal to 0, this goes away, and there's only one root to this polynomial-- one real root-- which is minus 1. We actually know the answer when lambda is equal to 0. If I make lambda a little bit bigger than 0, there will be a root which is close to minus 1. In fact, minus 1 is a good initial guess for that root, and my non-linear equation solver ought to converge very quickly to that value. If I choose lambda to be small, then it should be in that region of local convergence of my Newton-Raphson method, and I very quickly should find the next root. So I can step lambda along. I use the lambda equals 0 root as an initial guess for the next root, and I keep increasing lambda from 0 all the way up to 1, and when lambda is equal to 1, I'm going to find the root to the initial problem I was looking at. Does that makes sense? So I'm continuing from one set of solutions to the next. I vary this parameter continuously, and for previous values of that parameter, they serve as initial guesses for the next solution. Here's what a code that does that would look like. I define my lambdas. We don't know good ways to define the lambda. I've got to figure that out for myself, but here lambda is a vector that goes from 0 to 1, in 1/100 increments. Maybe that's too fine, but that's OK. And my initial guess for the solution when lambda equals 0 will be minus 1, and I loop over all the values of lambda, and I use f0, which is the one-dimensional non-linear equation solver in Matlab, to find the roots of this polynomial function-- x cubed minus 2 lambda x plus 1. Here's the solution and that becomes my initial guess for my next solution. So this loop is going to go round and round until I get lambda equals 1, and the solution that results there should be the solution to the initial problem I was interested in. Here's what that looks like graphically. I started with lambda equals 0. The root there serves as an initial guess for lambda equals 1/100, and so on, and these roots converge eventually, to something that's close to minus 1 and 1/2. So the actual value is a little bit bigger or a little bit smaller than minus 1.6. Here's one of the roots. So rather than having to have a precise initial guess, I transformed my problem from an easy to solve one to a more difficult to solve one, and I used the path along that transformation to give me rapid convergence to better and better initial guesses to the problem that I was interested in. Make sense? There's another way of doing this too. You don't have to go from 0 to 1. Let's go from lambda equals something big back down to 1 instead. So if lambda's big, this term dominates the polynomial equation. This is the biggest term in the polynomial equation, and it can either balance x cubed, or it can balance 1. It can't balance against both of them, it turns out. That So if we have this term large, and it balances against 1, then when lambda's large, an approximation for the root will be 1 over 2 lambda. If x equals 1 over 2 lambda, and these are the two largest terms in the equation, then these two will cancel, and f will be very close to 0. So that's one root when lambda is very large. If I balance these other two terms against each other, I'll see there are two other roots, which are plus or minus the square root of 2 lambda. So let's start with a large lambda instead, and these as initial guesses for the roots-- and let's trace them back to lambda equals 1. When lambda equals 1, we recover the problem we were interested in before. So here, I start with three different initial guesses, and I trace them back, and I'm actually able to find all three roots of the initial equation. So there's something close to minus 1 and 1/2, something close to 1/2, and something close to 1. So in one go, I found all three roots using continuation. Is this clear so far? Sam. STUDENT: Is that one go or is it three separate [INAUDIBLE].. PROFESSOR: That's an excellent question. So I had to initiate this process with three separate guesses, but it turned out in this case, all three guesses converged to three distinct roots in the end. it's one go in the sense that there's one trajectory that I had to follow, one passive lambda as I had to go through, but I had to solve three times to follow each of these three paths. There's actually no reason to stop at lambda equals 1. I stopped there because that's the solution to the problem. I'm interested in, but there's actually something richer to see here. If I keep decreasing lambda, eventually these two branches combine with each other, and then they jump down, and they follow the original branch that I had that went between minus 1 at lambda equals 0, and this root about minus 1.6 at lambda equals 1. So the paths that these trajectories follow can be very, very complicated, actually. A little bit unwieldy or unpredictable. But oftentimes, they can result in predictions of all three roots or multiple roots associated with the system of equations. Yeah. STUDENT: Going back to the first [INAUDIBLE] solve from lambda equals [INAUDIBLE] PROFESSOR: Zero to one? STUDENT: Zero to one. So then would you find only one solution there? PROFESSOR: Yes. So going from 0 to 1, I found one solution. Going from infinity-- 10 is big enough. 10 to 1, I was able to find three. I didn't know beforehand that I'm going to be able to do that. It happened to be the case that we could with this by construction. This is a particular problem where it works out that way. Other questions? So this is neat. Like we've done before with iterative methods, we turn hard to solve problems into a sequence of easier to solve problems. Good initial guesses converge fast with Newton-Raphson. We have a sequence of better and better initial guesses, which all converge very quickly to eventually, the solution that we're looking for evaluated at lambda equals 1. So this is one way of getting good initial guesses. You have a sequence of good initial guesses. Make sense? So we're just here. We had this polynomial problem and we injected lambda in a particular place. We injected it there because I knew it was going to work. I knew the problem could be transformed in this way in order to work out and distinguish this. It's not always obvious how to transform a problem to make it look this way, but oftentimes we have physical problems in which there's a parameter that we can do continuation with that is obvious. So think about fluid mechanics problems. Those are nonlinear partial differential equations. So we're trying to solve say, the Navier-Stokes equations. Rather than jump in and try to solve that right away, we might do continuation. We might find a solution at very small Reynolds numbers, where the equations are almost linear, exact solutions are known in that case, and we might progressively step the Reynolds number up, and use our solution set small Reynolds numbers as initial guesses for increasingly large Reynolds numbers until we get to the one that we're interested in. You can imagine doing that in mass transfer problems where you vary the Peclet number. In multicomponent phase equilibria problems, this might happen by varying temperature or pressure until we get to the temperature or pressure that we're interested in. If we have reaction equilibria problems, we might vary the different reaction rates until we get to the right balance of reaction rates to determine the equilibrium compositions. So they're often natural parameters in the problem that we're free to adjust. We can transform from easy to solve problems to hard to solve problems in a continuous fashion. It's not always going to work out the way that I showed. It's not always going to be so nice and neat. It may be quite a bit more complicated, and sometimes there are problems for which there isn't a natural parameter available to vary. In those cases, there's another strategy one can employ. It's called homotopy. This is the transformation from one problem into another problem in a continuous fashion. So oftentimes, we're seeking a set of roots. x star, which are a function of a parameter lambda, and they're the roots of an equation that looks like this. So they're the roots of this function, h, which is a linear superposition of a function f and a function g. When lambda equals 0, h is equal to g, and so the roots of h are the roots of g. And when lambda equals 1, h is equal to f, and so the roots of h are the roots of f. So we might transform-- we make very lambda continuously between 0 and 1 in a construction like this. So we might transform the roots from the roots of g into the roots of f. Maybe the roots of g are easy to find, but the roots of f are hard for us to find. There's a smooth transformation though, from g to f. When we have those sorts of smooth transformations, there's a chance things will be well-behaved, or at least where we encounter bad behavior-- and I'll show you what I mean by bad behavior in a minute-- there's a reliable way to interpret what's going on there. So we vary lambda in small increments from 0 to 1, and the solution x star for some lambda i is used as the initial guess to find the roots at some lambda, which is a little bit bigger than lambda i-- our next iterate in lambda. Does this idea make sense? Where do f and g come from? So this is a big question-- f is the problem you want to solve, typically, and g is some auxiliary problem. It can be arbitrarily chosen. There's nothing to tell you why one g should be better than another one, but oftentimes, we choose them in a physical way. So there will be a physical connection between the f problem and the g problem, and I'll show you that right now. So here's an example. We looked at this example before. We want to find the roots of the van der Waals equation of state. It's written in terms of the reduced pressure, the reduced molar volume, and the reduced temperature, and let's find those roots for a given pressure and temperature. So this is a function f of the molar volume equals 0, and we're looking for these three roots here. There's a root down here, which is the molar volume is low. Not a lot of volume per number of moles, so maybe this is a liquid like root. Out here, the molar volume is high. We have a lot of volume, the number of moles and materials. This is a vapor like root, and then there's an unstable root in between them. So three roots to find. We'd like to be able to find them in one go, or we'd like to have some reliable mechanisms for finding them. Homotopy is one way to do this. So let's construct a new function h of the molar volume, which is lambda times f. We want to find the roots of f plus 1 minus lambda times g. And as g, let's use something physical. So let's let g be a function that represents the ideal gas equation of state. So PV is nRT, but I've written everything in terms of the reduced pressure, molar volume, and temperature of the van der Waals equation of states. So you pick up this factor of 8/3 there. But this is the ideal gas equation of state. Its roots will be the roots of the ideal gas equation of states. I'm going to transform from something that's easy to solve. I know how to find the molar volume here, there's just one, to something that's hard for us to solve. It has three roots-- the roots of them van der Waals equation of state. So when lambda equals 0, I'll get the roots associated with the ideal gas, and then lambda equals 1, I'll get the roots associated with the van der Waals equation. Yes? STUDENT: If you're starting with lambda as 0 and you get the roots of g, you only get one root, but you're trying to find two roots [INAUDIBLE].. So how does that work? PROFESSOR: I'll show you. Here's what the process might look like. I've got to start with a reduced temperature and pressure. Here's my initial guess for the molar volume. It's the root of our ideal gas equation. Here's the definition of f, and the definition of g, and the definition of h. l times f plus 1 minus l times g, and I'm going to loop over all of my lambdas, solve the equation subject to some initial guess, and update my initial guess every time I update lambda. That's fine. That's the code you. Can implement the code if you want or play around with it. So what happens if I try to carry this out? So I'm going to vary lambda between 0 and 1. I start with my initial ideal gas guess for the root, and as lambda grows to 1, the molar volume shrinks until I get to lambda equals 1, and I've found one root of my equation. One root, good. There's not necessarily any reason to stop at lambda equals 1. That's the root I wanted, but there's no reason I have to stop there. So I could continue. If I keep pushing lambda up higher and higher, eventually I'll get to a point where all of a sudden the molar volume jumps down to a different place. This is the result of the algorithm. I just told lambda to go bigger. But there will be a discontinuity or a cusp here that I jump off of, and I find myself on a different solution branch, which is sort of crazy. So now I can do something clever. I found myself on a different solution branch. Why not try decreasing lambda along that solution branch instead? Maybe I'll find a different solution, and it turns out that's what happens. All right, so i jump down to this other solution branch, and now I try decreasing lambda from 2 back down to 1, and sure enough, I found the vapor molar volume, and now I found the liquid molar volume. There's no reason to stop and lambda equals 1. So if I keep decreasing lambda, eventually I'll hit another cusp, and I'll jump back up to the original solution branch. So I do my homotopy. I vary lambda from 0 to 1, and I find one root, but if I keep going, I might identify a cusp, and jump off that cross. I might reverse my direction with the homotopy, and I could find another root. I might hit another cusp as well. Now, when I hit a cusp, maybe it's best not to jump off the cusp, but instead to reverse direction at the cusp, and try to trace it out the other direction. And if I do that, if I get right to this last point, and I try to reverse direction again, I can trace out one more solution path, and I can find the third root to the equation. So in one go, if I'm paying attention to how my roots are changing-- if they change by too much, I might switch the direction with which I change lambda, and I could find all of these roots at once. So three roots in one go. Yes. STUDENT: Is this like trial and error? How do you know when to change? PROFESSOR: This is a wonderful question. So in this case, you can do it by trial and error. You can detect these jumps. That's even better. You look for jumps in the value of the solution that are of sufficient magnitude, and when you detect them, you can reverse the direction. There's an even more systematic way to do this, which is called arclength continuation. Give me one second, and I'll give you slide on arclength continuation, which is a systematic way of doing exactly this process of tracing out this path in the solution plane. Roots versus homotopy parameter. These cusps here, are sometimes called turning points, and they have a special property, which is the determinant of the Jacobian of the homotopy-- the homotopy function-- is going to be equal to 0. The Jacobian is going to be singular right at these cusps. So you can detect them by tracking what the Jacobian is doing right. The Jacobian is going to be non-singular at most of these other points, but when you hit one of these turning points, the Jacobian will be singular. There isn't a well-defined direction for the homotopy to proceed along. It hits this cusp where the slope is infinite in this plane. There isn't a well-defined direction to step along here. The question was-- OK, it seems a little ad-hoc. You've got this 2D plane, and it's easy to guess when you jump from one solution to another, and trace the path back. That's fine. There's a systematic way of doing this, which is called arclength continuation, which says I know that there's some curve in this solution plane. There is some curve here. I've shown it to you graphically. Let's try to parametrize that curve in terms of a measure of the arclength. Call it s, the distance along the curve. So the roots are a function of lambda, which can be a function of the distance along the curve, and lambda is a function of the distance along the curve, and there is a constraint which says s has to satisfy an arclength formula. This is the formula for the arclength in this plane of roots versus lambda. The derivative of this function with respect to s squared plus the derivative of lambda with respect to s squared equals 1. That defines s as a measure of length along this curve. You learned arclength formulas in the context of integration at one point. This is the differential version of that same formula. So I've got a differential equation. Well, it's actually a differential algebraic equation. This is an algebraic constraint that needs to be satisfied for all values of s, and it depends on how the roots change with s. We'll learn how to solve differential algebraic equations in the two sections from now. So we don't quite know how to solve something like this yet, but if we know how to solve those sorts of equations, then it's clear that we can trace out this entire curve in the solution plane, and then go back and try to find the places where the curve intersects lambda equals 1. So if we follow this path by solving this equation from s equals 0 out to some very large s, then we should be able to find all of the roots that are connected to the path. Yes. STUDENT: If there's a turning point that [INAUDIBLE] then does that imply multiplicity? PROFESSOR: Yes. So it's a real problem. That turning point singularity is actually built into this arclength formula as well. This is the derivative of x with respect to s squared, which takes the direction you're moving in out of the problem. This is a length, not a direction. So when we get to these turning points, the solution methods for these equations are going to have to know that I was headed in a downward direction first. I can get to this turning point, and then the question is which direction do I move in from the turning point? Do I go up or do I go down? Which way does the solution path change? The slope isn't defined there, so it's quite challenging to figure out exactly how to proceed. But it is what you say. It's like the solution has multiplicity at that point. Curves are tangent in the solution plane. It's a great question. OK, so that's the notion of arclength continuation. We may talk about this later when we talk about differential algebraic equations, and we'll talk about some equivalent sets of equations that we know how to solve, but they have to satisfy this constraint that s is a measure of arclength along the curve. But there you go. You can find all the solutions, at least all of the solutions connected to this path defined by the homotopy equation. You can find all of them in one go using some clever methodologies. Oftentimes, we're not just worried about these turning points or cusps, we encounter bifurcations in the solution. So we may have a path that has one root along it, and then that path may split, and there may be more than one root. Here's a simple example. Find the real roots of x cubed minus rx. If r is negative, there will always be one real root, and it'll be x equals 0. If r is equal to 0, there will still be one real root, it's 0. That root will have multiplicity 3 in that case. And if r becomes positive, I'll suddenly have three real roots instead of one. So as I vary the parameter r from a negative number to a positive number, I'll go from one real root to three. In the solution plane, that will look like this. Here are the roots as a function of r. r is less than 0. All the roots are equal to 0. When I hit r equals 0, I have multiplicity associated with the root, and the solution bifurcates, and there will be three possible solutions now. If I was trying to do a homotopy problem where I very r continuously from some negative number to some positive number, I'll hit this point, and I'll want to follow all three paths to find solutions out here at positive r. But I can detect those points. Just like I was said before, at these weird turning points or bifurcations where we have multiplicity in the roots, the Jacobian in is going to be singular. The Jacobian of the homotopy is going to be singular. Its determinant will be equal to 0. So I can detect these points by checking the determinant of the Jacobian. And when I find one, I want to do something smart to try to follow these paths. Your homework this week has a problem like that. You're trying to find all the azeotropes of this ternary vapor liquid equilibrium system, and you'll have a homotopy parameter that you change, and it'll bifurcate. OK, so as you change the parameter, you'll go from one path to two, and then those two paths will split again. You'll be splitting from a single component solution into a two component azeotrope, into a three component azeotrope. And you'll want to detect where these bifurcations occur, and try to follow solution branches off of those bifurcations, and there's some good hints in the problem for how to do that. So occasionally, we'll encounter problems that switch from having one solution to having many solutions. You've seen how this happens in the discontinuous sense with these turning points. Here, for a given value of lambda, all of a sudden it's clear that not only is there one solution here, but there's another solution down there. My path has to choose. Do I want this solution or the one below? I jump off this turning point. That's a sort of bifurcation. These are continuous bifurcations. They're often easier to take care of. So the bifurcations in a homotopy let us find multiple solutions. Every time we detect a bifurcation, if we just track those solution branches, we split our algorithm up to track each of the branches separately, they'll terminate at different roots to the homotopy equation. And when we get to the value of lambda equals 1 in our homotopy, we're there. We found multiple roots to the original equation. These are often of great physical interest. Do you know what sort of things these bifurcations represent? One occurs in the van der Waals equation. So if I start at very high temperatures and a given pressure, the van der Waals equation will have how many real roots? One-- a gas. And then as I turn the temperature down, there is a propensity for liquefaction of that gas phase. The formation of a liquid in coexistence with the vapor. So I'll go from having one root to having three roots all of a sudden, and there will be this bifurcation point where that transition happens in a continuous fashion. STUDENT: Can I ask you a question? PROFESSOR: Yes. STUDENT: [INAUDIBLE] PROFESSOR: Yeah. STUDENT: So as you early put it out that the turning point, the determinant of the Jacobian [INAUDIBLE] including lambda is singular. But here you're writing j of x. Is that [INAUDIBLE] PROFESSOR: Excuse me. This is a typo. So there should be a little h here. This is the determinant of the Jacobian of the homotopy equation at that root. Where there's a bifurcation, the Jacobian will be singular. I apologize. That's a little bit-- STUDENT: And that lambda. PROFESSOR: And that lambda. STUDENT: Just to make it clear, you have n state variables. Like x is how many unknowns you have, and then you're adding another unknown to lambda, another parameter. This Jacobian has n plus 1 dimension. This is also [INAUDIBLE] Is this correct? PROFESSOR: Well, no. Actually, the Jacobian with respect to x only is also singular at these turning and bifurcation points. STUDENT: [INAUDIBLE] as well. PROFESSOR: Yes. It's also true what you said, that if I have the Jacobian, and I take the derivatives with respect to lambda, that will also be singular, but it will be singular because the subspace, which is the derivatives with respect to x, gives you a singular component. The condition is the same as here. So it's the Jacobian of h, taking the derivatives with respect to x-- all the different x's of h. The determinant of that matrix at a particular lambda will be equal to 0, and the value of lambda will be the lambda at which you have this turning point, or which you have a bifurcation. Here's what it looks like in a two-dimensional sense. So I'll have my homotopy equation where I'm trying to find the roots of h. Say it's a two-dimensional-- x is dimensions. So it has x1 and x2, and these curves represent h1 equals 0 and h2 equals 0, and they intersect at one point for a given value of lambda. There's the root. Now I increase the value of lambda. The homotopy equations change. So these curves change shape, and all of a sudden I go from crossing curves to tangent curves, and we know tangent curves in the plane will have a Jacobian which is singular. And this is the bifurcation point. These two curves become tangent, and as I go to a bigger value of lambda, they'll change shape again, and I'll now have multiple roots. I'll bifurcate from one solution to two, or one solution to three, but the bifurcation will occur through this transition where I have tangent curves, and we know there, the determinant of the Jacobian is 0. There's a singularity under those conditions. Does that makes sense? OK, so a couple of things about the practice, and I'll give you a simple example you can try out, which I think is an interesting one, if pathological. So in practice, it's hard to hit this bifurcation point exactly. And I'm telling you that you should be detecting these branches. You should follow these branches, but clearly, I've got to know where this point is in order to follow the branches. That's a problem. If the determinant of the Jacobian is 0 at the bifurcation point, then it's going to change from positive to negative as I cross the bifurcation point with lambda. Actually, I don't necessarily have to find the point where the determinant of the Jacobian is equal to 0. What I really should do is try to track the sign of the determinant of the Jacobian, and see when it goes from positive signed to negative signed. And when it does that, it will have crossed through one of these bifurcation points or one of these cusps. That's the point at which I should stop, and say, OK, there's possibly other solution branches to follow. So as I step from my smaller lambda to my bigger lambda, the sign changes. I want to step in some different directions than the direction I was heading before, and those different directions will trace out the other solution branches. It turns out those different directions belong to the null space of the Jacobian. So if you find vectors that are very close to the null space of the Jacobian, those directions are the directions you should go to find these other solution branches. So it's very interesting. Yeah. STUDENT: Can you have a Jacobian that goes from positive down to 0 and [INAUDIBLE]? PROFESSOR: That may be possible. You won't encounter that in the problem that you're doing. That would be a very difficult problem to do branch detection on. Let's Suppose that was the case and you want to find exactly where that bifurcation point is. These bifurcation points are of physical significance. In the van der Waals equation of state, that bifurcation point is the critical point. It's the critical temperature at which that phase separates. Maybe I want to know that exactly. Well, all I need to do is solve an augmented system of equations. So let's solve a system of equations, which is our original hematology equation equal to 0, and determinant of the Jacobian of this homotopy equation equal to 0. This is if h had dimension n. This is now n plus 1 equations, and let's solve it for n plus 1 unknowns. Let's solve it for x and for lambda. x has dimension n, and we add one more unknown-- the value of the homotopy parameter at which this bifurcation occurred. So we can find that point exactly. So there may be cases where this happens where it doesn't continuously transition from positive to negative. It may hit 0 and go back up, but that's a really challenging case to handle, but you can handle it. You can find precisely the point at which this bifurcation occurs by solving these augmented equations. It's just another non-linear equation to solve. Yes. STUDENT: I'm not sure about the previous slide. So in the previous slide, why is the [INAUDIBLE] PROFESSOR: Let me go back to here. So let's let the blue line here, be the place where the first element of h is equal to 0. So this is all the values of x at which the first element of our homotopy top equation is equal to 0. And the red curve here, let's let that be all the values of x at which the second element of h is equal to 0. And so h itself, is only equal in 0 where both of these curves are equal to 0, and that's their intersection. So they intersect here, and this star is the solution. It is the root of this non-linear equation up here. Now, I increase lambda. I make lambda a little bit bigger. That changes h. h is a function of lambda. So h changes, and these curves change shape in the plane. And at that bifurcation point, these two curves will become tangent to each other. There will be one root, but in this solution plane, the two curves will be tangent. And we saw before that when we get these sort of tangency conditions or the equivalent in higher dimensions, the Jacobian of our function we're trying to find the roots of will be singular. It's like a multiple root. And when we increase lambda further, curves change shape again. I'm not changing the red one because that's complicated. I just changed the blue one for you as a graphical illustration. So the blue one changes shape, and now I've gone from having one solution to having three. So in order to go from one solution to three, in a continuous fashion, I have to go through this tangency state, and that defines the bifurcation point. Does that make sense? Here's a simple example you should try out. So I'll present it to you, and then you should try to solve it in Matlab. You actually know the solution, geometrically, so it's not too hard. So here's a function. The first element to that function, it's a function of two unknowns-- x1 and x2. The first element to that function is an equation for a circle of radius r with center at minus 3 minus 1, and the second element is an equation for a circle of radius r with center at 2, 2. So there's two circles. We want to find the radius where the circles just touch, and that point is a bifurcation point. If r is perfect, these two just touch. If r is a little too big, the two circles overlap, and there are two solutions to this problem. And if r is a little too small, there are no solutions to this problem. So the perfect r for touching is a bifurcation point. It has all the properties of a bifurcation point, just like we discussed. It's a multidimensional equation and it bifurcates. You know the solution to this problem, geometrically. You could work it out in a couple of minutes, but you could also solve it using these augmented equations. So we know that point at which the circles just touch, the value of r at which those circles just touch, will be the place where f of x is equal to 0 at which these equations define a circle, and at which the Jacobian of these equations-- the derivatives of f with respect to x, it's determinant is also equal to 0. So we want to solve this system of three equations for three unknowns. The values of x1 and x2, where the circles just touch, and the value of r, the radius of these circles at which they just touch. That would find the bifurcation point. That will find the touching of the circles. Can you see that, geometrically? All these bifurcation problems work exactly that way. There's some homotopy parameter. In this case, that's r. But there's some parameter we're varying as we pass through that bifurcation. We'd like to know the values of our unknowns and the value of this parameter at that bifurcation point or at that critical point. So you find this point by solving the augmented equations. We have a non-linear equation-- three nonlinear equations for three unknowns, and you can do that with a Newton-Raphson iteration. It's a non-linear equation. We know Newton-Raphson is the way to solve these things. To do Newton-Raphson, that means we need to compute the Jacobian-- this is confusing. We've got to compute the Jacobian of these augmented equations, the derivatives of these augmented equations with respect to our augmented unknowns. So that will be the derivatives of f with respect to x, the derivatives of f with respect to r, the derivatives of the determinant of the Jacobian with respect to x. That's this gradient here. The derivatives of the determinant of the Jacobian with respect to r. This is our augmented Jacobian. The augmented Jacobian times the Newton-Raphson step. That's equal to minus the function we're trying to solve evaluated at the previous iteration. Solve the system of equations, you'll get your Newton-Raphson step, and you can find the solution to the augmented equations. It'll converge quadratically. You should commit yourself, you can figure this out. Don't try to compute all of these derivatives. That's complicated. You're going to make a mistake. I'm going to make a mistake if I try to do that. What do we do instead to compute the Jacobian? How do we compute these derivatives? What's that? Finite difference, right. Don't play games with it. Try to use finite difference to compute the elements of this instead. You know the exact solution too, so you can use that as an initial guess, or something close to the exact solution as an initial guess to see if you can find this bifurcation point. So homotopy bifurcation, there are ways to get good initial guesses by continuously changing our problem. We get a succession of problems that will converge faster because we have good initial guesses, and that will give us a reliable convergence, hopefully, to one of the roots of the equation we're after. If we're able to pick up solution branches, we can track these cusps, or find these bifurcation points, we might be able to find multiple solutions. There are problems for which depending on your choice of g in this homotopy function, maybe some of the solution branches are disconnected from each other. Boy, that's crazy and complicated. We're not going to worry about those sorts of cases, but they can pop up and happen. But the point is there are a robust strategies you can employ to find multiple roots of these equations or multiple minima in optimization problems using precisely this technique. You'll have a chance to practice this on this week's homework assignment, and I think you should take a look at this problem. It's simple. It looks complicated , but if you can understand this from, then you'll understand everything you need to know about bifurcation and homotopy. Any questions? Yes. STUDENT: At a bifurcated point, we go from having one root to three roots, per se. And you said that when that occurs, we need to follow each of these solutions, right? PROFESSOR: Yes. STUDENT: And you said that we need to step in a different direction. So what does that exactly mean? In a 2D sense, we have this stepping forward or stepping backwards. PROFESSOR: This is a wonderful question. So let me go backwards here. Here's the bifurcation point, and when we go through that bifurcation point, there are now going to be three solutions. And I'm going to need some initial conditions for my solver to find each of these three solutions. I can't give them all the same initial condition. I'm going to try to track all three of these branches with different initial conditions. If I give them all the same initial condition, and my algorithm meets the minimum threshold for what an algorithm should be, they're all going to converge to the same root. So that's useless. So I need to give them different initial conditions. So what are those supposed to be? In your problem, I'll give you explicit directions for how to find those. It turns out those initial guesses should be perturbed in directions that belong to the null space of the Jacobian. So I want to perturb from that one solution, which was here, in a direction that's parallel to the tangency of those curves. Those two curves were tangent with each other. In the other solutions, they're going to sit off in the direction that's tangent to those curves. So I've got to find the vector that's in the null space, or very, very close to the null space, and that will tell me how to get better initial guesses for picking up these other solution branches. It's not trivial. It's actually quite challenging. You'll have a chance to practice. It's part five of five parts on this problem. The first four, they're relatively straightforward and easy. Part five is a little bit tricky. You get explicit directions, and the TAs know that that's the most difficult part of that problem. So we'll help you through it. It's something interesting to think about trying to do. You may find you're only able to pick up some of these solution branches, but not all of them. That's OK. There are ways to pick them all up if your algorithm is designed well enough. You may just pick up some of them, and that may be the result you turn in. You can look at the solution to figure out how the other ones are picked up as well. But you're looking at the direction along these tangent curves. That's a vector in the null space of the Jacobian. And if you perturb in those directions, then you can trace out these other roots so you can find the bifurcation. It's a wonderful question. You'll get a chance to practice on it. OK, last question. STUDENT: [INAUDIBLE] PROFESSOR: Yes. STUDENT: Is this supposed to be jrx? PROFESSOR: That's the Jacobian of f. That's the derivatives of f with respect to x. STUDENT: Not including r? PROFESSOR: Yes. Not including derivatives with respect to r. Just derivatives with respect to x. So this is the Jacobian of f. The derivatives of f with respect to x. The determinant of that will be 0 at the bifurcation point. It'll turn out that-- this will be the last thing and then I have to get off the stage here. It'll turn out that this matrix is also singular at the bifurcation point. It will be singular because the subspace j is singular. So it'll turn out to be singular as well. Don't worry about it. The condition for the bifurcation point is that the determine of the Jacobian is equal to 0, and that point is a root of the equations you're looking at. OK, thank you.
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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: This is the first of two lectures on spectroscopy and dynamics. Now, I'm a spectroscopist, and so this is the core of what I really love. And there are a lot of questions about, well, what are we trying to do? And you heard two lectures from Professor Van Voorhis, and he talked about ab initio calculations-- electronic structure calculations-- where you can get really close to the exact answer. And it's really a powerful tool. And it gets you the truth. But it gets you so much truth that you don't know what to do with it. And the same thing is true for spectroscopy. You get a spectrum. It contains a huge amount of information. And you can take lots and lots of spectra. But what are we trying to do? When I was a graduate student, we were doing a unique super high resolution spectroscopy. And so we thought what we were generating was excellent tests for quantitative theory. And if, in those days, I went to a lecture by a theorist, they were saying, we're generating stuff to check against experiment. And it's a circle, and that's not what we're trying to do. The theory and the experiment are just the beginning. And you can start from either extreme. What we want is, how does it all work? What is going on? Can we build a picture, which is intuitive and checkable and predictive, so that we can say, oh, yeah. If we want to know something, we can get to it this way. And the purpose of this lecture is to not provide a textbook view of spectra, but to give you a sense of how you can get to stuff that challenge your intuition. What we want to do as scientists is to be surprised. We want to do a good experiment or do a good calculation. And we want to find that the result is not what we expected. And we can figure out why it's not what we expected. And that's never conveyed in any textbooks. Now, this lecture is based on my little book of lecture notes. And I have a number of copies of it. And if people have a strong wish to have a copy, you can have one. I can give it to you. And a lot of this lecture is based on the first chapter of this book. But many of the topics are developed throughout the book. OK. So this is a two lecture sequence, and the first half will cover this stuff on this board. And it's in the notes, so you don't have to copy all this stuff. I just want you to see where we're going. So I've already talked a little bit about experiment versus theory. They complement each other. We can use theory to devise an experiment, which is path breaking. Or we can use an experiment to challenge the theorist to calculate a new kind of thing. And I hope that some of these things, those ideas, present themselves in this lecture. And it's really important that if there's something that you don't understand or don't capture the importance of, you should ask me a question. I really want to talk about what it's all for. OK. So I'm going to talk about spectra, and it will be what kinds-- rotation, vibration, electronic, and other ramifications. And going from atom to diatomic to polyatomic to condensed phase. Each step along this path leads to new complexities and new insights. Then I'll talk about, OK, we got a spectrum. What do we expect to be in the spectrum? Well, one of the things that's important is the transition selection rules or transition rules. Selection rules correspond to an operator-- eigenvalues of an operator-- that commutes with the exact Hamiltonian. And those correspond to symmetries. And there are propensity rules like, OK, which transitions are going to be strong and which are going to be weak? And a beautiful example of propensity rules are based on the Franck-Condon principle. And the Franck-Condon principle is one of the first keys you use to unlock what's in a spectrum, or what a molecule is doing. Because it's the first level of complexity that is presented to you in the spectrum. What are the vibrational bands, and how do we assign them, and what are they telling us? There is a very different kind of information in an absorption spectrum, because it's always from the lowest electronic state, lowest vibrational level. And so there's a simplicity, because of a kind of state selection. And in emission, it's a very different ballgame. Because in the gas phase, the emission is from many different levels. In the condensed phase, it's not. Why? OK. And then we get to dynamics. And the main thing I want to do in these two lectures is to whet your appetite for dynamics. And there are many kinds of dynamics ranging from a simple two level quantum beat to intramolecular vibrational redistribution. Which you can understand by perturbation theory to the strange behavior of electronically excited states losing their ability to fluoresce not because the molecule breaks, but because the bright state-- that's an important concept-- the bright state is something. It's not an eigenstate. It's a special state that we understand well. It's one of the things that we build perturbation theory around. The bright state mixes into an enormous number of dark states. And the molecule forgets that it knows how to fluoresce because the different components, different eigenstates dephase. And that is a beautiful theory. And when I was a graduate student, this theory of radiationless transitions-- Bixon-Jortner theory-- was just created and many people didn't believe it. They thought molecules, big molecules, they're really easy to be quenched by collision and the loss of the ability to fluoresce or the absence of fluorescence was somehow collision related as opposed to a physical process. So there's lots of good stuff. And some of the good stuff is Ahmed Zewail's Nobel Prize where he claims-- and that's why he got the Nobel Prize, because people believed that claim. Now, I'm not saying it's wrong. But part of getting famous is to have a package, which you can sell. And he sold the daylights out of it. And he calls it clocking real dynamics in real time. And it's basically wave packets. But they're wave packets doing neat stuff. And for example, one way a molecule can lose the ability to fluoresce is because the molecule breaks. And what is the mechanism by which a molecule breaks? Does the bond just simply break or is there some motion that precedes that? And what Zewail did was to show what are the motions that lead the molecule into the region of state space where the bond breaks. And, of course, if you want to manipulate molecules, you either want to get to those regions or avoid them. And so there's all sorts of insight there. OK. Now this is what I believe. That if you understand small molecules, you will see examples of everything you need to know to deal with almost any dynamical process in chemistry. Now, this is certainly an exaggeration, but this has been my motto for years. And so I really stress the small molecules. And it's not that small molecules are really hard. They're really beautiful, and they do enough so that you can anticipate what you need to deal with bigger molecules. So let's begin. OK. So what is a molecule? As chemists, we would never think of a molecule as a bag of nuclei and electrons. We wouldn't think of it as a bag of atoms either. We believe in chemical bonds. This is an important thing. It's not a conserved quantity. Bonds can break. But we believe that bonds tell an important story. And so almost all of our pictures for complicated phenomena are based on the-- I hesitate to use the word sanctity-- but the importance of bonds. OK. We start-- I'm going to erase this, because I've made my point, and it's embarrassing to keep emphasizing my secret motto. But it is true. OK. We have the Born-Oppenheimer approximation. And this is very important, because we can't solve a three body problem. We can solve a two body problem. But we have molecules, which are consisting of nuclei and electrons. And this Born-Oppenheimer approximation enables us to separate the nuclear part of the problem from the electronic part of the problem. Because these two things move at very different velocities. And so it's a profound simplification. We get potential energy curves or potential energy surfaces. And that is the repository of essentially everything we want to know. If we know the potential surface, we can begin to do almost anything. And certainly for a big molecule, it's not just a simple curve like this. If you have N atoms, there's 3N minus 6 vibrational modes. And well, that sounds terrible. But even for this, you have essentially an infinite number of vibrational levels and an infinite number of rotational levels. And so if you have a polyatomic molecule, you have 3N minus 6 infinities of infinities. So you're not wanting to get everything. You want to generate enough information to be able to calculate anything you want. And sometimes, you make approximations and you're not sure that those approximations are good. And you want them to break. You want to discover something new. So the Born-Oppenheimer approximation, we go from clamped nuclei calculation where the-- since the nuclei moves slow compared to the electrons, well, let's not let them move at all. And then we build a perturbation theory picture where we let them move. And we can deal with that because we understand vibrations and rotations. OK. So we have a potential energy surface. And there are things that we can anticipate about a potential energy surface. And LCAO-MO theory enables you to say a lot of important things about the potential energy surface. So it provides a qualitative framework. And so from molecular orbital theory-- and this is not what Professor Van Voorhis talked about. This is the baby stuff. And we don't expect to get the exact answer. But we do expect to be able to explain trends. Trends within a molecule and between related molecules. So this provides a framework for expectations. And there are things that we get like bond order. And we talk about orbitals that are bonding, non-bonding, and antibonding. And this comes directly out of the simple ideas. Recall when we had atom with a hydrogen, the hydrogen doesn't make pi bonds. And so they're pi orbitals for the atom A, which have nothing to interact with. And they're usually non-bonding. So we have these sorts of things. We have spN hybridization. And this is just telling you if a molecule wants to make the maximum number of bonds, you do something. And if it's sp cubed, you have four tetrahedrally arranged bonds. And if it's sp cubed, sp squared is planar with 120 degrees. These things tell you something about geometric expectations. Now, molecules don't follow the rules exactly, but they come pretty close. And so if you have some reason to believe that a particular hybridization is appropriate, then you have certain expectations for the geometry and how that's going to present itself in the spectrum. So bond order's related to internuclear distance and vibrational frequencies. Sp hybridization has to do with geometry, and all of these things are really important. So we have a potential surface. And let's say this is a boldface thing, implying that there are 3N minus 6 different displacement coordinates. This potential encodes the normal modes. What's a normal mode? Well, it's a classical mechanical concept. And it basically corresponds to situations where all of the atoms move at the same frequency in each normal mode. And these normal modes each has an expected frequency and expected geometry. Because if it's a polyatomic molecule, you have not just two things moving. They'll always be moving at the same frequency. But you have three or four or 100. And OK. So you learn about the shape of the potential and the force constants and so on. Now, we have the rotational structure. Now, molecules are not rigid rotors. But it's useful to think about molecules as rigid rotors to develop a basis set for describing rotation. And perturbation theory enables us to describe the energy levels of a non-rigid vibrating rotor. And it's straightforward. It may be ugly, but it tells you how you take information from the spectrum and learn about, say, the internuclear distance dependence of molecular constants like a spin orbit constant. Or some hyperfine constant. Or just the rotational constant. And it's a simple thing. And I'm toying with the idea of using this sort of problem on the exam. And I'm not sure whether I did on the second exam. But if I did, you'll have a chance to redeem yourself. OK. OK. Then in the gas phase, nothing much happens. You can have collisions, but the time between collisions can be controlled by what the pressure you use. And so you can sort of think about the gas phase as something where the molecules are isolated. And another way of saying that is the expectation value of the Hamiltonian in any state is time independent. The Hamiltonian is energy. Energy is conserved. And unless there are collisions, energy will be conserved. And so. But in the condensed phase, you have lot of collisions. They're very fast. And so one big difference between the gas phase and the condensed phase is energy is not conserved. And it's not conserved at different rates for different kinds of motions. And you want to understand that. OK. So now let's talk about the kinds of spectra. We have rotational spectra. And that usually is in the micro region of the spectrum. And it requires that the electric dipole moment be not equal to zero. I'll talk about this some more in a minute. We have vibration. And vibrational spectrum is in the infrared. And the requirement for vibration is that the dipole moment-- which is a vector quantity if you don't have a diatomic molecule-- changes with displacements, each of the normal modes. And so we have a molecule like CO2. CO2 does not have a dipole moment at equilibrium. It's a linear molecule, symmetric. But it can do this and that is a change in dipole moment. It can do this and that produces a new dipole moment. Or produces a dipole moment. And it can do this, which does not. So you have different modes, which are infrared active and infrared inactive. And it's related to this dipole moment. Now notice, I put a vector sign on it. It corresponds to motion, directions in the body frame where the dipole moment changes. When you do this, the dipole moment is perpendicular to the axis. When you do this, it's along the axis. And so there is stuff-- really a lot of stuff-- just looking at what vibrational modes are active. And then there's electronic. And that's mostly in the visible and UV. But the electronic spectrum could be in the X-ray region even. But mostly, molecules break when they get outside of the ordinary UV region. And so there's not much there. OK. What's needed. Well, what's needed is the electronic transition moment. Let's call this e1, e2. Going from two different electronic states, there is an electric dipole transition moment, which is not equal to zero. So H2, which has no vibrational spectrum and no rotational spectrum has an electronic spectrum. Everything has an electronic spectrum. OK. Now, a diatomic molecule. Here is sort of a template for everything a diatomic molecule can do. Now, they're cleverer than this. But this is sort of in preparation for dealing with greater complexity. And then we can have up here. OK. So this is the electronic ground state. And normally, if you have a diatomic molecule, you can predict what is the electronic ground state and how is it going to look-- how is its potential curve going to look relative to the excited states. That's something you should be able to do using LCAO-MO theory. OK. So this is the ground state. This is the repulsive state. And usually, the ground state correlates with ground state of the atoms. And so here we have a repulsive state, which also correlates with the ground state of the atom. So this is an excited state, and that correlates with an excited state of the atoms. And this is AB plus an electron. So that's one way the molecule breaks. And this dotted curve represents Rydberg states They're Rydberg states converging to every rotation vibration level of this excited state. And it can be complicated. But it's beautiful, because I know the magic decoder for how do you deal with Rydberg states. Because there's a lot of them, but they're closely related to each other. And we can exploit that relationship in guiding an experiment. And so here, now we have a curve crossing between a repulsive state and a bound state. And that leads to what we call predissociation. So the vibrational level of this state-- which would normally be bound above the curve crossing-- are not bound. And that's encoded in the spectrum too. And so for more complicated molecules, they're going to be these curve crossings or surface crossings. And we want to know how do we deal with them, and what is a diatomic-like way of dealing with them. And the important thing is at that internuclear distance where the curves cross, then you could be at a level-- starting at a level on this state-- the bound state. And it will have the same momentum at the crossing radius as the repulsive state. And that's where it goes. That's where it leaks out. I mean, we're normally used to thinking about processes-- non-radiative processes, all kinds of processes-- as an integral overall space. But because the momenta are the same on the two curves at this radius, they can go freely from one to the other. The molecules can go freely from one to the other. We get a tremendous simplification. And this is something that is always ignored in textbooks and is a profound insight. Because you know exactly where things happen and why they happen. And so you can arrange the information to describe what's going on at this point. And this is where semi-classical theory is really valuable. Because not only do you know that you have stationary phase at this point. But you know what the spatial oscillation frequency is. Because the spatial oscillation frequency is H-- or the wavelength-- is h over p. And you know what the momentum is at this point. And so you know where the nodes are. How far the nodes are apart and what the amplitude is here. And so it tells you exactly what you want to know in order to describe this non-radiative process. And my belief is that almost all of the complex things that molecules do happen at a predictable region in coordinated space. And you can get the information you need to understand them, because all of a sudden, the molecule is behaving in a kind of classical way. And we're entitled to think locally rather than globally. OK. I'm going very slowly. We may get through most of what I planned to talk about today. I have 11 pages of notes, and this is-- I'm on page three. So maybe we'll take off. OK. So how do you do spectra? Well, in the old days, you had some light source like a candle, and you had a lens, and you had an absorption cell. And there would be some sort of a spectrometer here. Could be a grading. It could be a prism. It could be anything. But it's something that says, OK, I looked at the selected wavelengths at which the gas in this cell removed light from the continuum. And then you have a detector, which in the old days was a photographic plate. But it could be a photo multiplier, and you're looking at the spectrum by scanning the grading or something like that. And so what you would get is some kind of a record where you have dark regions corresponding to where there has been no absorption and, well, actually it would be the other way around. There'd be bright regions, because there'd be no exposure of the emulsion on the plate and dark regions where there-- yes-- where the light hits. OK. So but we're much cleverer than this. And we can do all sorts of wonderful things. And again, I've been around for a long time and lasers were just beginning to be used when I was a graduate student. And I was one of the first small molecules spectroscopists to use lasers. But not as a graduate student. I wanted them, but we had such-- lasers were so terrible in the region between 1965 and 1971 when I was a graduate student. And so lasers were things to be admired, but hardly to be used. But one of the crucial things was dye lasers. Because these guys are monochromatic, and they can be tuned and tuned continuously over a wide region of the spectrum. And so that's way better than a candle or a light bulb. Because it's monochromatic, and you're asking one question at a time as you tune the laser. Now, lasers enable you to do many kinds of experiments. You can simply tune the laser through a series of transitions, and you get fluorescence every time the laser tunes through a transition. And if one laser is good, two lasers are better. And so you can do all sorts of things like suppose this spectrum is really complicated. And you want to be able to simplify it. And so you can do a double resonance experiment where you tune this laser to one line and then you tune this laser through a series of transition. That spectrum is going to be simple, and it's going to be telling you who this was. And there's just no end of tricks. And often, instead of detecting the fluorescence, you tune the laser. Let's do this. And so starting here is some kind of a continuum, ionization continuum. And so you have this photon being used twice. One here and one to take it above the ionization limit. And so you have an excitation, which you do want to know how strong it is. And so you might monitor the fluorescence. But you don't know who this is, and you find out by tuning this. But you would detect the excitation here by subsequent ionization. It's easy to collect ions. Every ion you produce, you can detect. Every photon you produce, you can't detect. Because you have a solid angle consideration and photo multipliers are not perfect. And so ionization detection is way more sensitive. So you can do that kind of thing. You can also do-- this is a kind of sequential excitation. You could imagine doing an experiment where you have an energy level here, and you have a laser, which is not on resonance. That's a coherent process. It uses the oscilltor strength at this level to get to here. And that's related to many other kinds of current experiments. And that's neat. Now, we're recording spectra, and we need to know what the rules are. And so there are certain transitions that are allowed and certain transitions that are forbidden. Now, I talked about the transition requirements for rotation, vibration, and electronic. But let's just talk about the electronic spectrum, because the other two are simple. The transition operator is equal to the sum over electrons of e times r sub i. It's a one electron operator. And that means if we have wave functions which are Slater determinates of spin orbitals like 2s alpha. This one electron operator can only have a non-zero matrix element if the two states differ by one spin orbital. That's a big simplification. And so one can actually use this to selectively access different kinds of states by designing an experiment. But the important thing is that for electronic transitions, we have the selection rule delta so is equal to 1. Not 2. Not 0. And now, the operator doesn't have any spin involved with it. And so that means delta s equals 0. You did not change, you did not go from a singlet state to a triplet state. And the only way you can get from a singlet state to a triplet state is if the triplet state is perturbed by a singlet state. So this picture I drew where I had a repulsive state crossing through a bound state, it might have been that one of those was a triplet. And as a result, and the other is a singlet, and you have spin orbit interactions, and you get extra states, extra lines appearing. But you get this wonderful selection rule. There is also for the electric dipole that we have plus to minus parity. Now what's parity? I don't like talking about parity, because the useful definition leads to complexity. But basically, parity corresponds to the symmetry, the inversion symmetry in the laboratory frame. Now, you say, well, a molecule doesn't have any inversion symmetry. But space is isotopic. And so you can go from a left handed to a right handed coordinate system. And that's what happens when you invert space. And so you can classify levels according to whether they're odd or even with respect to space inversion. This is close to the truth. This is close to all you need to know unless you're actually going to do stuff with the parity operator. But it's a useful way of saying, OK, I put parity labels on things. I learned how to do that, and that's enough. Now, you follow selection rules where good quantum numbers are conserved. Or they change in a way that you predict based on the way you did the experiment. A good quantum number, I remind you, is the eigenvalue of an operator that commutes with the exact Hamiltonian. There are very few rigorously good quantum numbers. But if a molecule has any symmetry, group theory tells you a bunch of things that commute with the Hamiltonian, and it gives you symmetry labels. And that's very important in inorganic chemistry where you have either molecules with symmetry or molecules with atoms with ligands in a symmetric arrangement. And since the transition is on the center atom usually that you can classify them using group theory as allowed or forbidden. So if we have an electronic transition, the easiest thing to observe is vibrational bands. If you have a relatively low resolution spectrum, you're going to see vibrational bands. You have to work harder to see the rotational transitions in each vibrational band, but you get an enormous amount of qualitative information just looking at the vibrational bands. Because the vibrational bands encode the difference between the ground state potential and the excited state potential. Now, this is a universal notation. Ground state is always double prime. Upper state is always single prime. Very strange, but that's the way it is. And so if these potentials are different, the vibrational bands encode the difference. And this comes from the Franck-Condon principle, which says nuclei move slowly, electrons move fast. The transition is an instantaneous process, as far as the nuclei are concerned. And so there's no change in nuclear coordinates, and there is no change in nuclear momentum. This is what's in all the textbooks, and nobody ever talks about this, because we don't really normally know or think about momentum. But we do know what it is. We do know what the operator is. And we know that kinetic energy is related to the momentum squared over 2 times the mass. So what is this? This means transitions are vertical. In other words, if we have a pair of electronic states, we draw these vertical lines. Not slanting lines. This means momentum is conserved. And this, here at this vertical point, we have this much momentum. And, well, there's the same amount here. Now usually, this just means-- the delta p is equal to 0-- means that of all the strong transitions, turning point to turning point transitions are the strongest. Because at a turning point, the vibrational amplitude is large. But there's more to it than that, because there are secondary maxima in the vibrational transition intensities. These correspond to stationary phase between the initial state and the final state. And so in addition to the strongest transitions, you get other transitions that you can explain by this delta p equals 0. Now, you don't know anything when you start. You know something maybe about the ground state. And this could be a polyatomic molecule if we're just looking at one mode. And suppose we have an excited state where the vibrational frequency is the same. In other words, there is no change in bonding character. And so what you end up getting is just the zero to zero transition in absorption. Or if you have many vibrational levels up here, you see v to v, delta v equals 0. Now, you could have a bound state, and it's usually true that the excited state is less bound. And so you would have-- the Franck-Condon active region corresponds to turning point to turning point in the lower state. So I shouldn't draw this. Let's draw a vertical transition. Well, I missed. Let me just start again. So we have an excited state, and we have a ground state. Ground state is bound. And this is v equal 0. And so we go from here to there. So if the excited state is less bound, it's a larger inner nucleus and smaller vibrational frequency. We have many vibrational levels accessed by the Franck-Condon principle. And if we have, in the other sense, we have an excited state, which is more bound than the ground state then the Franck-Condon region is narrower. Because this wall is nearly vertical. You have more transitions when it's displaced this way. And this branch of the potential is nearly much flatter, and you have fewer transitions here. So the vibrational pattern tells you qualitatively from the get go whether you're going to a more bound or a less bound state. That's very useful information. Now, you want to go further. You want to say, oh, well, I'm observing a bunch of vibrational levels in the excited state. What are their quantum numbers? Are we seeing the v equals zero level? How do we know what vibrational level we're observing? Or levels? And we can calculate Franck-Condon factors. But you can do many things. So I mean, if you have a situation like this, you might not get to the lowest vibrational level. But if you have a situation like this, you probably will get to the lowest vibrational level. So one way to get vibrational numbering is seeing the vibrational pattern terminate. That's a good sign it's v equals 0. If it terminates abruptly, it's v equal 0. But if it terminates slowly, you're not sure. But there are other really wonderful things about this. And you can do isotope separation. Isotope shifts. Since the vibrational frequency is the square root of k over mu, we can change and reduce mass. That changes which vibrational bands you observe, and it changes it in a quantitative way, and so you can often use that to tell. Another thing you can do is to now. If you have an excited state, the vibrational wave function is going to look like this. And there's a node here, and a node here, and node here, node here, node here. And so if we're looking at the vibrational progression observed from such a level, the intensities will have minima corresponding to how many nodes there are in the excited state. And the number of nodes is the vibrational quantum number. So there are lots of ways of doing it. And then there is something else. If you observe at moderately low resolution a vibrational band in an electronic spectrum, it will have the peculiar shape like that or like that. This is called a band head. And it corresponds to the fact that because the rotational constants in the upper and lower state are different, one of the branches-- you have delta J equals plus or minus 1, 0. And the delta J plus 1 is called the R branch and minus 1 is called the P branch. And delta J of 0 is called the Q branch. Now, these two guys are such that depending on the sign of the difference in B values, you get ahead on the high frequency side or ahead on the low frequency side. Well, actually, it's the other way around. But it tells you then the rotational constant, which is also a signal of how bound the state is. The rotational constant-- the shading of these band heads-- confirms your vibrational assignment. And typically, if you have a smaller vibrational frequency, you'll also have a smaller rotational constant. It's not always true. And it's always interesting when it doesn't happen. And so the vibrational bands have this asymmetric shape unless the rotational constant upstairs and downstairs is the same. And then you have a branch going this way and a branch going this way. And a gap in the middle that might be filled with some Q branch lines. And the Q branch lines tell you, oh, yeah, that was a transition where the lambda, the projection of L on the internuclear axis changed. And if it doesn't have this Q branch, it'll tell you this delta lambda equals 0. All sorts of stuff. And if you have no sign of band heads, but just this double hump structure, it tells you that the rotational constant is about the same as in the ground state. And it tells you also that the vibrational frequency is expected to be about the same. I better stop. So as soon as you go to polyatomic molecules. Instead of having just one vibration, you have 3n minus 6. And so 3n minus 6 downstairs, 3n minus 6 upstairs, that looks bad. But only some of the vibrational modes correspond to a distortion from-- a difference in geometry between the ground state and the excited state. And so most of the vibrational modes correspond to identical frequencies. And what we call that is Franck-Condon dark. Because the only transitions that are allowed are delta v equals 0 for that mode. And so only the modes that correspond to a change in structure appear as long progressions. And so there are only a few, and so the spectrum of a polyatomic molecule is not too much different from a diatomic molecule, because of the small number of Franck-Condon active modes. But you look closer and you see big differences. OK. So well, we'll talk more about this on Friday. And we might go to three lectures on this, because this is really-- you know the whole point of this course is to be able to understand how molecules talk to us.
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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: This recitation on mass spec is supposedly associated with reactive oxygen species. So [INAUDIBLE],, which happens all the time in this course, because we can't describe all the techniques as we go along. So what I'm going to do is just give you a two second overview of what you need to think about to put the paper you read into the big picture. I don't think the paper-- the paper also explains it. And this week we're going to focus on the mass spec paper, which is mostly sort of trying to figure out the technology, and then the next week is focused on the biology. And so the major unsolved problem-- so everybody and his brother is using mass spectrometry as a tool nowadays. There has been a revolution in mass spectrometry. [INAUDIBLE] The instrumentation is cheaper. The mass spectrometric methods have just really taken off. And people didn't even know who mass spectrometrists were, but they're starting to win major prizes, because it's revolutionized what we can do. I was talking to somebody yesterday, and they just got a mass on a protein that's 3.3 million. How do you get a protein into the gas phase that's 3.3 million? Right. Doesn't that sort of blow your mind? Anyhow, it's been a revolution. And we're going to be looking at-- this module seven, which is on reactive oxygen species, and we've been talking about the question of homeostasis. And so one of the things with these reactive oxygen species is they are used by us to kill bacteria, viruses, or parasites. But now, in the last five years or so, everybody's focusing on the fact that here are these reactive oxygen species that play a key role in signaling, which is everywhere, and the signaling process we're going to be looking at next time and is alluded to in this particular paper is epidermal growth factor receptor and epidermal growth factor. There are hundreds of these proteins that have receptors that are involved in growth and epidermal growth factor receptor [INAUDIBLE] of successful cancer therapeutics. So it's interesting what happens up here, what happens down here, how do you control all of that, and people are studying this. So we've already seen cystine is unique. And if you have a reactive oxygen species, and we'll see that the reactive oxygen species we'll be looking at, when we're going to be looking at a number is actually superoxide. So that's one electron reduced oxygen, which in the presence of protons can rapidly disproportionally give[?] oxygen gas to hydrogen peroxide. And hydrogen peroxide can react with cystines to form sulfenic acids, which is the subject of the paper you had to read. And so the question is how prevalent is this, and the question is, is this important and interesting in terms of regulation inside the cell? And so the key issue is-- even cystines aren't all that stable, you know, if you have proteins with cystines, and you let it sit around for a long time, you could form-- and the protein's concentrated, you could form disulfides. It's not a straightforward reaction, but you can form disulfides. The question is if you had hydrogen peroxide inside the cell, which you do, can you form sulfenic acids, and do they have a consequence biologically? OK, and that's the question we're going to address next time. And so the issue is this is unstable. So if you want to develop a method to look for this species, and you start cracking open cells, and you start working it up, what happens is this falls apart and reacts and gets destroyed. And an example of this is the area of DNA therapeutics and DNA drug interactions, therapeutics that interact with DNA. For decades, you see lesions on your DNA. How do you determine what the lesions are? Mass spec has been a major method to look at that. Almost all the lesions in the early days were complete artifacts of the analytical chemistry to work them up. They had to get them into some form that you could stabilize the lesion and then analyze it. And what was happening because they weren't careful enough and quantitative enough, they changed it to something else. And so the data in the early years was all completely misinterpreted. So the issue in this paper is that other people had developed this, and Kate Caroll has taken this on. Can we have a way of derivativizing this [? mentally ?] inside the cell, because if you disrupt-- if you disrupt this by cracking open the cells and trying to purify things, it undergoes further reaction, and what this can undergo further reaction to is SO2 minus, sulfonic acids or SO3 minus. Sulfinic acids and sulfonic acids. OK. And it turns out this reaction is also reversible with hydrogen peroxide a lot of people are looking at that at this stage is irreversible. Anyhow. So the question is can you develop methods to look at all of these things. And in fact Tannenbaum, who was in the chemistry department, but also [INAUDIBLE],, he is looking at nitrosation of SHs, again forming a reactive species, and he's developed new methods sort of like Carroll has to try to specifically look at these modifications. And in the end, what you want to do, and this is the key, you might be able to detect this-- the question is, is this interesting? So you have to have a way to connect this back to the biology inside the cell. And that's what the second paper is focused on. So what we're doing today is simply looking at the technology that's been developed to try to get a handle, how do you look at sulfenylation, you're not really focusing on the biology of the consequences. And so what we're using is mass spec. And we're using a method of mass-- how many of you have done mass spec? So if you know something and I say something wrong, you should speak up, because I'm not a mass spec expert. And actually, I've got a whole bunch of information from, say, the Broad, and I thought it was not very good. So we need a way of trying to figure out that you're going to see-- there's hundreds of variations on the theme. I'm going to give you a very simplified overview of what things you need to think about. And so if I say something that you don't agree with, tell me. OK, so when looking at mass spec-- this didn't exist when I was your age-- using soft ionization methods, and what does that mean? It means that you don't want your molecules to crack. So the issue is that what mass spec is about-- so really looking at mass spec, and the key issue of what you wind up looking at is mass to charge. OK. So m over z. OK, so the problem is how do we get something charged enough so that the mass is small enough so that you can see it, taking a look at the mass analyzer, which is going to be part of all mass spectrometers. OK. So there are two different ways you could change the mass to charge. You could dump an electron in. And if you dump an electron in, that produces radical species, which can then fragment. We want to avoid that. That's not soft ionization methods. But how can we control this? The way we can control this is dumping in protons. So what we do is we can control it by adding protons or by subtracting protons. And we'll see that the different methods we're going to be looking at, we'll see there are two main methods that most of you have probably heard about your classes. One is electrode spray ionization, so ESI. And I think, if you're in Brad's lab, they have a lot of these. Yesterday's class had people that had used these, but really didn't know much about what's inside the machine. So this is the kind of thing I think your generation, if you're going to use this as a tool, need to roll up your sleeves and understand a lot more about what's going on, and MALDI, maser MALDI. Matrix Assisted Laser Desorption-- it will become clear why it's called that in a minute. So these are the two methods. And what we do is we can protonate, so that we can move this into the analyzer range, where we can actually read it. So what we'll see is the analyzer-- I'm going to show you sort of what the three parts of a mass spectrometer are-- can only read 1,000 to 2,000 daltons. OK, so if you look at your protein, much, much bigger. So you're going to have to stick a lot of charges on there to be able to see anything. So that's the whole thing, and the question is, how do you do it by one method or by using the other method? OK, so all mass specs have sort of the same components. And you can go to websites. The Broad does have a website, and what the Broad will tell you is what all these spectrometers are, but I don't think they do a particularly good job telling you what's useful for what, and why it's useful, which is, I think, what you need to use if you're only going to use it fleetingly and then move out. So you have a source. So you have an inlet. How do you get your sample from the liquid phase or the solid phase into the gas phase? OK, so that's going to be that. And so what is the distinct ionization method? And we will see that there are many ways that you can ionize, and we're just going to briefly look at in a cartoon overview of how this happens. And then so once you ionize it, it needs to move from the source. So you need to have ion movement into the analyzer. So this is the mass analyzer. And this becomes important. And we will see in a second that there are many methods to do the mass analysis, mass to charge analysis, and then after you do this, you have a detector. And then, furthermore-- and I think this is a big part of it now, if you're doing wholesale anything, you have to have a really sophisticated method of data analysis. And so that's the other thing that I get frustrated about all the time. So you see people-- I mean, people do experiments where they've spent-- last year, somebody spent three months trying to get all the proteins out of a cell, 10,000 proteins out of the cell by mass spectrometry right. Now, because the technology is changing, they can do it in four days. But what do you do with all this information? And how do you use this information in a constructive way, and how do you know if it's correct or not? So those are the kinds of things. I think if you're going to use this-- I think everybody is going to be using this technology. You need to educate yourself about how to look at this. OK, so that's what the issue is. And so we have a source, an analyzer, and a detector. OK, so this is just a cartoon of that, which describes this in more detail. And I think he put this on the web. I think he put the PowerPoint on the web. I was doing this at the last minute yesterday. So it's different from the handout I gave you that's written out. This is a PowerPoint. OK, so you can go back and look at this, but one of the other things I wanted to say is that sometimes when you analyze your mass, you want to analyze it further, and that was true-- many of you might not have caught it, but that was true in the analysis that was carried out in this paper. Did anybody recognize that you had to analyze this using more than one mass spec? Did you look at the data carefully enough? So also you probably didn't read the supplementary information, which also is critical to think about. I mean, if you want to look at the methods, you need to get in there and roll up your sleeves and look at them. So we're going to see that the methods that people often use is they don't look at the whole protein, but they degrade it down into pieces. So then you can find here a whole bunch of pieces, OK. But that doesn't tell you anything. The mass does tell you something. It might tell you whether it's sulfenylated or hopefully, you can distinguish between any other modification, but it doesn't tell you the location of the sulfenylation. And so you can do a second method. So you could have some other gas. There are many ways to do this that you bring this in to now take a peptide. So you pick one mass charge. You throw in something that's going to degrade it by fragmentation, and then I'll show you in a minute we understand what kind of-- using certain methods, we understand the fragmentation patterns, which actually allow you to sequence the amino acids. And the reason I'm bringing that in is when I first got to MIT, Klaus Biemann was in the lab, and I did many experiments with him. And these are the first experiments that were done to sequence peptides by mass spec as opposed to doing Edman sequencing, which the mass spec was actually better, and there are pluses and minuses, but I noticed from looking at the literature, people were still using the same method that he developed. So this is just a cartoon. And it just shows you that there are many ionization methods. We're focusing on these two, FAB, fast atom bombardment. We didn't have any of these when I was your age. Fast atom bombardment was something I used a lot because I've worked on DNA drug interactions, and it allows you to look at nucleic acids. And a lot of these other methods don't. I mean, we're focused on proteomics in this particular paper, and then mass analyzer. So you have time of flight. I think Brad's lab has MALDI time of flight. So what does that mean? You've got a long tube in here, and what happens is you have mass to charge, and they're different sizes. And so the smaller ones fly faster. They don't want to keep away from the walls, but the smaller ones fly faster than the bigger ones. So that helps you differentiate between all the ions you're actually looking at. I guess somebody just told me you guys just got a new quadrupole ion trap. Anyhow, if you want to look at this, I have notes on all these things. But I think this is something you'd have to study in detail. And so while I have pictures of them all and how you can differentiate one from the other, I think it doesn't really mean that much to me, because I don't know enough about the physics of how they were designed. I mean, this really has revolutionized what you can do. OK. So that's the components of all mass spectrometry. What I want to do is very just briefly look at the ESI and then look at the MALDI and then show you what the issues are in general, and then we'll focus right in on the paper, and the recitation I did on Thursday, we didn't quite get through all of it. We got through most of it, but then we'll continue next week and also attach this to the biology, which is the second paper, the nature chemical biology paper also written by the Carroll group. ESI. So that's the one we want to look at next. And so that's up there. This is a cartoon of how this works. So what do you do, and how do you do this? So the first thing is you have your protein of interest, which I'll call the analyte, because we want to charge. Lots of times you put it under more acidic conditions, pH 6 or something, 6 1/2, depends on the protein. So you get more charge states. And if you're trying to look at something big, you need a lot of charges on there to get it into this mass range of 1 to 2,000 to be able to see it using this method, and apparently what you do here-- can you see this? Have you done this? Can you see this capillary? Can you look at what's going on? AUDIENCE: I don't think so. JOANNE STUBBE: OK. So I was just wondering, because I haven't ever. So it's all closed off. It's in a box, and so you can't-- there's not like a thing where you can watch what's going on? AUDIENCE: Not that I've seen. JOANNE STUBBE: OK, because I think it's sort of amazing. How do you get this huge protein and solution into the gas phase? Right. I mean, that, to me, is like mind boggling, OK? I mean, these guys were geniuses. And you know, there's been a number of Nobel Prizes for this, but I wouldn't have a clue how to do something like that. So what you do is apparently, you put it down a capillary and then you spray it out, and then you have to-- so what you get at the end of this, this plume of spray, apparently you've got a lot of the analytes and a lot of solvent molecules, and then the goal is during this process, to get into the analyzer is to get rid of-- to separate all the analytes mixed together into a single analyte and remove all the solvent. OK, so that's the goal. And apparently, according to the people that were here yesterday, this is taken from, I think, sort of one of the papers that was first out. This is the way they did it in the old days. I don't know if they still do it this way, but the goal is really, to get a single analyte with no solvent on it. OK, and so the question is, how do you do this, and the chamber they had was at atmospheric pressure, and then they had a potential and pressure gradient, which allowed it to get into the mass, before the mass analyzer. So you start here with the initial spray, and then as you go farther, you remove some water molecules. You finally get to the place where you've removed enough water molecules that all these positively charged species come together, they're incredibly unhappy. And then they fragment apart. I mean, that's the way they describe it. It sounds reasonable. So you get smaller and smaller till eventually, you get to a place where you have an analyte that you can look at specifically and the water has been removed, and that's what you look at. OK. So again, we need to be in the range of 1 to 2,000. So that's the way these things work. Although, I think, again, how you get to looking at the single ions I think in different mass spectrometers. And so what the issues are, I think, are shown here, and this is the beauty of this methodology. So if you have a protein of 10,000 molecular weight, you couldn't see it, because the mass analyzer is limited. So you have to go all the way down to eight charges on it to be able to see it. And then, you divide that by that, and you get-- what do you have? You have to do some corrections, but you get something that's this size. OK, but you can see it now because of all the charges on it, but the beauty is if you add more charges, you get another peak. And you get another peak. And it all has the same information, and it just differs by the number of charge. So you have all this information. You can use that-- all these informations together to give you a very accurate mass on this system. So this method by analyzing all the data, and now the computers do this, I guess, routinely can give you a very accurate mass. So if you look at this printout, it doesn't look like that. This is what it looks like. And what do you think's going on there? So we look at mass charge, and we're in the range of 1,000 to 2,000 daltons. And then what is this all-- what is all of these peaks associated with? Anybody got a clue? AUDIENCE: Isotopes. JOANNE STUBBE: Yeah, so isotopes. So where are we seeing isotopes before? So these are mostly stable isotopes. We spent recitation two and three looking at radio isotopes. OK. I would say, you know, radioactivity is pretty important. Stable isotopes are extremely important to mass spectrometry. So if you get into this, you're going to be able-- you'll see that being able to label things with different kinds of stable isotopes is key to really deconvoluting the complexity when you're looking at a whole proteome and thousands of peptides. We're getting down-- it becomes very complicated, and you have to be able to compute what you expect based on the normal natural abundance isotopic distribution. So that's the key thing. So we look at the normal isotopic distribution. And if you look at that, I think in the next one, I show you an example of that. So what are the isotopes-- you probably can't read this here, but if you pull out your computer, you'll see this. So we have C12, C13. OK, we have hundreds of amino acids with carbons. So you have C12 and C13. C12 is 99%. C13 is 1%. That's an actual abundance. OK, so every one of these has different natural abundance. We know what they are. In fact, if you're an organic chemist, you can measure isotope of x using a mass spectrometer, if you have something that's really accurate, which we do. I've measured a lot of C13 isotope effects, using a mass spectrometer, based on differences in natural abundance and changes. Yeah. AUDIENCE: [INAUDIBLE] JOANNE STUBBE: The what? AUDIENCE: The natural abundance of deuterium? JOANNE STUBBE: Yeah, I think it's up here. So it's up here. I think it's-- let's see, 3%. Yeah, protons deuterium 3%. AUDIENCE: Would you expect a huge distribution from that? JOANNE STUBBE: You see isotope effects on everything. You see-- if you do mass spec, I mean, this is something I think that's not appreciated, and you have a linker with deuteriums in it, and even if you chromatograph it, you change the chromatographic properties based on the deuterium, and so you might think it's migrating here, and it doesn't. It has an isotope effect on how it migrates. So yeah, you need to pay attention to all of this stuff. OK, and it seems like a small amount, but the beauty is that it is a small amount, but it's incredibly informative, and we have very powerful computers that can allow us to do the analysis. So we do have protons. You see deuterium used. You saw deuterium used in this paper you read today. They did CD3 and CH3's. OK, you can also see the tritium. OK, that's much smaller. I don't know what the ratio is, but you can look at it. But you also-- this one is also incredibly important and is widely used in proteomics-- N14 and 15, and people do isotopic labeling. So they might see N15 labeled lysine or arginine or deuterated lysine or arginine. And why do you think they would deuterate the lysine or the arginine or N15 label it? What do we know about lysine and arginine in terms of thinking about proteins and analysis of proteins? What do you think about lysine and arginine? You've seen it several times over the course of this semester, and you probably saw it in 5.07. AUDIENCE: [INAUDIBLE] JOANNE STUBBE: What? AUDIENCE: The protons will exchange? JOANNE STUBBE: Well, now as you put it-- no. So that that's true if it was on a hydrogen and a nitrogen, it would exchange, but they put the deuteriums in on carbon, so they're not exchanging. OK, so why that would happen in any amino acid, why lysine and arginine? And the reason is that almost all-- and this was also done in this paper, you don't work on the huge protein. You cleave it to pieces. And you cleave it into pieces, and where you cleave is with trypsin, which is the major-- you've seen this used now over and over again. That's a major thing you use because it cleaves next to basic amino acids. So these become really important in labeling experiments, if you read much mass spec data, or if you look at Alice Ting's work, everything is N15 and deuterium labeled, and lysine and arginine to try to make sure they have coverage of the whole proteome, which is what her lab actually looks at. OK. So we have isotopic labels, and we can take advantage of these, and we can calculate what the distribution should look like, OK, of the isotopes should be, depending on what the-- we know what the sequence is. We know what the abundance is. And so you can calculate the whole mass spec. So let's see. So there's going to be a number of things that we want to do, and what we're going to be describing today and the next time is a "workflow." These are the words that people use all the time, and "platform." And what we're trying to do in the case of the Carroll papers is simply look at whether the protein is modified or not. But as with most post-translational modifications, do you think this is going to be 100% modified? No. In fact, it's only partially modified. That adds to the complexity of understanding whether the biology is interesting or not, so what you have then is something that's modified and something that's more non-modified. So then the question is, how do you tell how much is modified and how much is non-modified? If this enhances the rate only a factor of two, and this is 99.8%, of this, are you ever going to be able to see an effect of this modification? That's the question that you have to focus on, and everybody and his brother is doing experiments like this. We will see in a second, hundreds of post translational modifications, and the question is what are they doing in terms of thinking about the biology of the system. OK, so what's the platform? What's the platform we're going to use? So there are two ways you can look at this. So we have a protein that has been modified. You're going to-- if you had a huge protein, and you only had a single OH on it, even if it was 100%, and the protein was, say, 300,000 molecular weight, you might not be able to see it. You need to do a calculation to see whether you could see it or not. If you have a small protein of molecular weight 30,000, or whatever-- I think the 22,000 or 23,000 like glutathione peroxidase, used in this paper, you could see it. So you could look at the protein directly. But how else could you do this? You would enrich. If you were doing this in the whole cell, you would want to separate this away from everything else. OK, so to do that, you want to be able to have a way to stabilize this, OK, and that's what this paper is all about, and then not only to stabilize it, but to separate the stabilized form out. So where does this happen? And in this particular cartoon, where do you see post translational modifications? Probably the most popular one is phosphorylation. So we have signaling cascades in kinases. And in fact, if you look at the epidermal growth factor receptor, it's a tyrosine kinase, and it gets phosphorylated and is regulated. And this sulfenylation is supposed to be on top of the phosphorylation. So you have multiple post-translational modifications that can affect activity. So Forest White, for example, in BE, works on kinase signaling cascades. And so he's developed a method, as have others, to be able to pull phosphorylated proteins out of a crude gemisch. OK. So, you know, if you look at this, here he's got iron bound to a phosphate and bound to some bead. So the iron's bound to some chelate around the bead, just like your nickel affinity column, which then binds to the protein. But this raises the issue that I was discussing in class, which I spent a lot of time on over and over again, but you need to think about, do you think these bonds are tight, how tight do you think those bonds are? What do you need to think about for this kind of analysis to work? It's the same thing with nickel affinity column that you talked about when you were looking at purification of proteins. AUDIENCE: It has to be stable enough. JOANNE STUBBE: It has to be stable enough. That's the key. So you have to undergo ligand exchange. It's got to-- if you didn't have-- when you start, you don't have phosphorylated form of your protein around. You have nothing. You have water there. OK, so the waters have to undergo exchange, so the phosphate can then bind, but it's an equilibrium, and so up and down the column is coming off and on. Yeah. AUDIENCE: [INAUDIBLE] JOANNE STUBBE: It could. I mean, so it's a question of what out competes what. It's a question of relative Kds. So what you have to do is study all of this to figure out how to optimize this, how did they arrive at this? Probably somebody did a lot of studies. OK. This is a new method. I don't know how new it is, but it's a method I don't know that much about, again, of pulling phosphates out. So that's one way. So you have-- so you usually have an affinity purification. And if we look at the Carroll paper, what she does in the next paper is she's going to figure out a way-- she's derivatized, she's made a dimedone derivative, which stabilizes the sulfenic acid, and then she attaches something to it that's going to allow us to affinity purify that. We'll come back and talk about that later. So what are they using over here? They're using-- this is-- if you look at histones that get acetylated or methylated, they have an antibody that's specific for the acetylated lysine, so they use antibodies to pull something out. So that's a method-- the second way of pulling things out are using antibodies. That's quite frequently used. And what did they use in this paper? Did it detect the modified sulfenic acid? Does anybody remember? Did you read the paper carefully enough? AUDIENCE: Like, a anti-dimedone antibody? JOANNE STUBBE: Yeah, so they use an antidimedone antibody. OK, so that becomes really critical that you know that your antibodies are actually working effectively. So we have antibodies, and then, another thing that people are interested in this department, the Imperiali lab, is sugars. We have sugars everywhere. OK, we don't really understand the function of these sugars. We understand some of them, but it's amazingly complex. And what we have are proteins called lectins, and any of you heard Laura Kiessling talk, maybe undergraduates wouldn't have done this, but she discovered a new lectin and discovered the basis, the structure the sugar that binds to this lectin. And so you can selectively move that type of sugar. Again, it's an equilibrium. So they're coming off and on, but it binds, hopefully, enough so that the other stuff washes through, and you enrich in the protein of interest. So these are sort of some of the tricks that are actually used. We're going to see, in the case of the Carroll paper, next time we use click chemistry to make something with a biotin on it, because biotin you all know can bind to streptavidin, which has pluses, and it has minuses, but it allows you to pull things out more easily, because the interaction is so tight. So you could do this-- the workflow could be on the intact protein, or it could be on peptides. OK. And so the bottom half of this graph shows what happens after you treat this with trypsin. So with trypsin, and you're always cleaving next to lysine or arginine. So the C terminus of your protein is always a lysine or an arginine. And you can find that more easily if you deuterate or N15 label it. That's what people routinely do in the [? Broad. ?] And then you have, I think this is the most amazing thing, so you have a protein. And then you have an HPLC column. Have any of you done HPLC? And so do you think-- you could have a protein of 300,000 molecular weight, and look at the separation of your peptides. But if you look at any one of these things, do you think it's pure? So it's not pure. So every one of these peaks, if it's 300,000 molecular weight, you can calculate-- the reason people use trypsin is-- does anybody know why use trypsin, besides that cleaves at lysines in its specifics? Why do people use trypsin as a thing to cleave a big thing down into a little thing? AUDIENCE: What's the rationale for cleaving it? [INAUDIBLE] JOANNE STUBBE: So the rationale for cleaving it is just to make it smaller and easier to analyze. That's the rationale for cleaving it. So a peptide, a small peptide. But the question is, how big is the small peptide that's easy to analyze? And so that's the rationale. It gives you a distribution of peptides that's pretty good, that are all accessible to mass spec methods. So I don't know what the distribution is, but you know, people have done that calculation. And so almost always the peptides fly, whereas if you use other things, and you have something much bigger, it might not get ionized in the appropriate way or in a quantitative way, and you completely miss it. So the trypsin has been most successful. But each one of these little peaks is not one peak. You'll see when you put it into the mass analyzer, and if you read this paper carefully, you will see they got multiple mass charge species, which then they associated with specific peptides, OK. They know the sequence of their protein. And then they always use tosyl phenyl chloro ketone. Why do they use that? Anybody have an any idea? So in the experiments where they're doing the trypsin cleavage, they put in tosyl phenyl chloro ketone. Anybody know why? OK. No good. This is something that-- so tosyl phenyl chloro ketone is an alpha halo ketone. So it's activated for nucleophilic attack, and what you do is you have an acylated N terminus and an aromatic, and that's specific for chymotrypsin, like proteases. And so what this does is that covalently modifies the active site of chymotrypsin, and kills chymotrypsin. If you choose the wrong time to cleave with trypsin, you don't start getting cleavage next to hydrophobics, which then makes the analysis of the peptides much more complex. So the analysis of the peptide, a lot of people have done a lot of peptide chemistry, and I was telling this story before. I always go off on tangents. But Stein and Moore won the Nobel Prize. Maybe this is what you do when you get old, but Stein and Moore won the Nobel Prize, you know, in the 1950s, the 1950s, for separating amino acids. Do you know that they had a three story column of Dowex that was composed of anion exchange Dowex and cations? It was all polystyrene backbones of anion and cation polystyrenes, to be able to separate the amino acids. OK. And when you do that, of course, it gets stuck on the resin. Your recovery's out of the bottom of this chromatography. You need tons of stuff to put on the column in the first place. And this is what's happened. I mean, you have a little tiny HPLC column that has huge number of theoretical place that allows you amazing separations. I mean, again, the technology is sort of mind boggling, what you can do now. OK, so what you're doing here is then you're just asking the question, if you have a post-translational modification, x, you can either look at the entire protein. And so you could probably tell it was modified, but telling the location of the modified location, you can't, or you can treat it with trypsin. And then you get, again, with trypsin, you have little pieces. And one of these little pieces will have an x on it. And then you can define it. And then if you want to do sophisticated analysis, you can hit it-- use a second mass spectrometer, and actually sequence this. OK. So I think the next one just briefly goes to MALDI. And MALDI-- so Matrix Assistant Laser Disorption-- have any of you ever done that? OK. So where do you do that? Do you do that in [INAUDIBLE] lab? AUDIENCE: No, in the undergrad lab. JOANNE STUBBE: Oh, OK, because this is Brad's new thing. OK. OK, so you're looking at peptides. OK, so what do you use as the matrix? AUDIENCE: We used some aromatic acid. I don't remember. JOANNE STUBBE: OK. So you probably used sinapinic acid. AUDIENCE: [INAUDIBLE] JOANNE STUBBE: OK. So this is so-- you're using a different one still from this one-- this is-- I don't know. I got this idea somewhere. I don't know. So when I've done this-- I did do this maybe 10 years ago-- I've looked at a lot of peptides. We went through five or six of them before we found one that really worked well. So I don't know how state of the art has become, you know what it is. But the other one in the book that I got this from was, again, an acid. And so what is the idea? So the first thing you have to do is you have to ionize. So the way you do that is you mix your matrix and solution with your protein of interest, your analyte, then you evaporate it. So you have a solid on a little plate. And then you use a laser beam at 337 nanometers. And the light is absorbed by whatever the matrix is and causes you to have a plume of material. This is, again, amazing to me that the protein goes into the gas phase. And then, you have to go through this, go into the analyzer. Did you do time of flight? OK, so you have time of flight. So you guys know what it is then. And in the end, you do detection. So, again, the protocol is the same, but the method is different, and this is widely used and easy really easy to use nowadays. So the issue then is this is what you face when you're looking at a whole proteome. So you just can't calculate the mass of all the proteins from the gene sequences. Why? Because almost every single amino acid in your proteins are modified. So that adds complexity to all of this. So de-convoluting the mass spec becomes more complicated. So this just shows you, you don't need to look at this, but if you look at cystine, you could form disulfides. You can attach a prenyl group, an isoprene group on it. You can attatch palmitic acid on it. You can sulfenylate it. You can nitrosate it. So you have many, many modifications of the amino acids that are chemically reactive and involved on catalysis, and then not only involving catalysis, they are involved in regulation. So that then adds to the complexity of trying to deconvolute what the mass spec, I think, is actually telling you. And then, sorry, it went backwards. And so then what that does is tells you-- whoops. I'm just completely discombobulated here. OK, so what that does is that, again, you're just adding different masses on to all of these amino acids. The problem is that you have modified, and you have unmodified. And the question is what's the distribution? OK, and so if you have a very non-abundant protein, and most of it's unmodified, it's going to be much harder to find. So these are just things you need to think about, and your technology to look needs to be extremely well worked out, so that when you look and you don't find something, you know what the lower limits of detection are. So here we are at our system. Now we're into the Carroll paper, and so what we're looking at is sulfenic acids, degenerated by hydrogen peroxide. We'll see-- do you think that's a fast reaction, hydrogen peroxide with a cystine? Anybody have any intuition? I think these reactive oxygen species you're going to find are not so intuitive about the chemical reactivity. I'll give you a table with what we think we know in general. But I think it's not so intuitive. If you look at the rate constants for reaction of a hydrogen peroxide with a cystine it's 1 per molar per second, really slow. OK. So then the question you have to ask yourself, so this was something that was debated in the literature for 15 years. Is this so slow that this could never happen inside the cell? Because I just gave you a second order rate concept. So we have two molecules interacting at the concentration, this could be high. This is really low. You can calculate the rate constant for the actual reaction. It's really, really slow. OK, so we'll see that there are some proteins, peroxiredoxins that are in humans, are there in quite high levels that can increase this rate to 10 to the fourth per molar per second. So there's a huge rate increase but you need to think about all this kinetic stuff to really understand if this modification can happen inside the cell. Otherwise, well, if it can't happen, why are you wasting your time looking for it? Which is what a lot of people are doing scientifically. OK, so let me see what the next-- OK. So now we're into making a reagent that can specifically modify this, or specifically modify this. OK. So the reagent that they chose-- she didn't invent this reagent-- was dimedone. And this reagent specifically interacts with sulfenic acids. It doesn't react with the free cystine. So you've got to study all of this. And if you're going to use this as a reagent inside the cell, you want it to be fast. You don't want to take 30 hours to do the reaction. You want it to be over fast, and you want it to happen at pH 7. So how do you think this reaction works? Where's the most reactive part of this molecule? AUDIENCE: Those two protons? JOANNE STUBBE: So two protons. So this these have low pKas, so you can easily form the enolate. Depends on the details, the experimental details. And now you have this, and what you end up with is this molecule. And so the question is, does this go in 5%? You need something that goes in quantitative yield at pH 7, rapidly. OK, we're going to come back and talk about what the issues are, because the issues are even harder if you want this region to work inside the cell. OK, we're doing this on glutathione peroxidase, which is what he's using as a model to see if all of this stuff works. OK. So what you really want to do if you're thinking about regulation in the end, is you want to know how much is in each form, and you know, if you read hundreds of papers published on methods trying to figure this all out, but what she did in this case, was she developed a second reagent with an iodo group. OK. And as you can see, what is the product of the reaction? The product of the reaction is the same as the product of this reagent. But this reagent does not react with sulfenic acid. OK, so you get no reaction. So how does this reaction work? What do you think? The what? AUDIENCE: SN2. JOANNE STUBBE: So it could work by an SN2, but the way probably works is it attacks the iodine. So you form-- this is probably the mechanism from what's been done in the literature. So you attack this, and you form this, which then gets attacked by the enolate. So it doesn't really matter what the mechanism is, but the key thing is for this to react-- if you're interested in a mechanism, which I am, it does matter what it is. So the key thing is now you have the same reagents. So how could you ever use it attached? How could you ever use it to distinguish sulfenylation from a cystine. So what did they do in this paper? AUDIENCE: [INAUDIBLE] JOANNE STUBBE: Yeah, so they put the deuterated form on this. So what they did then was in this paper, so you got to keep these straight, if they see deuteriums present, so they made this deuterium label, and this protonated so now you have a mass difference of 6. OK. And in the system, they're using glutathione peroxidase, which has three cystines in it. And one of the cystines is more reactive than the other two, but for proof of concept they mutated two of these cystines into serine initially, so you only had a single reactive cystine, but then they went back and studied the whole protein. OK, so let me just introduce you to this, and then we'll come back and talk about this next time. Let me just do one more thing. OK, so here is the difference in mass between these two species. So this is what you're looking at. If they start out with deuterium labeled dimedone, the peak that they observe is going to be associated with sulfenylation, and if they start out with the protonated material, the peak they observe is going to be associated with the [INAUDIBLE] group. OK, so that's the idea. And then what they did was they simply took their protein, and they have, in this case, 50 micromolar of their protein, and then they increase the concentration of hydrogen peroxide. They don't really talk very much about how they design the timing, but they use, you know, two equivalents. So they use variable amounts of hydrogen peroxide. And what you can see is the maximum amount. So now what you're using, we talked about this before, but we're using anti dimedone antibodies for the detection. And here, they're starting with no hydrogen peroxide. So you don't see any dimedone derivative, and then you increase the concentration. But you get to the highest concentration here that they looked at. So it's 100 micromolar versus 50 micromolar in the protein they used. But what did this immediately tell you? Did any of you look at this data very carefully? What is this? This guy here is associated with a [INAUDIBLE] group that is only reacted with iododimedone, so if you got 100% yield, what does that tell you? This tells you the maximum amount of material you're going to observe. So if you look at this peak, and you look at that peak, you can't do this by eyeball. You need to do this quantitatively. The phosphor images or methods that allow us to do this quantitatively. What do you see? AUDIENCE: It's not at the max. JOANNE STUBBE: Yeah, it's not at the max. And so what we'll do next time-- so these are sort of controls, and the question is how effective is this reagent, and if you start hanging stuff off of your dimedone over here, are you going to change the rate of modification? Can it get into the active site where this SOH actually is, these are the kinds of things we're going to talk about next time when we look a little bit more at the details of the reaction with this, and you should look at the reaction with gap dehydrogenase, which is another control enzyme they ended up looking at it, because what they do is address what the issues are that you're going to encounter when you get into something real that you care about. And that's much more complicated. OK so that's it.
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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. BOGDAN FEDELES: Hello and welcome to 5.07 Biochemistry online. I'm Dr. Bogdan Fedeles. This video is about pyridoxal 5 phosphate, or PLP, an essential metabolism cofactor derived from vitamin B6. All animals are auxotrophic for PLP, meaning they need to supplement their diet with vitamin B6 in order to survive. PLP is one of the most ancient cofactors, and surprisingly, it can catalyze chemical transformation, such as a transamination even without an enzyme. PLP is actually involved in a staggering number of biochemical transformations. This video summarizes the most important reactions involving PLP that you will see in 5.07, and will also show you how to write the complete curved arrow mechanisms for these transformations. Let's talk about PLP-catalyzed reactions. As we just mentioned, PLP is the cofactor derived from vitamin B6. This cofactor is very important for a number of reactions. In this course, we're going to look particularly at the transamination reaction. This is a crucial reaction for the metabolism of all amino the acids, and we're also going to encounter this reaction when we replenish the intermediates in the TCA cycle, what we call anaplerotic reactions. And we're also going to see PLP involved in reactions in the malate-aspartate shuttle that transfers redox equivalents, reducing equivalents, between mitochondria and cytosol. Let's take a look at the structure of vitamin B6, also known as pyridoxine. This is the molecule that we ingest when we get our daily vitamin supplement. Now in the body, this gets oxidized to form intermediate called pyridoxal. Notice the aldehyde group here, which is going to be the business end of the molecule. Now, the active co-factor, PLP-- it's actually the phosphorylated version of pyridoxal. This requires one molecule of ATP and the enzyme pyridoxal kinase. And we get PLP, or pyridoxal 5 phosphate. The name PLP comes from the initials as outlined here. Now, this nitrogen on the pyridine ring tends to be protonated because it's pKa, it's close to physiological pH, between 6 and 7. Now, for the rest of this presentation we're going to be abbreviating this phosphate group as such, and throughout the course. Now, a related molecule is pyridoxamine 5 phosphate, which we'll see, it's an intermediate in the mechanism of PLP-catalyzed reactions. Also known as PMP. Now, notice the PMP has an amine group here which replaces the aldehydic group, which is the business end of the molecule. Now, in all reactions with PLP, this co-factor is actually covalently bound to the enzyme that uses it. Typically, there's a lysine in the active site of the enzyme. As you remember, lysine has an amine group, and this can form a Schiff base with the aldehyde. The reaction proceeds in two steps. First, we form a tetrahedral intermediate. As such. And then, we form the Schiff's base. So this will be the enzyme-bound PLP. And this is where all the PLP-catalyzed reactions start. Let's take a closer look at the transamination reaction. Transamination reaction occurs between an amino acid and an alpha keto acid. We have here amino acid 1, where we highlighted the amine group, and alpha keto acid 2. As you can see, there's a keto group next to the carboxyl. Now, in a transamination reaction catalyzed by PLP, the amine group moves from the amino acid to the keto carbon of the alpha keto acid. And we obtain a new alpha keto acid, and a new amino acid. So in effect, the PLP-catalyzed transamination reaction facilitates the transfer of the amine group from an amino acid to an alpha keto acid. Now, this reaction actually occurs in two steps. In the first step, the amino acid transfers the group to the co-factor itself. So if you remember from the previous slide, the PMP contains an amino group, and that will actually contain this amino group that was taken from the amino acid 1. Now in the second step, the PMP will transfer its amino group to a different alpha keto acid to generate a new amino acid. Now, there are enzymes for virtually every single amino acid that can accomplish this first transformation, where by using PLP, to transfer the amine group and form an alpha keto acid and PMP. Now, in the second part of the reaction, however, the alpha keto acid 2 is typically alpha keto gluterate or oxaloacetate. So in this case, not any alpha keto acid can function. Alpha keto gluterate or oxaloacetate. Now, let's take a look at an example. For example, glutamate. It's going to be our amino acid. And oxaloacetate is going to be our alpha keto acid. And in a PLP-catalyzed transformation, we will obtain the alpha keto acid corresponding to glutamate, which is alpha keto glutarate, and the amino acid corresponding to oxaloacetate, which is aspartate. This enzyme that catalyzes this transformation is in fact ubiquitous, and it's found both in the liver and the muscles, and it is in fact a-- we can call it, depending and the product, we can call it aspartate, transaminase or glutamate oxaloacetate transaminase. In fact, if we find this in the bloodstream, this enzyme acts as a biomarker. And it tells us about some damage that might have occurred in muscle or liver, which forced the cells to spill out their contents. This biomarker is-- you'll often see as SGOT-- serum glutamate oxaloacetate transaminase. And this is just one of the biomarkers that are measured in blood tests that tells us about heart disease or liver disease. Let's take a look at the mechanism of the transamination reaction. And in particular, we're going to take a look at part one, which as we discussed before, the amino acid reacts with PLP to form an alpha keto acid and PNP. Here is our co-factor PLP, covalently bound to the lysine in the active side of the enzyme via a Schiff's base. And here is our amino acid starting material. So in the first step, the lysine that forms the Schiff base with the co-factor is going to be replaced by the amine functionality of the amino acid, and it will form a new Schiff base with the co-factor. This starts with the amine group attack on the pyridoxal carbon to form a tetrahedral intermediate. And following an additional proton transfer, the lysine can be kicked off to form the new Schiff base. So far, we have started with the Schiff's base corresponding to the PLP bound to the enzyme and we now obtain a co-factor forming a Schiff base with the incoming amino acid 1. So this portion of the mechanism is called transamination, because we're starting with one amine and we're forming a different amine. Now, let's take a look at the alpha proton attached to the alpha carbon, which we're going to highlight here. This proton is now in between two carbonyl-like groups. Here is the carboxyl group and here is the Schiff base group. So it becomes acidic enough that it can be removed by an active side base, for example the lysine in the active site. This will generate a carbanium alpha carbon. This carbanium can only form because it is resonance stabilized. And indeed, the PLP link system-- it's highly conjugated, and it's a good electron sink. For this carbanium, we can write, in fact, many different resonance structures. Let's take a look at one of them. This symbol denotes resonance structures. Notice in this structure that the positive charge on the pyrodine nitrogen is now gone, and highlights the fact that this is a good electron sink. And the ring now looks more like a quinone. That's why we call this a quinoid structure, or intermediate. This quinoid structure shows us a glimpse into how the reaction will proceed, because the alpha carbon now-- it's doubly bonded to a nitrogen, which anticipates how this alpha carbon will become a keto group and a product of the reaction will be an alpha keto acid. What happens? The quinoid structure can be re-protonated, but at a different place. For example, on the aldehydic carbon of pyridoxal. To highlight that these facts-- these structures are in fact resonance structures, we're going to put them in brackets. So let's take a look at what happened in this past couple of steps. So we had an alpha proton that was fairly acidic, it was able to be removed by the active site lysine. And then this proton came back to a different position. So all that's happened in just a couple of steps was a proton transfer. Now, looking at this intermediate, we can see that in fact the Schiff's base or the imine of the PMP form of the co-factor and the alpha keto acid corresponding to amino acid 1. So via a hydrolysis reaction, these two can come apart. So in the first step, an activated water molecule attacks alpha carbon, forming a tetrahedral intermediate. And one more proton transfer and we're going to kick off the pyridoxamine form of the co-factor. And notice we obtain PMP and the alpha keto acid corresponding to the amino acid 1. This last step is, in fact, just a hydrolysis reaction of a Schiff base. Now, let's take a look at the second part of the PLP transamination reaction. The part one left us off with formation of PMP. So in this second part, PMP will react with a new alpha keto acid to regenerate PLP and a new amino acid. In fact, this part of the mechanism-- it's the exact reverse of part one. Here is PMP and our alpha keto acid. In the first step, we're going to form-- as we've gotten used so far-- to a new imine between the keto group of alpha keto acid and the amine group of PLP. As usual, first we're going to get a tetrahedral intermediate. And one more proton transfer, and we can kick off the OH group to form the imine. Now, this portion of the reaction is, in fact, imine formation, which is the reverse of the hydrolysis step that we saw in part one. Now, as you remember, there is an active site lysine which can act as a general base. And it's going to de-protonate one of these two protons on the pyridoxal ring. The reason that we can form this carbanion here is because this negative charge is delocalized throughout the entire ring system. And let's show one important resonance structure. Which is none other than the quinoid structure we saw before. Just as before, the protonated lysine can now donate proton on a different position. For example, the alpha carbon of the alpha keto acid. As you can see here, now the proton is on the alpha position. And now where this starts to look more like the Schiff's base formed by an amino acid with the PLP version of the co-factor. So from here on onwards, we're just going to substitute the PLP-- the amino acid bound to the PLP-- with the active site lysine in the transimination reaction that we saw before. So first, the lysine can attack this carbon, forming a tetrahedral intermediate. And then, following some proton transfer, we can kick off the amino acid and generate the Schiff's base corresponding to the PLP co-factor bound to the enzyme. So here we have PLP, enzyme bound, and the new amino acid 2. We mentioned PLP is a very versatile co-factor, so let's take a look what other reactions besides transamination can PLP catalyze. One interesting reaction, used especially by bacteria, is a racemization. This involves taking an L amino acid, for example L alanine, and converting it via a PLP-catalyzed reaction to D alanine. This is an important reaction for bacteria, because they incorporate the alanine into the cell walls, which make it very hard to recognize by the immune system, and makes it very hard to digest by the host proteases. Let's take a look at a key intermediate in the PLP-catalyzed reaction. As before, L alanine is going to react with the PLP bound to the enzyme, and it's going to form an amine. Here we're highlighting the stereochemistry of the alpha hydrogen. And here is our active site lysine. As we've seen before, this alpha hydrogen is acidic enough that it can be removed by the lysine. And it's going to form a carbanion at this position. Now, this carbanion, as we've seen before, is able to form because the charge is, in fact, delocalized through the entire system of the pyridoxal ring. So for this structure, we can write a number of resonance structures, which we're not going to mention here. Now, this carbanion can be re-protonated. And here we had a-- the hydrogen was pointing up on the top of the plane of the page, but we can re-protonate it from the bottom, and that will change the stereochemistry of this carbon. See if that re-protonation happens from the bottom, we will obtain the Schiff base corresponding to the D alanine. So by being able to generate this carbanion at the alpha position, the PLP reaction and co-factor allows the inversion of the configuration and the alpha carbon, converting L alanine to D alanine. Now, another interesting reaction that requires PLP is de-carboxylation. Here we're looking at an amino acid-- for example, glutamate. In a PLP-catalyzed reaction, you can lose this CO2 group and form this molecule, which is called gamma aminobutyric acid, or GABA. This is, in fact, a very important neurotransmitter and inhibitor, a neurotransmitter that is required in the brain. Let's take a look how this de-carboxylation is catalyzed by PLP. As always, have we seen so far, the glutamate will react with PLP bound in the active site of the enzyme, to form a Schiff base. Here is the Schiff base. Like that. Now instead of de-protonating at the alpha position, the CO2 is activated to leave. Because it will leave behind the carbanion. Such as that. And as we've seen before, a carbanion formed at this position can de-localize throughout the entire pyridoxal ring, and therefore it stabilize and it can exist long enough. And we're not going to draw, but there-- you can imagine, there are a number of different resonance structures. Now, this carbanion-- it gets protonated quickly by a general acid, and will generate this structure, which is just a Schiff base corresponding to gamma aminobutyric acid with PLP. And now, from here, a transimination where the active site lysine will remove the PLP and free up the GABA products of the reaction. In this video we talked about PLP-catalyzed reactions. PLP is the co-factor that comes from vitamin B6. Here's vitamin B6, what we call pyridoxine, which is the molecule that we find in our vitamin pills. Now, in the body, pyridoxine formed pyridoxal, which is activated to form pyridoxal 5 phosphate, or PLP. And typically when it reacts, PLP is found as a Schiff's base bound in the active site via a lysine. PLP is very important for transamination reactions, which are essential for the metabolism of all amino acids. We have seen the transamination reaction where an amino acid 1 reacts with alpha keto acid 2 and the PLP catalyzed reaction forms an alpha keto acid 1 and amino acid 2, essentially transferring the group-- the amino group from the amino acid to the alpha keto acid. The reaction occurs in two steps, where first the amino acid is transferred to PLP to form PMP. Then PMP then transfers this group-- the amino group-- back to a alpha keto acid to generate a new amino acid. As we saw, the mechanism of transamination involves multiple steps. The first step is a transimination reaction where the Schiff's base that's formed between the active site lysine and the PLP becomes a Schiff's base between the incoming amino acid and PLP. Next, we have a proton transfer, which is allowed by the ability of the PLP ring to stabilize a negative charge, and move the proton from the alpha position to somewhere on the PLP ring via a quinoid structure. And finally, that the resulting Schiff base is hydrolyzed to generate PMP and an alpha keto acid. In the second part of the reaction, PMP now reacts with alpha keto acid to form a new Schiff base, and then the proton transfer happens in reverse, via, again, a quinoid structure, to generate the Schiff base corresponding to the PLP and the new amino acid 2. Which, via a transamination reaction will generate amino acid 2 and the PLP enzyme bound. Finally, we mentioned that PLP is very versatile, and it can catalyze other reactions, such as racemization, for example, switching the configuration of the alpha carbon from L alanine to D alanine, or de-carboxylation, generating alpha aminobutyric acid, or GABA, an important neurotransmitter from glutamate.
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https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/8.01sc-fall-2016.zip
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OK. Well, I like to lay in the hammock in the summer. And this is our version of that hammock. So we're having one rope, or perhaps two ropes, hanging between two trees here. And they span an angle theta on each side here. And we want to know what is the tension in this rope at the midpoint and here, right where it is attached to the tree. And here we already have the free-body diagram. I just drew this piece of the rope here that's hanging on this tree here. And we made an imaginary cut. We just cut it in the middle, which means gravity is obviously acting on it. It's hanging through. But it does so with m/2 here for our half rope. We know that there is a tension here at the midpoint, Tmid. And up here, we have a tension at that end point that goes under this angle here. I just use an ordinary Cartesian coordinate system with the i-hat direction going in the x direction and j-hat going in the y direction. So all we need to do is we need to apply Newton's second law and do an F equals ma analysis to figure out what these tensions are. So let's apply Newton's second law, the F equals ma analysis, to figure out what the tensions are at the midpoint and at the end over there. So we have F equals ma. And we'll have to very carefully separate the components here. Let's start with i-hat. We have Tmid minus Tend, but of course we have only the projection of Tend, so this is Tend sine theta. And since this rope is just hanging there, the acceleration is 0. In the j-hat direction, we have minus m/2 g and then plus Tend. And here we have the cos theta component, and that is also 0. So we see from here pretty much immediately that Tend equals m/2 g over cos theta. And we can stick this one here into the i-hat equation. So we'll get Tmid equals m/2 cos theta g sine theta. And that's nothing else but mg/2 over 2 tangent theta.
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https://ocw.mit.edu/courses/5-310-laboratory-chemistry-fall-2019/5.310-fall-2019.zip
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SARAH HEWETT: All right. I guess we can get started. So today, we're going to talk about what's significant in laboratory measurement and how to take measurements in the lab, how to do calculations with the lab, and how to do some of the data analysis when you have a lot of quantitative data, which you will have in the lab coming up when we do the Ellen Swallow Richards or the Charles River Lab. One thing I wanted to point out-- so there's hand-outs in the back. We're going to be going through how to do the calculations and how to do all the statistics. And there's an example problem that we're going to work through as we go in the back. So if you want to make sure you have a copy of that, that might be helpful. And one of the things that was pointed out this week in the lab is that in the lab, there is a typo left over from when we were in the old labs in Building 4. And so this Wednesday and Thursday, the TA session that it says, the help session about how to write the ferrocene lab report is going to be in the lab, not in 4:00, 4:30 as it says in the lab manual. So just come to lab as normal. The beginning of the lab period will be a quiz on the ferrocene lab, so anything procedure wise, calculations, reactions, all that good stuff, things that you should know, having done the ferrocene lab and hopefully having started to write up your report a little bit. And then after the first 20 minutes or so, it's going to be the quiz. And then your TAs will give a little bit of a lecture about what they're looking for in the ferrocene reports. So that's a good time to ask any questions about-- if you're unsure about where to write anything, how to write anything up if you have anything that you've already written and you want them to look at. They will be in the lab to help you out with that. And we have some more office hours, too, that got posted to Stellar. And I will have a slide about them at the end of this lecture, too, for more help. But, yeah, day four of ferrocene-- come take the quiz, then get some help on your reports. It is in the lab. You don't have to go anywhere else. Just come to lab as normal. OK, so if we're going to be talking about measurement today, then we need to talk about what things you should be thinking about if you're going to measure something. So if you're going to measure the length of this piece of wood with this ruler, how long would you be able to say this piece of wood is? AUDIENCE: There's no units. SARAH HEWETT: There's no units. Yeah, that's a problem. What else is wrong? AUDIENCE: It's not long enough. SARAH HEWETT: It's not long enough. So what's the minimum we could say about this piece of wood? AUDIENCE: [INAUDIBLE] SARAH HEWETT: It's longer than 1 something. So this is not a great measuring tool. What about this? Any better? Slightly better. So now, what could we say about the length of the piece of wood? AUDIENCE: Between 1. AUDIENCE: Between 1 and 2. SARAH HEWETT: It's between 1 and 2, probably a little closer to 2. But we can't really say too much more about it than that. And we still don't have any units. So this is just to get you thinking about what we need to be considering when we are choosing how we're going to measure something and when we were recording measurements, what information we can reasonably get from the measuring tools that we're using. So to talk about measurements a little bit more, we're going to need to talk about some terminology and get this out of the way, so that we're all using the same language. And you've probably seen these terms before in your other science classes in high school and middle school, going way back. But just as a quick refresher, so we're all on the same page, precision is how close repeated measurements are to one another. So if you measure the same thing a bunch of times, you would like the results to be the same. And you've probably seen these bulls-eye diagrams where if the results are all really close together, that is very precise. And you can measure that with the standard deviation. So the standard deviation gives you an idea of relatively how close your points are to each other. Accuracy is how close a measurement is to the true value or the value that you actually are trying to measure. So if you are throwing darts and you are trying to get the bulls-eye, then if you have high accuracy, then you have them all in the bulls-eye. If you have a high accuracy and high precision, then all of your darts always go to the center. You can have high accuracy where you're always around the target, but your points are scattered. And then you can have low accuracy and low precision where you're just all over the place. We measure accuracy by the percent error. So it's the error relative to the result that you're trying to get. And the answers don't need to be close to each other to be accurate. Like I said, they can be spread out, but still all close to the target value. So these are some terms that you probably heard there a little bit, used differently in chemistry sometimes than you may use them colloquially. So we just want to go over that. More terminology-- absolute error is how far away a measurement is from the true or accepted value. So if you are trying to measure, let's see, with this here, pipette, this pipette is designed to measure 10 milliliters. And you can look at it. And it has a tolerance of 0.04 milliliters. So that's an absolute error. It is how far away your measurement will be from the true value if you're trying to measure 10 and you measured 10.04. Then your absolute error is 0.04 milliliters. So you just subtract what you got from what you intended to get. The relative error is how far a measurement is from the true value, but it's relative to the quantity that you're trying to measure. So if you've measured 0.04 or 10.04 and you intended to measure 10, then your absolute error is the 0.04. To get your relative error, you'll make it a percent and divide it by the 10 that you intended to measure. So now, you have a 0.4% error. One way to think about this is if you have an absolute error of 1 milliliter when you're measuring out 2.5 liters of material, that's pretty small. But if you have an absolute error of 1 milliliter when you're measuring out 2.5 milliliters of material, then that becomes a much bigger problem. So depending on the quantity that you're measuring, sometimes it's a little bit more useful to report the relative error than the absolute error and vice versa, depending on the application. If you say, yeah, it had an error of 1 milliliter, and your friends are all saying, oh, you did such a great job, and then you say, I was only trying to measure 2 milliliters, then not so great. Other types of error are random error and systematic error. So random error is things that we can't really control for or identify. And it causes data to fluctuate relatively uniformly around a mean value. And that's one of the things that's given on the uncertainty in the measuring apparatus that you're using. So this pipette measures 10 milliliters, plus or minus 0.04 milliliters. That is a random or indeterminate error. The data will fluctuate around 10 milliliters within 0.04. You know that. And you can control for that using statistical analyses, which we'll talk about in a little bit. Systematic or determined error is error that has a known cause. And that could be from faulty or poorly calibrated instruments, human error, or chemical behavior inside reactions. So if you intended to get 5 grams of product and you got less than 5 grams, then you know that there is some error associated with your experiment. And it could be because maybe some of your iron oxidized. Or maybe you spilled some of your ferrocene. Or maybe you were not tearing the balance every time you used it. So that'll introduce error into your measurement if you were using the equipment improperly. Or if you're reading from a burette and you're always reading in the wrong direction or you didn't read to enough significant figures, that's systematic error. And so that's not necessarily from the glassware itself. That's from how you were using it. And these are types of error that you can talk about in your discussion section of your lab report. So you could talk about the error that is associated with the measurements that you can control and that you cannot control. So the last term that we need to talk about in terms of when we're making measurements or doing things in the lab is, what does it mean for something to be significant? So you may have heard of the term "significant" figures or this is a significant result. What does it mean for a number or measurement to be significant? And what makes a measurement that we take significant in the lab? Yeah? AUDIENCE: Ideally, significant means all the numbers that you're certain about plus one estimate or one final. SARAH HEWETT: Exactly. Yeah. So a number is significant. And you report the significant figures-- are in a measurement, all of the numbers that you're certain of, plus the first uncertain digit. So a lot of times, you'll have some sort of-- you won't have measurements every-- maybe tenth of a milliliter or 1 milliliter, depending on the size of the glassware that you're using. So you can be certain of some of the measurements. And then you're uncertain of others. So in our piece of wood, we were certain that it was bigger than 1, but we were uncertain of the next digit. So we could estimate that second digit. And that counts as significant. A quick review of significant figures and how to use them. All non-zero digits in a number are significant. Zeros in between non-zero digits are also significant. Zeros to the left of a non-zero digit are not significant. Zeros to the right are only significant if there's a decimal place. And you can avoid ambiguity in that with scientific notation. So if we just go through these really quickly, how many significant figures does this have? Four. AUDIENCE: Three SARAH HEWETT: Three. AUDIENCE: Six. SARAH HEWETT: Six. AUDIENCE: Two. SARAH HEWETT: Two, maybe. AUDIENCE: Maybe? SARAH HEWETT: As written, yeah, you would probably call it two, because the zero is to the right. But if you wanted the zero to be significant, then you could write it like this. And now, how many significant figures do we have? AUDIENCE: Three. SARAH HEWETT: Three. And if the zero is not significant, then to avoid people being confused about it, you could write it like this. And we have two. So that's one way you can use scientific notation to get around ambiguity. Yes? AUDIENCE: Yeah, I have a question. So for 590, if someone put a decimal point after the zero-- would it be three significant figures, or would it be two? SARAH HEWETT: So that's a question that I've always gotten. I think, just conventionally speaking, if you wrote something as like this, maybe. More conventionally correct would be to write it as 5.90 times 10 to the 2. Because you don't usually have a decimal place if there's nothing after it. And then if you wrote it like this, though, then you have four significant figures. And that's more than you want. So just go with the scientific notation, instead of having a hanging decimal place. Because that's not the typical way to write numbers. When you're doing math with significant figures, if you're adding and subtracting, then your final answer should have the same number of decimal places as the number you started with that has the smallest number of decimal places. When you are multiplying and dividing, the answer should be rounded to the same number of significant figures as the number you started with that has the least number of significant figures. So when you are multiplying and dividing, you want to count up your sig figs. And when you are adding and subtracting, then you pay attention to your decimal places. So if we go back to our piece of wood, then using that ruler that we had earlier, we can estimate. Maybe this is 1.8. So we have gradations every one unit. We still don't have units. Whatever this is. And then you can estimate this, one digit pass that. And then our uncertainty is our gradations divided by 2. So we know that we are within 0.5 of the correct answer. Because we can estimate that it's within this half of our interval. If we add more gradations to our ruler-- so if we have markings every 0.1, still no units, then we can say with certainty that our piece of wood is 1.567 bigger than 1.7 and smaller than 1.8. So it's 1.7 something. You can estimate that last decimal place. And then your error gets smaller to your uncertainty in that measurement. This is important in the lab. Because you need to make sure that you're choosing the correct apparatus for the amount of uncertainty that you want in your measurement. So if you see something in your lab manual that says measure 1.2 grams, then this balance is totally fine. Because the balances-- if you've never noticed at the top, they have the uncertainty written on there. So the uncertainty is in this last decimal place here. So if you want 1.2 grams, then you can't get 1.2. And then your last digit is the uncertain one. Should always write down all of the digits that are on the balance screen. But when you're going to calculate uncertainty or do an error propagation, then you know that the last digit is where your uncertainty is. This is plus or minus 0.01 grams. So we have three different types of balances in the lab. So you want to make sure that you're choosing the one that has the correct number of decimal places for the quantity that you are trying to measure. And the same thing goes for glassware. So the uncertainty in the glassware is frequently reported on the glassware itself. And there are standard tolerances for various sizes and grades of glassware. So if you've ever seen glassware that's volumetric, it may say either an A or a B on it. So the A is the highest standard. And then B is a slightly higher tolerance. And there are standard things of those that you can look up. But it's usually written on the glassware. And so this is an example of the 10-milliliter pipette that I was showing you earlier. So you can look at the pipette. And then on the top part here, it'll say 10 milliliters. And this one says plus or minus 0.04. This pipette says plus or minus 0.02. So that one is calibrated a little bit more precisely. These are also 10-milliliter pipettes. And you can look on the top part here to figure out what their tolerances. These are both 0.06. And one of the other things that you need to make sure when you are using different types of glassware is to figure out where the markings end. So these are both 10-milliliter pipettes. And they are designed to measure 10 milliliters within 0.06 milliliters. This one-- I don't know if you can see it from here. Probably not, but maybe if I hold it up here. The markings-- and they go from 0 to 10. And it ends here. And there are no more markings at the tip. So that means the only time you're measuring something is when your liquid is within where the markings are. So if you're going to measure 10 milliliters, you need to start your liquid here, end it at 10. And don't go below that, because that quantity is not calibrated. There are some pipettes like this one that are also designed to measure 10 milliliters, but these have gradations all the way to the end of the pipette. So to get 10 milliliters out of this one, you would fill it up to here and then drain it all the way out to the bottom. So you want to make sure that you check your glassware ahead of time to figure out how much of it you can use to measure liquids and what you get if you empty it all the way out, versus measuring by difference. If you look at your graduated cylinders, your graduated cylinders are different sizes. And the markings come in different levels of precision. So this has markings every half of a milliliter. And its tolerance is listed at the top as plus or minus 0.3. The 50-milliliter graduated cylinder has markings every 1 milliliter. And at the top, it'll tell you that its tolerance is plus or minus 0.5 milliliter. So again, that's the 1 milliliter divided by 2 rule. So sometimes that holds. And sometimes it's a little bit different. Volumetric glassware always has a smaller tolerance, so this is to 250-milliliter volumetric flask. And on here, it will tell you that its tolerance is plus or minus 0.12. So you should always look at these. And when you are making measurements in the lab, when you're writing things down in your lab manual, you should always write down what the uncertainty in the measurement that you're making is. So whether it's on one of the balances or a volume, always make sure that you write down what the uncertainty of the glassware that you're using is. If you look at something like a beaker, this also has a tolerance on it. And it's plus or minus 5%. So that's a lot when you are looking at a 300-milliliter beaker. That's going to be a lot more uncertain than any of these other measuring implements. So it's good to know before you choose, what piece of glassware you going to use to measure something, how precise you want your answer to be, and how much uncertainty is acceptable in the measurement that you are making. With burettes-- so this is a 50-milliliter burette. And it doesn't have a tolerance listed on it, but it has markings every tenth of a milliliter. So what would be uncertainty in the burette? Point? So if the markings are every-- a 50-milliliter burette. The markings are every 0.1 milliliters. So the uncertainty is-- AUDIENCE: 0.05. SARAH HEWETT: 0.05, so plus or minus 0.05 milliliters. In a 10-milliliter burette, the markings are every-- let's see. I think they're every 0.2. No. What are these? 0.05. Yeah, these are every 0.01 in a 10-milliliter burette. So what would the uncertainty be here? AUDIENCE: 0.005. SARAH HEWETT: Yeah. So make sure that you are looking at what glassware you're using, and you record the correct number of significant figures. Because it changes based on the type of burette and even within different types of glassware where they are, made different with varying levels of quality. So we can make one measurement. And then we can figure out what the uncertainty is in that one measurement. But what happens if we take multiple measurements and have to do math with them? So if you look at your worksheet here, this is an example of something that you might do in the lab where you have done a series of titrations to calculate the concentration of some sort of HCl solution. So the first question is, what is the uncertainty associated with a 50-milliliter burette? We already figured that out. And then if we were going to calculate the volume dispensed in each titration trial with the correct error, how would we do that? So one of the ways that you could think about adding up the error in these measurements is just straight up, adding the error. So for this first example, we have 12.52 milliliters. This chalk is no good. And then do volume by difference. We have 0.24 milliliters. And each of these has an uncertainty of plus or minus 0.05 milliliters. So if we added up the error in these measurements, then we would get 0.1 milliliters. But it is probably pretty unlikely that both of these measurements were off by the full 0.05 milliliters. So this would tend to overestimate the amount of error in your final answer. So what we do is we propagate the error. And when you're adding and subtracting, you use the absolute errors. Because you should be adding quantities that have the same units. So units will be consistent throughout your calculation. And when you do this-- did anybody subtract this out? I don't know if anyone has a calculator. Got one? AUDIENCE: Yeah. SARAH HEWETT: Great. So if we subtract these two numbers and we get 12.2 milliliters. And then how do we propagate the error for this calculation? So our error is going to be the square root of the squares of the errors of each individual measurement. So each one is-- so if anyone has done that-- or you do it in your head. Why not? The answer that you get is 0.070710678. So that's a lot of extra decimal places. And what we really care about in uncertainty is, where is the first decimal point that is uncertain? What is that first digit that we don't know for sure? And so we always round an error to the first significant figure. So the error associated with these measurements is just 0.07 mL. So to right this perfectly, you would write it as 12.28 minus 0.07 milliliters. And you can calculate the volumes for each of those trials. And then the error for all of them is going to be the same. Because the error for each of your direct measurements is going to be 0.05. So then we can talk about what happens to the error in a calculation when you are multiplying or dividing something, which is this next part. So if you're going to use this value to calculate the concentration of the HCl in this fake experiment, then you will have to do some multiplication and division. And you'll be using different quantities that have different units. So you can't just use the absolute error, because they have different units. So if you have your error with your burette reading, that'll be in milliliters. If you have your error with your concentration, that's in molarity. So we can't use the absolute error. So we'll have to use the relative error, which, if you remember, is the absolute error divided by the measurement that you've made. So you want to take a second and do out that calculation for how to calculate the concentration of the HCl. Then we can do it all together in a second. And we'll just do it for the first trial. And then if you guys want to have more practice, you can do it for the others afterwards. So if we're going to set up this calculation for the concentration of the HCl solution, where would we start? It's the first thing that we have to do or to calculate in this. Yeah? AUDIENCE: Figure out how much sodium hydroxide we have. SARAH HEWETT: Yeah. In what units? AUDIENCE: Moles. SARAH HEWETT: Moles. So we can figure out how many moles of sodium hydroxide we have by taking our volume. And then we have our units, our concentration return, moles per liter. So we need to turn this into liters. And then we can use our molarity to get moles. I guess we could do that as moles of NaOH in 1 liter. And then we can go from moles of NaOH to moles of HCl. What's our mole ratio? AUDIENCE: One to one. SARAH HEWETT: Too easy. One to one. Yeah. 1 mole. And then we have moles of HCl. And then the last step is to divide by our volume of HCl. Yes, which in this case is 0.0100. Yeah, liters. And that should give you, if anybody did it out, 0.60172 molar, hopefully. So that's our concentration. And now, we have to figure out what the error associated with that concentration is. So we will look at the error involved in each of these terms. So we have our error associated with our burette measurement. So our first one is going to be the 0.07 over 12.28 squared plus-- what's the next error that we have? AUDIENCE: [INAUDIBLE] SARAH HEWETT: Which one? No, say it. AUDIENCE: The last one. SARAH HEWETT: Last one. Yeah, we have error associated with this measurement, right? And the error associated with that is 0.02 milliliters. So you have 0.02 milliliters over your 10 milliliters. And then what's the last error that we have? Yeah? AUDIENCE: Concentration of NaOH. SARAH HEWETT: The concentration of NaOH. The 0.04. So you have 0.04 over 0.49. That's how you will calculate your error for the concentration of the HCl. And if you do this out, you get 0.08. So then, how would we report these two together? That's the thing from before. So what is this? What type of error is this? Is it an absolute or a relative error? Relative. So we calculated it from all of the relative errors. So this is a relative error. And it's also-- you could think of this as 8%. So in order to get it back into a standard error that we can report with our 6.60172, then we have to multiply these two together. And if we go back to our sig figs from before-- in this calculation, how many significant figures should our final concentration answer have? AUDIENCE: Two. SARAH HEWETT: Two. So this has two sig figs. So we'll round this to 0.60. Then you can multiply that by your 0.08 relative error. And your final answer should look something like this, way down here-- 0.60 plus or minus 0.05 molar. Any questions about calculating errors and doing error propagation? So this is why it's important to make sure that you have the correct tolerances and the correct errors associated with all of your measurements. Because when you go to do all of your calculations and write your lab report up, it will be very hard to figure out your error if you are missing some. OK. So now that we know how to deal with the various independent measurements, what happens when we make the same measurement more than once? And this goes back way-- probably to middle school when we talk about the mean or the average of a sample. So in statistics land, they call all of the possible values that you could measure of something the population. And it's really hard to know the true mean of the population, unless you have access to the entire population and you've taken all possible measurements, which is pretty much impossible, especially in the case of 5.310. We're not going to be making all of the possible measurements there are for each quantity that we measure. So in the lab, we can take the mean of a subset of the population. We call that the sample and then calculate our sample mean. So population mean, if you ever see in statistical literature, is denoted as mu. And then our sample population is denoted as the x bar, where you just add up all of your measurements, divide by the number of measurements you took. And that's your mean. So we calculate an average or a mean for the HCl solution on the back. If you didn't do the calculations from before, the concentrations that you should get are 0.60, 0.61, 0.61, and 0.64. So if you add all those up and divide by 5, then your mean concentration is 0.615 or rounded to the correct number of significant figures-- 0.62 molar. The other thing that you can calculate is the standard deviation of your measurements. And so remember, that's a measure of precision or how close all of your measurements are to one another. So small standard deviation means that all of your measurements are very close together, very precise. And to calculate the standard deviation, again you could calculate the standard deviation of an entire population. It's the sigma up there. And then you need to know your population mean. But in the lab, we are taking a smaller subset of the population. So we have to calculate a sample standard deviation, which is denoted as S. And the major difference here is that instead of dividing by N, which is the number of samples, you divide by N minus 1, so the number of measurements you took, minus 1. So to calculate a standard deviation, you calculate each of your measurements, subtract it from the sample mean, square it, add those all up, divide by N minus 1, and then take the square root. And you can do this on a scientific calculator, a graphing calculator. Excel does it for you, any statistical program that you want. You don't have to calculate this by hand in your lab manual or in your lab report. You could just show the formula for it. And then you can calculate it using any sort of statistical program. So if we calculate the standard deviation for our concentration of the HCl solution, then you should get-- not going to do this all out by hand right now, because we don't have a lot of time. But if you wanted to try it out later or check your answer for somewhere else, it should be 0.1732. It keeps going. So the standard deviation is typically rounded to the same number of significant figures as the measurements that you are using to get the standard deviation. You'll see it reported pretty frequently like that. So we can report our standard deviation as 0.017. Since we are taking a mean that is not the population mean, we know that we may not have exactly-- the mean of your sample will probably not be exactly the same as the population mean. But you can figure out how good of an estimate your mean is, of the overall population mean, by calculating the standard error of the mean. And that is done by taking the standard deviation of your sample and dividing by the square root of the number of data points that you have. So if you want to get a better and better estimate of the true mean of your population, then you can keep increasing the N value. And eventually, it'll make your standard error smaller to a point, since it's the square root. Once you get a big enough number of N, then incrementally, you get less and less benefit from adding more and more data points. But you can also make your standard error smaller by decreasing your standard deviation. And so if you can make more precise measurements, then that will also help improve the quality of your mean and how accurate it is toward the actual population mean. All of this is leading up to doing statistics. And all of statistics is essentially based on the Gaussian distribution or the normal distribution. And you've probably seen this before. And this graph is an axis of where the x-axis is the number of standard deviations away from the mean. And then the y is the probability of finding a measurement in that space. So you can see that the highest probability of all of your measurements is going to be very close to the mean. So if you're taking measurements accurately, they should be very close to the mean. And then as you get further and further away, there is less probability that some measurement you take will be three standard deviations away from the mean and so on. So this number Z is calculated by subtracting your number from the mean and dividing by the number of standard deviations. So it's just essentially the number of standard deviations away from the mean-- your measurement is. Oh, back up. We cannot go back. There we go. One way that we can use this is to figure out confidence levels of the measurements that we take or the means that we calculate. And so you can come to an error under the curve analysis here-- or an area under the curve analysis. And so for all measurements that are within one standard deviation, plus or minus from the mean, that accounts for 68% of the area there. If you go within two standard deviations, you can stay within 95% confidence that your mean is within that area. And then within three standard deviations, plus or minus, so 99% chance that your mean is within that area. So we can use this when we are calculating the accuracy or how confident we are that have measured the population mean with the samples that we have taken in the lab. We can do this because of the central limit theorem, which states that the distribution of sample means in a population will constitute a normal distribution, even if the population itself is not normally distributed. So if you take a bunch of samples or take a bunch of measurements in the lab, then you can use the normal distribution to statistically analyze them, even if you don't know the distribution of the population itself. So we can use that information to calculate confidence levels of a mean that we calculate. So what that means is we can calculate how confident we are that our mean falls within a certain range of the actual population mean. And you can say it with different levels of confidence. You could say I'm 99% confident. And you'll get a wider range or 95% confident. And you'll get a little bit of a smaller range. And the range of the values is what we call the confidence interval. And you can calculate a confidence interval for the actual population mean. And that is your standard mean, that Z square root we were talking about, times the standard deviation of the population, divided by the square root of N. To use Z as in the previous slide, you need to have a big enough sample size that your estimates of the mean and the standard deviation are a good enough substitute for the actual sigma, the population standard deviation. That's really challenging in lab situations. And in 5 through 10, we will not ever have a big enough sample size for that to be the case. So as an alternative, we can use the t statistic. So to calculate a confidence interval for a mean that you calculate in lab, you can take your mean, plus or minus the t statistic, times your sample standard deviation, over the square root of N. And you can calculate t statistics using this formula if you have all of this information. Or you can get a table of t values for varying confidence levels and varying degrees of freedom, which is N minus 1, which is the number of measurements that you took, minus 1. So if we're going to calculate a confidence interval for the mean of our HCl concentration, then we will have something like-- we need our average, plus or minus ts, square root of N. And you'll notice that this standard deviation over the square root of N is the standard error of the mean. So if you calculate that, then it'll make it slightly easier to calculate your confidence interval. Yeah. And so to do that, we're going to need some t table, which gives you the t statistics. Our average of the mean was 0.62, plus or minus. So the t value-- if we have four measurements, then our degrees of freedom is going to be 3, so N minus 1. And then we want to use a two-tailed t-test. Because we don't know which way our measurement is going to vary. So the two-tail means that it could be higher or lower than the actual mean. So you want to calculate one that has a value of 0.05. So that corresponds to the 95% confidence interval. And 3-- so 3.182 is our t value and then times 0.017. And what you get, if you do that out, is-- so when you're going to report your 95% confidence interval, you would add this number and subtract this number from your mean and then report it as a range in parentheses, so 0.593,0.647. And so that's your 95% confidence interval. And so what that's saying is that there is a 95% chance that the actual population mean lies between these two numbers. You are 95% sure that the actual mean is in there. So in your lab reports, when it says to calculate a 95% confidence interval, that is what we are looking for. Other issues you may run across in your data is if you take repeated measurements of the same quantity. You may have outliers. And sometimes you can look at your data and say, wow, there's definitely an outlier here and other times-- but even if you can do that, it's nice to be able to mathematically demonstrate that it is, in fact, an outlier. And so to avoid subjectivity in whether you're tossing out data points left or right, just to make yourself look better, you can use the Q-test to help decide whether a value can be kept in a data set or should be rejected as an outlier. And the way that the Q-test works is you take the absolute value of the result that you're worried about, the questionable value. Subtract it from its nearest neighbor, so whatever the next closest value that you measured was. And then divide it by the spread of the data, which is just your highest value that you measured, minus your lowest value that you measured. And then you have to look at another table of standardized Q values. And if the Q that you calculate is greater than the Q in the table for the number of data points in the confidence level that you want, then you can reject that point as an outlier at that confidence level. So the last question on here is not really related to the first whole set of problems. But if you took the following measurements for a concentration of an NaOH solution, are there any outliers in that data set? So what's our Q calculation for this data set? Which result might be questionable? 2.86. Yeah, that seems to be a little bit higher than everybody else. So our questionable result is 2.86 minus-- AUDIENCE: 2.52. SARAH HEWETT: 2.52 is its next closest neighbor divided by-- how do we calculate the spread of our data? AUDIENCE: [INAUDIBLE] SARAH HEWETT: So 2.86, which is our max. And our minimum is 2.38. So if you do that out, you get 0.34 divided by 0.48. So our Q value is 0.708. Now, if we look at a table of values for our Q-test-- so we can either calculate it at 90% confident that it's an outlier, 95% confident that it's an outlier, and all the way down. So if we wanted to calculate at 95% confidence level that this point is an outlier, then we had how many data points in this data set? 6. So we'd look at 6. 95%. Our Q value is 0.625. So 0.708 is greater than 0.625. So is it an outlier or not? The Q that you calculate is bigger than the Q in the table. Then it is, indeed, an outlier. And we can confidently say that that does not belong in our data set. Ooh, we're working on it. Come on. So you can see that if anything had changed, like if we had had fewer data points, then it would not be an outlier. Or if you wanted a higher confidence level, then it also would not have been an outlier. So it's a little more strict, how far away from the rest of the points it has to be before you can call it an outlier. Last thing we're going to talk about is doing the least squares regression. You'll make a bunch of calibration curves when you are doing different measurements with UV-Vis spectroscopy. So in UV-Vis, which you'll talk about in the next couple of labs, the absorbance of a sample at a certain wavelength, how much light it absorbs is related to its concentration. And this is a calibration curve with the concentration of bovine serum albumin protein versus absorbance, which you guys will be making a similar one in the catalase lab coming up. So it assumes that one variable is known. So that's our concentration, that we have known concentrations, and that all of the variation in the y-axis is linearly related to our x values. So that's one assumption that you have to have before you are going to make a least squares linear regression. The way that you can calculate all of the values in a least squares regression-- which the whole idea of is to get an equation for a line in the form of y equals mx plus b. So that's the point of all this. And you'll have different points. And you want to calculate a straight line through all of your points. So some of the terminology that is associated with this is the residual value. And so that's your vertical distance between a point and the line of best fit. So this is your residual value. Each point has a residual value. And the way that the line of best fit is generated in the "least squares method" is they take the square of all of these residual values and try to minimize it. And so you can calculate various quantities if you take all of the x values, subtract them from the average x and square them. Then you'll get the Sxx. You could do the same thing with the y values and then the x and the y values. The slope of the line is this value over this one. To get the y-intercept, you take the average y, subtract it from the slope and the average x. And then the r squared. This was a misprint in your slide hand-outs. It should be 1 minus this quantity in the hand-outs. The 1 is not on there. And so the R squared value, which you've probably seen before, is the coefficient of determination. And that is a measure of how well this best fit line can explain the variation in y in a linear fashion. So essentially, how well do your points fit a linear relationship? Is that a good way to explain your data? So the best way to generate this is in Excel. You could calculate all those quantities by hand if you want to. But Excel will do it for you or your other favorite mathematical software program. You graph your points as an x-y scatterplot. Then you can right click the data points. And it'll say add trend line. And then you can have options to display the equation and the R squared on the graph. So you'll get your y equals mx plus b equation there and your R squared value there. So you can see that this line is-- this is not a very linear set of data. So your R squared, ideally, is close to 1. And this is close-ish to 1, but that is definitely not a linear relationship. You can get it higher than that. And we'll talk about that more when you guys make your calibration curves for the bovine serum albumin. The Coomassie blue dye is only linear in certain regions of the calibration curve. So you'll have to calculate different lines for different pieces of the curve, depending on what your concentration is. The last thing that we can do to get more information out of a set of linear data is to use the LINEST program in Excel. And if you've never done this before, you type in your data as a series of x and y points. And then instead of graphing it-- well, you can graph it, too. But in addition to graphing it, you highlight a 2 by 5 area of empty cells. And while that's highlighted in the first cell, you type in equals LINEST. And then it'll prompt you to highlight your y values, then highlight your x values. And then you can type true and true. And what that'll do is if you type false, it'll set the y-intercept to zero. And we don't want to force our lines through zero. So you will have your slope. And then if you press Control-Shift-Enter, even on a Mac-- so on a Mac, press Control, not Command. Press Control-Shift-Enter. And it'll give you this array of data. Once you get this array of data, don't try to change any of these values or Excel will get mad at you, so just leave it. And this is the key to what each of the cells is telling you, because it won't give you this information. You can look it up online, or it'll be on these slides. So your first thing-- it'll just give you your slope and your intercept. And then it'll give you-- these are the two that we care about the most, are the standard deviation of the slope and the standard deviation of the y-intercept. So sometimes in this course, you'll make graphs where the slope is representative of a certain quantity. Or the y-intercept is representative of a certain quantity if you're graphing any equation in linear form. So then you'll have the standard deviations of each of those measurements. Then it'll also give you your R squared and a bunch of other information that you want if you're doing different statistical tests.
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https://ocw.mit.edu/courses/5-08j-biological-chemistry-ii-spring-2016/5.08j-spring-2016.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. ELIZABETH NOLAN: We're going to get started and what we'll do today is continue with fatty acid synthase. Because that's the paradigm for these macromolecular machines, like the PKS, and then we'll go over the logic of polyketide synthases. So we left off last time with this discussion about some molecules that will be involved and in particular thioesters, and I asked about the alpha H. So just going back to introductory organic chemistry, what are the properties of this atom here? AUDIENCE: [INAUDIBLE] acidic. ELIZABETH NOLAN: Yeah. OK, right. So this is acidic. So if you have-- OK? So what that means is if there is a base that can deprotonate that, we can get an enolate. OK, and this is the type of chemistry that's going to be happening with the thioesters that are used in fatty acid synthase and also polyketide synthase. And just to rewind a little bit more, if we think about carbon-carbon bond forming reactions in nature, which is what's happening in fatty acid biosynthesis and in polyketide biosynthesis, effectively, nature uses three different types of reaction. OK, so one is the aldol, two are the Claisen, and three [INAUDIBLE] transfer. OK, and so we're going to see Claisen condensations in FAS and PKS biosynthesis. And then after spring break, when Joanne starts with cholesterol biosynthesis, that will involve [INAUDIBLE] transfers. And hopefully, you've seen aldol reactions sometime before within biochemistry here. OK? So we need to think about just what the general Claisen condensation is that we're going to be seeing here and the consequences of this acidic proton. So also just keep in mind, rewinding a little more, nature uses thioesters not esters, and so the alpha H is more acidic. The carbonyl is more activated for nucloephilic attack. And there's some resonance arguments and orbital overlap arguments that can guide those conclusions, if you wish to do them here. OK. So let's imagine that we have a thioester. We have a base. OK, that's going to be [INAUDIBLE],, which is going to get us to here. So this is our nucleophile, and what you'll see coming forward is an enolate. So imagine we have that, and we add it with another thioester, and here's our electrophile. What do we get? We get formation of a beta-keto thioester, which is the Claisen condensation product. OK, you have two thioesters. OK? So effectively, this acyl thioester is doubly activated, so it can be-- did I lose it? Oh no, problems. Sorry about that. It can be activated as an electrophile at the C1 position, so next door to the sulfur. And it can be activated as a nucleophile at the C2 position here. So this is the general chemistry that's going to be happening by FAS and PKS in terms of forming carbon-carbon bonds between monomers here. OK? So in fatty acid synthase, we have two monomer units. OK? So we have acetyl-CoA and malonyl-CoA. Acetyl-CoA is the starter unit, sometimes called unit 0, and then malonyl-CoA is the extender. And so recall that in fatty acid biosynthesis, each elongation event adds two carbons, and if we look at malonyl-CoA, we have three here. Right? So there's decarboxylation of malonyl-CoA to generate a C2 unit, and there's details of that in the lecture 15 notes. And SCoA is coenzyme A, here, and there's some information as to the biosynthesis of these starter and extender units in the notes. We're not going to go over that in lecture here. So in terms of using these monomers to obtain fatty acids, first what we're going to go over are the domains in FAS. And so we can consider domains that are required for extension of the fatty acid chain and then domains that are required for tailoring of that effectively to reduce the carbonyl, as shown. And we're going to go through these, because what we're going to find is that with polyketide biosynthesis, the same types of domains are used. So this logic extends there. OK. So first, we have domains required for elongation of the fatty acid chain by one two-carbon unit. OK. So these include domains that may be abbreviated as AAT or MAT, and they can be grouped as AT and stand for acetyl or malonyltransferase. OK. We have an Acyl Carrier Protein, ACP, and this carries the growing chain between the domains of fatty acid synthase. And so in recitation this week, you're going to see how these domains move around and talk about the length of this acyl carrier protein. We also have the ketosynthase. So what the ketosynthase does is it accepts the growing chain from the acyl carrier protein, and it catalyzes the Claisen condensation with the next monomer. And what we'll see is that this ketosynthase uses covalent catalysis, and via a cysteine thiolate residue. So these are the key domains required for elongation of the chain. OK? And then what we also need are domains required for tailoring, and just to clarify, I'm defining domain here as a polypeptide with a single enzymatic activity. So domains can be connected to one another, or they can be standalone in different types of synthases, but domain means polypeptide with a single enzymatic activity. So what are the domains required for tailoring? And these work after addition of the C2 unit to the growing chain. So first, there's a ketoreductase. And as indicated, what this enzyme does is it reduces the carbonyl of the previous unit to an OH and uses an NADPH H plus. We also have the dehydratase here, and this forms an alpha, beta-alkene from the product of the ketoreductase action. And then we have an enoyl reductase that reduces this alpha, beta-alkene, and this also requires NADPH H plus here. And then some fatty synthases use a domain called a thioesterase for chain release, and that's noted as TE. And we'll see thioesterases in the PKS and in our PS sections here. So one comment regarding the acyl carrier protein, and then we'll just look at the fatty acid synthase cycle and see how these domains are acting. So in order for the acyl carrier protein to carry this growing chain, it first needs to be post-translationally modified with what's called a PPant arm. And that arm provides the ability to have these monomers, or growing chains, linked via a thioester. And so just to go over this post-translational modification, so post-translational modification of acyl carrier protein with the PPant arm. OK. If we consider apo acyl carrier protein, and apo means that the PPant arm is not attached. There's a serine residue. An enzyme called the PPTase comes along, and it allows for post-translational modification of this serine using CoASH, releasing 3', 5'-ADP to give ACP post-translationally modified with the PPant arm. OK? And we'll look at the actual chemical structures in a minute. What I want to point out is that throughout this unit, this squiggle, some form of squiggle here, is the abbreviation for the PPant arm. OK? And this is very flexible and about 20 angstroms in length. So what does this actually look like? So here we have CoASH. So PPant is an abbreviation for phosphopantetheine, here, this moiety, and here's the 3', 5'-ADP. And so effectively, what's shown on the board is repeated here. Except for here, we're seeing the full structure of the phosphopantetheinylated acyl carrier protein. So this squiggle abbreviation indicates this post-translational modification onto a serine residue of the ACP. Just as an example of structure, so here is a structure of acyl carrier protein from E. coli. It's about 10 kilodaltons, so not very big, and we see the PPant arm here attached. OK? So if we think about fatty acids biosynthesis, we can think about this in three steps, better iterated. OK. So first we have loading, so the acyl carrier proteins need to be loaded with monomers. Sometimes, this step the reactions are described as priming reactions. We have initiation and elongation all grouped together here and, three, at some point, a termination. OK? So we've thought about these before from the standpoint of biological polymerizations. So what about the FAS cycle? Here's one depiction, and I've provided multiple depictions in the lecture 15 notes. Because some people find different cycles easier than others, but let's just take a look. So this charts out the various domains-- the starter and the extender and then the chemistry that occurs on these steps. And so what needs to happen is that there needs to be some loading and initiation where the acetyl-CoA is loaded onto an acyl carrier protein. So that's shown here via transferase here, and then, from the acyl carrier protein, this monomer is loaded onto the ketosynthase. If we look here, we have one of our extender units, the malonyl-CoA, and the CO2 unit that gets removed during decarboxylation, as shown in this light blue. OK? We need to have this extender unit also transferred to an acyl carrier protein via the action of an AT. So we see lots of the CoA. Here we have the acyl carrier protein with the PPant arm. It's not a squiggle here. It is the next one with this malonyl unit loaded. There's a decarboxylation, and what do we see happening here? We have a chain elongation event, so Claisen condensation catalyzed by the ketosynthase between the starter and the first extender to give us this beta-keto thioester. So once this carbon-carbon bond is formed to give us the beta-keto thioester, there's processing of the beta carbon via those tailoring domains-- the dehydratase and the enoyl reductase. And so we see reduction of the beta ketone here, we see formation of the alkene, and then we see reduction to get us to this point. And so this cycle can repeat itself until, at some point, there's a termination event. And in this case here, we see a thioesterase catalyzing hydrolytic release of the fatty acid chain. This is the depiction you'll see in recitation today, or saw before. And I guess what I like about this depiction is that you see color coding separating the elongation and the domains involved in elongation with then the processing of the beta ketone here and then termination. OK. So we get some fatty acid from this. And so where we're going to go with this overview is looking at the polyketides and to ask what similar and different in terms of polyketide biosynthesis? And so where we can begin with thinking about that is asking what are the starters and extenders? And so these are the starters and extenders we saw for fatty acids, and here are the starters and extenders for polyketides, so very similar. Right? We just see that there's some additional options, so we also have this propionyl-CoA here. In addition to malonyl-CoA as an extender, we see that methylmalonyl-CoA can be employed. So what are the core domains of the PKS? They're similar to those of FAS, and we'll just focus on the PKS side of this table. So this is a helpful table when reviewing both types of assembly lines. So the core means that every module, which I'll define in a moment, contains these domains. So we see that there's a ketosynthase, an acyltransferase, and a thiolation domain. So this thiolation domain is the same as the acyl carrier protein. So there's different terminology used, and within the notes, I have some pages that are dedicated to these terminologies. OK? So for PKS, here, we have the ketosynthase, we have acetyltransferase, and then we have this T domain which equals acyl carrier protein here. OK? So then what about these tailoring domains that were required to produce the fatty acid? What we see in polyketide biosynthesis is that those domains are optional. So one or more of these domains may be in a given module. So that's an overview, and then we'll look at an example of some domains and modules. So we're going to focus on type 1 polyketide synthases. And in these, what we're going to see is that catalytic and carrier protein domains are fused, and they're organized into what we'll term modules. So a module is defined as a group of domains that's responsible for activating, forming the carbon-carbon bonds and tailoring a monomer. So there is an individual module for every monomer within the growing chain. And the order of the modules in the polyketide synthase determines the functional group status, and that functional group status is determined by whether or not these optional domains are there. OK? How do we look for modules? The easiest way is to look for one of these thiolation or ACP domains. So each module has one of these. So you can count your number of T domains, and then you know, OK, there's 7T domains, so there's 7 monomers, for instance. So each Claisen condensation is a chain elongation and chain translocation event. Keep in mind, the starting monomer-- so whether that's acetyl-CoA or propionyl-CoA-- does not contain a CO2 group. So there's no decarboxylation of the starting monomer, but decarboxylation of malonyl-CoA occurs, like in fatty acid synthase, and if that's the case, it provides a C2 unit. And if methylmalonyl-CoA is the extender, this decarboxylation provides a C3 unit because of that methyl group. So key difference, as we just saw, in fatty acid biosynthesis, we have complete reduction of that beta-keto group in every elongation cycle because of these three tailoring domains-- the KR, DH, and ER. In PKS, what can happen is that reduction of this beta-keto group may not happen at all, or it may be incomplete in each elongation step. So what that means is that polyketides retain functional groups during chain elongation. And if you look back at some of the structures that were in the notes from last time, you'll see that, in terms of ketones, hydroxyls, double bonds, et cetera. And also, the other point to note is that there can be additional chemistry, and that these assembly lines where polyketide synthases, non-ribosomal peptide synthatases can contain what are called optional domains. So these are additional domains that are not required for formation of the carbon-carbon bond or amide bond in non-ribosomal peptide synthases. But they can do other chemistry there, so imagine a methyltransferase, for instance, or some cyclization domain. So how do we show these domains and modules? So typically, a given synthase is depicted from left to right in order of domain and bond-forming reactions here. So let's just take a look. So if we consider PKS domains and modules, we're just going to look at a pretend assembly line. OK? So this I'm defining here as an optional domain. So in this depiction, going from left to right, each one of these circles is a domain, so a polypeptide with a single enzymatic activity. Note that they're all basically touching one another which indicates in these types of notations that the polypeptide continues. It's not two different proteins, but we have one polypeptide here. I said that there's modules, and we can identify modules by counting T domains. So here, we have three T domains. So effectively there's three modules. So we have a module here, we have a module here, and we have a module here. What do we see? Two of these modules have a ketosynthase, so that's the domain that catalyzes the Claisen condensation. We have no ketosynthase here, in this first module. Why is that? We're all the way to the left. This is effectively our starter or loading module. So the propionyl-CoA or acetyl-CoA will be here, as we'll see, and there's nothing upstream to catalyze a condensate event with. So there's no KS domain in the starting module here or loading module. OK. So this is often called loading or starter. So if we think about these optional domains for a minute and think about how they work. If we go back to fatty acid synthase, and let's just imagine we have this species attached. We have the action of the KR, the dehydratase, and the ER to give us the fully-reduced species. Where here, we have a CH2 to group rather than the beta-ketone. So what happens in PKS in terms of the different optional domains? So we could have this and have full reduction. We can imagine maybe there's no enoyl reductase. So the module has the ketoreductase and the dehydratase but no enoyl reductase, and so as a result, this polyketide ends up with a double bond here. OK? What if we have nobody dehydratase, like this? OK. We just work backwards from the FAS cycle. We'd be left with this OH group at the beta position. Right? And if we have none of them, so no ketoreductase, dehydratase, or enoyl reductase, the beta-ketone will be retained, here. So what this also means is that you can just look at some polyketide and assess what the situation is from the standpoint of these optional domains. So let's just take an example. If we have three cycles of elongation, and let's imagine we had an acetyl-CoA starter plus three malonyl-CoA. So what do we end up with? Let's imagine our chain looks like this. What do we see? So two carbons are added during each elongation cycle to the chain here, and we can see those here, here, here, and here. OK? So a total of four C2 units, one from the starter and then three from these three extenders. And then we can look at what the functional group status is and say, OK, well here, we have no ketoreductase. And here, there was ketoreductase action, but there's no dehydratase. And here, what do we see? We see that there was a reduction of the beta-ketone and then the action of the dehydratase, but we're left at the alkene, so no enoyl reductase. Right? So just looking, you can begin to decipher in a given module what optional domains are there. So what we'll do is take a look at an actual PKS assembly line and then look at the chemistry happening on it here. These are just for your review. This is a polyketide synthase responsible for making this molecule here. So D-E-B or DEB is a 14-membered macrolactone. It's a precursor to the antibiotic erythromycin here, and this is the cartoon depiction of the polyketide synthase required for the biosynthesis of this molecule. So what do we see looking at this polyketide synthase? So it's more complicated than this one here, but the same principles apply. And what we'll see is that it's comprised of three proteins. There's seven modules, so one loading or starter module and six elongation modules, and there's a total of 28 domains. OK? And I said before, the placement and the identity of these domains dictates the identity of the growing chain. So let's take a look. So first, how do we know there's three proteins? We know that in this type of cartoon because we end up seeing some breaks between different domains. So here, for instance, the AT, the T, the KS, et cetera, they're all attached to one another in the cartoon. That means it's all one polypeptide chain, but this one polypeptide chain has many different enzymatic activities in it, because it has different domains. When we see a break-- so for instance here this T domain and this KS domain are not touching one another. That means we have two separate proteins. So this T domain is at the terminus of DEBS 1, and DEBS 2 begins with this ketosynthase. OK? Likewise, we have a break here, between the T domain and this ketosynthase. So three proteins make up this assembly line, and so when thinking about this, these proteins are going to have to interact with each other in one way or another. And so there's a lot of dynamics in protein-protein interactions happening here. How do we know there's seven modules? And remember each module is responsible for one monomer unit. We count the T domains, so we have one, two, three, four, five, six, seven T domains. So like the acyl carrier proteins of fatty acid synthase, these T domains will be post-translationally modified with a PPant arm. And that PPant arm will be loaded with the acetyl-CoA or methylmalonyl-CoA or malonyl-CoA monomers. We have a loading module. So the loading module has no ketosynthase, because there's nothing upstream over here for catalyzing a carbon-carbon bond formation event. And then we see modules one through six, so sometimes the loading module is module zero. We see that each one has a ketosynthase, so there'll be carbon-carbon bond formation going along this assembly line. And we see that the optional domains vary. So for instance, module one has a ketoreductase as does module two. Look at module four. We see all three domains required for complete processing of that beta-keto group here. Here, only a ketoreductase, and here only a ketoreductase. OK? So just looking at this, you can say, OK well, we'll have an OH group here, here. Here we have complete processing. Just ignore this. It's in lower case, because it's a non-functional reductase domain. It's not operating as annotated here. So what happens? So again, there's post-translational modification of this T domain, so it has a serine. The serine gets modified with the PPant arm, as shown here, and we use that squiggle depiction, as I showed for the acyl carrier protein of FAS. So post-translational modification of these T domains has to happen before any of the monomers are loaded onto this assembly line. And these PPant arms allow us to use bioesters as the linkages and through the chemistry I showed earlier. So here, what we're seeing in this cartoon, going from here, this indicates that the T domains are not post-translationally modified. And here, we see the assembly line after action of some [? phosphopentyltransferase ?] loading these arms. OK? So each T domain gets post-translationally modified. What happens next? We have loading of monomers. And we'll look at module zero and one on the board and then look at how the whole assembly line goes. AUDIENCE: Do you ever get selected post-translational modification of the T domains and if so, does that facilitate different modules being like on or off, so to speak? ELIZABETH NOLAN: I don't know. I don't know in terms of the kinetics, and say, does one T domain get loaded by a PPTase before the other? These enzymes are very complex, and there's a lot we don't know. But that would be interesting, if it's the case. I wouldn't rule it out, but I just don't know. One thing to point out too, these assembly lines are huge. So this is something we'll talk about more the next time, as we begin to discuss how do you experimentally study them? But some are the size of the ribosome for the biosynthesis of one natural product. And what that means, from the standpoint of in vitro characterization, is that often you just can't express a whole assembly line, let alone say one protein that has a few modules. So often, what people will do is individually express domains or dye domains and study the reactions they catalyze in their chemistry there. And so it would be very difficult even to test that in terms of in vitro. Is there an ordering to how the T domains are loaded? And then there's question too, do you even know what the dedicated PPTase is? So there's some tricks that are done on the bench top to get around not knowing that, which we'll talk about later. So back to this assembly line to make DEB. So we're just going to go over the loading module and module 1 and look at a Claisen condensation catalyzed by the KS. And this chemistry pertains to the various other modules and other PKS. So we have our AT domain and our thiolation domain of module 0, and then we have the ketosynthase, the AT domain, the ketoreductase, and the T domain of module 1. OK. I'm drawing these a little up and down just to make it easier to show the chemistry. So sometimes you see them straight, sometimes moved around here, but it's all the same. So we have these PPant arms on the two T domains. So what happens now, after these have been post-translationally modified? We need the action of the AT domains to load the monomers onto the PPant arms here, so action of the AT domain. So what do we end up with? In this case, the starter is a propionyl-CoA, so we can see that here. And we have a methylmalonyl-CoA as the extender, that gets loaded, and I'm going to draw the cysteine thiolate of the ketosynthase here. So what happens next? We need to have decarboxylation of the methylmalonyl-CoA monomer to give us a C3 unit. And it's C3 because of this methyl group, but the growing chain will grow by two carbons. And then we need to have transfer of this starter to the ketosynthase. So the ketosynthase is involved in covalent catalysis here. So what happens, we can imagine here, we have attack, and then here, we're going to have the decarboxylation. We have chain transfer to the ketosynthase, and here, decarboxylation leaves us this species. OK? OK. So now, what happens? Now, the assembly's set up for the Claisen condensation to occur which is catalyzed by the ketosynthase. Right? So what will happen here? You can imagine that, and as a result, where do we end up? I'll just draw it down here. And what else do we have? We have a ketoreductase. So this ketoreductase will act on the monomer of the upstream unit, and that's how it always is. So if there's optional domains in module 1, they act on the monomer from module 0. If there's optional domains in module 2, they'll act on the monomer for module 1. OK? So we see here now we have reduction of the ketone from module 1 to here via the ketoreductase. OK? So if we take a look at what's on the PowerPoint here, what we're seeing is one depiction of this assembly line to make DEB indicating the growing chain. OK? So as we walk through each module, we see an additional monomer attached. So the chain elongates, and then you can track what's happening to the ketone group of the upstream monomer on the basis of the optional domains here. If we look in this one, which I like this one because they color code. So they color code the different modules along with the monomer, and so it's pretty easy to trace what's happening. So for instance, here we have the loading module, and we have the starter unit in red. And here we see that it's been reduced by the ketoreductase of the upstream blue module. Here, we have the green module, here is its monomer, and we see its ketoreductase acted on the blue monomer from module 1, et cetera here. So I encourage you all to just very systematically work through the assembly lines that are provided in these notes, and it's the same type of chemistry over and over again. And if you learn the patterns, it ends up being quite easy to work through, at least the simple assembly line. So as you can imagine, complexity increases, and we'll look at some examples of more complex ones as well. So where we'll start next time with this is just briefly looking at chain release by the thioesterase. And then we'll do an overview of non-ribosomal peptide biosynthesis logic and then look at some example assembly lines. So we have the exams to give back. I'll just say a few things. So the average was around a 68, plus or minus 10, 11, 12 for the standard deviation. I'd say, if you were in the mid 70s and above, you did really well. If you're into the low 60s, that is OK, but we'd really like things to improve for the next one. In terms of the exam and just some feedback-- and I'll put feedback as well in the key which will be posted later today or early tomorrow. There wasn't one question that say the whole class bombed, so that's good. There were a few things for just general improvement, and I want to bring this up, so you can also think about it in terms of problem sets. One involves being quantitative. So there's certainly qualitative trends and data, but there's also quantitative information there, and that can be important to look at. And one example I'll give of that involved question one. If you recall, there was an analysis of GDP hydrolysis and an analysis of peptide bond formation. And quantitative analysis of the peptide bond formation experiments will show that all of the lysyl-tRNAs were used up in the case of the codon that was AAA. Whereas, some of those tRNAs were not used up when the codon contained that 6-methyl-A in position one. Right? And if you linked that back to the kinetic model along with the other data, what that indicates is that proofreading is going on. Right? Some of those tRNAs are being rejected from the ribosome there. So that was one place where quantitiation, a fair number of you missed that. And another thing I just want to stress is to make sure you answered the question being asked. And where an example of that came up was in question one with the final question asking about relating the data back to the kinetic model. And so if a question asks that you really do need to go back to the model which was in the appendix and think about that. So many of you gave some very interesting answers and presented hypotheses about perhaps the 6-methyl-A is involved in regulation and controlling like the timing of translation. And that's terrific and interesting to think about, but it wasn't the answer to the question. Right? Which was to go beyond the conclusions from the experiments with GTP hydrolysis and formation of that dipeptide, and ask how can we conceptualize this from the standpoint of the model we studied in class? And then just the third point I'll make is related to question two and specifically to GroEL. But the more general thing is that if we learn about a system in class, unless there's compelling data presented in a question to suggest the model is something other than what we learned or its behavior is something other than what we learned, stick with what you know. So in the use of GroEL, the idea in that experiment was that, if you recall, this question was looking at these J proteins and asking, how do J proteins facilitate disaggregation? Right? And so a GroEL trap was used that cannot hydrolyze ATP, which means it's not active at folding any polypeptide. But the idea there is that these J proteins end up allowing monomers to come out of the aggregate, and then GroEL can trap and unfold the monomer to prevent reactivation. And so a number of people came to the conclusion that GroEL was binding that aggregate somehow in its chamber. And what we learned about GroEL is that its chamber can't house a protein over 60 kilodaltons. Right? We saw that in terms of the in vitro assays that were done looking at what its native substrates are. Right? So always go back to what you know, and then you need to ask yourselves, are the data suggesting some other behavior? And if that were the case, like what is your analysis of those data there? So please, even if you did really well, look at the key and see what the key has to say. And if you have questions, you can make an appointment with me or come to office hours or discuss with Shiva there. OK?
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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 MEHRAN KARDAR: OK, let's start. So in this class, we focused mostly on having some slab of material and having some configuration of some kind of a field inside. And we said that, basically, we are going to be interested close to, let's say, phase transition and some quantity that changes at the phase transition. We are interested in figuring out the singularities associated with that. And we can coarse grain. Once we have coarse grained, we have the field m, potentially a vector, that is characterized throughout this material. So it's a field that's function of x. And by integrating out over a lot of degrees of freedom, we can focus on the probability of finding different configurations of this field. And this probability we constructed on the basis of a number of simple assumptions such as locality, which implied that we would write this probability as a product of contributions of different parts, which in the exponent becomes an integral. And then we would put within this all kinds of things that are consistent with the symmetries of the problem. So if, for example, this is a field that is invariant on the rotations, we would be having terms such as m squared, m to the fourth, and so forth. But the interesting thing was that, of course, there is some interaction with the neighborhoods. Those neighborhood interactions we can, in the continuum limit, inclement by putting terms that are proportional to radiant of m and so forth. So there is a lot of things that you could put consistent with symmetry and presumably be as general as possible. You could have this and the terms in this coefficients of this expansion would be these phenomenological parameters characterizing this probability, function of all kinds of microscopic degrees of freedom, as well as microscopic constraints such as temperature, pressures, et cetera. Now the question that we have is if I start with a system, let's say a configuration where everybody's pointing up or some other configuration that is not the equilibrium configuration, how does the probability evolve to become something like this? Now we are interested therefore in m that is a function of position and time. And since I want to use t for time, this coefficient of m squared that we were previously calling t, I will indicate by r, OK? There are various types of dynamics that you can look at for this problem. I will look at the class that is dissipative. And its inspiration is the Brownian motion that we discussed last time where we saw that when you put a particle in a fluid to a very good approximation when the fluid is viscous, it is the velocity that is proportional to the force and you can't ignore inertial effects such as mass times acceleration and you write an equation that is linear in position. Velocity is the linear derivative of position, which is the variable that is of interest to you. So here the variable that is of interest to us, is this magnetization that is changing as a function of time. And the equation that we write down is the derivative of this field with respect to time. And again, for the Brownian motion, the velocity was proportional to the force. The constant of proportionality was some kind of a mobility that you would have in the fluid. So continuing with that inspiration, let's put some kind of a mobility here. I should really put a vector field here, but just for convenience, let's just focus on one component and see what happens. Now what is the force? Presumably, each location individually feels some kind of a force. And typically when we had the Brownian particle, the force we were getting from the derivative of the potential with respect to the [? aviation ?] of the position, the field that we are interested. So the analog of our potential is the energy that we have over here, and in the same sense that we want this effective potential, if you like, to govern the equilibrium behavior-- and again, recall that for the case of the Brownian particle, eventually the probability was related to the potential by e to the minus beta v. This is the analog of beta v, now considered in the entire field. So the analog of the force is a derivative of this beta H, so this would be our beta H with respect to the variable that I'm trying to change. And since I don't have just one variable, but a field, the analog of the derivative becomes this functional derivative. And in the same sense that the Brownian particle, the Brownian motion, is trying to pull you towards the minimum of the potential, this is an equation that, if I kind of stop over here, tries to put the particle towards the minimum of this beta H. Now the reason that the Brownian particle didn't go and stick at one position, which was the minimum, but fluctuated, was of course, that we added the random force. So there is some kind of an analog of a random force that we can put either in front of [? mu ?] or imagine that if we added it to [? mu, ?] and we put it over here. Now for the case of the Brownian particle, the assumption was that if I evaluate eta at some time, and eta at another time t2 and t1, that this was related to 2D delta function t1 minus t2. Now of course here, at each location, I have a noise term. So this noise carries an index, which indicates the position, which is, of course, a vector in D dimension in principal. And there is no reason to imagine that the noise here that comes from all kinds of microscopic degrees of freedom that we have integrated out should have correlation with the noise at some other point. So the simplest assumption is to also put the delta function in the positions, OK? So if I take this beta H that I have over there and take the functional derivative, what do I get? I will get that the m of xt by dt is the function of-- oops, and I forgot to put a the minus sign here. The force is minus the potential, and this is basically going down the gradient, so I need to put that. So I have to take a derivative of this. First of all I can take a derivative with respect to m itself. I will get minus rm. Well, actually, let's put the minus out front. And then I will get the derivative of m to the fourth, which is [? 4umq, ?] all kinds of terms like this. Then the terms that come from the derivatives and taking the derivative with respect to gradient will give me K times the gradient. But then in the functional derivative, I have to take another derivative, converting this to minus K log Gaussian of m. And then I would have L the fourth derivative from the next term and so forth. And on top of everything else, I will have the noise who's statistics I have indicated above. So this entity, this equation, came from the Landau-Ginzburg model, and it is called a time dependent Landau-Ginzburg equation. And so would be the analog of the Brownian type of equation, now for an entire field. Now we're going to have a lot of difficulty in the short amount of time that is left to us to deal with this nonlinear equation, so we are going to do the same thing that we did in order to capture the properties of the Landau-Ginzburg model qualitatively. Which is to ignore nonlinearities, so it's a kind of Gaussian version of the model. So what we did was we linearized. AUDIENCE: Question. MEHRAN KARDAR: Yes? AUDIENCE: Why does the sign in front of the K term change relative to the others? MEHRAN KARDAR: OK, so when you have the function 0 of a field gradient, et cetera, you can show that the functional derivative, the first term is the ordinary type of derivative. And then if you think about the variations that are carrying m and with the gradient and you write it as m plus delta m and then make sure that you take the delta m outside, you need to do an integration by part that changes the sign. And the next term would be the gradient of the phi by the [? grad ?] m, and the next term would be log [INAUDIBLE] of [INAUDIBLE] gradient of m squared and so forth. It alternates and so by explicitly calculating the difference of two functionals with this integration evaluated at m and m plus delta m and pulling out everything that is proportional to delta m you can prove these expressions. So we linearize this equation, which I've already done that. I cross out the mq term and any other nonlinear terms, so I only keep the linear term. And then I do a Fourier transform. So basically I switch from the position representation to the Fourier transform that's called m tilde of q, I think. x is replaced with q. And then the equation for the field m linears. The form separates out into independent equations for the components that are characterized by q. So the Fourier transform of the left hand side is just this. Fourier transform of the right hand side will give me minus mu r plus K q squared. The derivative of [INAUDIBLE] will give me a minus q squared. The derivative of the next term L q to the forth, et cetera. And I've only kept the linear term, so I have m tilde of q and t. And then I have the Fourier transforms of the noise eta of x and t. They [? call it ?] eta tilde of q [? m. ?] So you can see that each mode satisfy a separate linear equation. So this equation is actually very easy to solve for any linear equation m tilde of q and t. If I didn't have the noise, so I would start with some value at t [? close ?] to 0, and that value would decay exponentially in the characteristic time that I will call tao of q. And 1 over tao of q is simply this mu r plus K q2 and so forth. Now once you have noise, essentially each one of these noises acts like an initial condition. And so the full answer is an integral over all of these noises from 0 tilde time t of interest. Dt prime the noise that occurs at a time t prime, and the noise that occurs at time t [? trime ?] relaxes with this into the minus t minus t prime tao of q. So that's the solution to that linear noisy equations, basically [? sequencing ?] [INAUDIBLE]. So one of the things that we now see is that essentially the different Fourier components of the field. Each one of them is independently relaxing to something, and each one of them has a characteristic relaxation time. As I go towards smaller and smaller values of q, this rate becomes smaller, and the relaxation time becomes larger. So essentially shortwave wavelength modes that correspond to large q, they relax first. Longer wavelength modes will relax later on. And you can see that the largest relaxation time, tao max, corresponds to q equals to 0 is simply 1 over [? nu ?] r. So I can pluck this to a max as a function of r, and again in this theory, the Gaussian theory, we saw only makes sense as long as all is positive. So I have to look only on the positive axis. And I find that the relaxation time for the entire system for the longest wavelength actually diverges as r goes to 0. And recall that r, in our perspective, is really something that is proportional to T minus Tc. So basically we find that as we are approaching the critical point, the time it takes for the entirety of the system or the longest wavelength [? modes ?] to relax diverges as 1 over T minus Tc. There's an exponent that shows this, so called, critical slowing down. Yes? AUDIENCE: In principal, why couldn't you have that r becomes negative if you restrict your range of q to be outside of some value and not go arbitrarily close to the origin. MEHRAN KARDAR: You could, but what's the physics of that? See. AUDIENCE: I'm wondering if there is or is that not needed. MEHRAN KARDAR: No. OK, so the physics of that could be that you have a system that has some finite size, [? l. ?] Then the largest q that you could have [? or ?] the smallest q that you could have would be the [? 1/l. ?] So in principal, for that you can go slightly negative. You still cannot go too negative because ultimately this will overcome that. But again, we are interested in singularities that we kind of know arise in the limit of [INAUDIBLE]. I also recall, there is a time that we see as diverging as r goes to 0. Of course, we identified before a correlation length from balancing these two terms, and the correlation length is square root of K over r, which is again proportional to T minus Tc and diverges with a square root singularity. So we can see that this tao max is actually related to the psi squared over Nu K. And it is also related towards this. We can see that our tao of q, basically, if q is large such that qc is larger than 1 and the characteristic time is going to be 1 over [? nu ?] K times the inverse of q squared. And the inverse of q is something like a wavelength. Whereas, ultimately, this saturates for qc what is less than 1 [? long ?] wavelengths to c squared over [? nu ?] K. So basically you see things at very short range, at length scales that are much less than the correlation length of the system, that the characteristic time will depend on the length scale that you are looking at squared. Now you have seen times scaling as length squared from diffusion, so essentially this is some kind of a manifestation of diffusion, but as you perturb the system, let's say at short distances, there's some equilibrium system. Let's say we do some perturbation to it at some point, and that perturbation will start to expand diffusively until it reaches the size of the correlation length, at which point it stops because essentially correlation length is an individual block that doesn't know about individual blocks, so the influence does not last. So quite generally, what you find-- so we solved the linearized version of the Landau-Ginzburg model, but we know that, say, the critical behaviors for the divergence of the correlation length that is predicted here is not correct in three dimensions, things get modified. So these kind of exponents that come from diffusion also gets modified. And quite generally, you find that the relaxation time of a mode of wavelength q is going to behave something like wavelength, which is 1 over q, rather than squared as some exponent z. And then there is some function of the product of the wavelength you are looking at and the correlation length so that you will cross over from one behavior to another behavior as you are looking at length scales that are smaller than the correlation length or larger than the correlation length. And to get what this exponent z is, you have to do study of the nonlinear model in the same sense that, in order to get the correction to the exponent [? nu ?], we had to do epsilon expansion. You have to do something similar, and you'll fine that at higher changes, it goes like some correction that does not actually start at order of epsilon but at order of epsilon squared. But there's essentially some modification of the qualitative behavior that we can ascribe to the fusion of independent modes exists quite generally and universal exponents different from [? to ?] will emerge from that. Now it turns out that this is not the end of the story because we have seen that the same probability distribution can describe a lot of different systems. Let's say the focus on the case of n equals to 1. So then this Landau-Ginzburg that I described for you can describe, let's say, the Ising model, which describes magnetizations that lie along the particular direction. So that it can also describe liquid gas phenomena where the order parameter is the difference in density, if you like, between the liquid and the gas. Yet another example that it describes is the mixing of an alloy. So let's, for example, imagine brass that has a composition x that goes between 0 and 1. On one end, let's say you have entirely copper and on the other and you have entirely zinc. And so this is how you make brass as an alloy. And what my other axis is is the temperature. What you find is that there is some kind of phase diagram such that you get a nice mixture of copper and zinc only if you are at high temperatures, whereas if you are at low temperature, you basically will separate into chunks that are rich copper and chunks that are rich in zinc. And you'll have a critical demixing point, which has exactly the same properties as the Ising model. For example, this curve will be characterized with an exponent beta, which would be the beta of the Ising mode. And in particular, if I were to take someplace in the vicinity of this and try to write down a probability distribution, that probability distribution would be exactly what I have over there where m is, let's say, the difference between the two types of alloys that I have compared to each other over here. So this is related to 2x minus 1 or something like that. So as you go across your piece of material close to here, there will be compositional variations that are described by that. So the question is, I know exactly what the probability distribution is for this system to be an equilibrium given this choice of m. Again, with some set of parameters, R, U, et cetera. The question is-- is the dynamics again described by the same equation? And the answer is no. The same probability of distribution can describe-- or can be obtained with very different dynamics. And in particular, what is happening in the system is that, if I integrate this quantity m across the system, I will get the total number of 1 minus the other, which is what is given to you, and it does not change as a function of time. d by dt of this quantity is 0. It cannot change. OK. Whereas the equation that I have written over here, in principle, locally, I can, by adding the noise or by bringing things from the neighborhood, I can change the value of m. I cannot do that. So this process of the relaxation that would go on in data graphs cannot be described by the time dependent on the Landau-Ginzburg equation because you have this conservation here. OK. So what should we do? Well, when things are conserved, like, say, as the gas particles move in this fluid, and if I'm interested in the number of particles in some cube, then the change in the number of particles in some cube in this room is related to the gradient of the current that goes into that place. So the appropriate way of writing an equation that describes, let's say, the magnetization changing as a function of time. Given that you have a conservation, though, is to write it as minus the gradient of some kind of a current. So this j is some kind of a current, and these would be vectors. This is a current of the particles moving into the system. Now, in systems that are dissipative, currents are related to the gradient of some density through the diffusion constant, et cetera. So it kind of makes sense to imagine that this current is the gradient of something that is trying-- or more precisely, minus the gradient of something that tries to bring the system to be, more or less, in its equilibrium state. Equilibrium state, as we said, is determined by this data H, and we want to push it in that direction. So we put our data H by dm over here, and we put some kind of a U over here. Of course, I would have to add some kind of a conserved random current also, which is the analog of this non-conserved noise that I add over initially. OK. Now, the conservative version of the equation, you can see, you have two more derivatives with respect to what we had before. And so once I-- OK. So if we do something like this, dm by dt is mu C. And then I would have [INAUDIBLE] plus of R, rather than by itself. Actually, it would be the plus then of something like R m plus 4 U m cubed, and so forth. And we are going to ignore this kind of term. And then there would be high order terms that would show up minus k the fourth derivative, and so forth. And then there may be some kind of a conserved noise that I have to put outside. OK. So when I fully transform this equation, what do I get? I will get that dm by dt is-- let's say in the full space, until there is a function of qnt is minus U C. Because of this plus, then there's an additional factor of q squared. And then I have R plus kq squared plus L cubed to the fourth, et cetera. And then I will have a fully transformed version of this conserved noise. OK. You can see that the difference between this equation and the previous equation is that all of the relaxation times will have an additional factor of q squared. And so eventually, this shortest relaxation time actually will glow like the size of the system squared. Whereas previously, it was saturated at the correlation length. And because you will have this conservation, though, you have to rearrange a lot of particles keeping their numbers constant. You have a much harder time of relaxing the system. All of the relaxation times, as we see, grow correspondingly and become higher. OK. So indeed, for this class, one can show that z starts with 4, and then there will be corrections that would modify that. So the-- yes? AUDIENCE: How do we define or how do we do a realization of the conserved noise, conserved current-- conserved noise in the room? MEHRAN KARDAR: OK. AUDIENCE: So it has some kind of like correlation-- self-correlation properties, I suppose, because, if current flowing out of some region, doesn't it want to go in? MEHRAN KARDAR: If I go back here, I have a good idea of what is happening because all I need, in order to ensure conservation, is that the m by dt is the gradient of something. AUDIENCE: OK. MEHRAN KARDAR: So I can put whatever I want over here. AUDIENCE: So if it's a scalar or an [INAUDIBLE] field? MEHRAN KARDAR: Yes. As long as it is sitting under the gradient-- AUDIENCE: OK. MEHRAN KARDAR: --it will be OK, which means this quantity here that I'm calling a to z has a gradient in it. And if you wait for about five minutes, we'll show that, because of that in full space, rather than having-- well, I'll describe the difference between non-conserved and conserved noise in fully space. It's much easier. OK. So actually, as far as what I have discussed so far, which is relaxation, I don't really need the noise because I can forget the noise. And all I have said-- and I forgot the n tilde-- is that I have a linear equation that relaxes your variable to 0. I can immediately read off for the correlation, then this-- a correlation times what I need the noise for it so that, ultimately, I don't go to the medium is the potential, but I go to this pro-rated distribution. So let's see what we have to do in order to achieve that. For simplicity, let's take this equation. Although I can take the corresponding one for that. And let's calculate-- because of the presence of this noise, if I run the same system at different times, I will have different realizations of the noise than if I had run many versions of the system because of the realizations of noise. It's quantity and tilde would be different. It would satisfy some kind of a pro-rated distribution. So what I want to do is to calculate averages, such as the average of m tilde. Let's say, q1 at time T with m tilde q2 at time t. And you can see already from this equation that, if I forget the part that comes from the noise, whatever initial condition that I have will eventually decay to 0. So the thing that agitates and gives some kind of a randomness to this really comes from this. So let's imagine that we have looked at times that are sufficiently long so that the influence of the initial condition has died down. I don't want to write the other term. I could do it, but it's kind of boring to include it. So let's forget that and focus on the integral. 0 to t. Now, if I multiply two of these quantities, I will have two integrals over t prime. All right. Each one of them would decay with the corresponding tau of q. In one case, tau of q1. In the other case, tau of q2. Coming from these things. And the noise, q1 at time to 1 prime, and noise q2 at time [? t2 prime. ?] OK. Now if I average over the noise, then I have to do an average over here. OK. Now, one thing that I forgot to mention right at the beginning is that, of course, the average of this we are going to set to 0. It's the very least that is important. Right? So if I do that, clearly, the average of one of these in full space would be 0 also because the full q is related to the real space delta just by an integral. So if the average of the integral is 0, the average of this is 0. So it turns out that, when you look at the average of two of them-- and it's a very simple exercise to just rewrite these things in terms of 8 of x and t. 8 of x and t applied average that you have. And we find that the things that are uncorrelated in real space also are uncoordinated in full space. And so the variance of this quantity is 2d delta 1 prime minus [? d2 prime. ?] And then you have the analog of the function in full space, which always carries the solution of a factor of 2 pi to the d. And it becomes a sum of the q's, as we've seen many times before. OK. So because of this delta function, this delta integral becomes one integral. So I have the integral 0 to t. The 2t prime I can write as just 1t prime, and these two factors merge into one. It comes into minus t minus t prime over tau of q, except that I get multiplied by a factor 2 since I have two of them. And outside of the integral, I will have this factor of 2d, and then 2 pi to the d delta function with 1 plus 2. OK. Now, we are really interested-- and I already kind of hinted at that in the limit where time becomes very large. In the limit, where time becomes a very large, essentially, I need to calculate the limit of this integral as time becomes very large. And as time grows to very large, this is just the integral from 0 to infinity. And the integral is going to give me 2 over tau of q. Essentially, you can see that integrating at the upper end will give me just 0. Integrating at the smaller end, it will be exponentially small as it goes to infinity as a factor of 2 over tau. OK. Yes? AUDIENCE: Are you assuming or-- yeah. Are you assuming that tau of q is even in q to be able to combine to-- MEHRAN KARDAR: Yes. AUDIENCE: --2 tau? So-- OK. MEHRAN KARDAR: I'm thinking of the tau of q's that we've calculated over here. AUDIENCE: OK. MEHRAN KARDAR: OK. Yes? AUDIENCE: The e times the [INAUDIBLE] do the equation of something? Right? MEHRAN KARDAR: At this stage, I am focusing on this expression over here, where a is not. But I will come to that expression also. So for the time being, this d is just the same constant as we have over here. OK? So you can see that the final answer is going to be d over tau of q 2 pi to the d delta function q1 plus q2. And if I use the value of tau of q that I have, tau of q is mu R plus kq squared, which becomes d over mu R plus kq squared and so forth. 2 pi to the d delta function q1 plus q2. So essentially, if I take this linearized time dependent line of Landau-Ginzburg equation, run it for a very long time, and look at the correlations of the field, I see that the correlations of the field, at the limit of long times, satisfy this expression. Now, what do I know if I look at the top line that I have for the probability of distribution? I can go and express that probability of distribution in full mode. In the linear version, I immediately get that the probability of m tilde of q is proportional to a product over different q's, e to the minus R plus kq squared, et cetera, and tilde of q squared over 2. All right. So when I look at the equilibrium linear as Landau-Ginzburg, I can see that, if I calculate the average of m of q1, m of q2, then this is an equilibrium average. What I would get is 2 pi to the d delta function of q1 plus q2 because, clearly, the different q's are recovered from each other. And for the particular value of q, what I will get is 1 over R plus kq squared and so forth. Yes? AUDIENCE: So isn't the 2 over tau-- [INAUDIBLE]? MEHRAN KARDAR: Yes. Over 2. Yeah. OK. OK. I changed because I had 2d cancels the 2. I would have to put here tau of q over 2. I had 2d times tau of q over 2. So it's d tau. And inverse of tau, I have everything correct. OK. So if you make an even number of errors, the answer comes up. OK. But you can now compare this expression that comes from equilibrium and this expression that comes from the long time limit of this noisy equation. OK So we want to choose our noise so that the stochastic dynamics gives the same value as equilibrium, just like we did for the case of a Brownian particular where you have some kind of an Einstein equation that was relating the strength of the noise and the mobility. And we see that here all I need to do is to ensure that d over mu should be equal to 1. OK. Now, the thing is that, if I am doing this, I, in principle, can have a different noise for each q, and compensate by different mobility for each q. And I would get the same answer. So in the non-conserved version of this time dependent dynamics that you wrote down, the d was a constant and the mu was a constant. Whereas, if you want to get the same equilibrium result out of the conserved dynamics, you can see that, essentially, what we previously had as mu became something that is proportional to q squared. So essentially, here, this becomes mu C q squared. So clearly, in order to get the same answer, I have to put my noise to be proportional to q squared also. And we can see that this kind of conserved noise that I put over here achieves that because, as I said, this conserved noise is the gradient of something, which means that, when I go to fully space, if it be q, it will be proportional to q. And when I take its variants, it's variants will be proportional to q squared. Anything precisely canceled. But you can see that you also-- this had a physical explanation in terms of a conservation law. In principle, you can cook up all kinds of b of q and mu of q. As long as this equality is satisfied, you will have, for these linear stochastic equations, the guarantee that you would always get the same equilibrium result. Because if you wait for this dynamics to settle down after long times, you will get to the answer. Yes? AUDIENCE: I wonder how general is this result for stochastic [INAUDIBLE]? MEHRAN KARDAR: OK. AUDIENCE: But what I-- MEHRAN KARDAR: So what I showed you was with for linearized version, and the only thing that I calculated was the variance. And I showed that the variances were the same. And if I have a Gaussian problem of distribution, the variance is completely categorizable with distribution. So this is safe. But we are truly interested in the more general non-Gaussian probability of distribution. So the question really is-- if I keep the full non-linearity in this story, would I be able to show that the probability of distribution that will be characterized by all kinds of moments eventually has the same behavior as that. AUDIENCE: Mm-hmm. MEHRAN KARDAR: And the answer is, in fact, yes. There's a procedure that relies on converting this equation-- sorry. One equation that governs the evolution of the full probability as a function of time. Right? So basically, I can start with a an initial probability and see how this probability evolves as a function of time. And this is sometimes called a master equation. Sometimes, called a [INAUDIBLE] equation. And we did cover, in fact, this in-- next spring in the statistical physics and biology, we spent some time talking about these things. So you can come back to the third version of this class. And one can ensure that, with appropriate choice of the noise, the asymptotic solution for this probability distribution is whatever Landau-Ginzburg or other probability distribution that you need most. AUDIENCE: So is this true if we assume Landau-Ginzburg potential for how resistant? MEHRAN KARDAR: Yes. AUDIENCE: OK. Maybe this is not a very good-stated question, but is there kind of like an even more general level? MEHRAN KARDAR: I'll come to that, sure. But currently, the way that I set up the problem was that we know some complicated equilibrium force that exist-- form of the probability that exist. And these kinds of linear-- these kinds of stochastic linear or non-linear evolution equations-- generally called non- [INAUDIBLE] equations-- one can show that, with the appropriate choice of the noise, we'll be able to asymptotically reproduce the probability of distribution that we knew. But now, the question is, of course, you don't know the probability of distribution. And I'll say a few words about that. AUDIENCE: OK. Thank you. MEHRAN KARDAR: Anything else? OK. So the lesson of this part is that the field of dynamic or critical phenomena is quite rich, much richer than the corresponding equilibrium critical phenomena because the same equilibrium state can be obtained by various different types of dynamics. And I explained to you just one conservation law, but there could be some combination of conservation of energy, conservation of something. So there is a whole listing of different universality classes that people have targeted for the dynamics. But not all of this was assuming that you know what the ultimate answer is because, in all cases, the equations that we're writing are dependent on some kind of a gradient descent conserved or non-conserved around something that corresponded to the log of probability of distribution that we eventually want to put. And maybe you don't know that, and so let me give you a particular example in the context of, let's say, surface interface fluctuations. Starting from things that you know and then building to something that maybe you don't. Let's first with the case of a soap bubble. So we take some kind of a circle or whatever, and we put a soap bubble on top of it. And in this case, the energy of the formation-- the cost of the formation comes from surface tension. And let's say, the cost of the deformation is the changing area times some sigma. So I neglect the contribution that comes from the flat surface, and see if I make a deformation. If I make a deformation, I have changed the area of the spin. So there is a cost that is proportion to the surface tension times the change in area. Change in area locally is the square root of 1 plus the gradient of a height profile. So what I can do is I can define at each point on the surface how much it has changed its height from being perfectly flat. So h equals to 0 is flat. Local area is the integral dx dy of square root of 1 plus gradient of h squared, minus 1. That corresponds to the flat. And so then you expand that. The first term is going to be the integral gradient of h squared. So this is the analog of what we had over there, only the first term. So you would say that the equation that you would write down for this would be all to some constant mu proportional to the variations of this, which will give me something like sigma Laplacian of h. But because of the particles from the air constantly bombarding the surface, there will be some noise that depends on where you are on the surface in time. And this is the non-conserved version. And you can from this very quickly get that the expectation value of h tilde of q squared is going to be proportional to something like D over mu sigma q squared, because of this q squared. And if you ask how much fluctuations you have in real space-- so that typical scale of the fluctuations in real space-- will come from integrating 1 over q squared. And it's going to be our usual things that have this logarithmic dependence, so there will be something that ultimately will go logarithmically with the size of the system. The constant of proportionality will be proportional to kt over sigma. So you have to choose your D and mu to correspond to this. But basically, a soap film, as an example of all kinds of Goldstone mode-like things that we have seen. It's a 2-dimensional entity. We will have logarithmic fluctuations-- not very big, but ultimately, at large enough distances, it will have fluctuations. So that was non-conserved. I can imagine that, rather than this, I have the case of a surface of a pool. So here I have some depth of water, and then there's the surface of the pool of water. And the difference between this case and the previous case-- both of them can be described by a height function. The difference is that if I ignore evaporation and condensation, the total mass of water is going to be conserved. So I would need to have divided t of the integral dx d qx h of x and t to be 0. So this would go into the conserved variety. And while, if I create a ripple on the surface of this compared to the surface of that, the relaxation time through this dissipative dynamics would be much longer in this case as opposed to that case. Ultimately, if I wait sufficiently long time, both of them would have exactly the same fluctuations. That is, you would go logarithmically with the length scale over which [INAUDIBLE]. OK, so now let's look at another system that fluctuates. And I don't know what the final answer is. That was the question, maybe, that you asked. The example that I will give is the following-- so suppose that you have a surface. And then you have a rain of sticky materials that falls down on top of it. So this material will come down. You'll have something like this. And then as time goes on, there will be more material that will come, more material that will come, more material that will come. So there, because the particles are raining down randomly at different points, there will be a stochastic process that is going on. So you can try to characterize the system in terms of a height that changes as a function of t and as a function of position. And there could be all kinds of microscopic things going on, like maybe these are particles that are representing some kind of a deposition process. And then they come, they stick in a particular way. Maybe they can slide on the surface. We can imagine all kinds of microscopic degrees of freedom and things that we can put. But you say, well, can I change my perspective, and try to describe the system the same way that we did for the case of coarse grading and going from without the microscopic details to describe the phenomenological Landau-Ginzburg equation? And so you say, OK, there is a height that is growing. And what I will write down is an equation that is very similar to the equations that I had written before. Now I'm going to follow the same kind of reasoning that we did in the construction of this Landau-Ginzburg model, is we said that this weight is going to depend on all kinds of things that relate to this height that I don't quite know. So let's imagine that there is some kind of a function of the height itself. And potentially, just like we did over there, the gradient of the height, five derivatives of the height, et cetera. And then I will start to make an expansion of this in the same spirit that I did for the Landau-Ginzburg Model, except that when I was doing the Landau-Ginzburg Model, I was doing the expansion at the level of looking at the probability distribution and the log of the probability. Here I'm making the expansion at the level of an equation that governs the dynamics. Of course, in this particular system, that's not the end of story, because the change in height is also governed by this random addition of the particles. So there is some function that changes as a function of position and time, depending on whether, at that time, a particle was dropped down. I can always take the average of this to be 0, and put that average into the expansion of this starting from a constant. Basically, if I just have a single point and I randomly drop particles at that single point, there will be an average growth velocity, an average addition to the height, that averages over here. But there will be fluctuations that are going [INAUDIBLE]. OK but the constant is the first term in an expansion such as this. And you can start thinking, OK, what next order? Can I put something like alpha h? Potentially-- depends on your system-- but if the system is invariant whether you started from here or whether you started from there-- something like gravity, for example, is not important-- you say, OK, I cannot have any function of h if my dynamics will proceed exactly the same way if I were to translate this surface to some further up or further down. If I see that there's no change in future dynamics on average, then the dynamic cannot depend on this. OK, so we've got rid of that. And any function of h-- can I put something that is proportional to gradient of h? Maybe for something I can, but for h itself I cannot, because h is a scalar gradient. If h is a vector, I can't set something that is a scalar equal to a vector, so I can't have this. Yes? AUDIENCE: Couldn't you, in principle, make your constant term in front of the gradient also a vector and [INAUDIBLE]? MEHRAN KARDAR: You could. So there's a whole set of different systems that you can be thinking about. Right now, I want to focus on the simplest system, which is a scalar field, so that my equation can be as simple as possible, but still we will see it has sufficient complication. So you can see that if I don't have them, the next order term that I can have would be something like a Laplacian. So this kind of diffusion equation, you can see, has to emerge as a low-order expansion of something like this. And this is the ubiquity of the diffusion equation appearing all over the place. And then you could have terms that would be of the order of the fourth derivative, and so forth. There's nothing wrong with that. And then, if you think about it, you'll see that there is one interesting possibility that is not allowed for that system, but is allowed for this system, which is something that is a scalar. It's the gradient of h squared. Now I could not have added this term for the case of the soap bubble for the following reason-- that if I reverse the soap bubble so that h becomes minus h, the dynamics would proceed exactly as before. So the soap bubble has a symmetry of h going to minus h, and so that symmetry should be preserved in the equation. This term breaks that symmetry because the left-hand side is odd in h, whereas the right-hand side of this term would be even in h. But for the case of the growing surface-- and you've seen things that are growing. And typically, if I give you something that has grown like the tree trunk, for example, and if I take the picture of a part of it, and you don't see where the center is, where the end is, you can immediately tell from the way that the shape of this object is, that it is growing in some particular direction. So for growth systems, that symmetry does not exist. You are allowed to have this term, and so forth. Now the interesting thing about this term is that there is no beta h that you can write down that is local-- some function of h such that if you take a functional derivative with respect to h, it will reproduce that term-- just does not exist. So you can see that somehow immediately, as soon as we liberate ourselves from writing equations that came from functional derivative of something, but potentially have physical significance, we can write down new terms. So this is actually-- also, you can do this, even for two particles. A potential v of x1 and x2 will have some kind of derivatives. But if you write dynamical equations, there are dynamical equations that allow you to rotate 1 x from x1 to x2. That kind of term will never come from taking the derivative. So fine. So this is a candidate equation that is obtained in this context-- something that is grown. We say we are not interested in it's coming from some underlying weight, but presumably, this system still, if I look at it at long times, will have some kind of fluctuations. All the fluctuations of this growing surface, like the fluctuations of the soap bubble, and they have this logarithmic dependence. You have a question? AUDIENCE: So why doesn't that term-- what if I put in h times that term that we want to appear and then I vary with respect to h? A term like what we want to pop out together with other terms? MEHRAN KARDAR: Yeah, but those other terms, what do you want to do with them? AUDIENCE: Well, maybe they're not acceptable [INAUDIBLE]? MEHRAN KARDAR: So you're saying why not have a term that is h gradient of h squared? Functional derivative of that is gradient of h squared. And then you have a term that is h Laplacian-- it's a gradient of, sorry, gradient of h. And then you expand this. Among the terms that you would generate would be a term that is h, a Laplacian of h. This term violates this condition that we had over here. And you cannot separate this term from that term. So what you describe, you already see at the level over here. It violates translation of symmetry in [? nature ?]. And you can play around with other functions. You come to the same conclusion. OK, so the question is, well, you added some term here. If I look at this surface that has grown at large time, does it have the same fluctuations as we had before? So a simple way to ascertain that is to do the same kind of dimensional analysis which, for the Landau-Ginzburg, was a prelude to doing renormalization. So we did things like epsilon expansion, et cetera. But to calculate that there was a critical dimension of 4, all we needed to do was to rescale x and m, and we would immediately see that mu goes to mu, b to the 4 minus D or something-- D minus 4, for example. So we can do the same thing here. We can always move to a frame that is moving with the average velocity, so that we are focusing on the fluctuations. So we can basically ignore this term. I'm going to rescale x by a factor of b. I'm going to rescale time by a factor of b to something to the z. And this z is kind of indicative of what we've seen before-- that somehow in these dynamical phenomena, the scaling of time and space are related to some exponent. But there's also an exponent that characterizes how the fluctuations in h grow if I look at systems that are larger and larger. In particular, if I had solved that equation, rather than for a soap bubble in two dimensions, for a line-- for a string that I was pulling so that I had line tension-- the one-dimensional version of it, the one-dimensional version of an integral of 1 over q squared would be something that would grow with the size of the system. So there I would have a chi of 1/2, for example, in one dimension. So this is the general thing. And then I would say that the first equation, dhy dt, gets a factor of b to the chi minus z, because h scaled by a factor of chi, t scaled by a factor of z, the term sigma Laplacian of h gets a factor of b to the chi minus 2 from the two derivatives here-- sorry, the z and the 2 look kind of the same. This is a z. This is a 2. And then the term that is proportional to this non-linearity that I wrote down-- actually, it is very easy, maybe worthwhile, to show that sigma to the 4th goes with a factor of b to the chi minus 4. It is always down by a factor of two scalings in b with respect to a Laplacian-- the same reason that when we were doing the Landau-Ginzburg. We could terminate the series at order of gradient squared, because higher-order derivatives were irrelevant. They were scaling to 0. But this term grows like b to the 2 chi, because it's h squared, minus 2, because there's two gradients. Now thinking about the scaling of eta takes a little bit of thought, because what we have-- we said that the average of eta goes to 0. The average of eta at two different locations and two different times-- it is these particles that are raining down-- they're uncorrelated at different times. They're uncorrelated at different positions. There's some kind of variance here, but that's not important to us. If I rescale t by a factor of b, delta of bt-- sorry, if I scale t by a factor of b to the z, delta of b to the zx will get a factor of b to the minus z. This will get a factor of b to the minus d. But the noise, eta, is half of that. So what I will have is b to the minus z plus d over 2 times eta on the rescalings that I have indicated. I get rid of this term. So this-- divide by b to the chi minus z. So then this becomes bh y dt is sigma b to the z minus 2. Maybe I'll write it in red-- b to the z minus 2. This becomes sigma to the 4, b to the z minus 4. And then lambda over 2. This is Laplacian of h. This, for derivative of h, this is Laplacian of h. This term becomes b to the chi plus z minus 2, gradient of h squared. And the final term becomes b to the chi minus d minus z over 2. AUDIENCE: [INAUDIBLE]? MEHRAN KARDAR: b to the minus chi-- you're right. And then, actually, this I-- no? Minus chi minus d over 2 minus d over 2 [INAUDIBLE], That's fine. So I can make this equation to be invariant. So I want to find out what happens to this system if I find some kind of an equation, or some kind of behavior that is scale invariant. You can see that immediately, my choice for the first term has to be z equals to 2. So basically, it says that as long as you're governed by something that is diffusive, so that when you go to Fourier space, you have q squared, your relaxation times are going to have this diffusive character, where time is distance squared. Actually, you can see that immediately from the equation that this diffusion time goes like distance squared. So this is just a statement of that. Now, it is the noise that causes the fluctuations. And if I haven't made some simple error, you will find that the coefficient of the noise term becomes scale invariant, provided that I choose it to be z minus d over 2 for chi. And since my z was 2, I'm forced to have chi to be 2 minus d over 2. And let's see if it makes sense to us. So if I have a surface such as the case of the soap bubble in two dimensions, chi is 0. And 0 is actually this limiting case that would also be a logarithm. If I go to the case of d equals to 1-- like pulling a line and having the line fluctuate-- then I have 2 minus 1 over 2, which is 1/2, which means that, because of thermal fluctuations, this line will look like it a random walk. You go a distance x. The fluctuations in height will go like the square root of that. OK? So all of that is fine. You would have done exactly the same answer if you had just gotten a kind of scaling such as this for the case of the Gaussian Model without the nonlinearities. But for the Gaussian Model with nonlinearities, we could also then estimate whether the nonlinearity u is relevant. So here we see that the coefficient of our nonlinearity is lambda, is governed by something that is chi plus z minus 2. And our chi is 2 minus z over 2. z minus 2 is 0. So whether or not this nonlinearity is relevant, we can see depends on whether you're above or below two dimensions. So when you are below two dimensions, this nonlinearity is relevant. And you will certainly have different types of scaling phenomena then what you predict by the case of the diffusion equation plus noise. Of course, the interesting case is when you are at the marginal dimension of 2. Now, in terms of when you do proper renormalization group with this nonlinearity, you will find that, unlike the nonlinearity of the Landau-Ginzburg, which is marginally irrelevant in four dimensions. du by dl was minus u squared, this lambda is marginally relevant. d lambda by dl is proportional to plus lambda squared. It is relevant marginality-- marginally relevant. And actually, the epsilon expansion gives you no information about what's happening in the system. So people have then done numerical simulations. And they find that there is a roughness that is characterized by an exponent, say something like 0.4. So that when you look at some surface that is grown, is much, much rougher than the surface of a soap bubble or what's happening on the surface of the pond. And the key to all of this is that we wrote down equations on the basis of this generalization on symmetry that we had learned, now applied to this dynamical system, did an expansion, found one first term. And we found it to be relevant. And is actually not that often that you find something that is relevant, because then it is a reason to celebrate. Because most of the time, things are irrelevant, and you end up with boring diffusion equations. So find something that is relevant. And that's my last message to you.
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https://ocw.mit.edu/courses/7-016-introductory-biology-fall-2018/7.016-fall-2018.zip
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BARBARA IMPERIALI: Now, I want to talk today about one small thing before we move on to signaling, because it really kind of completes the work that we talked about with respect to trafficking. So I popped this question up last time, and it seemed like there weren't quite enough sort of people leaping to give me an answer. But let's just take a look at the big picture of things, as it's always good to do. Because this will also get me to one other topic, which is protein misfolding. So at the end of the day, what really defines where a protein is, what it does, is defined by its sequence. But you always want to remember that a protein sequence is defined by its messenger. The messenger is defined by the pre-messenger. Yes, there may be splicing events that really cause changes in localization. But the pre-messenger includes the content. And then what defines that is the DNA. There's certain aspects of regulation at the epigenetic level that we don't talk about barely in this course. But I want you to make sure that you realize at the end of the day, what the protein is, how it folds, is defined originally by the sequence of the DNA, although a long way along here. The post-translational modifications that we started talking about last time are defined by the protein sequence, which all the way is defined by the DNA-- so, so much of protein function. And there's one more aspect of proteins that's defined by the DNA sequence, and that's whether a protein folds well, or perhaps, in some cases, misfolds. And that's the thing I want to talk about very briefly today. Because I think that this captures the picture. So let's just go over here and write misfolded proteins, which, just like everything else, largely end up being dictated by the DNA. Because whether a protein folds faithfully into a good structure or misfolds can be a function of the protein sequence. So there could be mutations in the protein that ultimately end up that the protein misfolds and forms either a misfolded tertiary structure, or even worse, adopts an aggregated form that causes a lot of damage within cells and outsides of cells. So I want to talk just briefly about the processes that we have-- it's just one slide-- to deal with misfolded proteins. So when a protein is translated, it almost starts folding straight away, especially large proteins. A fair amount of a protein may have already emerged from the ribosome and started folding, even when the whole protein isn't made. The sequence ultimately ends you up with a well-folded protein. But if the protein does not fold fast enough, or there is a mistake in this, which might be caused intrinsically by the primary sequence, if there's a mistake in that-- so slow folding or incorrect folding-- then you will end up with a protein that's partially folded within the context of a cell. We especially encounter misfolded proteins when we are overexpressing proteins in cells, because you're just making one of a type of protein really quickly, and it doesn't have a chance to adopt its faithful structure. So there are proteins within the cell that helped sort of protect the folding process early on, to allow the protein to have enough time in singular, not with a lot of copies of itself around that are misfolded, to adopt a folded structure. And these proteins are called chaperones. I don't know if you guys are familiar with the term chaperone. It was a term that was heavily used in the sort of 18th and 19th century. A chaperone used to be an aunt or someone who you would send out with your beautiful young daughter to chaperone her, to protect her so she didn't get bothered by those mean men out there. So chaperones were-- that was the original definition of the chaperone. And it's kind of interesting that the chaperones are now proteins that help folding or protect against misfolding. How do they do this? Generally, a protein will fold poorly if it's very, very-- if it's quite hydrophobic. And hydrophobic patches are exposed in aggregate. So let's say, you have a protein, and there's a lot of copies, but they're not folded. If you have things that are hydrophobic that would normally end up tucked inside the protein, if the protein hasn't folded in its good time, these will just start to form aggregates, sort of associating with each other. It's just a physical phenomenon. If you put something that's got a lot of hydrophobic faces on the outside, this will start forming an aggregated bundle, and not a nicely folded protein at all. What the chaperones may do is in part hold the partially folded protein. So let's just think of this big jelly bean as a chaperone until things start to adopt a favorable state. But sometimes it's just too much. The chaperone cannot handle the flux of protein. So the protein ends up being recognized as misfolded. And then it gets tagged as a misfolded protein, and it gets taken to a place in the cell for disposal. So if you are unable to fold, there is a tagging process. And I mentioned it last time. It's a process known as ubiquitination. This is also a post-translational modification, but it's one that occurs on poorly folded proteins. And I'm going to describe to you that system, because the ubiquitination is the flag, the signal, or the tag, to take this protein to the great shredder, basically. And so what does the-- what's a paper shredder-- I like the analogy with a paper shredder. So here's a fellow who's got too much on in his inbox. So he just sends it straight to the shredder. It's a little bit about too much misfolded protein being made. So instead of sort of waiting to deal with the paperwork, you just send it straight to the shredder. And the proteasome is the cellular shredder that actually breaks proteins up into small chunks, and then digests them out. So think of the proteasome as a shredder, which chops up proteins into small pieces, mostly into short peptides that are 8 to 14 amino acids in length-- fairly small. Short peptides won't cause a problem in aggregation, and will then be further digested. Now, if you've got this shredder sitting around in the cell, it's like having a paper shredder on all the time. You've got a-- there's a risk things may end up in there without meaning to be. So for things to be tagged for shredding, they go through what's known as the ubiquitin system. So the first thing to get into that is-- and it's only then that proteins are tagged for shredding up or chopping up by the proteasome. So as you can sort of tell by its name, it's got protease function, but it's a large, macromolecular protease, with lots and lots of subunits that are important to cut that polypeptide into smaller pieces. But because many of your proteins may be partially folded or misfolded, they first have to be unfolded. So the ubiquitin is the signal to send proteins to the proteasome, where the second action is protease activity, and the first action is unfolding. So what I show you on this picture is the barrel structure of a proteasome. Let me explain the components of it. The red component of the proteasome is a multimeric ring that uses ATP and starts tugging apart the protein that you need to destroy. But it will only do that if the protein becomes labeled for destruction by the ubiquitin system. And I am showing you here a massively simplified version. Let's say this is a misfolded protein. It gets tagged with another protein. It's a really little protein known as ubiquitin. I've shown you the three-dimensional structure here. And using ATP, you end up managing to put a ubiquitin chain on the protein that's going to be destroyed. That is a post-translational modification that is a tagging for destruction. If the protein is not tagged, then it's not going to be chewed up. That makes sense. You don't want to be chopping up proteins in a cell with wild abandon. Once the ubiquitin chain is on here, the protein will bind to the unfoldase part of the proteasome, and with ATP, it will just stop tugging the rest of the residual structure apart to thread the protein down into the blue part of the barrel. It's a little hard to see it like this, but it's literally, the proteasome, are four concentric rings. Let me see. I hope my artwork is going to be good enough. Well, that's an unfoldase. And so is that. And then in the center, there is a protease. And each of these components is multimeric, having six or seven subunits. So it's a huge structure. It has a sedimentation coefficient of 20S, that entire structure. I don't know if you remember when I talked about ribosomes, they were so big we didn't tend to talk about them by size. We talked about them by sedimentation coefficient. And the large and small subunits of the ribosome, the eukaryotic one, just to remind you, were 40S and 60S. So just remember that S stands for Svedberg. It's a sedimentation coefficient unit. It describes how fast approaching precipitates. So once the protein has been labeled with ubiquitin, it binds to the unfoldase. And then the single strand feeds into the center core, which is two sections of protease. So it's feeding in here. It sees the protease activity. And then it's just short pieces of protein are spit out of the proteasome. Once these are really little pieces of peptide, they're readily digested by proteases within the cell. And you can recycle the amino acids, or you can do other things with these small pieces of peptide. They actually end up sometimes being sent for presentation on the surface of the cell by the immune system. And you may hear a little bit more about that later. So the proteasome-- oh, I apologize-- this should have been 26S-- has a molecular weight that's very large-- 2,000 kilodaltons. That's why we refer to it by its sedimentation coefficient. So this machinery is very important to get rid of misfolded or aggregated proteins to destroy them. Now, does-- are people aware of the sorts of diseases that can result from misfolded proteins? Has anyone been reading the news much about certain types of diseases, particularly in neurobiology? Anyone aware of those? Yeah. AUDIENCE: Was it mad cow disease? BARBARA IMPERIALI: Which one? AUDIENCE: Mad cow. BARBARA IMPERIALI: Yes. Mad cow. So there are a variety of neurological disorders, and mad cow is one of them. Creutz U-T-Z feldt-Jakob. But Alzheimer's disease is another one. Pick's disease is another. There are a wide variety of neurological disorders that result from misfolded proteins, both inside the cell and in the extracellular matrix, forming these tangles that are toxic to the neurons, causing them to no longer function, and then resulting in many of these neurological disorders. The ones I've described to you, I've mentioned to you here. I know many of you are familiar with Alzheimer's disease. Mad cow disease is a variant of a particular protein misfolding disease that was first noted in cattle. And they basically just fell down, dropped down. And it was in some cases ascribed to-- the contagion with the disease is ascribed not to a virus or to a microorganism, but literally, to misfolded proteins causing the formation of more misfolded proteins. So these are all collectively designated as prion diseases. I think you'll have read that term. And it's a particular kind of disease that the infectious agent isn't a living system-- not a virus, not a microbe, a fungus, a protozoan, but rather a protein, where it's misfolded structure nucleates the formation of more misfolded structure that leads to the disease. So I grew up in England during the years where there was a lot of mad cow disease in England. And even though I'm a vegetarian in the US for 30 years, I can't give blood in the US, because I lived in England during the time when there was a lot of mad cow disease. And this can be dormant for a long, long time before it suddenly takes over. So there's restrictions on blood donation in certain cases. And it's because it's not something you can treat with an antibiotic, you can treat with an antiviral. It's literally traces of badly folded protein that can nucleate the formation of more badly folded protein, that can lead to the diseases. These were-- there's particular instances of some of these diseases in tribes where there's pretty serious cannibalism, and eating your sort of senior relative's brains was considered to be something-- an important act of respect. And there was this transfer of some of these prion-type diseases through cannibalism as well. So eating contaminated meat, be it a cow, be it your grandparents, whatever, it's something that actually is-- it's a serious transmissible disease. And it's really-- in the situations where it can be sort of related back to contaminated meat are one thing. But there are variations in the case of Alzheimer's, where the sequence of proteins may dictate that they don't fold well, or they're not post-translationally modified properly, so they end up as misfolded proteins. So these are often genetically linked disorders, some of the things like Alzheimer's. And once again, remember that goes all the way back to the DNA, which might, in some cases, trigger the misfolded disease. So it's a fascinating area, and there's a tremendous amount to be studied. Because of the aging population, these diseases are piling up, and we need to mitigate the causes of the disease, and find ways, for example, to slow down. If there are these fibrils of protein that are misfolded, can we maybe inhibit that formation with some kind of small molecule inhibitor to mitigate the symptoms of the disease? So it's a very, very active area, because almost every-- many, many neurological disorders seem to be coming down to misfolded proteins. So let's move on now to signaling. All right. So we're going to spend two lectures on-- the remainder of this lecture plus the next lecture. And what I want to do in this lecture is introduce you to some of the paradigms, the nuts and bolts, the mechanics of protein signaling. And then in the next lecture, I'm going to show you examples of how all the characteristics that we define signaling by get represented in signaling pathways within cells. So I'm going to give you all the moving parts, and then we'll move forward to see how the moving parts might function in a physiological action, such as a response to something particularly scary, or as a trigger to do-- for the cells to do something different. So let me take you, first of all, to a cartoon-like image of a cell. And we're going to just take from the very simplest beginning. But then this topic will get quite complex, as you see. But that's why I think it's important to reduce the process of protein signaling down to simple aspects of it that we can really recognize, even in much more complicated pathways. So in protein cellular signaling, this is a complex system of communication that governs all basic activities of the cell. There are no cells that don't do signaling. Bacteria and eukaryotic cells may do signaling slightly differently. But they still do have an integrated correlated system that's responsible for triggering functions of the cell through a series of discrete steps. So protein signaling can be dissected into three basic steps, where you, first of all, receive a signal. And we're going to talk about what that signal is. What's the nature of that signal, is it small molecule, large molecule? Where is the signal? Where does it act? Then the next step is to transduce the signal. And finally, you have an outcome, which is a response. So we're going to talk about each of these components in order to understand flux through cellular signaling pathways, and how they work to give you a rapid response to a necessary signal. All right. So in this cartoon, let's just, for example, think about what if we want to trigger cell division? We might have a signal, which is the yellow molecule-- a small molecule, large molecule. We'll get to that later. There's a cell here, where on the surface of the cell is a receptor. And that would be the entity that receives the signal. So in the first step in the process, there's a buildup of a concentration of a signal. And it occupies the receptors on the surface of the cell, and in some cases, inside the cell. We'll talk about a bifurcation there. But really, a lot of cellular signaling is dominated by cells coming-- by signals coming from outside the cell. What happens upon this binding event is the transduction. If you bind to something on the outside of the cell, as a consequence, you might have a change on that structure. If it crosses the membrane, you might have a change on that same molecule structure that's on the inside of the cell. So that's why it's called transduction. You're transducing a soluble signal from outside, binding that signal to the cell surface receptor. And the cell surface receptor is responding in some way. And there are two principle ways in which we respond to extracellular signals, and we'll cover them both. The next event that might happen is through the change that happens to the intracellular component of the receptor. There might be a change, a binding event, another step occur within the cell. And as a function of that, you get a response. All right? So it's really-- thinking it in these three components is a good way to kind of dissect out the beginnings of the complication. And then what we'll be able to do is really start to see what kinds of molecules come in? How are they received? How is the signal transduced? And what's the ultimate outcome with respect to a response? Everyone OK with that? All right. Now this is what you have to look forward to. So we give you something with three moving parts, and suddenly we show you something with sort of, you know, 100 moving parts. And cell biologists very, very frequently look at these maps of cells, where what they're looking at with each of these sort of little acronyms or names, all of these are proteins, where they have been mapped out through cell biology and cellular biochemistry to be existing in certain components of the cell. And what has also been mapped out very frequently is, who talks to who? So the fact that JAK S might interact with STAT 35 and so on. So much of this was worked out through cell biology and biochemistry, and also by genetics. So Professor Martin has talked to you about identifying a player in a complex system by genetics. Let's say you have a cell that fails to divide. You might perhaps screen or divides unevenly, or has some defect in cell division. You might be able to pick out a particular player. Now, the key thing I want to point out to you with this cell is what's on the outside of the cell and runs across the membrane, and might have the chance, the opportunity, to have both an extracellular receptor and an intracellular function. And those key proteins are things like receptor tyrosine kinases. And we're going to talk about all of these in a moment. G protein-coupled receptors, and various other cell surface receptors-- so all of these-- anything that spans a membrane has the opportunity to be an important component of a signaling pathway. Because what you're routinely trying to do is have your signal recognized on the outside of the cell by something that spans the membrane. The signal will bind to that. And then you will have an intracellular response. So that's breaking it down. That's why proteins that are made through the secretory pathway that we talked about in the last lecture, that go through that endomembrane system, and end up being parked in the plasma membrane are so important. Other proteins that actually get secreted through that pathway are also important. What do you think they may be important for? Let's say you've made a protein within the cell. It goes through all the system. It doesn't stay parked in the cell membrane. It actually gets released from the cell. What might that be doing? AUDIENCE: [INAUDIBLE] BARBARA IMPERIALI: Yeah. Exactly. So that endomembrane system that I described to you, that pathway is great for making receivers. And it's great for making signals. And that's really what can sort of fuel the functions of cells. OK. So in systems biology, you may have heard this term quite frequently. Systems biology is research that helps us understand the underlying structure of signaling networks. So a lot of people who have common interests in engineering, computational analysis and cell biology, might bring in data to allow them to make models of cellular systems, to understand flux through signaling pathways. So they may make fundamental measurements about the concentrations of some components within the cell. And then try to say, OK, I know based on everything I've measured that this is a dominant pathway for gene regulation. And I could control this pathway by different-- by sort of different types of interactions. In this cellular system, I also show you another component, which is the nucleus. And when we discuss and describe specific cell signaling networks, in some cases the signaling network may involve receiving a signal, undergoing a variety of changes in the cytoplasm, but then a change that eventually results in a protein going to the nucleus. And oftentimes, those proteins that run into the nucleus are transcription factors that then trigger DNA replication or transcription. And then promote activities. So this is how you think about it. When you think of cellular signaling, it's really about, what does the signal need to do? And what's the pathway that I follow to get there? So all of those are membrane proteins. So now let's look at the canonical aspects of signal transduction. So the first-- and I'm going to rely on these little cartoons. But I want, both in this lecture and the next, to really show you where these recur in so many systems. So to that purpose, I want to talk about the characteristics. So the first critical characteristic is a signal and it's specificity. So a signal will be something that comes from outside of the cell. It could be a hormone that's produced in the hypothalamus and sent to another organ. But the most important thing about the signal is that that signal, which binds to a receptor in a cell membrane, is specific for a particular receptor, and the different signal won't blind to the same receptor. You have to have faithful signal specificity to trigger the right function. So if it's a hormone, it's got to be the hormone that you want to trigger the receptor, not a related but different-looking structure. If it's a small protein, you want it to be the exact one that binds with high specificity to a receptor. So what that means is if something is binding-- if a small molecule is binding to a protein on the surface of a cell with high specificity and high affinity, it means that even at a low concentration, it will make that binding contact. But all the other small molecules that are around won't crosstalk into that triggering that interaction. So we have high specificity, and we gain that specificity through macromolecular interactions, just like the ones we talked about in biochemistry. So if we have a small molecule or a protein bind to the receptor, it's making all those hydrogen bonding, electrostatic, noncovalent types of interactions with high specificity, so that a low concentration of the signal molecule is efficient for binding to the receptor to trigger the function. The next characteristic is amplification. Now let's put some lines between these guys. Now, with all the signaling pathways that you're going to see, we're going to be looking where in a pathway you get amplification. Very commonly, you might have a response that's just the result of a single molecule binding a single receptor. But at the end of the day, you might want a large response. You might want to make a lot of ATP. Or you might want to replicate all of the genome. So you need some kind of amplification where in a sense you're turning up the volume on your signal. And you need to do that rapidly. So frequently in signaling pathways, you go through a cascade of reactions where the signal might affect an enzyme. But once you make that enzyme active, it might work on many, many copies of another enzyme. And then each of those may work on even more copies. So that's what I mean by amplification, where at some stage you've generated a molecule that can result in the cascade of a reaction. So we often refer to these as cascades. So if you're Spanish-speaking, cascada. You want to think about a waterfall coming from just a single molecule of water. You're getting a large increase in your signal as a result of amplification. The next feature or characteristic of signaling is feedback. At the end of the day, if you're signaling, I got to make some ATP. I got to run out of the woods. I'm getting chased. At a certain stage, you need to stop all of the process occurring. So feedback is just a negative feedback loop that might slow down some of those steps that are involved in amplification. So for a pathway, you only want the pathway turned on for a prescribed amount of time. And then you want to be able to say, I'm done with that whole pathway. I don't need to keep churning through all those enzymes. It's time to stop that. And that usually occurs through negative feedback. And remember, we talked about negative feedback when we were talking about enzyme catalyzed pathways. So feedback is very often some kind of negative feedback, which suppresses a series of transformations, perhaps through a product of those transformations acting as an inhibitor on an early step. And then finally, the other component of a signaling network-- if you think of signaling networks as electronic structures, you have integration. So that's the last characteristic feature. And let me go back to that big circuit diagram quickly to show you an example of integration. So if you look at this signaling pathway, all these signaling steps are not single. You just have a signal come in, and you end up, for example, in the nucleus. But rather, other components may have crosstalk within one pathway, and start out either amplifying or turning down a particular signaling pathway. So these are networks. They're not pathways. They're networks that interact and communicate, all to amplify signals or turn down signals. So integration is an important part of signaling, because you're often dealing with the integrated function of a number of pathways to get a particular response. And that actually ends up being one of the situations where sometimes a particular enzyme may look like a perfect target for a therapeutic agent. But if you don't take into account the integration steps, you may be trying to deal with a-- you may think you're dealing with a single pathway, but you're, rather, dealing with crosstalk with a lot of other pathways. And what often happens in a cell is there's compensation from other pathways. Is everybody following? Any questions here about this? So what I want you to think about is that it's just amazing what is orchestrated to have even the simplest functions in the cell, how many interacting components there may be. Specificity, amplification, feedback, and integration-- all right, so let's talk briefly about types of signals and how we name them, where they come from, in order to make sure we're all on the same page with respect to the language that's used. So now, signals may take different molecular forms. They may be small molecules, for example, an amino acid or a phospholipid-- just something little. Alternatively, they may be proteins. They may be carbohydrates. They might take different forms in terms of their molecular structure. But we tend to describe signals by where they come from. So what I've shown you here is a picture from the book that just describes how we refer to certain signals. So there are four different terms-- autocrine, juxtacrine-- and I'm going to just give you a little hint to how to remember these terms-- paracrine, endocrine. OK. So these don't tell you anything about the molecule. They tell you about where it's come from. So an autocrine signal is a signal that may come from a cell, but it's signaling to itself. So it may produce a component that's released. And so it's producing this through a secretory pathway. It's a release, and it stays in the vicinity of the cell. So the self is self-signaling. So whenever you see something auto, you just want to say, oh, that means it's coming from the same cell where the signal occurs. Let's move to the next one, which is paracrine. I'm going to talk about jux-- and that's usually from a nearby cell, not a cell that's in contact-- definitely a different cell. So paracrine is-- we would always call nearby. And endocrine is completely from somewhere else, so perhaps coming through the circulatory system. One cell may release an endocrine signal. It may weave its way through the vascular system, and then target a cell. So endocrine is always from a distance. And juxtacrine is the only one that's a little odd. It's really from cells that actually are in contact with each other. So it's not self-signaling within a cell. It's not a cell that's nearby but pretty close. It's actually physically making a contact. And so that's the last terminology there. So hopefully, I can get this calcium wave to show you. This is just a video of juxtacrine to signaling. I just want you to sort of keep an eye on things. It's usually a cell. What you're observing here is a dye that lights up in the presence of calcium flux. It's called Fura-2. And so when you stare at these for long enough, what you can notice is that when one signal will often come from an adjacent cell right near it-- so there are long prostheses. You're not looking at the entire cell, but they're definitely-- for example, this little duo down here, they keep signaling to each other. And that's just a juxtacrine signaling, because the cells are in the contact. So that just shows you the difference there. If it was autocrine, you just have a single cell responding. If it's paracrine, they would be at more of a distance to each other. I hope that imagery-- this is from a website in the Smith Lab at Stanford. OK. And then the last thing I want to give you an example, there are many, many hormones in the body that undergo endocrine signaling, and so one example I thought I would tell you about, you all know that insulin is made in the pancreas. It's an important hormone for regulating glucose levels. And it's actually-- functions at the muscle level. So insulin is an example of an endocrine signal, because it travels a distance from where it's made in the body to where it functions in the body. All right. Now-- so we've talked about the types of signals. Let's now move to the types of receptors. Now, we cover both the intracellular and the cell surface receptors. But we really will focus a lot on the cell surface receptors. I just want to give you a clue that not all signaling is cell surface. So what I've shown you here is a cartoon where you see signaling, where a signal comes from outside the cell. It goes into the cell and triggers a change. And then the majority of the time we'll talk about these receptors that are in the plasma membrane. And they have an outside place where the signal binds, and they trigger a response inside. And it's only very specific signals that are able to signal intracellularly, that is, to cross the membrane to get inside the cytoplasm to do the triggering. What kinds of molecules can cross the membrane easily? We talked about that before, when we talked about getting across that barrier. Yeah? Nonpolar. OK. So you can look at a-- you can stare at a molecule, and if it's very polar or pretty large, it's not going to be able to sneak through a membrane. So something like a steroid molecule, a large, greasy molecule, can definitely make that transition. And so those are the only types of signals that we can really do inside the cell, because they can get across the cell. Many, many other signals have to go through this-- bind to the outside of a cell, and transduce a signal to the inside of the cell. So one very typical signal that can bind to an intracellular receptor is a steroid. So remember when I talked to you about these lipidic molecules, things like testosterone and cortisol. These are very hydrophobic molecules. So they literally can cross from the outside of the cell without a transporter. So for example, the hormone cortisol. And when that functions, it just-- an amount of it becomes available, for example, in the bloodstream. It crosses the cell, and it binds to an intracellular receptor. Once it binds to that intracellular receptor, this disengages a different kind of chaperone protein that's keeping it stable. Once it's found, it can then go into the nucleus and trigger transcription. So this is the one example of an intracellular receptor that we'll talk about. I just wanted to show you a little bit about the steroid receptors. These are molecules that are very-- these are macromolecules, proteins that are very-- quite a complex structure. But they can literally-- and I'll show you the picture at the beginning of the talk next time-- they can literally engulf these proteins. So once the steroid is bound to that, it completely changes shape. And that's what enables the change for it to be triggered and sent to the nucleus. Now, the key types of receptors that we'll focus on, though, are the cell surface receptors. And there are three basic classes of molecules that occur in the plasma membrane that are critical for cellular signaling. They are the G protein-coupled receptors, the receptor tyrosine kinases, and then you will talk in the lecture 22 about ion channels, and how they perform a receptor function. So the membrane proteins, first of all, I want to underscore their importance. 50%-- they comprise 50% of drug targets, the receptor tyrosine kinases and the G protein-coupled receptors. The G protein-coupled receptors have this 7 transmembrane helix structure, which spans a membrane. This would be the outside of the cell, and the inside of the cells-- so there's signals going across there. The receptor tyrosine kinases are another important type of receptor. They are dimeric proteins that in the presence of a ligand dimerize, and then cause intracellular signaling. Once again, they cross the plasma membrane from the outside to the cytosol. And then lastly, there are the ion channels, which also may cross the plasma membrane. And when you think about these classes of proteins, there's a tremendous amount to be learned with respect to their functions. And they are so important to understand their physical functions in the body, because they really represent the place, the nexus, where signaling happens in the cell. So I want to briefly show you a picture of a GPCR. It's a 7 transmembrane helix structure. You can see it here. There are about 30% of modern drugs actually target the GPCRs. And here, I'm just going to show you the structure of a GPCR. Those are the 7 transmembrane helices. If you stretch them out, that's about the width of a membrane. That's typical of a signal that would bind to that kind of receptor. This is a chemokine. It's a small protein receptor. So you can see that structure and how it would go from one side of the membrane to the other. In it's bound state, the chemokine binds to the 7 transmembrane helix receptor through kind of a clamping action. The magenta is the chemokine. The blue and the green space filled parts are actually what holds the chemokine. And if you look at it where the membrane would be, you can see how you can transduce a signal from one side of the membrane to the other, by the binding of the magenta molecule to the outside of the cell, to those loops outside the cell. That would have a significant perturbation to the biology and chemistry of what's going on on the inside. So next class, we'll talk about pathways that are initiated by these G protein-coupled receptors, and what that terminology means. OK.
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https://ocw.mit.edu/courses/7-013-introductory-biology-spring-2013/7.013-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. HAZEL SIVE: I want to discuss with you today a very topical and interesting question, which is the notion of stem cells. In fact, I'm going to discuss two things, the first of which is another concept that you need, following on from the concepts we had right at the beginning of lecture. I feel like this microphone-- The first is a concept that you need in addition to the ones we started lecture with, and then we'll talk about stem cells. So today, we'll talk about potency, and then we'll talk about stem cells. Potency, along with fate, determination, and differentiation, is one of those terms that you need to know and you need to understand in order to understand stem cells. Potency refers to the number of possible fates that a cell can acquire, number of possible fates open to a cell. And this is a very important concept of development because, in general, potency decreases with age, and decreases as different parts of the organism become specialized. So in general, potency decreases with age. But I will put in here, and we'll explore this more in a moment, except for some stem cells. And we haven't defined a stem cell yet, but we will. What kinds of potencies are there? There's the big one, totipotent, where a cell can become all fates. And there's really only one cell that can do this in the normal animal, and is the zygote. And in most animals, even as the zygote becomes just two cells or a few cells, that full potency is lost. And cells instead, in the embryo, are multi or pluripotent, which means that they can acquire many fates, but not all fates. Embryonic cells, especially in the early embryo, and many stem cells can also become, are also multipotent or pluripotent. And then as time progresses-- is that a hand up? Yes, sir. AUDIENCE: How do you reconcile the fact that human cells, we can separate them. Even at the eight cell stage. HAZEL SIVE: That's a great question. The question is how do I reconcile what I'm telling you with the fact that you can get identical sextuplets, or octuplets actually? It's a good question. That's true. In different animals, the very early embryonic cells are sometimes totipotent up to a while. OK? And so for example, in armadillo, here's a piece of, you know, fact for your back pockets. In the armadillo, the eight-cell embryo almost always splits into eight single cells, each of which becomes a baby armadillo. OK? So those cells are totipotent. In mice, even at the two cell stage, the two mouse cells are probably not equivalently potent and they're not totipotent. So very, very seldom-- almost never-- get identical mouse twins. OK? So it's one of these generalities. And if you ask me, you get the specifics are a bit different. As cell fate restriction continues, cells can become bipotent, or unipotent, whereby one or just two fates are open to them. And so if we look, taking this concept, let's now start the lecture about stem cells. You're going to need this concept. Stem cells, I'll point out, whatever they are, got almost 3,000 hits yesterday on Google News. This is way below baseball, which got 45,000. I checked. But still, you know, as science topics go, stem cells are really up there. And they're on the covers of magazines, over and over again. And we'll talk more about why that is. Here's a diagram-- it's not on your handouts-- that I drew for you. Let's not dwell on it. But let's now move on to Topic Number 2, which will fold in this concept of potency and the concepts of fate, determination, and differentiation, and talk about stem cells. And let's do, as is our custom, let's define what a stem cell is. I think that a stem cell can be defined as a cell of variable potency that has the capacity to self-renew. Cells of variable potency that can self-renew. They can make more of themselves. Despite the hype, despite covers of Time Magazine and almost every front page of every newspaper across the world, stem cells are normally found in our bodies. And normally, as we'll explore, they're used for organ maintenance and repair, organ maintenance and repair. But the thing, you know, that has everyone fired up is that you can somehow harness these cells for therapeutic purposes. And that you can repair what the body cannot, by being clever, and using the power of these cells as they normally have it, or as you can give it to them. And so there's this question of therapy and therapeutic stem cells, where the idea, again, is that you would repair a damaged organ by introducing somehow, injecting or otherwise introducing, extra or somehow special stem cells-- which I am going to abbreviate heretofore as SC-- by introducing extra stem cells into a damaged body. Does this work? It does work. It works for the hematopoietic system, as in bone marrow transplants. And it also works for skin cell transplants. Well, let's just put skin cells. OK. Skin stem cells can be grown from your own skin. And in the case of burn victims, this has really saved countless lives. The original technology began to be developed here at MIT by Professor Howard Green, who is now at Harvard. But the idea is to take your skin and grow it on something like gauze or some kind of some solid support, and then to cover a burn patient with layers of support on which there are some stem cells. And these stem cells will help fill in the holes in the skin left by the burn. Normally when a wound heals, as I'm sure you've noticed, it heals from the sides. The only way a wound can heal is from the side. And if it's a big wound, it can take a very, very long time to heal. And you can get infections and so on while the healing process is going on. So seeding the inside of a wound with stem cells that can start the skin regeneration process and seal up the body against infection, that's been incredibly useful. And we'll talk more about bone marrow transplants in a moment. OK. We previously talked about this process by which cells decide, are undecided initially, they decide what they're going to become. And then they differentiate into their final function. Stem cells fit into this litany somewhere between the commitment stage and the differentiation stage. And in this diagram, these multiple arrows are there for a reason. There are multiple steps between commitment and differentiation. And somewhere along the way, a group of cells with capacities we'll talk about, leaves this lineage and sits around and waits, partially determined, so that it can go on and make more differentiated cells when they're needed. And I've added on there the potency timeline, decreasing with age, of the developing animal. But let's diagram the notion of stem cells on the board. Stem cells generally divide slowly. Here's one. It's a variable potency. It may be multipotent. It may be bipotential. And it is somewhat committed, which is a slightly difficult concept. Because last time we talked about committed versus uncommitted. But now I'm telling you something can be somewhat committed. And that gets to this multiple arrows there were as cells progress in their fate decisions, they change their molecular signature and they really do become closer and closer to a cell that's made a decision. But it's kind of like, you know, if you're weighing up going to med school or going to graduate school in bioengineering, you know, you have decided that it will be one or the other, but you haven't decided which. You're somewhat committed. And then when you make the decision to go to graduate school, you have now become committed. OK? So the cell is doing the same kind of notion there. There's your stem cell, variable potency. Under the correct stimulus, that stem cell will divide to give rise to two different cells. One is another stem cell. And the other is something we'll call a progenitor. The progenitor is more committed than the stem cell. The progenitor cell is going to go on and divide, usually a lot. Progenitors divide rapidly. And their progeny will eventually go on and differentiate into one or more different kinds of cells, maybe a stripey cell, and a spotted cell type, and a cell type with squiggles. And so here are the differentiated cell types. And the number of different differentiated cells that comes out of this process is a reflection of the potency of the stem cell. OK? So here you've got these progenitors. The idea is that these progenitors will have similar potency. But as I'll show you, there's a whole variation on this. But here the number of differentiated cell types, the number of cell types reflects potency of the stem cell. OK? This kind of diagram is called a lineage diagram. It tells you what-- not only what the final fate of the cell is, it tells you something about the progress towards that final fate. So a lineage we can define as the set of cell types arising from a stem cell or a progenitor. Let's talk about the discovery of stem cells, because this is something that really was pivotal in helping understand whether or not there was some way that the body normally repaired itself. It was clear that during early development, there was lots of cell division and lots of changes. Cell types were formed and organs were formed. But it really wasn't clear in the adult how much repair there was, how much turnover of tissues there were, and really what the whole dynamic process of maintaining the adult was. And the discovery of stem cells came about because people looked to see how long cells lived. And what they found was found using a turnover assay that measures the half life of cells. And they found that in almost all organs, in fact, probably in all organs, cells did not live forever. They turned over. They died. And they were replaced by new cells. And this turnover assay implied that there was some kind of replacement. And the cells doing the replacement were called stem cells. You find this by a pulse/chase assay, which we'll go over on your handout in a moment. And what was found was really variable for different organs. Firstly, all organs, about, show cell turn over. Red blood cells have a half life of about 120 days. There are a lot of red blood cells in your body. And in fact, that implies that there are about 10 to the seventh new red blood cells made a day. In your intestine, the half life of cells is three to five days, in the small intestine. And the hair on your head has a half life of about four years. So it's variable for different kinds of cells. If you look at your first handout, it diagrams a pulse/chase assay where a cell population is labeled with a nucleotide analog. It's a normal nucleotide, but it's got a bromine added to it. And it acts like deoxythymidine, gets incorporated into DNA, and you give just a short pulse of this nucleotide analog. So only some of the cells get labeled. And you only give it for a short time. So you get a labeled cell population. And then you stop the labeling by adding lots of unlabeled thymidine, and that's called a chase. And you follow the cells over this long chase period. And then you can watch and see what happens to those cells that you initially labeled over a very short time. And so in this example, I've got four cells initially labeled. Over time, they're only two cells left. And if you measure the time from going from four cells to two cells, you can get to the half life of that cell population. OK? You can also-- if you don't have a handout for this, just look on the screen-- you can also follow the labeled cell population and see what those cells become. And you can see that they go on to differentiate as particular cell types. So this is a kind of way of labeling the lineage of these cells. And that is useful, too. This was the theory behind stem cell definition. But what is a stem cell look like, and how do you isolate one? It turns out that that's really difficult. So isolation and assay in the adult stem cells are very, very rare. And that is one of the issues with using stem cells for therapy. There are very few of them and they're hard to isolate. Hematopoietic stem cells comprise about 0.01% of the bone marrow, which is where the stem cells reside, and where the precursors of your whole blood and immune system reside. The way that this was dealt with was through a really clever technique that has the acronym of FACS. I'll give you a slide in a moment. It stands for Fluorescence Activated Cell Sorting. We'll go to a slide in a moment so we don't have to spell it out. And the idea behind FACS is that you label stem cells. And you might be guessing what to label them with. But you label them usually via their cell surface proteins with some kind of tag, often an antibody tag. And then you can use that tag to make them different colors. And you can then sort them, cell by cell, through the Special Fluorescence Activated Cell Sorter. Sort individual cells, and then you can assay individual cells or small groups of them for stem cell properties. Let's look at a slide of how the FACS, the Fluorescence Activated Cell Sorter works. No, let's not. I've really gotten ahead of myself here. And I'm going to go back because I want to show you this. Hold on to that thought and let's go back to this notion of a pulse/chase assay. I forgot that I had this here. This is really important. This is a pulse/chase assay of intestinal cells. And so your small intestine lies inside your belly. And if you look at its anatomy, it contains many tubes whose surface is thrown into folds to increase the surface area for food absorption. And if you look at these folds, they are very closely packed. So you get a huge surface area increase. And the cells of these folds that are doing the food absorption turn over every three to five days. Those are the ones I was talking about. So if you blow up one of these folds, which is called a villus, there's the part that's sticking up into the cavity, into the lumen of the small intestine. And then there's this part that kind of dips down into the lining of the intestine, which is called a crypt. The stem cells lie somewhere at the base of the crypt. It's not exactly clear where. But the idea is that somewhere at the bottom of this crypt there are these stem cells that under specific conditions will start to proliferate. And they will move up into the villus and replace the villus cells that are dying. You can monitor this by doing a pulse/chase experiment. So here's the crypt. And they've given, in this experiment, a pulse of thymidine has been given. And you're looking kind of at time 0, right after the pulse of thymidine has been given. Here the cells, the black cells are in the crypt. They're labeled. And then if you look over time, you can see that these black cells move away from the crypt. They're moving up into the villus. And here they are actually right on top of the villus, replacing the cells that were turning over. So that's a really beautiful demonstration of a pulse/chase assay. OK. Now we can go to our Fluorescence Activated Cell Sorter. The idea is that you take a mix of cells-- you don't have this on your handout, just look on the screen-- the cells are labeled with fluorescent antibodies. And you put them into a reservoir and droplet generator. And the cells drop out of this reservoir, one at a time. And as they drop out, they go past a laser. And you can tune the laser to whatever wavelengths you want. It excites the cells. And if the cells emit in the particular wavelength you're interested in, the detector will detect that. And then it actually gives a charge to the cell that it is the correct fluorescence. And as the cells are dropping down, the cells of the correct color are deflected by an electric charge. And different color cells can be collected into different flasks. OK? This really works. It's a fantastic machine. You can collect cells about, you know, you can collect millions of cells an hour. It's pretty quick. But you do it one cell at a time. And in that way, you can isolate cells, which have got stem cell properties. OK. So you've used your FACS machine. You've got cells that look as though they've got stem cell properties. And now, let's look at the assays that may be used. And there are three assays that you should know. One is a repopulation assay to test stem cellness. And this is a transplant assay where you're transplanting test cells, test stem cells, into the adult. And you have removed from the adult, endogenous cells that might be competing with those stem cells. That would include the stem cells. We'll go through a slide in a moment. I'm going to list them here. Another one is an in-vitro induction assay, where you are going to take isolated cells and you're going to treat them with various inducers, various signaling molecules, and you're going to test and see what fates those cells can acquire. And a third assay is called an embryo incorporation assay, where you are going to take cells that may be stem cells, and you're going to test them in a chimeric embryo. Let's go through your next slides to discuss each of these points. Bone marrow transplants resulted in a Nobel Prize in 1990 for E. Donnall Thomas and Joseph Murray. It's a technique that saved millions of lives, and here how it works in a mouse. You take the mouse and you irradiate the mouse to destroy the bone marrow and the stem cells associated with the bone marrow. If it's a person, to destroy the diseased burned bone marrow. And then you replace that bone marrow with normal bone marrow to either try to make the person better, or in this case, to test something about stem cells. The irradiated mouse or person would die, but the normal bone marrow will cause the mouse to live. And if you've put stem cells into that mouse, you can start getting them out of that mouse whose life you have saved, and isolate more stem cells. Those kinds of assays led to the definition of the hematopoietic stem cell. Here it is. It's a pluripotent stem cell that gives rise to all of these different kinds of cells, the immune cells, and all of the blood cells. It's a very, very powerful stem cell. And the ability to actually make this diagram and say that there was a single cell that gave rise to all these different lineages was because of a titration assay where you could take these putative hematopoietic stem cells that were difficult to isolate-- and still haven't been isolated in their purity-- but you can titrate them down. And you can introduce what you think is one stem cell, 10 stem cells, 100 stem cells, and so on, into an irradiated mouse, and ask how many stem cells does it take to repopulate the entire blood system and immune system. And it turns out, you have to mix these cells with carrier cells, otherwise it doesn't work. But it turns out that one cell can repopulate the entire hematopoietic system, which is really extraordinary, and led to the diagram that I showed you in the slide before. OK. Here's another assay. This is an in-vitro induction assay. And the idea here is that you start off with something, which you think might be stem cells, by various criteria. And then to test what these cells can do, you put them into plastic tissue culture dishes, and you add some nutrients and so on to allow the cells to divide. And then you add some inducers. And you remember a couple of lectures back, inducers are just ligands for various signaling systems. You might add fibroblast growth factor to this one, and retinoic acid to that one, and then you ask what happens to the cells. They will go on, in general, to differentiate into different cell types. And depending on what they differentiate into, you can say something about the potency of these putative stem cells. You can't test if they're really stem cells, but you can say something about their potency. You can do a similar experiment, but in a whole mouse. The mouse is made from, the embryo is made from a part of the very early embryo called the inner cell mass. And you can inject labeled, putative stem cells into an early mouse embryo, into this inner cell mass part of the embryo, put it into a mother, a recipient mother, and then ask what comes out, what kind of embryo comes out of that process. And if you see that the baby that comes out of this chimeric embryo has got a green liver and green ears and green whiskers, you'll know that these cells that you put into the chimeric embryo, that you made the chimeric embryo with, had the capacity to give liver, ears, and whiskers. OK? So this is a powerful assay to, again, look at the potency of cells, not, in this case, the stem cellness of cells. One of the things about stem cells is that you only want them to work when you want them to work. If you cut yourself, in the normal process of keeping your liver the right size, in the normal process of keeping your heart muscle correct, you want your stem cells to be working and keeping everything homeostatic. You don't want them to be dividing out of control, because then you'll get cancer. And so something has to control what stem cells do and when they do it. And this is a question of regulation and the notion of a stem cell niche. Stem cells are kept quiescent, usually in G0 that we talked about in the cell cycle lecture. And they're kept quiescent by signals from the surrounding cells. So their cell-cell interaction, and by signals from the surrounding cells. And those are given a special name by stem cell biologists. They're not that special, but they're given a special name. They're called niche cells, or the niche. OK? They're just surrounding cells that are secreting signals. On some kind of activation-- you cut yourself, your organ normally needs repair-- things change. So on activation, by some kind of environmental input-- local or less local-- niche cells induce the stem cells to divide. And they do this-- and they induce the stem cells to divide into a stem cell plus a progenitor, which then goes on to do all the things that I diagrammed that progenitors do, on the first board. And the niche cells do this because they have changed their signaling. This is yet another use, or a very related use, of the notion of cell-cell signaling in controlling life. Here's a diagram. Here are niche cells. You don't have this. Just look on the screen. Surrounding cells, maintaining stem cells quiet. When there's an environmental input, the niche cells change. They activate the stem cells to divide and form more stem cells and progenitor cells. One really fantastic example of the niche, and the interaction between stem cells and the niche, is in the hair. This is from my colleague Elaine Fuchs at Rockefeller, who over many years has figured out that in the hair follicle-- this is the hair sticking out of the skin-- in the hair, there's a small group of cells on one side of the hair shaft called the bulge cells, and these bulge cells are the stem cells. Her investigators isolated these bulge cells and did the following experiment to show that they were hair stem cells. There's a kind of a mouse called a nude mouse that has a very bad immune system. So it's useful for immune system experiments. But it also has no hair. And you can do a transplant into this mouse of these bulge cells. And you get little tufts of hair growing where the rest of the mouse is nude. And you've done a careful experiment where you've labeled the cells you have transplanted in so that you can show that they actually came from the transplanted tissue and not from the mouse, the nude mouse tissue. And you can do that because you labeled them with GFP and they're green. So this the green hair. And it shows you that these bulge cells are stem cells. During the life of a hair-- your hair has a life, we discussed four years on your head-- during the life of a hair, the bulge cells and the niche cells are in different places. And depending on whether they're touching or far apart from one another, there's induction of hair growth or not. So at a particular stage of the hair cycle-- this is called the hair cycle, look on the bottom of the screen here-- the bulge cells and a group of cells called the dermal papilla that lies right at the bottom of the hair shaft are touching one another. And at that point, a particular signaling pathway called the Wnt pathway is activated. And these dermal papilla cells tell the bulge cells to start dividing and start making a new hair shaft. After that's happened, the bulge cells move away from the dermal papilla cells. Here they are during growth, the growth period. You see the bulge and the dermal papilla cells are no longer touching. At this point, the stem cells become quiescent. And there are enough progenitor cells in the hair shaft to give you formation of the hair. And then the stem cells remain quiescent until the next hair cycle starts and they get in contact with the dermal papilla again and start making new hair. This is a really beautiful story that's shown us quite clearly how the niche cells can control these stem cells. All right. So let's spend the last minutes talking about therapeutics. Here's the dream. You know, the dream is that you have a stem cell population for every organ in the body, including things like limbs, such that if your limb gets severed or your heart becomes really diseased or your spinal cord is injured and you can't walk anymore, that you can just inject into a patient the correct stem cells and everything gets repaired. That's the dream. And that's really the holy grail of what thousands and thousands of investigators are going after. And it's given precedence by bone marrow transplants, which really are very successful. Turns out that it's kind of tough. It's tough because these adult stem cells are really rare. The hematopoietic stem cells are special because they're kind of liquid. They're single cells. They're not attached to anything. And they are relatively easy to identify. But other stem cells are very difficult. So the idea is that you inject stem cells to repair a damaged organ. You need stem cells of the correct potency, otherwise you're not going to repair the specific organ. But adult stem cells are very rare. And so the quest has been to find some kind of substitute for adult stem cells. And those substitutes come in two flavors. One are embryonic stem cells, abbreviated, ESCs. Embryonic stem cells are cells that grow from the inner cell mass of an early mammalian embryo. And you can grow from them groups of cells that will keep growing in the laboratory for a long time and have variable potency. So the embryonic stem cells are derived from the inner cell mass of an early mammalian embryo-- human in the case of human. And you grow out so-called ESC lines, which means that these cells grow continuously in culture. And each embryonic stem cell line has a unique potency and can be used to do different things, in terms of theoretical repair. If you look on your next handout, the idea is that you take this inner cell mass, you plate the cells as single cells, they grow and grow and grow. And if you treat them with various inducers, you can get them to become heart or neuron progenitors, inject those into animals and make them better. There are some issues with embryonic stem cells. And the problems are twofold. One is ethical in that you have to harvest human embryos. You have to obtain and harvest human embryos to get these human stem cell lines. And currently, you're really not allowed to do much human embryo work to obtain human stem cells. But the second, even if you were, is that these cells are non-autologous. They do not match the person into which you're putting them. They do not match the immune system of the recipient. And so they'll be rejected. And it's the same as an organ transplant. You have lots of problems of rejection. So the latest thing that is very exciting and wonderful and potentially might be very useful, is the use of things called IPS cells, in which you convert adult cells into stem cells. And you do so, as I'll tell you, by adding some transcription factors to them. The advantage of these is that they are self cells. You could do them from yourself. The disadvantage is that they're really not proven. And there is still, I will just say, lots of problems. But there are a lot of people, including some of the top investigators in the world, working on these IPS cells. So look at your last handout. We will remove our interesting calendar reminders there. Adult differentiated cells, which are unipotent, can be treated with three or four transcription factors. And finding these transcription factors is the key. And once you express these transcription factors in these adult cells, like magic, they become stem cells. It really was like magic. You can test the potency of these stem cells in the same way we discussed. And Professor Yamanaka in Japan and Professor Jaenisch here have shown that these are really very powerful cells. And those are the promise of the future. And we'll stop there and meet on Friday.
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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. BOGDAN FEDELES: Hi, everyone. Welcome to 5.07 Bio Chemistry Online. I'm Dr. Bogdan Fedeles. I'm going to help you work through some more biochemistry problems today. I have here question 2 of Problem Set 8. Now, this is the question I put together to get you thinking about the electron transport chain. As you know, the electron transport chain is a fundamental redox process through which we convert the chemical energy of the covalent bonds into an electrochemical gradient. This electrochemical gradient is like a battery, and it can be used inside the cell to generate, for example, ATP, which is the energy currency of the cell, or it can be dissipated to generate heat. We're going to see both of these modes in action in this problem. Now in most organisms, the electron transport chain helps to transfer electrons all the way to molecular oxygen. However, in this problem, we're dealing with an organism that lives deep inside the ocean where the atmospheric oxygen is not available. And it turns out this organism transfers its electrons to sulfate. Sulfate is the final electron acceptor. Part A of this problem asks us to write the order of the electron carriers as they would function in an electron transport chain for this organism. Now, for a number of redox processes, the problem provides a table with the electrochemical reducing potentials, as you see here. Now, I've selected the ones that are mentioned in the problem, and I put them into a smaller table here. As you can see, we're dealing with cytochrome A, B, C, C1. This is the flavin mononucleotide. This is the sulfate, the fine electron acceptor, and ubiquinol. Now, on this column here we have the redox potential, which are the electrochemical reduction potentials denoted by epsilon, or e0 prime. Now, e0, as you know from physical chemistry or physics, denotes the electrochemical potential in standard conditions. However, in biochemistry, we use the e0 prime notation to denote that the pH is taken into account, and it's not what you would expect, like of hydrogen ion's concentration equals 1 molar, but rather it's a pH of 7. The hydrogen ion's concentration equals 10 to the minus 7. So therefore, these numbers are adjusted to correspond to pH 7. The electrochemical potentials we see in this table are reduction potentials, and they tell us how easy it is to reduce a particular species. Therefore, the higher the number, the easier it is to reduce that particular species and the more energy the reduction of that species will generate. Therefore, the electron transport chain will go from the species that hardest to be reduce towards the species that are easiest to be reduced. Therefore, the order of the electron carriers will be from the ones that have the lowest reductive potential to the ones that have the highest reductive potential. So now if we're going to sort all these electron carriers in order of their potential, we're going to get the following order as you see here. So the electrons are going to flow from the flavin into the coenzyme Q, and then the electrons are going to flow coenzyme Q to cytochrome B, and then Cytochrome C1, C, A, and sulfate. And as you can see, flavin has a negative reduction potential. It's like the hardest to be reduced. And the next one is ubiquinol. It's barely positive. And then the highest number is sulfate 0.48 volts. Now, let's take a closer look how the electrons are going to be transferred through this proposed electron transport chain. In the first reaction, here we have the flavin, I've written the flavin adenine dinucleotide, FADH2, the reduced version, is going to be converted to the oxidized FAD version of it. And in this redox reaction, we're going to use the coenzyme Q, the oxidized version and reduce it in the process. So the electrons get transferred from FADH2 to coenzyme Q. Now, in the next reaction, the reduced version of coenzyme Q is going to get oxidized back to coenzyme Q and in the process cytochrome B is going to go from its oxidized form to its reduced form. Now, this process continues with every single step, every single electron carrier up until we get to the sulfate where the reduced form of the cytochrome A will donate its electrons to the sulfate, and sulfate would get reduced to its reduced form. It's called sulfite. So if we were to draw how the electrons move through this chain, the electrons are going to start at FADH, and then they're going to be transferred to coenzyme Q in the reduced form. And then coenzyme Q is going to pass it to the cytochrome B. That's going to be in its reduced form. And then cytochrome B is going to pass it to cytochrome C1, and then cytochrome C, cytochrome A, and finally, they're going to end up in sulfite. Another thing to notice here is that except for the initial flavin and the final electron acceptor, sulfate, all the other intermediates get regenerated. So we go from the oxidized version to the reduced version and back to the oxidized version. So all these electron carriers are going to be sufficient only in catalytic amounts. So the only thing that gets consumed is the FADH2 and the sulfate. These are two reactants. And we get in this reaction FAD and sulfite. What we just said will help us segue into the Part B of the problem, which asks us to calculate how much energy do we get by converting one molecule of FADH2 and one molecule of sulfate into FAD and sulfite, respectively. Now as we pointed out here, only the FADH2 and sulfate are consumed in this reaction. All the other electron carriers are recycled and regenerated in the course of the electron transport chain. In order to calculate the energy, it's useful first to write the half reaction of the redox processes. Here are the two half reactions of this redox process. FADH2 gets oxidized through FAD and donates its two electrons. And the epsilon, or e0 prime is minus 0.22 volts. Now, this is the potential from the table, and that's a reduction potential. The equation as written is an oxidation, and therefore, the potential that we need to take into account is the minus of this one. Sulfate is then going to accept the two electrons and going to get reduced to the sulfite and water. And the electrochemical potential for this is 0.48 volts. So now when we add these two together, we get the overall process where FADH2 gets oxidized by sulfate to generate FAD and sulfite. And the electromotive force is just the mathematical sum of these two keeping in mind that this has to be taken as a negative sign. Because, again, as written, this is an oxidation and this the potential for the reduction reaction. So electromotive force is actually 0.7 volts. Now, we can easily convert from the electromotive force to a delta g0 prime value, and the relationship is written here, delta g0 prime. It's minus nF delta e0 prime and is the number of electrons in the process as we see here. Two, F is the Faraday's constant and delta e0 prime is going to be the electromotive force. And if we go through the number crunching, we get a delta g0 prime minus 135 kilojoules per mole. Notice because it's a negative number that means there's a spontaneous process as written. And as you know, the negative delta g will correspond to a positive electromotive force. Now, we're just one step away from calculating how much ATP we can produce with this energy. As you know, we generate ATP out of ADP and phosphate, and this is the reaction that's catalyzed by ATP synthase. And it takes about 30.5 kilojoules per mole to form ATP out of ADP and phosphate. Therefore, the 135 kilojoules per mole that we generated from 1 mole of FADH2, it's going to be enough for about 4 molecules of ATPs. This is in contrast, which was the normal processes that use oxygen as their final electron acceptor where out of one FADH2 molecule, will generate at most 2 molecules of ATP. So in some ways, sulfate is actually a better electron acceptor and can give us more energy. Part C of these problem deals with a culture of this microorganism in the lab. And we're adding to this culture dinitrophenol, a compound we're told has a pKa of about 5.2. So let's explore what happens to the electron transport chain of the organism when we add dinitrophenol. Here I put together a cartoon representation of the electron transport chain of our organism. So as you can see here, this is the extracellular environment. This is the outer membrane. This is the inner membrane where we have all these complexes I denoted here with these rectangles of the electron transport chain. And FADH2, for example, is going to donate its electrons. They're going to be passed along all the way to sulfate. And in the process, protons are going to get pumped into this intermembrane space. Now, these protons can be used in the ATP synthase as they travel back into the intercellular space. Their energy can be used to convert ADP and organophosphate to ATP as we just discussed in Part 2. Now, to this organism, we said we're going to add dinitrophenol. Here is the structure of dinitrophenol. And we're told the pKa of this proton, right here, the pKa is about 5.2. When this compound diffuses through the membrane, it's going to go through this intermembrane space, which has a very low pH and also in the intercellular space in the cytosol, which has a much higher pH. So because pKa 5.2, it's a relatively low, much lower than 7, pKa, in the intermembrane space where it's more acidic, it's going to be protonated. So we can write, for example, dinitrophenol OH in equilibrium with dinitrophenol O minus plus a proton. Now, because here we have a lot of protons, this equilibrium will be shifted to the left. That is the protonated form of dinitrophenol. However, here in the cytosol, the NPOH, it's going to be in the same equilibrium O minus plus H plus. But because the pH is fairly high, that is there are not a lot of protons, this equilibrium is going to be shifted to the right. This equilibrium is going to be shifted to the left. So now look what happens. So because this equilibrium has shifted to the left, it's going to keep soaking up a lot of these protons. Then the neutral dinitrophenol molecule is going to diffuse through the membrane as such and enter the intercellular space to cytosol where it's going to be deprotonated. The equilibrium is shifted to the right. So in effect, dinitrophenol is going to carry the protons from the intermembrane space inside the cell. Now it's going to do that in parallel with the protons that are going to be flowing through the ATP synthase to generate ATP. So in effect, we're discharging this battery where the concentration of protons is basically our electrochemical gradient. It's going to be discharging the battery without producing ATP. So as you know, if you short circuit a battery, the battery is going to heat up because you're discharging an electrochemical gradient. Similarly, dinitrophenol, by taking these protons from the intermembrane space and bringing them inside into the intercellular space, it's going to be generating heat. Therefore, we can answer Part C by saying that the medium in which these cells are growing is going to heat up when we add dinitrophenol to it. The processes described in this problem are fairly universal. Now, in eukaryotes, like more evolved organisms, they would happen in the mitochondria. Now, if you look back at this diagram, if this was the double membrane of the mitochondria, this would be the inside of the cell that contains the mitochondria, this would be the intermembrane space, and this will be the inside of the mitochondria or the mitochondrial matrix. Similarly, by adding a compound like dinitrophenol, who can dissipate the electrochemical gradient in the mitochondria and cause the cell to heat up. In fact, this process is actually used by a number of organisms to generate heat instead of chemical energy, or ATP. For example, the brown fat cells in newborns in mammals have a special protein that allows to dissipate this electrochemical gradient in the mitochondria to generate heat. Another good example is the seeds of many plants. When they germinate, they actually generate a lot of heat that can be used to melt the ice or the snow around them. That's why some of the plants can start growing even before the snow has melt in the early spring. I hope that working through this problem will help you understand better the inner workings of an electron transport chain and how it can convert the chemical energy of chemical bonds into an electrochemical gradient, which can then be used to generate high energy compounds like ATP. Or it can be dissipated to generate heat.
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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 one more thing let's do with this operator. So we're getting accustomed to these operators, and these permutation operators can also act on operators themselves. So that's important. So consider the action on operators. So for example, an operator B that acts belonging to the linear operator some V, when acting on V tensor V, we define two operators, B1 and B2. And you define them in an obvious way, like B1 acting on Ui 1 tensor Uj 2. OK, B as an operator knows how to act on every vector on the vector space capital V. So when you say 1, you're meaning that this operator acts on the first Hilbert space. So this is equal to B times Ui 1 tensor Uj 2. So it just acts on the first state. How does it act? Via B, that is an operator, in the vector space V. Similarly, if you have B2 of Ui tensor Uj 2 you have Ui 1 tensor BUj 2 to OK. So these are operators that act either on the first state or in the second state. So the permutation operators can do things to these operators, as well. So we can ask a question, what is P21B-- should I start with one? Yes, one-- P21 dagger. Remember, when you ask how an operator acts on an operator, you always have the operator that you're acting with come from the left and from the right. That is the natural way in which an operator acts on an operator. You can think of this thing as your operator is being acted upon as having surrounded by a [INAUDIBLE] and a [INAUDIBLE].. And then when the states transform, one transforms with U, one transforms with U dagger. So always the action on an operator is with a U and a U dagger. So if you ask how does the permutation operator act on B, you don't ask generally what's the product of P times B. You ask this question. This is the question that may have a nice answer. Then we'll see that there's other ways of doing this. So we want to investigate this operator. So what I can do is let it act Ui, Uj. So what do we get? We get P21B1. Now, P21 dagger, we saw that it's Hermitian anyway, so it's just P21. And now it acts on Ui 1. I'll put the j here, and Uj 2. So I let up the P21 on that state, and that the moves the i's and the j's. Now, B1 acts on the first Hilbert space. So now we have P21 and we have BUj 1 and tensor Ui 2. Now, P21 is supposed to put the second state in position one and the first state in position two. So this is Ui 1 BUj 2. I could put this thing-- BUj 2. And then you see, oh, this term is here. So this is nothing else than B2 acting on the same state of the Ui Uj, which means-- I guess I could use this blackboard-- that P21 B1 P21 dagger is B2. So it has moved you. The operator used to act on the first particle. Two and one changes the first particle with the second. It moved it into the other one. Similarly, you could do this also. Would not be a surprise to you that P21 B2 P21 dagger is equal to B1. And you don't have to do the same argument again. You could multiply this equation by P21 from the left and P21 dagger from the right. These things become one and one, and the operators remain on the other side and gives you this. So this second equation comes directly from the first. You don't have to go through the arguments. So what is the use of this thing? You may have a Hamiltonian, and you want to understand what it means to have a symmetric Hamiltonian. And these operators allow you to do that. So for example, you may have an operator O 1,2. What is an operator O 1,2? It's an operator build on things that act on one or act on two. So if you want to imagine it, it could be an O that depends on the operator A acting on the first label, an operator B acting on the second label, an operator C on the first label, an operator D on the second label. Could be a very complicated product of those operators acting on all kinds of labels. Now suppose you act with P21 O 1,2 P21 dagger. Now, the great advantage of having a P and a P dagger acting on a string of operators is that it is the same as having a P and a P dagger acting on each one. Remember, if you have like P and P dagger, and it's a unitary operator on ABC, it's the same as PAP dagger, PABP dagger, PCP dagger. It's like acting on each one. So when you have this P21 P21 dagger acting on this, it's as if each one of those is surrounded by a P21 P21 dagger. So each label one will become a label two, and each label two will become a label one. And therefore, this operation is going to give you O2,1 for an arbitrary operator acting on these two labels. Now, it may happen that the operator is symmetric if O is symmetric. By that, we mean O 2,1 is equal to O 1,2. If that happens-- if that happens-- then from this equation you would have P21 O 1,2 P21 dagger is O1,2 is itself. And you could multiply by a P21 from the right, giving you P21 O 1,2 equal O 1,2 You're multiplying by a P21 from the right that cancels this P21 dagger. P21. And there you see that an operator is symmetric if it commutes with the permutation operator. So if always symmetric, this is true, and this is true, and then finally, P21 with O 1,2 commutator is 0. Oops. Too low. Let me see. It's a commutator. It's 0. So that's basically how you manipulate these operators on this Hilbert space.
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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 material from hundreds the MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR BAWENDI: Last time you talked about the first law of thermodynamics. And you talked about isothermal expansion, the Joule expansion. You saw a very important result. which is that for an ideal gas, the energy content is only dependent on the temperature, nothing else. Not the volume, not the pressure, it just cares about the temperature. So, if you have an isothermal process for an ideal gas, the energy doesn't change. q plus w is equal to zero for any isothermal process. And you also saw that du then could be written as Cv dT for an ideal gas always. This is not generally true. If you have a real gas and you write du is Cv dT, and your path is not a constant volume path, then you are making a mistake. But for an ideal gas, you can always write this. And this turns out to be very useful to remember. OK, now most processes that we deal with are not constant volume processes. So energy, which has this wonderful property here, du is Cv dt for constant volume process, which happens to be equal to d q, constant volume, because there's no change. there's no work if you've got a constant volume process. So du here is a very interesting quantity, because it's related to the heat that's going in or out of the system under constant volume process. But as I said, we're not operating usually in a constant volume environment. When I flail my arms around I generate work and heat. This is not a constant volume process. If I'm the system, what's constant when I do this? Anybody have an idea? What's the one function of state? I'm the system, the rest are the surrounding. What's the one function of state that's constant when I'm doing all my chemical reactions to move my arms around? Temperature? STUDENT: Pressure? PROFESSOR BAWENDI: Pressure, right. Pressure is constant. What is the pressure at? One atmosphere, one bar. So the most interesting processes are the processes where pressure is constant. When I had have a vial on bench top, and I do a chemical reaction in the vial, and it's open to the atmosphere, the pressure is constant at one atmosphere. When you've got your cells growing in your petri dish, the pressure is constant at one atmosphere, even if they're evolving gas, pressure is constant. So we'd really like to be able to find some sort of equation of state, or some sort of rather function of state that's going to relate the heat going in or out of the system with that function of state, because this isn't going to do it. du only relates to the heat under constant volume. And the heat is a really important thing to know. How much heat do you need to put into a system, or how much heat is going to come out of a system when something is happening in the system? All right, this is a really important quantity to know. Your boiling water or whatever, you want to know how much heat do you need to boil that amount of water under constant pressure? And this is where enthalpy comes in. You've all heard of enthalpy. H we're going to write it as the function of temperature and pressure. And the reason enthalpy was invented was exactly for that reason, because we need some way to figure out how to relate the heat coming in or out of a system under a constant pressure process. Because it's so important. And I should add and also under reversible work, where the external pressure is equal to the internal pressure. OK, so we're going to define enthalpy as u + pV, these are all functions of state here, So H is a function of state, and we're going to see now how this is, indeed, related to the heat flow in and out of the system. If you have a constant pressure, reversible work process. Let's take a system. Under constant pressure T1, V1, going to a second -- this is the system, so let me write the system here. And it's more dramatic if the system is a gas, p, T2, V2, And let's look at what happens to these functions of state, to H to u under this transformation. OK, so let's look at delta u. Delta u is q plus w. That's the first law. And this is a constant pressure path, so now I can write, this q is actually q under constant pressure. Little p means the path is a constant pressure path. And I'm doing reversible work. So that w is minus p, dV where p is the pressure inside the system, minus p delta V. Rearrange that, delta u is plus p delta V is equal to q p. All right, so this is the heat flowing in or out of the system, and these are all functions of state. This depends on the path. It tells you right here, the path is constant pressure. These don't depend on the path, right. V doesn't care how you get there. u doesn't care how you get there. In this case, p is a constant because the path is constant. So we can bring the p inside, delta u plus delta p V it's q p. Take this delta outside again, delta of u plus p V is equal to q p. And there you have it. There is the H right there. The u plus p V. Delta H is equal to q V. And this is the reason why enthalpy was invented, and why it's so important. Because we want to know this. So this for a finite change. If you want to have an infinitestimally small change, you end up writing dh is dq sub p. It's not always equal to the heat. It's only equal to the heat if your process is constant pressure reversible work. OK, so this is the kind of, this is the kind of concept that needs to be branded into your brain, so that if I come into your bedroom in the middle of the night and I whisper to you delta H, you know, you should wake up and say q p, right? Heat under constant pressure reversible work. This should become second nature. This is where the intuition comes from. This is why people right tables and tables of delta H's. Why you have delta H's from all these reactions, because this is basically the heat, and the heat is something we can measure, we can control. We can figure out how much heat is going in and out of something. This is what we're interested in. OK, so last time you looked at -- any questions on this first? Yes. STUDENT: [INAUDIBLE] from the T delta V to the delta p here? What was the reasoning behind that? PROFESSOR BAWENDI: p is constant here. It's constant pressure. OK, so now, last time you looked at the Joule expansion to teach you how to relate derivatives like du/dV. du/dV under constant temperature. du/dT under constant volume. You use the Joule expansion to find these quantities. Now these quantities were useful because you could relate them. The slope of changes, with respect to volume or temperature of the energy with respect to quantities that you understood, that you could measure. We're going to do the same thing here. So if we take as our natural variables for enthalpy to be temperature and pressure, and we have some sort of change in enthalpy, dH, and it's going to be related to changes in temperature and pressure through the derivatives dH through the slope of the enthalpy in the T direction, keeping pressure constant, dT plus the slope of enthalpy in the pressure direction, keeping the temperature constant, dp, and these are knobs that we can turn. We can change of temperature. We can change the pressure. These are physical knobs that are available to us as experimentalists. And so when we turn these knobs on our system, we want to know how the enthalpy is changing for that system. Because eventually they will tell us maybe things about how heat is changing further on. OK, but in order to relate turning these physical knob to this quantity here, which we don't have a very good feel for, we've got to have a feel for the slopes. If I keep the pressure constant. I change the temperature, what does that mean? What is dh/dT? If I keep the temperature constant, and just change the pressure, dH is going to change, but how is it going to change? What does this mean in terms of something I can physically understand? That's the program now for the next few minutes. What are these quantities? What is dH/dT as a function, keeping pressure constant, what is dH/dp, keeping temperature constant? All right, let's start with the first one, dH/dT, keeping the pressure constant. And we're going to look at a reversible process to help us out, but the result is going to be independent of whether or not we have a reversible process or irreversible process. Constant pressure, that means dp is equal to zero. So for reversible process, constant pressure, what do we know? This is already branded in your brain, right? Reversible process, constant pressure dH=dq. So we can write that down, dH=dq, constant pressure. That's by definition of enthalpy. That's why we created enthalpy. What else do we know? Well we can go look up here, looking at the differential, there are no approximations here. This is just an equality. I have a constant pressure process. This term here is equal to zero. That means that dH is also equal to dH/dT, constant pressure dT. All right, so now I've got more dH/dT under constant pressure. dH is equal to this, and it's also equal to this. All right, these two are equal to each other. Now, I know how to relate the heat flow to temperature change, through the heat capacity. dq constant pressure is that heat capacity, and I have to tell you the path for the heat capacity. So it's C sub p, the heat capacity under a constant pressure path, dT, all right? So these two are equal to each other. So, these two are equal to each other as well, which tells me that this derivative, dH/dT constant pressure is Cp. So now I have my first of my two slopes, in terms of something that's related to my system, the heat capacity of the system. Something I can measure and I can tabulate, and when I turn my dT knob here I know what's going to happen to the enthalpy. So this is the first, this is the one. So it's very similar to what we saw with the volume and the energy, where du/dV under constant temperature was equal to Cv in this case here, right. and you're going to find that there's a lot of these analogies between energy and enthalpy. You just change volume to pressure and basically you're looking at enthalpy under a constant -- anything that's done at a constant volume path with energy, there's the same thing happening under constant pressure path for enthalpy. So you can guess the answer usually that way. OK, so now we have the other one, dH/dp constant temperature. How do we relate this to something physical? Well, it's going to be an experiment, very much like the Joule experiment. The Joule experiment was a constant energy experiment, right. Here we're going to have to find a constant enthalpy experiment, and that is going to be the Joule-Thomson experiment. That's going to extract out a physical meaning to this derivative here. OK, the Joule-Thomson experiment. This is going to get us dH/dp constant temperature. What is this experiment? You take a throttle valve, which consists of some sort of porous plug between two cylinders that is insulated. There is insulation here. Insulation on the bottom. It's like a, think about a tube, a large tube, insulated tube, with a bunch of -- a frit inside here, which prevents flow of gas, which slows down the flow of gas from one side to the other. It's a blockage in this tube here. Then you put two pistons, one on that side here, and one on this side here, and the external pressure here, we're going to call that p1. The external pressure here, we're going to call that p2, and we're going to do it slowly enough that the pressure on this side of the cylinder is in equilibrium with the external pressure, and the pressure on this side of the cylinder is in equilibrium with this pressure. But not so slowly that these two are in equilibrium with each other. So this is restricting the flow, so there's some sweet spot here. When I'm pushing slowly enough here that the pressure here is equal to that one, but not so slowly that the air flow from here to here is fast, compared to how fast I'm pushing. You got the picture here? Any questions on that? All right, then as I push through, I'm going to start with all of my gas on this side, and at the end, I'm going to have all the gas on the other side. Let me first ask you this, is this a reversible or in irreversible process? Right, let me add one more piece of data here which I said in words, but which is actually important to write down before doing the problem. Is this a reverse -- any guesses? How many people vote for that this is a reversible process? I've got one vote back there, two votes, three votes, four votes. Anybody else? How many people think this is irreversible? It's about a tie, and everybody else doesn't now. All right, I'm going to give you ten seconds, fifteen seconds to make up your mind. You're not allowed to be on the fence here. You've got to decide, all right? This is, you can talk to your neighbors, you know, do a little bit of thinking. And I'm going to give you ten seconds to figure this out, what your vote is for that. All right, let's try again. How many people vote that this is reversible? That looks like a majority to me. Irreversible? Let's look at the show of hands? All right, so this is the majority here. Good thing physics doesn't work on the rule of the majority, otherwise we'd be in big trouble. Wow, let's walk through that. I'm sorry to say that this is the wrong answer. OK, why is that the wrong answer? Well, just think, you know, think about it. You're pushing through here. p1 is greater than p2. What does it mean for a process to be in equilibrium or reversible? It means at any point you can reverse the direction of time, and it will look fine, right. So now I'm pushing on this plug here, with p1 greater than p2. I'm pushing, I'm pushing, I'm pushing. p1 is greater than p2. Then I want to reverse direction of time. I want the arrow of time to go so that the gas goes from p2 to p1. p2 is less than p1. Is that going to work out? If p2, the pressure in p2, is less than the pressure in p1, is the gas going to want to go from p2 to p1 and the whole thing reverse back? You've got to put more pressure on one side than the other if you want to push that gas through the throttle, right? So this is where the time scale issue comes into play. I said let's do this slowly enough that this p1 is in equilibrium with this p1, but not so slowly that this pressure is equivalent to that pressure. If I do it really, really slowly, so that everything is reversible, well I won't be able to do it, because p1 and p2 are different. But suppose that I fix my pistons here, with p1 greater than p2, and I don't touch it, eventually p1 will be equal to p2. I'll come to some sort of equilibrium. So this is a system which is out of equilibrium. If I stop, if I move slowly, if I move more slowly, then these two will want equilibrium. So this is an irreversible process. The Joule-Thomson experiment is irreversible. OK, important -- if you are part of this group here, think about it and make sure you understand that. All right, so this is the experiment. Now what are we doing with that? The initial state, let's look at what we are doing. So initially, we're going to have our piston, there's the plug sitting here. Our piston on the right side here fully out, and the piston on my right side, your left side, fully inside. There's no gas on that side here. So there's p2 sitting here. There is p1 sitting here, and all of the gas is sitting on that side of the plug. Then after we're done with the experiment, we'll have transferred all of the gas from one side to the other. p1 here. p2 sitting here. And there's going to be some volume V2 and some volume V1, but are not necessarily the same. Especially since the pressures are different. we don't know yet about temperature so I don't know what to say about these volumes because I don't know what the temperatures' are going to do. OK, so let's go through this and see what we would do which is to calculate the heat and the work. This is well insulated. So, what is the -- let's do the first one here. What's q for this process here? Anybody? STUDENT Zero. PROFESSOR BAWENDI: Zero, right. This is an adiabatic process. It's well insulated. Heat is not going in or out, adiabatic. q is equal to zero. So all we need to find out is the work now. Let's divide it up into the two sides, the work going on on the left hand side, my left hand side or your left hand side, and the work going on on the right hand side. So let's first look at the left hand side. OK, so w, first of all, work is being done to the system on the left hand side here. I'm pressing on the gas. So I expect that to be a positive number. The pressure is constant, p. The V goes from V1 to zero. So we write down p1, V1. On the right hand side, the work, let's call this left hand side, let's call this right hand side. Here there's an expansion going on, so the system is doing work to the external world. This piston is being brought out, so we expect the work to be negative, negative. And we start out with zero volume. We end up with a volume of V2, and the external pressure is constant to p2. Minus p2, V2. Minus p delta V. So the total work is the work from the left hand side plus the work on the right hand side, which is p1 V1 minus p2 V2. Which I can rewrite as minus delta pV. Delta pV is p2 V2 minus p1 V1. It's the pressure volume multiplied together at the final state, minus pressure volume from the initial state, with the minus sign here because it's the negative of that. All right, what is delta u? delta u is q plus w. q is zero. delta u is just w. So this is also just delta u. delta u is minus delta pV, for the process. OK, delta H is delta of u plus pV. By definition, that's how we define enthalpy up here. H is u plus pV. Delta H is delta of u plus pV, which is equal to delta u, plus delta pV. Now delta u is minus delta pV. So I have minus delta pV plus delta pV. This is equal to zero. So this irreversible process, this Joule-Thomson process, is a constant enthalpy process. Delta h for this process is equal to zero. Adiabatic q equal to zero. It's also delta H which is zero. The two didn't necessarily follow, because remember, delta H is dq so p is only true for a reversible constant pressure process. This is an irreversible process. So a priori, this was not necessarily true. It turns out after you do all the math, it turns out to be delta H equals zero. All right, so this is the experiment. How do we go from that experiment to the terms that we're trying to get, these slopes. Remember, we're trying to get delta H, we're trying to get dH/dT constant pressure and dH/dp constant temperature. OK, these are the two things were trying to get here. OK, so let's write down, what we know here. I'm missing something. Oh, we already know one thing. We already know this guy here. We already did that. OK, dH/dT constant pressure is Cp. That was easy one. So we already know that. So now we can write or differential dH as Cp dT plus dH/dp, constant temperature, dp. Now we want to find out what this guy is here. Now for this experiment, this is a constant enthalpy experiment for the Joule-Thomson experiment, this is equal to zero. So I can rearrange this to get this dH/dp in terms of things that I can either measure, like the heat capacity, or that I have control of, like dT and dp. So in this case, dH/dp constant temperature is minus Cp dT/dp, and this is under constant temperature, no, not constant temperature. Whatever the experi -- for that the experiment is. For that experiment, the constraint, so we need a constraint here, right, we need a constraint here. Right? We need a constraint here. The constraint isn't constant temperature because the temperature is going to be changing. It's not constant pressure, because we have a delta p going on. It's not constant volume either. The constraint is the constraint of the experiment, and the constraint of the experiment is that the enthalpy is constant. So the constraints we have here, is the constant enthalpy. It's the constant enthalpy process that we're looking at. This we can do experiments on. It's tabulated in books, and this we can measure in the experiment. Delta p here is the change in pressure from the left side to the right side, and we can put a thermometer, measure the temperature before the experiment, and measure the temperature after the experiment. So this is something we can measure. So now we have this derivative, in terms of physical quantities, things that we can measure. Things that we can relate to the properties of the substance that we're doing the experiment on. So this is basically delta T and delta p and the Joule-Thomson experiment. And so Joule and Thomson did these experiments, and they measured lots of gases, and they found that, in fact, this was something that they could measure. Sometimes it was positive, sometimes it was negative, and it was an interesting number. And so they defined them, after many experiments, the limit of this, delta T delta p and the limit of delta p goes to zero as the Joule-Thomson coefficient. So, basically dT/dp, constant enthalpy is equal to mu, by definition, Joule-Thomson, where mu Joule-Thomson is the Joule-Thomson coefficient, just like you saw last time eta sub j was the Joule coefficient for dT/dV under constant energy. So there's, again, total analogy here between what we're doing with enthalpy to what you did last time with energy, replace p with T and H with u. Flip those two and you get the same thing. OK, so now we have dH/dT is equal to Cp, and we can also write, then, dH/dp under constant temperature is equal to minus Cp mu Joule-Thomson. We have our two derivatives in terms of physical quantities, which is going to allow us, then, whenever we have a change to go back and, when we have a change where we adjust the temperature and the pressure, we'll be able to know what the enthalpy change is. OK, now let's take two cases. Let's first start talking about ideal gases. The last time you saw that for an ideal gas, the energy only cared about the temperature. It didn't care what the volume was doing. du/dV under constant temperature was equal to zero for an ideal gas. And by analogy, we expect the same thing to be true here, because enthalpy and energy have all this analogy going on here. So let's look at an ideal gas. So for an ideal gas, we saw that u was only a function of temperature. We also have the equation of state for an ideal gas, pV = nRT. We can write our definition of enthalpy, h is u plus pV. This only depends on the temperature. pV= nRT. So we have u only depends on temperature plus nRT. The only valuable now on this side is temperature. Pressure and volume have dropped out. So enthalpy, for an ideal gas, only cares about temperature. Pressure has dropped out of the picture completely here. So there is no p dependence here. H for an ideal gas is only a function of temperature. This is not true for a real gas, fortunately, but it's true for an ideal gas. So for an ideal gas then, dH/dp under constant temperature, that has to be equal to zero. Because temperature is constant H only cares about temperature. and that's equal to zero. And if that's equal to zero, that means that the Joule-Thomson coefficient for an ideal gas is also equal to zero. We're going to actually prove this later in the course. Right now, you're taking it for granted. Right now we told you Joule did all these experiments and he found out that for an ideal gas, that the limit in and ideal gas case was that the eta J was equal to zero. Therefore, from experiments, u is only a function of temperature for an ideal gas, and therefore from these experiments, we come out with delta H dH/dp is equal to zero. The Joule-Thomson coefficient is equal to zero. Later we are going to prove it exactly. but right now you're going to have to take it for granted. So, if the Joule-Thomson coefficient is equal to zero, just like we wrote, du = Cv dT for an ideal gas, we're going to have dH = Cp dT for an ideal gas as well. dH is Cp dT. This term goes away. dH = Cp dT. That's the only thing that's left behind. So if you know the heat capacity, you know the change in temperature, you know what enthalpy is doing for an ideal gas. This needs to be stressed that this is the ideal gas case. Now regular gases, real gases, fortunately as I said, don't obey this. This is important because we use this all the time for when, when we do technology. One example of Joule-Thomson coefficient being not equal to zero for a real gas that you've like experienced is if you take a bicycle pump, take a bicycle pump, and you're pumping up your tire, you're working pretty hard, so you're getting hot. But if you touch the valve going into your tire, which basically measures the temperature of the air going into your tire, that is getting hot, right. So if you've got to pump that tire really a lot, then you're going to you're going to really feel a lot of heat there. The compression of the, basically it's an adiabatic compression. You're taking the air inside of the pump and you're compressing it. You're doing it so fast that there's not enough time for heat to come out of the gas that's inside the pump towards the walls of the pump. So your time scale it just fast enough that this is basically an adiabatic compassion. Your compressing it really fast, all right, so you're changing the pressure. You're changing the pressure, and the temperature is going up. dT/dp is positive. dT/dp is positive. dT/dp is positive, well that's mu JT. dT/dp is mu JT. So for a real gas like air, this is a positive number. It's not zero. Air is not an ideal gas. It's one simple example of -- The fact that mu JT is not zero for real gases is how we are able to liquify things like hydrogen and helium, by compressing them and pushing them through a nozzle, and the expansion through the nozzle cools the gas. Right, in this case here it wouldn't happen if it was an ideal gas. Or in many kinds of gas refrigerators where you push a gas through a nozzle close to room temperature, what you find is that the gas coming out on the other side under lower pressure is cooler than the gas that went through on the other side. Real refrigerators actually work with liquids that go into gases so use the latent heat of the liquid, so it doesn't really work like the Joule-Thomson expansion. So this is real. This is real, unlike the Joule coefficient which is very small so that most gases have tiny Joule coefficients. So if you do a Joule experiment, you hardly measure a temperature change. With real gases, here you do actually measure it. You can feel it with your finger on your bicycle tire. OK, so we're going to see this using a Van der Waal's gas. Let's look at a Van der Waal's gas and see what happens in the Van der Waal's gas. Any questions, first? What we've been talking about, the Joule-Thomson experiment, constant enthalpy process? OK, so let's take our Van der Waal's gas. Remember the equation of state for Van der Waal's gas is not pV is equal to nRT, but p plus the attraction term. And then V minus the excluded volume term is equal to RT. Two parameters, this is the attraction between two atoms or molecules in the gas phase. This is the repulsion, not the repulsion. This is the fact that we occupy a finite volume in space, because they're little hard spheres in this molecule. OK, in a few weeks, you're going to find out that we can calculate dH/dp from this equation of state, and you're going to find out that dH/dp from that equation of state is proportional to b minus a over RT. This is going to be probably a homework at some point to do this. For now, let's take it for granted. Let's take it for granted that we know how to calculate this derivative from an equation of state like this. But now we're going to use that. OK. So since dH/dp under constant temperature is proportional to minus mu JT, then we have that dT/dp under constant enthalpy than it is related to the negative of this, a over RT minus b All right, this is how the temperature changes when you change, when you have pressure changing. So when you do an expansion or compression, my adiabatic compression of my bicycle pump is dT/dp. All right. Or when I do an expansion of hydrogen or helium at low temperature, through a Joule-Thomson experiment, when I want to liquify hydrogen or helium. I want to cool a gas with a Joule-Thomson experiment, what temperature do I have to be at? So this tells you that you have to be careful what temperature you're at, because depending on how high, how big this temperature here is, you could either be, have a negative dT/dp, if this first term is small enough, meaning if temperature is very high, then you end up with a negative term. If the temperature is very small, then one over the temperature is large, and the first term wins and you have a positive number. So there's some special temperature which is going to depend on the gas where the first term is going to be equal to the second term, where Joule-Thomson is zero, or it's going to behave like an ideal gas. So when that is the case, we're going to call that temperature the inversion temperature or T inv. We call that inversion because on one side you end up cooling if you compress. And on the other side of that temperature you end up heating if you compress. OK, so there's some temperature, t inversion minus b where the gas behaves like an ideal gas. The Joule-Thomson coefficient is zero and that inversion temperature, you can solve for it. Conversion temperature is equal to a over R times b. OK, so when you are at that temperature, everything looks like an ideal gas, as far as the enthalpy changes are concerned. Now if you're at the temperature which is higher than the inversion temperature, in that case here, a over RT is small compared to b, and this is going to turn out to be negative. So if you had a high temperature, this a small compared to b. If you're negative, which means that dT/dp at constant H is less than zero. So that means that if you compress something it's going to cool. The temperature rises when the pressure drops, right? Or in this case here, if I do my Joule experiment delta p is negative, p2 is less than p1, that means that delta T is positive, right? So in this experiment here, this side is going to heat up. So for materials where T inversion is low, lower than room temperature, then you would end up heating up in this expansion. You're basically expanding the gas from one side to the other and the expansion causes the temperature to rise. If T is less than T inversion, you have the opposite case, and dT/dp is greater than zero. So in this experiment here, delta p is less than zero. You need to have this whole thing greater than zero. So delta T is less than zero as well. So if you're below the inversion temperature and you do the Joule-Thomson experiment, you're going to end up with something that's colder on this side than that side. Ideal gas would be the same temperature. But now, so this is where the refrigeration comes in. So if you take a gas, and you're below the inversion temperature and you make it go through this irreversible process, the gas comes out colder from that side than that side. So the work that you're doing to expand, to go through this experiment, ends up cooling the gas. OK, for most gases, T inversion is much greater than 300 degrees Kelvin. Much greater than room temperature. For more most gases, if you do this experiment at room temperature, you end up cooling the gas, and you can cool it measurably, which is why also, your bicycle pump, you know, you push down, you compress, this is an expansion here. You're expanding from this side to that side. Bicycle pump you are compressing the gas, which is the opposite, you end up heating up the air in the bicycle pump in your compression. There are two exceptions to this rule that most gases have a T inversion which is greater than 300 degrees Kelvin, and that's hydrogen. The T inversion for hydrogen turns out to be 193 degrees Kelvin, and the T inversion for helium turns out to be 53 degrees Kelvin. And so when, there's a lot of liquid helium that's being used on campus. We use a liquid helium. And so in order to make a liquid helium, you can't take helium at room temperature and do this, because if you did, you would just heat it up, because the room temperature is above the inversion temperature, so Joule-Thomson would heat up the helium. So you need first to take the liquid helium and cool it below 53 degrees Kelvin before you can do the Joule-Thomson to cool it even further to make liquid helium. So you have to do it in stages. You take your room temperature liquid helium, and you cool it with liquid nitrogen to 77 degrees Kelvin, the new, you're not quite there yet, unfortunately right? Then you take hydrogen you cool it would liquid nitrogen to 77, then you can use your hydrogen gas. Do a Joule-Thomson hydrogen gas which you first cool with liquid nitrogen. Liquid nitrogen, 77, that's below 193, so you can do Joule-Thomson on hydrogen, cool the hydrogen to below 53, then use your cold hydrogen to cool the helium, and then you can do the Joule-Thomson on the helium to cool it further until it liquifies. So that's what you do when you make liquid helium. All right, any questions on this lecture or any of the concepts that we talked about? The last thing that we're going to do today then is to look at a relationship which is going to turn out to be very useful. It's a relationship for ideal gases which relates the heat capacities at constant pressure and constant volume. Cp = Cv + R. This is very useful because often you just have lots of tables of Cp's but sometimes you want to know what the energy change is going to be for the ideal gas, and you know that du is Cv dT not Cp dT. So, you need to know what Cv is, and if this is true always, then there's a very easy way to go from one to the other. We're going to do it two ways. Today we'll do the first way, and then next time we'll do the second way. The first way is just to turn the crank on the math, and the second way is to do a little bit more imaginative about the process. OK, let's just turn the crank on the maths. What do we know about the Cp and Cv? Well Cp, we already know how to relate it to dH/dT through the slope of the enthalpy, and we also related Cv to the slope of the energy with respect to temperature under constant volume. And we also know that H is u plus pV. Right, H is u plus pV. So we're going to take the derivatives of both sides of this equation here, by, with respect to temperature, keeping pressure constant. So we have dH/dT keeping pressure constant, is du/dT keeping pressure constant. Got to keep track of what's being constant, kept constant, plus dPV/dT, keeping pressure constant. dH/dT keeping pressure constant, that's just Cp, and then we have du/dT keeping pressure constant. The p is constant here, comes out of the equation, so we have p and then we have dV/dT well it's an ideal gas, an ideal gas for dV/dT for an ideal gas is equal to R over p because pV is equal to RT, right. So dV/dT is Rp. So dV/dT is times R over p, the p's are going to cancel out here. OK, so now it's very tempting at this state to say, oh there's the answer right here. du/dT is Cv. There I have it Cp is equal to Cv plus R, right? But even though it looks like that's the right way to do it, it's actually not right because it turns out to be right by sort of by accident. But here you've got pressure constant. du, this is du, not H here. du/dT is only equal to Cv when the volume is constant, not when the pressure is constant. So if you're going to turn the crank on the math correctly, you're going to have to change this p into a V somehow. Because this isn't Cv mathematically speaking. We don't know what it is yet. In order to change this from a p to a V, you have to use the chain rule. So let's use the chain rule. And then we'll be done. OK, so u is actually a function of temperature and volume, which in this case here could be a function of pressure and temperature. So if we want du/dT under constant pressure, you have to use the chain rule. There's the pressure sitting right here. It's going to be du/dT under constant volume, plus du/dV dV/dT under constant pressure. All right, chain rule. du/dT constant pressure is the direct derivative with respect to temperature here, which is sitting by itself under constant volume, keeping this constant but there is temperature sitting right here too. That's where that term comes from, du/dV dV/dT. Now, for an ideal gas, du/dV under constant temperature is equal to zero. It doesn't care what the volume is doing. It only cares what temperature is. If temperature is constant, there's no change in energy. For an ideal gas, this is zero. It's not zero for a real gas. Right, so this whole term disappears and for an ideal gas, it turns out that du/dT constant pressure is equal to du/dT constant volume, but this is equal to Cv for an ideal gas. It wouldn't be true for a real gas, and this is a common mistake that people make for real gases to equate this. This is only true for an ideal gas. Since it's true for an ideal gas, then we can go ahead and replace this with Cv, and then we have Cp with Cv plus R, which is what we were after. OK, next time we'll do the other way of getting to the same answer.
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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 where were we last time? So we had derived the Fermion Lagrangian which is at the top of the board, and really the only difference between this and what we talked about earlier is that we had to perform the multipole expansion. So we distinguish between collinear momenta that are scaling like a collinear momentum and momentum that are scaling like an ultrasoft, and only the collinear pieces were showing up in here, and same with the collinear gauge field. It was only showing up in here, whereas both were showing up in here. There's only a single type of derivative here, because ultrasoft and collinear momenta are the same size in the end component. But they're different in these other components, so we had these labels in this label operator that picked out the collinear momenta. So we could define collinear covariate derivatives, if you like, with this subscript n, whereas this D is like a full derivative that involves both types of gauge field. And then at the end of lecture, it was kind of rushed, but I was talking about the collinear gluon Lagrangian, and I said there is a set of replacement rules that we could make to effectively do the same thing that we did here. And now I've just written out for you with that Lagrangian is with those replacement rules. So curly D is basically just taking all the derivatives and taking only the leading order pieces. So we take the collinear pieces in this component and this component. We take the ultrasoft pieces in this component. And then if you wrote the original Lagrangian as a commutator of two covariant derivatives, you'd just replace it by the leading order pieces, and that's what the leading order action will be. Then, we had to think about gauge fixing. Since this is the collinear gluon Lagrangian, we should think about collinear gauge fixing. This here as a general covariant gauge fixing with parameter tau, and then there's a corresponding ghost term. And in this Lagrangian, the usual way it would look, this would be i partial, and then I said that, because we don't want the collinear gauge fixing term to break ultrasoft gauge invariance, we're going to turn that i partial into a covariant derivative under-- we're going to include this piece here at the n dot a ultrasoft to make it ultrasoft gauge invariant at lowest order. OK? So that's really the only-- other than just doing this most naive thing by just replacing derivatives by covariant derivatives, and you might think, well, I'll just keep this as a partial derivative. Since it's to doing gauge fixing, I don't need to make it covariant, but we do want to make it covariant under the ultrasofts, and that's why I write this curly D ultrasoft. OK. So that gives you the leading order Lagrangian. Once you put together this with what you want for the ultrasoft, then you would have the full leading order Lagrangian, and the ultrasoft part is actually very simple. So for the ultrasoft part, we just take a full QCD action for the ultrasoft field. So this is just q ultrasoft bar i D slash ultrasoft q ultrasoft. And likewise for the gluon piece, it's just QCD, and we would do the gauge fixing without thinking about any complications, like these ones, and these are just involving ultrasoft fields, so just QCD, just ultrasoft fields. OK? So the only place-- and you take everything here together-- the only place that the ultrasoft and collinear fields talk to each other in the Lagrangian is in this single component n dot partial, n dot a ultrasoft. And that comes about basically because of the power counting, that this is really the only place that these two things can interact. And we'll talk about the implications of that later on, but that's the leading order Lagrangian. So once you have this piece, and this piece is pretty straight forward with this additional complication of worrying about what gauge symmetry means, which we'll talk more about later. But we had to be careful not to break it, when we introduced this term, because we wanted to in some sense have a gauge fixing both for the collinear gluon and a separate gauge fixing for the ultrasoft gluon, and that's why we wanted to do this. And then this piece here is simple. It's just QCD, because it's, if you'd like, it's the lowest energy mode, and it doesn't know about any of the complications that we had for the collinear modes. OK? So any questions about this so far? OK. So everything that we did in deriving these actions is tree level. All the steps that we did were tree level. So you can ask, if I start to do loops, will there be some Wilson coefficient that shows up somewhere here? Will I generate some new operators that I don't see here? Those are reasonable questions, and that's what we're going to address next. So to go further, we'll use symmetries, and we're actually going to consider three different symmetries, gauge symmetry, which I've been promising you for a while, reparameterization invariance, and spin symmetry. Where I'll put a question mark by this last one, because we need to answer the question whether there even is a spin symmetry. These two here, this number one and number two, will turn out to be quite important. Number three is not so important. So reparameterization invariance here will be like reparameterization invariance in HQET, except now it's different. We've introduced the parameters and then n bar, and we'll have to see what kind of symmetries we have with respect to that choice of basis factors that we made. But it otherwise will be analogous to our discussion of HQET. So let's actually first dispense with the one that in some sense is the least important, this number three. So first, let's revisit our spinners a little bit. So if I put together the information that we talked about when we derived the equation at the top of the board there for the Lagrangian, we worked out at a tree level we have this formula that relates the fields. So from that formula, if we just project onto the spinner pieces, we can write down a formula that relates the spinners. So throwing away the gauge fields, we have in momentum space the u of p, the spinner, would be related to whatever spinner we have for this cn field which I call Un by that formula. And then if you take this formula, you also see that if I hit it with a projector, n slash n bar slash over 4, if I hit the u with a projector, it's going to kill this piece. Because the n slash n bar slash can be pushed through the p perp slash, then n bar squared is 0, so that kills that second piece. So we also have this formula. And then once you have this formula, you have the formulas that we wanted for that. OK? But this actually, this spinner here, this Un is not exactly the same as the spinner that we talked about earlier. So let me come back. Let me come to that in a minute. So first thing you might consider is whether, when I take this cn field, and I take the Lagrangian up at the top of the board, do I just get the collinear propagator that we talked about? And indeed, you do. It's kind of obvious for the momentum-dependent parts, and really you might only worry about the spin. And so if you consider this formula, and you consider the sum over spins of u u bar which is what's going to appear in the numerator of the propagator for the Fermion when you're driving the Fermion propagator, you get a sum over the physical spins. Then, from this formula, this is a projector on something you know how to do the spin sum for which is the full theory spinner. And that spin sum is p slash, so this is like p slash sandwiched between projectors. And you could work out that that's exactly the numerator that we had before after a little bit of Dirac algebra. So that part works as expected. If we take this Lagrangian, and we work out what the propagator is, we get exactly the propagator we got from expansion. So quantizing lc0 gives us that propagator. But the situation is not quite the same for the spinner. And in some sense, this is not-- this point is not absolutely crucial, but actually, there's a little simplification I want to do in order to discuss the spin symmetry, and in order to do that at this point, it's important to understand this point. So that I'm going through this in some sense, because then it'll be very easy to discuss what spin symmetry we have in the theory. So this guy is actually not equal to our expanded spinner which putting in some normalization I could write like this. So this is what we got by expanding, something like this which is very simple where this u is equal to 1, 0 or 0, 1. in the Dirac representation. But that's actually not what we get if we just take the formula up here and use this. So let's see what we do get. So if you use the formula up there, then you have the following. So here's the full theory spinner with a conventional normalization. So this thing and this thing here is the projector. OK? So you could work out what that product is, and it turns out that you can write it in the following way which is very closely related but not precisely the same as what we had before. So you can write it in what looks like the same form as we have over here, except this curly U, curly capital U, is a more complicated object. And it's just whatever I get by multiplying these two things out which turns out to be something I can write in that form. So it's some two-component spinner, but it's got momentum dependence, unlike our simple 1, 0 and 0,1. But everything we said about Un really depended only on the fact that it could be written in this form in terms of some two-component spinner-- the fact that n slash on it was 0, the fact that it had projection relation. So these formulas here, if you have a formula like these, these formulas here are true. OK? So actually, it would be true whatever spinner we have there. So why should I want this U twiddle spinner and rather than the U spinner? That actually has to do with this reparameterization invariant, so it'll become clear when we talk about that. But these extra terms in U relative to those for the other guy, the simple U, actually ensure the proper structure under reparameterizations. And basically, it'll become clear in a moment, but basically, if we wanted to get this U spinner, we should have a slightly different projector, which I'll call Pn prime. So we could have used a different projector which is this one, and then we could have come up with another projector which was the Pn bar projector which would satisfy that the sum is 1, if we wanted. And this projector here, when acting on the full theory field, would have given something that would have been proportional to this combination over here. OK? So you just have to believe me. I don't want to go through the algebra, or you can check it yourself. But this projector here is not invariant under that symmetry of reparameterization variance. When we talk about RPI, it'll be clear why we want a projector which is this projector and not the slightly different projector which has this extra n slash over 2. OK. But nevertheless, the important point that I wanted to emphasize is really that we have this way of decomposing the spinner, the true spinner, for our field cn in terms of a two-component object, U twiddle. OK? So we are able to do that. And if you want to talk about spin symmetry of the theory, it's easier if you use a two-component notation. So the reason that I wanted to go through this is really to have on the board this equation, which I can then use to motivate writing down to two-component version of cn. And once I have a two-component version of SCET, then it's very easy to see what the spin symmetry is. So let's write down a two-component version of our collinear quark Lagrangian. You can do the same thing, of course, in HQET right down the two-component version rather than a four-component version. If you have a four-component version, it has this projection relation. If you have the two-component version, the projection relation is built in. The reason to consider the four-component version is if you want a couple this object to four-component fields, like the ultrasoft field, then it's, of course, a nice thing to have a four-component version, but some things are easier in two components. So take cn, and write it as follows, where this phi n has two components. And I've set things up, so the dimensions of cn are equal to the dimensions of phi n. OK? So I can take that formula, plug it into our SCET Lagrangian, and then I can, using the Dirac representation for the gamma matrices, write out a Lagrangian for this phi n. So that requires doing a bit of algebra, which I will take you through. And then we get an equivalent Lagrangian but in terms of this field, and it looks as follows. OK. So it's almost independent of spin but not quite. There's this sigma 3 sitting there. If sigma 3 wasn't there, it'd be like HQET, where you have an SU2 symmetry. The fact that sigma 3 is there means you don't have an SU2 symmetry. And really, all you have is a U1 symmetry, and that U1 really corresponds just to helicity. So in four-component notation, that U1 would be the following. It's projecting a spin operator onto perpendicular indices anti-symmetric in both of them, and in the two components, that just becomes a sigma 3. So obviously, if we do an exponential rotation with respect to sigma 3, sigma 3 commutes with sigma 3, so does the identity. And so we could have a rotation of this guy by that, and that's the only spin symmetry that you have is this helicity. And so because of the coordinates we're using, this corresponds to the spin along the direction of motion which is the three direction, if you like, which sometimes we denote by just saying it's along the direction n which is then more independent of how we pick our axes. So this symmetry here is actually related to what you would call the chirality in QCD. So it's not really a new symmetry. So this is just related to the chiral rotation. So if you look at gamma 5 times c and gamma 5 in our representation would be 0, 1, 1 0. And then if I write it out in terms of this two-component thing, then that is just giving me 1 over root 2, and it's swapping up and down. And so that just means that phi n has gone to sigma 3 phi n. So multiplying by gamma 5 is actually the same as multiplying by sigma 3 in the two-component notation. OK? So this is not really new symmetry, and actually all the usual things that you would say about chiral rotations in QCD would apply here too. So chiral rotations, of course, are not exact chiral rotations are broken by masses. Chiral rotations are broken by non-perturbative effects. You can worry about anomalies. They are useful in perturbation theory for quantifying operators, and that remains true here, but if the collinear fields were non-perturbative, then you should worry about those other things as well. OK. So spin symmetry, there's not really anything new there to talk about, but along the way, we saw we could write SCET in two-component form which is kind of nice. So let's talk about something that's more important which is gauge symmetry. Is there any questions about the spin before we talk about gauge symmetry? AUDIENCE: So the extra term you got in determining the theta n bar, that is why you had the minus n slash [INAUDIBLE] and why you don't have the full SU2? PROFESSOR: Yeah. I looked at it once, but it's hard for me to remember the answer. I think not, but I think if you do the other version, then you just get something more complicated here but still breaks SU2. Yeah. AUDIENCE: Is it-- I'm just looking for [INAUDIBLE].. PROFESSOR: So they're two disconnected fact. The thing that I wanted to motivate that I could use this formula, and I wanted to be honest with you about where that came from, and that's why I told you this other fact. But I could have jumped right here and said, remember, and glossed over it, and that would have been fine for this discussion. So they're in some sense two disconnected things, at least in my mind. AUDIENCE: Do you have some intuition for what theta n bar is? PROFESSOR: For theta n bar? AUDIENCE: [INAUDIBLE] PROFESSOR: Oh. Yeah. It's really just saying-- my intuition for it is really just that, when you're producing energetic particles, it's what we did at the beginning. When you're producing energetic particles, these are the spin components that you have at lowest order, and that leads to, if you like-- the fact that you have this form, and you have this projection relation is kind of non-trivial, even though you don't have a spin symmetry. There is something to it. It's not like a SU2 symmetry of the spin, but it does, for example, when you go and look at form factors, it does lead to non-trivial relations for those form factors. So I don't call it a symmetry. I don't think of it as a symmetry, because it's not a formal group theory statement of Lagrangian, but I think it's basically-- well, OK. Let me not speculate any further. It caused some confusion in the literature actually as well. All right. So what about gauge symmetry? So we're doing a non-abelian gauge theory. A gauge transformation is something like this. And in order to just talk about gauge symmetry, what we need to do is we need to talk about the gauge symmetries that are related to the two type of gluons that we have. We have a collinear gluon. We have an ultrasoft gluon. You know that the field acting like the connection for that gauge symmetry. So there should be, in some sense, a gauge symmetry associated to the ultrasoft gluon and a separate one associated with collinear gluon, if the meaning of these things as gauge fields is going to have any sense to it. So rather than have just this set of gauge transformations that we can make in QCD, we should have some more complicated structure of gauge symmetry in the effective theory, because we have a more complicated structure of gauge fields. So in order to talk about U's, we're going to want to divide them into camps. And the useful thing to think about is that you want to have U's that, first of all, leave you inside the effective theory. And really what this means is that you want talking about gauge symmetry to commute with your discussion of talking about power counting. So by way of example of this, say you had a gauge transformation that was large, that made a formula like that one, that i partial on it was scaling like q. OK? So this would be a very large momentum associated to that spacetime dependence. Then, if you talked about the gauge transformation which would be U times c, even if this guy has collinear momentum scaling, the new guy would not because of this fact. OK? So if we want to talk about gauge symmetries, we want to talk about symmetries that take us to the fields that we've already decided to have a certain power counting. And we don't want the gauge symmetry to mess that up, so we're going to demand that that's true. So given that, we can think about dividing up the gauge symmetry into pieces, into pieces that have different scaling in terms of how big the transformation are, how the coordinates are behaving. And it's convenient for this discussion to divide them into three parts. A global transformation, where this guy is not spacetime dependent, a collinear transformation, which I'll call Uc, where i partial on Uc scales like a collinear momentum, and then an ultrasoft U Us, where the i partial scales like an ultrasoft momentum. And basically, what we want to do is we want to connect this gauge transformation to the gauge field for the collinears and this transformation to the gauge field for the softs. OK? So one complication that we have to deal with is this momentum labels that we were talking about. Because the way that we distinguish between these types of momenta and these types of momenta was dividing things into these labels and residuals. Gauge transformations are something that's very nice in position space but not so nice in momentum space, and that's because gauge transformations are local transformations. Local means local in x not in P. So what is simple multiplication in position space becomes convolutions in momentum space. So if you go over to this hybrid notation that we developed, then you would have the following. At this field, if I do a transformation, it goes into something like that. Where the way that you should think of what this sum over q is in the following way. But if you have a transformation in position space that looks like that, then in a momentum space, that simple product is convolution. OK? So this is one dimension, and you'd have that correspondence. And so this complication that we have here just correspond to the complication that you'd have for any gauge symmetry that you want to write in momentum space. But we can be slick about this and introduce a notation and then dispense with it, because it's really not a technical complication. It's just a notational complication. So let's do that in the following way. Let's let Uc which is a matrix with two indices be defined by this Uc of bl minus ql. So I'll say define a matrix. Then we can use a matrix notation for the convolutions, where we sum over repeated indices, and that takes care of this notational complication. So what is meant by this is that the Pl, ql entry of this matrix is a number, and that number is given by this thing that we call Uc Pl minus ql. OK. So then the formula up here could just be written in terms of that matrix as a kind of matrix multiplication. OK. So what are the transformations once we adopt that? Let me write them down and then talk about them. So cn of x goes to Uc of x cn of x, where you should understand this thing is a matrix that has two indices, like this, also dependent on spacetime. And I just dot over the repeated indices in this thing which is like a vector. So this is like a matrix, and this like a vector. That makes it look like it's the usual notation which is the point of introducing this notation. So for the collinear gauge field, it's going to be basically that it, as we said over here, that it becomes the gauge field of that transformation, and so that's what this would be. That would be the standard gauge transformation, and then there's one little complication here, which is I wrote curly D ultrasoft, the ultrasoft derivative, the thing that has the n dot a ultrasoft field in it. So the reason that I wanted to do that is because I'm thinking here of the ultrasoft field. When I'm thinking about collinear fields, I'm thinking of the ultrasoft field as a background field. And if I was to do a background field gauge transformation, then I would have a covariant derivative here under the background. So this n dot a ultrasoft is like a background field to the collinear gauge field. So this, if you want to make an analogy of what this is like, it's like the quantum gauge transformation of a field that has a background. If you look up how background field gauge works and background field transformations, you'd have the same thing. You'd have something covariant under the background, and that's what we're going to do here as well. Yep? AUDIENCE: Kind of a silly question. So UC of x is a matrix. So what exactly UC of p [INAUDIBLE]?? PROFESSOR: Well, no. This is just a number. So just think of this as like this is some vector, and this is some other vector. The difference of these are some label on this thing. Right? And for each value of this, I get a different number, and then for each value it's based on, I get a the different number, but this thing is just a number. And I'm saying think of it like a matrix in the sense of, because we are thinking of these as discrete-- AUDIENCE: But isn't UC [INAUDIBLE]?? PROFESSOR: Oh, yeah. Yeah. That's another matrix. It's an orthogonal matrix. It's also a matrix in that space. Yeah. Yeah. Yeah. So here, I'm defining a matrix in a new space which is this momentum space, and that's just really my way of making transformations in momentum space look like as simple as they do in position space by just having a matrix notation for the convolutions. AUDIENCE: So that number is essentially a matrix. PROFESSOR: It's a matrix, yeah, in that space. Good point. What about the ultrasoft fields? Well, ultrasoft fields shouldn't transform under collinear gauge transformations because of exactly the same logic that I gave you a minute ago about why I shouldn't be able to inject a hard momentum like this one into the effective theory. If I want my ultrasoft fields to remain ultrasoft, then I can't let them transform, because that would spoil their power counting. So q ultrasoft under this Uc transformation just goes to q ultrasoft, and A ultrasoft has to go to A ultrasoft. So these things should not be touched by that transformation. AUDIENCE: So why cannot make the ultrasoft and the collinear talk to the-- PROFESSOR: Well, we don't want that. You could think of trying to develop effective theory, where they would mix. Right? But whenever you have a Lagrangian that mixes two fields, the first thing you try to do is to force them not to mix. So it's better, say I set the theory up so that they mixed. The first thing I would try to do to that Lagrangian is make field redefinitions to go to some form of the Lagrangian that wouldn't mix. OK? So I prefer, if you like, to develop the one that doesn't mix right from the start, and you could view it that way. So you could always think of I'm going to tell you like two sets of transformations. You could always think of formulating linear combinations of those transformations. Say these are the fundamental ones, not the ones you wrote down but these other ones. Right? And then by some manipulations, one could go to the ones I'm talking about. All right. So now, the other set of transformations that we want are these ultrasoft ones, and for that type of transformation, we don't have the same type of problem with the collinear. We can't say the collinear fields don't transform, because that type of transformation doesn't spoil the collinear field being collinear. We can always add ultrasoft momentum to it. So these fields cn and An will transform like background fields-- sorry, like quantum fields in a background. So from the point of view of the cn fields which are the shorter distance fields, you can really think of like what we're doing here is exactly like what you would do if you had quantum and background gauge transformations. And the physics of thinking of the background as a longer wavelength mode is exactly the right way of thinking about how these fields which are shorter wavelength think about the ultrasoft fields. So what that means is that they transform like matter fields in the appropriate representation. OK? So with that in hand, we can write down how things transform under the ultrasofts. A Fermion just transforms like that on the left, and this, unlike our notation over there, this is one number for all entries in the vector. This guy doesn't have any indices. OK? So this is like transforming all the terms in the vector by the same overall color matrix, overall number in the momentum matrix base. So this guy should transform like an adjoint, and you can read the adjoint as a U on each side. So there's two ways of writing the transformation here. You could write it as like this. If you wrote down an adjoint transformation, then this guy would have two indices A and B, and you'd have a transformation like that, where this is in the adjoint. But if you want to write it in terms of the U ultrasoft that's e to the ita, which is in the fundamental, then you write it like this, and those are equivalent. So why are they equivalent? So they're equivalent because-- well, OK. Maybe I'll tell you later. Yeah? AUDIENCE: I'm always confused about why n dot is a background field to A ultrasoft, because it actually has the same size of momentum, so its wavelength is the same size. PROFESSOR: The wavelength is not the same size. The component's power counting is the same size, but the n dot An field carries momentum which have P squared which is much larger. AUDIENCE: Right. P squared [INAUDIBLE] wavelength, right? PROFESSOR: So I'm talking about the wavelength which is related to that P squared. So you think about the n dot An field carries a large P minus momentum still. Right? Even though it has a small-- it carries in order a 1 P minus momentum. So if it's confusing, think about just the P minus momentum. The n dot An field, right, has P minus, which is the what are lambda to the 0. So just let's think about wavelength with respect to P minus. AUDIENCE: That's definitely true, yeah. PROFESSOR: Right. Whereas the n dot A ultrasoft field carries P minus, which is what are lambda squared, so just thinking about those two. AUDIENCE: You were saying conflating [INAUDIBLE] PROFESSOR: Yeah. Yeah. All right. We have room here to squeeze the final transformation. So the final transformations are just what you'd expect, that the ultrasoft fields transform as regular gauge fields. So let me just squeeze that in. All right. So there's the regular transformations that you'd have for a gauge field. Yeah? AUDIENCE: So I just want to clarify. These two analogies for background field transformations, they're not disconnected. Right? PROFESSOR: They're not disconnected in the sense-- that's right. Yeah. So if you take the background field notation, and you just really think about that, then this is how you would transform the quantum field into the background gauge transformation. And this is how you would transform the quantum field under the quantum gauge transformation. Yeah. That's right. Yeah. AUDIENCE: So even if you made these to separate the gauge transformation for just [INAUDIBLE] they still have the same quantum numbers? PROFESSOR: Yeah. That's right. AUDIENCE: Are they going to meet at some level? PROFESSOR: We're going to make sure they don't. AUDIENCE: OK. PROFESSOR: Yeah. In some sense, we really want-- the picture is this one that we had earlier, where we really want the collinear fields to live up here, so we had these hyperbolas. Right? We want this guy to live up here and this guy to live down there in momentum space. So this was P minus. This is P plus. So we want them to live in different places. They have the same quantum numbers, except we want them in momentum space to live in different places and to describe different degrees of freedom. Right? So everything that we did here is exactly related to this picture, that we want to talk about this guy as having larger momentum, talk about gauge transformations that shuffle things around with components that have large momentum versus small momentum. All right. So we're going to hypothesize that these gauge transformations are the fundamental symmetry of the effective theory. And I tried to motivate them physically by this analogy with background field gauge and thinking about a longer wavelength mode with respect to a shorter wavelength mode which is in this picture exactly the fact that this hyperbola is higher than this one. Maybe the hyperbola should stay on the board. OK? So they multipole, where the multipole expansion that we did between these fields and the statement that this is a longer wavelength mode is exactly the fact that there's a separation between these guys. So we'll take these to be the fundamental transformations that do not get corrected by power corrections. And I'm not claiming that that's necessarily a unique way of getting to the effective theory. You could think of there might be an example in the literature of where people were thinking of other gauge transformations. These are the ones that people talk about now, but in the early days, I think, there is an example in the literature, where people are thinking of other transformations, and you could prove that they could be connected to these ones. But we'll just take the ones that are the most useful from the start and not worry too much about what we could have done. All right. So there's no mixing in the power counting, and that's something that we like. Set up that way. If there was a mixing, if we connected things that were different sizes in lambda, then you could say, well, that the power counting, that the gauge transformation induces power suppressed terms. And you could end up with some situation where you're trying to connect things at different orders, and we don't have that here. And that's because, if we want to think about this thing as a symmetry, it should really be a symmetry of the leading order action that shouldn't require some other sub-leading terms necessarily to compensate for it, at least for a gauge symmetry like this. We'd like to think that what the meaning of gauge symmetry is and the redundancy encoded in gauge symmetry is all something that you're talking about at the order of the leading order action. And you're not talking about redundancies in fields in sub-leading order, et cetera. OK. So let's do an example, now that we have this, of how it comes in. And I want to come back to our example of our heavy-to-light current and talk about gauge symmetry there. If you just took the tree level result that we had in that case, then we had something that just involved these two guys, and this hv field was ultrasoft. This guy was collinear. So if you did a Uc transformation, then under our rules, what you'd find is that the cn transforms, and this guy doesn't, the h doesn't. So that would not be gauge invariant. But actually, we saw that this also wasn't the lowest order current. There was this additional thing, the Wilson line showed up, and that Wilson line is going to fix this issue. It's going to make it gauge invariant. So we have to ask, how do Wilson lines transform under gauge symmetries? And that Wilson line in position space, I called it something like this. I can't remember if I used a twiddle or a bar. We'll use a bar here for the position space Wilson line. So in general, in just a regular gauge theory, the Wilson line between two points transforms at the endpoints. So you get U of x. Then, you get a U dagger of y. So if you like, what we would say for this Wilson line is that we have a U of x on one side and a U dagger of minus infinity on the other side. Minus infinity is really long wavelength physics, and so we have to worry a little bit about the overlap between our different types of gauge transformations. I divided them into three camps-- collinear, ultrasoft, and the global, so let me do the following. To avoid some double counting with what I called U global, let's just simply take, since U global acts the same everywhere, let's just simply take Uc dagger at minus infinity to be 1. So this is at some particular spacetime point. We take it to be 1, and this U global will transform at that point. We then know there's no overlap between Uc and U global. OK? Then, that's enough to ensure that. So with that, our guy under a collinear gauge transformation just transforms on one side. If you like, what's happening is that the coordinate here corresponds to this very large momentum, and we're stretching ourselves out to long distance physics which is corresponding to the smaller momentum. So when I think about transformations that should be collinear, I want them to be associated to the x. And I want to associate other types of transformations to what's going on at infinity which is really a spacetime of its own, which is this ultrasoft spacetime that's sitting at the collinear infinity. So that's why I want, actually, it to look like this. AUDIENCE: [INAUDIBLE] just with the collinear fields? PROFESSOR: Yeah. AUDIENCE: OK. PROFESSOR: Yeah. So this guy here is a path-ordered exponential, and it just involved the n bar guy. Yeah. So that's what it was in position space, and in momentum space, they called it W, and it was a sum over a bunch of things. I'm not going to-- we had this formula. Let me not write it all out again. As a short form for that formula, we can write the following formula which is given our notation that we've developed a useful way of thinking about it. So if we formally expand out this exponential, and we just keep the order of that expansion, where the P bar acts on all the fields to the right, that will correctly reproduce this denominator, the sum over permutations is something that we still have to do. It was 1 over n factorial, k factorial in here. M factorial with this notation. OK? But this is like a little shorthand for that messy formula. AUDIENCE: [INAUDIBLE] PROFESSOR: Yeah? AUDIENCE: I think it's related to the gauge transformation equation before. PROFESSOR: Yeah. AUDIENCE: How do you get the i epsilon prescription? [INAUDIBLE] PROFESSOR: So that actually is related to the fact that, when you look at the-- for this Wilson line with the i epsilon, it's not related to the gauge transformation at infinity, but it's related to the fact that the path goes that way. So that when you do the Fourier integral, the fact that it goes from minus infinity to x will give the i0. OK? But for these Wilson lines here, if you think about what's happening at the place where the i0 would actually matter, that's exactly the 0 bin. So it's going to be a little more-- Yeah. So you don't really care. AUDIENCE: OK. PROFESSOR: Yeah. I can only say that-- we'll explain what I just said to Ilia more when we come to it later on. OK. So this field here has-- I said, it's a momentum space, but then I wrote x. So really what I mean by momentum space is momentum is in the label momentum space. There's some guy, and the labels are these labels on the fields here. Right? So there's still an x, and that x corresponds to the residual coordinate that we were using to do the multipole expansion. So in some sense, this isn't simply the Fourier transform of that, because of that. It's still knows about the separation between the large momentum and the soft momentum from that multipole expansion. So this dependence on x allows us to think about the multipole expansion on these Wilson lines. It encodes all the residual momenta, and the fields in the Wilson line carry a residual momenta. So we're just putting our notation to use here. And if you really strictly want to connect to that formula up there, then you would take the residual momenta to be 0. And then the Fourier transform with respect to just the large momentum Pl minus gives a line which is just depending on that one single coordinate. So if we just had a coordinate which is conjugate to the large momenta, which is the thing that's showing up here are the large momenta, then that would be the analog of what we were talking about up here. Where in general, this thing also has a dependence on the residual x, which we'll make use of in a minute. OK. So let's think about, so I told you how this Wilson line transforms under Uc see. How does it transform under U ultrasoft? So under Uc, again using this matrix-type notation, it's just Uc. So this is a matrix multiplication in momentum space. What about under an ultrasoft transformation? Well, under an ultrasoft transformation, this should be a derived-- in some sense, under a collinear transformation, it was a derived quantity, because we knew of the gauge field transform. The thing that gave us the transformation of the Wilson line is just the transformation of the gauge field. OK? So that's what led to, actually, this form, once we put in this fact. So under the ultrasoft transformation, we just take the transformation that we have for the collinear gauge field, and then we derive how the Wilson line should transform under the ultrasoft gauge symmetry. OK? So how does this guy transform? This guy transforms by putting U's on the left and the right that are at x. But every field in this exponential has the same, U U dagger, multiply it, think about expanding that out. All the U U daggers that are next to each other cancel. You just get an overall U U dagger that comes right outside the exponential. So this guy is just transforming with U ultrasoft of x. Wilson line back again, U ultrasoft dagger of x, and that's because this Wilson line with our notation is really a local thing with respect to the coordinate x. All the fields sit at x. OK? So it just transforms like that, because that's how the gauge field inside it transforms. So in some sense, both of these are derived from the transformation of An, but it's easier to think about how a Wilson line transforms and then think about how it should transform under collinear than just go through the An route and impose this condition. OK. So let's come back now to our full current which was this, that had the Wilson line in it. And now if you ask how it transforms under Uc, you see what's going to happen. You get Uc dagger, Uc, and the ultrasoft guy doesn't transform, and then these cancel. So it's invariant under the transformations that have momenta of order of the ultrasoft. It'll have a momentum of what are the linear scale. There's two objects in this thing that have momenta that can be of that size. The gluons, they're in W in this collinear quark field. Both of those things transform, but that transformation cancels between the two things. AUDIENCE: [INAUDIBLE] PROFESSOR: Whoops, that's important. Thank you. And then under an ultrasoft transformation, we can also transform this guy, and then everybody's transforming. Gamma, of course, has not got color indices, and again, it's gauge invariant. OK? Missed a gamma. Gamma's not so important. All right? So this current that we derived earlier is actually invariant under two different types of symmetries which are really momentum space splitting of a standard gauge symmetry into pieces where we have a large momenta and pieces where you have these larger momenta, collinear momenta and pieces where we have a smaller momenta, this ultrasoft momenta. But both of these types of momenta are momenta that we're allowing inside the effective theory. Yeah? AUDIENCE: Just trying to remember that the ultrasoft scale was the same for the [INAUDIBLE].. PROFESSOR: Yeah. The momentum of the ultrasoft-- the ultrasoft momentum of the collinear quark, you pulled out mv, and then the remaining part was ultrasoft in the discussion for this particular thing. All right. So it's the Wilson line, the moral of this is that what we were doing before, where we're integrating out those collinear gluons, is actually connected to gauge symmetry. Because without that Wilson line, we wouldn't have a result that's invariant under the type of gauge symmetry we've been talking about. So this Wilson line carries these n collinear gluons, and those gluons would be the gluons that in the full theory would give you a gauge symmetric result. They would combine. They would be the attachments to the heavy quark that combined with attachments to the light quark to give you a result that's invariant, if you looked at word identities, for example. But in the effective theory, these are just put into a Wilson line, because that's the leading order way that they can show up, but still we have a gauge invariant answer. So the Wilson line is really needed for the gauge symmetry. And finally, the ultrasofts, as can be clear from the notation, the ultrasofts can just as well include the global transformation, if you like. So we don't really need three. We could put the global back with the ultrasofts. Then, we just have two. All right. So that's our symmetry. That's an example of how it works. Now, let's talk about covariant derivatives. So gauge symmetry ties together regular derivatives with the gauge field, and here, what it ties together at the following things. So because of the treatment of the A ultrasoft field as a background, it actually ties together both of these things. And then it does what you expect in a collinear sector. It ties together the derivative with the gauge field. And if you're acting on ultrasoft fields, then there's an ultrasoft covariant derivative which is just like that. So now, you can ask the question if I just had this gauge symmetry, which is going to force me to use covariant derivatives-- which I was already thinking ahead to when I wrote down the Lagrangian earlier. Right? We already wrote it in terms of these kind of objects, so it seems like we're good to go. So you can ask the question if I just have power counting, the cn that we decided we're going to use, and gauge symmetry, is that enough? Am I done? Do I have everything I need for the leading order Lagrangian? So power counting told us that we had this type of derivative, and that we could have two of these perp derivatives. OK? So those are the two type of terms we were getting. There's no other order lambda squared field structures that would have the right dimensions and that could do the job. So these are really the only two that we have to think about. So you could try to think about maybe I can have something with some additional n bars floating around, but really, it boils down-- since you need to get a lambda squared, you need either two perp derivatives. You need two of them, because you still have a rotation symmetry around the direction of motion, or you could have an n dot D. But there is one operator that is not rolled out, by gauge symmetry alone, and that is this operator. When we wrote the operator, we had D perp slash D perp slash. But nothing stops me from doing something similar in some way by just taking the D perps and contracting them, like that. That's a different operator than the D perp slash D perp operator that we got before. Oops, only one of those. OK? So this operator is a different operator. In a priori, you could say, well, do loop connections generate this operator which is different than the D perp slash D perp slash operator. The answer is no, actually, but so far what we've done, it's not enough to see that. So we need actually this other symmetry that I alluded to, the reparameterization invariance, and that will actually rule out this guy. So what's RPI? Well, when we formulated this theory, we needed a direction for the collinear line or for the collinear particles. And then we needed an auxiliary vector n bar, and we formulated the whole set up in terms of this n and n bar. But having vectors like that that we write down, that breaks Lorentz invariance in the same way that specifying v mu in HQET breaks Lorentz invariance. So if you say that m mu nu is the set of Lorentz transformations, the usual six generators that are anti-symmetric, then the ones that are broken are these ones. And there's five of them, and the one that's not broken is this one which corresponds to rotations about the three, axis with the axis specified by n. OK? So those rotations would act in the components of these guys that are 0, if you like. So there's no issue there, and these guys are the guys that are connected to the pieces that were non-0 in general. So there's going to be a larger, because there's five things here and because there's two vectors, it's going to be a larger set of reparameterization symmetries than in HQET. First, talk about just the n and n bar themselves. There's three types of reparameterization invariance that would leave n squared equals 0 and bar squared equals 0 and n dot n bar equals 2. And those were the formulas that we were using over and over again really. Everything else was just convention. OK? So what are those three types? Well, we could take n, and we could change it by some amount in the perp direction and leave n bar unchanged. And since n bar dot something perp is 0, that would satisfy this, or we could do the opposite. That would be type two. So don't change n, and change n bar. And so this is two transformations there's two things specified by this perp guy and two here. And then there's one more which is called three, and this one let me write it like this. It's a simultaneous transformation of these two guys, where I just do a multiplicative factor. Well, a multiplicative factor is not going to change the fact that's the square of something 0. The place where the normalization comes in is this n dot n bar, and if I just rescale them both by an opposite amount, then that remains satisfied as well. So I formulated this in terms of an infinite-- I'm going to think about this is an infinitesimal transformation for delta perp, infinitesimal for epsilon perp. I could also expand this exponential and think of it as infinitesimal, but the finite one's easier. So let's just think of the finite transformation in the last one. Now, this is an effective theory. So whenever we think about transformations, we should think about power counting. We just spent a lot of time thinking about that for gauge symmetry, and we should think about power counting here too. Did you have a question? AUDIENCE: [INAUDIBLE] PROFESSOR: Yeah. So what I was just about to talk about, how big things can be. So the power counting that's the right power counting for these guys, so these things are infinitesimal or finite, but they also have some counting in a different space which is the power counting space. And the right power counting for these guys is as follows. So this means I can make arbitrarily large transformations in a power counting sense of type two and type three. There's no constraint. That's what this means, so that maybe that's easier to swallow. This one, there is a constraint, and the reason that you can think of there being a constraint, I can give you a simple example, and then you'll see how you would derive the other things as well. So think of n dot P. If you transform n dot P, that becomes n dot-- think of this is n mu, P mu. That becomes n mu, P mu plus delta perp dot P perp. So this is order lambda squared, so this better be order lambda squared. But this momentum here is order lambda, so you need to have a power counting for the delta perp that makes it of order lambda. In order to keep the transform guy the same size as the guy you started with, because I don't want to consider transformations that would take me away from my power counting, that would violate it. So imposing that this thing is of order lambda squared says that this is true. If you go through the same logic for the other guys, the n bar component was order lambda 0, so you can make the epsilon perp of order lambda 0. It doesn't cost you anything, because there's no constraint. You're not making something that would mess up the power counting. All right. So what does this correspond to physically? So type three is actually pretty simple. Type three is like a boost along-- if you think about our back-to-back vectors, type three is like the analog of a boost, but you're transforming it in the passive sense of transforming the axis. And what the outcome is is very simple. It just implies that any operator that has an n must have a corresponding n bar sitting next to it, effectively. So I'll give you some examples. So really, if you have-- let's just say it this way-- if you have an n bar mu in the numerator, then you either have a corresponding n in the numerator, or-- well, you have an n bar in the denominator. Since we could have n bars in the denominator and that was showing up in some places. Those are the two possible ways of compensating for the transformation. So it's just like a simple counting that you have. So if you look at our lc 0, we had various terms. One of them was n bar slash 1 over i n bar over D. So here, we have an n bar in both the numerator and the denominator, and that compensates for the transformation. And then in another term, we had an n bar slash n dot D, and then that again is invariant in this type of transformation. What about type one and two? So type one is the following. So think about these guys as in this kind of language they would be back to back. We're trying to describe the physics in this cone for the collinear particles, and n was the vector that was pointing inside that cone. What this is saying, type one, is that I can rotate that guy, and as long as I'm not rotating it by too much-- i.e. I'm rotating it by a small amount of order lambda, which you can think of as staying inside the cone-- then I can describe everything equally well by some other vector that lives in that cone, and I can decompose the momenta and the modes in terms of that vector, and it will work equally well. OK. So that's what type one is physically doing, and so you're not making a large transformation here, because you want to still be inside the cone. You still want something that's pointing in the collinear direction. Type two is related to the fact that this n bar vector was just an auxiliary vector that we used to decompose things. We didn't really care about it. It didn't have a strong physical motivation. It was just needed because we're using light cone coordinates. So I can make a very large transformation of that guy. So here's a large transformation, and I can use some other guy. And we gave you an example earlier, which was 3, 2, 2, 1 as a possible value for the n bar, and that's something that you could think of getting to by a finite type two reparameterization transformation. OK? So that's the picture for what these transformations are, and we will finish up talking about them next time and will show that actually this additional term that we could write down in the Lagrangian is actually ruled out reparameterization symmetries.
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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 consider an example of motion, in which we want to use our energy concepts. So suppose we have a ramp which is circular, radius r. And we have an object here. And now, here's a surface a certain distance, d, that has friction, where the coefficients of friction is non-uniform. So we'll write it mu-0 And we'll write it as mu-1x. We'll take x equals zero. And then here, there's a spring and a wall. And here there is no friction. Now as we drop, assuming we have enough height here, h, so that this block will slide across the friction, make it all the way across and start compressing its spring. What I'd like to find is how much the spring has been compressed. OK, now how do we analyze this? Well the key is to use our energy principle, where we have w-external equals the change in mechanical energy. And the key is to, what we're going to do is the tool that we're going to use is what we call energy diagrams for the initial. So what does this mean? We want to choose, we want to first identify the initial and final states that we're referring to. So, in our picture, here is our initial state. And I'll draw the final state in when the spring has been compressed a distance, x-final. So I drew it on my diagram. So we have initial and final states. And now what we want to do is choose reference points, zero-points, zero-point for each potential function. And show that on our diagram. So for the potential energy of gravity, here, if we chose this to be y, then this y equals 0. And u at y equals 0 is 0. And we'll call that the zero point for gravity. And the zero point for the spring is when this x, now I'm going to call here x equals 0. So, let's call this a variable, I can call it anything I want. I'll call it u-final. And this is where u is zero. So u-final just measures the stretch of the spring. And so at y equals 0 is 0. And u spring little u equal 0, 0. So now I can identify my energies. So let's talk about the initial energy is all gravitational potential. That's mgy. It's starting at rest. And e-final well, here, this is the distance where it comes to rest, also. So there's no final kinetic energy. There's no gravitational potential energy, because we're on the surface at y equals 0. But our spring has been compressed by 1/2 k little-u-final square. Now, in terms of our external work equals the change in mechanical energy, we have now identified the right-hand side using these tools of the energy diagrams. And I can write my description u-final squared, minus mgy. Now what I have to do is think about the friction force, f kinetic friction, as the object moves. This friction force is non-uniform. If we were to draw n, and mg, then our friction force here is equal to the integral minus the integral from x equals 0, to x equals d. Equals 0 to x equal d, of the friction force, which is equal to the coefficient of friction, ukndx. Now that 0 to d, notice that our coefficient of friction is varying. And I chose it intentionally to show you that friction is really an integral. So we have mu-0 plus mu-1 xmgdx. Let's put our integration variable in there. And these are just two separate integrals. The first one is easy. It's minus mu-0 mgd. And the second one is -mu-1 mg. And we're just integrating x-prime, dx prime between 0 and d. So that's simply d-squared over 2. And that's equal to ku-final squared minus mgy. Let's write that as y-initial. And we're starting it at y-initial equals h. And so, even though this is complicated, I can now solve for how much this spring has been compressed with a little bit of algebra. And so I'm just going to bring a bunch of terms over to the other side. And take the square root of 2, divided by k of mgy-initial. Minus mu-0 mgd. Minus mu-1 mgd-squared over 2. And that's how much the spring is compressed. Notice what I did not do was divide this up into a bunch of different motions. I picked an initial state. I picked a final state. I drew my energy diagrams with my zero points. I described the key parameters of initial and final states, y-i and u-final. I defined the initial mechanical energy, the final mechanical energy. And then applied the work energy, work mechanical energy principle. I had to integrate the friction force because it was non-trivial and solve for how much the spring has been compressed.
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https://ocw.mit.edu/courses/5-61-physical-chemistry-fall-2017/5.61-fall-2017.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation, or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. ROBERT FIELD: Last time, we talked about photochemistry. And the crucial thing in photochemistry is that the density of vibrational states increases extremely rapidly-- extremely, extremely-- and especially for larger molecules. And this enables you to understand intermolecular vibrational redistribution, intersystem crossing, internal conversion. The matrix elements are different in each of these cases. But what happens when the vibrational density of states gets large is that you have a special state, which is a localized state or something that you care about that you think you understand-- a bright state-- which is different from the masses. And because of the density of states, that bright state or that special state, gets diluted into an extremely dense manifold of uninteresting states. And as a result, the system forgets whatever the bright state wanted it to tell you. And so you get faster decay of fluorescence, not because the population goes away, but because the unique thing-- the bright state, the localized state-- the thing that you can prepare by excitation from the ground state, which is always localized, its character is dissipated. And so you get what's called statistical behavior. Now I have a very strong feeling about statistical behavior, and that it's mostly a fraud. But the current limit for statistical behavior is when the density of states is on the order of 100 per wave number. That's the usual threshold for statistical behavior. As we know more, as we develop better experimental techniques, we're going to push that curtain of statistical behavior back. And we'll understand more about the short-time dynamics and whatever we want to do about manipulating systems. So statistical just means we don't know, but it's not what we expected. And it's usually boring. OK. Today I'm going to talk about the discrete variable representation, which is really weird and wonderful thing, which is entirely inappropriate for a course at this level. But I think you'll like it. So in order to talk about the discrete variable representation, I'll introduce you to delta functions. And you sort of know what delta functions are. And then I'm going to say, well for any one-dimensional problem-- and remember, you can have a many-dimensional problem treated as a whole bunch of one-dimensional problems. And so it really is a general problem. But if you have a one-dimensional potential, you can obtain the energy levels and wave functions resulting from that one-dimensional potential. Regardless of how horrible it is, without using perturbation theory, which is should make you feel pretty good. Because perturbation theory is labor intensive before you use the computer. The computer helps you, but this is labor free because the computer does everything. At the beginning of the course, I said, you cannot experimentally measure a wave function. And that's true. but if you can deal with any potential and have a set of experimental observations, energy levels, which you fit to an effective Hamiltonian, you can generate the wave functions associated with the effective Hamiltonian directly from experimental observations. So it's not a direct observation, but you get wave functions. And you can have whatever complexity you want. OK, so let's begin. Delta functions. So this is not the Kronecker delta. This is the delta function that is a useful computational trick. And it's more than that. We write something like this. OK, this is a delta function. And it says that this thing is non-zero when x is equal to xi. And it's really big. And it's 0 everywhere else. And as a result, you can say that we get xi delta x xi. So that's an eigenvalue equation. So we have the operator, x, operating on this function. And this function has the magical properties that it returns an eigenvalue and the function. So this is more than a mathematical trick. It's an entry into a form of quantum mechanics that is truly wonderful. So this is part one. Part two will be DVR-- discrete variable representation. And the issue varies. Suppose we have a matrix representation of an operator. Well, suppose we wanted some function of that operator. How do we generate that? And there are lots of cases where you care about such a thing. For example, you might want to know about something like this-- e to the ih t over h bar. This would tell you something about how a system propagates under the influence of a Hamiltonian, which is not diagonal. This is kind of important. Almost all of NMR is based on this sort of thing. Also-- well, we'll get to it. So extend DVR to include rotation. And then the last part will be-- can be determined for even the most horrible situations, like a potential that does something like this, or something like that. Now we don't want to have a continuum. So I put a wall here. But this is a summarization from an unstable isomer to a stable isomer. This is multiple minima. Anything you want, you would never want to perturbation theory on it. And you can solve these problems automatically and wonderfully. OK, so let's just play with delta functions. And a very good section on delta functions is in Cohen-Tannoudji, and it's pages 1468 to 1472, right at the end of the book. But it's not because it's so hard, it's just because they decided to put it there. And so we have notation-- x xi. It's the same thing as x minus xi. So basically, if you see a variable at a specified value, it's equivalent to this, and this thing is 0 everywhere except when x is equal to xi. So the thing in parentheses is the critical thing. And it has the property that if we do an integral from minus infinity to infinity, some function of x-- dx-- delta x xi dx. We get f of xi. Isn't that wonderful? I mean, it's a very lovely mathematical trick. But it's more than that. It says this thing is big when x is equal to xi. It's 0 everywhere else. And it's normalized to 1 in the sense that, well you get back the function that you started with, but at a particular value. So it's infinite, but it's normalized. It should bother you, but it's fantastic and you can deal with this. It's an eigenvalue equation. I mean, you can say, all right, we have delta x xi xi delta x xi. And we can have delta functions in position, in momentum, in anything you want. And they're useful. So suppose we have some function, psi of x, and we want to find out something about how it's composed. And so we say, well, we have c of xi j delta x minus xi dx. This is the standard method for expanding a function. OK, what we want is these expansion coefficients. And you get them in the standard way. And that is by-- normally, when you have a function, you write it ck j the function-- I'm sorry. So here we have an expansion of members of a complete set of functions. And in order to get the expansion coefficients, you do the standard trick of integrating phi j star, psi j dx. OK, so this is familiar. This is not. But it's the same business. In fact, we've been using a notation incorrectly-- the Dirac notation. We normally think that if we have something like k and psi k of x, they're just different ways of writing the same thing. But the equivalent to the Schrodinger picture is really x psi. This is a vector, not a function. This is a set of vectors. And this is how you relate vectors to functions. We just have suppressed that. Because it's easier to think that the wave function is equivalent to the symbol here. It's not. And if you're going to be doing derivations where you flip back and forth between representations, you better remember this, otherwise it won't make sense. This is familiar stuff, you just didn't realize you understood it. OK, and so now, often, you want to have a mathematical representation of a delta function. And so this thing has to be localized and it has to be normalized. And there are a whole bunch of representations that work really well. One of them is we take the limit, epsilon goes to 0 from the positive side, and 1 over 2 epsilon e to the minus x over epsilon. Well this guy, as epsilon goes to 0, this becomes minus infinity. And it's 0 everywhere except for x is equal to 0. And so if we wanted to put x minus xi, then fine. Then it would be 0 everywhere except when x is equal to xi. So that's one. That's a simple one. Another one is limit as epsilon goes to 0 from the positive side of 1 over pi times epsilon over x squared plus epsilon squared. This also has the properties of being an infinite spike at x equals 0. And it's also Lorentzian. And it has the full width. It'd have maximum-- or a half width it would have-- a full width at half maximum of epsilon. If you make the width as narrow as you want, well that's what this is. OK, there are a whole bunch of representations. I'm going to stop giving them because there's a lot of stuff I want to say, and you can see them in the notes. OK, so what if we have something like xi, that just is the same thing as saying x minus xi is 0. You can produce a delta function localized at any point by using this trick. OK, there are some other tricks that you can do with delta functions, which is a little surprising. And that's really for the future. Delta of minus x is equal to delta of x. It's an even function. Derivative of a delta function is an odd function. Delta-- a constant times x is 1 over the absolute value of that constant, delta x. A little surprising, but if you have x, which-- I better not say that. OK, we have now another fantastic thing-- g of x. So what is the delta function of a function? Well it's a sum over j of the g ex and x of j times delta x minus xj. This is something where you're harvesting the zeros of this function. OK, when this is 0, we get a big thing. So the delta function of our function is a sum over the derivative-- the zeros of g of x. And at times, it's kind of neat. OK, there's also stuff in Cohen-Tannoudji on Fourier transforms of the delta functions. I'm not going to talk about that. But one of the things is, if you have a delta function and x, and you take the Fourier transformer with it, you get a delta function of p, momentum. Kind of useful. It enables you to do a transform between an x representation and a position representation and a momentum representation. It's kind of useful. OK, now we're going to get to the good stuff. So for every system that has a potential, and where the potential has minima, what is the minimum potential? What is the condition for a minimum of the potential? Yes? AUDIENCE: The first derivative has to be 0 with respect to coordinates. ROBERT FIELD: Right. And so, any time you have a minimum, the first term in the potential is the quadratic term. And that's the same thing as a harmonic oscillator. The rest is just excess baggage. I mean, that's what we do for perturbation theory. We say, OK, we're going to represent some arbitrary potential as a harmonic oscillator. And all of the bad stuff, all of the higher powers of the coordinate get treated by perturbation theory. And you know how to do it. And you're not excited about doing it because it's kind of algebraically horrible. And nobody is going to check your algebra. And almost guaranteed, you're going to make a mistake. (Exam 2!) So it would be nice to be able to deal with arbitrary potentials-- potentials that might have multiple minima, or might have all sorts of strange stuff without doing perturbation theory. And that's what DVR does. So for example, we want to know how to derive a matrix representation of a matrix. So often, we have operators like the overlap integral or the Hamiltonian. So we have the operator, the S matrix, or the Hamiltonian matrix. And often, we want to have something that is a matrix representation of a matrix. For example, if we're dealing with a problem in quantum chemistry, where our basis set is not orthonormal, there is a trick using the S matrix to orthonormalize everything. And that's useful because then your secular equation is the standard secular equation, which computers just love. And so you just have to tell the computer to do something special before, and it orthonormalizes stuff. And you can do this in matrix language. And if you're interested in time evolution, you often want to have e to the minus I h t over h bar. And that's horrible. But so we'd like to know how to obtain a matrix representation of a function of a matrix. Well, suppose we have some matrix, and we can transform it to be a1, a n, 0, 0. We diagnose it. And if a is real and symmetric, or Hermitian. We know that the transformation that diagonalized it has the property that t dagger is equal to the inverse. And we also know that if we diagonalize a matrix the eigenvectors that-- they say, if you want the first eigenvector, the thing that belongs to this eigenvalue, we want the first column of t dagger. And if you want to use perturbation theory instead of the computer to calculate t dagger, you can do that, and you have a good approximation. And you can write the vector, the linear combination of basis functions, that corresponds to t dagger as fast as you can write. And the computer can do it faster. So if the matrix is Hermitian, then all of these guys are real. And it's just real and symmetric. Well then these guys are still numbers that you could generate, but they might be imaginary or complex. Suppose we want some function of a matrix. So this is a matrix, this has to be a matrix too. It has to be the same dimension as the original matrix. So you can do this. Well we can call that f twiddle. But we don't want f twiddle. And so if we do this, we have the matrix representation of the function. So this is something that could be proven in linear algebra, but not in 5.61. You can do power series expansions, and you can show term by term that this is true for small matrices, but it's true. So if your computer can diagonalize this, then what happens is that we have this-- we can write that f twiddle is the-- So this is the crucial thing. We've diagonalized this, so we have a bunch of eigenvalues of a. So the f twiddle is just the values of the function at each of the eigenvalues. And we have zeros here. And now we don't like this because this isn't the matrix representation of f. It's f in a different representation. So we have to go back to the original representation. And so that's another transformation. But it uses the same matrix. So if you did the work to diagonalize a, well then you can go back and undiagonalize this f twiddle to make the representation of f. And so the only work involved is asking the computer to find t dagger for the a matrix. And then you get the true matrix representation of this function of a. Is it useful? You bet. Now suppose a is infinite dimension. And we know that this is a very common case. Because, even for the harmonic oscillator, we have an infinite number of basis functions. But what about using the delta function? We also have an infinite number of them. So what do we do? We can't diagonalize an infinite matrix. So what we do is we truncate it. Now the computer is quite happy to deal with matrices of dimension 1,000. Your computer can diagonalize a 1,000 by 1,000 matrix in a few minutes, maybe a few seconds depending on how up to date this thing is. And so what we do is we say, oh, well let's just take a 1,000. So here is an infinite matrix. And here is a 1,000 by 1,000 block. That's a million elements. The computer doesn't care. And we're just going to throw away everything else. We don't care. Now this is an approximation. Now you can truncate in clever ways, or just tell the computer to throw away everything above the 1,000th basis function. That's very convenient. The computer doesn't care. Now there are transformations that say, well you can fold in the effects of the remote basis states, and do an augmented representation. But usually, you just throw everything away. And now you look at this 1,000 by 1,000 matrix. And you get the eigenvalues. And so this might be the matrix of x, or q if we're talking in the usual notation. This is the displacement from equilibrium. Well, it seems a little-- so we would like to find this matrix. Well, we know what that is. Because we know the relationship between x and a plus a dagger are friends. And so we have a matrix, which is zeros along the diagonal, and numbers here and here, and zeros everywhere else. It doesn't take much to program a computer to fill in as many one-off the diagonal, especially because they're square roots of integers. So that's just a few lines of code, and you have the matrix representation of x. Now that is infinite. And you're going to say, well, I don't care. I'm going to just keep the first 1,000. We know that we can always write-- so we have v of x, and this is a matrix now. And it's an infinite matrix. But we say, oh, we just want v of x to the 1000th, and we'll get v of 1,000. We have a 1,000 by 1,000 v matrix. And we know how to do this. We diagonalize that. Then we write at each eigenvalue of x, what v of x is at that eigenvalue. And so now we have a v matrix, which is diagonal, but in the wrong representation for us. And then we transform back to the harmonic oscillator representation. And so everything is fine. We've done this. And we don't know how good it's going to be. But what we do is we do this problem. So we have the Hamiltonian, which is equal to the kinetic energy, I'm going to call it k, plus v. We know how to generate the representation of v in the harmonic oscillator basis by writing v of x, diagonlizing x, and then undiagonalizing-- or then writing v at each of the eigenvalues of x, and then going back to the dramatic oscillator basis. We know k in the harmonic oscillator basis. It's just tri-diagonal, and it has matrix elements delta v equals 0 plus minus two. So now we have a matrix representation of k, which is simple, add a v which is-- computer makes it simple. Add any v you want, there it is. So that's a matrix, the Hamiltonian. You solve the Schrodinger equation by diagonalizing this matrix. And so you have this h matrix, and it's for the 1,000-member x matrix, and you get a bunch of eigenvalues. And so then you do it again. And maybe use a 900 by 900, or maybe you use a 1,100-- you do it again. And then you look at the eigenenergies of the Hamiltonian, e1, say, up to e100. Now if you did a 1,000 by 1,000, you have reasonable expectation that the first 100 eigenvalues will be right. And so you compare the results you get for the 1,000 by 1,000 to the 900 by 900, or 1,100 by 1,100. And you see how accurate your representation is for the first 100. Normally, you don't even care about the first 100. You might care about 10 of them. So the computer is happy to deal with 1,000 by 1,000s. There's no least squares fitting, so you only do it once. And all of a sudden, you've got the eigenvalues, and you've demonstrated how accurate they are. And so depending on what precision you want-- you can trust this up to the 100th, or maybe the 73rd, or whatever, to a part in a million, or whatever. And so you know how it's going to work. And you have a check for convergence. So it doesn't matter. So the only thing that you want to do is you want to choose a basis set where we have-- x is the displacement from equilibrium. OK, so this is the equilibrium value. This is the definition of the displacement. And so you want to be able to choose your basis set, which is centered at the equilibrium value. You could do it somewhere else, it would be stupid. It wouldn't converge so well. And you want to use the harmonic oscillator, k over u. You have a couple of choices before you start telling the computer to go to work. And you tell it, well I think the best basis that will be what works at the equilibrium-- the lowest minimum of the potential, and matches the curvature there. You don't have to do that. But it would be a good idea to ask it to do a problem that's likely to be a good representation. And all this you've done. You get the energy level. So what you end up getting-- So you produced your Hamiltonian. And since it's not an infinite dimension Hamiltonian, we can call it an effective Hamiltonian. It contains everything that is going to generate the eigenvalues. And we get from that a set of energy levels-- e vi-- and a set of functions. So these guys are the true energy levels. And these are the linear combinations of harmonic oscillator functions that correspond to each of them. Who gives that to you? And so then you say, well, I want to represent the Hamiltonian by a traditional thing. Like, I want to say that we have omega v plus 1/2 minus omega x equals 1/2 squared, et cetera. Now we do a least squares fit of molecular constants to the energy levels of the Hamiltonian. And there's lots of other things we could do, but-- so we say, well in the spectrum, we would observe these things, but we're representing them as a power series in v plus 1/2. Or maybe in-- where we would have not just the energy-- the vibrational quantum number-- but the rotational constant. We could say the potential is v of 0 plus b x j j plus 1. Well, that means we could extend this to allow the molecule to rotate. And we just need to evaluate the rotational constant as a function of x, and just add that to what we have here. Another thing, you have t dagger and t for our problem. And you have, say, the 1,000 by 1,000, and maybe the 900 by 900 representations. You keep them. Because any problem you would have, you could use these for. So you still have to do a diagonalization of the Hamiltonian, but the other stuff you don't have to do anymore. Now maybe it's too bothersome to store a 1,000 by 1,000 t dagger matrix. It's a million elements. Maybe you don't, and you can just calculate it again. It takes 20 minutes or maybe less. And so this is a pretty good. OK I've skipped a lot of stuff in the notes, because I wanted to get to the end. But the end is really just a correction of what I said was impossible at the beginning of the course. We have in the Schrodinger picture H psi is equal to E psi, right? That's the Schrodinger equation. So this wave function is the essential thing in quantum mechanics. And I also told you, you can't observe this. So it's a very strange theory where the central quality in the theory is experimentally inaccessible. But the theory works. The theory gives you everything you need. It enables you to find the eigenfunctions, if you have the exact Hamiltonian. Or it says, well we can take a model problem and we can find the eigenvalues and wave functions for the model problem. We can generate an effective Hamiltonian, which is expressed in terms of molecular constants. We can then determine the potential. And we can also determine psi. All of them. So what I told you was true, but only a little bit of a lie in the sense that you can get as accurate as you want a representation of the wave function, if you want it. And DVR gives it to you without any effort. And so it doesn't matter how terrible the potential is, as long as it is more or less well behaved. I mean, if you had a potential, which-- Even if I had a v potential-- the discontinuity here-- you would still get a reasonable result from DVR. What it doesn't like is something like that, because then you have a continuum over here. And the continuum uses up your basis functions pretty fast. I mean, yeah, it will work for this. But you don't quite know how it's going to work, and you have to do very careful convergence tests, because this might be good. But up here, it's going to use a lot of basis functions. So for the first time in a long time, I'm finishing on time or even a little early. But DVR is really a powerful computational tool that, if you are doing any kind of theoretical calculation, you may very well much want to use something like this rather than an infinite set of basis functions and perturbation theory. It's something where you leave almost all of the work to the computer. You don't have to do much besides say, well what is the equilibrium value? And what vibrational frequency have I got to use for my basis set? And if you choose something that's appropriate for the curvature at the absolute minimum of the potential, you're likely to be doing very well. Now other choices might mean you don't get the first 100, you only get the first 50. But you might only care about the first 10. Or you could say, I'm going to choose something which is a compromise between two minima, and maybe I'll do better. But it doesn't cost you anything in the computer. Your computer is mostly sitting idly on your desk, and you could have it doing these calculations. And there is no problem where you can't use DVR. Because if you have a function of two variables, you do a two-variable DVR. It gets a little bit more complicated because, if you have two DVRs, now you're talking about a million-- 1,000 by 1,000-- two of them, and couplings between them. And so maybe you have to be a little bit more thoughtful about how you employ this trick. But it's a very powerful trick. And there are other powerful tricks that you can use in conjunction with quantum chemistry that enable you to deal with things like-- I've chosen a basis set, which is not orthonormal. And my computer only knows how to diagonalize an ordinary Hamiltonian without subtracting an overlap matrix from it. And so if I transform to diagonalize the overlap matrix, well then I can fix the problem. And so there is a way of using this kind of theory to fix the problem, which is based on choosing a convenient way of solving the problem, as opposed to the most rigorous way of doing it. OK, so I'm hoping that I will have a sensible lecture on the two-level problem for Wednesday. I've been struggling with this for a long time. And maybe I can do it. If not, I'll review the whole course. OK, see you on Wednesday.
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https://ocw.mit.edu/courses/10-34-numerical-methods-applied-to-chemical-engineering-fall-2015/10.34-fall-2015.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. JAMES SWAN: OK. Let's go ahead and get started. We saw a lot of good conversation on Piazza this weekend. So that's good. Seems like you guys are making your way through these two problems on the latest assignment. I would try to focus less on the chemical engineering science and problems that involve those. Usually the topic of interest, the thing that's useful to you educationally is going to be the numerics, right. So if you get hung up on the definition of a particular quantity, yield was one that came up. Rather than let that prevent you from solving the problem, pick a definition and see what happens. You can always ask yourself if the results seem physically reasonable to you not based on your definition. And as long as you explain what you did in your write up, you're going to get full points. We want to solve the problems numerically. If there's some hang up in the science, don't sweat it. Don't let that stop you from moving ahead with it. Don't let it make it seem like the problem can't be solved or there isn't a path to a solution. Pick a definition and go with it and see what happens, right. The root of the second problem is trying to nest together two different numerical methods. One of those is optimization, and the other one is solutions of nonlinear equation, putting those two techniques together, using them in combination. The engineering science problem gives us a solvable problem to work with in that context, but it's not the key element of it. OK, good. So we're continuing optimization, right. Move this just a little bit. We're continuing with optimization. Last time we posed lots of optimization problems. We talked about constrained optimization, unconstrained optimization. We heard a little bit about linear programs. We started approaching unconstrained optimization problems from the perspective of steepest descent. OK, so that's where I want to pick up as we get started. So you'll recall the idea behind steepest descent was all the unconstrained optimization problems we're interested in are based around trying to find minima, OK. And so we should think about these problems as though we're standing on top of a mountain. And we're looking for directions that allow us to descend. And as long as we're heading in descending directions, right, there's a good chance we're going to bottom out someplace and stop. And when we've bottomed out, we've found one of those local minima. That bottom is going to be a place where the gradient of the function we're trying to find the minimum of is zero, OK. And the idea behind steepest descent was well, don't just pick any direction that's down hill. Pick the steepest direction, right. Go in the direction of the gradient. That's the steepest descent idea. And then we did something a little sophisticated last time. We said well OK, I know the direction. I'm standing on top the mountain. I point myself in the steepest descent direction. How big a step do I take? I can take any size step. And some steps may be good and some steps may be bad. It turns out there are some good estimates for step size that we can get by taking a Taylor expansion. So we take our function, right, and we write it at the next iterate is a Taylor expansion. About the current iterate, that expansion looks like this. And it will be quadratic with respect to the step size alpha. If we want to minimize the value of the function here, we want the next iterate to be a minimum of this quadratic function. Then there's an obvious choice of alpha, right. We find the vertex of this quadratic functional. That gives us the optimal step size. It's optimal if our function actually is quadratic. It's an approximate, right. It's an estimation of the right sort of step size if it's not quadratic. And so I showed you here was a function where the contours are very closely spaced. So it's a very steep function. And the minima is in the middle. If we try to solve this with the steepest descent and we pick different steps sizes, uniform step sizes, so we try 0.1 and 1 and 10 step sizes, we'll never find an appropriate choice to converge to the solution, OK. We're going to have to pick impossibly small step sizes, which will require tons of steps in order to get there. But with this quadratic estimate, you can get a reasonably smooth convergence to the root. So that's nice. And here's a task for you to test whether you understand steepest descent or not. In your notes, I've drawn some contours. For function, we'd like to minimize using the method of steepest descent. And I want you to try to draw steepest descent paths on top of these contours starting from initial conditions where these stars are located. So if I'm following steepest descent, the rules of steepest descent here, and I start from these stars, what sort of paths do I follow? You're going to need to pick a step size. I would suggest thinking about the small step size limit. What is the steepest descent path in the small step size limit? Can you work that out, you and your neighbors? You don't have to do all of them by yourself. You can do one, your neighbor could do another. And we'll take a look at them together. OK, the roar has turned into a rumble and then a murmur, so I think you guys are making some progress. What do you think? How about let's do an easy one. How about this one here. What sort of path does it take? Yeah, it sort of curls right down into the center here, right. Remember, steepest descent paths run perpendicular to the contours. So jumps perpendicular to the contour, almost a straight line to the center. How about this one over here? Same thing, right? It runs the other way. It's going downhill 1, 0, minus 1, minus 2. So it runs downhill and curls into the center. What about this one up here? What's it do? Yeah, it just runs to the left, right. The contour lines had normals that just keep it running all the way to the left. So this actually doesn't run into this minimum, right. It finds a cliff and steps right off of it, keeps on going. Steepest descent, that's what it does. How about this one here? Same thing, right, just to the left. So these are what these paths look like. You can draw them yourself. If I showed you paths and asked you what sort of method made them, you should be able to identify that actually, right? You should be able to detect what sort of methodology generated those kinds of paths. We're not always so fortunate to have this graphical view of the landscape that our method is navigating. But it's good to have these 2D depictions. Because they really help us understand when a method doesn't converge what might be going wrong, right. So steepest descent, it always heads downhill. But if there is no bottom, it's just going to keep going down, right. It's never going to find it. Oh, OK. Here's a-- this is a story now that you understand optimization. So let's see, so mechanical systems, conservation of momentum, that's also, in a certain sense, an optimization problem, right. So conservation of momentum says that the acceleration on a body is equal to the sum of the forces on it. And some of those forces are what we call conservative forces. They're proportional to gradients of some energy landscape. Some of those forces are non-conservative, like this one here. It's a little damping force, a little bit of friction proportional to the velocity with which the object moves instead. And if we start some system like this, we give it some initial inertia and let it go, right, eventually it's going to want to come to rest at a place where the gradient in the potential is 0 and the velocity is 0 on the acceleration is 0. We call that mechanical equilibrium. We get to mechanical equilibrium and we stop, right. So physical systems many times are seeking out minimum of an objective function. The objective function is the potential energy. I saw last year at my house we had a pipe underground that leaked in the front yard. And they needed to find the pipe, right. It was like under the asphalt. So they got to dig up asphalt, and they need to know where is the pipe. They know it's leaking, but where does the pipe sit? So the city came out and the guy from the city brought this. Do you know what this is? What is it? Do you know? Yeah, yeah, yeah. It's a dowsing rod. OK, this kind of crazy story right, a dowsing rod. OK, a dowsing rod. How does it work? The way it's supposed to work is I hold it out and it should turn and rotate in point in a direction that's parallel to the flow of the water in the pipe. That's the theory that this is supposed to work on. I'm a scientist. So I expect that somehow the water is exerting a force on the tip of the dowsing rod, OK. So the dowsing rod is moving around as this guy walks around. And it's going to stop when it finds a point of mechanical equilibrium. So the dowsing rod is seeking out a minimum of some potential energy, let's say. That's what the physics says has to be true. I don't know that flowing water exerts a force on the tip of the dowsing rod. The guy who had this believed that was true, OK. It turns out, this is not such a good idea, though, OK. Like in terms of a method for seeking out the minimum of a potential, it's not such a great way to do it. Because he's way up here, and the water's way underground. So there's a huge distance between these things. It's not exerting a strong force, OK. The gradient isn't very big here. It's a relatively weak force. So this instrument is incredibly sensitive to all sorts of external fluctuations. The gradient is small. The potential energy landscape is very, very flat. And we know already from applying things like steepest descent methods or Newton-Raphson that those circumstances are disastrous for any method seeking out minima of potential energies, right. Those landscapes are the hardest ones to detect it in. Because every point looks like it's close to being a minima, right. It's really difficult to see the differences between these. Nonetheless, he figured out where the pipe was. I don't think it was because of this though. How did he know where the pipe was? What's that? STUDENT: Where the ground was squishy? JAMES SWAN: Where the ground was squishy. Well yeah, had some good guesses because it was leaking up a little bit. No, I looked carefully afterwards. And I think it turned out the city had come by and actually painted some white lines on either side of the street to indicate where it was. But he was out there with his dowsing rod making sure the city had gotten it right. It turns out, there's something called the ideomotor effect where your hand has very little, you know, very sensitive little tremors in it. And can guide something like this, a little weight at the end of a rod to go wherever you want it to go when you want it to. It's like a Ouija board, right. It works exactly the same way. Anyway, it's not a good way to find the minimum of potential energy surfaces, OK. We have the same problem with numerical methods. It's really difficult when these potential energy landscapes are flat to find where the minimum is, OK. So fun and games are over. Now we got to do some math. So we talked about steepest descent. And steepest descent is an interesting way to approach these kinds of optimization problems. It turns out, it turns out that linear equations like Ax equals b can also be cast as optimization problems, right. So the solution to this equation Ax equals b is also a minima of this quadratic function up here. How do you know? You take the gradient of this function, which is Ax minus b, and the gradient to 0 to minima. So Ax minus b is 0, or Ax equals b. So we can do optimization on these sorts of quadratic functionals, and we would find the solution of systems of linear equations. This is an alternative approach. Sometimes this is called the variational approach to solving these systems of linear equations. There are a couple of things that have to be true. The linear operator, right, the matrix here, it has to be symmetric. OK, it has to be symmetric, because it's multiplied by x from both sides. It doesn't know that it's transpose is different from itself in the form of this functional. If A wasn't symmetric, the functional would symmetrize it automatically, OK. So a functional like this only corresponds to this linear equation when A is symmetric. And this sort of thing only has a minimum, right, when the matrix A is positive and definite. It has to have all positive eigenvalues, right. The Hessian right, of this functional, is just the matrix A. And we already said that Hessian needs all positive eigenvalues to confirm we have a minima. OK? If one of the eigenvalues is zero, then the problem is indeterminate. The linear problem is indeterminate. And there isn't a single local minimum, right. There's going to be a line of minima or a plane of minima instead. OK? OK, so you can solve systems of linear equations as optimization problems. And people have tried to apply things like steepest descent to these problems. And it turns out steepest descent is kind of challenging to apply. So what winds up happening is let's suppose we don't take our quadratic approximation for the descent direction first. Let's just say we take some fixed step size, right. When you take that fixed step size, it'll always be, let's say good for one particular direction. OK, so I'll step in a particular direction. It'll be good. It'll be a nice step into a local minimum. But when I try to step in the next gradient direction, it may be too big or too small. And that will depend on the eigenvalues associated with the direction that I am trying to step in, OK. How steep is this convex function? Right? How strongly curved is that convex function? That's what the eigenvalues are describing. And so fixed value of alpha will lead to cases where we wind up stepping too far or not far enough. And there'll be a lot of oscillating around on this path that converges to a solution. I showed you how to pick an optimal step size. It said look in a particular direction and treat your function as though it were quadratic along that direction. That's going to be true for all directions associated with this functional, right. It's always quadratic no matter which direction I point in. Right? So I pick a direction and I step and I'll be stepping to the minimal point along that direction. It'll be exact, OK. And then I've got to turn and I've got to go in another gradient direction and take a step there. And I'll turn and go in another gradient direction and take a step there. And in each direction I go, I'll be minimizing every time. Because this step size is the ideal step size. But it turns out you can do even better than that. So we can step in some direction, which is a descent direction, but not necessarily the steepest descent. And it's going to give us some extra control over how we're minimizing this function. I'll explain on the next slide, OK. The first thing you got to do though is given some descent direction, what is the optimal step size? Well, we'll work that out the same way, right. We can write f at the next iterate in terms of f at the current iterate plus all the perturbations, right. So our step method is Xi plus 1 is Xi plus alpha pi, right. So we do a Taylor expansion, and we'll get a quadratic function again. And we'll minimize this quadratic function with respect to alpha i when alpha takes on this value. So this is the value of the vertex of this function. So we'll minimize this quadratic function in one direction, the direction p. But is there an optimal choice of direction? Is it really best to step in the descent direction? Or are there better directions that I could go in? We thought going downhill fastest might be best, but maybe that's not true. Because if I point to a direction and I apply my quadratic approximation, I minimize the function in this direction. Now I'm going to turn, and I'm going to go in a different direction. And I'll minimize it here, but I'll lose some of the minimization that I got previously, right? I minimized in this direction. Then I turned, I went some other way, right. And I minimized in this direction. So this will still be a process that will sort of weave back and forth potentially. And so the idea instead is to try to preserve minimization along one particular direction. So how do we choose an optimal direction? So f, right, at the current iterate, it's already minimized along p, right. Moving in p forward and backwards, this is going to make f and e smaller. That's as small as it can be. So why not choose p so that it's normal to the gradient at the next iterate? OK, so choose this direction p so it's normal to the gradient at the next iterate. And then see if that holds for one iterate more after that. So I move in a direction. I step up to a contour. And I want my p to be orthogonal to the gradient at that next contour. So I've minimized this way, right. I've minimized everything that I could in directions that aren't in the gradient direction associated with the next iterate. And then let's see if I can even do that for the next iteration too. So can it make it true that the gradient at the next iterate is also orthogonal to p? By doing this, I get to preserve all the minimization from the previous steps. So I minimize in this direction. And now I'm going to take a step in a different direction. But I'm going to make sure that as I take that step in another direction, right, I don't have to step completely in the gradient. I don't have to go in the steepest descent direction. I can project out everything that I've stepped in already, right. I can project out all the minimization I've already accomplished along this p direction. So it turns out you can solve, right, you can calculate what this gradient is. The gradient in this function is Ax minus b. So you can substitute exactly what that gradient is. A, this is Xi plus 2 minus b dotted with p, right. This has to be equal to 0. And you can show that means that p transpose A times p has to be equal to 0 as well. You don't need to be able to work through these details. You just need to know that this gives a relationship between the directions on two consecutive iterates, OK. So it says if I picked a direction p on the previous iteration, take how it's transposed by A, and make sure that my next iteration is orthogonal to that vector, OK. Yeah? STUDENT: So does that mean that your p's are all independent of each other, or just that adjacent p is? K, k plus 1 p's are? JAMES SWAN: This is a great question. So the goal with this method, the ideal way to do this would be to have these directions actually be the directions of the eigenvectors of A. And those eigenvectors for symmetric matrix are all orthogonal to each other. OK? And so you'll be stepping along these orthogonal directions. And they would be all independent of each other. OK? But that's a hard problem, finding all the eigenvectors associated with a matrix. Instead, OK, we pick an initial direction p to go in. And then we try to ensure that all of the other directions satisfy this conjugacy condition, right. That the transformation of p by A is orthogonal with the next direction that I choose. So they're not independent of each other. But they are what we call conjugate to each other. It turns out that by doing this, these sets of directions p will belong to-- they can be expressed in terms of many products of A with the initial direction p. That'll give you all these different directions. It starts to look something like the power iteration method for finding the largest eigenvector of a matrix. OK? So you create a certain set of vectors that span the entire subspace of A. And you step specifically along those directions. And that lets you preserve some of the minimization as you step each way. So what's said here is that the direction p plus 1 is conjugate to the direction p. And by choosing the directions in this way, you're ensuring that p is orthogonal to the gradient at i plus 1 and the gradient i plus 2. So you're not stepping in the steepest descent directions that you'll pick up later on in the iterative process. OK? So when you know which direction you're stepping in, then you've got to satisfy this conjugacy condition. But actually, this is a vector in n space, right. This is also a vector n space. And we have one equation to describe all and components. So it's an under-determined problem. So then one has to pick which particular one of these conjugate vectors do I want to step along. And one particular choice is this one, which says, step along the gradient direction, OK, do steepest descent, but project out the component of the gradient along pi. We already minimized along pi. We don't not have to go in the pi direction anymore, right. So do steepest descent, but remove the pi component. So here is a quadratic objective function. It corresponds to a linear equation with coefficient matrix 1 00 10, a diagonal coefficient matrix. And b equals 0. So the solution of the system of linear equations is 00. We start with an initial guess up here, OK. And we try steepest descent with some small step size, right. You'll follow this blue path here. And you can see what happened. That step size was reasonable as we moved along the steepest ascent direction where the contours were pretty narrowly spaced. But as we got down to the flatter section, OK, as we got down to the flatter section of our objective function, those steps are really small. Right? We're headed in the right direction, we're just taking very, very small steps. If you apply this conjugate gradient methodology, well, the first step you take, that's prescribed. You've got to step in some direction. The second step you take though minimizes completely along this direction. So the first step was the same for both of these. But the second step was chosen to minimize completely along this direction. So it's totally minimized. And the third step here also steps all the way to the center. So it shows a conjugate direction that stepped from here to there. And it didn't lose any of the minimization in the original direction that it proceeded along. So that's conjugate gradient. It's used to solve linear equations with order n iterations, right. So A has at most n independent eigenvectors, independent directions that I can step along and do this minimization. The conjugate gradient method is doing precisely that. Doesn't know what the eigendirections are, but it's something along these conjugate directions as a proxy for the eigendirections. So it can do minimization with just n steps for a system of n equations for n unknowns. It requires only the ability to compute the product of your matrix A with some vector, right. All the calculations there are only depended on the product of A with a vector. So don't have to store A, we just have to know what A is. We have some procedure for generating A. Maybe A is a linear operator that comes from a solution of some differential equations instead, right. And we don't have an explicit expression for A, but we have some simulator that produces, take some data, and projects A to give some answer, right. So we just need this product. We don't have to store A exactly. It's only good for symmetric positive definite matrices, right. This sort of free energy functional that we wrote or objective function we wrote only admits symmetric matrices which are positive definite. That's the only way it will have a minimum. And so the only way a steepest descent or descent type procedure is going to get to the optimum. But there are more sophisticated methods that exist for arbitrary matrices. So if we don't want symmetry or we don't care about whether it's positive definite, there are equivalent sorts of methods that are based around the same principle. And it turns out, this is really the state of the art. So if you want to solve complicated large systems of equations, you know Gaussian elimination, that will get you an exact solution. But that's often infeasible for the sorts of problems that we're really interested in. So instead, you use these sorts of iterative methods. Things like Jacobi and Gauss-Seidel, they're sort of the classics in the field. And they work, and you can show that they converge on are lots of circumstances. But these sorts of iterative methods, like conjugate gradient and its brethren other Krylov subspace methods they're called, are really the state of the art, and the ones that you reach to. You already did conjugate gradients in one homework, right. You used this PCG iterative method in Matlab to solve a system of linear equations. It was doing this, right. This is how it works. OK? OK. OK, so that's country ingredients. You could apply it also to objective functions that aren't quadratic in nature. And the formulation changes a little bit. Everywhere where the matrix A appeared there needs to be replaced with the Hessian at a certain iterate. But the same idea persists. It says well, we think in our best approximation for the function that we've minimized as much as we can in one direction. So let's choose a conjugate direction to go in, and try not to ruin the minimizations we did in the direction we were headed before. Of course, these are all linearly convergent sorts of methods. And we know that there are better ways to find roots of non-linear equations like this one, grad f equals zero, namely the Newton-Raphson method, which is quadratically convergent. So if we're really close to a critical point, and hopefully that critical point is a minima in f, right, then we should rapidly converge to the solution of this system of nonlinear equations just by applying the Newton-Raphson method. It's locally convergent, right. So we're going to get close. And we get quadratic improvement. What is the Newton-Raphson iteration, though? Can you write that down? What is the Newton-Raphson iteration that's the iterative math for this system of non-linear equations, grad f equals 0? Can you work that out? What's that look like? Have we got this? What's the Newton-Raphson iterative map look like for this system of non-linear equations? Want to volunteer an answer? Nobody knows or nobody is sharing. OK, that's fine. Right, so we're trying to solve an equation g of x equals 0. So the iterative map is Xi plus 1 is Xi minus Jacobi inverse times g. And what's the Jacobian of g? What's the Jacobian of g? The Hessian, right. So the Jacobian of g is the gradient of g, which is two gradients of f, which is the definition of the Hessian. So really, the Newton-Raphson iteration is Xi plus 1 is Xi minus Hessian inverse times g. So the Hessian plays the role of the Jacobian, the sort of solution procedure. And so everything you know about Newton-Raphson is going to apply here. Everything you know about quasi-Newton-Raphson methods is going to apply here. You're going to substitute for your nonlinear. The nonlinear function you're finding the root for, you're going to substitute the gradient. And for the Jacobian, you're going to substitute the Hessian. Places where the Hessian is, the determent of the Hessian is 0, right it's going to be a problem. Places where the Hessian is singular is going to be a problem. Same as with the Jacobian. But Newton-Raphson has the great property that if our function is quadratic, like this one is, it will converge in exactly one step. So here's steepest descent with a fixed value of alpha, Newton-Raphson, one step for a quadratic function. And why is it one step? STUDENT: [INAUDIBLE] JAMES SWAN: Good. So when we take a Taylor expansion of our f, in order to derive the Newton-Raphson step, we're expanding it out to quadratic order, its function is quadratic. The Taylor expansion is exact. And the solution of that equation, right, gradient f equals 0 or g equals 0, that's the solution of a linear equation, right. So it gives exactly the right step size here to move from an initial guess to the exact solution or the minima of this equation. So for quadratic equations, Newton-Raphson is exact. It doesn't go in the steepest ascent direction, right. It goes in a different direction. It would like to go in the steepest descent direction if the Jacobian were identity. But the Jacobian is a measure of how curved f is. The Hessian, let's say, is a measure of how curved f is. Right? And so there's a projection of the gradient through the Hessian that changes the direction we go in. That change in direction is meant to find the minimum of the quadratic function that we approximate at this point. So as long as we have a good quadratic approximation, Newton-Raphson is going to give us good convergence to a minima or whatever nearby critical point there is. If we have a bad approximation for a quadratic, then it's going to be so good, right. So here's this very steep function. Log of f is quadratic, but f is exponential in x here. So you got all these tightly spaced contours converging towards a minima at 00. And here I've got to use the steepest descent step size, the optimal steepest descent step size, which is a quadratic approximation for the function, but in the steepest descent direction only. And here's the path that it follows. And if I applied Newton-Raphson to this function, here is the path that it follows instead. The function isn't quadratic. So these quadratic approximations aren't-- they're not great, right. But the function is convex, right. So Newton-Raphson is going to proceed downhill until it converges towards a solution anyways. Because the Hessian has positive eigenvalues all the time. Questions about this? Make sense? OK? So you get two different types of methods that you can play with. One of which, right, is always going to direct you down hill. Steepest descent will always carry you downhill, right, towards a minima. And the other one, Newton-Raphson, converges very quickly when it's close to the root. OK, so they each have a virtue. And they're different. They're fundamentally different, right. They take steps in completely different directions. When is Newton-Raphson not going to step down hill? STUDENT: [INAUDIBLE] What's that? STUDENT: [INAUDIBLE] JAMES SWAN: OK, that's more generic an answer than I'm looking for. So there may be circumstances where I have two local minima. That means there must be maybe a saddle point that sits between them. Newton-Raphson doesn't care which critical point it's going after. So it may try to approach the saddle point instead. That's true. That's true. Yeah? STUDENT: When Hessian [INAUDIBLE].. JAMES SWAN: Good, yeah. With the Hessian doesn't have all positive eigenvalues, right. So if all the eigenvalues of the Hessian are positive, then the transformation h times g or h inverse times g, it'll never switch the direction I'm going. I'll always be headed in a downhill direction. Right? In a direction that's anti-parallel to the gradient. OK? But if the eigenvalues of the Hessian are negative, if some of them are negative and the gradient has me pointing along that eigenvector in a significant amount, then this product will switch me around and will have me go uphill instead. It'll have me chasing down a maxima or a saddle point instead. That's what the quadratic approximation of our objective function will look like. It looks like there's a maximum or a saddle instead. And the function will run uphill. OK? So there lots of strengths to Newton-Raphson. Convergence is one of them, right. The rate of convergence is good. It's a locally convergent, that's good. It's got lots of weaknesses, though. Right? It's going to be a pain when the Hessian is singular at various places. You've got to solve systems of linear equations to figure out what these steps are. That's expensive computationally. It's not designed to seek out minima, but to seek out critical points of our objective function. Steepest descent has lots of strengths, right. Always heads downhill, that's good. If we put a little quadratic approximation on it, we can even stabilize it and get good control over the descent. Its weaknesses are it's got the property that it's linearly convergent instead of quadratically convergent when it converges. So it's slower, right. It might be harder to find a minima. You've seen several examples where the path sort of peters out with lots of little iterations, tiny steps towards the solution. That's a weakness of steepest descent. We know that if we go over the edge of a cliff on our potential energy landscape, steepest descent it just going to run away, right. As long as there's one of these edges, it'll just keep running downhill for as long as they can. So what's done is to try to combine these methods. Why choose one, right? We're trying to step our way towards a solution. What if we could craft a heuristic procedure that mixed these two? And when steepest descent would be best, use that. When Newton-Raphson would be best, use that. Yes? STUDENT: Just a quick question on Newton-Raphson. JAMES SWAN: Yes? STUDENT: Would it run downhill also if you started it over there? Or since it seeks critical points, could you go back up to the [INAUDIBLE].. JAMES SWAN: That's a good question. So if there's an asymptote in f, it will perceive the asymptote as a critical point and chase it. OK? And so if there's an asymptote in f, if can perceive that and chase it. It can also run away as it gets very far away. This is true. OK? The contour example that I gave you at the start of class had sort of bowl shape functions superimposed with a linear function, sort of planar function instead. For that one, right, the Hessian is ill-defined, right. There is no curvature to the function. But you can imagine adding a small bit of curvature to that, right. And depending on the direction of the curvature, Newton-Raphson may run downhill or it may run back up hill, right? We can't guarantee which direction it's going to go. Depends on the details of the function. Does that answer your question? Yeah? Good. STUDENT: Sir, can you just go back to that one slide? JAMES SWAN: Yeah. I'm just pointing out, if the eigenvalues of h is further negative, then the formula there for alpha could have trouble too. JAMES SWAN: That's true. STUDENT: Similar to how the Newton-Raphson had trouble. JAMES SWAN: This is true. This is true, yeah. So we chose a quadratic approximation here, right, for our function. We sought a critical point of this quadratic approximation. We didn't mandate that it had to be a minima. So that's absolutely right. So if h has negative eigenvalues and the gradient points enough in the direction of the eigenvectors associated with those negative eigenvalues, then we may have a case where alpha isn't positive. We required early on that alpha should be positive for steepest descent. So we can't have a case where alpha is not positive. That's true. OK. So they're both interesting methods, and they can be mixed together. And the way you mix those is with what's called trust-region ideas, OK. Because it could be that we've had an iteration Xi and we do a quadratic approximation to our functional, which is this blue curve. Our quadratic approximation is this red one. And we find the minima of this red curve and use that as our next best guess for the solution to the problem. And this seems to be working us closer and closer towards the actual minimum in the function. So since quadratic approximation seems good, if the quadratic approximation is good, which method should we choose? STUDENT: Newton-Raphson. JAMES SWAN: Newton-Raphson, right. Could also be the case though that we make this quadratic approximation from our current iteration, and we find a minimum that somehow oversteps the minimum here. In fact, if we look at the value of our objective function at this next step, it's higher than the value of the objective function where we started. So it seems like a quadratic approximation is not so good, right. That's a clear indication that this quadratic approximation isn't right. Because it suggested that we should have had an minima here, right. But our function got bigger instead. And so in this case, it doesn't seem like you'd want to choose Newton-Raphson to take your steps. The quadratic approximation is not so good. Maybe just simple steepest descent is a better choice. OK, so it's done. So if you're at a point, you might draw a circle around that point with some prescribed radius. Call that Ri. This is our iterate Xi. This is our trust-region radius Ri. And we might ask, where does our Newton-Raphson step go? And where does our steepest descent step take us? And then based on whether these steps carry us outside of our trust-region, we might decide to take one or the other. So if I set a particular size Ri, particular trust-region size Ri and the Newton-Raphson step goes outside of that, we might say well, I don't actually trust my quadratic approximation this far away from the starred point. So let's not take a step in that direction. Instead, let's move in a steepest descent direction. If my Newton-Raphson step is inside the trust-region, maybe I'll choose to take it, right. I trust the quadratic approximation within a distance Ri of my current iteration. Does that strategy makes sense? So we're trying to pick between two different methods in order to give us more reliable convergence to a local minima. So here's our Newton-Raphson step. It's minus the Hessian inverse times the gradient. Here's our steepest descent step. It's minus alpha times the gradient. And if the Newton-Raphson step is smaller than the trust-region radius, and the value of the objective function at Xi, plus the Newton-Raphson step is smaller than the current objective function, it seems like the quadratic approximation is a good one, right. I'm within the region in which I trust this approximation, and I've reduced the value of the function. So why not go that way, right? So take the Newton-Raphson step. Else, let's try taking a step in the steepest direction instead. So again, if the steepest ascent direction is smaller than Ri and the value of the function in the steepest descent direction, the optimal steepest descent direction or the optimal step in the steepest ascent direction is smaller than the value of the function at the current point, seems like we should take that step. Right? The Newton-Raphson step was no good. We've already discarded it. But our optimized steepest descent step seems like an OK one. It reduces the value of the function. And its within the trust-region where we think quadratic approximations are valid. If that's not true, if the steepest descent step takes us outside of our trust-region or we don't reduce the value of the function when we take that step, then the next best strategy is to just take a steepest ascent step to the edge of the trust-region boundary. Yeah? STUDENT: Is there a reason here that Newton-Raphson is the default? JAMES SWAN: Oh, good question. So eventually we're going to get close enough to the solution, all right, that all these steps are going to live inside the trust-region ring. Its going to require very small steps to converge to the solution. And which of these two methods is going to converge faster? STUDENT: Newton-Raphson. JAMES SWAN: Newton-Raphson. So we prioritize Newton-Raphson over steepest descent. That's a great question. Its the faster converging one, but its a little unwieldy, right. So let's take it when it seems valid. But when it requires steps that are too big or steps that don't minimize f, let's take some different steps instead. Lets use steepest descent as the strategy. So this is heuristic. So you got to have some rules to go with this heuristic, right. We have a set of conditions under which we're going to choose different steps. We've got to set this trust-region size. This Ri has to be set. How big is it going to be? I don't know. You don't know, right, from the start you can't guess how big Ri is going to be. So you got to pick some initial guess. And then we've got to modify the size of the trust-region too, right. The size of the trust-region is not going to be appropriate. One fixed size is not going to be appropriate all the time. Instead, we want a strategy for changing its size. So it should grow or shrink depending on which steps we choose, right. Like if we take the Newton-Raphson step and we find that our quadratic approximation is a little bit bigger than the actual function value that we predicted, we might want to grow the trust-region. We might be more likely to believe that these Newton-Raphson steps are getting us to smaller and smaller function values, right. The step was even better than we expected it to be. Here's the quadratic approximation in the Newton-Raphson direction. And it was actually bigger than the actual value of the function. So we got more, you know, we got more than we expected out of a step in that direction. So why not loosen up, accept more Newton-Raphson steps? OK, that's a strategy we can take. Otherwise, we might think about shrinking instead, right. So there could be the circumstance where our quadratic approximation predicted a smaller value for the function than we actually found. It's not quite as reliable for getting us to the minimum. These two circumstances are actually these. So this one, the quadratic approximation predicted a slightly bigger value than we found. Say grow the trust-region, right. Try some more Newton-Raphson steps. Seems like the Newton-Raphson steps are pretty reliable here. Here the value of the function in the quadratic approximation is smaller than the value of the function after we took the step. Seems like our trust-region is probably too big if we have a circumstance like that. Should shrink it a little bit, right? We took the Newton-Raphson step, but it actually did worse than we expected it to do with the quadratic approximation. So maybe we ought to shrink the trust-regional a little bit. And you need a good initial value for the trust-region radius. What does Matlab use? It uses 1. OK. It doesn't know. It has no clue. It's just a heuristic. It starts with 1 and it changes it as need be. So this is how fsolve solves systems of nonlinear equations. This is how all of the minimizers in Matlab, this is the strategy they use to try to find minima. They use these sorts of trust-region methods. It uses a slight improvement, which is also heuristic, called a dogleg trust-region method. So you can take a Newton-Raphson step or you can take a steepest descent step. And if you found the steepest descent step didn't quite get you to the boundary of your trust-region, you could then step in the Newton-Raphson direction. Why do you do that? I don't know, people have found that it's useful, right. There's actually no good reason to take these sorts of dogleg steps. People found that for general, right, general objective functions that you might want to find minima of, this is a reliable strategy for getting there. There's no guarantee that this is the best strategy. These are general non-convex functions. These are just hard problems that one encounters. So when you make a software package like Matlab, this is what you do. You come up with heuristics that work most of the time. I'll just provide you with an example here, OK. So you've seen this function now several times. Let's see, so in red covered up back here is the Newton-Raphson path. In blue is the optimal steepest descent path. And in purple is the trust-region method that Matlab uses to find the minima. They all start from the same place. And you can see the purple path is a little different from these two. If I zoom in right up here, what you'll see is initially Matlab chose to follow the steepest descent path. And then at a certain point it decided, because of the value of the trust-region that Newton-Raphson steps were to be preferred. And so it changed direction and it started stepping along the Newton-Raphson direction instead. It has some built in logic that tells it when to make that choice for switching based on the size of the trust-region. And the idea is just to choose the best sorts of steps possible. Your best guess at what the right steps are. And this is all based around how trustworthy we think this quadratic approximation for objective function is. Yeah, Dan? STUDENT: So for the trust-region on the graph what Matlab is doing is at each R trust-region length it's reevaluating which way it should go? JAMES SWAN: Yes. Yes. It's computing both sets of steps, and it's deciding which one it should take, right. It doesn't know. It's trying to choose between them. STUDENT: Why don't you do the Newton-Raphson step through the [? negative R? ?] JAMES SWAN: You can do that as well, actually, right. But if you're doing that, now you have to choose between that strategy and taking a steepest descent step up to R as well, right. And I think one has to decide which would you prefer. It's possible the Newton-Raphson step also doesn't actually reduce f. In which case, you should discard it entirely, right. But you could craft a strategy that does that, right. It's still going to converge, likely. OK? OK. I've going to let you guys go, there's another class coming in. Thanks.
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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 our example is a well-known one. But it's important to understand it. And it's a good example. It's a familiar one. It's the hydrogen molecule ionized. And we spoke a little about it last time. So let's do that now. So H2 plus, hydrogen ion. So what is this system? Well, it is two protons sharing one electron. That's basically it. It's the fact that if you have a hydrogen atom-- this is hydrogen-- you could bring another proton and form a bound state. That's an important thing in order to think about it-- you see, you can ask what is the bound state energy of your system? It's the energy you need to dissociate your system. But this system, if you have two protons-- proton, proton-- and one electron, it can be dissociated in many ways. It can be dissociated by just totally separating it and destroying it. But it can be dissociated or liberated by just removing the proton. So this is already a bound state. So if you think of energies, you have the bound state the energy of hydrogen here, hydrogen itself, and the H2 ion-- H2 plus-- must be lower because you could imagine taking the proton out and somehow this system is a configuration in which this proton is captured here. So you should get lower energy than the hydrogen atom energy. So that's the hydrogen ion. And that should be still more energy than if you bring another electron, which is the H2 molecule. You bring another electron. This should bind this and the other electron and lower its energy even more. So how does one solve for this hydrogen ion? Well, as we discussed, you first solve for the electronic wave function when the protons are separated at distance r. So we said fix the lattice. So here it is. The lattice is fixed. It's a separation capital R. And now I'm supposed to find the electron wave function. Easy? No. It already is hard. We spent several lectures-- you've been studying this three times already, maybe, in quantum mechanics, the hydrogen atom, one proton, one electron. To protons, one electron is much harder. I don't think there's a simple analytical way to solve it. So already at this step we have trouble solving it. What is the electron, Hamiltonian in this case? The electron Hamiltonian is just one Laplacian here minus h squared over 2m for the electron variable, and then the potential that is the interactions of the electron with the two nuclei. So if we call this distance r1 and this distance r2, that is the whole electron Hamiltonian. Now, don't think of r1 and r2 as you have two variables, two positions. No, there is just one position. The dynamical variables for this electron, this p and r for this electron-- r is the position of the electron. It happens to be that r is equal to r1. And if you know-- so r1 is really the length of r, the position of the electron. And r2 is the length of capital R, if we wish, all the way here, minus r. r is the vector here, as well. So there is r1 and r2, but it's not two electrons or two variables. It's just two distances that depend on the single position of the electron. And this is the momentum operator of the electron. OK, so you can't solve this. That's life. It can't be solved. But we can do something and we can do something variational. We can try to find some kind of approximation for the state of electrons and use that. So there is an [INAUDIBLE] that you could try, a variational wave function, for even the electron. You see, in our argument in this lecture, we did the variation and approximation for the nuclear degrees of freedom or for that thing. But even for the electron wave function, I can't solve them. So I have to do a variational method. Now, these variational methods have become extremely sophisticated. People do this with a series of Eigenfunctions and find answers that converge with 15 digits accuracy. It's just unbelievable what people can do by now with this. It's a very nice and developed field. But we'll do a baby version of it. So this baby version is going to stay. My wave function for the electron as a function of position and as a function of the separation is going to be simple. It's going to be a sum of ground state wave functions. It's going to be a number-- a, that's just a number-- times the ground state wave function based on the first proton plus the ground state wave function based on the second proton. So these psis, or psi0 of r is the ground state wave function of hydrogen pi a 0 cubed e to the minus r over a0. This is called-- people have given it a name, even though it's-- you would say it's not that original to put some wave function like that. This is a simple approximation. And the technique is called LCAO technique. And this calls for Linear Combination of Atomic Orbitals. A big name for a rather simple thing. It has some nice things about this wave function. This system is invariant. This molecule is invariant. And they're taking a reflection around this thing changing the first proton and the second proton. So that's the symmetry of your Hamiltonian. And it's the symmetry of [INAUDIBLE] with the Hamiltonian. So you can demand that your wave functions have that symmetry. And this is nice here because if you change r1 and r2, the wave function is invariant. So that's a very nice thing about this wave function. It shows-- you see, you can put in the variational method anything and it will still give you some answer. But if you put something that mimics the real wave functions and the real wave function of this system is going to have a symmetry under the exchange of the two protons. They're identical. So this lattice that you've created has a symmetry. So the electron configuration has to respect that symmetry. So that's very nice. Now, this is electronic wave function. So I ask you, is this wave function better where the protons are far away or when the protons are very close to each other? Give you a minute to think about it. When is this approximation or this wave function better? If the distance between the protons is very little or the distance between the protons is very large? You're right in that when the two things collapse, this looks like a true wave function because this is a solution when there's a single nucleus and a single electron. But this is the solution for hydrogen. And it gives you the ground state of hydrogen. But when the two protons collapse, it has become helium nucleus. It has two protons. So it doesn't have the right decay rate. It doesn't have the right Born radius. On the other hand, when they're far apart, it does have the right thing. So this is very good. It's excellent when the things are far away. But it's not great when they're close together. So here you go. You have to normalize this wave function. Even that takes some effort. It turns out that the value of a for normalization-- the value of a is 1 over 2 1 plus a constant i. And that constant i is e to the minus capital R over a not times 1 plus r over a not plus 1/3 of r over a not squared. Wow, so very funny. It's not so easy to even calculate the normalization of this thing. But that's the normalization. And then recall that what we have to do is just put the electronic Hamiltonian inside this wave function, this phi of r. And this is going to give us an energy which is approximately the electronic energy as a function of r. So you have to evaluate the Hamiltonian-- I'm sorry, that Hamiltonian there-- should have a square here. And now we evaluate this. We have to go even more into gross and get what is the potential energy contributed by the electronic cloud. So here is the function that you get. So our plot here I'll call-- this calls for a variable x, which is going to be the separation divided by a not. This is the separation between the protons divided by a not. So here is going to be the electronic energy as a function of x. And here is x. And all right. So as x goes to 0, means the nuclei are going on top of each other. That's the place where the wave function is not all that great. And that's not so good. It turns out that here you get minus 3 Rydbergs. Remember, the Rydberg was e squared over 2 a not. And it's about 13.6 ev. It's just the ground state energy of the hydrogen. You do get that the energy due to the electronic configuration is negative and it goes to this value. And then it starts to grow quadratically and goes up. And I want to know, in your opinion, what's going to be the next asymptote? So what is it going to asymptote as x goes to infinity? Let let's make the analogy. The right answer is it actually stops here. And you'll remember this problem that you've solved many times, probably. If you have a-- suppose a square well. You have a wave function that is like that. Now, if you have two square well-- and it has a ground state energy. If you have two square wells like that and you ask, what is the ground state energy? The ground state energy is roughly equal to the ground state energy for a single square well because what happens is that the wave function goes like that and then like-- well, actually, like that. So yes, the wave function spends equal time. But it's the same energy as if there would be a single square well. So here it is. The protons are very far away in the symmetric state, the ground state. The electron is half the times here, half the times there. But the energy is the same energy as if it would be in either one. So this is an intuition that you may have from 804 or 805. So here it is. That's what it does. That's good. In fact, if you ride the ee of x over Rydbergs, it behaves here like minus 3 plus x squared plus dot, dot, dot. So it starts growing and then asymptotes with an exponential there. OK, but what did we say was the Born-Oppenheimer approximation was the idea that then you have dynamics of the nuclei based on-- so Born-Oppenheimer tells you that the Hamiltonian for the nuclear degrees of freedom is given by h. This h effective is capital P over 2m squared plus the nuclear nuclear potential, plus this electronic energy. So this electronic energy we already calculated. The nuclear nuclear potential, in this case, is the proton proton potential, v m n is e squared over r. It is repulsive. You see, this-- if you wanted to minimize the electronic energy, you still don't get anything. You get a system that collapses to zero. That's certainly not the molecule. But this n n is this, so we can write it as e squared over 2a0 and then put a 2 over r over a 0 to get all these numbers nicely. So the nvv is a Rydberg times 2 over x. So in terms of x there, vnn is a 2 over x-- oops-- a 2 over x potential like that. So now you have the possibility of getting a stable grounding, so the potential for the nucleons. So the total potential for the nucleons is the sum of these two potentials. And how does it look? Well, some of the two potential-- so ee of x plus vnn of x. Let's divide by Rydberg as a function of x, goes-- here is minus 1. And it goes more or less like-- OK. Let me try to get this right. 1, 2, 3, 4, 5. OK. It goes up here, down, crosses the minus 1 line, and moves to a minimum here. And then it goes like that. Pretty much it's something like that. It's a rather little quadratic minimum. It's rather flat. And these are the numbers you care for. This is the minimum. Maybe it doesn't look like that in my graph too well. But it's a very flat minimum. And r over a0 at the minimum is, in fact, 2.49. And e over Rydberg at the minimum is minus 1.297. Those are results. Now, you have a potential. You could calculate the quadratic term around the potential. And then you get an approximate oscillation Hamiltonian that those would be the nuclear vibrations. You could calculate the frequency of the nuclear vibrations and the energy of the nuclear vibration. It's a very simple but nice model. You can calculate everything pretty much about this molecule and compare with experiments. So how does it do the comparison with experiment? It does OK. It doesn't do very well. So let me tell you what's happening. So again, we do get this system. Happily, the minimum was below minus 1. The minus 1 was the situation in which you have a hydrogen atom and a free proton. So there is a bound state energy but the bounds state energy is not conceptually right to think of it as this big energy. It's just this little part here in which this is the extra binding that you can have when you bring a proton near to the hydrogen atom. So this extra part-- so the bound state energy-- I'll write a couple of words here and we'll stop. The bound state energy is E. This E that we determined here at the minimum is minus 1 Rydberg minus 0.1297 Rydbergs which is minus 1 Rydberg minus 1.76 ev. So this is what you would call the bound state energy of this system because it's the least energy you need to start to dissociate it-- not to dissociate it completely, but to start to dissociate it. 1.76. True value is 2.8 ev, so not great, but sort of order of magnitude. Another thing that you can ask is, what is the separation between the nuclei in this ion? How far away are they? And this gives a prediction that it's about 2.49 a0. So separation is r equal 2.49 a0, which is 2.49 times 0.529 angstroms. It's about 1.32 angstroms. And through value's 2.8. Experimental value, the value of r, our experiment is not that far. It's 106 angstroms. So it's 20% there, 25% there. But it gives you a way to think about this system. You get all the physics. You understand the physics. The Born-Oppenheimer physics is that the electron energy in the fixed lattice is a potential term for the nuclei. Added to the nucleus nucleus repulsion gives you the total potential term for the nuclear degrees of freedom. You can find the stable situation. And if you want to do quantum mechanics and vibrations of the nuclei, well, do the harmonic oscillator associated with this thing. The nice thing is you can complicate these wave functions a little and get better and better answers. And it's fun. It's things that can be done numerically. And you have a remarkably powerful tool to understand these things. So with this, we'll close our whole chapter in adiabatic physics. And next time, we will begin with scattering.
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https://ocw.mit.edu/courses/5-08j-biological-chemistry-ii-spring-2016/5.08j-spring-2016.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. ELIZABETH NOLAN: We're going to end the unit on synthesis today. And the focus of today's lecture will really be looking at one system in detail and the types of experiments that are done to elucidate a biosynthetic pathway for a non-ribosomal peptide. And so just to recap from last time, if we think about studying assembly lines in lab, and we're thinking about this for a non-ribosomal peptide synthetase, what needs to be done? So first, it's necessary to, typically, overexpress and purify domains, didomains or modules. And so on Monday, it came up that often, these proteins are enormous and it's not possible or feasible to express entire modules, or entire proteins that have multiple modules. So oftentimes, people will look at individual domains, or didomains, which are smaller and more amenable, to overexpression in an organism like E. coli. Then it's necessary to assay for A domain activity. So we're called the A domains through the adenylation domains. And the question is, what monomer is selected and activated? And so the ATP-PPi exchange assay comes up here. There needs to be assays for loading of the T domain, or carrier protein, with the monomer. Assay for peptide bond formation, which is the condensation domain. And then often, some assay for chain released by the thioesterase domain. OK, so assay for TE activity. Chain release. And so in terms about of thinking about these T domains, we learned that these T domains need to be post-translationally modified with the Ppant arm, which means we need an enzyme called a PPTase. And so in many cases, we don't know what the PPTase is for a given gene cluster. And what's done, often in the lab, is that a PPTase from B. subtilis, named Sfp, is used in order to post-translationally modify the T domains with the Ppant arm. So there is a serine residue in these T domains that gets modified. We looked over that in a prior lecture. So this one is very useful. And if you don't know the enzyme to use, people will use recombinant Sfp And just recall, we have the T domain. There's a serine moiety. We have a PPTase. That's going to stick on the P pant arm here. So we call this apo, holo, and then the amino acid, or aryl acid monomer, in the case of NRPS gets loaded here via a thioester. And so Sfp can be used to get us here. And even what people have done is make modified analogs, where there's some R group. So you can imagine using chemical synthesis to load a monomer, or even some other type of group that, for some reason, you might want to transfer here. And this Sfp is very promiscuous and it can do that. And so the take-home here is if you need a PPTase, overexpress, purify, and utilize Sfp. Here's just an example for review, where we have a carrier protein, so a T domain, and we have the PPTase activity here, Sfp, attaching this Ppant arm. And here, it's described with an R group. And just to give you an example of possibilities here, there have been many reports of CoA analogs being transferred to T domains by Sfp. And these can range from things like an isotope label to peptides to steroids to some non-ribosomal peptide derivative, or a fluorophore. So this has been used as a tool. And you might ask, why is this possible? And if we just take a look at the structure of Sfp from B. subtilits with coASH and magnesium bound. What we see is that this end of the coA is extended out into the solvent. And at least in this structure here, it's not interacting with regions of the protein. So you can imagine that it's possible to attach some group, even a bulky group, here and be able to transfer it there. So where we're going to focus the rest of the lecture is on an assembly line responsible for the biosynthesis of a natural product called enterobactin, and this is a siderophore. And so in thinking about this, what I would just like to first note is that when we talk about these assembly lines, we can group them into two types, which are non-iterative and iterative assembly. And so what does this mean? So we've seen examples of non-iterative assembly last time on Monday with the ACV tripeptide and the vancomycin synthetase. So in these non-interative assembly lines, effectively, each step has its own module. So each carrier protein, T domain, each condensation or catalytic domain, is used only once as the chain grows. And we see the chain passed along from module to module here. So also, the PKS we looked at for synthesis of DEB is one of these non-iterative assembly lines. So in contrast, in the example we're going to look at today with the enterobactin synthetase is an iterative assembly line, and this is similar to what we saw in fatty acid synthase. So in these iterative assembly lines, effectively, only one module is employed over and over again. So you can have the same carrier protein and same catalytic domain used for multiple cycles of chain elongation. And that's what we saw in fatty acid synthase, where there are multiple cycles in addition of a C2 unit via the same domains. And so what we're going to see today is this type of iterative assembly is responsible for the synthesis of this molecule here. So first, just an overview of building blocks. And then we'll talk about why organisms want to make this molecule, and then focus on the biosynthetic logic and experiments. So this molecule, enterobactin, is produced from two monomers. So we have 2, 3 dihydroxybenzoic acid, or DHB, and we have serine here. And there is a two-module assembly line responsible for the synthesis of this natural product. And that assembly line is shown here. So we see that there's three proteins, EntE, EntB, and EntF. We have an initiation module, elongation module, and this TE domain for termination. So overall, three separate proteins, two modules, and seven domains. So this NRPS is quite small. And this is an example of a non-ribosomal peptide that's produced by E. coli. So E. coli makes this molecule, as well as some other gram-negative bacteria. So this is iterative. We have three of each of these monomers, yet only two T domains here, so imagine one responsible for each. So before we get more into this biosynthetic logic, let's just take a moment to think about why this molecule is produced. So this is a case where we actually have very good understanding about why an organism is producing a natural product. And this actually gives a segue into JoAnne's section on metal homeostasis, which will come up after cholesterol after spring break. So many bacteria use non-ribosomal peptide synthesis machinery in order to make chelators in order to acquire iron. And that's because iron is an essential nutrient and it's actually quite scarce. So if you imagine an organism in the soil, maybe it needs to obtain iron from a rock. Somehow it needs to get iron from our pool, and concentrations are very tightly regulated, and most iron is tightly bound. And we can also think about this from a standpoint of solubility, so simple KST type things. We all know that iron 3, which is the predominant oxidation state in aerobic conditions, is very insoluble. So our cars rust up here in the Northeast because they sit outside on the road in the winter, and that's no good. So we can think about 10 to the minus 18 molar. And then if we think about free ion in human serum, for instance, the concentration is even lower because there's inherent toxicity associated with free iron. And you'll hear about that from JoAnne in more detail later. So these organisms have a predicament because for metabolism, they need iron on the order of micromolar concentrations. So how does some organism obtain micromolar iron when in environments where, say, that's 10 to the minus 24 molar? And there's a number of strategies that come up, but one of the strategies is the biosynthesis of non-ribosomal peptides that act as metal scavengers and metal chelators. And so I just show you two examples here. And we have enterobactin, which we're going to focus on today. And this is really just a wonderful molecule. Yersiniabactin-- and I put this up here, in part, because there were some questions about those cyclization domains in the bleomycin gene cluster, that we looked at that assembly line on Monday. And this is another example where cyclization of cysteine residues occurs in order to give the final natural product via those modified condensation domains here. So if we think about enterobactin for a moment longer, what happens, effectively, this molecule can bind iron 3 with higher affinity. And the iron bound form is shown here. So these aryl acids, these catechol groups, provide six oxygen donors to the iron center to get a structure like this. So in terms of the organism in production, what happens when these organisms are confronted with iron limitations? So essentially, they're starved for essential nutrients. They'll turn on biosynthesis. So they'll express the enterobactin synthetase, which will allow for production of enterobactin. So this is happening in the cytoplasm. So we have those three proteins that comprise the assembly line that use the HDML serine to produce the natural product. And then in addition to that biosynthetic machinery, the organism needs to also express and use a whole bunch of transport machinery. So what happens is that this natural product is exported into the extracellular space. So this is a gram-negative organism, so it has an inner membrane and an outer membrane. And it's in the extracellular space that enterobactin will scavenge iron 3. So there's formation of the coordination complex, shown in cartoon here. And then there's a dedicated receptor on the outer membrane that will recognize that iron bound form and bring that into the cell. And then through transport and through breakdown of the natural product, this iron can be released and then used. So iron is a co-factor of many types of proteins and enzymes here. So a whole lot is going on. We're going to focus on the biosynthetic part. And so in thinking about this, from the standpoint of a non-ribosomal peptide synthetase, what's something interesting? So in the examples we saw last time, we had the ACD tripeptide, the vancomycin synthetase. These assembly lines are only forming peptide bonds, so we saw formation of amide bond. If we take a look at enterobactin and we think about the monomers coming from, what do we see? So this has some C3 symmetry. And we can see that it's comprised of three of these DHB serine monomers, so 1, 2, 3. And effectively, there's formation of amide bonds between DHC and serine. So it's shown and throughout here. But there's also ester linkages formed. So this ring here is often called a trilactone, or a macrolactone, and somehow, these three esters need to be formed. So how is the enterobactin synthetase doing this? So if we look at an overview of different enterobactin synthetase, the gene cluster, what do we learn? So the first point to make is that there are actually six proteins required. So you've seen three so far, in terms of the assembly line. So we have an A, B, C, D, E, and F. And A, B, and C are required for the biosynthesis of this aryl acid building block here, this DHB. And then this is a case, rather unusual, where the PPTase was identified, and we're going to talk about that more as we go through the experiment. So I just told you about using Sfp if you don't know what to do. This was the case where the researchers were able to identify the dedicated PPTase for the assembly line. So that's EntE. And then we have B, E, and F that provide an iterative assembly line that yields the natural product, as shown here. OK, so also just note that B is coming up twice. We're seeing it here and we're seeing it here. So that should bring up a question, what's going on with this enzyme? And then we'll address that as we move forward. So in terms of thinking about this synthetase, we'll do an overview and then look at the experiments. So we have an A domain and B. We have a protein here that has a T domain and an IDL domain that we'll get back to. This is EntE, and then here we have EntF. And then we have our PPTase. So effectively, here, we can have our initiation. Here, we have elongation. And here, termination. So what is the overview, in terms of what happens for A domain activity? Loading of the T domains and peptide bond formation. So for the overview, we'll first consider getting a monomer on to EntB. So EntB has a T domain. And that has a serine. The serine needs to be modified under the PPTase EntD. Holo EntD. We put the Ppant arm. And then what we'll see is that EntE is the A domain that's responsible for activating DHB and transferring that monomer to EntB. So then in terms of EntF and getting the two domains of EntF loaded, it's going to be loaded with L-serine. And so here, you have EntF, again, focusing on the T domain. Again, we have that action of EntD to give us the holo form with the Ppant arm. And then we see that, in this case, the A domain is within the same protein. So the A domain of EntF is going to activate L-serine and transfer that to the T domain. So we have EntF, A domain to get us here. So then what about peptide bond formation? So we see the C domain, condensation domain as an EntF. And so what we can imagine is that we have our EntB loaded with the aryl acid monomer plus EntF loaded with L-serine. And what's going to happen? The C domain of EntF is going to catalyze formation of the amide bond here to give us EntB plus EntF, effectively, with DHA serine attached. So this gives us some insight, just this overview, in terms of how the amide bond is formed and pretty much follows what we saw for the ACV tripeptide and vancomycin biosynthesis for the heptapeptide that forms its backbone. So a question we have at this stage is, well, we see in that structure, in addition to these amides, there's also esters. How are those formed? And then what assays are needed? And so first, we're going to think about formation of the ester linkages, and then we'll launch into the experiment. So let's take a look at this assembly line. So we have EntE, the A domain, EntB, this didomain. That has the T domain. And here's EntF. And we see in this cartoon, the T domains are already modified with the P pant arm. And here is the serine residue of the TE domain that, ultimately, accepts the chain. So what happens? If we take a look, so we saw this on the board, EntE becomes loaded with dihydroxybenzoic acid. EntF becomes loaded with serine. The condensation domain catalyzes this formation of an amide bond between two monomers. And then what happens? We see transfer of this DHB serine unit to the TE domain here. And then we can imagine these two domains being loaded with monomers again. And what happens? What do we think about this? Effectively, formation of one amide bond transferred to the TE domain. Formation of another amide bond. And look. The second moiety here is transferred to the TE domain, to the initial monomer, via this ester linkage. This is really unusual behavior for a TE domain. And what happens again? We see this happen again, so we get this linear trimer of enterobactin, effectively. And then what happens? Chain release to form the macrolactone here. We have this group that can come around here. So what is the hypothesis? The hypothesis that was put forth by the researchers is that in this assembly line, effectively, this thioesterase is serving as a waiting room. And it's allowing these DHB serine monomers to wait around and remain attached, such that these esters can be formed. And somehow, it senses this appropriate size, this linear trimer, and then catalyzes chain release, as shown here. AUDIENCE: Does it mess up? ELIZABETH NOLAN: Does it mess up? AUDIENCE: Yeah, does it always give you a 3 under [INAUDIBLE] circle? ELIZABETH NOLAN: Yes, to the best of my knowledge. What's very interesting is that-- so this is worked out as Chris Walsh's group. Recently, Alison Butler's lab at Santa Barbara has discovered an enterobactin analog that has an additional unit in it here. So it looks like there's other thioesterases around that serve as waiting rooms and can accommodate different ring sizes. But this one will just give this size. And that's a very interesting question, just in terms of, how is this thioesterase doing that? We need more structural understanding to get at that. In addition, these are just some overviews that I've put in the notes, other depictions of this process and the waiting room hypothesis from the literature. So we're going to look at the experiments that were done to study this. And I really, 1, like enterobactin, so I got excited about this molecule as an undergraduate, actually. But beyond that, why I like to present on this system, in terms of experiments, is that many firsts came from it, and it really serves as a paradigm for many, many other studies. So if we just consider the various firsts that came from the studies of the enterobactin synthetase, 1, it was the first siderophore synthetase to be studied, and there's hundreds of siderophores out there and many have been investigated since this one. It's the first example of a siderophore synthetase to be characterized for the Ppant arms. This was the first identification of a dedicated PPTase for one of these assembly lines. And the first identification of the thioesterase domain that has this behavior of forming this cyclo-oligomer. And the first identification of an aryl carrier protein, so this T domain that carries DHB here. And in terms of the experiments we'll go through, these experiments that were devised in this system have been generalized across many, many assembly lines and the methods are still routinely used today. But a major difference I want to point out is that today, we have so many microbial genomes sequenced that a lot of work is driven by bioinformatics here, in terms of identifying that NRPS. So often, the gene cluster may be identified well before the natural product is ever isolated. And this is a case where the natural product isolation occurred first, so that was done well, well before here. And this is a case where we know how to get the organism to produce this natural product. You starve the organism of iron and it will start to make it, in many instances, for other natural products produced by these assembly lines. We don't know how to get the organism to actually make the molecule in a laboratory setting there. So there's some interesting work being done about that. Some actually recent work out of Northeastern, actually trying to grow organisms in soil-like environments and seeing what they can be provoked to produce. If you're curious, I'm happy to give you references. OK, so where are we going to start, in terms of characterization of this synthetase here? We're, more or less, going to follow the logic outlined up here for this. So here's the cartoon of the players. And the first order of business is that it's necessary to characterize the adenylation domains. So we need to ask, what the monomers are selected and activated? And we have two adenylation domains to consider, so EntE and the A domain of EntF. So what was done? For EntE, where we'll start, this protein was purified from E. coli and characterized here. And so how was it characterized? It was characterized by ATP-PPi exchange, like what we saw for the amino acyltransferase synthetase characterization. And so what was observed is that when EntE was combined with dihydroxybenzoic acid, ATP and radiolabeled PPi, there was incorporation of the P32 label into ATP. So that indicates formation of this adenylate intermediate. And resulted in the conclusion that EntE is the A domain that activates this aryl acid monomer, so this chemistry, which should be very familiar at this stage based on our discussions in the translation unit. So what about EntF and its A domain? So again, we're working with E. coli proteins. EntF was purified from E. coli. And again, this ATP-PPi assay was performed. And so in this case, what was observed is that when EntF was incubated with L-serine, ATP, and radiolabeled PPi, there was incorporation of the radiolabel into ATP, which indicates that EntF, its A domain is responsible for that activation of L-serine. And so you can imagine in each set of experiments, the researchers also tried the other monomer, and in the case of EntF, would have seen no ATP-PPi exchange with DHB. And likewise, for EntE, if they tried with L-serine, there would be no exchange. You'd want to see that, in terms of making a robust conclusion here for that. So that's good. Now, the next question is we need to get these monomers to these T domains here. And so that's the next step, is to study the T domain. And something you all need to appreciate about the time of this work, there wasn't a whole lot known about PPTases. There wasn't Sfp that you could borrow from your lab mate, or maybe you've expressed 100 milligrams for yourself and you could get that Ppant arm on here. And so JoAnne may want to elaborate, but there were a lot of effort to try to figure out, what is going on here? JOANNE STUBBE: And graduate students had no thesis. Because they couldn't get any activity of any of the enterobactin genes. ELIZABETH NOLAN: Yeah. JOANNE STUBBE: Until it was discovered what was going on. ELIZABETH NOLAN: Right. So this was a major, major effort, undertaking, and discovery here. And so they couldn't find activity, and that's because these two domains needed to be modified and they weren't modified. But some little detective work here. So in the analysis of purified EntF, what analysis of this purified protein had revealed, in some instances, with substoichiometric phosphopantetheine. And so is that a contamination or is that meaningful? In this case, it was a meaningful observation that, when chased, proved to be very hopeful. It suggested that maybe there's a T domain here that's modified. That's something we can infer from this. So if this Ppant arm is attached to EntF, how does it get there? And if we rewind and think about what was going on at the time, it was only shortly before that the PPTase for the acyl carrier protein and fatty acid synthetase was discovered. So that might beg the question, is it possible that this enzyme also modifies EntF, if you don't know much about its substrate scope? And so that hypothesis was tested and it didn't pan out. So if EntF was incubated with ACPS from fatty acid biosynthesis and coASH, there was no product formation. There was no transfer of the Ppant arm to here. Yeah? AUDIENCE: Was it obvious the fatty acid synthesis-- was there [INAUDIBLE] synthesis at the time, or did it have a name? ELIZABETH NOLAN: I don't think it had a name, but I defer to JoAnne, who was on the thesis committee, because this is really the first one. AUDIENCE: Were the analogs of mercury obvious at the time? ELIZABETH NOLAN: No, it's really discovery work at this stage. The question is, is there a possible lead from somewhere? And if you try it, what will happen? And really, there is no clue as to what is this modification and that design involved. But if you see an enzyme with activity in one system, maybe it will be active with another. Maybe not. And in this case, it didn't work, but it was something certainly worthwhile to try. So then what was done? So there is a search for another PPTase, and this was done using BLAST. And what BLAST, this bioinformatics, revealed was the identification of that enzyme EntD. So here, we have this EntD, the PPTase here. And so EntD was overexpressed and purified. And so in this case, a histag was used, affinity column purification. And the question is, what happens if we incubate EntD with EntF and coASH? And so in these experiments, radiolabeled coASH was used, and radio labels are commonly used to look for transfer of either Ppant arms or monomers, as we'll see as we go forward, to these domains. And so the question is, will we see transfer of the radiolabel to EntF in the presence of EntD and coASH? And so here are the results from the experiment. So we have formation of holo EntF, as monitored by the radiolabel, versus time. And so what's done, the reaction is run for a given time point. The reaction is quenched with acid to precipitate the proteins. And then you can imagine measuring radioactivity in the pellet. coASH will remain in the soluble fraction and then protein in the pellet here. You can imagine control assays with EntD included there. And so what do we see? So as I said before, that we tried the acyl carrier, ACPS from fatty acid synthesis. There's no reaction. But look. When EntD is present, we see transfer of this Ppant arm to the protein here. So this was a really exciting result at the time. We have a new enzyme, a new activity, this post-translational modification there. And this opens the door to further studies. If you can get the Ppat arm on, then we can look at loading the monomers here. So what's the next step? We have EntF loaded. We're also going to want to try to load EntD-- EntB, excuse me-- here. But of course, you need to know some more about EntB. And so let's think about that. I'll also note, just noted here, the next step is to look at loading of L-serine onto this moiety here, as drawn. And you can think about how to do that experimentally. So what about EntB? This was another mystery, in terms of experimental work and exploration. And so initially, EntB was purified and characterized for its activity that led towards the biosynthesis of the BHP monomer. So this ICL domain is involved in the series of reactions that give DHB. On a historical note, it was thought there was another protein. And this protein was named EntG that was thought to be required for enterobactin biosynthesis. And EntG would be the T domain that is for the aryl acid. So effectively, it would be this T domain, or aryl carrier protein for dihydroxybenzoic acid. But the problem was they couldn't find a gene for EntG. And so as it turned out, what more detective work showed is that this EntG is actually just the N-terminus of EntB here. So they realize that EntB has another role, another function, and that in addition to having this function and the synthesis of dihydroxybenzoic acid, because of this domain at the N-terminus, it's also the carrier protein for this monomer here. So how is this sorted out here? What we can do is just take a look at a sequence alignment. And so this is from one of the papers about all of these explorations. And effectively, what we're taking a look at are known [INAUDIBLE] phosphopantetheinylation sites, the proteins. So something is known about fatty acid synthesis and some other carrier proteins here from different organs. And so effectively, if we just look at this region of the alignment, we see this serine residue with an [INAUDIBLE] And this happens to be serine 245 of EntB. So this led to the hypothesis that maybe this serine 245 towards the C-terminus terminus of EntB is the site where the Ppant arm is attached here. And so this means that some experiments are needed to show that EntB has this carrier protein, or T domain, and that it can be modified with the Ppant arm. And it was predicted EntD would do this. And also, that once modified with the Ppant arm, the aryl acid can be transferred to EntD. So if we just think about EntB for a minute, So have the N-terminal domain. Here's the C-terminal domain. Here's the ICL domain. Here's the T domain for an aryl carrier protein. So amino acid 1, 285. This is 188. It's not quite drawn to scale. So we serine 245 around here, which is the serine of interest for post-translational modification with the Ppant arm. And so what was done is that pathways were performed, where EntB was incubated with EntE and radiolabeled coASH, like what we saw for the studies of EntF. But they made a few additional constructs. So they considered full length EntB, so as shown here. They considered an EntD variant where with C-terminal 25 amino acids were deleted. So you can imagine, they just put a stop codon in and leave the last 25 amino acids. So the serine is still there, but a bunch of the C-terminal residues aren't there. And then they also considered a variant of EntB where they deleted this entire N-terminal domain. And so the question is, what are the requirements? Well 1, does this reaction work? Does EntD modify EntB with the Ppant arm? And then if yes, what are the requirements? So is the N-terminal domain needed? Are these C-terminal residues needed? And so these are the gels that come from this experiment. And so what we're looking at on top are total proteins, so Coomassie staining. And on the bottom, we're looking at radioactivity. And the idea here is that we want to track the radiolabels. So in lane 1, we have assays with full length EntB. In lane 2 with the C-terminal truncation. And in lane 3, deletion of the N-terminal domain. OK, so the question is, what do we see from these data here? And so these give us a sense as to where the individual proteins run on the gel. And here, we're looking at the radioactivity. So what do we see? In lane 1, you see a huge blob of radioactivity. This isn't the most beautiful gel, actually. Nevertheless, there's much to learn. So what do we see? We see radioactivity associated with EntB. That's really good news. We see transfer of this radiolabeled Ppant arm. What about lane 3? So what do we see there? AUDIENCE: Also a lot of radioactivity. ELIZABETH NOLAN: Right. We have a lot of radioactivity. We're looking at the construct that only has this C-terminal domain. So what does that tell us? AUDIENCE: It's shorter. That's why it moved down the gel further. ELIZABETH NOLAN: Right. So that's why it has a different migration on the gel. But in terms of seeing the radioactivity, what did we learn? Is this region of the protein essential or dispensable? We don't need this N-terminal domain in order for EntD to modify EntB. So we're seeing that. What about in the middle? AUDIENCE: The deleted region is important [INAUDIBLE] ELIZABETH NOLAN: Right. We see very little radioactivity here. Basically, almost nothing, especially compared to these spots. So deletion of those C-terminal amino acids is detrimental, and so that region is important. So maybe there's protein-protein interaction going on, or something with information that's important. So from here, we learn that EntD transfers the Ppant arm to EntB. The ICL domain is not essential for this, but the C-terminus of this region is here. So now what? We've got in here via the action of EntD. Can we get attachment of the monomer? And so our hypothesis is that EntE, which we saw EntE activate DHB to form with the adenylate, it will also transfer this moiety to EntB. So in this case, what was done, again, we're looking at use of a radiolabel. In this case, the radiolabel is on the DHB lane. And this is an important point. In order to see this, we cannot have radiolabeled Ppant arm, in this case, because that would give you a big background. So they're going to prepare EntB with the Ppant arm unlabeled. We know that will work based on the prior study. And now, we repeat that with unlabeled coASH. And then ask, if we incubate total EntB with EntE, ATP, and radiolabeled DHB, do we see transfer of the radiolabel to this protein here? JOANNE STUBBE: Let me ask a question. This will be a question during class. Can you do this experiment with tritiated CoA and C14-labeled serine, based on what you know about radioactivity? We actually discussed a similar situation. AUDIENCE: Would it last longer [INAUDIBLE] JOANNE STUBBE: Did you go back and look at the lifetimes? Is it infinite compared to any experiments? So it's not lifetimes. Do you have any ideas? AUDIENCE: I mean so tritium, the energy, the particle released, is much lower than the energy of C14. JOANNE STUBBE: Right. So the difference is in the energies. We talked about this. You can tune the scintillation counter. So you measure tritium to C14. So people that do enzymology for a living often use tritium in C14 at the same time. And you can quantitate, if you do your experiments right. It's a very powerful tool together, actually. ELIZABETH NOLAN: So another option, the non-simplistic approach. So basically here, if we're looking at the four lanes, again, we're looking at total protein. We're looking at radioactivity, and can consider the overall reaction, and then a variety of controls. OK, and I want to move forward to get through the rest of the slides and we just have a few more minutes, but you should work through this gel and convince yourself that there is transfer of this radiolabel in the presence of EntD And this was done with unlabeled EntD. OK, so what about peptide bond formation? We have the T domains loaded with the monomer. Can we see activity from the C domain, in terms of the formation of an amide bond. And so this experiment requires a lot of components. So what is the experiment? To look at whether or not EntF catalyzes condensation. Basically, we can incubate EntE, holo EntE, holo EntF, ATP, and these monomers. And what we want to do, in this case, is look at transfer of radiolabeled DHB to serine-loaded EntF. And I guess I got a little ahead of myself on the prior slide. So this is a case where if you add C14 labels in both of your monomers, you'd have a big problem. So key here is to use unlabeled serine and radiolabeled DHB so you're not getting a big background. And an important point to make here in these experiments is that we're looking for detection of a covalent intermediate. So effectively, having this guy here attached to EntF. And so the radiolabel is here. So that's what we're looking for, not the final product, and that's indicated by having the gels. So what do we see? We have the total protein and then the autoradiograph. And so we have holoEntF, EntE, holoEntB. And the question is, where do we see radiolabels transfer? And if we look at lane 3, where we have the proteins, ATP, serine, and radiolabeled DHB, what we observe is that there is some radioactivity here, which is indicative of a covalent intermediate. And again, you should work through these gels and work through the different conditions and make sure it makes sense to you what's seen in each one. So finally, at that stage, the activities for all of these different domains have been found. And so the question is, in the test tube, can we actually get enterobactin biosynthesized, which is going to rely on this TE domain. So the idea is if we incubate everything together, similar to what was done before, can we detect the actual small molecule, rather than this intermediate attached to EntF here? And so the way this was done was by monitoring the reaction by HPLC using reverse stage chromatography. And so here, we have all of the reaction components. Here, we see standard. So in enterobactin, this is the linear trimer, the linear dimer, a monomer. Here's the DHB substrate. And the question is, over time, do we see enterobactin formed? So you can imagine quenching the reaction, taking the soluble component, which should have this small molecule, and looking at each POC. And where you should just focus at the moment is here. So the enterobactin peak, what we see is that over time, there's growth in this peak. You can imagine doing something like LC-MS analysis to confirm it is the species you expect here. So this is full in vitro reconstitution of a non-ribosomal peptide synthetase in the test tube, and it really paved the way for many, many additional experiments related to these types of biosynthetic machines. And so with that, we're going to close this module, and I hope you all have a great spring break. And I leave you in the good hands of JoAnne starting the 28th here for lecture.
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https://ocw.mit.edu/courses/8-591j-systems-biology-fall-2014/8.591j-fall-2014.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu PROFESSOR: Today what we're going to do is we're going to talk about some of the big ideas in class. Today what we're going to do is talk about pattern formation in biology, some of the different mechanisms that are employed in development and other contexts in order for organisms to figure out where to put things. Now before I get into that too much, I wanted to make couple administrative type announcements. So first, we have graded the exams and we'll hand them back at the end of class today. If I try to forget then please somebody remind me. What we're going to do is we're going to start by just talking about diffusion a bit more. I think that there's a fair range of different maybe experiences thinking about diffusion and what it can do for you. So we'll work on our intuition a little bit. And then I'll say something about the reading that you did from [INAUDIBLE] book about robust mechanisms for pattern formation in the context of development. So simple diffusion with degradation leads to these exponential profiles that are not robust to changes in the concentration or production rate of the morphogen. Whereas if you have self enhanced degradation then this leads to power law fall off it's more robust to the concentration of this morphogen. Then what we're going to do is transition to these turning patterns, reaction to fusion systems where you can have a really surprising effect, whereby in a well mixed situation these chemicals or proteins or reactants, they might just reach a stable state. However, once you add diffusion, then you can start getting pattern formation, which is a very funny thing because diffusion is normally something that's smooths out profiles. Normally we think about diffusion as being something that removes patterns and indeed that's generally what happens. But following from this class of work from Alan Turing at the end of his life, he showed that it's actually in principle possible to have the emergence of patterns from diffusion. And we'll talk about some of the ways that could possibly happen. So Turing patterns are already a surprising case where diffusion leads to patterns. There's another interesting phenomenon where if you add another source of noise that-- again, you think noise, demographic noise, the random creation destruction of individuals or chemicals or proteins. Normally, we think this should also kind of be a force for removing patterns. But there's been a number of pieces of work over the last decade showing that in some cases, demographic noise can actually lead to patterns either in space or in time. We're going to talk more about this in the context of these so-called noise induced predator-prey oscillations. But since we're talking about patterns here I will tell you something about how such noise can enhance the formations of patterns in the context of the Turing mechanism. And then we'll end by talking some about recent work of how E. coli find the center of their cell when they want to divide. So you can imagine this is not an obvious thing to figure out how you might do. And we'll talk about the so-called Min system that E. coli use to find the center of the cells so they know where to septate, where to cut off and make two cells. So I just want to start by making sure that when we talk about diffusion, we're all talking about the same thing. Now, hopefully for those of you that have not been thinking about diffusion so much recently, you did read the notes that Alexander Van Oudenaarden put together for Systems Biology a few years ago. I think it's useful, but in addition to the math, as maybe many of you know I'm a huge fan of graphical representations of ideas. So I just want to make sure that we all agree on some of the basic ideas of how, for example, we get flux in the context of diffusion. So imagine that we have a one dimension system where there's some chemical with concentration c. We can imagine, for example, it starts out looking like this. This is a concentration profile as a function of position x where we have maybe a box of length l. Concentration of some chemicals of function of position x in some space. Now I just want to make sure that we're all thinking about the same thing. So it starts out some concentration of c1, over might be some c2. Linearly here. I just want to make sure we all agree on the magnitude of the flux at some different points. So, in particular, we can think about A, B, this is capital C. So where is the magnitude of the flux largest? A, B, C. D corresponds to them being the same. E as don't know. Are there any questions about the question? Yes. AUDIENCE: On the y-axis, you're plotting concentration? PROFESSOR: Yes, this y-axis is the concentration is the function of position x. All right, I'll give another five seconds to think about it. All right, do you need more time? Let's vote. Ready three, two, one. OK, great. So I'd say a vast majority here are agreeing that it's going to D. The flux is going to the same everywhere. Now the flux-- can somebody remind us? We'll say the flux of J. And what is it going to be? So there's a minus D times the change in c with respect to x. This was derived in the notes. So this is highlighting that what leads to a flux is the change in a concentration with respect to position. And so it doesn't matter that the concentration is higher here than here, we have the same flux. And the flux is in which direction? Left, right, up, down? Three, two, one. Left, so the flux is here. Minus sign-- like always, it's hard to remember from equations whether they're plus minus signs but you should be able to just remember you can draw something like this and say OK, this is a positive DCDX, so flux is going to be in the negative direction, right? Now, what will be the change in the concentration with respect to time at this point right now? So this is change in the conservation at point b with respect to time. Is it greater than 0, equal to 0, less than 0, or can't determine? Does everybody understand the question I'm trying to ask? Change in the concentration at this point with respect to time, at this time that's wrong. I'll give you 5 seconds. Ready? Three, two, one. OK. And so this, again, it's pretty good but not everyone. So it is going to be b. Can somebody explain why it's b here? Yes. AUDIENCE: There's as much flux coming in as coming out on the other side? PROFESSOR: That's right. So it's true that there's a net flux here coming to the left. But from the sample, if we want to know whether there's a change in the concentration at that point, we need to know what's the net number of particles moving to the left verses the net particle number there that are coming in from the right. Now of course, on this left face-- if I draw a little box here-- are all the particles crossing this position? Are they all moving to the left? No. So the idea is that all the particle motion is random. And indeed, there's only a slight access of particles moving to the left as compared to the right. In the sense that, if you look at the concentration there, the concentration is rather large, it's only a little bit larger to right than to the left of this plane. Which means that there's only a slight excess of particles defusing to the left as coming to the right, but that leads to a net flux of particles crossing this plane to the left. So where will the concentration be changing? At this point here will the concentration be changing with respect to time? Ready three, two, one. No. So well does it change at all ever anywhere in this example? Is this a steady state profile, yes or no? Ready three, two, one. AUDIENCE: No. PROFESSOR: No. What should be the steady state profile? AUDIENCE: Flat. PROFESSOR: Flat, OK. So how does that come about if the concentration is not changing any? Yeah. AUDIENCE: It'll only change at the edges. PROFESSOR: It's only changing the edges initially. Now, it's important to note that it's not that the concentration profile is not going to start looking like this. So that will not actually be how it-- eventually it's going to be flat. But it's going to smooth out on the edges. Do you guys understand what I'm trying to say? It's not that it's a line that kind of goes like this, but instead we do end up getting curvature. One more of these, and then we'll consider ourselves to be expert diffusionists. All right, so let's say the concentration as a function of position looks something like this. What I want to know is, where is DCDT maximal? We have some different points. We have A, B, C, D. Five seconds. Ready three, two, one. All right, so we finally got a lot of disagreement. I like it. Turn to your neighbor, you should be able to find somebody that disagrees with you. What is it that determines? AUDIENCE: [ALL DISCUSSING] PROFESSOR: All right, did you guys all agree? You all agree on it? OK let's go ahead and reconvene. And let me see, maybe they were forming the domains of some sort. All right, ready. Wait, he says, no, no. All right, OK, one question. AUDIENCE: Is this instantaneous in some sense? PROFESSOR: Yeah. AUDIENCE: Exactly at the inset. PROFESSOR: All right, so I guess what I would say is that this is concentration profile at some time as a function of position in time. If you want to know the concentration profile some time delta t later, right? t plus delta t. Well then, it's going to be the concentration we had before then a little bit. Plus delta t times the derivative with respect to time. Right, so this is the concentrate file at this time. And I want to how much is it going to change in the next delta t? And I'm assuming, of course, that I'm not adding any particles or taking any particles away. So just do the diffusion. All right, let's see where we are. Ready three, two, one. OK. All right so it's we got domain, definitely some domains. All right. Well, it's going to end up being A, as I think the majority of the group is now saying. And why is it a? AUDIENCE: Because of the second derivative. PROFESSOR: Second derivative, right? So just what we said before is that, if we wanted to know the change of concentration with respect of time, at this point that was zero. Because the flux particle was leading that point and the flux coming in were the same. And we can also think about the second derivative. So this indeed just from fix, like I said, was first law/ we got a D, so this gives second derivative c with respect to x squared. So that's a 2. What this is saying is that, to determine how rapidly this concentration is going to change at a particular point with respect of time, we need to know the curvature of the concentration with respect to position. So many people were saying c here. And actually, this is where the concentration will remain exactly constant over this next delta t. And of course, if we want to know about what the concentration is going to do for a long time, then we have to do something more subtle. But if we want to how the concentration is changing right now, then we just look at the curvature. And indeed, the curvature here is maximal. So concentration DCDT at this point, is it going to be greater or less than 0? Ready three, two, one. PROFESSOR: Less than 0, right? Now, it's true that only the curvature at a particular point and matter is to know how the concentrations going to change over this next delta t. But if you want to know about what's going to happen over longer periods of time, than you have to worry about more of the global structure. And what's true is that this local maximum may quickly go away. So you don't have to wait very long before this thing gets smooth and then DCDT will-- the magnitude will go to 0. And sometimes this big hump will stick around longer, but because of the curvature, this has the maximal magnitude of DCDT. And it makes sense that the concentration is going to go down because the net flux here is to the right and to the left. And so there's a net flux, particles leaving, concentration is going to come down. Now are there any questions about what we've said? If you find this discussion confusing, I would strongly encourage you to get together with a friend and just draw some random lines, curves, and just make sure you understand where diffusion is going to pull things and so forth. No questions? All right, I just want to say a few things about the so-called French flag model in development. This is perhaps the most famous. Does everybody know what the French flag looks like? What's that? All right. There are three stripes and some orientation. No, but the idea here is just that a simple way to specify a structure across one axis is to have a defusing chemical or generally a protein. And then based on the concentration of that protein or morphogen, you can just have the tissue read out what it's supposed to be, or what it's supposed to do. So the idea here is that, you have some morphogen that starts maybe on one end of an embryo and then diffuses. So we end up with some curve that tells us about the morphogen concentration as a function of position. This is a functional position. So this is along the embryo. Now the general challenge in the context of development is asking-- we started this embryo with many different cells, they're all genetically identical, and they all start out maybe without any positional information. They don't know where they are. But what that means is they don't know they need to develop into a head or a tail or something in between. Now what often happens is that you can have a deposition often by the mother so you have a maternal deposition of, say, RNA that leads to expression of some morphogen on one end and then it diffuses. And based on the concentration you can say what part of the body you should be developing into. So the idea is you just have these thresholds that might be some M1, M2. All right, now all the cells over here know that they should maybe develop into a head. Over here, they're going to develop into the mid body. And over down here it could be the low body, for example. So just from the concentration that a cell feels at a particular location. So if you had a cell right here, you just say OK the concentration of morphogens between M1 and M2 and then from that you know what l you're supposed be. Now if you have no degradation within the body. Let's just imagine you say, that at one end is 0. We're told morphogen concentration is M0. And let's say that the other end, we know the morphogen concentration is going to be 0. Let's say there's a lot of degradation at an end. If there's no degradation in between, what should the profile look like given these? So, no. All right. So just diffusion with these boundary conditions. I'm going to draw some options and hopefully you can close your eyes and imagine what it should be. All right, we know that starts out M0 over L is equal to 0. Is it closer to A, B, C, or D? Do you understand what I'm asking? Yes. AUDIENCE: That is a steady state. PROFESSOR: This is the steady state concentration profile of the morphogen given there's no degradation in the interior. Ready three, two, one. All right. So at least a majority are agreeing it's going to be B. Although, there's a significant minority that's saying A. This is why I'm bring this up because, we're so used to seeing exponential profiles and think that you should always be getting them. And I want to be clear why it's an exponential profile. That's the thing we start with, and for example this model in [INAUDIBLE] book. Noori's book, you get exponential profiles as a result of first order degradation. Whereas if you don't have any degradation in the interior, then we say, OK, DCD-- well, do we want to use C or M now? We'll use M, just because now we're thinking very particularly for morphogen. All so DMDT. We want to know the steady state profile. We set this thing equal to 0. Integrate twice. Right, so we're going to get-- morphogen profile should just be some Ax plus B, OK? My line's kind of crappy, I'm sorry. However, when we have first order degradation, then we get these exponential profiles. Are there any questions about--? AUDIENCE: Is it possible to have [INAUDIBLE]? PROFESSOR: Yeah, right. What does this require? AUDIENCE: The morphogen is generated into halves. X equals 0 and then-- PROFESSOR: Degraded it at-- AUDIENCE: Disappearing. PROFESSOR: Yeah. Right. That's what I'm saying, is that when I said that at the end it was a 0, this means that we have degradation at this point. Any time that the morphogen reaches the end of the embryos, it gets degraded. So it's not degradation inside if it's only degradation here. And this of course a situation where you have constant flux throughout the embryo, constant rate of production of the morphogen, constant rate of degradation over here. But when you have this first order degradation-- if we're told that instead-- well, maybe. So, if in addition to having diffusion, you also have this first order degradation. Now remember M is a function of both x and t, right? And this is the situation where we get exponential profiles. Now in Murray's book, he often discusses this where the boundary condition is that you have some concentration M0 at one end. And then we get an exponential profile here. What was the characteristic length scale? This is something we should be able to figure out. You should, of course, be able to solve the equations, but you should be able to figure it out from our favorite approach of dimensional analysis. What's the units of things here? So units of D is what? Length squared over time. Units of alpha? So this is a morphogen concentration over a time. This is a morphogen concentration. So this thing has to be one over time. So the length scale is then going to have to go as the square root of D over L. And indeed, if you solve it, it's exactly equal to that, but this dimensional analysis just told us that it had to scale like that. So this is telling us about a characteristic length of which this morphogen profile is going to fall off. This is the characteristic length L. Incidentally, you'll see many cases of things that look like this. So if there's first order rate that's something is either degraded or is uptaken or whatnot, then together with diffusion, you'll get an exponential profile with a [INAUDIBLE] length scale that's given by this ratio. And you'll see this in many different contexts of, for example, nutrients going into a biofilm. If you have cells picking up nutrients then it's not that it's being degraded, but it's being imported. Then you'll have a similar process. This is a nice, fine situation except for what was the problem that [INAUDIBLE] pointed out? It's not robust. It's not robust against what? Initial condition. Yes, but I think we have to be a little more specific on that. Right, it's not robust to changes in M0. Incidentally, in most context, I think for mathematical simplicity, it's useful to just have a banner condition where M0 is constant. But in general, is that what the mother is going to be fixing necessarily? What would the mother typically be fixing? What was that? AUDIENCE: Production rate. PROFESSOR: Production rate. How do we figure out what the production rate is in this situation? That's right, the flux ware. That's right. The production rate is the flux right at or just to the right of 0. Incidentally, where's the flux maximal here? Is it maximal at A, B, or C? Ready three, two, one. You can do it verbally. Fun. A. Right. And that's because the profile steepest here and then shallows out. So flux is decreasing here. And that's because some of the morphogen is being degraded. So indeed, we can figure out the production rate. Right. This is equal to the flux, which is equal to minus DCDX at that location. Right, so the morphogen profile M is a function of x is some M0e to the minus x over L, where L is given by this character of lengthscale, right? So if we ask what's DMDX? Well, it's going to be-- we get a minus M0 over L. Evaluated at x equal to 0 just makes that go away, so we end up with this. So you figure out what flux is, right? Yeah, so, it's just an extra couple terms, it's fine. But that would be what would typically be constant because there might be some RNA deposit there or if it's actually transcription off of the genes at in the cells in this location and then it's based on whether there's one or two copies and so forth. So this is all fine, except that this thing is not robust. It changes in M0 or this production rate. Whereas in many cases, what we seek experimentally is that if we, for example, have the production rate, then the profile that we see out here is somehow remarkably similar. Yes? AUDIENCE: [INAUDIBLE]. PROFESSOR: Right. Yeah, OK. So I'd say that in many cases, it's not that it's directly this M squared business. But there are a number of cases in Drosophila, where they actually do see this. In many cases, it was a self enhanced degradation but not that it was directly, it was via binding to a receptor. So this was seen in patch lists together. I want to remember which were the models that-- I'm worried I'm always doing it get so it's like patchless and frizzled, or-- are there any Drosophila people here that came up with these crazy names? All right, yeah, so I always got confused which are the morphogens and which are the receptors in these situations. But I'd say that it is something that's observed in number of different contexts, at different stages of development, both in Drosophila and-- they see this in the wing as well as in the body. And so in different stages I think it is signature. But I'm very much not a developmental biologist. All right so the proposal that [INAUDIBLE] suggests as a way to make this thing more robust is to somehow change this term. Now, the goal in all this was to have some degradation rate. And he writes it as F is a function concentration M. In principle, this thing could be a function of the position x, but he doesn't want that. And why was it that he didn't want that? Yeah. AUDIENCE: Because we already know [INAUDIBLE]. PROFESSOR: Yes, right. So the whole goal is that we want to start with some situation where we don't know where we are as a function of position x. And then from that get some spatial information. So we really want something that's not a function of x in there. Because then we've already solved the problem that we're trying to address. And so one proposal is indeed just to have something that-- and this is very much a phenomenological approach-- which is to have this thing be self enhanced degradation. So the idea is that here if we have the morphogen profile that looks something like this. There are multiple ways you can have something that looks like this. So if along the way to degradation, if the morphogen actually binds the receptor, but then activates the degradation. And this is seen, for example, in Drosophila. This is Hedgehog and Patched. So R is-- in this case, you have the morphogen is diffusing outside the cells. It binds to the receptor, and then the receptor activates imported degradation of the morphogen. So it's not that the morphogen is directly leading to its own degradation, but it's via some network here. And there was another one that they talk about, which is that, when the degradation is mediated by 2. These look like this, and this is in context. And this is again Drosophila, but you could probably guess what this thing does. And so it is a wing development as well. If you write down an actual model of one of these things, and you could play with it and see where you're going to get something that's going to look like an M squared type term. But the nice thing about this phenomenological approach is it kind of gives you a sense of the kinds of effects that you should be looking for that will enhance the robustness of the pattern. Now the key thing in this whole discussion to remember is that if you have a profile here, if you change, it's just somehow translationally invarion in the sense that since there's no information about where you are here, these patterns can kind of slide over. So if you change the boundary condition M0, that's just equivalent to sliding this over some distance and you could figure out where everything has to go. And that's true not just for the exponential profile, but for any of the profiles. And that's just because there's no information about what's going on once you're inside the embryo. Can somebody give the intuitive explanation for why some sort of self enhanced degradation like this might make the pattern more robust? AUDIENCE: Decrease faster [INAUDIBLE]. PROFESSOR: Decrease faster when M is big. That's right. And so the idea is that if you change the M0 here, for example, you double it, then with a self enhanced degradation you kind of quickly come back to where you were. So there's some sense that if the pattern quickly comes down, then you can get it being more robust. In particular, this profile for large x in the proper regime, this thing falls off as some A a over x squared. And indeed, one of the problems in Murray's book is that if this is just degrading as M to some power n, then you can calculate what happens here. But this is some sort of power law fall off. So of course, it's falling off rapidly here, but not as rapidly as if it were an exponential. And this is just like in the case of the power law distributions of networks that we were talking about before, that for the exponential fall off, then it's just very, very unlikely that you'll find some node with many edges. Similarly here, you quickly fall to zero here as compared to a power law distribution. So that means that you can specify patterns out further in some sense. Are there any questions about this basic idea and where it comes from? I think the discussion in [INAUDIBLE] book is actually reasonable. So I don't really want to spend too much time on it. In the context of the eldar model that was from the last section of that chapter. One thing I just want to highlight is that it's useful to be able to look at these models, and from the equations that you see-- for example, the supplemental material of a paper-- make sure you are very clear about what assumptions they've made in getting to this model. And so maybe we'll just spend a few minutes talking about this. So this is again a Drosphil embryo two, two and a half hours after fertilization where at that stage, you really have the embryo that looks like a-- we're trying to understand the patterning around the front back. So we think about diffusion along in one dimension, but it's really around this radial direction. And in particular, where this is trying to understand patterning in the dorsal region. And what we have is a situation where we have this protease inhibitor. And so the protease distributed uniformly. Inhibitor starts out out there, and then we have the morphogen that somehow starts out in here. And we want to end up with a situation where the morphogen profile in here is kind of well defined. So where you just have the morphogen at the center of this dorsal region. And then we have diffusion of these inhibitors coming in and then the morphogen coming out. And the question is, how can we make sense of this? And if you look at the equations that they write down-- So we have diffusion of the inhibitor. This is the complex of the inhibitor and the morphogen-- complex c. And so this is a case where they wrote down what they basically knew about the biologists system. And then they asked, well, what range of parameters would lead to a robust pattern of the morphogen in here, if you vary things such as the total concentration inhibitor, the morphogen or the protease? So we have diffusion of each of these things, the inhibitor, the morphogen complex. We're not going to worry about the protease, because this thing is uniformly distributed. You can argue about whether that's the right thing to do, but that's based on the model. So we have diffusion of each of these guys. Can somebody say what's going on here? What are these things trying to capture? John? AUDIENCE: [INAUDIBLE]. PROFESSOR: So this is just saying, OK, the complex is formed as a result of binding of the inhibitor and the morphogen. It's proportional to concentration those two things, at that particular position at the particular time. And just remember that any time that you create something, if you have binding at two things together, then you have to consider these things going away. And it's not they're being degraded, it's just that they're forming this complex. So in these equations you just have to be very careful about keeping track of which things are really going away and which things are just changing form. Yeah? AUDIENCE: [INAUDIBLE]. PROFESSOR: Right. You're saying why that there's not a minus term? So this is an assumption. And I think this is a kind of thing that you have to be clear about in any of these, because that's always going to occur at some rate. And the assumption is that this is dominant. And then just be clear in words, what is this thing? AUDIENCE: [INAUDIBLE]. PROFESSOR: Right. The protease. And you're saying cleaving inhibitor. AUDIENCE: Yeah, cleaving the compacts by degrading the [INAUDIBLE]. PROFESSOR: Right. So at a rate proportional to the Proteus concentration-- the complex concentration-- this complex goes away. Now the question is we have to figure out where it went or what happened, right? Well we have a minus alpha c here but then we have a plus alpha c here. But this isn't the morphogen right. This is at the rate that the complex somehow is being disappeared by the protease, the morphogen is appearing. But then, we don't have the same term here. And that's how you know that what the protease is doing is it's degrading the inhibitor as part of the complex. So by looking at these equations, you can actually figure out what are the assumptions that have been made of what the biology is that's being captured here. But in many cases, there are assumptions of what's big and what's small. And then what's this thing here? Somebody else that has not yet-- do these equations look familiar? Yes, please. AUDIENCE: It's the protease [INAUDIBLE]. PROFESSOR: That's right. It's the protease degrading the inhibitor when the inhibitor is not down to the morphogen. Now of course, what the authors did next is they searched numerically across parameter space over [INAUDIBLE] magnitude, and they found that for some parameter regime there was a robust pattern of the morphogen developed in here. There's no reason for you to necessarily remember which thing it was. But what they found is that if this term went away and this term went away, then you end up getting a robust profile of the morphogen against changes in the overall morphogen inhibitor concentration. Now this is pretty weird, I would say. I don't know, do you think-- is this one weird? Weirdness is in the eye of the beholder. But I would say it's not so shocking. Can somebody say verbally what this would correspond to? AUDIENCE: The protease doesn't degrade the inhibitor on its own. PROFESSOR: That's right, the protease doesn't degrade the inhibitor on its own, only when it's part of the complex with the morphogen. And independently there was experimental data already indicating that was true. So you could have if you want to just written the model without that, because there was already evidence for that. But as I said, this is a pretty weird thing. And so, this would be saying that, for some reason, diffusion of the morphogen on its own is very small. In particular, small compared diffusion of the morphogen when it's a complex. Now can somebody say why I might think that's weird? You don't have to believe it's weird. AUDIENCE: You'd think bigger is better. PROFESSOR: That's right. Exactly. This is saying, well, when it's complex, it's bigger, it's somehow diffusing faster. So this would not happen do the simple Stokes type drag. This would have to be biology. Somebody's biology is allowed to do anything that doesn't violate the laws of physics. Indeed, later there was experimental evidence of something like this was actually happening. And you could imagine having either from sort of active transport type dynamics for the complex, or from binding type dynamics of the morphogen. I think the key thing to take from this example is just that, in general, we would like biological function-- biology would like biological function to be robust to things that are often fluctuating or varying. And that's one way to try to make guesses about what might be happening in the system. This sort of computational exercise is not all a proof that this has to be what's happening. They wrote down a particular model-- it could have been that there are other terms they're not aware of and so forth. But it's at least a way of generating hypotheses that you can go and test. And quite generally, I'd also say that its essential for all of us as consumers of models to be able to look at some of these equations and figure out what it is that they've assumed. I want to move on to this idea of pattern formation via the reaction diffusion or Turing-type patterns. And I think it's really important to start by just acknowledging that this is really a surprising finding, that it's even possible for this to happen. Because this is a situation where you have a couple chemicals-- or proteins, or reacting elements-- and if you just mix them, if you have them in some well mixed tube, they reach some steady state. Whereas somehow if you allow diffusion, then you can get these patterns. And I would say based on my intuition at least, I would not have thought that this would be possible. And this really just is because if you can just imagine, you start with some profile. Now there's some structure here. But diffusion's going to actually cause this to come down, and this to come up. So diffusion acts to remove these spatial patterns. But somehow in some prime regimes, if you have coupled reactions that are activating and inhibiting each other, then you can actually get these spatial patterns. And as you read about, there's some idea that you need to have a local activation and a global inhibition. But ultimately the mathematical-- you always have to go and think about more carefully about the math, and would be indicated just by those words. This is at least a way to guide your thinking a little bit. But ultimately what matters are around this stable what would be the stable fixed point, you have to actually look at these derivatives and so forth. And the derivatives can be subtle I would say. Just because you think of something as activator inhibitor doesn't mean that it's going to have that role around the fix point that you're studying. So this is just a caution. What I want to do is just give you, for example, one example of a simple model that does experience these Turing patterns. And in an appropriate regime. And this Levin-Segel model of pattern formation. And this was published in 1976, and it was actually meant as a model of predator-prey interactions in ecology. We'll talk significantly more about predator-prey interactions in a few weeks. But I just want to write down what they said. And this is a simplified model of their stuff. So it's actually trying to think about some plankton herbivore before interactions. And this is derivative with respect to time. We're going to follow the nomenclature of a paper by Butler and Goldenfeld a few years ago, because they're the ones who thought about the demographic noise enhancing the patterns. PROFESSOR: All right, so we have [? phi ?] and [? phi. ?] First of all, who's eating whom? All right, we're going to say it verbally. Who is the predator? Ready three, two, one. AUDIENCE: Herbivore. PROFESSOR: Herbivore. Herbivores are not what they used to be. So indeed, the herbivore benefits from the presence of the plankton. So nobody cares about plankton, I guess. So this corresponds to the predator-prey type interaction. So there's some death rate. So let's see, the herbivore or the predator. And then what you see is there's these two growth terms for the plankton. So there's this term, which would be kind of simple, exponential growth. And then this term, which is actually some sort of super exponential growth. There's some sense in which the plankton benefit each other. And this was originally introduced because of something called predator satiation. But it's just a general reflection of the fact that in many cases, individuals benefit from the presence of other individuals. And in particular, this is known as the ally effect, and we'll spend time talking about this in a few weeks as well in the context of populations and ecology. For now, I just want to use this as an example of a model that gives you these Turing patterns. So if you want to, you can go ahead and ask, well, what happens if we just have a well mixed situation? If everything is not a function of x, but it's still can to be a function of time. You can solve these equations, and you can find what the steady states are equal to. So there is indeed a steady state stable coexistence of the predator and the prey in a well mixed situation. However, if you then go and you analyze the stability of different spatial modes, what you'll find is that in some situations, particular wavelength or wave vectors become unstable. And it's just over some range of wavelengths, and that corresponds to the wave length of the Turing patterns that you'll see. If you're curious about these things in more depth, I encourage you to attend Mehran Kardar's class Statistical Physics in Biology, he is an expert on these topics and I think is a wonderful clear lecturer. Now there's going to be some condition for these things leading to Turing patterns. Now from your reading, do you think it's going to be mu--? So this is Turing patterns require-- I'll give you 30 seconds to think about what. Do you need more time? I'm not sure where we are. Maybe another 15 seconds just to-- Let's see where we are. Ready three, two, one. OK, we're all over the place. I think we're uniformly distributed between A, B, and C. So there's an idea that you're supposed to have so-called local activation and global inhibition. Why don't we turn to a neighbor and see if you can figure out what's going on. OK, why don't we reconvene, I'm curious where you're thinking is. Let's go ahead and vote. Ready three, two, one. OK, so we have many, many C's. The idea that some mu has to be much less the nu. Yes. I want to make sure. All right, mu is telling us about the diffusion coefficient of the prey. Nu is the diffusion coefficient of the predator. So what you want is to have local activation. And the thing that's activating itself is the plankton. So you can see that that's like, for example, psi squared term. And from the stand point of a spatial situation, you can say, all right, well, let's say we have some region with a lot of this prey. Well, it's able to activate itself, so it kind of comes up. And as it does that, it is creating more of the predator, phi. But because the predator has a larger diffusion coefficient-- well, it's obviously going to grow here. But it will also diffuse away. And that's this global inhibition of neighboring regions. Now of course, these are just words, you have to take it a little bit of a grain of salt. But if you can actually do this calculation analytically, and derive in this model the condition for Turing patterns to emerge as a function of everything. And for example, if you have b 1/2, p 1. So for unity type parameters, then the requirement to get Turing patterns is that nu over mu is the greater than 27.8. The key thing though is that this thing is much larger than 1. So a general thing that emerges in these Turing patterns is that you need this so-called inhibiting type partner to have a much larger effective diffusion coefficient than the activating partner. Now the problem with this is that this thing is indeed much larger than 1, which limits the overall usefulness of this is a mechanism. Because if it really is simple diffusion, how much bigger would the-- let's say, low Reynolds number regime, these are just molecules experience diffusion. How much bigger does the-- this is so confusing. So this is big and this is little, right? So how much bigger does the activator-- they're too many things to keep track of here. It's because I don't have anything written down. So that's the activator. So this one's the activator, and this one's the inhibitor. And the inhibitors has to diffuse much more than the activator. So the inhibitor has to be much smaller than the activator. OK good, so how much bigger does the activator have to be than the inhibitor for this to be true? Does it scale as this 28 or 28 squared or 28 cubed? In terms of radius, how much? Scales linearly, remember? Einstein equation, you guys had fun thinking about on the exam. KT over gamma. This would require, for example, the activator to be 30 times bigger in terms of radius. And the inhibitor, in order to have this emerge just as a result of diffusion, simple diffusion. And that's just not a typical thing for just a range of protein sizes. But you could imagine putting up various ways to make this happen. But there is a very nice development that I mentioned, which is that there's another surprising aspect in these problems, which is that in many cases, demographic noise can lead to patterns-- -- or maybe you might call them quasi-patterns. But things that, for all intents and purposes, look like some of this pseudo-periodic. So there's a pseudo-Turing pattern, maybe. And indeed, if you actually do this simulation with those parameters where you do the explicit demographic fluctuations-- i.e. you take into account that this corresponds to a prey giving birth. This corresponds to a predator eating a prey, random events just like what we've done in the context of a Gillespie simulation. Then what you find is that with demographic fluctuations or demographic noise, then the condition here nu over mu has to be only 2.48. And indeed in Butler and Goldenfeld, they derive these things using field theoretic approaches. There was a 2011 paper and also 2008 or 2009. The nice thing here is that what you see is that in the presence of these demographic noise that will be there, the difference in diffusivity that are required is not nearly as large as when you're thinking about the mean field equations. Are there any questions about this before we talk about center fighting in E. coli? Yeah. AUDIENCE: [INAUDIBLE]. PROFESSOR: OK, now you're thinking about the actual plankton-herbivore. I think there are many things to say. One of them would just be that maybe the patterns that you see between actual plankton and actual herbivores is not due to the Turing mechanism. For the Turing mechanism to be at play to generate spatial patterns, it requires that this inhibitor do something equivalent to diffusion more rapidly. It has to have to move away more rapidly. AUDIENCE: [INAUDIBLE]. PROFESSOR: Right. It could be on the plankton only move around on the currents, whereas the herbivore actually has direct motions. I have to say, I don't actually know enough about plankton to know whether this is at all-- That base statement I thin is true, but I don't know how the numbers work out. Yeah. AUDIENCE: [INAUDIBLE]. PROFESSOR: Yes. Both in this case and the simple predator-prey oscillations. One thing we're going to see the predator-prey populations later is that you can also get this sort of oscillations in time. So this could be in time or this could be a function of space. But in both cases, it's really that you have a resonant enhancement in some areas. So that the demographic fluctuation somehow excite all frequencies or all wavelengths. But then what happens is that, in this case, if you look at the eigenvalues-- OK, I want to be close to 0, though. If you look at how the modes decay-- so the eigenvalue's function of a wave number-- what you see is that you end up getting things that look like this. So whereas a Turing pattern is when particular wave length or wave modes actually becomes unstable, and then you get mean field type patterns. And in this situation what happens is that you're just close here, so then you excite all the wave numbers or wavelengths. And then some of them take a long time to go away, so then they build up and then that's the resulting patterns that you see. So what I want to do is just for the last 10 minutes to talk about this Min system. Because it's I think it's a beautiful example of a combination of Systems Biology questions together with reaction diffusion systems, and also some beautiful in vitro experiments that I've quite appreciated. So the question is-- imagine yourself, you've gone to all this work right, so now you're nice and long. So you'd like to divide into two cells. And the question is, how do you know where to divide? So you imagine that you're some E. coli cell. You're maybe six microns long, everything is great. And the question is, you want to divide? Where would you like to divide if you're an E. coli cell? Middle. OK, so what you'd like to do is go here. And indeed, what happens is that there's the so-called z-ring. There's a a pseudo-polymer protein that forms a ring around here, then it constricts itself and pinches off the membrane. And that's how you get two cells. So there's this formation of this z-ring that constricts and that's this cell division event. The question is how do you know where to put it? And that's what this Min system is for. And it's, of course, like everything in bacterial genetics, it was identify by mutations. So the Min system it's called that because these mutants formed so-called mini cells. If you, instead of dividing the center where you might have two copies of the genome here. If instead you divide over here, then what your going to end up with is a very long cell with two copies of the genome. You're going to end up with this mini cell without any DNA. And those mini cells, are they going to do well for the long term? AUDIENCE: No. PROFESSOR: Right. They actually can survive for a little bi. And people have argued that maybe that could be useful for synthetic biology because they still make proteins, but they're not going to go take over the world because they don't have any DNA. But these are the so-called mini cells that happen if you have mutations in The min system. Now, I think the three players are MinC, MinD, and MinE. So I think that MinA and MinB ended up not actually existing or something, in the sense that they identified the mutants, but then they were wrong about something. So the three that actually ended up being involved in the actual Min system are C, D and E. So this MinC prevents formation of the z-ring. Now MinD and MinE are kind of the amazing guys. So this guy binds to the membrane and recruits MinC. Whereas MinE-- what it does is it pulls MinD off the membrane. So it binds to MinD and ejects from the membrane. And what has been seen by doing imaging in live E. coli cells with fluorescently labeled MinD and MinE is that there are remarkable oscillations from pole to pole. And indeed, that doesn't require MinC, so MinC is somehow following the others. What happens is that MinD binds over here, then MinE binds and pushes it off. And then the MinD comes over here, and then MinE pulls it off, and it goes back and forth. The period is minutes, maybe. So it's a really remarkable thing that happens in vivo. And the idea of what's happening is that if MinD comes here and then it comes here, on each of the edges, then it doesn't hang out in the middle. And that means that it's only in the middle where MinC can then bind-- oh, sorry. Because MinC is following MinD, then only in the middle can you form the z-ring, because that's where MinC is not. Right. It's wonderful to do things in live cells because that's the native context and so forth. But I think that in some cases, it's also wonderful if you can take purified components and recapitulate interesting behaviors in vitro. Because then at least you know what is sufficient to generate a particular kind of dynamical behavior. And because this system had been proposes as a model system for reaction diffusion mechanism to get this behavior. And so there's this paper by Martin Loose, in Petra Schwille's lab in-- where was it? In Dresden. And what they did is they took a supported lipid bilayer-- so a membrane on glass-- and then they added purified components of MinD and MinE that were fluorescently labeled. And then they saw amazing patterns, these amazing reaction diffusion waves traveling along. Now on the first day of lecture, I actually showed you what some of those things look like. I didn't want to use up the projector again, but maybe I can pull up this movie because it is kind of fabulous. So maybe you can't see these things very well. But this is a MinD and MinE and an overlay of the two, imaged on a two dimensional membrane. And the movies are just amazing. When you see this, you think that it's a simulation. It's so incredible watching these things go. After class, you can come up, I can show you the paper, and you can look at the movie in more depth. But they're really amazing patterns, and it's patterns that you would predict from some sort of Turing type reaction diffusion mechanism. So there are few things that are may be worth saying in this business. So these are fluorescently labeled MinD, MinE. They started out with MinD that was uniform. And then they added MinE, and over the time scale of about an hour they started seeing these sorts of patterns. And as maybe you expected from the in vivo behavior, what you see are these things where MinD looks like this, whereas MinE looks like that. And then the wave travels here to the left. Because the MinE is ejecting the MinD from the membrane, causing this whole thing to move. So this is a situation where you have both of these proteins in the liquid. All right, so in buffer, as well as on the membrane and they're coming on and off and so forth. And in this situation, it's wonderful because they can do all sorts of things like control the concentration of MinE, and see that the velocity of these waves changes as you change concentration in MinE. So this is a really experimentally trackable system. And then you can ask what kind of model would lead to that sort of behavior? Such waves, do you think that it requires ATP? Just for fun, we can vote yes or no, it's all right. Ready three, two, one. AUDIENCE: Yes. PROFESSOR: Yeah. So indeed it does require ATP. And that's a general feature of these Turing type patterns is that there is a non-equilibrium structure formation. Now one nice thing you can do in this sort of system is you can ask, well, what happens if I photo bleach a particular area? So I come in and I photo bleach. So now the profile looks like-- I locally deplete the fluorescence here. Question is, will this move together with the traveling wave, or does it stay fixed? And in these mechanisms, do you think should this locally depleted area, should it move or should it stay where it is? Do you understand the question? Move, yes/no. Ready three, two, one. AUDIENCE: [INAUDIBLE]. PROFESSOR: It actually doesn't move. Depleted area doesn't move. And this is just a reflection. These waves are not the result of individual molecules moving together with the wave. A wave like what you see in the ocean, or on a stream, or what not. So the motion is a result of the individual molecules coming and going and communicating with each other, but it's not a reflection of actual molecules having directed motion. Because indeed, there's no mechanism in these models for directed motion. What you see is that in all these models, the only thing that is a function of position is diffusion terms. So it's all random diffusion, but then you get global motion.
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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: One active learning strategy that I use in 595 are called mud cards. And they are a really easy activity to use. Anyone can implement them, and it's a really low barrier. So I like to introduce that early in the class so that students can see that it doesn't necessarily disrupt the class flow. It doesn't really take a lot of prep. It doesn't take a lot of resources. So a mud card is, at minimum, is an index card. And at the end of every class, I ask them to write down on the index card what they might want to know more about, what they're still confused about, or something that they found particularly compelling or interesting. I ask them to specify that so that I know if they just say, active learning, I don't know whether they thought it was really interesting or whether they thought it was confusing. So I asked them to specify. But we use we use that because-- and then I also talk to them about why it's so effective. They're able to anonymously identify what they understood or what they're still confused about. I can go back to my office afterwards, I can look through the cards. I can sort them out really quickly, even in a large class, and I've used them in a class of 100 people, even though 595 is just around 15. And generally speaking, the cards fall in three categories. You get three main problems. And so you know that maybe you didn't do a very good job of helping students understand a particular topic, or that students are really interested and want more information about something. I know that really easily, really quickly. So I can make a decision to either get more resources, write something up that I can post on our course website, or I can come to class the next time and say, hey, looks like most of you were confused about X. Let's talk a little bit more about that. And I can prepare ahead of time and be totally ready to go in with multiple explanations or multiple examples of whatever it is people are confused about. I looked over some of the mud cards we had from last time and there were some really good questions, some really good points. Someone asked if all those learning theories that we discussed were equally valid. They get they get targeted feedback based on what they don't understand. And it's really easy. It takes two minutes at the end of class. I have a colleague who calls them tickets to leave, meaning he doesn't let people leave the class until they've handed him an index card with something on it. But they're very, very effective. I believe they were first used in the Aero-Astro department here at MIT, but they're used all over the place now and they're extremely, extremely effective. For instructors that are thinking about using mud cards in their classes, I would say the biggest issue or the biggest concern would be to make sure that you use them early in the semester and that you use them often. If you use them one day and then you don't use them for another two weeks or even another week, students really won't get in the pattern of filling them out and filling them out thoughtfully. And they won't see the utility, because you really need to come back the next time with useful information saying, I looked at these cards. I understand you don't get this. Let me help you. And that act really encourages students to keep filling them out, and it's just a win-win for everyone.
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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 we're just picking up where we stopped last time. So we are beginning to discuss how we are going to sort of do a geometrical approach to physics, using a more general set of coordinates now. So we began talking about how things change when I discuss special relativity, so for the moment keeping ourselves just at special relativity. We, by the way, are going to begin lifting our assumptions that it is simply special relativity fairly soon. But to set that up, I need to start thinking about how to work in more general coordinate systems. So we're going to do it in the simplest possible curvilinear coordinates. So it's basically just going from Cartesian coordinates in the spatial sector to plane polar coordinates. One of the things which I have emphasized a few times, and I'm going to continue to hammer on, is that these are a little bit different from the curvilinear coordinates that you are used to in your past life. In particular, if I write out the displacement, the little vector of the displacement element in the usual way, I am using what's called a "coordinate basis," which means that the vector dx is related to the displacement, the differential of the coordinates, by just that thing contracted with all the basis vectors. And so what that means is I have a little displacement in time, which looks normal. Displacement in radius, which looks normal. Displacement in the z direction, which looks normal, and a displacement in an angle, which does not. In order for this whole thing to be dimensionally consistent, that's telling me e phi has to have the dimensions of length. And that is a feature, not a bug. Last time, we introduced the matrix that allows me to convert between one coordinate system and another, so just basically the matrix-- it's sort of a Jacobi matrix. It's a matrix of partials between the two coordinate systems. And this idea that things are-- they look a little weird. So the way I did that was I didn't actually write it out, but I did the usual mapping between x, y and r and phi, worked out all of my derivatives. And sure enough, you've got something that looks very standard, with the possible exception of these r's that are appearing in here. So notice the elements of this matrix. These do not have consistent units-- again, feature, not bug. This guy is basically just the inverse of that. This is the matrix that affects the coordinates in the opposite direction. And notice in this case, you have some elements where their units are 1 over length. So let's just continue to sort of catalog what some of the things we are going to be working with look like in this new coordinate representation. And this will lead us to introduce one of the mathematical objects that we are going to use extensively as we move forward in studying this subject. So what I want to do next is look at what my basis vectors look like. So what I want to do is characterize what my e r and my e phi look like. And these are going to look very familiar from your intuition, from having studied things like E&M in non-Cartesian coordinates. So your e r is just related to the original Cartesian basis vectors, like so. And if you like, you can easily read that out by performing the following matrix multiplication on the original Cartesian basis vectors. Your e phi perhaps looks a little wacky. So you can see the length coming into play there. A good way to think about this is if your intuition about basis factors-- I have to be careful with this language myself-- your intuition about basis vectors is typically that they are unit vectors. These are not unit vectors. They do form a nice basis, but they are not unit vectors. In particular, the basic idea we're going to go with here is that e phi, it's always going to sort of point in the tangential direction. But no matter where I put it in radius, I want that vector to always sort of subtend the same amount of angle. In order to do that, its length needs to grow with r. So that's where that's a little bit different from your intuition. And there's a very good reason for this, which we will get to, hopefully, well before the end of today's class. So last time, when we first began to talk about tensors a couple of lectures ago, the first tensor I gave you-- so confining ourselves just to Cartesian coordinates-- was the metric, which was originally introduced as this mathematical object that came out of looking at dot products between basis vectors. It's essentially a tensor that allows me to feed in two displacements and get the invariant interval between those displacements that comes out of that. I am going to continue to call the dot product of two basis vectors the "metric." But I'm going to use a slightly different symbol for this. I'm going to call this g alpha beta. In the coordinate representation that we are using right now, so in plane polar coordinates, this becomes-- you can work it out from what I've written down right here. This is just the diagonal of minus 1, 1 r squared 1. So this equals dot here. This is-- I'll put PPC for plane polar coordinates under that. And then using that, you know that you can always work out the invariant displacement between two events. It's always going to be the metric contracted with the differential displacement element. And this is going to be minus dt squared plus dr squared plus r squared d phi squared plus dz squared. That, I hope, makes a lot of sense. This is exactly what you'd expect if I have two events that are separated in plane polar coordinates by dt, dr, d phi, dz. This is what the distance between them should be. So the fact that my basis vectors have this slightly annoying form associated with them, it all sort of comes out in the wash here. Remember at the end of the day, if we think about quantities that are representation independent-- and that's going to be the key thing. When you assemble scalars out of these things, the individual tensor components, they can be a little bit confusing sometimes. They are not things that we measure. They are not things that really characterize what we are going to be working with. And we really want to get into the physics, unless we're very careful about it. This is something you can measure. And so sure enough, it comes out, and it's got a good meaning to it. Let me just wrap up one last thing before I talk about sort of where we're going with this. So just for completeness, let me write down the basis one forms. Just as the basis vectors had a bit of a funny form associated with them, you're going to find the basis one forms likewise have a bit of a funny form associated with them. And the way I'm going to get these-- and so these are going to be the Cartesian basis one forms-- basically, I'm not carefully proving all these relations at this point, because you all know how to do that. I'm just using line up the indices rule. And when you do that, you get this. And likewise, your basis one form for the axial direction, I'll just write down the result. It's going to look like this. So the key place where all of this-- so right now, these are all just sort of definitions. Nothing I've done here should be anything that even approaches a surprise, I hope, just given the you guys have done-- the key thing that's probably new is all this garbage associated with coordinate bases, this extra factors of r and 1 over r that are popping up. But provided you're willing to sort of swallow your discomfort and go through the motions, these are not difficult calculations. The key place where all of this really matters is going to be when we calculate derivatives of things. It'll turn out there is an important rule when we talk about integrals as well a little bit later, but let's just say that. So for now, we'll focus on derivatives. So all the derivatives that we've been looking at so far, we have, indeed, done a couple of calculations where we've computed the derivatives of various vector valued and tensor valued quantities. And it was helped by the fact that all the bases, when I work in Cartesian coordinates, are constant. Well, that's not the case now. So now, we need to account for the fact that the bases all vary with our coordinates. So let me just quickly make a catalog of all the non-trivial-- there's basically four. In this one, where I'm just doing plane polar coordinates, there are four non-trivial derivatives we need to worry about. One of them actually turns out to be 0. So the radial derivative of the radial unit vector is 0. But the phi derivative of the phi unit vector is not. you go and take the phi derivative of this guy, and you basically get-- take the phi derivative of this, you're going to get this back, modulo factor of radius. So I can write d e r d phi as e phi over r. If I take the derivative of e phi with respect to r, I get e phi back, divided by r. So the simplest way to write this is like so. And finally, if I take the phi derivative of the phi unit vector, I get e r back, with an extra factor of r thrown in. And a minus sign. So we're going to see a way of doing this that's a little bit more systematic later, but I want to just keep the simple example, where you can just basically by hand calculate all the non-trivial derivatives easily. Of course, there's also a t unit vector and a z unit vector. But they're constants, so I'm not going to bother writing them out. All the derivatives associated with them are equal to 0. So let's imagine now that I have assembled some vector. So I have some vector field that lives in this spacetime. And I'm using this basis. And so I would write this vector with components v alpha. And let's let the-- so this is going to be a curvilinear coordinate system, so this will be plane polar coordinates being used here, plane polar coordinate basis vectors. And what I would like to do is assemble the tensor that you can think of essentially as the gradient of this vector. So let's begin by doing this in a sort of abstract notation. So the gradient of this guy-- this is sort of ugly notation, but live with it. Following what we have been doing all along, what you would want to do is just take the root of this whole thing. It's going to have a downstairs component on it. So attach to it the basis one form. If you prefer, you can write it using the d notation like I have there, but I just want to stick with the form I wrote in my notes. Looking at this the way I've sort of got this right now, I can think of, if I don't include the basis one forms here, this should be the components of a one form. So this should be a kind of object. So let's just expand out that derivative. Let's write it like this. So you just-- we haven't changed calculus. So when I do this, I'm going to basically use the old-fashioned Leibniz rule for expanding the derivative product of two things. Here's the key thing which I want to emphasize-- in order for this whole thing to be-- for this to be a tensorial object, something that I couple to this basis one form, the sum of these two objects must obey the rules for transforming tensors. But the two objects individually will not. So this is an important point which I'm going to emphasize in slightly different words in just a few moments again. This is one of the key things I want you to get out of this lecture, is that when I'm taking derivatives of things like this, you've got to be a little bit careful about what you consider to be components of tensors and what is not. Now as written like that, this is kind of annoying. So my first object has a nice basis vector attached to it. My second object involves a derivative of the basis vector. However, something we saw over here is that derivatives of basis vectors are themselves proportional to basis vectors. So what I'm going to do is introduce a bit of notation. So let me switch notation slightly here. So the beta derivative of e alpha can be written as-- in general, it can be written as a linear combination of basis vectors. So what we're going to do is define d-- I want to make sure my Greek letters are legible to everyone in the room here. So let me write this nice and clearly. d beta of e alpha, I'm going to write that as capital gamma mu beta alpha e mu. This gamma that I've just introduced here in this context is known as the Christoffel symbol. Fact I'm calling this a symbol, it's got three indices on it. You might look at it and go, ooh, smells like a tensor. Be a little bit careful. In much the same way that these two terms are not individually components of a tensor, but their sum is, this guy individually is actually not a component of a tensor, but when combined with other things, it allows us to assemble tensors. So for our plane polar coordinates, there are exactly three non-zero Christoffel symbols. So gamma phi r phi is equal to 1 over r, which is also equal to gamma phi phi r. Gamma r phi phi is minus r. And you can basically just read that out of that table that I wrote down over there. All the others will be equal to 0. Now from this example, this is what it makes it smell like every time you introduce a new coordinate representation. You're going to need to sit down for an hour and a half, or something like that, and just work out all the bloody derivatives, and then go, oh, crap, and read out all the different components of this thing, and assemble them together. There actually is an algorithm that we will get to at the end of this class that allows you to easily extract the Christoffel symbols provided you know the metric. But right now, I just want to illustrate this thing conceptually. The key thing which you should know about it is that it is essentially the-- I almost said the word "matrix," but it's got three indices. It's a table of functions that allows me to relate derivatives of basis vectors to the basis vectors. So before I go on to talk about some of that stuff, let's take a look at the derivative a little bit more carefully. So the derivative of the vector-- so let's basically take what I've written out up there. I'm going to write this as the beta derivative of vector v. And I can write that as the beta derivative of e of v alpha-- so the first term where the derivative hits the vector components. And then I've got a second term where the derivative hits the basis. I'm going to write this like so. This is sort of annoying. One term is proportional to e alpha, one is proportional to e mu. But notice, especially in the second term, both alpha and mu are dummy indices, so I'm free to relabel them. So what I'm going to do is relabel alpha and mu by exchanging them. As long as I do that consistently, that is totally kosher. And when I do that, I can factor out an overall factor of the basis object. This combination that pops up here-- so we give this a name. And this is a combination which, by the time you have finished this semester, if you don't have at least one nightmare in which this name appears, I will not have done my job properly. This shows up a lot at this point. This is called the "covariant derivative." And it shows up enough that we introduce a whole new notation for the derivative to take it into account. I'm going to call this combination of the partial derivative of v and v coupled to the Christoffel symbol-- I'm going to write this using the, if you're talking LaTeX, this would be the nabla operator. So I made a point earlier when we were talking about derivatives a couple of weeks ago that we were reserving the gradient symbol for a special purpose later. Here it is. So whenever I make a derivative that involves the gradient symbol like this, it is this covariant derivative. And the covariant derivative acting on vector components, it generates tensor components. Partial derivative does not. And what I'm going to do, just in the interest of time-- it's one of those calculations that's straightforward but fairly tedious-- I have a set of notes that I meant to make live before I headed over here, but I forgot. I have a set of notes that I'm going to show on the website by this evening which explicitly works out what happens when you apply the coordinate transformation using that-- it's been erased-- when you use that L matrix to construct the coordinate transformation between two representations. If you try to do it to partial derivatives of vector components, basically what you find is that there's an extra term that spoils your ability to call that-- it spoils the tensor transformation law, spoils your ability to call that a tensor component. So the partial on its own doesn't let you. You get some extra terms that come along and mess it all up. On next p set, you guys are going to show that if you then try to apply the tensor transformation law to the Christoffel symbols, you get something that looks tensorial, but with an extra term that spoils your ability to call it tensorial. There's a little bit of extra junk there. But the two terms exactly conspire to cancel each other out so that the sum is tensorial. So part one of this will be notes that I post to the website no later than this evening. Part two, you guys will do on the p set. So just saying in math what I just said in words, if I do this, like I said, you basically will eventually reach the point where what I am writing out right now will become so automatic it will haunt your dreams. Wait a minute, I screwed that up. It's so automatic I can't even write it properly. Anyhow, something like that will-- modulo my typo-- that should become automatic. And the key thing which I want to note is that if I take these guys, and I attach the appropriate basis objects to them, this is an honest-to-god tensor. And so this derivative is itself an honest-to-god tensor. A typical application of this, so one that will come up a fair bit, is how do you compute a spacetime divergence in each coordinate system? So suppose I take the divergence of some vector field v. So you're going to have four terms that are just the usual, like when you guys learned how to do divergence in freshman E&M in Cartesian coordinates. You get one term that's just dv x dx, dv y dy, et cetera. So you got one term that looks just like that, and you're going to have something that brings in all of Christoffel symbols. Notice the symmetry that we have on this one. Actually, there is Einstein summation convention being imposed here. But when we look at this, there's actually only one Christoffel symbol that has repeated indices in that first position. So when I put all this together, you wind up with something that looks like this. So go back and check your copy of Jackson, or Purcell, or Griffith, whatever your favorite E&M textbook is. And you'll see when you work in cylindrical coordinates, you indeed find that there's a correction to the radial term that involves 1 over r. That's popped out exactly like you think it should. You have a bit of a wacky looking thing with your phi component, of course. And let me just spend a second or two making sure. It's often, especially while we're developing intuition about working in a coordinate basis, it's not a bad idea to do a little sanity check. So here's a sanity check that I would do with this. If I take the divergence, I take a derivative of a vector field, the final object that comes out of that should have the dimensions of that vector divided by length. Remembering c equals 1, that will clearly have that vector divided by length. That will clearly have that vector divided by length, vector divided by length, explicitly vector divided by length. That's weird. But remember, the basis objects themselves are a little weird. One of the things we saw was that e phi has the dimensions of length. In order for the vector to itself be consistent, v phi must have the dimensions of v divided by length. So in fact, when I just take its phi derivative, I get something that looks exactly like it should if it is to be a divergence. Let's move on and think about how I take a covariant derivative of other kinds of tensorial objects. This is all you need to know if you are worried about taking derivatives of vectors. But we're going to work with a lot of different kinds of tensor objects. One of the most important lectures we're going to do in about a month actually involves looking at a bunch of covariant derivatives of some four-indexed objects, so it gets messy. Let's walk our way there. So suppose I want to take the derivative of a scalar. Scalar have no basis object attached to them. There's no basis object. When I take the derivative, I don't have to worry about anything wiggling around. No Christoffel symbols come in. If I want to take the covariant derivative of some field phi, it is nothing more than the partial derivative of that field phi-- boom. Happy days. How about a one form? The long way to do this would be to essentially say, well, the way I started this was by looking at how my basis vectors varied as I took their derivatives. Let's do the same thing for the basis one forms, assemble my table, do a lot of math, blah, blah, blah. Knock yourselves out if that's what you want to do. There's a shortcut. Let's use the fact that when I contract a one form on a vector, I get a scalar. So let's say I am looking at the beta covariant derivative of p alpha on a alpha. That's a scalar. So this is just the partial derivative. And a partial derivative of the product of something I can expand out really easily. So using the fact that this just becomes the partial, I can write this as a alpha d beta p alpha plus p downstairs alpha. So now what? Well, let's rewrite this using the covariant derivative. Pardon me a second while I get caught up in my notes. Here we are. I can write this as the covariant derivative minus the correction that comes from that Christoffel symbol. Pardon me just a second. There's a couple lines here I want to write out very neatly. So when I put this in-- oops typo. That last typo is important, because I'm now going to do the relabeling trick. So what I'm going to do is take advantage of the fact that in this last term, alpha and mu are both dummy indices. So on this last term that I have written down here, I'm going to swap out alpha and mu. When I do that, notice that the first term and the last term will both be proportional to the component a alpha. Now, let's require that the covariant derivative when it acts on two things that are multiplied together, it's a derivative, so it should do what derivatives ordinarily do. So what we're going to do is require that when I take this covariant derivative, I should be able to write the result like so. It's a healthy thing that any derivative should do. So comparing, I look at that, and go, oh, I've got the covariant derivative of my one form there. Just compare forms. Very, very similar, but notice the minus sign. There's a minus sign that's been introduced there, and that minus sign guarantees, if you actually expand out that combination of covariant derivatives I have on the previous line, there's a nice cancellation so that the scalar that I get when I contract p on a, in fact, doesn't have anything special going on when I do the covariant derivative. So I'm going to generalize this further, but let me just make a quick comment here. I began this little calculation by saying, given how we started our calculation of the covariant derivative of a vector, we could have begun by just taking lots of derivatives of the basis one forms, and assembling all these various tables, and things like that. If you had done this, it's simple to find, based on an analysis like this, that if you take a partial derivative of a one form, that you get sort of a linear combination of one forms back. Looks just like what you got when you took a partial derivative of the basis vector, but with a minus sign. And what that minus sign does is it enforces, if you go back to a lecture from ages ago, when I first introduced basis one forms, it enforces the idea that when I combine basis one forms with basis vectors, I get an identity object out of this, which is itself a constant. If you are the kind of person who likes that sort of mathematical rigor, some textbooks will start with this, and then derive other things from that-- sort of six of one, half a dozen of the other. So we could go on at this point. And I could say, how do I do this with a tensor that has two indices in the upstairs position? How do I do this with a tensor that has two indices in the downstairs position? How do I do it for a tensor that's got 17 indices in the upstairs position and 38 in the downstairs position? The answer is easily deduced from doing these kinds of rules, so I'm just going to write down a couple of examples and state what it turns out to be. So basically, imagine I want to take the covariant derivative-- let's do the stress energy tensor-- covariant derivative of T mu nu. So remember, the way that the Christoffel got into there is that when I looked at the derivative of a vector, I was looking at derivatives of basis objects. Well, now I'm going to look at derivatives of two different basis objects. So I'm going to wind up with two Christoffel symbols. You can kind of think of it as coming along and correcting each of these indices. I can do this with the indices in the downstairs position. Guess what? Comes along and corrects all them with minus signs. Just for completeness, let me just write down the general rule. If I am looking at the covariant derivative of a tensor with a gajillion upstairs indices and a gajillion downstairs indices, you get one term that's just a partial derivative of that guy, and you get a Christoffel coupling for every one of these. Plus sign for all the upstairs, minus sign for all the downstairs. That was a little tedious. You basically just, when I give you a tensor like that, you just kind of have to go through. And it becomes sort of almost monkey work. You just have to rotely go through and correct every one of the indices using an algorithm that kind of looks like this. Oh, jeez, there's absolutely a minus sign on the second one. Thank you. I appreciate that. So the way that we have done things so far, and I kind of emphasized, it sort of smells like the way to do this is you pick your new coordinate representation, you throw together all of your various basis objects, and then you just start going whee, let's start taking derivatives and see how all these things vary with respect to each other, assemble my table of the gammas, and then do my covariant derivative. If that were, in fact, the way we did it, I would not have chosen my research career to focus on this field. That would suck. Certainly prior to Odin providing us with Mathematica, it would have been absolutely undoable. Even with it, though, it would be incredibly tedious. So there is a better way to do this, and it comes via the metric. Before I derive what the algorithm actually is, I want to introduce an extremely important property of tensor relationships that we are going to come back to and use quite a bit in this course. So this is something that we have actually kind of alluded to repeatedly, but I want to make it a little more formal and just clearly state it. So this relationship that I'm going to use is some kind of a tensor equation, a tensorial equation that holds in one representation must hold in all representations. Come back to the intuition when I first began describing physics in terms of geometric objects in spacetime. One of the key points I tried to really emphasize the difference of is that I can have different-- let's say my arm is a particular vector in spacetime. Someone running through the room at three-quarters the speed of light will use different representations to describe my arm. They will see length contractions. They will see things sort of spanning different things. But the geometric object, the thing which goes between two events in spacetime, that does not change, even though the representation of those events might. This remains true not just for Lorentz transformations, but for all classes of transformations that we might care to use in our analysis. Changing the representation cannot change the equation. Written that way, it sounds like, well, duh, but as we'll see, it's got important consequences. So as a warm-up exercise of how we might want to use this, let's think about the double gradient of a scalar. So let's define-- let's just say that this is the object that I want to compute. Let's first do this in a Cartesian representation. In a Cartesian representation, I just take two partial derivatives. I've got a couple basis one forms for this. So I've got something like this. The thing which I want to emphasize is that as written, in Cartesian coordinates, d alpha d beta of phi-- those are the components of a tensor in this representation. And the key thing is that they are obviously symmetric on exchange of the indices alpha and beta. If I'm just taking partial derivatives, doesn't matter what order I take them in. That's got to be symmetric. Let's now look at the double gradient of a scalar in a more general representation. So in a general representation, I'm going to require these two derivatives to be covariant derivatives. Now, we know one of them can be very trivially replaced with a partial, but the other one cannot. Hold that thought for just a second. If this thing is symmetric in the Cartesian representation, I claim it must also be true in a general representation. In other words, exchanging the order of covariant derivatives when they act on a scalar should give me the same thing back. Let's see what this implies. So if I require the following to be true, that's saying-- oops. So let's expand this out one more level. So now, I'm correcting that downstairs index and over here. So the terms involving nothing but partials, they obviously cancel. I have a common factor of d mu of phi. So let's move one of these over to the other side. What we've learned is that this requirement, that this combination of derivatives be symmetric, tells me something about the symmetry of the Christoffel symbols itself. If you go back to that little table that I wrote down for plane polar coordinates, that was one where I just calculated only three non-trivial components, but there was a symmetry in there. And if you go and you check it, you will see it's consistent with what I just found right here. Pardon me for just a second. I want to organize a few of my notes. These have gotten all out of order. Here it is. So let me just use this as an opportunity to introduce a bit of notation. Whenever I give you a tensor that's got two indices, if I write parentheses around those indices, this is going to mean that I do what is called "symmetrization." We're going to use this from time to time. If I write square braces, this is what we call "anti-symmetrization." And so what we just learned is that gamma mu alpha beta is equal to gamma mu alpha beta with symmetrization on those last two indices. We have likewise learned that if I contract this against some object, if these were anti-symmetric, I must get a 0 out of it. So that's a brief aside, but these are important things, and I want to make sure you have a chance to see them. So trying to make a decision here about where we're going to want to carry things forward. We're approaching the end of one set of notes. There's still one more thing I want to do. So I set this whole thing up by saying that I wanted to give you guys an algorithm for how to generate the Christoffel symbols. The way I'm going to do this is by examining the gradient of the metric. So suppose I want to compute the following tensor quantity-- let's say is g the metric tensor, written here in the fairly abstract notation. And this is my full-on tensor gradient of this thing. So if you want to write this out in its full glory, I might write this as something like this. But if you stop and think about this for just a second, let's go back to this principle. An equation that is tensorial in one representation must be tensorial in all. Suppose I choose the Cartesian representation of this thing. Well, then here's what it looks like there. But this is a constant. So if I do this in Cartesian coordinates, it has to be 0. The only way that I can make this sort of comport with this principle that an equation that is tensorial in one representation holds in all representations-- this leads me to say, I need to require that the covariant derivative of the metric be equal to 0. We're going to use this. And I think this will be the last detailed calculation I do in today's lecture. We're going to use this to find a way to get the Christoffel symbol from partial derivatives of the metric. There's a lot of terms here and there's a lot of little indices. So I'm going to do my best to make my handwriting neat. I'm going to write down a relationship that I call "Roman numeral I." The covariant derivative in the gamma direction, G alpha beta-- you know what, let me put this down a little bit lower, so I can get these two terms on the same line. So I get this thing that involves two Christoffel symbols correcting those two indices. This is going to equal 0. I don't really seem to have gotten very far. This is true, but I now have two bloody Christoffel symbols that I've somehow managed to work into this. What I'm trying to do is find a way to get one, and equate it to things involving derivatives of the metric. So this is sort of a ruh-roh kind of moment. But there's nothing special about this order of the indices. So with the audacity that only comes from knowing the answer in advance, what I'm going to do is permute the indices. Then go, oh, let's permute the indices once more. So I'll give you guys a moment to catch up with me. Don't forget, these notes will be scanned and added to the web page. So if you don't want to follow along writing down every little detail, I understand, although personally, I find that these things gel a little bit better when you actually write them out yourself. So those are three ways that I can assert that the metric has no covariant derivative. They all are basically expressing that same physical fact. I'm just permuting the indices. Now there's no better way to describe this than you sort of just stare at this for a few moments, and then go, gee, I wonder what would happen if-- so stare at this for a little while. And then construct-- you know I have three things that are equal to 0. So I can add them together, I can subtract one from the other. I can add two and subtract one, whatever. They should all give me 0. And the particular combination I want to look at is what I get when I take relationship one and I subtract from it two and three. So I'm going to get one term that are just these three combinations of derivatives, gamma. And I get something that looks like-- let me write this out and then pause and make a comment. So I sort of made some lame jokes a few moments ago that essentially, the only reason I was able to get this was by knowing the answer in the back of the book, essentially. And to be perfectly blunt, for me personally, that's probably true. When I first wrote this down, I probably did need to follow an algorithm. But if I was doing this ab initio, if I was sitting down to first do this, what's really going on here is the reason I wrote out all these different combinations of things is that I was trying to gather terms together in such a way that I could take advantage of that symmetry. A few moments ago, we showed that the Christoffel symbols are symmetric on the lower two indices. And so by putting out all these different combinations of things, I was then able to combine them in such a way that certain terms-- look at this and go, ah, symmetry on alpha and gamma means this whole term dies. Symmetry on beta and gamma means this whole term dies. Symmetry on alpha and beta means these two guys combine, and I get a factor of 2. So let's clean up our algebra. Move a bunch of our terms to the other side equation, since it's a blah, blah, blah equals 0. And what we get when we do this is g mu downstairs gamma is equal to 1/2. What we're going to do now is we will define everything on the right-hand side-- I've kind of emphasized earlier that the Christoffels are not themselves tensors, but we're going to imagine that we can nonetheless-- we're not going to imagine, we're just going to define-- we're going to say that we're allowed to raise and lower their indices using the metric, in the same way you guys been doing with vectors and one forms and other kinds of tensors. So let's call everything on the right-hand side here gamma with all the indices in the downstairs position, gamma sub gamma alpha beta. And then this is simply what I get when I click all of these things together like so. If you go and you look up the formulas for this in various textbooks that give these different kinds of formulas, you will typically see it written as 1/2 g upstairs indices, and then all this stuff in parentheses after that. When you look things up, this is the typical formula that is given in these books. This is where it comes from. So I need to check one thing because it appears my notes are a little bit out of order here. But nonetheless, this is a good point, since we've just finished a pretty long calculation, this is a good point to introduce an important physical point. We're going to come back to this. We're going to start this on Thursday. But I want to begin making some physical points that are going to take us from special relativity to general relativity. So despite the fact that I've introduced this new mathematical framework, everything that I have done so far is in the context of special relativity. I'm going to make a more precise definition of special relativity right now. So special relativity-- we are going to think of this moving forward as the theory which allows us to cover the entire spacetime manifold using inertial reference frames. So we use inertial reference frames or essentially, Lorenz reference frames, and saying that Lorentz coordinates are good everywhere. We know we can go between different Lorentz reference frames using Lorentz transformations. But the key thing is that if special relativity were correct, the entire universe would be accurately described by any inertial reference frame you care to write down. And I will probably only be able to do about half of this right now. We'll pick it up next time, if I cannot finish this. The key thing which I want to emphasize is, gravity breaks this. As soon as you put gravity into your theory of relativity, you cannot have-- so we will call this a global inertial frame, an inertial frame that is good everywhere, so "global" in the mathematical sense, not "global" in the geographic sense-- not just earth, the whole universe, essentially. As soon as we put in gravity, we no longer have global reference frames and global inertial reference frames. That word "inertial" is important. But we are going to be allowed to have local inertial frames. I have not precisely defined the difference what "local" means in this case, and I won't for a few lectures. But to give you a preview as to what that means, it's essentially going to say that we can define an inertial coordinate system that is good over a particular region of spacetime. And we're going to have to discuss and come up with ways of understanding what the boundaries of that region are, and how to make this precise. So the statement that gravity breaks the existence of global Lorentz frames, like I said, it's a two-part thing. I'm going to use a very handwavy argument which can be made quite rigorous later, but I want to keep it to this handwaving level, because first of all, it actually was first done by a very high-level mathematical physicist named Alfred Schild, who worked in the early theory of relativity. It's sort of like he was so mathematical, if it was good enough for him, that's good enough for me. And I think even though it's a little handwavy, and kind of goofy in at least one place, it gives a good physical sense as to why it is gravity begins to mess things up. So part one is the fact that there exists a gravitational redshift. So here's where I'm going to be particularly silly, but I will back up my silliness by the fact that everything silly that I say here has actually been experimentally verified, or at least the key physical output of this. So imagine you are on top of a tower and you drop a rock of rest mass m off the top of this tower. So here you are. Here's your rock. The rock falls. There's a wonderful device down here which I label with a p. It's called a photonulater. And what the photonulater does, it converts the rock into a single photon, and it does so conserving energy. So when this rock falls, the instant before it goes into your photonulater, just use Newtonian physics plus the notion of rest energy. So it's got an energy of m-- mC squared, if you prefer, its rest energy-- plus what it acquired after falling-- pardon me, I forgot to give you a distance here-- after falling a distance h. So that means that the photon that I shoot up from this thing-- let me put a few things on this board. So the instant that I create this photon, this thing goes out, and it's going to have a frequency omega bottom, which is simply related to that energy. This photon immediately is shot back up to the top, where clever you, you happen to have in your hands a rerockulater. The rerockulater, as the name obviously implies, converts the photon back into a rock. Now, suppose it does so-- both the photonulater and the rerockulater are fine MIT engineering. There are no losses anywhere in this thing. So there's no friction. There's no extra heat generated. It does it conserving energy, 100%. What is the energy at the top? Well, you might naively say, ah, it's just going to go up to the top. It's going to have that same energy. It might just have it in the form of a photon and omega b. There will be some frequency at the top. And your initial guess might be it's going to be the same as the frequency at the bottom. But if you do that, you're going to suddenly discover that your rock has more energy than it started out with, and you can redirect it back down, send it back up. Next thing you know, you've got yourself a perpetual motion machine. So all you need to do is get your photonulater and your rerockulater going, and you've got yourself a perpetual motion machine here. I will grant that's probably not the weakest part of this argument. Suppose you had this. I mean, you look at this. If technology allowed you to make these goofy devices, you would instantly look at this and say, look, if I am not to have-- let's just say I live in a universe where I'm fine with photonulaters. I'm fine with rerockulaters, but damn it, energy has to be conserved. I am not fine with perpetual motion machines. If that's the case, we always fight perpetual motion. We must have that the energy at the top is equal to the energy this guy started with. When it sort of gets back into-- imagine that your rerockulater is shaped like a baseball catcher's mitt. You want that thing to just land gently in your mitt, and just be a perfectly gentle, little landing there. And when you put all this together, going through this, taking advantage of the fact that if you work in units where you've put your c's back in, there will be a factor of g h over c squared appearing in here, what you find is that the frequency at the top is less than the frequency at the bottom. In other words, the light has gotten a little bit redder. now I fully confess, I did this via the silliest argument possible. But I want to emphasize that this is one of the most precisely verified predictions of gravity and relativity theory. This was first done, actually, up the street at Harvard, by what's called the Pound-Rebka experiment. And the basic principles of what is going on with this right now-- I just took this out to make sure my alarm is not about to go off, but I want to emphasize it's actually built into the workings of the global positioning system. Because this fact that light signals may travel out of a gravitational potential, they get redshifted, needs to be taken into account in order to do the precise metrology that GPS allows. Now, this is part one, this idea that light gets redder as it climbs out of a gravitational field. Part two, which I will do on Thursday, is to show that if there is a global inertial frame, there is no way for light to get redder as it climbs out of a gravitational potential. You cannot have both gravity and a global inertial reference frame. That's where I will pick it up on Thursday. So we'll do that. And we will then begin talking about how we can try to put the principles of relativity and gravity together. And in some sense, this is when our study of general relativity will begin in earnest. All right, so let us stop there.
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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/20-020-introduction-to-biological-engineering-design-spring-2009/20.020-spring-2009.zip
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>> Sally: What are you working on? >> Dude: I've got a great new idea. I'm going to make cocoa beans that reproduce as fast as viruses. That way, we'll have enough cocoa beans to make as much chocolate as we want! >> Sally: You know how much I like chocolate! Explain your project a little more. How are you going to engineer this yummy idea? >> Dude: Well viruses reproduce really fast so I thought I could take that part from a virus and put it into a cocoa plant. >> Sally: Plants are pretty safe to work with but viruses can be dangerous. Which virus were you thinking about using? >> Dude: Well I read that Ebola virus has a short sequence that... >> Sally: ...EBOLA!?! Ebola is a bio-safety level FOUR agent. >> Dude: Well the worst are level five, so I thought it was safe enough. >> Sally: Level five? There is no bio-safety level five. You'd need lots and lots of training to work in any level three or four facility... ...and you'd have to wait a few years since they don't let minors in those labs. >> Dude: DD: But there was this show I saw on TV where... >> Sally: What you see on TV and in movies can be very deceiving. But let's talk about all the bio-safety levels. If we're going to make your viral cocoa beans, we need to come up with a way that keeps everyone safe, inside the lab and outside too. >> Dude: OK. I do know that biosafety is a combination of safe lab practices and techniques, wearing the right safety equipment, and working in a lab designed to contain the experiment. But with biosafety level 1 that just means a room with screened windows and a door that closes. That doesn't sound like much. >> Sally: That's because BL1 experiments use only well-characterized microbes that don't make a healthy person sick. But remember, there's no eating or drinking in the lab since you wouldn't want to eat your experiments. >> Dude: When do I get to wear a lab coat like yours? >> Sally: If you're working at biosafety level 2, you'll have to wear it, and gloves, too, since some BL2 agents can be infectious. BL2 labs have impervious benches and most of the work is done in biological safety cabinets. Lots of BL2 workers also get immunizations if they'll be working with certain agents. >> Dude: I hate getting shots. >> Sally: Well, not all level 2 agents require immunizations. Really, the rules for BL1 and BL2 are pretty basic because the agents have minimal risk associated with them. But let's think for a minute about what biosafety you'd need for your mash-up of a virus with a cocoa plant. >> Dude: Well maybe more protective clothes, like the workers in bio-safety level 3 labs. They wear full suits and maybe masks or something. It's nothing I can't handle. Imagine me in a space suit! Very cool! And think about how much chocolate we'll have! >> Sally: Earth to Dude! Level three agents are very serious, potentially lethal, and level four agents, like Ebola virus, have such a high lethal risk that there are only a few places in the world where they are studied. Restrictions for working with these agents are very tight. Everything entering or leaving BL3 and BL4 labs has to be decontaminated. Sometimes the lab has to be in its own building to control foot traffic. I know I don't want to work where there are double door entries and airlock rooms. >> Dude: I think that sounds like fun... >> Sally: ...If you are so excited about the protective clothing, we can always be overly cautious in a level 1 or 2 facility. But can't we find another way to make more cocoa beans? >> Dude: OK. >> Sally: Maybe we can find another virus that's not so dangerous and we can keep working on your chocolate idea. Let's start looking in here.
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https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/8.06-spring-2018.zip
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PROFESSOR: So so far even though these things look maybe interesting or a little familiar, we have not yet stated clearly how they apply to physics. We've been talking about vector spaces, V, for a particle. Then V tensor N. And we've looked at states there. We've looked at permutation operators there. Symmetric states there. Antisymmetric states there. What is missing is something that connects it to quantum mechanics. And that is given by the so-called symmetrization postulate. So it's a postulate. It's something that you technically can't derive. Therefore, you postulate. You can say it's an extra axiom even, quantum mechanics. You would say also that probably there's no other way to do things. So in some sense, it's forced. But I think it's more honest to admit it's an extra postulate. So here it goes. I'll read it first. Then I'll write it. So it says the following. If you have a system of N identical particles, arbitrary states in V tensor N are not physical states. The physical states are the states that are totally symmetric. In that case, the particles are called bosons. Or totally antisymmetric, in which case the particles are called fermions. That's basically it. The arbitrary state that is neither symmetric nor antisymmetric, and all that is not a physical state of a system of identical particles. So it's stated here. In a system with N identical particles, physical states are not arbitrary states in V tensor N. Rather, they are totally symmetric. In parentheses, they belong to SymN of V. In which case, the particles are said to be bosons or they are totally antisymmetric. Are totally antisymmetric. They belong to anti NV. In which case, they are said to be fermions. All right. So this is really pretty fundamental. It's a statement that we began our discussion by saying we have an electron at this up and another electron at this down. Is it the plus minus state or is it the minus plus state? And we said it's neither one. You cannot declare those two to be equivalent. And that's what this says. It's not arbitrary states in this. But they have to become totally symmetric, if they're bosons or totally antisymmetric if they are fermions. So we have to make a few comments. And that's what will keep us busy for the next 20, 25 minutes. And to understand this and see what it implies and how we use it. So here it is. The statement we have made is a statement of fact in three dimensions. There's no particles that we like and we study. There are further possibilities in worlds of lower dimension two, three dimensions, space, time, or so two dimensional space, that allows for further kinds of statistics that are interesting. But this is the general statement. So the first comment is this is the general statement. But further statistics, other types of space, exist in two dimensions, not in three, where we live. We say three in the Galilean ways. 3 plus 1 time. But this is spatial dimensions. This is the statistical definition of bosons and fermions. But then comes the spin statistics theorem. That applies then you understand with quantum mechanics. And that's a deep theorem that ends up telling you that particles with spin 0, 1, 2 integer are bosons. And particles with spin 1/2, 3/2 [INAUDIBLE] are fermions. So it's a great achievement of quantum field theory associating the statistical properties with the spin. It's a deep connection. It's valid for elementary particles or composite particles. The list of particles is not that big. Particles of spin 1, we know a lot of them. A photon is spin 1. Gluons are spin 1. The w's, the c's are spin 1 particles. Particles spin 0. We didn't know an elementary one for a while. We finally know the [INAUDIBLE] is there, spin 0. Particles of spin 2. Just [INAUDIBLE] along. Higher spin unknown. It may exist. String theory has them. All kinds of theories postulate them. They may exist. They may not. Spin of 1/2, fractional spin. Well, all the matter particles, quarks, muons, leptons, neutrinos are spin 1/2. Spin 3/2 is hypothetical as an elementary particle. If it exists, it would be called the gravitino. There are composite particles. You could [INAUDIBLE] with quarks, or with other particles, mesons, baryons. You can get several spins. Not with elementary particles. Many time unstable particles. So anyway, those are very interesting things. Now next fact is that you can go from elementary particles to composite particles. So if you have, for example, the hydrogen atom. That this has a proton and an electron. And you want to figure out if it's a boson or a fermion. Well, you think of another hydrogen atom here, p and e. And you write a wave function that involves the protons, the first proton, the second proton, the first electron, and the second electron. That's a little bit of a funny notation. This would correspond to maybe the coordinates of the first proton, the second proton, the first electron, the second electron. Since protons are identical particles that are fermions, the wave function must change sign if P1 and P2 are exchanged. There would be a minus sign. It's antisymmetric. Since the electrons are fermions, it should antisymmetric by exchanging this 2. So it's also antisymmetric under this exchange. So finally, if you exchange this item with this item, you must exchange the two protons and the two electrons at the same time. Two minuses and you get a plus. Therefore, the hydrogen atom is a boson. So these statistical properties build up. So an object built with a number of bosons and a number of fermions will be a boson or a fermion, depending on those numbers. So that's a nice thing. So the hydrogen atom is a boson.
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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 extend our concept of momentum to a system of particles. Again, we need to choose a reference frame. So we'll have a ground frame. And let's consider N particles. Now when we have a lot of particles, we need some type of notation. So let's use the symbol j. And it will goes from 1 to N. And then our arbitrary j particle will be moving. This particle will have mass mj. And it will be moving with a velocity vj. Now recall in our system, we have many other particles. We can call that one 1. This is one n. We have lots of different particles in the system. And this just represents an arbitrary particle in that system. And the momentum of the jth particle is just the mass, mj, times the velocity, vj. And again, we're assuming some fixed reference frame. So the total momentum of this system, we now have to add up the momentum of all the particles, all the way up to the nth particle. Now, when we make a sum like this, there is a standard mathematical summation notation, which we'll write like this. We'll do the sum, this capital sigma sin of j goes from 1 to j goes to N of the momentum of the jth particle. And that represents the sum j goes from 1 to n of mj vj. And this is what we call the momentum of the system. This is a vector sum. And now let's see how Newton's second law applies to the momentum of the system. Suppose that acting on our particles-- for instance, here's our jth particle-- we have a force Fj acting on the jth particle. Then we know that from Newton's law that the force will be also the sum of the forces on all of the particles, F1, F2, plus dot, dot, dot, plus FN. So once again, we can write this as a sum j goes from 1 to N of the force on the jth particle. And that's the force on the summing over all the forces on all the particles in the system. But now, we can apply Newton's second law. So Newton's second law is the statement that the force on the jth particle causes the momentum of the jth particle to change. And when we write that now, the total force on the system, j goes from 1 to N, is just the sum of the change in momentum. Because every single term-- let's just look at that. T1 plus dP2/dt plus dot, dot, dot, plus dPN/dt, that's what we mean by the sum. We can rewrite this as d/dt of P1 plus P2 plus P3 plus dot, dot, dot, plus PN. And what we see is that the total force is the derivative of the sum j goes from 1 to N of the momentum. But recall, this sum we've defined as the momentum of the system. So our conclusion is the total force causes the momentum of the system to change. Now so far, all we've done is we've recast Newton's second law in this form. Our next step is to analyze the forces on the individual particles we have and apply Newton's third law. So we'll do that next.
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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, let's get started. Why doesn't everyone take 10 more seconds to answer this clicker question here. All right. So, good. It looks like just about everyone is able to go from the name of an orbital to the state function. That's important. And we're actually going to get a little bit deeper in our clicker questions here, since when you do your problem-set it won't be quite this straight forward that you'll be answering this kind of question, but actually you'll be thinking about how many different orbitals can have certain state functions or certain orbital names. So let's go to a second clicker question here and try one more. So why don't you tell me how many possible orbitals you can have in a single atom that have the following two quantum numbers? So let's say we have n equals 4, and n sub l equalling negative 2. How many different orbitals can you have that have those two quantum numbers in them? And this should look kind of familiar to some of the problems you may have seen on the problem-set if you started it this weekend. So this should be something you can do pretty quickly, so let's take 10 more seconds on that. All right. OK, looks like we got the majority, which is a good start, but we having discrepancy on what people are thinking. So, let's go through this one. So, what we're saying is that we have n equals to 4, and m sub I being equal to negative 2. If we have n equals 4, what is the highest value of l that we can have? STUDENT: 3. PROFESSOR: OK. We can have n 4, l 3, and then, sure, we can have m sub l equal negative 2 if l equals 3 What's the second value of l that we can have? 2. OK. So we can have this orbital here. What about l equals 1, can we have this? No, we can't. Because if l equals 1, we can not have m sub l equal negative 2, right, because the magnetic quantum number only goes from negative l to positive l here. So that means it's not possible, if we've made these stipulations in the first place, to have a case where l equals 1. So this means we can only have 2 different values of l. We already know our value of m. So now we're just counting up our orbitals, an orbital is completely described by the 3 quantum numbers. So we end up having 2 orbitals here. All right, so hopefully if you see any other combination of quantum numbers, for example, if it doesn't quickly come to you how many orbitals you have, you can actually try to write out all the possible orbitals and that should get you started. All right. So today we're going to finish up our discussion of the hydrogen atom. We'd started on Monday talking about radial probability distributions for the s orbitals. We'll finish that up, and then we're going to move on to talking about the p orbitals. We'll start with talking about the shape, just like we did with the s orbitals, and then move on to those radial probability distributions and compare the radial probability at different radius for p orbital versus an s orbital. And once we do that, we're actually going to move on to multi-electron atoms. So, you might have noticed that we will have spent about 6 and 1/2 lectures just getting to the point where we have only one electron, so we're only up hydrogen so far. And you might have kind of been projecting ahead and thinking if we keep up at this pace, pretty much we would only get to carbon by the end of the semester. So I will assure you that we will not be spending 6 lectures per atom as we move forward, and in fact, what we're going to find is that by taking all the principles we've learned about the hydrogen atom and applying that to multi-electron atoms, but making a few of changes and making a few modifications to take into the fact that we're going to have electron-electron repulsions going on, we'll be able to think about any multi-electron atom using these same general ideas, and the Schrodinger equation ideas that we came up with and have looked at for the hydrogen atom. And this is really good news because it's good to get passed carbon. I'm an organic chemist, so I love carbon, it's one of my favorite atoms to talk about, but it would be nice to get to the point of bonding and even reactions to talk about all the exciting things we can think about once we're at that point. So, one we finish our discussion of how we think about multi-electron atoms, we can go right on and start talking about these other things. All right. So, let's pick up where we left off, first of all we're still on the hydrogen atom from Monday. And on Monday what we were discussing was the solution to the Schrodinger equation for the wave function. And we also, when we solved or we looked at the solution to that Schrodinger equation, what we saw was that we actually needed three different quantum numbers to fully describe the wave function of a hydrogen atom or to fully describe an orbital. We didn't just need that n, not just the principle quantum number that we needed to discuss the energy, but we also need to talk about l and m, as we did in our clicker question up here. We also talked about well, what is that when we say wave function, what does that actually mean? And first we discussed the fact that well, in terms of a classical analogy, we don't really have one for wave function, we can't really think of a way to picture wave function thinking in classical terms. But we do have an interpretation for wave function squared. And when we take the wave function and square it, that's going to be equal to the probability density of finding an electron at some point in your atom. And that's useful in terms of seeing a general shape, but if we're actually interested in thinking about how far away that electron is from the nucleus, you can see that instead of talking about probability density, which is the probability per volume, instead it would be much more useful to talk about something called radial probability distribution, or in other words talking about the probability of finding the electron at some distance, which we define as r, from the nucleus in a spherical shell that is just infinitesimally small, infinitesimally thin at distance or at a thickness that we'll call a d r here. So, basically what we're saying is if we take any shell that's at some distance away from the nucleus, we can think about what the probability is of finding an electron at that radius, and that's the definition we gave to the radial probability distribution. And we can look at the formula that got us here. This is the radial probability distribution formula for an s orbital, which is, of course, dealing with something that's spherically symmetrical. It's somewhat different when we're talking about the p or the d orbitals, and we won't go into the equation there, but this will give you an idea of what we're really talking about with this radial probability distribution. So, what we can do to actually get a probability instead of a probability density that we're talking about is to take the wave function squared, which we know is probability density, and multiply it by the volume of that very, very thin spherical shell that we're talking about at distance r. So if we want to talk about the volume of that, we just talk about the surface area, which is 4 pi r squared, and we multiply that by the thickness d r. So if we take this term, which is a volume term, and multiply it by probability over volume, what we're going to end up with is an actual probability of finding our electron at that distance, r, from the nucleus. So, the example that we took on Monday and that we ended with when we ended class, was looking at the 1 s orbital for hydrogen atom, and what we could do is we could graph the radial probability as a function of radius here. And when we do that we can see this curve, this probability curve, where we have a maximum probability of finding the electron this far away from the nucleus. And we call that most probable radius r sub m p, or most probable radius. And what is discussed is that for a 1 s hydrogen atom, that falls at an a nought distance away from the nucleus. And remember, a nought, that's just the Bohr radius, it's a constant -- that's all we need to worry about. We talked about the Bohr model and how that told us an exact distance. It was a classical model, right, so we could say the electron is exactly this far away from the nucleus. We can not do that with quantum mechanics, the more true picture is the best we can get to is talk about what the probability is of finding the electron at any given nucleus. And the most probable one here is that a nought. The other thing that we looked at, which I want to stress again and I'll stress it as many times as I can fit it into lecture, because this is something that confuses students when they're trying to identify, for example, different nodes or areas of no probability. In an orbital is remember that this area right here at r equals zerio, that is not a node. We will always have r equals zero in these radial probability distribution graphs, and we can think about why that is. At first it might be counter-intuitive because we know the probability density at the nucleus is the greatest. So the probability of having an electron at the nucleus in terms of probability per volume is very, very high. But remember that we need to multiply it by the volume here, the volume of some sphere we've defined. And when we define that as r being equal to zero, essentially we're multiplying the probability density by zero. So that's why we have this zero point here, and just to point out again and again and again, it's not a radial node, it's just a point where we're starting our graph, because we're multiplying it by r equals zero. So, we can look at other radial probability distributions of other wave functions that we talked about. We talked about the wave function for a 2 s orbital, and also for a 3 s orbital. So, let's go ahead and think about drawing what that would look like in terms of the radial probability distribution. So what we're graphing here is the radius as a function of radial probability. And for a 2 s orbital, you get a graph that's going to look something like this. So, again, we're starting at zero. We have one node here, and we can again define that most probable radius. And it turns out that for a 2 s orbital, that's equal to 6 times a nought. So when we think about what it is that this radial probability distribution is telling us, it's telling us that it is most likely that an electron in a 2 s orbital of hydrogen is six times further away from the nucleus than it is in a 1 s orbital. So another way to say that is, in a sense, if we're thinking about the excited state of a hydrogen atom, the first excited state, or the n equals 2 state, what we're saying is that it's actually bigger than the ground state, or the 1 s state of a hydrogen atom. And when we say bigger, remember this is not a classical description we're talking about. We are talking about probability, but what we're saying is that most probable radius is further away from the nucleus. So we can also look at this in terms of the 3 s orbital. And in this case, we have a graph that looks something like this. So you can draw that into your notes. And again, we can define what that most probable radius is, that distance at which we're most likely to find an electron. And in the case of the 3 s orbital, that's going to be equal to 11 . 5 times a nought. So again, what we're saying here is that it is most likely in the 3 s orbital that we would find the electron 11 and 1/2 times further away from the nucleus than we would in a around state hydrogen atom. And I just want to point out here in terms of things that you're responsible for, you should know that the most probable radius for a 1 s hydrogen atom is equal a nought. And you should know that a 2 s is larger than that, and a 3 s is even larger, and of course, hopefully as we go to 4 and 5, you would be able to guess that those are going to get even larger. But you're not responsible for knowing specifically that it's 11 . 5 times greater. You just need to know the trend there. Another thing to point out in these two graphs is that we do have nodes, and we figured out last time, we calculated how many nodes we should have in a 2 s orbital. And in terms of radial nodes, we expect to see one node. And how many nodes do you see in the 3 s orbital? two, good. I'm glad to hear that no one counted this r equal zero as a node. So we expect to see two nodes right here in the 3 s orbital. And we can calculate that with the formula that we used, which was just n minus l minus 1 equals the number of nodes. Or we could just look at the radial probability distribution itself and see how many nodes there are. So if we're looking at these two situations here and we're actually thinking of them from a more classical standpoint, which is natural for us to do because we live our lives in the every day world, not thinking about things on the atomic size scale all the time, most of us. So, for example, if we were to look at this 3 s orbital here, you might have the question of how this can be, because we're saying that, for example, we have probability of having an electron here, an electron can also be way out at this radius here. But what we're saying is there's a node here, so that there's no probability of finding an electron between those two points. So you can think of it, if we were to just think of it as a straight line that we were going across, essentially what we're saying is that we're getting from point a to point c without ever getting through point b. So, that can be a little bit confusing for us to think about, and when it's a very good question you might, in fact, say well, maybe there's not zero probability here, maybe it's just this teeny, teeny, tiny number, and in fact, sometimes an electron can get through, it's just very low probability so that's why we never really see it. And in fact, that's not the answer. The answer is, in fact, there is zero, absolutely zero probability of finding a electron here. So basically we're saying yes, we can go from point a to point c without ever going through point b. That might seem confusing if you're thinking about particles, but remember we're talking about the wave-like nature of electrons. So, the quantum mechanical interpretation is that we can, in fact, have probability density here and probability density there, without having any probability of having the electron in the space between. And you can think about that if you think about a standing wave, for example, where you can have amplitude at many different values of x, so an amplitude at many different distances, but you also have areas where there is a 0 amplitude. So, remember this makes sense if you just think of it as a wave and forget the particle part of it for right now, because that would be very upsetting to think about and that's not, in fact, what's going on, we're talking about quantum mechanics here. Yes. STUDENT: [INAUDIBLE] PROFESSOR: Oh, I'm sorry. So it's n minus l minus 1. So here we have 3 minus l equals 0, because it's an s orbital, minus 1, so we have two radial nodes here. OK. So let's actually go to a clicker question now on radial probability distributions. So I mentioned you should be able to identify both how many nodes you have and what a graph might look like of different radial probability distributions. So here, what I'd like you to do is identify the correct radial probability distribution plot for a 5 s orbital, and also make sure that it matches up with the right number of radial nodes that you would expect. All right. Let's take 10 more seconds on that, this should be a quick identification for us to do. All right. So it looks like 82% got the correct answer here. So, you should know that there's four radial nodes, right, we have 5 minus 1 minus l -- is there a question? STUDENT: [INAUDIBLE] PROFESSOR: It is very difficult for me to draw graphs on the computer. That's a good point, I'm sorry. This was my best attempt at hitting zero and not having the graph go down there. I'm not the most gifted at drawing on the computer. So yes, it should be zero at zero, but I made the line too thick. So, assuming -- if anyone got it wrong because of that, that's my apologies, that's my fault. But you should see that there are four radial nodes here since we have a 5 s orbital. And also that we know that the zero does not count as a node, if per se I actually had managed to hit zero in drawing that, so the correct answer would be the bottom one there. So, you should be able to generally identify and draw the general form of these radial probability distributions. Obviously we don't expect you to know exactly what the distances are, but you should be able to compare them relatively. Yes? STUDENT: [INAUDIBLE] PROFESSOR: No, they actually don't, and when you graph it all out. You can see this if you look at some examples in your book, actually. So this doesn't fall, for example, at 6 a nought, but that's a really good question. And the trend always is that the probability gets smaller with each of the peaks as you're drawing them. All right. So we can switch back to our notes. So we got our clicker question set there. And so now we can move on to thinking about p orbitals, we now have two ways to talk about p orbitals. We can talk about the wave function squared, the probability density, or we can talk about the radial probability distribution. So when we talk about p orbitals, it's similar to talking about s orbitals, and the difference lies, and now we have a different value for l, so l equals 1 for a p orbital, and we know if we have l equal 1, we can have three different total orbitals that have sub-shell of l equalling 1. So we can have, if we have the final quantum number m equal plus 1 or minus 1, we're dealing with a p x or a p y orbital. Remember, we don't do a one-to-one correlation, because p x and p y are some linear combination of the m plus 1 and m minus 1 orbital. And if we talk about m equals 0, we're looking at the p z orbital. And the significant difference between s orbitals and p orbitals that comes from the fact that we do have angular momentum here in these p orbitals, is that p orbital wave functions do, in fact, have theta and phi dependence. So they do have an angular dependence that we're talking about. And what I'm showing here is not on your notes, if you're interested you can look it up in your book. This is a table that's directly from your book, and what it's just showing is the wave function for a bunch of different orbitals. I mentioned last time that there was this list in your book. And what I want to point out here is this angular dependence for the p orbitals for the l equals 1 orbital. So, first, if I point out when l equals 0, when we have an s orbital, what you see is that angular part of the wave function is equal to a constant. So, remember we can break up the total wave function into the radial part and the angular part. When we look at this angular part, we see that it's always the square root of 1 over 4 pi, it doesn't matter what the angle is, it's not dependent on the angle. In contrast when we're looking at a p orbital, so any time l is equal to 1, and you look at angular part of the wave function here, what you see is the wave function either depends on theta or is dependent on both theta and phi. So we do, in fact, have a dependence on what the angle is of the electron as we define it in the orbital. So what this means is that unlike s orbitals, p orbitals are not spherically symmetrical -- they don't have the exact same shape at any radius from the nucleus. And these shapes of p orbitals probably do look familiar to you, most of you, I'm sure, have seen some sort of picture of p orbitals before. So what I want to point out about them is that they're made up of two nodes, and what you can see is that nodes are shown in different colors here and those are different phases. Sometimes you see this written when you see p orbitals, one is written as plus, one is written in minus. That's not a positive and negative charge, that's actually a phase, and that arises from the wave equation. Remember when we have waves we can have positive or a negative amplitude. When we talk about p orbitals the phase of the orbital becomes important once we talk about bonding, which hopefully you were happy to hear at the beginning of class we will get to soon. And it turns out that when you constructively have two p orbitals interfere, and when I say constructively, I mean they're both either positive or they're both the negative lobes, that's when you got bonding. Whereas if the phases where mismatched, you would not get bonding. So, that's going to be important later when we get to bonding, but just take note of it now, we have two nodes, each with a separate phase -- or we have two lobes, excuse me, each with a separate phase. And when we look at this, it's actually split by what's called a nodal plane, which is pointed out in light orange here on this picture, but what we just mean is that there is this whole plane that separates the two lobes where there is absolutely no electron density. So, the wave function at all of these points in this plane is equal to zero, so therefore, also the wave function squared is going to be equal to zero. So, if we say that in this entire plane we have zero probability of finding a p electron anywhere in the plane, the plane goes directly through the nucleus in every case but a p orbital, so what we can also say is that there is zero probability of finding a p electron at the nucleus. So, again we can use these probability density plots, which are just a plot of psi squared, where the density of the dots is proportional to the density, the probability density, at that point. So what we can say is look at each of these separately, so if we start with looking at the 2 p z orbital, the highest probability of finding an electron in the 2 p z orbital, is going to be along this z-axis. We can see that right here. And in terms of thinking about the phase of this p orbital, the phase is going to be positive anywhere where z is positive. So we would say we have a positive phase here and a negative phase there. Remember, that's going to become important when we talk about bonding, we don't need to worry about it too much right now. We can also think about where the nodal plane is in this p z orbital, so how would we define the nodal plane here? What would the nodal plane be? So, it's the x-y plane, you can see there's no electron density anywhere there. And similarly, actually, if we're looking at our polar coordinates here, what we see is it's any place where theta is equal to 0 is what's going to put up on the x-y plane. So another way to define the nodal plane is where theta is equal to 90 degrees. So let's look now at the 2 p x orbital. This is the probability density map, so we're talking about the square here. The highest probability now is going to be along the x-axis, so that means we're going to have a positive wave function every place where x is positive. What is the nodal plane in this case? Um-hmm. So, it's going to be the y z nodal plane, or in other words, we can say it's any place where phi is equal to 90 degrees. So you can see if you take phi, and you move it over 90 degrees, we're right here in the y z plane. Anywhere where that's the case we're going to have no probability density of finding an electron. And finally, we can look at the 2 p y, so the highest probability is going to be along the y-axis. It's going to be positive in terms of its wave function or in terms of its phase anywhere where y is positive. And the nodal plane's going to be in the x z plane, or again, anywhere where phi is going to be equal to 0, that takes us to the x z plane. So, let me get a little bit more specific about what we mean by nodal plane and where the idea of nodal plane comes from, and nodal planes arise from any place you have angular nodes. So we talked about radial nodes when we're doing these radial probability density diagrams here. You can also have angular notes, and when we talk about an anglar node, what we're talking about is values of theta or values of phi at which the wave function, and therefore, the wave function squared, or the probability density are going to be equal to zero. So, you remember from last time radial nodes are values of r at which the wave function and wave function squared are zero, so the difference is now we're just talking about the angular part of the wave function. And, in fact, these are the only two types of nodes that we're going to be describing, so we can actually calculate both the total number of notes and the number of each type of node we should expect to see in any type of orbital. And our equation for total nodes is just the principle quantum number minus 1. And when we talk about angular nodes, the number of angular nodes we have in an orbital is going to be equal to l. So that's why we saw, for example, in the p orbitals we had one angular node in each p orbital, because l is equal to 1 there. And we talked about the equation you can use for radial nodes last time, and that's just n minus 1 minus l. You can go ahead and use that equation, or you could figure it out every time, because if you know the total number of nodes, and you know the angular node number, then you know how many nodes you're going to have left. So you don't really have to memorize that. So, let's go ahead and just do a few of these. They're pretty straight forward to do and it gives us an idea what kind of nodal structure we can expect it an orbital. So for a 2 s orbital, how many total nodes will we have? Yup, I heard one, so 2 minus 1, one total node. Angular nodes, we're not going to have any of those, we'll have zero, l equals 0, so we have zero angular nodes. And in terms of radial nodes, we have 2 minus 1 minus 0, so what we have is one radial node. So, what you find with the s orbital, and this is general for all s orbitals is that all of your nodes end up being radial nodes. That has to be the case because l equals 0 for s orbitals. Let's look now at a p orbital, so how many total nodes do we have here? Yup, so one total node, 2 minus 1 is 1, and that means since l is equal to 1, we have one angular nodes, and that leaves us with how many radial nodes? Yup, zero radial nodes. So, for a 2 p orbital, all the nodes actually turn out to be angular nodes. So, let's have you try one more, if we can switch over and talk about a 3 d orbital. So, I'm asking very specifically about radial nodes here, how many radial nodes does a hydrogen atom 3 d orbital have? So, you can go ahead and take 10 seconds on that. All right. So most of you got that, though there is this little sub-set we have thinking that we have one, so let's actually write this out here. So if we have a 3 d orbital, we're talking about n minus l minus 1, what is n equal to? What is l equal to? OK. and 1 is equal to 1. So, it turns out that we have zero nodes that we're dealing with when we're talking about a 3 d orbital. OK. So we should be able to figure this out for any orbital that we're discussing, and when we can figure out especially radial nodes, we have a good head start on going ahead and thinking about drawing radial probability distributions. We did it for the s orbitals, we can also do it for the p, we can do it for the d. All we have to figure out is how many nodes we're dealing with and then we can get the general shape of the graph here. So, let's actually compare the radial probability distribution of p orbitals to what we've already looked at, which are s orbitals, and we'll find that we can get some information out of comparing these graphs. So if we draw the 2 p orbital, what we just figured out was there should be zero radial nodes, so that's what we see here. The other thing that I want you to notice, is if you look at the most probable radius, for the 2 s orbital it's actually out further away from the nucleus than it is for the 2 p orbital. So what we can say here is that the 2 p is less than or smaller than the 2 s orbital. So think about what that means, we're, of course, not talking about this in classical terms, so what it means if we have an electron in the 2 p orbital, it's more likely, the probability is that will be closer to the nucleus than it would be if it were in the 2 s orbital. We can also take a look the 3 s, which we have looked at before, and we figured out that that should have two radial nodes. We can look at the 2 p, which should have one radial node, and we just figured it out for the, excuse me, for the 3 p has one radial node, and for the 3 d here, we should have zero radial nodes, we just calculated that. So again, what we see is the same pattern where the most probable radius, if we talk about it in terms of the d, that's going to be smaller then for the p, and the 3 p most probable radius is going to be closer to the nucleus than it is for the 3 s most probable radius that we're looking at. So, there are 2 different things that we can compare when we're comparing graphs of radial probability distribution, and the first thing we can do is think about well, how does the radius change, or the most probable radius change when we're increasing n, when we're increasing the principle quantum number here? So, from going from the shell of n equals 2, let's say, to the shell of n equals 3. And what we find is we're going from about or exactly a 6 a nought here, to almost three times that when we're going from 2 s to 3 s. So we say if n increases, the orbital size is also going to increase. And when we talk about size, I'm again just going to say the stipulation that we're talking about, probability -- we're not talking about an absolute classical concept here, but in general we're going to picture it being much further away from the nucleus as we move up in terms of n. The other thing that we took note as is what happens as l increases, and specifically as l increases for any given the principle quantum number. So if we're keeping n the same, we look and what we saw was that size actually decreases as we increase the value of l. So, I want to contrast that with another concept that seemed to be opposing ideas, and that is thinking about not how far away the most probable radius is, but thinking about how close an electron can get to the nucleus if it's actually in that orbital. And what we'll find is that we actually see the opposite. So if we compare l increasing here, so a 3 s to a 3 p to a 3 d, what we find is that it's only in the s orbital that we have a significant probability of actually getting very close to the nucleus. So, if I kind of circle where the probability gets somewhat substantial here, you can see we're much closer to the nucleus at the s orbital than we are for the p, then when we are for the d. So, the size still for an s orbital is larger than for a d orbital, but what we say is that an s electron can actually penetrate closer to the nucleus. There's some probability that it can get very, very close the nucleus, and that probability is actually substantial. A kind of consequence of this is if we're thinking about a multi-electron atom, which we'll get to in a minute where electrons can shield each other from the pull of the nucleus, we're going to say that the electrons in the s orbitals are actually the least shielded. And the reason that they're the least sheilded is because they can get closest to the nucleus, so we can think of them as not getting blocked by a bunch of other electron, because there's some probability that they can actually work their way all the way in to the nucleus. So, this is a concept that's going to become really important. Soon when we're talking about multi-electron atoms, and I just want to introduce it here, that it is sort of opposing ideas that even though the s is the biggest and it's most likely that the electron's going to be furthest away from the nucleus, that's also the orbital in which the electron can, in fact, penetrate closest. All right. So I think we are, in fact, ready to move on to multi-electron atoms, and what happens is when we solved the relativistic version of the Schrodinger equation and we're discussing more than one electron, we actually have a fourth quantum number that falls out and that we need to deal with and this is called the electron spin quantum number. And I promise, this is the last quantum number that we'll be introducing. And this spin magnetic quantum number we abbreviate as m sub s, so that's to differentiate from m sub l. And when you solved the relativistic form of the Schrodinger equation, what you end up with is that you can have two possible values for the magnetic spin quantum number. You can have it equal to plus 1/2, and that's what we call spin up, or you can have it equal to minus 1/2, which is what we call spin down. So, there's two kind of cartoons shown here that give you a little bit of an idea of what this quantum number tells us. And this spin is an intrinsic quality of the electron, it's a property that is intrinsic in all particles, just like we would say mass is intrinsic or charge is intrinsic. Spin is also an intrinsic property. One way to think about it, if we want to use a classical analogy, which often helps to give us an idea of what's going on, is the spin of an electron, we can picture it rotating on its own axis. So that's kind of what's shown in these pictures here. So you can see, if it's spinning on its own axis in this direction we'd call it spin up, where as this way it would be what we call spin down. So, it turns out there's not actually a good classical analogy for spin, we can't really think of it like that, but if that helps give you an idea of what's going on here then it's valuable maybe to consider it spinning on its own axis, even though that's not technically what's exactly happening here. But the reason that I like that analogy is that it points out a very important part of spin, and that's the idea that it's a description of the electron. It is not dependent on the actual orbital. So we can completely describe an orbital with just using three quantum numbers, but we have this fourth quantum number that describes something about the electron that's required for now a complete description of the electron, and that's the idea of spin. So, we need to actually add on this fourth quantum number, and it's either going to be plus 1/2 or negative 1/2. So, we can talk a little bit actually, because it's an interesting story about where the idea of spin came from, and it was actually first proposed by two very young scientists at the time, George Uhlenbeck shown here, and Samuel Goudsmit who's here, and they're with a friend, and I can't remember who that is, but he did not have anything to do with discovering spin. And what you can see in this picture is that these are actually, they're pretty young guys in this picture. I think this is taken about two years after they discovered the fourth quantum number. So hopefully, you can picture yourself at this age in a similar situation with an anonymous friend and think this is something, kind of observations maybe you can make as well. And what they were doing when they discovered that there must be this fourth quantum number is they were looking at the emission spectrum of sodium, and, so specifically, they were looking at the frequencies, and if we think about the frequencies of sodium, it was already known at this time that you could calculate what those would be based on the difference in the energy levels -- this happened in about 1925. So they actually knew exactly what they were expecting. So, let's say they were expecting to see one certain frequency or one line in the spectrum at this point here. It turns out that what they actually observed, so this is the actual of what they saw, is if they were expecting their line at some given frequency, which I'll show by this dotted line here, what they actually saw was two lines, and one was just the teeniest tiniest bit of a higher frequency than what was expected, and one was just, just below what they expected. And if we're talking about things in spectroscopy terms here, this is what we call a doublet. So it's centered at this frequency that was expected, but it's actually split into two different frequencies. And this was an amazing observation that they made. They were totally surprised and excited, and they were thinking how could this happen, where did we got this split doublet from. And what they could come up with, what they reasoned, is that there must be some intrinsic property within the electron, because we know that this describes the complete energy of the orbital should give us one single frequency. And that the fact that it split into two was telling them that there must be some new property to the electron, and what we call that now is either being spin up or spin down. But at the time, they didn't have a well-formed name for it, they were just saying OK, there's this fourth quantum number, there's this intrinsic property in the electron. So, where the story gets kind of unfortunate, but also a little bit more interesting is the fact that well, they did publish what they observed, and they did write that up. They put it in a pretty low impact paper. I think it was in French, so it wasn't really hitting all the scientific community of the time in any major way. And they didn't put their explanation of what they thought was going on, it just sort of was observing what they saw. These were very young scientists, of course, so what you would expect that they would do, which makes sense, is go to someone more established in their field, because they have the completely radical revolutionary idea, let's just run it by someone before we go ahead and publish this paper that makes this huge statement about this fourth quantum number. So the person they chose to talk to, and I think it was just Goudsmit that went to him and discussed it is Wolfgang Pauli, which is shown here. So how many before five minutes ago had heard the name Goudsmit? All right, a couple. OK, cool. How about Pauli, like the Pauli exclusion principle? Hmm, OK. So, Pauli seems to be getting a little bit of fame that you'll see in a second here. So, it turns out they go and they discuss their idea with Pauli, and what Pauli tells them is that this idea is ridiculous, that it's rubbish, if they go ahead and try to publish this their scientific careers are ruined. They can pretty much pack it up and go home, because everyone's going to think they're ridiculous, no one will believe what they say, and it's a stupid idea anyway, is basically the gist of this conversation. And in chemistry, just like in any discipline, you have all types of scientists, but also all types of personalities, and unfortunately Pauli had a personality that was known for, first of all, being very arrogant, but also the very unfortunate trait of taking other people's scientific ideas as his own. And as the story goes, as Goudsmit was leaving and the door with slamming, Wolfgang Pauli was already writing down this idea into a scientific paper of the idea of a fourth quantum number. And, in fact, he did make some more strides, he was a brilliant thinker, maybe he put it more articulately than those two younger scientists could have. But now, it has come to light that they are the ones that do get credit for first really coming up with this idea of a spin quantum number, and it's interesting to think about how the politics work in different discoveries, as well as the discoveries themselves. You see that a lot with Nobel prizes, there's usually a nice little scandal, a nice little interesting story behind who else was responsible for the Nobel Prize-worthy discovery. So, here, Pauli came out on top, we say, and he's known for the Pauli exclusion principle, which tells us that no two electrons in the same atom can have the same four quantum numbers. So let's just think exactly what this means, and that means that if we take away function and we define it in terms of n, l and m sub l, what we're defining here is the complete description of an orbital. In contrast, if we're taking the wave function and describing it in terms of n, l, m sub l, and now also, the spin, what are we describing here? An electron. So now we have the complete description of an electron within an orbital. So, that's an important distinction to make -- what three quantum numbers tell us, versus what the fourth quantum number can fill in for us in terms of information. So what that means is that we're limited in any atom to having two electrons per orbital, right, because for any orbital we can either have a spin up electron, a spin down electron, or both. So, if we look at neon just as an example, neon has ten electron in it, and once we look at all the orbitals written out here, this is probably a familiar thing for you to look at, but it's important to think about why, in fact, we don't just put all ten electrons, why wouldn't they just want to go in that ground state, that lowest state, right? That should be the most stable, the lowest energy orbital for it to be in, and the reason they can't do that is because of the Pauli exclusion principle -- the idea that all of the electrons have to have a different set of four quantum numbers, so only two of them can have the same set of three quantum numbers here, because for m sub s, we're only left with two options. So let's try a clicker question and thinking about the Pauli exclusion principle. It might look a little bit similar to a question we just saw, but hopefully you'll find that it is, in fact, not the same question. So you can go ahead and take 10 seconds on that. OK, great. So, most of you recognize that there are four different possibilities of there's four different electrons that can have those two quantum numbers. Actually the easiest way is probably to bring this down here. And the next highest percentage of you thought that we still only had two. So, remember we solved this problem earlier in the class, but we were talking about orbitals. So there's two different orbitals that can have these three quantum numbers, but if we're talking about electrons, we can also talk about m sub s, so if we have two orbitals, how many electrons can we have total? Yeah. So we have two orbitals, or four electrons that can have that set of quantum numbers. So you'll notice in your problem-set, sometimes you're asked for a number of orbitals with a set of quantum numbers, sometimes you're asked for a number of electrons for a set of quantum numbers. So make sure first that you read the question carefully, and realize the difference that is between the two. So, that's where we'll end today. So on Friday, we'll start with talking about the wave functions for the multi-electron atoms.
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https://ocw.mit.edu/courses/5-95j-teaching-college-level-science-and-engineering-fall-2015/5.95j-fall-2015.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation, or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: And before we get into the topic, which is educational teaching with educational technology, I just want to go back to some of the previous assignments, a little bit of logistics, et cetera. So you should have feedback on all the assignments that you can be given feedback on. The assignment where you gave a presentation to a friend, I can't really give you feedback on. I looked over it. You all made nice observations, good pointed observations, and I hope that you can see how-- and I think I made this comment on the wiki-- that you can see how just your presence, and the choice of words that you make, can have a big effect. And that was the whole point of that assignment. So that's my feedback on that assignment. The previous one, the last one that I feel like I can give you substantive feedback on is the active learning, how to incorporate active learning. As I mentioned last time, you guys did such a great job of incorporating active learning in the first time you'd submitted that, where it was just creating a course outline, that there wasn't a lot extra to add. But I added comments whenever possible. One thing that I commented on at least two people's homework, the one about active learning, was the idea-- you mentioned you doing demos. So there's a very interesting piece of work out of Eric Mazur's group at Harvard about the use of demos. And that paper is called "The Crouch and Mazur Paper." And it's a optional reading in, I believe, it's the active learning session. So this is the active learning class, which is class 5, right? Am I right about that? Or it was 6? AUDIENCE: It was 6. PROFESSOR: 6? AUDIENCE: Yes, 6. PROFESSOR: 6. AUDIENCE: [INAUDIBLE] PROFESSOR: Oh, OK. So it's 5. Yeah, so that's a really interesting paper. Because I think sometimes we say, oh, I'm going to do a demo. Apriori, a demo isn't active. I mean, it's active for you. You're doing the demo. But it's not necessarily active for students. And if we enter into the enterprise of doing a demo thinking we're being-- well, thinking they're being active-- they might not be. And Croutch and Mazur did an interesting study where they looked at-- they were doing a freshmen mechanics class. And they had amasses on an air track, when you have like a massive 2M, and you slam into it with a mass of M. What happens to the mass of 2M? What happens to this mass of M that was coming in, that kind of stuff. And so that it would show the students these demos. Oh, look what happens. Here comes a mass of M at a velocity, da, da, da. And then when they asked students in the same class at the end of the same class period to say what happens when a mass of M collides with a mass of 2M, a mass of M going x miles an hour collides with the mass 2M, what happens, students got it wrong. They had just seen the demo. They had just seen the demo. And they still got it wrong. And I think that's a really profound finding. Because as experts, we think, of course, they're going to see this. It's going to be emblazoned into their brain. They'll never forget it. But they never learn it, basically. So it's more evidence for active learning for sure. But what they found out was they forced students to make a prediction, to discuss that prediction with somebody else before they saw the demo. And when they did that, the scores on these sort of post-demo quizzes, if you will, went way up. So it's a small tweak, but it's a profound one. Because you might really just be wasting your time if you show a lot of demos and you don't ask students to make these predictions beforehand. This goes back to the mathematician George Polya, who wrote in the '50s about asking people to make predictions before you gave them the answer. I'm not sure he had any data for it. But this was his thing. This is pretty good data suggesting that you really should ask students to make predictions before showing the demo. So a couple people brought that up. And I wanted to make sure everybody heard that. OK? If you can't find this reference, let me know. It might be hiding somewhere else. I'm pretty sure it's in this-- it is? OK. OK, so that was that. But overall, a great job on that assignment. And then I briefly looked over the assignment it was due today or whenever creating prompts at various levels of Bloom's Taxonomy or a taxonomy of your choosing. Again, I thought you did a great job. I'd like to hear a little bit of reactions to that assignment. Was it difficult? Was it easy? Do you see how it could be useful? Does it seem like a waste of time? AUDIENCE: I think it was very, very useful. Yeah, it lets you see what the students will go to when you [INAUDIBLE]. Otherwise, because it's like you are solving too. So you kind of feel and see what level to take them to. It's very good. PROFESSOR: Great. AUDIENCE: Otherwise, you don't know. You just [INAUDIBLE]. And then you don't know how [INAUDIBLE] PROFESSOR: Exactly. And I mean, it's happened to me when I first started teaching intro materials. [INAUDIBLE] was like, OK, they'll calculate the stress on this plane, and blah, blah, blah. And you're like, well, why are they doing that? What is that getting them? And it's just this reality check for you, so that you don't go off into the weeds sort of. So, yeah, [INAUDIBLE]? AUDIENCE: I think [INAUDIBLE] most of very good question that [INAUDIBLE] They're [? setting ?] very good questions and [INAUDIBLE] more than just one, [INAUDIBLE] three at the same time. So I think it's really very good. It's a good experience. PROFESSOR: Right. And also to make sure you're aware of that, and then also decide, well, if they don't remember the value of the universal gas constant in these units, but they do the apply part fine, they just use the-- so it's important, we always have to think about that when we grade questions. But I think by parsing it out like that, it makes you more cognizant of those issues that might come up, in terms of how the students respond to the question. AUDIENCE: I thought it was actually difficult to make a question that only satisfied one of-- like applied [INAUDIBLE] Because everything I was asking [INAUDIBLE] to try to apply, was also analyzed. Because I tried to-- I guess i could have just made one that did both. But I was sitting and I was like, no, OK, not this one. But really, I almost left it blank and said, I'm stuck. I finally come with something that I think just answered the [INAUDIBLE]. But it was interesting to think about. PROFESSOR: Right, right. And you know, in real life, you don't have to just have one. You can embed them together, but you know, recall and apply, or whatever. And I think for some topics, it might be easier to add to silo-ize it a little bit. Given the data, construct a whatever, so that's more of an apply. But there may be classes of subjects where you can't really do that quite the same way. Other reactions? AUDIENCE: I was also thinking of it in terms of grading. Like if you have a rubric or you have assigning the number of points based on which level you were asking. Like as opposed to say, OK, five points for each question, to say, OK, you get one point for the remembering and then you get 10 points for the creative. PROFESSOR: Right. Right. I would be careful-- I mean, I think that's a fine idea. I would be careful to make sure that students know that, so they don't kill themselves memorizing pi out to 25 digits or something, right? So if you do have these sort of, this is what I think is important, you'll need to specify that. I mean, and a rubric could incorporate that for sure. But you do want to make sure students know what your logic is. AUDIENCE: I think also there are some questions you said that you give the formulas to the equation. Just to make sure that the student [INAUDIBLE] PROFESSOR: Yes, yes. Right. And in most classes, that's totally reasonable. You know the joke I make about it's rare to real life that somebody comes in and says, I need this answer right now! Don't look up anything. I mean, it's kind of a weird pretense. So it's usually reasonable to just put the equations at the bottom, or write it in the problem statement. Anything else on that? OK, so today we're going to talk about teaching with educational technology. I had written on the website to bring a web-enabled device. If you didn't, I have an extra laptop. And we can share. It's shouldn't be a problem. As part of this, I hope that you'll be able to identify appropriate educational technologies to advance an intended learning outcome. And I want to get back to that in just a second. And maybe describe best practices and potential pitfalls of particular educational technology, so somewhat modest goals. I think it's a really interesting thing to think about technology-- and I didn't make this up. I heard this at a conference. So they said, when you walk into Home Depot or one of these big home improvement stores, if you happen to be blessed by having an employee come up to you-- because sometimes it's hard to find an employee in those stores-- what do they ask you? AUDIENCE: Is everything OK? Is there anything I can help you with today? PROFESSOR: Right. OK, is there anything I can help you with? If and you say, yeah. Yeah, you can. And then they say? Usually, it's, what are you working on? What are you trying to do? They don't say, what tool do you want? I mean, they might. But if they're good, they don't. And I think that's a word the wise when we think about educational technology. It's not, what looks really cool and fun, and what tool do I want? It's, what am I trying to build? What do I want my students to build? What am I trying to build? What am I trying to accomplish? And then I can step back and say, which of these technologies is going to help me do it? And it's easy to get wrapped up-- I mean, we're all kind of techie, geeky. I mean that is a compliment. But it's kind of fun. Like, oh, cool, look what this can you. And then we go walking around looking for something to do with it. And that might be OK, but you have to check your assumption. You have to check your assumptions about why you're doing it. Are you doing it just to be cool? Or are you doing it because it's really going to help student learning, or it's going to help advance the learning outcome? We can use educational technology as some kind of a scapegoat, but that's true for anything. It's true for that amazing exam question that you thought up in the shower. Like, I know what I'll do. They'll do this, and do that, and blah, blah. Maybe it's cool, but maybe it's not going to get you anywhere. Maybe it's not consistent with your learning outcomes. So it doesn't have to be technology. But often, technology is kind of pretty shiny. And we tend to gravitate toward it. But it's remembering that we're using it to advance some particular learning outcome, OK? So you guys did the readings. What are educational technologies? What does it mean? Can we come up with a definition? AUDIENCE: Yeah. I think from what I can understand, educational technology is a tool which, perhaps, the student and also have the teacher. So that means something that can enable inter-communication between the teacher and the student. And then the student and the student. PROFESSOR: OK, so we have a lot of ideas in that statement. So it's a tool. You said, it enables communication. AUDIENCE: Yeah, students connect. And then the teacher [INAUDIBLE]. PROFESSOR: And I think you had something else in there. Did you? Or no? Somebody else? [INAUDIBLE]? AUDIENCE: Education technology is like a tool to enhance students' learning-- enhancement of student's learning. And enhancement of instructors' teaching as well. AUDIENCE: The technology part is like a electronic device. PROFESSOR: So it has to have electricity? It has to involve electrons? No, that's everything. Wait, let's let Michelle-- what should we say about that? AUDIENCE: I think that's kind of correct to say it has to be an electronic device, right? Like if it's a clicker or if it's a-- [INTERPOSING VOICES] PROFESSOR: It must be electronic. That sounds so like 1950s. I think we had-- yeah, Gordon? AUDIENCE: I would just adjust it a little bit and add in front of the two [INAUDIBLE]. AUDIENCE: A kind of tool that's created [INAUDIBLE] by time or place. PROFESSOR: So it lends itself to asynchronicity? AUDIENCE: Yes, something like that. PROFESSOR: Most-- Anything else? AUDIENCE: I think it's also enhancer, easier use of materials. PROFESSOR: Facilitates-- AUDIENCE: Yeah, materials like. It has to [INAUDIBLE]. PROFESSOR: Right, a distribution, aids distribution? AUDIENCE: [INAUDIBLE] PROFESSOR: Did I hear something else? AUDIENCE: [INAUDIBLE] PROFESSOR: Efficient, that's a loaded word. Anybody have anything else? Or does somebody want to make a comment on this collection of attributes? AUDIENCE: [INAUDIBLE]. PROFESSOR: I don't think I have any cynics in this-- well, the cynics haven't said anything. AUDIENCE: I think all the comments are positive things. Like if you're doing it well what should happen. But they aren't necessarily what actually happens. Because most of the time, it's probably not done well. Using technology is not optimal. PROFESSOR: OK, that's one observation. It's sort of, you know, that rhyme, there was a little girl who had a little curl right in the middle her forehead. And when she was good, she was very, very good. And when she was bad, she was horrid. So it's a little bit like this, right? When it's good, it's very, very good. And when it's bad, maybe it's even worse than if you hadn't bothered. But-- Alex, you had something to say? No? AUDIENCE: Well, on the side of the cynics, I thought that students are often disinterested when they don't see the point, or don't see them using it later. It's not necessarily a definition of it. PROFESSOR: Right, right. No, but that's a very good point. And somebody had quoted this idea of an app, that if an app doesn't load in something like seven seconds, people stop using it. Like if you try to get an app to open on your phone, and you're like, what's going on here? Come on, 1, 2, 3, 4-- OK, forget it. And that's a similar idea, that if it doesn't look like it's going to help you-- and whatever that means, help, it's self-defined, help. But if it doesn't look like it's going to help you, you're not going to buy into it. Some other thoughts about this list? Personally, I don't think that to be an educational technology it has meet all these criteria at once. That's an important thing to note, that it might improve productivity. The old school learning management systems that you put up problem sets, and you put up-- the grade book was in one place, and that kind of thing. That might help your productivity. You're not looking around for stuff. But it's not necessarily promoting learning at any level. It's a delivery system, a materials delivery system. So it kind of does this. And it does this. But it might not really be part of the learning enterprise. It's a store. It's a repository. And that's fine. There's nothing wrong with that. But that's one type. If you're saying that it enables communication, well, that might be part of this. But it might be something else. If I say, all right, it's a wiki. And I want you guys to read what everybody else wrote, and to make a comment on at least one other persons' posting, well, I have facilitated communication between you in an asynchronous way. I didn't tell everybody to do it at noon on the 29th of October. But it does facilitate communication. I think this one is a loaded one, for sure. And it gets to Michelle's comment about if you did it right, it would do that. But if you did it wrong, it's likely to not do that at all. So these are attributes that you could use to describe some educational technologies. I do have a definition from the reading book. But before I put that up, any other thing you want to add? It's interesting to think about it, like does it have to be electronic? So I have had classes in the past where people argue, OK, well, a blackboard is an educational technology. I mean, you could do that without some sort of human-made product. And without it, I would have to be walking around speaking very slowly and hoping you took notes, or walking around with-- I don't know-- a piece of paper that I wrote on, and sharing it with you or something. So a blackboard does fit a lot of these categories, right? It's an efficient delivery system. At some level, it enables communication. It's a distribution of learning materials. So we want to think about whether it means it has to be electronic. And you may have guessed by now, there is not a single definition, so that's the other issue. Comments? Yes, go ahead. AUDIENCE: Just a comment. What about if it's not well-moderated [INAUDIBLE]? PROFESSOR: Yeah, that's the efficient one, that we've all gotten sucked into websites, or sucked into, perhaps, an online discussion, or commenting on something. And it might be interesting. You might learn. But it may not be efficient at all. Or you might just be on the screed end of things, where people are just either ranting or they're reinforcing misconceptions or whatever. So yeah, you do have to kind of make sure people are using it carefully. But that's true for virtually anything you do. Any other comments? AUDIENCE: And I think, as well, that the fact that these tools are valuable. A student can decide to stay home in the bad weather since he knows this is available. Then he can go [INAUDIBLE]. That could be good and bad. Because the student would continue to learn on his own and become a good student. And then he could miss out on some critical points as well. PROFESSOR: Right, right. But that implies some sort of like web-housed delivered repository of content, or of video lectures, or whatever. You could have a course that used clickers, let's say, that didn't even have a website, but they used clickers. So that course might be using educational technology, maybe in a very, very good way in the classroom, but they don't even bother with a website. So in that case, the student has to come to class if they want to experience that particular use of technology. I think you're right, that if you do have a website, or you do have lectures online, or you do have rich content online, then it does sort of, perhaps, make being in the classroom a little bit less necessary. I will say, if you employ active learning methods in class, group discussions, et cetera, that being in the classroom is, certainly, I would argue, is value added for the student. So it's not a replacement experience, but it might be better than nothing, let's say. It might not. Yes, Gordon. AUDIENCE: I found something written very interesting when it said that if you use clickers-- I would love to do that. To take attendance of the students, that it might [INAUDIBLE] the students from using technology. But it's only [INAUDIBLE]. It might actually be good to have technology they can use non-evasively to find out if the student is actually coming to your class. AUDIENCE: What if they gave their clicker to their friend? PROFESSOR: Yeah, it's true. AUDIENCE: One person will come with five friend's clickers, the clicker won't-- AUDIENCE: If you have like RFIB technology or something like that that can use to [INAUDIBLE] come to the class to know that, OK, this person is the one using [INAUDIBLE] [INTERPOSING VOICES] [INAUDIBLE] AUDIENCE: So in certain courses you have to do that. AUDIENCE: Yeah, you must do that. AUDIENCE: You have to show up? AUDIENCE: Yeah, you should show up. But if you decide it's difficult to take attendance, and if you have this wonderful way of getting the students to [INAUDIBLE]. AUDIENCE: [INAUDIBLE] PROFESSOR: Well, I don't know if there's an answer to that. For some students, it's never a problem. For some students, it's awesome. For some students, it's just more information. And it's just another path. And they're learning. And they're learning. And they're learning. And it's supplementing what they're learning in the class. It kind of depends on-- AUDIENCE: Taking attendance that Gordon is talking about, so is it [INAUDIBLE]? PROFESSOR: Well, yeah. And actually that's an important point is that I think I showed these slides about clicker use in 511 1. I showed when we did active learning. And the first year they used them, they used them more or less to take attendance. And students figured that out pretty quickly. They did all sorts of things, sharing clickers, clicking and then leaving, or coming in at the end and clicking, but not being there the whole time. And if you ask students, well, what did you think about the clickers, they're like, fu. This is crappy. This didn't help me. This is terrible. And probably it didn't. There's an attitudinal thing that says at some point, if you think it's not helping, it's not helping. So the next year, they made it a point to really incorporate it into the class, to ask good questions, rich questions, questions that had interesting and informative distractors. So the wrong answers told the instructors what the problems and misconceptions were. And this was made clear to the students. And in that way, if you ask students what they thought about the clickers, they were not at all resentful of the use of clickers. And they did not see them as a way of just taking attendance. They didn't think they were sort of baby or whatever. And they were therefore more effective. [? Rachel, ?] you had a comment. AUDIENCE: Yeah, I was just going to say going to 802. PROFESSOR: That's the E and M, Electricity and Magnetism freshman year. AUDIENCE: The class I took it in [INAUDIBLE] So we did all the online [? courseware ?] and then the idea was that you come to class and you just learn for two hours from the teacher using group discussions, and all these other things. And it was one of the worst classes I've ever [INAUDIBLE]. Because the professor wasn't prepared to do these two hour discussions, like group discussions, and group work, and all these different things that you could tell he wasn't used to doing. And it was just a giant mess. So I think they're really useful tools. But they also require the professor to really put something into it for the students to get something out of it. PROFESSOR: Right, right. And that's a consideration. For some things, you can adopt them. There's kind of a low bar for adoption and a low bar for solid use, maybe not innovative, maybe not groundbreaking, but a solid pedagogical use. For others, if you're going to use them effectively in the classroom, you're going to have to spend some time engaging with the technology, making sure you understand it, making sure you understand what you can do with it. Otherwise, it's not going to work. A couple stories, one is the idea of-- well, maybe I'll hold that one. I'll hold that one. I'll bring it up later. Remind me. Any other comments about this? So this is a definition that I got, I think from one of the [? reasons, ?] so technological processes and resources that are used, created, or managed for learning and/or improving performance. So that's crazy broad. And then in that definition, there's this technological, which gets around the electronic, but brings back the idea of well, a blackboard is probably a technology at some level, or at least it was at some point. So it's admittedly an incredibly difficult concept to pin down. And maybe at some point, you don't necessarily have to. You have to decide which slice of it you want to deal with. And so the other important thing is that it's-- and it doesn't have to be, but it can be-- an integral part. Or it should be well-integrated within the course, and or it could be a supplemental learning aid. So it could be a way of delivering content outside of class, or the idea of an enrichment of the material. And it could be something like hardware, really just like hardware. I'm going to come back to how we chop this up in just a second. Remember that we don't want to do any thing-- this is the idea about walking into Home Depot. We don't want to do anything without thinking first about our intended learning outcomes. What intended learning outcome is the technology going to advance, not, what am I going to do with this technology? And so in terms of categorization, this is something that I came up with just after grappling with things. So this is just my thing. And we're going to talk more about different ways to think about it. So this is just one slice of it. You could think about technologies that you use in the classroom, so that would be clickers, that would be using some sort of projection system, using a smart board, using some sort of simulations that you show students, et cetera, like virtual demos. It could be a way for students to engage with the content and with other students outside of the classroom. So it could be some sort of a wiki, some other collaboration tool, and/or it could be something that delivers the content to the student outside of class. And there may be an overlap. You may show a simulation in class, and then decide students should have access to it outside of class. So they're not mutually exclusive. And so the engagement with the content can include these collaboration tools, extra materials, content enrichment materials. And then the third one might be student assignments. So the idea, maybe you're going to say to students, OK, I want you to create a website for a company that you're going to create, and this parameters. And you want to make sure you're explicit about what the company sells, and how you deal with customers, or whatever. The assignment involves the technology. Go out and make an audio recording of people using a particular technology. Or if you're going to design something for use in some community, interview the people in the community, and submit this audio recording. So it could be that technology could be used in the student assignments. And you could back that all the way up to online problems as they use in MITx. And then for assessment and feedback, the idea that you can either use clicker questions, which would be for the formative feedback. Or you can use it as some sort of an online quiz, or some other way of giving feedback to the students. So that gets to the summative and the formative idea. And then the idea of just using a learning management system, just to administrate the course. That's one slice. There's another couple other very interesting ways to slice this up. And I have a handout here, because these are woefully tiny, these schematics. So if you just take one and pass it along. One is called the SECTIONS framework. And the other is called the backward Ed Tech Tool Flowchart. So these are different. These are not mine. And these are different ways of thinking about technology. So the first one is this Ed Tech chart. And I know you can't read that here. So that's why you have the handout. And I like it, because it kind of starts with the activity that you want students to use. So it starts with this, what do you want students to do? And that's what the purple box are, the student tasks. And then it kind of creates these sort of guiding questions that sort of reroute you. And then it gives you a suggestion for a tech tool. So I'm going to give you a few minutes just to parse that, just to engage with it a little bit. AUDIENCE: It's very beautiful. PROFESSOR: I like this one. I like this one. I think it's very useful. Any other comments or thoughts about it? Yeah, Gordon? AUDIENCE: I think it's [INAUDIBLE]. PROFESSOR: OK, Gordon doesn't get it. Somebody want to-- maybe nobody gets it. But do you want to explain it a little bit? Out loud, David, please. AUDIENCE: Yeah, [INAUDIBLE]. PROFESSOR: But can you explain to everybody? AUDIENCE: [INAUDIBLE] PROFESSOR: Oh, sure. AUDIENCE: OK, [INAUDIBLE] PROFESSOR: It's certainly not exclusive. I mean, it doesn't have everything on here. But it kind of gives you, I think, a way of thinking about, OK, I start with what I want students to do, which most of us have a pretty good handle on. And then it helps you just think backwards to the technology. [INAUDIBLE]? AUDIENCE: Yes, I just want to add that it's only about what you want to do, and how can you did it. PROFESSOR: Yes. AUDIENCE: That's what it's all about. And secondly, [INAUDIBLE] but what you want them to do, and how do you have them doing it, that's what [INAUDIBLE]. PROFESSOR: Yeah, Yeah. No, it is great. Other thoughts? And yeah, we'll get to get back to the idea of the taxonomy shortly. Other thoughts? Let's see, there's a second one, which is on the other side of your handout. And so these are just-- I know you can't see this. That's why you have the handout. These are just ideas. Again, there's no rules really. But depending on what you're trying to do, it just may help you start to think about how to get the right tool for the job. So it's called sections, because each of those boxes with the red outline, S-E-C-T-I-O-N-S. The S is student characteristics. The E is ease of use. C is cost, teaching and learning, interaction to promote active learning, organization, novelty, and speed. And this chart, I think-- I'm going to let you engage with it for just a minute. But it brings up the idea of the learning curve for the instructor, which you don't want to lose sight of, that you want to factor that into your choices about what technologies to use. So again, I'm going to shut up for a minute or so, so you can look at this. AUDIENCE: [INAUDIBLE]. PROFESSOR: Right, right. But what's the difference between this one-- I mean, there's many differences, obviously. But what's kind of a utility difference? AUDIENCE: One's a flow chart that points you to the answer. So it asks the yes or no questions. The other just gives you the-- then you have to select from a-- it doesn't guide you directly. PROFESSOR: Right. Exactly. It kind of helps you think about it. Oh, my gosh, I have no money. So that's certainly going to rule out a whole set of options for you. But it doesn't say, oh, you have no money, think about these. It's not a flowchart. It doesn't point you to an end result. But if you went through this exercise and answered these questions, I think you'd be in a better place, with respect to choosing the particular technology. Gordon? AUDIENCE: Yeah, also I think that second one is not [INAUDIBLE] if you want to use technology and you have the money, [INAUDIBLE] the things you can think about. Because if you don't think of everything [INAUDIBLE] PROFESSOR: Right. It's some things to just keep in your head when you're making these choices. It's not necessarily a formalism for making a choice. So one other slice of all this technology is kind of a classic, I would say classic, model, which is called the SAMR model, S-A-M-R. I know I'm throwing a lot of models at you. But you have to kind of find the slice that works for you. The SAMR model, I think, is very useful to keep in mind when you're deciding why you're gravitating to a particular technology. And it's this idea of, is the technology just a substitution? So back in the day, when everybody said, oh, courses need web pages. You have to have a course web page. Well, what was the course web page? It was an electronic version of your syllabus, basically. It had the name of the course. It had some picture. And then there was a link. And you could click on it, and the students got the syllabus. They got a text version of the syllabus, a soft copy of the syllabus. But then they printed it out and brought to class. So it was a one-to-one substitution. Instead of seeing the syllabus in a piece of paper, they see it online. It was complete substitution. It didn't do anything extra, except maybe made it a little easier for students not to lose the syllabus. But it didn't really have any functional difference. Then the next level-- that's the S. The A is that there's an augmentation. So there's a substitute, but with a little bit of an improvement. So maybe students could hand the homework in online. They're still handing homework in. They're still probably writing their homework out, scanning it, and handing it in. But it's a little easier. It's a little bit easier. Maybe you don't lose their homework, that kind of thing. So that's the augmentation. The next level will be a modification, so some sort of serious redesign of a task that you couldn't do before that technology. And then the final would be it just redefines the tasks that you do, the tasks that students do, in a way that's completely different. So I'm going to ask you to think about that for your own courses in about two minutes. There's a little bit of a schematic here that might help. These are sort a little bit baby. But the idea of substitution, so the idea of using a word processor instead of a typewriter, that was a substitutional technology. So then augmentation, you get spell check. Once you're in a computer, you get to have spell check, and formatting, and cutting and pasting. And that's different than what you can do on a typewriter. And then so modification, well, there was like an email, or you could put graphs and images right into the document. It changed sort of the way we viewed a whole document. And then the idea of having hypertext, or some sort of a web page that students navigated in a nonlinear, or a non-prescribed way, that would the redefinition, let's say. So that's some sort of baby examples of that. And then this next one, which is pretty fun, is there's this woman, Kathy Schrock. And if you Google Kathy Schrock, she comes up. She's got a ton of stuff. And this to get back to Bloom's Taxonomy there, is that she's taking Bloom's Taxonomy, remember, understand, apply, analyze, evaluate, create, and then she said, OK, well, here's some apps that can facilitate those things, facilitate the act of remembering, or facilitate the act of evaluating, or creating. So it's not mutually exclusive. It's not exhaustive. But it is important to remember that some of these tools are supporting remembering, or they're supporting some other level, understanding. But they're not necessarily at the highest level. Or they're not at the lowest level. It's good to be aware of that. Before I charge you with your assignment, I did want to demo a few other technologies, just so you can take a look at them. And there was that list in the reading, which I hope you checked out. I invited you to this-- well, I thought I invited you to-- oh, wait a minute. Huh. Hang on. Well, let's just start up a new one. So I can create this picture here. And I can share this, then, with you. And I can invite the board. And so this is the link. And if you have a device and want to log in there, you can check it out. And then you can write on it. Oh, somebody wrote on it. Somebody erased my picture. I don't know who that was. Oh. AUDIENCE: Is that a [INAUDIBLE]? PROFESSOR: U-Q-P-L. AUDIENCE: So that's different from the link that was in the wiki. PROFESSOR: I know. I just created a new one, because I didn't want to go look for the wiki link. AUDIENCE: But this is free? PROFESSOR: This is free, yeah. So we're working a little backwards here. I'm showing you the tool. And now I'm asking you what you might want to do with it, which is not necessarily best practice. But I have a use that I-- you can change the color. This pen here changes the color. No, something changes the color. Maybe not. Oh, somebody got blue. When I taught thermodynamics, we would often do sketches. I'd say, OK, sketch the Gibbs free energy versus temperature, or Gibbs free energy versus pressure. Tell me what the first derivative of g, with respect to t, is, things like that. Sketch the slope, sketch the curvature, et cetera. And so this kind of application is you could have students log in, do that kind of a sketch. And then people can evaluate it. It's anonymous, you can see. I can't tell who said what it, except David, who wrote his name. But I can't see who did what, so that it could work pretty well in a small class. If you had a class of 400 people, I'd be afraid. I'd be very afraid. But in a small class, you can come up with some interesting uses of it. The good thing about it is that you don't have students coming up to the board. It minimizes that motion, the sort of getting everybody unsettled and settled again. It's not a cure. I wouldn't say this redefines anything. In terms of the SAMR model, I'd say it's probably a modification, maybe an augmentation, maybe. I don't know. But it's certainly not and R. So I wanted to just demo that, because it's kind of fun. The other thing I wanted to do is, I think you guys saw this list? Did you see this list from the readings? This is just some things that I've compiled. It's kind of a mishmash. But there's some interesting things on here that you might want to explore, depending on your learning outcomes. And we've used Backchannel. Today's meet is pretty similar to Backchannel. Backchannel chat is pretty similar Backchannel. So those are some things that are pretty similar. We just looked at the web app. We've used Plickers. We've talked about Socrative, which is just a cell phone-based clicker. But there's a lot of interesting uses. And you might want to just play around with them. And I'll augment this list. So if you click it again from the wiki, you'll get a augmented list of things, a few things I want to add to it. So now, given your experience in life, and the readings, et cetera, is that I'd like for you to think about a technology that you haven't used for-- actually, what you want to first is think about a learning outcome. But you want to think about a learning outcome. You want to think about a technology that might advance that learning outcome. And then just jot down why you chose this technology, where does it fit in the SAMR model, how you would use it in your class, any kind of difficulties. And then there's a Google doc. I'm pretty sure everybody can access it. We can only hope. And if you can type your responses into that. So we can give you about five, seven minutes for that activity. And then we're going to do a lightning round based on it. So use whatever. Use the handouts. Use your knowledge, use, et cetera. And then somebody can tell me really quickly if they can get into the Google doc. Yeah, you guys are good. I was going to say, write your name down. But it doesn't really matter whether your name is up or not. So I want to make sure that you take a look at everybody else's some of the learning outcomes. We're going to do this lightning round where you're going to be talking to other people about the technology they chose, their learning outcome, they chose it, et cetera. And you're going to be trying to help your partner troubleshoot, or perhaps, implement the technology in a better way, solve some of their problems with the implementation process, or perhaps, think about a different technology altogether given their desired learning outcomes. So let's just take a minute, and either read it on your own device or read it up here, so that you kind of get a sense for what you're going to be talking about. And if you haven't read over it, please do that now. So there's some really interesting ideas up here, I think, and a broad spectrum of ambition. We had some pretty small-- like one assignment's worth of technology, perhaps. And then we also have kind of a whole course design. So that's another thing we didn't really talk about. I didn't narrow the scope in any way, or define the scope in any way. So it could be for a whole class. Or it could be for one particular assignment, or one particular class activity. And that's fine. That's OK. So did everybody get a chance to look over the other posts? AUDIENCE: I'll say, I've have had [INAUDIBLE] AUDIENCE: So it gives a follow up of all of these things is that when the AWW where everyone was drawing, I could draw here. But it was never showing up there. PROFESSOR: Interesting. AUDIENCE: Oh, yeah, my thing I drew never showed up. AUDIENCE: Yeah And then the other thing with this now, with the Google doc, I tried to edit it. And it doesn't let me click on it. It says, do you want to use the app? So then I had to download the app. And now I say, yes, use the app. And then it says something about the server not being recognized. But then if I go back and say, no, thanks, and I try to use it without the app, then I can't put anything in. PROFESSOR: Interesting. OK, so who had trouble? AUDIENCE: Yeah, I downloaded the app and used it instead. PROFESSOR: All right, did anybody that was using a smartphone not have trouble? AUDIENCE: I didn't have trouble. PROFESSOR: You didn't have trouble? AUDIENCE: No, I downloaded and use it. PROFESSOR: Oh, OK. And were you an Android, or Android? AUDIENCE: [INAUDIBLE] PROFESSOR: Android, iPhone, iPhone. Who else had trouble? It was more than just the two of you with this one? AUDIENCE: With this one. PROFESSOR: With this particular one? AUDIENCE: Yeah, [? Rachel ?] had problems with it. PROFESSOR: So for now, if you wouldn't mind, just-- let's see. We don't have to complete this right now. But maybe if there's a lull, just come on over here. And you can type your app in, just so we have a complete record-- your technology in. But that's interesting. It's a good thing to keep in mind that access might be an issue. I did try this out on two computers. But I tried it out on two computers. And I didn't try phone. AUDIENCE: [INAUDIBLE] Couldn't do. AUDIENCE: [INAUDIBLE]. PROFESSOR: Yeah, after is fine. So if we could get up. And anybody else that wasn't able to input their ideas, if you will do it later, that's fine. So we're going to do the lightning round. And if I could get you gentlemen to come around and fill in the back. Just fill in up to here. Great, so what we're going to do is I'm going to ask the row in the back, this row, to share their technology, their intended learning outcome, their concern. And then this row will give their advice. And it's two minutes for the whole process. And then we'll switch. OK? AUDIENCE: So what do I do? PROFESSOR: So you just have to tell me more or less what you wrote on the screen. OK, but hold on. I got to set the timer. OK, go. [INTERPOSING VOICES] PROFESSOR: Is going to share their ideas with the other side. [INTERPOSING VOICES] PROFESSOR: It's moment you've been waiting for. So we're going to do this two more times, this time and one more time, OK? All right. [INTERPOSING VOICES] PROFESSOR: Yeah? AUDIENCE: For the course which I want to teach using educational technology, so it was a big challenge, because the head of department wanted to say no. But then it continued because many of the professor already preferred it. PROFESSOR: Wow, so does [INAUDIBLE]. [INTERPOSING VOICES] AUDIENCE: Anybody can do anything. And you can reconfigure it. PROFESSOR: Alex, if you come around. And then, [INAUDIBLE], if you come around, please. Everybody slide down. AUDIENCE: We already went that way. PROFESSOR: Oh, you went that way. AUDIENCE: We already saw those people. PROFESSOR: So, [INAUDIBLE], if you come on over. [INTERPOSING VOICES] PROFESSOR: So I know in my conversations, I heard some really interesting ideas, some fascinating uses, and some incredible obstacles. I'll just point out David's obstacle is on the spreadsheet. But he wants to use his piece of software. And it costs $12,000 a year. So it sounds fabulous. It sounds really fabulous. And your department actually bought it for you? AUDIENCE: Yeah. Have it for all faculty [INAUDIBLE] PROFESSOR: Yeah, so that's great. But that's a huge hurdle. So that's an example of trade-offs that you might not always be able to make. So that was my interesting-- but I did hear some other wonderful ones too. So I hope that you take a look at it. You'll note that I always give you time after this lightning round. Because it's a very noisy kind of little crazy exercise. And I always like to give you a few minutes of silence afterwards to sort of process it a little bit. Because it can be a bit of an overload. So that's sort of the part of the design or the planning for this exercise. Anything anybody want to share about something they learned, or some issue or problem that was solved, or still unsolved? All right, I see some people are still typing. I'm going to go on. Feel free to keep adding to this. And again, please take a look at it. Because we only got so many pairings. And you can see virtually everything from the Google doc, assuming everybody gets in. So what I wanted to do is flash back a little bit to what we talked about the second class that we met. We talked about all these different learning theories, about behaviorism, cognitivism, constructivism. I think we've done some other exercise with these three learning theories before. But it's a good chance for you to reflect back, to do a little backward transfer, and to reinforce some of the concepts from the beginning of the semester, and to also think about how technologies might be-- so we talked about how technologies might fit at different levels of Bloom's, or supports different levels of Bloom's, but also, different technologies-- different anything you do in the classroom. But use of different technologies can actually be more behaviorist, cognitivist, or constructivist in nature. So I think it's useful for you to think about that just as a way of just getting a bit of a marker. About again, it's this idea about, what are we trying to do with this tool? And so I'm going to divide you up. And we have nine people left, so three groups of three, which works absolutely perfectly. Yes, three groups of three. And we're going to have the behaviorist group, the cognitivist group, and the constructivist group. What I want you to do is think about a particular technology that sort of would be consistent with this-- ooh, OK, one group will have four-- a particular technology that's consistent with that particular learning theory. And so this will take about maybe four minutes. And think about, perhaps, even think about a learning outcome that that tool advances, but at that level, so the behaviorist level, let's say. And then what I'm going to do is the activity is called a jigsaw. And so first, you're going to be in these homogeneous groups. So one group talks about constructivism. One group talks about behaviorism. And one group talks about cognitivism. And then I'm going to mix you up. So each group then has one cognitivist, one constructivist, and one behaviorist in it. So this is an active learning technique. And I wanted to model it for you. So let's just create some groups. [INTERPOSING VOICES] PROFESSOR: All right, folks. I don't know how far you got here. But now each one of you is going to be expert in your particular field. So I'm going to take-- and this is the constructivists. So here's three constructivists. I'd like one cognitivist and one behaviorist to find a constructivist, and to make a new group. All right, so now each one of you is the content expert for your particular theory. And just sort of explain what you came up with to your peers. So you'll just go around the table and talk about it. [INTERPOSING VOICES] AUDIENCE: They're still learning. But they're having an active roll in participating in when they're going to learn about, rather than just being passively taught that. AUDIENCE: And that's? AUDIENCE: Constructivist AUDIENCE: OK, so what technology [INAUDIBLE] AUDIENCE: Well, I think that a cool way to do it is to have computers set up throughout the room to have internet access. PROFESSOR: So the class is officially over. I mean, it's five of. I don't know if people have to go other places or not. So we can take this up beginning of class next time if there's things you want to finish up. I'd like to hear some of what you came up with for sure. So can we just plan to do that, and just tie up the exercise? Not the best of the best pedagogical strategy, to make you wait a week, but it is what it is, so. OK, thanks very much. And I look forward to seeing your posts on the assignment.
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https://ocw.mit.edu/courses/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: So this first lecture is a spirited but lighthearted introduction into the course. I usually never talk about technicalities first. I want to talk about the excitement of the field first. So I will talk about technicalities a little bit later. So when you take a course in AMO physics, the obvious question is, what is AMO physics? What defines our field? Actually, by far the best definition of what AMO physics is-- it is what AMO physicists do, which is defined by a community of AMO researchers. And this is really characteristic for our field. I will tell you in the next 10 minutes or so about these enormous dynamics of the field, how AMO physics has changed in a fraction of my lifetime. And what happened is, I felt, whatever I was doing-- and it turned out to be very different of what I did 10 and 15 and 20 years ago-- has stayed in the center of AMO physics. So myself and the whole community, we're moving, and took the field along. So therefore AMO physics is what gets AMO researchers excited. It's not a joke. It's the way it sort of happened. Well, a little bit more mechanically, we can define AMO physics. AMO physics is what is made out of the building blocks we have in AMO physics. So defined by the building blocks. And in AMO physics, we build systems out of atoms, or molecules, and then photons, or light, and in general, electromagnetic fields-- electric, magnetic fields, microwaves and all this. So in other words, everything which is interesting and we can put together with those building blocks, this is AMO physics now. And this may redefine AMO physics in the future. Now, historically this meant that those building blocks, atoms and molecules, were available in the gas phase. So you had gases of atoms, gases of molecules, and you studied them. And almost all of AMO physics was actually about individual particles, individual atoms, individual molecules, because in the gas phase at high temperature, pretty much, you learned that in statistical mechanics each particle is by itself. And the partition function of the whole system is just factorized into partition functions of the individual particles. Well, maybe with some exception-- occasionally, particles collide. And the field of collisions-- collisions between atoms and atoms, atoms and ions, atoms and molecules, molecules and photons-- this was widely started in AMO physics. Well, now the field has really moved away from just individual particles and collisions between two particles. What is in the center of attention is few body physics. And that of course takes us to entanglement. For instance, people study entanglement of eight ions in ion traps. And of course this is deeply related to quantum information science. Or if you want to go from few body to many body, this is now starting to overlap with condensed matter physics. And this is now widely studied in the field of [INAUDIBLE] atoms and quantum gases. Now, when we talk about many body physics, we get, of course, into overlap with condensed matter physics. In condensed, I would actually say now a larger fraction of AMO physics and condensed matter physics overlap strongly, that we speak the same language, we study the same Hamiltonian, a lot of theorists apply the same methods to topics coming from either field. However, often in technology, how the systems are studied, there's a different culture, different tradition, and still two different communities. And the overlap comes about because there are systems in nature which you can say-- they are natural two or few level systems. Of course, we've helped nature a little bit by engineering. Those systems are, for instance, quantum dots or NV centers in diamond. Or another overlap with condensed matter physics is that AMO physics is now-- one frontier of AMO physics is the optical control of mechanical oscillators. Micro cantilevers, membranes, tiny mirrors in cavities, they have mechanical motion. And the mechanical motion is strongly coupled to the photon field. Of course, for fundamental AMO physicists, a mechanical oscillator is nothing else than an harmonic oscillator. So one can say-- and this is sort of my third attempt in defining for you what AMO physics is-- AMO physics is almost defining itself by low energy quantum physics. So all of the quantum mechanics which doesn't take place at giga and tera electron volt, which takes place at low energy, this is AMO physics. Of course, maybe not all of it. Usually when it is in [INAUDIBLE] solution, it's biophysics, and it's distinctly different from AMO physics. Or if the solid state is involved, then it's more condensed matter physics. But I would say there is one part of solid state physics which is already becoming interdisciplinary with AMO physics. And this is when the control of the system is not done by wires and carbon probes but it is done by lasers. So if you have a solid state system and you use all of the methods you apply to atoms, you use electromagnetically-induced transparency, coherence, all those concepts, then AMO physicists feel at home and they don't care if the two level system or the harmonic oscillator is an ion oscillating in an ion trap or whether it's a small cantilever, the methods and concepts are the same. So therefore, one could say AMO physics is sort of the playground where we can work on extensions of simple systems which we understand and cherish. And of course, our exactly solvable problems are the hydrogen atom-- my colleague, Dan Kepner, would have said, maybe the hydrogen atom, if you understand the hydrogen atom, you understand all of atomic physics. I'm not so sure. I would actually say, in addition to the hydrogen atom, you have to know the two-level system. And of course, you have to understand the harmonic oscillator. So these are three paradigmatic Hamiltonians, and a lot of understanding of much more complicated systems really comes from taking the best features of those three systems and combining them. If you have questions or comments, this is an interactive class. Feel free to speak out or interrupt me. OK so, now we know, or we don't know, what AMO physics is. Let me now address-- how has AMO physics developed? And I mentioned to you that AMO physics has done breathtaking evolution in my lifetime, or even in the shorter part of my life, which is my research career. Well, traditionally, almost all fields in science started with observing nature. The pursuit of science was born out of human curiosity to understand the world around us. And atomic physicists, well, they started to observe atoms and molecules, usually in the gas phase, and what they were doing. And already there was some evolution, because original observations at low resolution were taken to a completely new level when high-resolution methods were developed, when lasers came along, when people had light sources which had fantastic resolution and eventually finer and finer details of the structure of atoms and interactions between atoms were resolved. But AMO physics is a field which has taken the pursuit of science much further. So there is not just observation of nature-- and I want to write that with capital letters-- there is CONTROL OF NATURE. And you maybe take it for granted, but you should really appreciate it, that controlling nature, having control over what you study, modify it, advance it, take it to the next level, is really something wonderful. It is completely absent in certain fields, like astrophysics. In astrophysics, all you can do is, you can observe. In atomic physics, we create the objects we can observe. So the control of nature, the control of our atomic physics system, developed in stages. The first kind of control was exerted about internal states. If you have an atom at thermal energies, it would only come in hyperfine states which are thermally populated, or molecules come in rotation states, and well your limited control was simply to raise and lower the temperature. But with the advent of optical pumping-- this actually happened already with classical light sources before the invention of the laser-- so with optical pumping, you can pump the internal population of molecules into, let's say, a single rotational state. So this is control over the internal Hilbert space. And this was actually rewarded with a Nobel Prize to Alfred Kastler in 1966. Of course, the next step after controlling the internal degrees of freedom is have control over the external degrees of freedom, and this means control motion. This was of course pursued by understanding the mechanical aspect of light. How do photons mechanically interact with atoms? This eventually led to laser cooling and Bose-Einstein condensation. And those developments were recognized with major prizes. Well there is more to it than controlling internal and external degrees of freedom. You can then also say, well, how much can we control the number of building blocks? And eventually AMO physics advanced to exert control onto single quantum systems-- single photons, single atoms. A single atom in a cavity exchanging a photon with a cavity thousands of times. So this control of single quantum systems was actually just recognized with a Nobel Prize a few months ago. Well, at this point I sometimes make the joke, we have gone from big ensembles in many, many quantum states to single photons, single atoms, in a single quantum state-- a single quantum state for many particles is Bose-Einstein condensation. So we have really gone down to single atoms, single photons, single quantum states. Well, what comes next? To have no atoms and no light in vacuum. Well, the vacuum has some very interesting properties. And if you talk to Frank Wilczek, the nature of the vacuum and dark energy is one of the big mysteries in physics and in science in general. But the study of that is definitely outside the scope of AMO physics. So what happens is, when we have gone down and have now control over the building blocks, now we can sort of go up again in a controlled way, create complexity by assembling a few photons, a few atoms into new entangled states, so we can now take our system into very different regions of Hilbert space. So what is defining now the control is, we want to use this pristine control over the building blocks to now put in something which hasn't existed naturally before, or when it existed it was completely obscured, completely hidden by thermal motion or by you can say homogenous broadening our lack of control. And the best buzzwords are here now entanglement and many-body physics. It's hard to capture that as in a diagram, but let me try that. I don't know actually what I'm drawing, but I think you get the message that this is sort of Hilbert space. And I have 2 x's. One is sort of entropy, high temperature, and the other one is complexity. And for quite awhile, people studied hot gases. So these are gases, there's a lot of entropy in it. The complexity is actually not particularly high. And everything is described by a statistical operator-- the density matrix. The pursuit of cooling, and actually, control-- gaining information about a system is also a way of cooling. If you know in which state the atom is, the entropy of the system is zero even if you haven't changed the state of the atom. So control and cooling, control measurement and cooling, has now taken us to the point where we have systems which have no entropy anymore. They are very, very well defined. And our goal now is to take these systems to much more complexity where wave functions become entangled and we have strong correlations in many-body systems. But this is now, maybe here, this is now described by wave functions. But here is the wave function of a single particle, and here we have highly correlated, highly entangled many-body wave functions. So at least for me-- but all predictions are notoriously incorrect when you look at them in a few years from now-- but for me, this is where the future of our field is moving, to get into interesting regions of Hilbert space where no person has been before. As an experimentalist, and with a lot of experimental graduate students in this room, I want to emphasize that a lot of those rapid developments of the field are driven by technology. So it's driven by technology advances. In the '50s and early '60s, people thought AMO physics is pretty much dead. Only a few people with gray hair continue what they have done and the field will eventually die. But then technological developments made it possible to do major conceptual advances. I've mentioned the conceptual advances-- let me now say a few words about the technology which has driven it. There was one phase of developing lasers which I experienced when I was a student. But those lasers were fantastic. They were very narrow already, very stable, but they were very expensive and very complicated. So if you had one laser, this defined your laboratory and then you studied a lot of things with the single laser you could afford. Well, you're probably not used to one big dye laser pumped with a big argon ion laser. It was a $250,000 investment for the lab. And of course, you couldn't afford a second laser. It required 50 kilowatts pf electrical power. And all this power had to be cooled away by gushing water through thick pipes. So it was an expensive undertaking but you could really do wonderful science with it. So what I've seen in the last 20 years is the proliferation of solid state lasers, starting with diode lasers and continuing until the present year. So now, in a lot of our laboratories here at MIT, we have 10 lasers. And we've stopped counting them, because adding a laser to the system is almost like adding an amplifier to a circuit or adding another circuit to a data acquisition system. But it's not just the simplicity of the lasers which we have now, the robustness-- to have 10 lasers in the lab, it's fine. Previously, if you had three lasers in the lab, you spent 90% of your time just keeping the lasers running. Those lasers, not so much continuous-wave lasers, but pulse lasers have also very, very different properties. Laser pulses got very, very short-- femtosecond or even attosecond. Shorter pulses means that the energy is now focused to a much shorter temporal window. Therefore, laser pulses of very, very high intensity-- if you focus a short pulse laser on an atomic system, you can easily reach an electric field of the laser which is stronger than the electric field of the atom. So in other words, if you have protons and electrons-- well, maybe the outer electrons, not get down to the single protons. But maybe an ionic core, and you've electrons. Now, you should first look at the motion of free electrons in the strong field of the laser and add the atomic structure as a perturbation. It really takes the hierarchy of effects upside down. So the appearance of high intensity lasers has given rise to a whole new field of atomic physics. Lasers got more precise. The invention of the frequency combs, recognized by the Nobel Prize in 2005, meant now that we can control laser frequencies at a level of 10 to the minus 17. And this has completely redefined precision metrology and has advanced the control over atoms and molecules I've mentioned before. Finally, another technical development which plays a major role in research being pursued here at MIT and elsewhere are the development of high finesse cavities. High finesse cavities in the microwave range-- then they're superconducting-- or high finesse optical cavities by having super mirrors. It is actually those super cavities which have enabled the study of single photon physics. Because after all, photons move away with the speed of light. And if you want to observe a photon in your laboratory, it has to bounce around zillions of times in order to have enough time for the photon to do something interesting. So sometimes a field at the frontier of science is defined by paradigms. If you want to explain to somebody why your field of interest is cool and exciting, you usually do it by picking a few really exciting examples. And I want to show you how, over the years, it has advanced. Definitely in the '50s and '60s, you would have mentioned that we understand now atomic structure of multi-electron atoms. Optical pumping just started. So these were flagship developments of AMO science. The cool thing to do in the '60s and '80s was, use the new tool, the laser, applied to atoms and do laser spectroscopy. Sub-doppler spectroscopy, sub-natural spectroscopy, resolving hyperfine structure-- wow. I mean, this was really exciting in those days. And well, the older people I have met, my teachers, my thesis adviser, these were people who started their research career before the laser was invented but then, as a young researcher, embraced this new tool and helped to redefine the field. Definitely in the '80s and '90s, the cool pictures were those of magneto-optical traps, atoms standing still and hovering around. So the new aspect where mechanical forces of light which led to laser cooling and trapped atoms. In the late '90s, of course, the excitement was about Bose-Einstein condensation. And it was really Bose-Einstein condensation which drove AMO physics from single atoms, maybe two atoms colliding, to many-body physics. It's always easier to analyze those things by looking backward, so if I'm now getting closer to the past, I have to be a little bit vague. But in the 2000s, I think hot topics were ultracold fermions and the study of entanglement and correlations. And what is the paradigm now or in the near future? Well, I think you have to help to define it. If you make an interesting discovery, this is what people will be pointing to and would say, this is what now defines AMO physics. Some candidates are, of course, if there is a major breakthrough in quantum computation-- let me put question marks here. In the field called atom science, we may actually do some progress towards topological states which have different symmetries and different properties. And another emerging frontier is micro-mechanical oscillators. The last couple of years, we just had the breakthrough for the first time. Mechanical objects were cooled to the absolute ground state. So at least for that community it was, but for many of us, Bose-Einstein condensation was 15 years ago. Any questions? Well, eventually we have to talk about this course. So I've told you about, at least, a snapshot of where AMO physics is, how it has developed, and on what trajectory it is. In this course I want to present you the concepts behind many of the major advances in the field. So over the years, quite often a topic was added to the course, because I felt, hey, that's getting really exciting, that's what people want to do in research, that's what graduate students want to do here. And then the subject was added and other subjects were dropped. I know in the '90s, I was teaching aspects of laser cooling, sub recoil laser cooling, which was the latest excitement. This year, I may mention it for 30 seconds. So the course has evolved. It wants to stay connected to what is exciting, what is hot, and what prepares you for research at the frontier. 8.422, the second part of the two-course cycle in the graduate course in AMO physics, is somewhat different, not radically different, but is somewhat different from part one, from 8.421. First of all, 8.421, 8.422 can be taken out of sequence. We alternate between AMO 1 and AMO 2. And whenever you enter MIT, you're probably in your second semester, take whatever we offer. So I expect-- let me ask you, who of you has taken AMO 1? Yes, statistically, it should be about half of the class. It can be taken out of sequence because that's the way how we've structured it. But to give you one example is, in 8.421 you really have to learn about hyperfine structure. You have to learn about atoms, you have to learn about Lamb shift and all that. So you have to learn what all these atomic levels are. And here in that course, in 8.422, I will say, here's a two-level system and then I run with that. And we do all sort of entanglement, manipulation of two-level systems. So it helps you if you know where those two levels come from, but you don't really need the detailed knowledge of atomic structure, for instance, to understand that. So this is why the different parts of the course are connected, but in terms of learning the material, somewhat decoupled. I've spoken to many students who said there was no problem in starting with 8.422. The only sort of critical comment I've heard is that taking 8.422 first and then 8.421 is anti-climactic. You see all this excitement in the modern physics, and then, eventually, you have to work on the foundation. Prerequisites for this course-- the course announcement said 8.05. It is actually 8.05 and 8.06. The main part of 8.06 which we really need here is perturbation theory-- time independent, time dependent perturbation theory, and this is usually covered in 8.06. However, I've had students who took the course without 8.06. If you're really determined and want to acquire certain things by self-study, you can follow this course. So the topics we will cover include QED. I really want to talk about light-atom interactions from first principles. Sure, 95% of what we're doing is just done by saying we have a matrix element, which maybe the dipole matrix element. But you really have to know what are the approximations, what are the conditions, which lead to the dipole approximation. And I want to do that from first principles, and we do that starting on Monday. So a discussion of light-atom interaction has two parts. One is the simple part-- excitation and stimulated emission. Because this can be simply described by a unitary time evolution, and you can do a lot, if not everything, by using Schroedinger's equation. Things get much more complicated and richer if you include spontaneous emission or, more generally, if you include dissipation. Then we talk about open systems. And for fundamental reasons, we need a formulation using the density matrix, a statistical operator, and a master equation. One major part of the course is discussion and [INAUDIBLE] derivation of the mechanical forces of light. This will include a discussion of important experimental techniques using those mechanical forces. So various simple and sophisticated methods of trapping and cooling. We will spend some time in not talking about atoms at all. We're just going to talk about photons, about single photons. We want to understand where the photon nature of light makes light very, very different from a classical electromagnetic wave. Also, it's not the focus, we will come across basic building blocks of quantum information science. Pretty much when atoms and photons interact, this is a fundamental quantum gate. And we'll talk about the many-body physics off quantum gases. So maybe it becomes clearer what we are covering by saying what we are not covering. And this tells you that there is at least some selection of topics. It's not that we talk about everything. We will not talk about the physics above 10 electron volts. We will not talk about collisions-- or maybe I should say, high-energy collisions. We will, of course, talk about nanokelvin collisions, which is the physics of the scattering links and some s-wave collisions which are really relevant to understand quantum gases. We're not talking about any advanced topic in atomic structure. All we do about atomic structure is done in 8.421. And if you want to graduate in atomic physics at MIT, yes, you have to understand atomic structure at the level of the hydrogen atom. And maybe know a little bit about a new phenomena-- when another electron enters, when electrons interact and that's a helium atom-- but we're not going beyond it. Let me just mention here that there is, of course, more interesting things in atomic structure. For instance, if you go to highly charged ions, you have QED effects. You can discuss very interesting correlations between two electrons in an atom. And you can have very relativistic effects if you have highly ionized atoms. If you have bare uranium, then the electron in the lowest orbits becomes relativistic. You can even see, if you scale the fine structure constant-- which has an e squared in it-- with the charge of uranium 92, well, 1 over 137 times 92 gives about 1, so you really get into a new regime of coupling for atomic physics. But we won't have time to talk about it. And this may be an omission, because there is really interesting work going on. We're not talking about high intensity lasers and short laser pulses. This choice maybe mainly determined by that the experimental program in the physics department is not overlapping with that. But of course, you know we have world class researchers and short pulse lasers in the electrical engineering department. Any question about the syllabus? Questions about what to expect in the next 12, 13 weeks? Well, then let's talk about some technicalities. I'll keep it short because all this information is available on the website. We run the website of the CUA server. So it is cua.mit.edu/8.422. When I say all the information is available, I have to say-- not yet. I realized yesterday night that [INAUDIBLE] has re-modified the server. I didn't have access to the server. I just got it an hour ago. What you will find under this URL is the website from when the class was taught two years ago. And actually, more than 90 percent of the information is the same. I always try to improve the course, but I know it will be more incremental changes and not dramatic changes. So if you look at the website, you'll already get a taste what the course is, but within the next day or two, we'll have updated information. I know, at least for most of you, I have contact information for MIT registration. But if you are a Harvard student or a listener just sitting in and you haven't registered for the class, I would ask you to send an email to our secretary, Joanna, [email protected], because I would like to have an email list for the class for corrections to the homework assignment or last minute announcements. And then we'll add you to the mailing list. The schedule of the class is, well, we meet at this time on Monday, Wednesday. And let me know disclose, I plan to occasionally teach on Fridays at the same time. Over many years, it has been my experience, if you have a class on Monday, Wednesday at one o'clock, you don't have another class Friday at one o'clock. Is that assumption correct? So within the next few days, I will tell you on the website which Friday I would like to teach. The reason is rather simple. Like every faculty member who's active in research, I have to go to funding agency workshops and conferences. I try to keep it to a minimum, but on average I will miss two or three classes. And instead of asking a [INAUDIBLE] or a graduate student to teach the class, I would like to teach the class myself. So I will have makeup classes on Fridays. OK, that's the schedule. Finally, talking about requirements, one requirement is homework. We have 10 wonderful problem sets with a lot of problems which I actually designed from current research. So you will actually recognize that a lot of problems are created based on some research papers which came out of my own group or other groups at MIT in the last few years. The good news is, there is no mid-term, there is no final, but there will be a term paper. And the term paper is due on the last day of classes. The teaching team are myself, Joanna Keseberg, our secretary-- that's also where you will be dropping off your homework-- and then we have assembled a wonderful team of five TAs. They are all advanced graduate students. And actually, I've picked TAs from all of the active groups here at MIT. From Martin Zwierlein's group, I. Chuang's group, Vladan Vuletic's group, and my group. Are any of the TA's around? Nick, Alexei, Jee Woo, Lawrence, and Molu. Each TA will be responsible for two weeks in the course, and the TA will indicate availability and office hours on the weekly homework assignment. Other details about the term paper, how long the term paper should be, the, kind of, what I regard as an honesty code, that you're not trying to get access to solutions from previous courses, all that is summarized on the website. And once I've updated the website, I'm sure you will read it. Any questions? OK, well, with that we will actually be ready to jump into the middle of the course and start with some heavy duty equations of QED. But no-- just joking. I think this would maybe spoil the introduction. What I actually like to do is, to make the transition from the introduction to the discussion of atom-photon interactions and quantum electrodynamics, I always like to start the course by giving you an appetizer by talking to you in a lighthearted but hopefully profound way about the system, which helps to showcase what are we doing in this course. And until a few years ago, I would have started with the simplest example of laser cooling, simply beam slowing or optical molasses, in the simplest possible picture, just to give you a taste of what we will be doing together during the semester. But as I said, AMO physics is moving along. And what I now want to use as an example which clearly synthesizes many aspects of this course are atoms in an optical lattice. So let me, in the last, next 20 minutes or so, make some connections to different topics of this course by using as a starting point a concrete, very simple system, but very, very rich and profound, and these are atoms in optical lattice. So the situation I want to use here is that you have two laser beams which interfere. And those laser beams form an optical standing wave. So next week we will learn how this electromagnetic wave interacts with atoms. So we have to put a few atoms into the system. And we will derive, from first principles, the QED Hamiltonian, which, after a lot of manipulation and eliminating complicated terms, will be the dipole interaction. But of course, each symbol here is an operator, and there's a long story behind that. In about two months, we will describe light-atom interaction with a formalism which uses a Bloch vector and the optical Bloch equations. There is a vector with three components which describes what is the state the atom is in. One component will tell us if the atom is in the upper or the lower state, whereas the other components tell us whether the dipole moment of the atom is in phase or out of phase with a driving electromagnetic field. Well, if there is part of the dipole moment which is in phase with a driving electric field, then the suitable expectation value defines a mechanical potential. And if the electric field is a standing wave, then we generate, through light-atom interaction, a periodic potential for the atoms. We will learn everything which we have to know about this potential. In a simple case, it is simply the Rabi frequency divided by the detuning of the electromagnetic wave. But we will find it very interesting to look at it from very different point of view. And maybe let me use this example to point out that I'm really a big friend of explaining the same physics from very, very different perspectives. And when we talk about optical standing waves, we will use a picture of a classical potential, like a mechanical potential, and the atom is just moving around. We will use a photon picture that every time the atom feels a force, photons are involved. You don't see them in the classical potential, but photons are behind it, and ultimately, the forces of the classic potential come from stimulated absorption emission of photons. I may go down to the microscopic level. And I may ask you, but in the end, it's an atom, but the atom consists of electrons. And the electrons are simply oscillating because it's driven by the electromagnetic field. It's actually something which most people are not aware of, but you can ask the question now-- is the force in the optical lattice on the atom, it's ultimately a force on the electron? Well, if you have a charged particle, you can have two kinds of forces-- an electric force or the Lorenz force. And I don't know if you would know the answer, but an atom is in an optical lattice, is the force just the [INAUDIBLE] the potential which the atom experiences, is that fundamentally due to the Coulomb force exerted by the electric field of the light, or is it due to the Lorenz force exerted by the magnetic part of the light? Who knows the answer? Who doesn't know the answer? OK, great. I was actually surprised when I derived it a few times for the first time. And it's not in the standard textbooks. So anyway, I hope, even for many of you who know already a lot about the subject, I hope I can add other perspectives for you. OK, so what I've discussed so far is that the standing wave of light creates now a periodic potential for the atom. And we will understand it from many, many different perspectives, from the photon type, quantum optics perspective, to the most classical description. So what immediately comes to my mind is that we now want to look at this periodic potential in two different situations. We can ask, well, whenever you have light, spontaneous emission is a possibility. And we may ask, what happens when spontaneous emission is not negligible? And we discuss that approximately in week nine of the course. Then what happens is, an atom has an excited and a ground state. And those in the excited and the ground state, do the atoms feel the periodic potential? Well, eventually, we have to generalize the notion of ground and excited state into [INAUDIBLE] states, and I will tell you all about it in a few weeks. But the situation can now be, when an atom is in the ground state, it has to move up the standing wave potential. Then it's getting excited with a laser. It has to move up again the standing wave potential. And then there may be spontaneous emission. So what I'm just describing here is a situation where the atom is mechanically moving up the potential. It's excited at the top of the potential. And it emits when it's on top of the potential. So the atom is doing mechanical work, and this is called Sisyphus cooling. This is the way, this is one method, one mechanism, of laser cooling which leads to the lowest temperatures in the laboratory before evaporative cooling is used. So that's some cool aspect we will come along by discussing motion in a standing wave, but taking into account that photons are emitted and really understanding when and where are the photons emitted. Well, the second situation is, of course, if spontaneous emission is completely negligible. Then we have a situation where all what matters is the potential, and we can completely forget that it's photons, it's quantum optics which has created this potential. We can simply use a classical potential in our description in our Hamiltonian. Now, again, we have two limiting cases. One case is when this potential, or optical lattice, is really deep. Deep means that atoms are sitting deep in the potential and they can oscillate around, but they cannot jump over to the next potential. So in this situation, you would say, well, that's boring. Nothing happens, the atom just stays put. But what is boring for some of you is exciting for some others, because these atoms are exquisitely controlled. They cannot collide, they cannot interact with other atoms. So these are really the most ideal situation you can imagine for atoms. Of course, if you have less than one atom per site. And this is the way how you want to prepare atoms for the most accurate interrogations. And you can build atomic clocks, optical atomic clocks, based on atoms insulated in such a lattice, which approach now 10 to the minus 17 accuracy. Well, if you drive a clock transition, if you take the atom from the ground to the excited state, you may face a situation I mentioned earlier that the periodic potential for the ground and excited state are different. And that would actually interfere with your clock, because the clock frequency depends now on what the lattice is doing to your atoms. However, and we discuss that in week four, there are what people call magic wavelengths, where you pick a certain wavelength for your optical lattice where the periodic potential is absolutely the same for ground and excited state. So that means you have, then, the perfect decoupling between the ticking of the clock, the internal structure of the atom and the mechanical motion of the atoms in the potential. And in that situation, you create a situation which has been studied since [INAUDIBLE], namely with trapped ions. If you have a single ion in an ion trap-- and ion trapping is pursued here at MIT in Ike Chuang's group-- you have just a single object completely isolated. And in the form of the aluminum ion, this has just been demonstrated to be the most accurate atomic clock in the world with 10 to the minus 17 accuracy. So anyway, with optical lattices, avoiding spontaneous emission using magic wavelengths, we can only engineer with neutral atoms what has been available with trapped ions but we can simultaneously have 10,000 copies and look at 10,000 atoms which are all identical copies, identical systems with each other. So you see already from that example that there will be intellectual overlap and synergy between talking about neutral atoms in optical lattices and talking about trapped ions and how they are cooled and how they are manipulated. And we'll talk about trapped ions at the end of the course, sideband cooling of trapped ions, which is week 12. OK, so this is the simple but pristine situation that we have a deep lattice. The other limit is now that we have a weak, or shallow lattice. And the new physics which now comes into play is that atoms can move around-- they can tunnel from side to side. So towards the end of the course, I think the last months, approximately week 10, I will give you a short summary of what some of you may have already learned in the solid state course-- namely, band structure of atoms in optical lattices, Bloch states, effective mass, et cetera. This has now become language of atomic physics, because there is a very clean and straightforward realization of this physics using cold atoms. What we are mainly interested in, in our research, is when we have atoms which tunnel in this optical lattice. And for bosons, if the interactions get strong enough, the Bose-Einstein condensate is destroyed and what forms is a Mott insulator. This is a phase transition. And for fermions, we have a crossover from a metal to a fermionic Mott insulator. And with that, we are already overlapping conceptually with condensed matter physics, because the Mott insulator is a paradigm-- one of these paradigmatic examples where you understand some deep physics. It's a paradigm of condensed matter physics where you have only a partially-filled band. Common sense, or undergraduate textbooks, would say, partially-filled band, this means you've a metal, you've a conductor. But because of the interactions of the atoms, the system is an insulator. So this is where, really, the many-body physics profoundly changes the character of your system. Any questions? Well then, let me add one last aspect to my introductory example of optical lattices. In a sort of flyover, I described to you what is the physics we encounter when we have an optical lattice switched on and the atoms move around or they don't move around, and both cases are interesting. But if you think you've explored everything, well, then you think harder and say, hey, there's another angle we can get out of it. And this is, we can take the optical lettuce and simply pulse it on, switch it on and off. Well, what happens is, and then it becomes a time-dependent problem. It becomes something where we can shape and control the wave function of atoms in time-dependent way. Let me give you one example. If we start with very cold atoms-- can be a Bose-Einstein condensate-- and then, for a short time, we switch on the lattice, afterwards, we observe that we have still some atoms at zero momentum. But now we have atoms which have a momentum transfer of plus minus 2 h bar k. You can understand that-- and this again exemplifies that we want to look at the physics from different angles-- this can be described that you have an two-photon transition from the ground state with zero momentum to the ground state with two-photon recoil. So you can understand it in the photon picture. But you can also understand it by saying you have some matter waves which are now exposed to a periodic potential. And you simply ask what happens to waves in a periodic potential? Well, that's the same what happens to optical waves when they encounter a grading, and that's the physics of diffraction. So in a nutshell, this happens when we use a pulse on an optical potential. And let me now finish by telling you that, again, in this situation, we discover the ambiguity or the two-sidedness of a lot of things we do in atomic physics. I've mentioned to you that the same experiment, the same experimental setup, cold atoms in optical lattice-- you turn up the lattice and you have the world's best atomic clock, atoms just isolated from each other. You turn down the lattice and you have an interesting condensed matter physics system. Let me just show you that the same two-sidedness of atomic physics-- precision and pristine control and interesting many-body physics-- happens when you pulse on optical lattices. So one application of those optical lettuces is atom optics. You have atoms and when they encounter an optical lattice, some atoms will just continue, will not be diffracted. They have zero momentums. Others are diffracted with a transverse momentum. You can then expose the atoms to a second zone where, eventually, momentum is transferred, and then the atoms come back together. So what we have here is an atom interferometer where atomic matter waves first go through a beam splitter, are reflected back towards each other, all by photon transfer, and then they're recombined. So this pulsed optical lattice acts as a beam splitter. It is the way how, today, the most precise atom interferometers are built. And this is used for precision measurement, for measurement of inertial forces, gravity, rotation, acceleration, and it is used for navigation and accurate observation of the changes of the gravitational field of the Earth. Since we just had a talk by Holger Mueller from Berkeley day before yesterday over at Harvard, he talked about the combination of atom interferometer and frequency combs, another development-- let me just mention that. If you derive those laser frequencies from an optical frequency comb, and with an optical frequency comb you can pretty much count the frequencies. So this laser is mode, I don't know-- 100,000 or whatever you are in the comb. By doing that, by combining it, you can actually build an atomic clock where-- let me just say, plus combs-- can give an atomic clock which is called the Compton clock because the frequency is now given by the rest mass of the atom, but divided by a big integer number. So using completely unrelated developments in AMO science which I've mentioned, the optical frequency comb, combining it with the physics of a pulse standing wave, you now have an atomic clock which is directly related to the energy of the rest mass. But finally, if you pulse on an optical standing wave and your object are not individual atoms, non-interacting atoms, your object is the Bose-Einstein condensate or atoms which strongly interact, then you're not transferring recoil to individual atoms, you're transferring momentum to a complicated many-body system. And this means what we are measuring is the dynamic structure factor. If you have atoms, you want to do spectroscopy, you want to know what energy levels are there, and then you know your atom. If you have a strongly-interacting system, you also want to know what energy levels are there. But each energy level in a homogeneous system or in a periodic potential is associated with momentum and quasi-momentum. So in other words, if you have a more complicated system, you want to figure out what are the possible states in terms of momentum and energy. And the optical standing wave, the pulsed optical standing wave is the way how we impart momentum and energy to a system. What I actually just told you is a story in my own research career. I was a post-doc with Dave Pritchard. He had trained at MIT in the '90s by a pioneer in laser cooling. And when we had Bose-Einstein condensates in the late '90s, Professor Pritchard and myself, we teamed up. I was the expert on Bose-Einstein condensation and he was the expert on atom interferometry. So just by sort of exploring things, we took Bose-Einstein condensates and we pulsed on a standing wave. What was on our mind was, hey, let's build an interferometer. But I'm more of a many-body physics person. I suddenly said, yes, but if we now change the momentum and the frequency of this standing wave, what happens? And I suddenly realized that this is a way to measure properties of a Bose-Einstein condensate in a way which hadn't been done before. So I realized-- and this is maybe the last thing I want to tell you today-- I realized in my own research and my collaboration with Dave Pritchard, that we'd built an experiment, and we just turned one knob at our experiment, and the following day we were no longer doing atomic interferometry, we were doing many-body physics. So this is, I think, what makes our field exciting. We are using the tools, the precision, and the control of atomic physics, which leads to the most accurate atomic clocks in the world-- atomic clocks, which are the most accurate in the world. And we are using those tools to do entanglement and many-body physics. And I think it's just a compelling combination. Anyway, that's an appetizer, that's an outlook over the semester. Do you have any questions about the last examples I gave you or the course in itself? OK. No homework assignment today. And we'll meet Monday at the same time here in this lecture hall.
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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: You first are facing the calculation of the energy eigenstate with some arbitrary potential. You probably want to know some of the key features of the wave functions you're going to calculate. So in fact, all of today's lecture is going to be devoted to this intuitive, qualitative insights into the nature of the wave function. So we will discuss a few properties that help us think clearly. And these are two of those properties. I want to begin with them. Then we'll do a third one that we have already used, and we will prove it completely. And then turn to the classical and semi-classical intuition that lets us figure out how the wave function will look. And that's a great help for you. Even if you're solving for your wave function numerically, you always need to know what the answer should look like. And it's ideal if before you calculate, you think about it. And you realize, well, it should have this t properties. And if you find out that those are not true, well, you will learn something about your intuition and see what was wrong with it. So we're talking about one dimensional potentials, time independent potentials. And a first statement that is very important, and you will prove in an exercise after spring break, and that is the fact that one dimensional potentials, when you look at what are called bound states, you never find degeneracies, energy degeneracies. And this is when x extends from minus infinity to infinity. You've seen already, in the case of a particle in a circle, there are degenerate energy eigenstates. But if the potential extends to infinity, there is no such things. Now what is a bound state? A bound state sounds like a complicated concept. But it is not. It really means an energy eigenstate that can be normalized. Now if an energy eigenstate can be normalized and you live in the full real line, that the wave function must go to 0 at infinity. Otherwise you would never be able to normalize it. And if the wave function goes 0 at infinity, the bound state is some sort of bump in the middle region or something like that. And it eventually decays. So this is bound by the potential in some way. And that's basically what we use to define a bound state. We'll take it to be that generally. So this is something, this property, which is very important, is something you will prove. But now we go to another property. We've emphasized forever that the Schrodinger equation is an equation with complex numbers. And the solutions have complex numbers. And suddenly, I wrote a few lectures ago a wave function was real. And I was asked, well, how can it be real? Well, we've discussed stationary states in which the full wave function, capital PSI, is equal to a little psi of x times the exponential of e to the minus i et over h bar. And there in that exponential, there is complex numbers on this little psi of x in front of that exponential, which is what we called basically those energy eigenstates. The e to the minus i et over h bar, it's understood that little psi of x is the thing we've been looking for. And this psi of x solves the time independent Schrodinger equation h psi equal e psi. And that equation has no complex number in it. So little psi of x can be real. And there's no contradiction. Because the full solution to the time dependent Schrodinger equation is complex. But here is a statement. With v of x real, the energy eigenstates can be chosen to be real. And the words can be chosen are very important here. It means that you may find a solution that is complex, but you need not stick to that solution. There is always a possibility to work with real solutions. And what is the way you prove this? This I will put this in the notes. You don't have to worry about the proof. You consider the Schrodinger equation for psi. And you show that psi star, the complex conjugate of psi, solves the same equation that psi solves. And therefore, if psi is a solution, psi star is a solution with the same energy. That part is very important. Therefore, if you have two energy eigenstates with the same energy, you can form the sum. That's still an energy eigenstate with the same energy. Even formed in difference, that's still an energy eigenstate with the same energy. And the sum of psi plus psi star is real. And the difference psi minus psi star, if you divide by 2i, is real as well. Therefore you can construct two solutions, the real part of psi and the imaginary part of psi. And both are solutions to the Schrodinger equation. So I've said in words what is the proof of the first line. It's that if you have a psi, psi star is also a solution. Therefore, psi plus psi star and psi minus psi star are solutions. So given a complex psi, then psi psi of x. Then psi real of x that we define to be psi of x plus psi star of x over 2. And the imaginary part of the wave function 1 over 2i psi of x minus psi star of x are both solutions with the same energy as this one has. So these are the two solutions. So far so good. You don't like to work with complex psi? No need to work with complex psi. Work with real psi. But here comes the second part of the argument, the second sentence. I want you to be alert that the second sentence is very powerful. It says that if you have a bound state of a one dimensional potential, more is true. There are no genuinely complex solutions in this case. Any solution that you will find, it's not that it's complex and then you can find the real and imaginary part. No, any solution that you will find will be basically real. And how can it fail to be real? It just has a complex number in front of it that you can ignore. So it is a very strong statement. That the wave function, it's not that you can choose to work it. You're forced to do it up to a phase. So how is that possible? How is that true? And here is the argument for the second line. If we're talking bound states, then these two are real solutions with the same energy. So now suppose these are bound states. There is a problem if there are two real solutions with the same energy. They would be degenerate. And property number 1 says there's no such thing as degenerate energy bound states. So they cannot be degenerate. So if you start with a complex psi, and you build these two, they must be the same solution. Because since there are no degenerate bound states, then psi, I will write it as psi imaginary, of x must be proportional to psi real of x. And both are real, so the only possibility is that they are equal up to a constant, where the constant is a real constant. You see there cannot be degenerate bound states. So the two tentative solutions must be the same. But that means that the original solution, psi, which is by definition the real part plus i times the imaginary part, is now equal to psi r plus i times c times psi r again, which is 1 plus ic times psi r. And that is basically the content of the theorem. Any solution is up to a number, just the real solution. So you're not going to find the real solution has non-trivial different real imaginary parts here. No, just the real solution and a complex number. Now if you want, you can just write this as e to the i argument of 1 plus ic times square root of 1 plus c squared psi r. And then it's literally the way it's said here. The wave function is proportional to a real wave function up to a phase. So that's a very neat situation. And therefore, you should not be worried that we are going to have to assume many times in our analysis that the bound states were trying to look for are real. And we plot real bound states. And we don't have to worry about, what are you plotting? The real part? The imaginary part? Many times we can just work with real things.
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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. The title here is the topic for our class today. We want to discuss formally the derivation Optical Bloch Equations, but usually when I teach you something, I have a general concept in mind and the concept right now here is, how can we get from unitary time evolution of a quantum system to rate equations anticipation. And this is sort of the subject of master equation open system dynamics, and this cartoon sort of tells you what you want to do. We have a total system, which is one part we are interested in, and the other one, often it has many, many degrees of freedom, and we don't want to keep track of them. But they have one Hamiltonian. But we are only interested in how our atomic system evolves, and we want to find an equation, which is no longer a Schrodinger equation. How does an initial density matrix describing our system develop with time? And as we will see, in general, it follows a master equation, and we want to discuss what are general principles of such equations. But it's very important here is that we are not keeping track of what happens in the environment, and that's associated here with a little bucket or trash can. Every result photons or such are measured, the environment is constantly projected on a measurement basis, and therefore, re-introduce probabilistic element into the part we don't observe mainly the part which characterizes the atoms. So for that we need the formalism of density matrix, and that's where we want to start. At the end of last lecture, I reminded you that the density matrix can always be written as an ensemble where you say you have a certain probability for certain rate function. This is always possible, but it is not unique. So each of you could actually create the same density matrix by preparing a number of quantum states with a certain probability and saying this is my density matrix. And also each of you has prepared different quantum states. If you sum them up in that same way, you get the same density matrix and therefore, all observables or measurements you will do on your ensemble in the future will be identical because the density matrix is a full description of the system. So at that level, it may look sort of very trivial that we have different unraveling. Unraveling means you look microscopically what is behind the density matrix, but in your homework assignment, you will all show that you can have a system which have very, very different dissipation mechanisms, but they're described by the same equation. So therefore, it's physically not possible to distinguish by just measuring the density matrix what causes a dissipation. Of course, if you know what causes a dissipation, fluctuating fields or collisions, you know more than the density matrix knows. Any questions about density matrix or the agenda we want to go through today? Just a quick reminder and since this was covered in 8421, and most of you know about it, I pre-wrote the slides. The density matrix is a time evolution. There is one which is sort trivial and covered in more elementary takes, and this is the Hamiltonian evolution, and I'm sure you've seen it. The time evolution, the unitary time evolution involves the commutative of the Hamiltonian with a density matrix, and later on, we want to specialize including dissipation, including the environment to the evolution of a tool evolutionary system driven by a monochromatic field And this the famous Jaynes-Cummings Hamiltonian, and we can characterize this system by a density matrix, but will be very important is that we distinguish between populations, the diagonal parts, and the coherences. And if you simply put this density matrix into this equation, you find the time evolution of the system, and we will refer to that result later on. Many of you have seen it in 8421. we can parametrize the density matrix for the two-level system. So we can parametrize the densities of the two-level system by a local vector, r, which is defined by this equation. And then the equation of motion is simply the rotation of a Bloch vector on the Bloch sphere, and it has a rotational axis, which is given by-- the undriven system rotates around the z-axis. This is just e to the i omega naught t, the normal evolution of the free system. But if you divide it with a monochromatic field rotation over the x-axis and the x-axis, of course, can take your two-level system, and flip it from the ground to the excited state. So it's just a reminder of the simple unitary time evolution, but now we want to add dissipation on top of it. And what I've decided that before I discuss with you Optical Bloch Equation and master equation in general, I want to give you a very simple model. I really like simple models which capture the essence of what you are going to discuss. So what I want to use is I want to use a beam splitter model I formulated for photons, but it would also immediately apply to atoms. And this model, what I like about it, it has all the ingredients of integration of the master equation we'll do later on without the kind of many indices and summations and integration, but it captures every single bit of what is important. And I usually like to present exactly solvable, simple models where you get it, and then I can go a little bit faster for the general derivation because you know exactly what the more complicated equations, what they are doing. So in other words, what I want to derive for you is we want to have the following situation. We have a beam splitter, and we know everything about beam splitters because we talked about them in the first part of the course, and we have a wave function, which is the input, and this is a photon. And we want to understand after the beam splitter, how has the system evolved. In general, it will be a density matrix, and what you want to find out is, what is the equation for the density matrix. Maybe this density matrix goes through the next beam splitter and then we want to know what comes out of it. And all we have to apply is the formalism we developed for the beam splitter earlier in the course. Of course, the beam splitter is not as harmless as it looks like. There is another part and another part one here brings in the environment, and for the environment, we will use the vacuum. That's the simplest environment. It's actually important environment because it is the environment we will use all the time when we discuss spontaneous emission. We send photons into the nirvana, into the vacuum and they disappear, and this is our modified. But the other one which is often not so explicit. If you send your photons away, you're not keeping track of what happens, but you could as well perform a measurement, and this is what we put in here. We say those photons hit a bucket or detector and measure them, we observe them. There's nothing else we do with them, so we can as well measure them, and this being immediately lead to the equation for the density matrix. So this is what you want to discuss, and it will have all the ingredients later on in a mathematically simple form for the derivation of master equation. So let's consider a similar photon for that. So the wave function, this is superposition of no photon and 1 photon, and the coefficients are alpha and beta. And just for simplifying notation, I pick alpha and beta to be real. So what do we expect to happen at the beam splitter? Well, there's a probability that the photon gets reflected. Probability to reflect the photon and therefore to observe the photon. This probability is, of course, beta square-- the probability that we have a photon to begin with-- and then the beam splitter, remember we categorized the beam splitter with angle sine theta cosine theta. Sine theta was the reflection amplitude, cosine theta, the transmission amplitude. So this probability, which I call P1 is the probability for reflection and for measurement. And now naively you would think what happens after the system has passed through the beam splitter, with a probability of P1, we've measured the photon. We know for sure there is no photon left. The system is in the vacuum state, but then you would say, well, maybe with probability 1 minus P1, we have not measured anything. Nothing has happened to the wave function, and that means the wave function just continuous. Well as we will see, this is wrong. We are missing something. What we are actually missing is that if you measure nothing, the wave function is not sine. The possibility that we could have measured something, changes the wave function. I will comment on that in much more detail in a few lectures down the road when I derive for you quantum Monte Carlo wave function. I will have a wonderful discussions with you about how does non-observation change a wave function. So we will talk about the physics behind it in some more detail. Right now, I don't want to get into this discussion. I simply want to use our beam splitter equation, so we can just take the beam splitter equation and apply it. So our output state is obtained by taking the operator for our beam splitter, and maybe you remember that the propagation for beam splitter was discovered by an operator, which had a dagger b dagger a in the exponent. a and b are the two input nodes. And the angle of the beam splitter, which interpolates between 0% and 100% reflection transmission is theta. And we're now looking for the output state of the total system. We're not performing the measurement yet, and this is now acting on the total system, which is the cross product of our photon system, of our system of interest. And the other input, which we call the environment or the vacuum, is 0. Well, look a few weeks back, we have done that all. The output state is, well, there was a probability, alpha, that we had no photon in the state psi. And if we have no photon in the state psi and no photon in the vacuum, this is the state 0, 0. What I denote here with this second place is the environment. And now we have one photon. We have exactly one photon with the amplitude beta, and this photon is split with cosine theta transmitted and with phi theta reflected. If you transmit it, we have 1, 0. If we reflect it, we have 0, 1. And again, this is the environment, and here is a photon in the environment. So let me just be clear that this is where the environment comes in. It is a vacuum state, and here, this is the output part for the environment. This is where we do the measurement. And I don't think it matters. I haven't really told you which is mode A, which is mode B. It doesn't matter, but one, let's say the environment is mode B, and the system evolves in mode A. As you can see, I'm using a new program, which has some nicer features in terms of handwriting, but it is a little bit rough in scrolling, so I sometimes have to scroll back and forth. So what is our output? Now, we have two possibilities. The environment is 0, or the environment is 1, and we perform a measurement. So we have to now go into a probabilistic description. So with probability P1, we have done a measurement, and our output state is now the vacuum state. With probability 1 minus P1, we have not detected anything in the vacuum, and therefore, our state of the system is alpha 0 plus beta cosine theta 1. Is alpha 0, so it is not beta 1 as naively would have assumed. It's not the original state. There is a cosine theta factor, which we got exactly from the beam splitter from the unitary evolution provided by the beam splitter. And since these state is no longer normalized, I have to normalize it by alpha squared plus beta square cosine square theta. So now we have done our measurement probability P1 to detect the photon. This projects the system into the vacuum state with the probability 1 minus P1. We have that state. Just one second. Scroll in the pictures. Write that down. In fact, millions that our system is now described by a density matrix with probability P1 and 1 minus 1 minus P1. With probability P1, we are in the vacuum state, and with probability 1 minus P1, we are in that state, the denormalized state psi naught, which I just hold down. Question? AUDIENCE: Have you considered theta to be some like a dynamical phase evolution system. It's very low order like when you expand it the first time it looks almost identical to [INAUDIBLE] quantum effect maybe. The environment is measuring the state in some way, and I mean, it's the lowest order now. PROFESSOR: Yeah, I Quite agree that random 0 is just an example of it. Pretty much, it's all the same. Yeah. What we do here is, I like the beam splitter because the beam splitter provides an exact formulation of the measurement process. You really can use a beam splitter to discuss what happens fundamentally when you perform a measurement. And the beams splitter is one typical implementation of that, but it has all the features you'll find in any measurement system. And especially what you observe here, let me just emphasizes is, the fact that we you do not make a measurement is changing the way function from the initial wave function psi to psi naught, we have a factor of cosine theta here. And that's also very general. A measurement perturbs, modifies your wave function no matter what the outcome of the measurement is. So let me write it down because we want to take it to the next level. So we have now found in terms of the beam splitter, angle theta, and the parameters of the initial state, alpha beta. We found the density matrix after the beam splitter. Yes. So what is the next step? Our goal is to derive the master equation for the density matrix, the time evolution of the density matrix. So since we want to discuss the time evolution, we want to find a differential equation. So what we want to figure out is, what is the difference between the output density matrix and the input density matrix. The input was, of course, pure state characterized by the matrix population alpha squared and beta squared of diagonal matrix element of alpha beta. The difference between the density matrix is can just calculate the difference. You can simplify things by applying some trigonometric identities, so this is an exact result, cosine theta minus 1. Here we have alpha beta cosine theta minus 1, and on the diagonal, we have cosine 2 theta minus 1 divided by 2. Anyway this is an intermediate result. We're interested in the differential equation. We want to sort of find out what happens when we observe, when we have the density matrix interacting with the environment all the time. And this can be simulated by beam splitters by using many beam splitters with a small degree of reflection. So we want to simplify this result now for the case of many beam splitters, and each of them has a small tipping angle, theta, and for later convenience, I defined theta to be gamma times delta t over 2. That's just my definition of theta. So what we have in mind now is that we start with the system psi, and we have many such beam splitters with an infinitesimal tipping angle. Each beam splitter has the vacuum at its input state. And we always perform the measurement. If I take the equation above, which I know you can't see anymore, we find a differential equation for the density matrix, which is we find an infinitesimal change delta over the density matrix, which looks like this. So all I've done is, I've used the equation above, and I've done a Taylor expansion in the small angle theta. And the reason why I brought in the square root, well, we get cosine theta. The first order Taylor expansion or the lowest order Taylor expansion from cosine is 1 minus theta squared. So I get the square root squared. So I get gamma, which appears here, and then I divide by delta t. So this is just an exact mathematical expression, and the next step is to form a differential equation. But before I do that, I want to emphasize the two features we are using here. They sort of enter automatically, but these are the two big assumptions we make when we derive a master equation. The first one is that we always have a vacuum state as the input. So in other words, the environment is always in the same state, which is a vacuum state, and this is sort of called a Bohr approximation. What it means is that we do a measurement here, but the vacuum is not changing. In other words, we are not overloading the vacuum with so many photons that suddenly the vacuum Is no longer in the vacuum state. Or in the case of spontaneous emission, the vacuum can just take as many photons as you dumping into it. They disappear so quickly that for all practical purposes, the environment stays in the vacuum state. So this is called the Bohr approximation. The environment is not changing. It has enough capacity you to be modified by the measurement process. And the second thing which is related is, the vacuum is always in the same state, and there are no correlations from here to here to here, there is no memory effect. Everything is completely uncorrelated. So the environment is uncorrelated. It has no memory. It's correlation function is a delta function, and this is called Markov approximation. So these are the two effects which are important. One is no memory for the environment, Delta function correlation, Markov approximation, and the two, of course, are related in the environment is all of this in the same state. So remember this is a change for the density matrix, and alpha beta where the original parameters of the density matrix for the input state. So I can now rewrite everything as a differential equation. The density matrix has a derivative for the diagonal matrix elements and for the coherences. Here we have plus gamma. Here we have minus gamma 0, 1, 1, makes sense because we conserve the trace. We have unity probability that we have a stellar system. And therefore the two diagonal matrix elements, the population, the sum of them cannot change with time. And for the coherence, we have gamma over 2. And if you're familiar with Optical Bloch Equations, which we derive next, we can say that these means if 0 is the ground state that the ground state changes because-- call it spontaneous emission from the excited state. This equation would say that the excited state decays with the rate gamma, and sometimes you may have wondered about that there are factors of two appearing, which also appears here that when the excited state decays with a rate gamma, we have a factor here for the coherences, which is gamma over 2. So what we have accomplished in contrast to let's say Einstein's equation with the Einstein a and b coefficient, which lead to rate equations for the population, we have now a new feature. We have an the equation for the coherences, and we find a decay of the coherences with half the rate as a decay of the population. Questions? So if you want you could rewrite this model for photon, which goes through beam splitters, undergoes measurement. You can rewrite it from atomic wave function and you measure whether the atomics in the excited and ground state and the equation for the measurement performed on the atom is exactly as the equation by which the beam splitter acts on the photon state. So what I've shown here it's very specific for a single photon because I could use simple equations, but everything is what you find in a much more general situation. So before I give you the general derivation of the master equation, let me talk about what we have learned from this example and what the general procedure is. The first thing is our goal is to find a differential equation for the density matrix of the system. I just remember. There was one thing I wanted to mention. In the previous derivation with the beam splitter, I started with a purer state, and the purer state developed into a statistical mixture, and this statistical mixture would then transform the next beam splitter into another statistical mixture. I derived the differential equation for you for the first state from the pure state to the statistical mixture. But if you would spend a few minutes, you could immediately show that you can start with an elementary density matrix, look how it evolves through the beam splitter, and you get exactly the same differential equation. So the general procedure is, we want a differential equation how the density matrix evolves with time. And this will be obtained by finding an operator which acts on the initial density matrix. This operator is not a unitary operator because we are performing measurements through the environment, even if you don't actually perform them. Once we dump something into the environment, it's out of our control and anybody could go and perform a measurement, and so we should assume that this measurement has been taken. It's one of those quantum mechanical things that you don't even have to care whether somebody does it. The environment does it for you. So this operator is called a Liouvillian operator. It's sometimes called-- and I haven't really traced down why-- it's called sometimes super operator. I know what superconductivity is, but I don't know what the super powers of this operator are, but that's just a name which you will find. The second thing which we have used is that the evolution of this system can be obtained from the time evolution of the total system by performing a trace over the degrees of freedom of the environment. This was exactly what we actually did when we said the system continues with probability P naught in one state and probability P1 in the other state. The operation which lead to this density matrix was exactly the partial trace. So this is the second general feature which we have to implement. Thirdly, if we could do one and two exactly, we would have an exact formulation for a small part of a quantum system no matter how complicated the environment is. In practice, we can solve the equations only when we make simplifying assumptions about the environment. One is, it is large, and more important, therefore, it's unchanging, and this is the Bohr approximation. And the second feature is, it has a short correlation time, tau c. In the beam splitter, I have made the assumption that there is no correlation between different beam splitters in the derivation, which I want to walk you through. Right now, you will see explicitly where the correlation times enter. And this is called the Markov approximation. And finally, this is number four, the whole possibility to derive a master equation hinges on the fact that we have different time scales which are very different. We are interested in the evolution of our system. We want to know how it relaxes, and this is on a time scale 1 over gamma. So we call this slow. We are interested in the variation of our system, the atomic system or the photon state, which passes through the beam splitter, and this time scale has to be much slower than the fluctuations of the environment. So therefore, if the environment has fluctuations, which in the beam splitter model where assumed to be 0, it was delta function of time, if that correlation time is much smaller than the time it takes for the system to relax and to evolve, that opens a window delta t, and this is the time scale of the master equation. Just to give you one example for the spontaneous emission, the correlation time, tau c, would be the time it takes the photon to disappear from the atom. And the photon has disappeared from the atom when it is one wavelengths away. So typically, the correlation time f the vacuum for spontaneous emission is one cycle of the optical frequency. It's very, very fast. Whereas typical decay times of the excited states, a nano second. It's six orders of magnitude slower, and this is what we describe. But on a time scale of a femtosecond of one optical cycle, the photon has not detached from the atom and it could go actually back to the atom. During that time, we talked a little bit about it when we did this diagrammatic discussions of resonance scattering for very, very early times. You don't have exponential decay because you cannot do the approximations where we approximated the kernel by something which was completely energy dependent. And so what happens at such short times, we encounter here again in such short times, we will not have a simple description of the system. So the last point, let me summarize. Our goal is that we describe the density matrix of the system, and we want to find the Liouville operator or some matrix which acts on it. And because we will integrate over time steps, which are larger than the correlation time of the system, we can also call it, it will be a coarse-grained evolution. Any questions about that? I really like the discussion, the derivation of the master equation, how it is presented in atom-photon indirection. But it is presented on more than 50 pages with many, many equations. So after giving you all of the principles, all of the concepts, I want to go with you now over those equations and point out how the principles, which we encountered with the beam splitter, how they are now implemented in a very general context. I will not be able to give you all of the mathematical aspects of it, but I think by now you know that the book atom-photon interactions are actually wonderful. You can get a lot of conception information out of it by looking at the equations without understanding every technical detail. So I would really encourage you, if there is something which piques your interest and I hope there will be things which you'll find very interesting, that you go to the book and read it. So I'm exactly following that actually I used copies of the book. So we have a Hamiltonian, which is describing the atomic system. It describes the reservoir and then there is an interaction. We keep it very general here, but you may always think well, the atom is your favorite two-level system. The environment is maybe the vacuum with all its possible modes, and the interaction is the dipole interaction or the a dot p interaction. So we start out with an equation, which is nothing else than Schrodinger's equation for the density matrix. The time derivation of the density matrix is commutative with the Hamilton. But it is often useful and you've seen it many times, to go to the interaction representation, that the time dependence due to the unperturbed part of the operator is absorbed in a unitary transformation, so therefore, this density matrix in the interaction representation evolves not with, h, because h naught is taken care of, it only involves due to the coupling between the two systems, between the system and the environment. So now this equation, we are interested in a time step delta t. And this time step, delta t remember, we want to coarse grain, will be larger than the correlation time of the reservoir, and you will see exactly where it comes about. So we want to now do one of those coarse-grain steps. We take this equation and we integrate from time t to time t plus delta t. So this is exact here. But now we want to iterate, and that means the following. We have expressed the time step in the density matrix by having the density matrix there. But now we can do in a first-order perturbation theory, we can do one step and we get the second order result by plugging the first order result into this equation. It's the same we have seen with our diagrams and such. We have an exact equation. It's useless unless we do something, and what we do is, we realize that we can iterate it because the part we don't know involves one more occurrence of the interaction potential. And when you plug the nth order solution in here, you get the n plus first order solution. And this is exactly what is done here. And I skipped a few equations here. This is what he's done here, number one. And number two is, we are interested in the system, not in the reservoir, so therefore we perform the trace over the reservoir. And the trace of the reservoir for the photon beam splitter mend, we say we have two possible states, we detect a photon or not. And for the system, we have now a density matrix which is probability P naught in one state, probability P1 in the other state. And this is exactly what the operator partial trace does. Remember also I want to really make sure that you recognize all the structures. The time evolution of a density matrix was a commutator with h, but in the interaction picture, it's a commutator with v. But since we are putting the first order result in here, the second order result is now the commutator of v with the commutator of v and o. It's just we have iterated one more time. So the sigma tilde, the density matrix for our system, tilde means in the interaction picture is now the partial trace over the reservoir of the total density matrix. Tilde means in the interaction picture. And the important part here is that it is exact. We have not done any approximation here. Any questions? Of course, now we have to make approximations because we cannot solve an interactive problem exactly. The first one is-- and what do we want to do in the end? We want to keep the first non-trivial term but to the extent possible, we want to factorize everything. We want to get rid of the entanglement of the environment in the system and only get sort of the minimum which is provided by the coupling. So this evolves as follows. The interaction we assume is a product of two operators. One operator acts on the system, one operator acts on the environment. So this could be the dipole acting on the atom, the vacuum field, e, interacting with the environment or it could be p dot a. Or maybe your system has a magnetic moment, m, and the environment consists of fluctuating magnetic fields. So we'll pretty much find in every kind of measurement that a measurement involves the product of two operators. One is an operator for your system and one is an operator for the reservoir of the environment. And so this is one thing we want to use, and now there is one thing which the moment we will set in our equations, there is one thing which will naturally appear. Let me scroll back. What we have here is the interaction operator, v, at two different times. So this means if something happens at different times and we integrate over times. This is a correlation function, a correlation function between v at the time t prime and the time t double prime and since the reservoir part of this interaction is the operator, r, so what we have here now is, we have a correlation between the operator, r, at two times, which characterizes the environment. And now comes an important approximation. You remember I said we want to assume that the environment has a very short correlation time. Whenever a photon is emitted, it appears dramatically fast. It disappears in one optical cycle, and the environment is sort of reset, it's back in the vacuum state. So this is now expressed here that this product over which we take the partial trace has a very short coherence time. And the fact is now the following. We are integrating over a coarse-grained step delta t, but this correlation function goes to 0 in a very short time. So therefore it will not contribute a lot. Let me write that down. What we are going to approximate is our total density matrix is now approximately factorizing in a density matrix describing the atomic system. Well, we describe the atomic system when we trace out the environment. We describe the environment when we trace out the atomic part. And if we now form the direct product, we are back to the total system, but we have factorized the total density matrix into two parts. What we neglect here is a part which cannot be factorized which is the correlated part of it. But what happens is, since we are integrating over time steps delta t and the correlation decay in a very, very short time, the result is that this complicated part, which we could never calculate, is smaller than the first part by the ratio of the time where the correlations contribute over the time delta t, the time step we are going to take. So this is a very critical assumption. There is a whole page or two in the book where an photon-atom interaction they discuss the validity of this assumption, but I've given you the physical motivation that we indicate over much larger time, and if this time is large and the correlation is lost for short time, they only contribute with this small parameter to the result. So in other words, this means after-- we have an interaction between the environment and the system. We write it down in second order, but the second order result is now we evaluated by factorizing the density matrix into our system in the reservoir. So that means in that sense if you factorize something, it looks as if it's not interacting, but the trick is the same. You write down something to first and second order, and once you have factored out the important physics, now you can evaluate the expression by using an approximation, which is now the approximation that the density matrix factorizes. So with that, this is the approximation that we have made that the correlation time is very short. And now we have a differential equation for the density matrix sigma, which describes or atomic system. We have traced out the degrees of the reservoir. And now we want to insert B.17. You probably don't remember what B.17 is. It says that the interaction operator is a product of a, the operator a for the atoms and r for the reservoir. So the reservoir part, tau prime tau double prime, gives a correlation function. This is the correlation function between the operator, r, at two different times and the part which acts on our system, the a part, is explicitly kept here. So this is now a general master equation. It tells us the time evolution of the density matrix in this form. It looks very complicated, but this is because it's very general. In order to bring it into an easier form, we want to now introduce a basis of states, energy eigenstates of the unperturbed system, and write down all of these operator into such a basis of states. But anyway you saw here how we had an exact equation, and the main approximation we made is that the operator acting on the environment has a very short correlation time. Any questions? Well, you're only a few minutes away from producing this result to Fermi's golden rule, which you have known for a long, long time. It's just we have made very general assumptions. You see sort of how the assumptions propagate, but now if you write it down for an energy eigenbasis, you will immediately see results you have probably known since your childhood. So we want to have energy eigenstates of the atomic operators, so this is sort of ground and excited state if you think about a two-level system. The previous equation, I have to go back to it. Our previous equation is a differential equation for the density matrix here, and here is the density matrix. So now we formulate this equation into an energy eigenbasis, and what do we get? Well, we get an equation for the matrix elements, and what matrix elements are important? diagonal matrix elements which are population of diagonal matrix elements which are coherences. So we pretty much take this equation, use the energy eigenbasis and look, what do we get for the populations and what do we get for the coherences. So the structure is now the following. That we have our matrix elements ab. There is one part which looks like a unitary time evolution. This is what comes from the Hamilton operator. This is sort of the-- we'll see that in a moment-- but this is the time evolution without relaxation and now we have something here which are generalized relaxation coefficients. And you will find if you go further above that those relaxation coefficients are directly related to the correlation function of the reservoir. So we can now specify what happens between populations. Population means that we have a differential equation, let's say between sigma aa the puller, and sigma cc, so we have a rate coefficient which connects the population in state A with the population in state C. And if you take this expression, you find several things. Well, you find Fermi's golden rule, in a generalized way-- that's always nice-- you find Fermi's golden rule. When you integrate over time, you often get data function, and you expect to get a delta function because of energy conservation. So you get that, of course, naturally. Secondly, we have second order matrix element, which you know from Fermi's golden rule, but now we have the following situation that the matrix element in Fermi's golden rule may actually depend on the state, mu, of the environment. So you have maybe 10 different possibilities for the environment, and Fermi's golden rule gives you spontaneous emission, which is different for those 10 states. And naturally, since we have performed the partial trace over the environment, we have all those rates weighted with the probability that the environment is in one of those states. So what you find here is a simple generalization of Fermi's golden rule. And if you look at the off-diagonal matrix elements, for instance, you want to know what is this rate, what is the rate coefficient, which gives you the time derivative of the coherence, and it's multiplied with the coherence. You'll find now that in general, this rate coefficient has a damping term, but it may also have an imaginary term. And I hope you remember when we played with diagrams, that we had something similar. There was something which we called the radiative shift. I called it the AC stock effect of a single photon, and here it is a level shift which comes because the environment interacts with your system and it shifts the levels a little bit. So in addition to just relaxation, spontaneous emission, and damping, there is also a dispersive part, a level shift, and it has exactly the same structure. Let me add that delta ab is the difference between the shift of state b and the shift of state a. And those shifts have exactly the same structure. You have to take the principle part of something which has 1 over the difference of energies, and we discussed that this has to be understood by somewhere adding an infinitesimal imaginary part and doing the right thing with complex function. It's actually related to Laplace time difference between Laplace transformation and Fourier transformation. So anyway what I find sort of beautiful is that we started with a most general situation. We perform the partial trace. We made one assumption of short correlation times, and a lot of things we have known about quantum system just pops out in a very general form here. Any questions about so far? Well, the coherences are, of course, more interesting than the population. Coherence is always something physicists get excited about it because it captures something which goes often beyond classical system that we have quantum mechanical coherences. And what happens is the coefficient here, which provides the damping of the coherence, just comes out of the formalism, has two parts. And one part is an adiabatic part and the other one is a non-adiabatic part. Well, and that makes sense. If you have two quantum states and there is a coherence, some phase between the two, the phase can get lost if you do a transition between the two states or one state undergoes a collision and is quenched. So you definitely have one part which is due to the fact that the quantum states or the population changes, and you find that there is this state-changing part, which is pretty much the sum of all the rate coefficient leading out of state, leading to the decay of state a and leading to the decay of state b. In other words, if you have a two-level system, which has a coherence and you have decay of the excited state and decay of the ground state, you would expect that those decay terms appear also in the decay of the coherence between the two levels and they do, and they appear with the correct factor of 1/2. But there is another possibility and this is the following. You can have no [INAUDIBLE] of the population of the state, but you can still lose the coherence. The model you should maybe make is that you have spin up, spin down. You are not perturbing the populations in spin up and spin down, but the environment provides fluctuating magnetic field. Then due to the fluctuating magnetic field, you no longer can keep track of the phase, and that means in your identity matrix the off-diagonal matrix elements decay. And we find that here this is the second part which in this book is called the adiabatic part, and the physics behind it is now pure de-phasing. So it's an independent way for coherences to decay independent of the decay of the population. Questions? Collin. AUDIENCE: Where does the Markov approximation come in? PROFESSOR: The Markov approximation is, so to speak, the delta function approximation, which would say that-- I mean, I introduced the correlation function between the reservoir operator and said the correlation time tau c is very, very short. The Markov approximation would actually state that it would actually say in a more radical way the correlation time is 0. And the Bohr approximation, the fact that the reservoir is unchanged came in when we said the total density matrix for the second order expression just factorizes. It factorizes into the environment, which is just this density matrix of the environment. It's not changed by the interaction with the system. And this is the Bohr approximation. We just use the same expression for the reservoir independent of the measurements the reservoir has done. Other questions? Yes, Nancy. AUDIENCE: [INAUDIBLE]. PROFESSOR: This is something very general. Thank you actually for the question. Whenever we have some damping of the population, the coherence is only damped with a factor of 1/2. One way to explain it in a very simple way is, that if you have an amplitude alpha excited and alpha ground, the population in the excited state is this squared. So you sometimes make the model that the amplitude decays with gamma over 2, but the population is-- because you take the product, decays with gamma. So you would say alpha e and alpha g both decay with half the rate, but the probability is for the total rate, and the coherence is the product of the amplitudes. So therefore when you look at the coherence, this decays with 1/2 gamma e. This decays with 1/2 gamma g, and this is what you get here, 1/2 gamma in state a, 1/2 gamma in state b. Whereas the probability to be in this state decays with twice that because the probability is squared. So you find that pretty much in any quantum mechanically corrects derivation, which you do about the decay of population coherences. Other questions? So we have done two things. We have done the very, very simple derivation using the beam splitter model where you may not even notice where I did the Markov approximation because I jumped from beam splitter to beam splitter and left all the correlations behind. Here in the most general calculation, you have seen exactly where it enters, but maybe now you have the full forest in front of you, and you don't recognize the trees anymore. So let me wrap up this lecture by now focusing on the system we want to discuss further on, namely a two-level system interacting with a vacuum through spontaneous emission. But I also want to make some generalizations. I want to give you some generalizations about what kind of environments are possible in quantum physics. Let me just see how I do that. So this part I actually owe to Professor [? Ikschuan ?] who wonderfully compiled that. So what I want to do now is, I want to call your attention to the operator form which, is rather unique. Remember when we did second order perturbation theory, we had sort of the commutator of v with the commutator of v and rho. This came from iterating the exact equation of motion for the density matrix. And you want to specialize that now to Jaynes-Cummings model. I mean, in the end, at least in this course, we always come back to the Jaynes-Cummings model because it captures a lot of what we want to explore. So the Jaynes-Cummings model in the rotating wave approximation is very simple. It raises the atom from ground to excited state and destroys a photon, or it does the opposite. So this is our simple interaction between our system, the two-level atom and our reservoir which is just the vacuum of all the modes. And you, again, recognize what I said in general. You usually always find it by linear form, an operator which acts on the modes, on the reservoir on the vacuum, and an operator which acts on the vacuum. Now, we want to make the explicit assumption that the initial state of the reservoir is the vacuum state. It's empty. And I want to show you what is the structure of the operators we obtain. And so if you put v into here, you have first the commutator with rho, which I write down here, and then we have to take another commutator with v. And the result of that is the following, that when it comes to relaxation pauses, based on the general structure of the time evolution of quantum mechanics, we have this double commutator. And the operator which, couples our system to the environment is erasing and lower an operator. I mean, the atom because it interacts with the environment either absolves the photon or emits the photon. But those operators, sigma plus and sigma minus, appear now always as products because we have two occurrences of the interaction, v. But if you look at the double commutator structure, the operator sigma plus sigma minus appears. This is just the general structure of this double commutator appears to the left side of the atomic density matrix to the right side of the atomic density matrix and then there is the coarse term where the atomic density matrix is in the middle of the two. So this is actually something which is very general and very important in the theory of open quantum system. What I'm discussing with you now is this famous Lindblad form. And the story goes like that. You want to know what are possible environments, not just empty vacuum. You can have fluctuating fields. You can have, you name it. But if you are saying that your environment interacts with your system through an operator, and our operator is now the operator sigma minus, which is a spontaneous emission, you need heavy t. The mathematical structure of a valid quantum field remains that if your system interacts with an environment by emitting a photon sigma minus, this is now the structure of the master equation. This is the structure of the time evolution of the density matrix. So the operator sigma minus and its and its Hermitian conjugate sigma plus, all of this have to appear in this combination. Yes, Collin. AUDIENCE: This is still in my interaction picture. Right? There's no dynamical phase evolution that we put back in there. PROFESSOR: Yeah. OK. What we do in general if the system is driven by a laser beam, for instance, Rabi oscillation, we simply add up the dynamics of the Rabi oscillation of the unitary time evolution to the time evolution done by the reservoir. AUDIENCE: So this form is always in the interaction picture? PROFESSOR: Well, this is, you would say, this form is what is the relaxation provided by the environment. And if you drive the system in addition with the coherent field, unitary time evolution, you would add it to it. So in other words, what I'm telling you here is that this is the general structure, and if you have a system which interacts with an environment in five different ways, with a dipole wand, with a magnetic moment and such, you have maybe five interaction terms and then you have to perform the sum over five operators, and here one of them is a sigma minus operator. So in other words, if you want to know what is the whole world of possibilities for quantum system to relax and dissipate with an environment, you can pretty much take any operator which acts on your system, but then put it into this so-called Lindblad form and you have a possible environment. And I mentioned I think last week that people are now in our field actively working on environmental engineering. They want to expose a system to an artificial environment and hope the system is not relaxing, let's say, to a broken ground state, but to a fancy correlated state. So what this Lindblad form, if the operators appears in this way, what it insures is the following. Just imagine if we have an equation, a derivative of the density matrix, which depends on the density matrix, you could write down a differential equation and say is it possible? Well, it has to be consistent with quantum physics. You have certain requirements. One requirement is that rho, the density matrix, always has to be the density matrix. The trace equal 1 has to be conserved. A density matrix always must have non-negative eigenvalues, otherwise, what you write down, it might be a nice differential equation, but quantum mechanically, it's nonsense. But now there is one more thing, which is also necessary. This time evolution of the system's density matrix must come from a unitary time evolution of a bigger system. So you must be able to extend your system into a bigger system, which is now the environment, and this whole system must follow a Schrodinger equation with a Hamilton as a unitary time evolution. And this is where it's restrictive. You cannot just write down a differential equation and hope that this will fill some requirement, and what people have shown is under very general assumptions, it is the Lindblad form which allows for it. So some operator always has to appear in this form. So often in this Lindblad form, you have an operator, which is called a jump operator, which is responsible for the measurement, which the environment does on your system. The jump operator is here, the operator which takes the atom from the excited to the ground state. With that you need a photon, and the photon can be measured. So often you can describe a system by a jump operator, and if the jump operator is put into this Lindblad form, then you have a valid master equation for your system. So let me wrap up. If you take now the definition of the raising and lowering operator, and you take the form I showed you, the Lindblad form, you'll find now this differential equation for your two-level system. And this is one part of the Optical Bloch Equation. Now, coming to Collin's question, if you include the time evolution of the classical field, this is a coherent evolution of the Bloch vector, which I showed at the beginning of the class, and we add this and what I wrote down at the beginning of the class. Then we find the famous Optical Bloch Equations in the Jaynes-Cummings model. So these are now the Optical Bloch Equations, and I hope you enjoy now after this complicated discussion, how simple they are. And it is this simple set of equations, which will be used in the rest of the course to describe the time evolution of the system. Just because I did some generalizations about the Lindblad equation, I copied that into the lecture notes from Wikipedia, and probably now you sort of understand what is the most general Lindblad equation. It has a Hamiltonian part and then it has jump operators like you sigma minus operator, but it has to come in the form that the jump operator and its complex conjugate, emission. conjugate, is on the left side, on the right side, and left and right of your density matrix. So this is the generalization I've mentioned. Yeah. with that I think with that we've derived the master equation, and on Wednesday, we will look at rather simple solutions, transient and steady-state solution of the Optical Bloch Equation. Any questions? One reminder about the schedule, this week we have a lecture on Friday because I will not be on town next week on Wednesday. And of course, you know today in a week, next week on Monday. It's [INAUDIBLE] day. So we have three classes this week, no class the following week, and then the normal schedule for the rest of the semester.
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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: OK. Back to cavity QED, back to the fully quantized radiation field, back to vacuum Rabi oscillation. Let me just recapitulate and sort of make the transition from this intense discussion about homework to the intellectually stimulating discussion about atoms and photons. So in the semiclassical description of the electromagnetic field, photons can only be emitted because we have a Hamiltonian with the semiclassical electric field. So if you don't drive the system with an electric field, you cannot stimulate the emission of photons. But we know this is not what happens. Photons are emitted into empty space, photons are emitted into a vacuum. And for that we needed a quantized description of the electromagnetic field. We did field quantization, and we have now our quantized Hamiltonian. And on Monday I started to discuss what is sort of the paradigmatic situation, the paradigmatic example, for how you should think about the vacuum and how you should think about emission of photons into the vacuum. And these are the vacuum Rabi oscillation described by the Jaynes-Cummings model. So the situation which I have in mind, or which you should have in mind, is an idealized situation, but it has been realized experimentally. And some of those idealized experiments were recognized with the Nobel Prize research of Haroche and Dave Wineland. So the situation is we have an atom, but it only talks to one mode of the electromagnetic field, and we make sure that the atom only talks to one mode of the electromagnetic field not by eliminating other modes; they exist. I mean, an atom can emit upwards and downwards. But we surround it with a cavity which has such a small mode volume, it has such a small volume, that the single photon Rabi frequency is huge, and therefore the emission into this one single mode dominates over the emission into all other modes. So this is a condition that the single photon Rabi frequency has to be larger than gamma. And, of course, we also have to make sure that the system is idealized so the loss of photons because of losses in the mirror, or finite reflectivity in the mirror, also has to be smaller. So that means for several Rabi periods we have a system which has only two parts, a two-level atom and one single mode of the cavity. So that's the system we have in mind, and we discussed the Hamiltonian. We saw that the Hilbert space of the atom is excited in ground state, the Hilbert space of the photons is spent by the [? flux ?] states, but what happens is-- so there's an infinite number of states, because of the infinite number of states of the photon field-- but what happens is the Hamiltonian couples only an excited state with n photons to a ground state with n plus 1 photons. So the whole Hilbert space is segmented now into just pairs of states labeled by the index n. So after so much work, we are back to a two-level system. And here is our two-level Hamiltonian. And, well, a two-level system does oscillations between the two levels. Rabi oscillations, no surprise. And this is what I want to discuss now. But the new feature is that these are really, well, these are now really two levels. Each of them is the combined state of the atom and the quantized radiation field. So now we have included in our two-level description the quantum state of the electromagnetic field. So first you should realize that this Hamiltonian is absolutely identical to spn 1/2 in magnetic fields. And you can recognize by [? comparing ?] this Hamiltonian, this matrix, to the matrices we discussed for spn 1/2 in the magnetic field, that this corresponds to the situation where this spn 1/2 had a transverse field in the x direction which caused a precession from spn up to spn down. And this x component of the field corresponds now to the single photon Rabi frequency times n plus 1. That's the off diagonal matrix element in this matrix. The thing which we have to discuss, and I will focus later, is that it depends on n. So for each pairs of state labeled by n, the photon number, we have a different off diagonal matrix element. But let's discuss first the most important and simplest case. Let's assume we are on resonance, and we want to assume that we have a vacuum. Then our Hamiltonian is simply this. And when we prepare the system in an initial state, which is an excited state with no photon in the vacuum, then we'll have oscillations to the ground state with one photon. These oscillations are exactly the oscillations we saw on the spn 1/2 system. We can just map the solution. I'm not really writing anything here. So what we obtain is the famous vacuum Rabi oscillations. where the probability to be in the excited state oscillates with the single photon Rabi frequency omega 1. I think there's a little bit of an ambiguity in language. Is it the single photon Rabi frequency? or is it the vacuum Rabi frequency? Because there's always the question about plus minus one photon because we start in the excited state without photon so you want to say it's a vacuum Rabi frequency. But then you have the ground state with one photon, and this photon is reabsorbed and then you may want to call it the one photon Rabi frequency. So I leave it to you, but it's called vacuum Rabi oscillation and this Rabi frequency is usually referred to as the one photon Rabi frequency because we obtained the Rabi frequency by calculating the electric field of a single photon. So the Rabi oscillations which we are observing now correspond to the periodic spontaneous emission and re-absorption of the same photon. There's only one photon which is spontaneously emitted and reabsorbed in a completely reversible coherent way, and the time evolution is unitary. So it's a periodic spontaneous emission and re-absorption of the same photon. This has been experimentally observed. Actually, let me back up. Experiments are done in the microwave regime. The leading groups are, well, in the older days, Dan Kleppner, Herbert [? Weidner, ?] and Serge Haroche. And this involves Rydberg atoms. Rydberg atoms in superconducting high q cavities. And those Rydberg atoms, because things scale with n and n squared, the principal quantum number, have a fantastically strong coupling to the electromagnetic field. And there is a homework assignment on Rydberg atoms in such cavities. The other example is in the optical domain. And this really involves the D line of alkali atoms. You drive them on the D line. rubidium and caesium are often used, and the work is enabled by the development of so-called supermirrors which have an extremely high reflectivity, and you can realize an excellent q factor. And the leaders in this field are Jeff Kimble and Gerhard Rempe at the Max Planck Institute. So let me just discuss an example taken from the optical domain. So the generic situation is that you have two mirrors which define a single mode cavity. Usually, you have a stream of atoms. Traditionally in atomic beams, then in some experiments in slowed atomic beams, more recently in atoms which are falling out off the mode, and only very recently single atoms with the help of other laser beams that are trapped inside the cavity. So they are streamed in such a way that only one or a few atoms are in the mode volume interact with a single mode of the cavity at a given time. And then you want to figure out what is now happening, and you have the probe laser, you send it through the cavity, and then you record the transmission with a photodiode. Yanosh? AUDIENCE: What is the mirror made of for the [INAUDIBLE]? PROFESSOR: The mirror is made of a glass substrate, but then you would [INAUDIBLE] the coating. And the mastering is really to put coatings on which are very pure, but then also I think using ion sputtering, you make sure that the coating is extremely smooth and does not have any surface irregularities which would scatter a tiny fraction of the light. I know there are some people in Ike's group and [INAUDIBLE] group who work with high q mirrors. What is a typical example for the reflectivity? Or the q factor you can reach? AUDIENCE: 5 [INAUDIBLE] and a finesse of, maybe, 500,000. And they're called superpolished mirrors. PROFESSOR: So finesse of about a million, and that means the mirrors have 99.9999% reflectivity. And the superpolishing, I think, that was the last step. People had controlled the materials, but then they found ways to make a super polish and avoid these even one part per million scattering by surface roughness. OK. So if you do that experiment, what would you expect? Well, it's a [INAUDIBLE] experiment so if you would scan the probe laser, and there is nothing in it, what you would expect is you would just expect a transmission peak at the cavity resonance. And if you tune much further, you get the next peak at the free spectral range. Let me just indicate that. So this is a case for 0 atoms in the cavity. If you put 1 photon in the cavity, you no longer-- sorry, 1 atom in the cavity, you're no longer probing a cavity, you're really probing a system, which is no longer the cavity by itself. It's an atom-photon system. It's a couple system. And we know it's described by our two-by-two Hamiltonian, and this Hamiltonian has two solutions. And the two solutions are split by the one photon Rabi frequency. So the two eigenvalues of our Hamiltonian are at plus minus omega 1 photon. So therefore, for n equals 1, we have a situation that we have two peaks split by the single photon Rabi frequency. Of course, I have assumed that great care has been spent to make sure that the cavity resonance is right where the atomic resonance is. So this is now for 1. If you have 10 atoms, remember the two-by-two Hamiltonian looks the same, but it has the square root n plus 1 factor. So neglecting the 1 roughly when we have 10 atoms in the cavity, it's square root 10 larger Rabi frequency. And therefore, we would expect that we have now a splitting of the two modes, which is square root 10 plus 1 larger. Actually, I didn't-- sorry, I have to collect myself now. I showed you that the Rabi frequency scales with square root n plus 1 in the photon field. But you should realize that everything is [? isometric ?] between photons and atoms. It's the complete coupling between photons and atoms. And if you would now look-- but I don't want to do it now-- if you would now look what happens if several atoms are present in the mode volume, you would also get a scaling which is n plus 1 in the atom number, because the atom coupled coherently. It is actually an effect of super-radiance, which we'll discussed later. So just take my word. You have the same scaling with the atom number. But I have to give you my word now, because in the experiment this is what people varied. AUDIENCE: If they had varied power or number of photons instead, like, we couldn't have drawn these same diagrams, right? Because then the top part of the Lorentzian changes. If you're changing the photon number then Lorentzians change. PROFESSOR: Say again? AUDIENCE: So, like, right now, yes, we are varying the number of atoms so we can talk about the splitting. But if we were varying the power or the number of photons instead, then each of the Lorentzians, their height would change. How would you draw this observation if you were changing the photon numbers? PROFESSOR: You know, I don't want to go into line shape. I would probably be a Lorentzian. I mean, all I want to discuss here is that we have a two-by-two Hamiltonian, which is split. And if we have one atom and one photon, it is split by the single photon Rabi frequency. If we have one atom and 10 photons, the atom can of course absorb and emit only one. As I derived on the previous page, we would have now a Rabi splitting, which is square root n plus 1, n being the number of photons. But if you would start in an empty cavity with 10 atoms in the excited state, because all the atoms are identical, they would spontaneously emit together, and then you would have 10 atoms in the ground state, and then you would have 10 photons. And so maybe this helps you. If you start with 10 atoms in the excited state, they do everything together. If you have 10 atoms in the ground state with 10 photons, and now you have 10 photons and it's clear the 10 photons lead to a Rabi frequency, which is proportional to the square root of 10 or to the square root of 11. So therefore, what you will observe is you will now observe a splitting of the single mode of the cavity which goes by the square root of n plus 1. I don't want to discuss the line shape and the [? strings, ?] I just want to sort of discuss, in a way, the eigenvalues of the Hamiltonian, and the eigenvalues are the positions of the transmission peaks with a cavity. And that has been observed. I mentioned the two leaders of the field are Gerhard Rempe and Jeff Kimble. Well, Gerhard Rempe, he did his Ph.D. In the same group at the same time as I did so I know him very well. Then he went and did post doc work with Jeff Kimble, and now is the Director of the Max Planck Institute. He has the world leading group in cavity QED. But this is sort of here the two leaders have a joint paper, which is the first observation of the vacuum Rabi splitting in an optical cavity. Of course, you can easily observe it if you have a strong atomic bean with many atoms, because then you have a good signal. And secondly, the splitting is large and easily resolved. So what they managed to do is they managed to throttle down the atomic beam that fewer and fewer atoms at the given time were in the cavity. And eventually they came down to the limit of one atom. That was an historic experiment. Of course, it's not perfect in the sense that you do not see the deep cut between the two peaks simply because, when on average you have one atom in the cavity, sometimes you have to atom in the cavity, and then you have a peak in the middle. So those experiments in those days were done only with average atom numbers and not with trapped atoms where you know for sure there's exactly one atom in the cavity. OK. So I don't show you an experiment, but let me just state that this sort of single photon Rabi flopping has been observed. You start with the cavity in the vacuum field, and you sort of see this oscillation to the ground state with one photon. But what I want to discuss now is the situation that we are not starting with an empty cavity. We are starting with a coherent field. You can also start with a thermal field so there are different experiments you can do. What would we expect now? So now the initial photon state is not the vacuum state, but the thermal state. If you have a microwave cavity and you heat it up a little bit, you have to cool it down to below 1 Kelvin. People use either helium-free chrio stats or dilution refrigerators, but if you warm it up a little bit, you have a few microwave photons in the cavity. Or, that's even more controlled, you can make the cavity ice cold, but then you inject a few photons from your synthesizer into it-- from your microwave generator-- and then you have a weak coherent field. But a thermal state or a coherent state. So what we then have is, OK, we would expect now a Rabi oscillation; however, the frequency for the Rabi flopping is now proportional to n plus 1. And we have our photon field in the superposition of flux states. So the fact that we have a superposition state implies now that the Rabi oscillations have a different oscillation frequency for the different [? tablets ?] of states labeled by n. And that leads to a dephasing. So that would mean that if you would look at the probability to be in the excited state-- just think about it. You have a wave function where the atom starts in the excited state, and the photon field is in a superposition. So now you have a two-component wave function which has different parts, and each part has a specific Rabi frequency. So you would have oscillations. Let's say there is a certain probability that the cavity is in the vacuum, and then that means that there is a component which oscillates at the vacuum Rabi oscillation frequency. But if you have a component in your coherent or thermal state which has two photons in it, then you have Rabi oscillations which are faster. And now you have to superimpose them all. And if you all superimpose them, and you find that very soon there is a damping and maybe a little bit of vigor, but you see at damping of the population in the excited state. Q [? 2 ?] dephasing. I'm just hesitating. I think I took this plot out of my notes, but I would expect now the damping should actually lead to a probability to be in the excited state of 1/2. So let me just try to correct that. So there is a little bit of oscillation, but then there is sort of a damping. And eventually, if you have only a small number of photon states, then there will be a time where you have sort of at least a partial commensurability. You have maybe five frequencies. You know, square root 5, square root 4, square root 3, square root 2. But then there is sort of a time where all these different frequencies have done an integer number of oscillations each, and then you get what is called a revival. And if you go to a large photon number, you have square root 100, square root 99, square root 88, the revival will happen at a later and later time and eventually at infinite times if you use a microscopic field. But for small coherent states, or thermal states, which only involve a few photons, you will get a revival phenomenon. And this has indeed been observed. This was actually the PhD thesis of Gerhard Rempe, and it shows the probability in the excited state. They had previously observed the Rabi oscillations at early times, but now the experiment had to be adjusted, I think by using slower atoms, to observe the longer time. And here, well, 1987 for the first time revivals have been seen. Let me dwell on that, or first are there any questions about what happens now? Atoms in the cavity to Rabi oscillations? And if the photon field is a superposition of only a few states due to this pseudo commensurability, you find times where you have revivals. I just worked out something this morning which I think is nice, because it will highlight how you should think about spontaneous emission. So let me discuss. It doesn't really matter, but I want to give you a specific example that we have a coherent state. A lot of you know what a coherent photon state is. For those who don't, it doesn't really matter for what I want to explain, and recover that in [? 8.4.22. ?] But if you have a laser or if you have a microwave generator, what comes out is a field which has a normalized amplitude of alpha, but your field is in a superposition state or [? flux ?] states. With these prefactors, I just wanted to give you an example. What I really just mean is that we have a coherent superposition of number states. We have prepared that. So now we have one atom in the excited state, it enters the cavity which has been prepared with the short pulse for a laser or microwave synthesizer in these state alpha. And now we want to discuss-- so this is at t equals 0-- and now I want to discuss what happens as a function of time. Well, we know that if you have one tablet, n, we have Rabi oscillations between the atoms in the excited state, and we have n photons. Or it has emitted the photon, and then we have n plus 1 photon and the cavity. But now, we have a superposition state, and we have amplitudes an. So I mean, that's what we get. And this includes everything. It includes everything a two-level atom does in a single mode of a cavity. And this is spontaneous emission, stimulated emission, and reabsorption. But I want to use that now to discuss with you the misconceptions about spontaneous emission. Colin? AUDIENCE: We're talking about just spontaneous emissions into the cavity? PROFESSOR: OK. I've singled out a single mode. But what happens is-- and you're just two minutes, 30 seconds, ahead of me-- that we had discussed vacuum Rabi oscillations or Rabi oscillations when we have n photons in the cavity. This was our two-level system, our Hamiltonian, and all we get is Rabi oscillations with the Rabi frequency omega n. And now we have to sort of do averaging. I'm now discussing that we have a coherent superposition of number states. Let's say, a pulse of coherent radiation, a coherent state, and this is what we get. You can now, if you want, put in a [? zillion ?] of other modes, have another sum over all the other modes you want. So I'm just doing the first step in discussing with you what will happen, but adding more and more modes will actually not change the structure of the answer and will be, of course, quantitatively a mess but conceptually not more complicated. So I want you to really look at that and realize where is the spontaneity of spontaneous emission. Where do you see any form of randomness associated with spontaneous emission in this expression? I don't see it. This is a wave function, and this time evolution is unitary. Everything is deterministic, and depending now how we choose our coefficient, there is even a revival. It's not dissipative that a photon is spontaneously emitted, and it's done. We saw in the single photon Rabi oscillation it can be reabsorbed, we saw in a slightly more complicated situation that there are at least partial revivals, and it now depends how long we wait whether revivals will take place or whether they will be complete revivals or partial revivals. But we don't need a revival in a coherent evolution, the coherent evolution can just go to a complicated wave function and it's still a single coherent wave function fully deterministically obtained form the Hamilton operator. Sometimes it pops into our eyes through a reversible oscillation or through revival, but we don't need that. So let me write it down but then explain you something. So it's unitary. There is no spontaneity at all. However, eventually we want to retrieve the classical limit. So if we would go to this situation that the average photon number is much smaller than 1, then the fluctuation in the photon field around the mean number are very small. For the coherent state the fluctuations are square root n. And then, we retrieve the limit of semiclassical Rabi flopping with the Rabi frequency omega r, which is-- I'm not consistent here with lower and uppercase [INAUDIBLE]. So it's uppercase or lowercase omega n, and this is square root n times the single photon Rabi frequency. And, of course, for a large number of photons, we can always make the approximation that we do not have to distinguish between n and n plus 1. So this is the ultimate limit if we would work in the limit of large photon numbers. So the way how you should look at it is the following. This system undergoes a time evolution to a state which is rather complicated. But if you make the number n large, this becomes approximately a state where you have simply-- you know what the rate of the semiclassical limit? In the semiclassical limit, we have a constant laser beam with constant electric field amplitude, e, and then we have driven Rabi oscillations between ground and excited state. So therefore, I don't want to show you mathematically, but in the limit of large n, you can approximate this complicated entangled wave function by the product of Rabi oscillations between ground and excited state times a coherent photon field. And the correction between what I just said and this complicated wave function is like 1 over n, because it's sort of a 1 over n approximation where we have neglected terms which the relative importance of them is 1 over n. So therefore, there are people who will say and who will tell you when we have an interaction of an atom with a coherent state, and let's just think in the number of n being large, that n times out of n plus 1, we have a coherent state. The atom does Rabi oscillation and what it does is it just emits photon into the coherent field and takes it back, like in semiclassical physics. But in one case out of n cases, or the rate 1 over n of the rate of the wave function is sort of fuzzy. It's not a coherent state; it's something much more complicated. And if you do not keep track of this complicated nature of the wave function and just do some simple measurement by, let's say, just measuring the phase of the electromagnetic wave by projecting onto a coherent state, then you would find that with the probability of n the system was just staying in a coherent state. And with a probability which is one part out of n, something else has happened, and your detector cannot capture the entanglement of that state. And this last part is what some people associate with spontaneous emission. I don't know. That's my view where the spontaneity in this process is. It's not a spontaneity in the time evolution. It's more a spontaneity if you do not care to detect this complexity, but map it back to a coherent state. And then with a precision which is 1 over n, you retrieve the semiclassical limit, but the difference between the semiclassical limit and the entangled wave function, this is what some people say is spontaneous because it's not captured by a single picture. I'm actually expecting some people to disagree with me, but this is sort of my view, what I'm sort of learning from the simple examples I've given to you. Since Ike is an expert on it, maybe, Ike, can I ask you the question is there actually a simple way to show that if I go to a large n limit that you can sort of really show that n parts out of n plus 1 is really described fully by the semiclassical limit and there is only a 1 over n fraction where we have to look at the more complicated wave function? AUDIENCE: I don't think there's a simple way to do it, but one can look at the equivalence of a [? and a factor ?] state, and they're only different by one photon number. PROFESSOR: It's sort of clear. I mean, everything is if you approximate n by n plus 1. If you don't care about the small difference, everything falls into place and is simple. But I was just wondering if one could show sort of in a more direct or more intuitive or maybe more quantitatively what is really the extra part beyond stimulated emission absorption into the coherent state. So what sort of really the nature of what people call the spontaneously emitted photon? AUDIENCE: I think that I don't the question, because I still argue that it's purely [? unitary ?] evolution even-- PROFESSOR: OK. AUDIENCE: For that system, and therefore, it's purely [INAUDIBLE] and nothing spontaneous is happening at all. PROFESSOR: OK. All right. Good. OK. Fine. What is next? I think this finishes our discussion on vacuum Rabi oscillations and revivals. I have now two topics in light atom interaction which you may not find in many textbooks, but it's my experience that they're really relevant. One is very conceptual. It's about the rotating wave approximation. And the other one is just the opposite, very technical. It's not really a new concept, but this is about saturation intensities and cross-section of an atom for absorption. The last things, cross-section for absorption and saturation intensity, that's what you need when you talk to atoms in the laboratory. These are the quantities in which we think intuitively about light atom interaction. So it's not involving any concept. I want to spend 20 minutes in introducing for you saturation, saturation parameter, cross-section, what's different between monochromatic light and broadband light. But before I do that, I have a few minutes on the rotating wave approximation. So let's call it rotating wave approximation revisited. Again, rotating wave approximation. And what I want to discuss can be discussed in the fully quantized picture, but also in the semiclassical picture. In the fully quantized picture, just a reminder, what we discussed earlier was that when we have the atomic raising and lowering operator and the photonic raising and blowing operator, we got four terms. And two of the terms are co-rotating, two are counter-rotating. But I can get exactly the same number of four terms in this semiclassical picture, and I want you to see both. But in the quantized picture, it's actually easier, because when you see a and [? a dega, ?] you know immediately one is absorption one is emission. So therefore let me explain to you what I want to tell you about the rotating wave approximation using the semiclassical picture, because then you immediately know how to apply it to the quantized picture. So what I want to bring in the here in addition to what we have discussed about light atom interaction, we had sort of a dipole Hamiltonian, is the fact that we have circular polarized light, left-handed and right-handed light, and I want to sort of use that and combine it with angular momentum selection roles, which as you remember we discussed after our discussion of dipole, quadrupole, and magnetic dipole positions. So I put now all those parts together and revisit the rotating wave approximation. So what I hope for is it tells you a little bit how selection roles, angular momentum, circular polarization, and semiclassical field which rotate in one direction how they are all connected. So I have to set up the situation by saying that we use as a quantization axis the direction k, which is either the direction of the propagation of the light beam or, in general, it's orthogonal to the polarization of the electric and magnetic field. And I can talk about an electric field driving an electric dipole transition. I can talk about a magnetic field driving a magnetic transition. It doesn't really matter. I will use [? Bsc ?] amplitude, but you can also immediately think electric dipole, and this field is linearly polarized. But I want to immediately decompose this field into right-handed and left-handed field. Or a linearly polarized field can be regarded as a superposition of a field which circulates this way plus one which circulates the other way. And ultimately, the message we will see is that if you have linearly polarized light, we always get counter-rotating term, we always have a [INAUDIBLE] shift and such. But if you use rotating fields or circularly polarized light, selection roles may actually lead to the result that there is no counter-rotating term at all. So this is eventually what I'm aiming for, and this will be the final point of the discussion. So the field which rotates in the right-handed direction where the rotating field is a superposition of x and y, or i and j. And one the rotates has a cosine omega t and one has sine omega t. I don't need to write down the left-handed part, because there's just a minus sign. Or this will actually become very handy. I do all the discussion for the right-handed part, but I can always obtain the expression for the left-handed part by replacing omega by minus omega. Which will mean that some emission process by the right-handed part will be an absorption process by the left-hand part. Be we'll see. You can change angular momentum by plus 1 by absorbing a right-handed photon or-- you'll see. We'll get there. So anyway, those signs will become important. Let me now take the above expression for the right-handed part and replace cosine omega t and sine omega t by e to the i omega t. So this was the i component, this was the j component. We divide by 2. Just to avoid confusion, I want to emphasize I've started with a real field. So I'm not using, as you often do in e and m complex field and the real fields are the real part, I have not started out by adding, you know, imaginary parts to the field. I've started out with a linearly polarized field in the x direction cosine omega t, and I've decomposed it into two real fields. One is right-handed, one is left-handed. Complex numbers only come because I want to use a complex exponential to replace cosine omega t and sine omega t. We are almost done with the decomposition of the field. I just wanted to-- we have now four terms, and I want to [? recoup ?] them. i minus imaginary unit j. i plus imaginary unit j. This is e to the i omega t. And this is e to the minus i omega t. So what have we done? Well, we've just started with linearly repolarized light, and I've rewritten the expression twice, and now we are looking only at one of the circular components, and in the end what we have is four terms. Well, that's also what we had in the fully quantized Hamiltonian, and we now want to identify what those four terms. Two will be co-rotating, two will be counter-rotating, but it's very helpful to analyze those terms. But there are two things we have to look at now. One is we have an e to the i omega t. And, well, it's probably a sign convention, but trust me, if you put that in the Schrodinger equation, it mean that you increase the energy of the atom if you drive it with e to the i omega t. You take it from a ground state to an excited state, which differs in frequency by omega, and therefore this means you have increased the energy of the system, and this corresponds to absorption. Whereas this one here means we take an atom from an excited state to a ground state, and this is the situation of stimulated emission. Remember, in selection roles we take our field and we multiply it with a dipole moment, electric, magnetic dipole moment whatever. But now we want to use also the spherical tensor decomposition of those dipole moments. It's a complicated word, but what it means is those terms are dotted with the dipole moment, and if you do it now component-by-component, we retrieve selection roles because this peaks out the-- let me just write it down-- the x plus y tensor component of the matrix element. And this corresponds to delta m equals plus 1. We change the angular momentum by one unit, and of course this term is then delta m equals minus 1. So we have done the work. What I want to do now is just map those terms into and energy level diagram. I like sort of pictorial representations, and each term becomes now a graphical [INAUDIBLE]. So let us assume we have a system, hydrogen is to p state. But let's say generally we go from a j equals 0 to j equals 1 state, which has three components. Now, I have set it up in such a way that-- oops, we need a little bit extra space. I've set it up in such a way that the states here, this is m equals 0, this is m equals plus 1, and this is m equals minus 1. So therefore-- let me just use color coding now-- this one here is delta m equals plus 1 so this one always moves to the right. It changes angular momentum by 1 so it can always move to the right, whereas the other one, delta m equals minus 1, moves to the bank. Absorption is e to the i omega t, always moves up. And stimulated emission moves down. So with that what happens is this term here transfers one unit of angular momentum and energy. So that would mean this term goes up here. It could go up here, if [? there were a ?] state. The other term-- let me use a green color-- is driving the process in the opposite direction. But now we have to also consider that you can go down here, and you can go down to a virtual state. A virtual state is just something which has the same wave function as a state, it just has an e to the i omega t, which is not-- it's a driven system. You drive it. You [INAUDIBLE] a state. You [INAUDIBLE] a state at the drive frequency, and it just means, in this case, this state has an oscillation e to the i omega t, which is very, very different from what a state which is populated would have, and this is what we call a virtual state. So in other words, what is possible is we have our three states, plus minus 1 and 0, but this is the spatial wave function including angular momentum. But we can now drive it by plus omega and minus omega, and therefore we can have it as virtual states pretty much at any energy we want. But this process here is not possible, because this would require to go to a state which has m equals 2, which does not exist. So now what I've shown here is if we would stock in the m equals 0 state, I've shown you the four terms, two are co-rotating and two are counter-rotating. If you neglect this virtual state which has a detuning of about 2 omega, or 2 resonance frequency of the atom, this is the rotating wave approximation. One term is responsible for absorption; the other term is responsible for a stimulated emission. But if I don't make the rotating wave approximation, I have those two extra terms. So this is only the right-handed light, and I want to sort of play a little bit with this concept. If I would take the left-handed light, I would add sort of four more arrows. Two more here and two more here. But let's just keep the situation as simple as possible. But I really sort of like that you write down right-handed, left-handed side, decompose it into its components, and each component is now in this diagram connected to an arrow where one direction is angular momentum, the other one is energy. So let me now talk about other energy diagrams. And this will lead to the answer. Well, can we create a situation where we have only two terms, which would be the simplest two-level system, can be directly realized without any rotating wave approximation a two-level system? So if we had two levels, which have only m equals 0 and m equals 1. So this would be the situation I just discussed with those two levels. So the only way how I can fit in this arrow is this one, and the diagonally downward arrow is that. So in this case, rotating wave approximation is not an approximation, it is exact. But some purists will actually say, hey, you can never realize that when you have an n equals plus 1 state. Then you always have an m equals minus 1 state. And then you have a virtual state down there, and then you get two more terms, which are the counter-rotating terms which I just showed above. So whole I would say if you have a neutron star which makes an infinitely high magnetic field, you can have a huge [INAUDIBLE] splitting between m equals plus 1 and m equals minus 1 and completely move one of the angular momentum states out of the picture. But, of course, in the rotating wave approximation we are neglecting off resonant terms at 2 omega, omega being an electronic excitation energy, so I'm really talking about Zeeman shifts here to eliminate the other state which may be comparable to electronic energies. So in principle, I can say this is my Hilbert space, and in this Hilbert space no rotating wave approximations is needed. But it's maybe an artificial Hilbert space. When I had a discussion with other people, we came up with the possibility of some forbidden transition. If you go from a doublet s to a doublet s state so all you have is a spin system which has 1/2 angular momentum plus 1/2 minus 1/2. And then you realize that the only way how you can fit in the orange arrow is in this way, and the green arrow in this way. So here you would have a situation where the rotating wave approximation is exact. But, of course, it's not an electric dipole transition; it's some sort of weaker conversation, which may be forbidden. I need two more minutes. I have discussed the case where we have quantized along a direction, I called it the k direction, and the polarization of the electromagnetic field was [? i and j ?] was perpendicular to it. So let me now discuss a case where we quantize along the polarization of the electromagnetic field, and you remember from our discussion on selection roles that this is pi light. So in this case, our magnetic or electric field is polarized along the i direction, and the real cosine omega t gets decomposed into e to the plus e to the minus i omega t. And we know already one term is absorption one is emission. And now, if I take my j equals 0 to j equals 1 system, pi light has a selection role of delta m equals 0. So now I have an arrow, which I want to be orange, which goes up. And a green arrow-- this is a great program. The only thing is you have to be very careful when you change color and press carefully. That's why sometimes the colors are not doing what I want. But here is green. But now, of course, with linearly polarized light we can always go down to a virtual state. We have now four terms. Two are rotating, two are counter-rotating. So therefore the quick conclusion of the last ten minutes is that there is the possibility that counter-rotating terms can be 0 for sigma plus sigma minus light due to angular momentum selection roles. But what we have also learned is if you have the m plus 1 state in there's an m minus state, if you have circularly polarized light and we drive a transition between two m states, the counter-rotating term does not come from the same set of two states, m equals 1. It involves m equals minus 1. So it's the other state which is maybe degenerate or only slightly [? split ?] by a magnetic field which is responsible for the counter-rotating terms. Anyway we have talked so much about rotating wave approximation and those terms, I just wanted to show you how it is modified if you use degeneracy p states and angular momentum. Any question? OK.
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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: We're going to continue our discussion of correlation functions, and in particular, g2 of 0. If you think I'm focusing a lot on one quantity, g2 of 0, that's correct. But I have to tell you, I have been involved in so many discussions with people who looked at g2 of 0 with both Einstein condensates when Bose-Einstein condensates in [INAUDIBLE] were discovered here at MIT by [INAUDIBLE] and collaborators. There was some mystery about, in essence, the g2 of 0 function. So a lot of physics actually goes into it. And I've learned a lot from my own research and through discussions. And in this unit, I want to summarize it. So the confusion sometimes comes because people use classical pictures, quantum pictures, and they may not see the relation. So let's continue the discussion. I want to go through four different views of the same physics that you will recognize that they're interrelated. The first one, just to repeat that, is if you have light which is Gaussian intensity fluctuations, light from random sources. I think if you would look at the twinkle of starlight, if you look at the light from a light bulb, because of the central limit theorem, you will find that the intensity that the electric field has a Gaussian distribution, and the intensity in exponential distribution. And if you then ask, what is g2 of 0, you'll find it's 2. Because for an exponential distribution, the average of the intensity squared is two times the intensity, the average of the-- what did I say. This average of the squared intensity is two times the average of the intensity squared. Before I forget it, one student asked me after class, I'm discussing g2 of 0, but what about g2 of tau? Well, my understanding is that almost all of the physics is in g2 of 0. And there is a simple number, 1 or 2, in those special cases. What happens is, if you have completely independent sources, if you completely independent statistics, then the probability of finding a second particle, a second photon, is independent of the first. So it means g2 of tau would be unity. This is the case for an uncoordinated system. But if you have g2 of 0 which is higher, this means correlation. This means some form of coherence. And usually, the correlated value simply decays to the uncorrelated value with a characteristic coherence time. And this characteristic coherence time can be inhomogeneous, technical. It can be limited by the Fourier transform of a light pulse, the monochromaticity, the bend which delta omega is 1 over tau. So it has all of the usual suspects for what limits the coherence time. So therefore in general, we're not learning something fundamental in looking at g2 of tau. We're just learning the usual spiel about correlation times and coherence times. The second angle I gave you with the brief interference is something which is deeply insightful. It shows you the following. If you have a plane wave, the intensity is constant. If you have two plane waves which interfere, you create density fluctuations, because the interference pattern can be constructive and can be destructive. So therefore, if the intensity fluctuates, then i square average is no longer i average square. And what happens is because of the nature of interference, because the average of cosine square kx is-- well, there are factors of 1/2 and 2 which just come out of cosine and sine square function. You get exactly g2 of 2. But that should already tell you something, which will be important for both quantum particles, and light, and the classic limit. The moment if you have light or particles in one single mode, you can't have any form of interference. Only two modes can interfere. And you will always get a g2 function of 1. But we'll come to that in a second. Any questions at that point? Yes. AUDIENCE: I thought that when we talked about the g2 function several weeks ago, you said that if we had a single mode thermal light, it would have the g2 function of 2. And that's therefore, all we were using for lasers when we used them in experiments is the fact that it's single mode, not the coherence in the g2 function. PROFESSOR: You have a good-- that's a good point, Cory. Let's just try to connect the two pictures. Slowly, slowly. What did we discuss? This question is great for me to repeat something. When we looked at black body radiation, I wrote down the partition function. And I said I'm only considering a single mode. And then, we have of course in thermal equilibrium a Boltzmann type, or Bose-Einstein type, distribution of finding certain numbers, number of photons populating the single mode. And that resulted in a g2 function of 2. AUDIENCE: It's not really single mode though, right? Because you're summing over many frequencies. PROFESSOR: No, it was single frequency. I have to think about it. That's a really good question. Right now, I would quickly say, in thermal equilibrium-- I'll give you partial answer, and when I think about it, I'll give you full answer on Friday at the beginning of the next class-- I think would be calculated where the thermal fluctuations in the intensity. And the intensity fluctuations, and those intensity fluctuations. Those intensity fluctuations result in a g2 function of 2. That's correct. Maybe what I want you to say here is, if you have two modes which have all maybe single occupation or constant occupation, and they interfere, then they create a spatial interference pattern if the phase between the two modes is random, which gives rise to additional fluctuation. So most likely, the two fluctuations are independent. One is the fluctuations because of the interference of two modes. But when I say the single mode has a g2 function of 1, I meant actually a single mode with constant amplitude. I'll double check and give you a more complete answer on Friday. Other questions? OK. The third view is classical versus quantum statistics. And I think this really shows that we need at least some quantumness to find correlations. Let's assume we have n particles. And we have N possible states. And if it's a classical system, let's assume we want to find one particle in a certain state. We have a big box and want to-- we have a big box, have a small sub-volume, and say what is the probability of finding one particle here? So this probability to find one particle is P1 is small n over big N. The probability to find two particles. If you have non-interacting classical particles, well, it's simply P1 squared. Independent classical particles, they don't care what they are doing. You grab into a sub-volume, which is maybe phase space cell. And therefore, your quantum state, you have probability P of finding one. But since each particle moves around independently, the probability to find two particles is just P squared. It's a little bit like if you toss two coins, one with your left and one with your right, and you ask what is the probability to find a head or tail, and you simply multiply the probabilities. But now, following reasonings which Bose and Einstein introduced, you want to use counting statistics, which takes into account the indistinguishability of particles. So if we go from the classical distribution to the distribution of indistinguishable particles, then the classical probability P 1 is reduced by n factorial, because we are not-- let me put it this way. The number-- I don't want to go through a mathematical argument. If you grab one particle, it can be the probability to find one particle is represented by a microconfiguration classically. And each configuration when you permute particles counts as independent. But if you have quantum indistinguishability, you're not counting permutations as an independent configuration. And therefore, you have a reduction by n factorial. If you look at the probability of finding two particles, the reduction is-- this is just the counting statistics n factorial by 2. So therefore, you will find that classically, that quantum mechanically for bosons, is two times. The probability to find two bosons in one quantum state is two times the probability squared to find one boson. So this is just counting statistics. This can be applied to conditions in a Bose-Einstein in a Bose gas. When you have inelastic collisions, spin relaxation, dipolar collisions, two body collisions, two-body collisions have a rate gamma 2, which is proportional to the probability of finding two particles at the same time at the same location. And that brings in the g2 function. Whereas three-body collisions, gamma 3, reflects the third order correlation function g3, which is defined in an analogous way. So if you now compare at the same density the two-body rate coefficient, like the same density means we are looking at the rate of two-body collisions. And we compare between two-body collision in a thermal cloud, which is a g2 factor of two, a thermal cloud of bosons. Whereas a Bose-Einstein condensate is a constant occupation of particles in one mode and has therefore a g2 factor of 1, exactly as [INAUDIBLE] of the Bose-Einstein condensate is for particles for meta waves, but the laser is for light. Whereas, if you look at three-body collisions, those who do experiments with Bose-Einstein condensates will know that usually the lifetime of Bose-Einstein condensates is limited by three-body collisions. Well, I didn't derive it, but actually yes, we have the Gaussian statistics for randomness. For Gaussian statistics, the n-body collision function has a factor of n factorial. So therefore, three-body collisions scale with 3 factorial, and two-body collisions with 2 factorial. So in other words, you have a thermal cloud, you have a Bose-Einstein condensate at the same density. But what matters for two-body collision is not the average density. It is the average of n square. And the fluctuations in the thermal cloud because of the g2 function are two times enhanced compared to Bose-Einstein condensate. And therefore, you find that at the same density, you have more loss. You have a higher rate of two and three-body collisions. Some of this was not-- became better understood soon after Bose-Einstein condensates were realized. The Boulder group studied three-body collisions. And for two-body collisions, myself in the post-doc clarified the situation that even the mean field energy, which everybody had measured before, requires two particles to interact with short range interaction is therefore proportional to g2. And therefore, when people had determined the mean field energy of a condensate without really knowing they had already determined the g2 function of Bose-Einstein condensates. OK, so this is pretty much the counting statistic which you do in undergraduate class when you derive the three different statistics-- Boltzmann statistics, Bose-Einstein statistics, Fermi statistics. And it gives you the result of that is also for bosons, you have the factor of 2. For fermions, of course, you get 0. The average between bosons 2 and fermions 0 is 1. And this is the classical statistics. Any questions? Then let me finish this discussion with a quantum mechanical description. If you describe the detection or collision of two particles in two modes. So you have mode. A mode operator, annihilation operator a1 you and a2. But if you are asking what happens if I measure two particles, you annihilate two particles with an operator a1 and a2. But then, because of the indistinguishability of particles, you have to consider an exchange term, which is a2a1. So whenever you detect two particles in two different modes, your signal is proportional to something related to a1a2 plus a2a1. This extra exchange term gives you an extra factor of 2 for bosons, which is exactly what appears in the g2 function. And, of course, if you have an exchange term for fermions, you get 0. And this, of course, leads to the antibunching, or the g2 value of 0 for fermions. However, and this is obvious, if you have a single mode, then the only operator to detect two particles is a1 times a1. And this has no exchange term. And this is the situation of the laser and the Bose-Einstein condensate. And that means they have a g2 function of 1. So Cory, this is another argument why single mode occupation has a g2 function of 1. But I have to reconcile it with your question. Right now, I think there may be difference between a canonical ensemble and grand canonical ensemble. In a canonical ensemble, we have [INAUDIBLE] the total particle number is fixed. But in a grand canonical it fluctuates. And maybe this leads to additional fluctuations for the case of a single mode chaotic light. But I will have to think about it, your question. But here is the same argument why if you occupy only a single mode-- if you look, for instance, at a Bose-Einstein condensate, you do not have the second exchange term. And the missing exchange term just propagates through the equations and gives you the g2 factor, the g2 function of 1. OK so let me sort of wrap it up. I've given you different angles at the g2 function. Some are simply classical intensity distribution. Some is simply interference. Another aspect is Poissonian counting statistics. My understanding and my interpretation is they look as different as they can possibly look, but they reflect the same physics. Because interference of to light only happens because light consists of photons. And photons are bosons. So the classical interference and the quantum mechanical counting of bosons lead not for random reasons, but lead for profound reasons to the same result. And chaotic light, which seems to be determined by just random fluctuation, well, the random fluctuations if you have single mode light, a thermal distribution of single mode photons, comes from random phases. And these random phases lead to random interference. And so we are back to interference. Finally, let me give you my view on the measurements of g2. When you look through the literature, you find the famous Henry [INAUDIBLE] experiment. In one of your homework assignments, you looked at a seminal experiment where people dropped atoms out of cold atom clouds and measured the g2 function for cold atoms. Sometimes it's confusing when you directly compare the two experiments. But let me try to give you a common description, or the common denominator, between all these experiments. You can say that all experiments to measure the two particle correlation function is about comparing the probability of finding two particles with a probability of finding one particle. And well, you restrict your measurement, your detection of particles, to either one quantum state, one mode of the electromagnetic field, one coherence length, or coherence volume, of light. Or if you use the semi-classical argument for the description of particles, you take one phase space cell. This is one phase space cell, one mode, one coherence volume, one quantum state, is sort of the different definitions of what a quantum state is, or what a wave packet is with Heisenberg limited uncertainty. You cannot define a particle in phase space to within better than the phase space volume of a quantum state, which is h square. Or is it h bar? I don't think there is a bar-- h square. And so this is how you can relate wave packets to quantum states to phase space volume. It's pretty much mathematically I could say in different basis sets, in different basis sets, it is one quantum state. But you can use wave packets. You can use time dependent description. But what I said here-- one quantum state, one mode, one coherence volume-- this is as well as you can define a particle or photon to be by fundamental reasons by Heisenberg's uncertainty relation. AUDIENCE: What do you mean by one coherence volume? PROFESSOR: What? AUDIENCE: What do we mean by one coherence volume? PROFESSOR: Well, if you have a laser beam which propagates, and it is TEM00, you would say the transverse coherence volume is the size of the laser beam but then the longitudinal coherence volume is given by the coherence length. Just envision you have a laser beam here. And the coherence volume where all of the photons are coherent is the area of the laser beam times the coherence length. And whenever you find a photon in this kind of volume, then it is in a Heisenberg Uncertainty limited or Fourier-limited state. It is in, if it's a wave packet and pulse laser, it's in a time dependent quantum state. But it is as coherent. It is fully coherent. So you have to figure out for your system what is this minimum-- what is the coherence volume, what defines the fundamental phase space cell of your system? And now you do the following. You are asking, what is the probability to find two particles? And what is the probability to find one particle? So if p is the probability to find one particle, then you have three options. p square, 2 p square or 0 to find two particles. One is the classical case of distinguishable particle. This is the case of bosons, bosonic atoms or photons. Collins Therefore includes the Hanbury Brown-Twiss experiment. And 0 is fermions. And the question is now different experiments, how do you define, with spatial resolution or temporal resolution, the coherence volume? And that can involve transverse coordination, temporal resolution, and whatever you can use to define the phase space cell or the coherence links, the coherence volume of a laser beam. Let me give you one example, which I think illustrates it. And this is the example of an atom cloud. If you have an atom cloud, all the particles which are in a volume of a thermal de Broglie wavelengths are coherent. The momentum uncertainty of particles in a thermal cloud is 1 over the thermal-- is the thermal momentum. And according to Heisenberg, the position uncertainty related with this momentum spread is just the thermal de Broglie wavelengths. So you can say that all the atoms in a cubic de Broglie wavelengths are coherent. All the atoms are in one semi-classical quantum state. So therefore, this is sort of atoms in one single mode. So the Hanbury Brown-Twiss experiment with atoms, or the measurement of the g2 function, could be defined as follows. You have an atom cloud. And if you had an electron microscope or some high resolution device, you grab into your cloud and ask, what is the probability for one particle. What is the probability for two particles? And you will find that p2 is 2 times p1 squared. If your volume is too big, you lose the factor of 2, because you average over uncorrelated volumes. Now, in your homework, you were looking at the question, how can I really grab into a cloud and just pick out atoms out of one phase space cell of effective size, lambda de Broglie cubed? And the way how it was done is that you take an atom cloud and you drop it and expand it. When the cloud expands, there is a mapping from momentum space into position space. Then you use some form of pinhole, which provides transverse coordination. You use a detection laser, which gives you temporal resolution. And well, this was part of your homework, but I just wanted to give you the bird's view on it. By controlling the transverse correlation and the temporal resolution of the detection, you create a situation that what you count only atoms which originated from one phase space cell in your cloud. So in other words, let's say experiment where without electron microscope, without submicron spatial resolution, you can literally grab into a cloud, capture a volume of the thermal de Broglie wavelengths, open your hand, and figure out how is the probability for two particles related to the probability of finding one particle. Any questions? All right. OK, good. So it's time for new chapter. So in this chapter, we want to look at interactions between light and atoms using Feynman diagrams. And I know this part of the course is a little bit formal. We are using exact solutions for the time evolution operator in quantum physics. But I'm not doing it to teach you sophisticated formalism. Well, it's interesting to learn it anyway. But it is this picture of really writing down the exact solution of the time evolution operator, which helped me to understand much better what virtual states are, what certain virtual photons are. So there are things we talk about it all the time. And the question is, how do you define a virtual state? What is the virtual emission of a photon? And the only way how I can really explain it to you is by showing you the equation and say look, the virtual photon is just a term in this equation. So that's the goal of this chapter. I don't want to overemphasize the mathematical rigor, but I want to really show you what it means to have virtual photons and what exactly virtual states are. So let me first motivate that by reminding you of two diagrams. In physics, we always like to draw something-- a few lines, a few doodles. And when we have a two level system, we have the two processes which are emission and absorption. Well, so this is easy as long as the light is in resonance. But if the light is not in resonance, we may ask what about this process? And since somehow the weekly line of the photon ends here, I may even put in a dashed line and say, this is a virtual state. What does it mean? Or we can even ask, is it possible-- I'm just playing with straight lines which are quantum state and wiggly lines-- is it possible that this process happens. That would actually mean that an atom in the ground state, just out of the blue, emits a photon. Can that happen? And I can again draw a dashed line and say, here's a virtual state. Or, we can say we start in the excited state. The atom is in the excited state. But can the atom-- its a two level system, so we don't include higher states-- but can a two level atom in the excited state absorb another photon? So let me just change the color of these photons to red, because I want to-- and this would remain black. All right. Good. So question is, are those processes possible? Yes or no? AUDIENCE: Yes. PROFESSOR: Yes, they are. And they have experimental, observable consequences. However, they look funny because something seems to be strange with energy conservation. And what we will see is that in the end, at the end of the day, that means if you let the system evolve for a long time, after a long time, energy has to be conserved. And therefore, we will always need a second photon to conserve energy. So those weirder processes, where photons are emitted by ground states out of the blue or excited states absorb other photons without-- also, there is no higher lying state. That looks a little bit strange. So what you will see is-- and I want to show you that mathematically-- if an atom emits a photon out of the ground state, you would say, where does the energy come from. But quantum mechanic allows us to violate energy conservation by an amount delta e for a time which is h bar over delta e. So for short movement, energy can be violated. But then, you need the second photon to reconcile energy conservation. You can violate energy only for short times. And if you would say, what does it really mean to violate energy conservation for short times? Well, I want to show you the equation, which exactly explains what it means. Next question, just to see. If you have an excited state, and you just say it's possible to absorb another atom, you are in the excited state, and you absorb another photon. And at least you said, yes this is possible. My question is, when you are in the excited state and absorb a photon, in which quantum state is the atom after absorbing a photon? AUDIENCE: Ground state. PROFESSOR: Pardon? AUDIENCE: Ground state. PROFESSOR: In the ground state. Yes. There is no other state. And the operator-- and I want to show you that-- when photons are exchanged, always transforms the ground to the excited state, because a dipole operator connects the ground to the excited state. So therefore-- but we'll see that all in the time evolution operator in the formal solution that this dash line is actually the ground state. And while it's not obvious, what is this dash line? AUDIENCE: The excited state. PROFESSOR: It is the excited state. The atom in the ground state emits a photon. Therefore, the atomic system has now lower energy, because a photon has been emitted. But the character of this state is now the excited state. So we will later see that this means that in a perturbation analysis, we are violating energy. We are violating energy, because the real excited state is omega photon plus omega atom, the resonant energy for the atom. The real excited state is omega photon plus omega atom higher. So you will actually find that in a perturbation analysis, this term has an energy denominator, which is of resonant by exactly this separation. And this term, because the real count state is here, the virtual state here which is the count state is here, has actually the same energy denominator as that state. And for this state, of course, this is something you've seen of resonant light scattering. The energy [? denominator ?] is there. So, OK. So I'll explain to you that we involve virtual states. So when we have virtual states, we want to clarify what is their energy. But we discussed that already in the discussion. And what is the wave function, ground or excited state. Yes. So just to repeat it, when we draw those diagrams, those diagrams show the energy of the atom, but taking into account the energy of the photon. In other words, if you have a ground state, and we absorb a photon, the photon has disappeared, and the energy of the atom is now the ground state and the photon energy. And this is a dashed line. Similarly, when an atom emits a photon in the ground state, the photon is emitted. Therefore, the atomic energy is lower than the ground state by the photon energy. And we draw the dashed line here. This is what we mean when we say that is the energy of the virtual state. And this is what we draw in those diagrams. By the way, yesterday we had the CUA seminar. And the speaker was actually talking about trapping atoms with quantum fluctuations. And he conceded explicitly Casimir forces, forces of the vacuum. And diagrammatically, forces of the vacuum come, because the ground state atom emits a virtual photon and reabsorbs. This is actually, I will mention it later, the same diagram is actually also the diagram which leads to the Lamb shift. These are all sort of when you have atoms in the lowest state, and they interact with the electromagnetic field. Well, if they interact with the vacuum, they cannot absorb photon. All we can do is emit a photon, and this leads to this diagram. But just to tell you that you have to be careful, yesterday's speakers actually used this process. But he drew this diagram. He drew it in the opposite way, which I think confused some people in the audience. The correct way is to draw it like this. The virtual state for vacuum fluctuations for the Lamb shift for the Casimir force is below the current state. OK, so we have-- let me just redraw those three diagrams. I want to now introduce the time evolution. So I want to use those. Seems I'm running out of space. So we want to introduce the time axis for-- so we had this diagram. We had this diagram. And we had-- this is a virtual state. OK. So those three diagrams correspond to the following situation. If time evolves from the bottom to the top, we can now draw the first process when atom in the ground state first absorbs and then emits a photon. And let's label the second photon in the purple color. So an atom starts out in the ground state. Then it reaches the time where it interacts the electromagnetic field. There is a photon which propagates from earlier times to this time, t equals t prime. At the time t prime, the photon disappears. And the atom, which was in the ground state, goes to the excited state. It may propagate in the excited state for a while. And then it emits the photon and is back into the ground state. So this is the temporal diagram for this process. For the next situation, we also start with an atom in the ground state. But now we have the situation that a photon is emitted by the ground state. As a result, the atom switches to the excited state. Strong violation of energy, but possible for short times. Then there is a real photon, which is now absorbed, absorbed by an atom in the excited state. And that takes us back down to the ground state. And finally, in the third scenario, we have an atom in the excited state. There is a real photon which is absorbed that switches the atom into the ground state. And then the atom in the ground state emits a photon. And as a result, it is back in the excited state. So let me just make a note that when a ground state atom absorbs a photon, this is the co-rotating term in the quantum description, which a lot of you have seen. The opposite process of emitting a virtual photon-- I shouldn't say virtual photon. It's a photon. What makes it virtual we will see later when we have the more accurate mathematical formulation. So this is a counter-rotating term. And we have used the rule that every photon, which with a real or virtual-- every photon which is interchanged with the atom absorbs or emitted-- changes the atomic state from ground to excited. So an atom goes from the ground to the excited state either by photon absorption or by photon emission. Both is possible. So whenever photon appears or disappears, it changes-- let's say we start in the ground state. We go to the excited state. And this is possible by photon absorption-- this is a co-rotating term-- or emission. What I've just said reflects that when we derived the dipole approximation, that the essential term of the dipole operator is of diagonal. The dipole operator is an operator between calm and excited state. Therefore, it has those two matrix elements. Whereas electric field operator is the equation and annihilation operator a and a dagger. Or which is very elegant if I use for the two level system a spin one half description. And I use sigma plus and sigma minus, raising and lowering operator. Sigma plus takes the ground state to the excited state. Sigma minus takes the exciting state to the ground state. Then in the fundamental atom light interaction, we have those four terms. And now you see that sigma plus takes the atom from the calm to the excited state. But sigma plus appears both with photon absorption and with photon emission. So therefore, the atom can-- this is what the quantum mechanical operator tells us-- go from the ground to the excited state either by photon absorption or by photon emission. Any questions? Yes. AUDIENCE: So if we were to look beyond the dipole approximation, would we see transitions that leave the [INAUDIBLE] state? PROFESSOR: Yes or no. What we need here is a bilinear interaction, which has an raising and lowering operator for the atom. And here it has a raising and lowering atom operator for the photon number. And it has the co-rotating term where you raise here and you lower there. You lower here and you raise there. But it has also terms where you raise the atomic excitation, and you raise the photon number by one, which of course violates energy. But the message I want to give you is this is possible for short times. And this is exactly what the quantum mechanical equations tell us. So the short answer to your question is, as long as we describe the system by the dipole interaction, or what we have even beyond-- I'm not using a perturbation argument. I'm writing down the operator. And this operator can now effect the system to all possible orders. So even the non-perturbative strong coupling limit will always have products of atomic raising operators, lowering operators with photon raising and lowering operators. The little bit longer answer is, some of what I'm telling you may be gauge dependence. If you work in the-- if you use, not the dipole description, but you use the p dot a Hamiltonian, this Hamiltonian has a e square term. The e square term allows the atoms to exchange two photons at the same time. So in other words, in the a square interaction, we may have a vertex point in time where the atom interact with the photon field where two photons are exchanged. And then the rules are different. So in that sense-- but this is quite often when we give a quantum mechanical pictures, which has terms, which nicely shows the time evolution. And you think wow, yeah, this is what happens. One photon at a time, the atom goes ground, excited, excited, ground state. And you have your picture. And it is fully consistent with the exact quantum mechanical result. You may be able to get the same result out of a different picture. That point will actually occur again when we talk, for instance, when we talk about density operators. A density operator is an average of our quantum states. But you can get the same density operator by averaging quantum states in different basis sets. So if you get one picture out of one specific basis set, it gives you the correct intuition, it's a correct description, but it is not the only possible description. [? Collin ?]. AUDIENCE: Maybe the thing, I don't know. AUDIENCE: I think what you're trying to get at is that picture works as long as you can limit yourself to a two level system. Because that's correct on a whole. Or is it like strong coupling? But as long as you can identify the two level system, or I think the dipole versus particle approximations [INAUDIBLE] an atom may have transition probabilities to others, like from an [INAUDIBLE] state. But-- AUDIENCE: Actually, I was just wondering if it was possible to have it like a transition matrix element to [INAUDIBLE] coupling the ground state to the ground state. PROFESSOR: Yes. The e square operator does that, because in the other gauge, which is pa, pa. p is in dipole operator. p is the momentum operator for the atoms. And the p operator will switch the atom from ground to excited state. a, the vector potential, involves a and a daggers. So the p dot a term does pretty much the same as the d dot e term, because it's a bilinear product of an operator which has an effect on the atoms and has an effect on the photons. And it means you always switch the quantum state in your atom and create or annihilate the photon. But the a squared term-- it's just a square, it doesn't have a p in front of it-- will allow an atom in without changing the state to create two photons, emit two photons. And then there's probably a cross term which means to absorb and emit a photon. You will actually, in one of your next homework assignments, you will actually look at the a square term. And I think it's a very educational problem. So you really see that you have two very different descriptions, but the results fully agree. But this is already an expert discussion. Why don't we for now just hang onto the dipole approximation. We have d dot e, and we just learn what is inside the dipole approximation, and what is an exact quantum mechanical description using the dipole approximation. But and then we can come back to the discussion, is everything we describe really real. And while the answer is, what is real in quantum physics is the final result, the intermediate results, you have to say that this is only one possibility to go to the final result. OK, so we want to-- let me just put in one more page. So what we want to do now is we want to do a calculation. It's a perturbative calculation, but we can take it to all orders. So therefore it's exact and general. We want to do a perturbative calculation of transition amplitudes. And what I am following here the discussion in atom photon interaction on those pages. And the idea is the following. We have the Hamiltonian for the atom. And then we have a Hamiltonian which describes the transverse field. And the interaction of the transverse field with the atoms. What we want to use is we want to use the interaction picture, which is often the nicest picture to describe the effect of the interaction between two systems. In the interaction picture, you are transforming from you Schrodinger type e function to wave function psi tilde. And those wave function psi tilde include already the dynamics of the Hamiltonian H naught. So that means that all operators in the normal Schrodinger picture become now operators in the interaction picture by canonical transformation, which involves the dynamics due to the unperturbed Hamiltonian. And so that's just a reminder what the interaction picture is. And we are now interested what happens in the interaction picture to an initial wave function psi of ti. This wave function already includes the time evolution due to H naught. So the only change now comes because we have interactions with the electromagnetic field. So this allows us now to focus only on the relevant interactions, the interactions we want to understand. And the formalism which is used in atom photon interaction focuses on the time evolution operator, which I'm sure all of you have seen. So the time evolution of our system, how the wave function involves from the initial to the final time, is described by the time evolution operator. I will come back to the wave function later, but in order to derive the time evolution of the wave function, it's at least convenient to focus first on the time evolution operator. So in other words, I want to show you a formal solution for the time evolution operator. And once we know the time evolution operator, we apply it to the wave function. And then we are talking again about the wave function. OK, since the derivation involves many equations and they are all printed in the full beauty in atom photon interaction, I decided to use pre-written slides here. Also, most of it is fairly elementary. The equations are complicated, but the concepts behind them are very, very simple. It's actually the beauty of using a kind of a time evolution operator. It's a little bit mathematically formal, but it allows us to express what is really going on in the solution in very, very simple terms. So the goal is to do a perturbative expansion for the time evolution operator. If we have no interaction potential v, the interaction picture, nothing happens in the interaction picture, and therefore, the time evolution operator is a unity matrix. But if you have a coupling term v, there is time evolution. And the time evolution can be treated in lowest or in higher orders in the perturbation. And we want to use now perturbation expansion, where we have correction terms, which is the first order, second order, and nth order correction for time evolution operator. So what we simply want to do is we want to do an iterative solution. We want to find an iterative solution for the time evolution operator. The next two lines is just a reminder. What is the differential equation for the time evolution operator? Well, we are really just talking about Schrodinger's equation. Schrodinger's equation for the wave function turns into differential equation for the time evolution operator. Schrodinger's equation in the interaction picture says that the time derivative of the wave function in the interaction picture has nothing to do H naught. H naught has already been absorbed in the definition of the wave function. So the time evolution in the interaction picture only comes from the interaction term. And now, if you write the wave function at time tf as the time evolved wave function from the initial time to the final time, then you take the derivative here. You actually take a derivative of the time evolution operator. And therefore, in one step, from Schrodinger's equation, you find a differential equation for the time evolution operator. In other words, this is now the operator equation for the time evolution operator. And it's nothing else than a 100% rewrite of the Schrodinger equation. Interrupt me if you have questions. This part should be-- we'll go through that for solely pedagogical reasons. And that means you should really understand it. So this equation can now be formally solved in the following way. Take this expression for the time evolution operator and insert it in this equation, and you find an identity. Of course, you haven't really solved it, because you have the time evolution operator expressed by the time evolution operator. This looks like a circular conclusion, which is just nonsense. But if you inspect it more closely, you observe that the time evolution operator here is expressed by the time evolution operator multiplied with v. And if you now think in an iterative solution that we want to express. We want to expand the time evolution operator in the different orders in the interaction parameter v. Then you find actually that if you are only interested in the first order of the interaction operator, you can use the [INAUDIBLE] order here, because you get one more power of v. So that's how you get the first order. If you want to know the second order of the time evolution operator, you can plug-in the first order on the right hand side. So therefore, you get an iterative solution. The first order solution is one. The [INAUDIBLE] order solution is the unity operator. The first order solution is by just putting the unity operator here. The second order solution is by taking the first order solution, plugging it in in here. And then you get two integrals over the operator v. So in that sense, we have formally solved the time evolution of the system in all orders. So this is what we get out of it. Yes, the equations are getting longer. But the structure is fairly obvious. So you saw that the-- let me just flashback. The first order term had a temporal integral over v. The second order term has two temporal integrals over v. And there's a time order in between tau 1 and tau 2. And the nth order term now involves n temporal integrals where the times are ordered in such a way. Now we want to go back to the Schrodinger equation. So we want to use this already fairly complicated expression for the time evolution operator and apply it to the initial wave function. And we want to calculate matrix elements between the initial wave function and some other phi f, some other wave function in a given basis. So we are also going back. I don't want to go through too many steps. Here, we're also going back to the original Schrodinger picture. So I've taken all the tildes off. But you know the step form the Schrodinger picture to the interaction picture was only taking out the Hamiltonian H naught. And the Hamiltonian H naught for eigenstates phi i and phi f, is simply a time evolution, e to the e ei with a certain time. So all what this transformation to the interaction picture gave us, it eliminated, at least for a short while, all of those phase factors, which are simply the evolution of the eigenfunction in the Schrodinger picture. So let me look at a typical term which we have right now. And you can't expect something simpler than this, because this is the general solution. It's an exact solution to all orders. It's a summation over terms to all orders. And eventually the summation of an infinite number of diagrams is an exact solution. So what we've got here is we have an integral over n times. And the n times are time ordered. This is the formal solution for the time evolution operator. And that means that we integrate overall times, but what it is under the integral sign is a product of the operator v at times tau 1, tau 2, tau 3, tau 4. Since we got back to the Schrodinger picture, what happens is between times tau 1, and tau 2, the particle simply evolves with the phase factor given by the Hamiltonian H naught. So the picture is the following. The time evolution operator in the interaction picture is a product over the operator v tilde at different times. If you simply plug it back into the above equation-- and you have to look at it for a little while to see that everything works out correctly. But what that means is then that if you go to the Schrodinger picture and write down the operator in the Schrodinger in the basis of unperturbed eigenfunction, that you have products of the operator v at times t1, tau 2, tau 3, they're time ordered. And because we have left the interaction picture, you also get the propagation with the phase factors given by the eigenenergies of the unperturbed Hamiltonian H naught. So therefore, you get just an infinite number of terms which have all the following structure, that the particle propagates in its eigenstate. It reaches a vertex where the interaction switches the particle from one state to the next. Then in the next state, which may now be the excited state, we have propagation in the excited state. Then the excited state is, again, exchanging a photon, has an interaction term, and is switched to-- well, for the two level system, it has to go back to the ground state. In a more general situation, it goes to another quantum state. So therefore, what we have exactly derived is that the propagation, or the time evolution, of the wave function is nothing else then many, many of those factors. And each of those factors can be represented by a diagram like this. So let me just write down what this diagram here means mathematically. It means that initially, we have a photon of energy epsilon, of polarization epsilon, wave factor k, and a certain energy. And we have particles in state a. That means that we get from the initial time to the time tau 1. And tau 1 is the time of the first vertex of the first photon exchange, that the wave function evolves. And it evolves with an energy of the unperturbed Hamiltonian, which is the atomic energy. And we have one photon h bar omega. At the time tau 1-- and this is when the diagram on the left hand side has a vertex-- we are now bringing in the interaction operator, which acts on the atom. It takes the atom from state A to state B. And it changes the state of the photon. In this case, a photon with a certain wave factor and polarization simply disappears. So this is now the time tau 1. The next Vertex is reached at time tau 2. And between the time tau 2 and tau 1, the system evolves according to the Hamiltonian H naught. And the energy is now, because we have no photon, is simply the energy of the atom in state B. Then we reach the next vertex. We start with a state B and no photon. The interaction switches, as now I assumed we have more than two levels. Two are state C. And we have now a photon, which may be different, k prime epsilon prime. Since we are now, for this given example, I've selected n equals 2, something which is second order in the perturbation. But that means now that between the time tau 2 and the final tie-in for the time propagation, that's now everything is done. Things propagate. And the unperturbed Hamiltonian H naught gives us now a phase factor, which reflects the energy of the atomic state, and the energy of the photon in mode k prime epsilon prime with energy omega h per omega prime. So in other words, I hope you see that at the end of this-- I mean, this is why people have Feynman diagrams. It's a complicated, involved mathematical formulation with multiple integrations. But at the end of the day, what we have derived is that the most general solutions to all orders in the probation is nothing else than a sum and integral over diagrams like this. So when I depict it here, the second order diagram, of course this has an exact mathematical meaning. And what we have to do is when we solve a quantum mechanical equation, we have to now allow those interactions, which happen at time tau 1 and tau 2, to happen at arbitrary times. So therefore, we have to sum over amplitudes by integrating over times tau 1 and tau 2. So on Friday-- I have to stop now. I think there's a seminar right now. On Friday, I will show you that with this description, we have actually captured everything I explained to you earlier-- those virtual states, the emission of virtual or real photons. So everything that which was maybe qualitative at the beginning of this section has now a precise, mathematical meaning. But that's what we do 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: This definition in which the uncertainty of the permission operator Q in the state psi. It's always important to have a state associated with measuring the uncertainty. Because the uncertainty will be different in different states. So the state should always be there. Sometimes we write it, sometimes we get a little tired of writing it and we don't write it. But it's always implicit. So here it is. From the analogous discussion of random variables, we were led to this definition, in which we would have the expectation value of the square of the operator minus the square of the expectation value. This was always-- well, this is always a positive quantity. Because, as claim 1 goes, it can be rewritten as the expectation value of the square of the difference between the operator and its expectation value. This may seem a little strange. You're subtracting from an operator a number, but we know that numbers can be thought as operators as well. Operator of minus a number acting on a state is well defined. The operator acts on the state, the number multiplies a state. So this is well defined. And claim 1 is proven by direct computation. You certainly indeed prove. You can expand what is inside the expectation value, so it's Q hat squared. And then the double product of this Q hat and this number. Now, the number and Q hat commute, so it is really the double product. If you have A plus B times A plus B, you have AB plus BA, but if they commute it's 2AB, so this is minus 2 Q hat Q. Like that. And then, the last term is the number squared, so it's plus Q squared. And sometimes I don't put the hats as well. And all this is the expectation value of the sum of all these things. The expectation value of a sum of things is the expectation value of the first plus the expectation value of the second, plus the expectation value of the next. So we can go ahead and do this, and this is therefore expectation value of Q squared minus the expectation value of this whole thing. But now the expectation value of a number times an operator, the number can go out. And this is a number, and this is a number. So it's minus 2 expectation value of Q, number went out. And then you're left with expectation value of another Q. And the expectation value of a number is just the number, because then you're left within the world of psi star psi, which is equal to 1. So here is plus Q hat squared. And these two terms, the second and the third, are the same really. They are both equal to expectation value of Q squared. They cancel a little bit, and they give you this. So indeed, this is equal to expectation value of Q squared minus expectation value of Q squared. So claim 1 is true. And claim 1 shows in particular that this number, delta Q squared, in the expectation value of a square of something, is positive. We'll see more clearly in a second when we have claim number 2. And claim number 2 is easily proven. That's another expression for uncertainty. For claim number 2, we will start with the expectation value of Q minus Q squared, like this, which is the integral dx psi star of x and t, Q minus expectation value of Q, Q minus expectation value of Q, on psi. The expectation value of this thing squared is psi star, the operator, and this. And now, think of this as an operator acting on all of that. This is a Hermitian operator. Because Q hat is Hermitian, and expectation value of Q is real. So actually this real number multiplying something can be moved from the wave function to the starred wave function without any cost. So even though you might not think of a real number as a Hermitian operator, it is. And therefore this whole thing is Hermitian. So it can be written as dx. And now you have this whole operator, Q minus Q hat, acting on psi of x and t. And conjugate. Remember, the operator, the Hermitian operator, moves to act on psi, and the whole thing [INAUDIBLE]. And then we have here the other term left over. But now, you see that you have whatever that state is and the state complex conjugated. So that is equal to this integral. This is the integral dx of the norm squared of Q hat minus Q hat psi of x and t squared, which means that thing, that's its complex conjugate. So this completes our verification that these claims are true, and allow us to do the last step on this analysis, which is to show that if you have an eigenstate of Q, if a state psi is an eigenstate of Q, there is no uncertainty. This goes along with our measurement postulate that says an eigenstate of Q, you measure Q and you get the eigenvalue of Q and there's no uncertainty. In particular, we'll do it here I think. If psi is an eigenstate of Q, so you'll have Q psi equal lambda psi, where lambda is the eigenvalue. Now, this is a nice thing. It's stating that the state psi is an eigenstate of Q and this is the eigenvalue, but there is a little bit more than can be said. And it is. It should not surprise you that the eigenvalue happens to be the expectation value of Q on the state psi. Why? Because you can take this equation and integrate dx times psi star. If you bring that in into both sides of the equation then you have Q psi equals integral dx psi star psi, and the lambda goes up. Since my assumption whenever you do expectation values, your states are normalized, this is just lambda. And by definition, this is the expectation value of Q. So lambda happens to be equal to the expectation value of Q, so sometimes we can say that this equation really implies that Q hat psi is equal to expectation value of Q psi times psi. It looks a little strange in this form. Very few people write it in this form, but it's important to recognize that the eigenvalue is nothing else but the expectation value of the operator of that state. But if you recognize that, you realize that the state satisfies precisely Q hat minus Q on psi is equal to 0. Therefore, if Q hat minus Q on psi is equal to 0, delta Q is equal to 0. By claim 2. Q hat minus Q expectation value kills the state, and therefore this is 0. OK then. The other way is also true. If delta Q is equal to 0, by claim 2, this integral is 0. And since it's the sum of squares that are always positive, this state must be 0 by claim 2. And you get that Q minus Q hat psi is equal to 0. And this means that psi is an eigenstate of Q. So the other way around it also works. So the final conclusion is delta Q is equal to 0 is completely equivalent of-- I'll put in the psi. Psi is an eigenstate of Q. So this is the main conclusion. Also, we learned some computational tricks. Remember you have to compute an expectation value of a number, uncertainty, you have these various formulas you can use. You could use the first definition. Sometimes it may be the simplest. In particular, if the expectation value of Q is simple, it's the easiest way. So for example, you can have a Gaussian wave function, and people ask you, what is delta of x of the Gaussian wave function? Well, on this Gaussian wave function, you could say that delta x squared is the expectation value of x squared minus the expectation value of x squared. What is the expectation value of x? Well, it would seem reasonable that the expectation value of x is 0. It's a Gaussian centered at the origin. And it's true. For a Gaussian it would be 0, the expectation value of x. So this term is 0. You can also see 0 because of the integral. You're integrating x against psi squared. Psi squared is even, x is odd with respect to x going to minus x. So that integral is going to be 0. So in this case, the uncertainty is just the calculation of the expectation value of x squared, and that's easily done. It's a Gaussian integral. The other good thing about this is that even though we have not proven the uncertainty principle in all generality. We've only [? multivated ?] it. It's precise with this definition. So when you have the delta x, delta p is greater than or equal to h bar over 2, these things are computed with those definitions. And then it's precise. It's a mathematically rigorous result. It's not just hand waving. The hand waving is good. But the precise result is more powerful.
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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. So. Last day we started looking at phase diagrams, and we looked at unary phase diagrams, one component systems, and one component system P versus T, pressure versus temperature. And this is the situation for water. Water is an exception, because it's got this negative solid equals liquid coexistence curve, but otherwise, you have large, single-phase regions here, which I've designated circle p equals 1, and then along these lines, we have equilibria. So this is liquid goes to vapor, this is solid goes to liquid, this is solid goes to vapor. And we know that for water, if we take the 1 atmosphere, if this is pressure in atmospheres, the 1 atmosphere isobar, then that would put this point at 100 degrees Celsius, and this point at 0 Celsius. And then we have the triple point, which is solid equals liquid equals vapor at 0.01 degrees C, and p equals, I think it was 4.58 millimeters of mercury. So that's the triple point. And we looked at a variety of other one-component phase diagrams, and I think we came to a pretty good understanding. And then up here, we have the supercritical fluids. So I've got a break in the line here. And for water, you reach supercriticality at 374 degrees Celsius, and up here at 218 atmospheres. So these are not phenomenal, these are not geological pressures at all. And so that's if we want to the decaffeinating of coffee and so on. And so what happens here, is you have a highly compressed vapor or a highly rarefied liquid, and you end up with solubilizing power of a liquid, but transport properties of a gas. So this stuff has really good diffusivity. So it can penetrate into interstices and do its work very quickly, to say nothing of the fact that you're up at 374 degrees Celsius or higher. So today what I want to do, is I want to look at two-component systems. So component number is circle c, c equals 2. So that means now I have to deal with pressure, temperature, and lowercase c is composition. I want to know how things vary with composition. So let me give you an example. Suppose I have two substances, A and B. So here's substance A, and I've got its one-component phase diagram. And in this case, I'm going to put solid-liquid vapor in the conventional setting, where the solid-liquid line is-- the coexistence curve has a positive slope. So this is A, and let's say I've got B over here, and it's got its solid-liquid coexistence curve, and so we've got its melting point, and whatever. So now the question I want to ask is, what happens if I mix A and B? And let's just say, here's the melting point of A, so this is at p equals 1 atmosphere. And here's the melting point of B. And I want to ask the question, how does the melting point of the mixture vary? So let's say I'm going to connect these, and I want to ask, what happens if I've got an AB mixture? So I'm going to put lowercase c here, which is concentration. So at the one end, I've got 100% A, and over here, I've got 100% B, and I know those melting points. Question is, what happens when I mix them? Does the melting point-- is it just a straight line? Is it a linear variation? Do you go through a local maximum? Do you go through a local minimum? Or do you go wild? I mean, what happens? So a question you might ask is, suppose I made a 50-50 alloy of 50% A and 50% B. If I knew the end-member melting points, could I predict this? Or if not, do I have an archive? Remember the phase diagram compendium is an archive. What is the value of the melting point at 50-50 or 75-25? So that's the question I want to ask. Now this is getting really messy, because now I have to plot pressure, and I have to plot temperature, and I have to plot composition. So this is really crazy, right? I've got pressure, and I'm going to have composition here, and I'm going to have to have now a third axis, aren't I? I'm going to have to have a third axis. I'm going to need a temperature axis. And I've got to make a three-dimensional drawing. And this is messy. But we're in luck, because 3091 is solid state chemistry. So as solid state chemists, we're more interested in what happens in the solid here. Now, I've drawn these lines with a slope, but they're really not to scale. It turns out that, you know, you can move 10,000 feet up or down in the atmosphere and have a huge variation in the boiling point because this line is shallow. But this line is virtually straight up and down. You have to go to geological pressures to change the melting point very much. And so, as a result, this is almost insensitive to pressure, to first order, right? So we can say solid equals liquid is almost insensitive to pressure, whereas liquid equals vapor is very sensitive very sensitive to pressure. But we don't care about this so much. So what I'm going to do is throw it away. I'm going to throw it away, and just say, let's look at just T versus C. Temperature versus composition on the strength of the fact that the melting point and all of these things are mildly dependent on pressure. All right. So now I'm going to give you three different types. There's many, many different types of binary phase diagram. So now we're going to be looking at C equals 2. So instead of a unary, this is going to be called binary. Binary phase diagrams. And they come in all shapes and sizes, but they can pretty much be classified in several bins. And I've made the classification. So classify binary phase diagrams according to their bonding. It all comes back electronic structure and bonding. Vary by bonding. Because bonding, then, ultimately indicates solubility, and that's what we're looking at here. We're going to look at how well A dissolves into B, and how well B dissolves into A. That dictates solubility. So it all comes full circle. So I've made this up. You're now going to find this in the book. So I just decided, for ease of teaching 3091, is I'm going to just give the different types. And so I call them type one, type two, and type three. You're not going to find that in the books. I made that up, because it's simple. So type one. What's type one binary phase diagram, according to me? What is the type one? It means complete solubility. So A and B are completely soluble in one another as solid and liquids. That's the first thing that you'll see on a type one phase diagram. And the second one is, a change of state is present. And you know the only change of state we care about here is solid goes to liquid. So we're going to show solid goes to liquid, and the thing is totally, as they say in California, totally soluble as liquids and solids. Now, here's some bonding rules that you can think about. So what would be the characteristics of two substances A and B that would give complete solid solubility? Well, one thing is they'd have to have identical crystal structures. That would heighten the chances of this. Identical crystal structures. So if they're both FCC metals, you've got a better chance of making an infinite variation in solution composition than if one is an FCC metal and the other is a BCC metal because FCC will just sit on the lattice site. But there's more fine structures. Second one is similar atomic volumes. So if we're going to have a substitutional solid solution, let's make sure we're replacing oranges with oranges, and not trying to put a grapefruit on a site that normally is occupied by a lemon, because there's going to be a size restriction there. And the third thing is, even if they have identical crystal structures and similar atomic volumes, you can still run into trouble with respect to complete solid solubility if you have a large variation in electronegativity. And I'll show you an example of that. So if you have a small difference in electronegativity, it means there's a low propensity for polarity, and ultimately no chance of electron transfer. So I think it's pretty cool. So I wrote those down, but was disappointed to learn that for metals, they were enunciated about 75 years ago by the British metallurgist Hume-Rothery. So I can't take credit for this. This is one name. It's one of these hyphenated British names. He's Sir da-da-da-da Hume-Rothery. So he has the Hume-Rothery Rule. And he called such systems that mix as binaries forming isomorphous, meaning they have the same structure. So let's take a look at the prototypical isomorphous phase diagram, and it looks like this. We're plotting temperature versus composition. We threw away the pressure coordinate. So I got pure A on the left, pure B on the right. Little c is concentration, so it varies from 100% A to 100% B. And the vertical axis is temperature. So one extreme, I've got the melting point-- this is a melting point of pure A. And at the other extreme, I've got the melting point of pure B. So that, we know. And now the question is, how does melting point vary as a function of composition? And for an isomorphous phase diagram, it looks like this: lens-shaped. Up here is all liquid. You see A and B mix in all proportions. And down here is all solid. And furthermore, it's an all-liquid solution. It's a mixture. And this is an all-solid solution. And then comes the piece that if you take enough thermodynamics, you'll be able to rationalize. I'm simply going to tell you without proof that when you have a multicomponent system, when c is greater than 1-- so we're in that situation. Now, c is 2. When c is greater than 1, it is impossible to move from one field, up here, this is-- I'd better put the label on it-- this is a single-phase field. So up here, p equals 1, because it's homogeneous liquid solution, p equals 1. When you're in pure materials, you can go from solid to liquid. But when you're in a two-component system, you cannot move from one single-phase field to a second single-phase field without moving through a two-phase field. And we'll get to that, what it means to move from one field of p equals 1 to another field of p equals 1 requires traverse or transit across a field of p equals 2. And so what's in here has to be the end member. So this must be liquid plus solid. So I'm going to call it slush. They have a slush in here. The other terminology that people give this is, it's lens-shaped, right? Looks like a lens. So let's call this lens shape. So we're going to use the Latin word for lens, which is lens. But we want to make it adjectival. So we're going to use the adjective-- what's the genetive form of lens? Lentis? So we will call this lenticular. This is lenticular. The phase diagram has a lenticular shape. Looks like a lens. And you know, just as over here, I showed you, every line represents a coexistence curve, right? The lines are all coexistence curves. These lines are coexistence curves. The line up here is liquid equals vapor, so this one must be the coexistence of the two things on either side. So what do I have here? I have liquid. Here I have slush. So I'm going to write that. So I'm going to write the equilibrium, liquid goes to liquid plus solid. And this line, this coexistence curve, it's called the liquidus line. This is the liquidus equilibrium, the liquidus coexistence curve, and so on. And so what's the liquidus? The liquidus is the lowest temperature. So you pick a composition. The liquidus is the lowest temperature at which you can have a single-phase liquid solution, OK? So liquidus equilibrium. And the liquidus is the lowest temperature at which all liquid is stable. You go below that temperature at that composition, you start making solid, because slush requires that there be some solid present. And then the lower line, it also is a coexistence curve. So on the one side, I've got solid. On the other side, I've got slush. So I'm going to write that one down. That solid goes to liquid plus solid. And that's called the solidus equilibrium, or that's the solidus line, the solidus coexistence curve, and the solidus is the complement. The solidus is the highest temperature at which you can have all solid present. So you pick a composition, and I'll tell you what the highest temperature is at which you'll have a single-phase solid solution. So solidus is the highest temperature at which all solid is stable. All right. Good. And I'm going to look at a few of these things. It's always fun to see if I'm on track with the Hume-Rothery rule. So here's copper nickel. They're both FCC metals. So we've got copper melting at about 1085, nickel melting at about 1455. And there's the phase diagram, lenticular phase diagram. So up here is all liquid, and then-- this is all metallurgy terminology. A solid solution of A and B, they call alpha. Metallurgist looks at that, goes oh, it must be a solid solution, OK? So there's p equals 2 in between, p equals 1, p equals-- Here's a ceramic system. This is nickel oxide, magnesium oxide. It's not metals, but they have identical crystal structures, very similar atomic volumes. They're ions, so we have nickel and magnesium substituting for one another on the cationic sublattice. So they must have very nearly equal sizes. If I look at this, and I see a lenticular diagram, it means they must have similar atomic volumes, which means the dominant defect in here must be Schottky, not Frankel because Frankel needs a big difference in atomic volume. If you have a big difference in atomic volume, we wouldn't have a lenticular phase diagram. So here's-- mag oxide melts at 2800 degrees centigrade. It's a great refractory. You can hold molten iron in it. And here's nickel oxide down here. All right. Here's an interesting one. This is gold nickel. Both FCC metals. Shade your eyes from the lower part. Just look at the upper part. It looks lenticular. It's what you'd expect, gold and nickel. They're both, you know, card-bearing metals. But you've seen already, with the cesium-gold, gold has a fairly high electronegativity. And what happens is that you get over here, at about 33 atomic percent of nickel in gold, you have an atomic ratio that allows to have something that starting to approximate electron transfer. So it's almost as though you have a lenticular phase diagram between pure nickel and this nickel-gold compound. But up in here, it's nice lenticular stuff. All right. What's the next one? Oh, yeah. So now I want to go in and I want to start talking about what goes on inside here. So what I'm going to do, is I'm going to blow this up. And I'm going to start at 40%, 40 weight percent nickel, and I'm going to say, what happens if we take a crucible that's 40 weight percent nickel and copper, and we cool it from all liquid at 1300, down to all solid at 1200, pausing in that slush zone at 1250? So we're going to take snapshots and say, what's the contents of the crucible look like at 1300, at 1250, and at 1200, and come out of it, have an appreciation for what all of this stuff means. So let's start our experiment. So we're going to have three crucibles here. And we're going to start with 40% nickel in copper, and that's going to be the experiment we'll perform. So I've got three crucibles. And so if I look at the phase diagram, here I am. This was 1300 degrees C, this was at 1250 degrees C, and then this was at 1200 degrees C. So at 1300, that's trivial. 1300 for this thing, it should be just all liquid. And maybe it's copper. I know at this temperature, it's going to be blinding white heat. So it doesn't matter. I could use this. But I felt compelled, it's one of those rare opportunities to use colored chalk. So I want this to sort of be copper-colored. The meniscus is going to look like this, because the liquid metals have a very high surface tension. They want to ball up. So you're going to have a meniscus looking like this, and this is all single phase, all liquid. And over here, I'm going to do the easy one. If we get down to 1200, it's all solid state, right? At 1200, I'm going to end up with something that's solid, and it's going to be polycrystalline, and these are all going to be grains of alpha. All solid. And I'm going to label all of these as alpha solid solution. They all have the same composition, which is 40% nickel in copper. And these are all 40% nickel. Now, according to this phase diagram, when we get down into the center there, something else happens. We end up in a two-phase regime. And that two-phase regime is a regime in which certain compositions are forbidden because that's an equilibrium. It doesn't allow us to have what's in between. I think the next slide actually shows this. So what I've designated as c2 is this 40%. So c2 is now going to park at that point in the middle. Whoops! Want to get into that. So here's where we are. We're in the center of that two-phase regime. So this is the solidus. This is the liquidus. And we're at 1250. t equals-- And we're at this value. We started at c2, which is 40%. But in this regime, 40% is forbidden. If you stop at 40% at 1250, this says that the stable phases are a liquid and a solid. But the liquid has less nickel in it, and the solid has more nickel in it. It's like if this is a solubility limit. In other words, if you started at pure copper, and you started adding nickel, you can keep adding nickel until you get to this concentration. You try to add any more nickel, it's like adding too much sugar to water. What do you have? The stuff just falls to the bottom of the cup, doesn't dissolve. That's this. It's the solid. Only instead of being pure nickel, it's still a copper nickel solution, but kind of rich. So I'm going to call this c star. It's the solubility limit on the liquid side. And this one here I'm going to call c star on the solid side. In other words, I could start from this side, and keep adding copper to nickel, and I make a homogeneous alloy until I get to this composition. If I try to add any more copper, I jump across. So we're here in the middle. But now, isn't there something strange here? It's saying I'm going to have a liquid that's nickel-poor and a solid that's nickel-rich. And I'm just putting up here what the diagram says. And we know that FCC metals, which is denser, the liquid or the solid? The solid. So down here I'm going to have alpha, OK, a bunch of grains of alpha solid solution, and up here, I'm going to have a liquid solution. And the composition here is c star liquid, and the composition here is c star solid. It's different composition from this. Here the composition is equal to c2, or the 40% nickel. So I've got disproportionation. But the total mass, I can't have sources or sinks. So I have to conserve mass. So I'm going to ask you to just do the algebra. You've got two equations and two unknowns. I know what the n-member concentrations are. They're given by the n's of this thing called the tie line. The tie line ties the two ends of the two-phase region together, and then we just do a mass balance. Well, fortunately, the metallurgists have thought about this for a while, so we don't have to go through and figure out, well, if two apples cost 15 cents-- we don't have to do that kind of thing. We can just invoke the lever rule, and it will tell us how much of the liquid that's nickel-depleted and solid that's nickel-rich add up to this. And so let's look at the lever rule. Basically what's going to happen here is that the stuff that started off as 40% nickel, 60% copper, that's my initial mix, it's going to break into two layers. It's going to break into a liquid layer, which, if you go to the phase diagram, looks like it's about 38% nickel and 68% copper, and that's what we're going to call c star liquid. And then over here it's 45% percent nickel and 55% copper. And that's what we call c star of the solid. And so it's just a lever. So there's c star. This is the c2 that we want. And we want to ask, what's the relative amount of the liquid phase and the solid phase? And why they call it a lever rule is, look, if you cooled something that was at this composition, it would be 100% liquid. And if you cooled something at this composition, it will be 100% solid. And so it's going to be a constant variation across here. So you use, you take this, and you know, and I want this amount, I'm taking this versus this, kind of thing. It's a lever construction. So we'll just put it down. So the percent of the liquid in the crucible at 1250-- and then this is a general thing. Whenever p equals 2, you'll use a lever rule. Percent of the liquid is given by this one. The composition of the solid at the end of the tie line minus the composition that you started at. This is your bulk initial concentration, divided by the length of that tie line. c star solid minus c star liquid. And then multiplied by 100%. So you get a percent, and the number here is going to be 45 minus 40 over 45 minus 32, and that works out-- going to multiply by 100, or else there will be complaints. And then this turns out to be 38%. So what that means is that if you take this amount of liquid here-- pardon me-- based on conservation of mass, if you drop and hold at 1250, 38% of that volume is now sitting here in the liquid, which means 62% of it by volume has turned into solid. And the composition here is different from the composition here, but if you add up all the nickel in the crucible and sum it versus all the copper in the crucible, you still end up with 40% net nickel. So this very powerful. Because what you can do here is you can separate metal. Because I started here with 40% nickel, 60% copper, and now I've got something-- the liquid face, I wrote here, the liquid phase is now 68% copper. So can you see that if I'm clever about this, I could use this as a technique for enriching. And if I'm thinking about recycling metals, maybe I need to know something about phase diagrams because it's a very simple way of concentrating impurities or concentrating the desirable stuff. Let's look at one other thing here that's really very interesting. Suppose I had something, instead of c2 at 40%, suppose I choose a c1. So this is 40% nickel. Now what if I take something that's 35% nickel, and I cool it down to 1250? What happens? If I park here at 1250 degrees, according to the phase diagram, that substance has to phase separate, and I'm going to end up with the same thing. I'm going to end up with liquid and solid. Now, isn't there a contradiction there? How is it that whether I started with 40% or 35%, I end up with the same end members? What's the missing piece? The relative amounts will be different! The relative amounts. Here, look. As I'm getting closer to the liquid, can you see that axiomatically, I'm going to make more liquid and less solid if I ended up with something over here? But at 1250, those are the end members. That's the only stuff that's going to be present. Ah! That's so good. I think I've got some other stuff here. Oh yeah. Oh, this was-- yeah. So p equals 2. Lever rule. Whenever you see p equals 2, you have two things you think of. p equals 2 means phase separation. And that phase separation means you're going to get different amounts, and the lever rule. And why I have this Manchurian Candidate-- it's this movie. I know there's a remake. The remake is horrible. The original, if you see the original, is about some fellows that were brainwashed. They were American prisoners of war brainwashed in North Korea and they're brainwashed to become assassins on cue. And the cue is the Queen of Diamonds. When someone shows the Queen of Diamonds, they just go into automatic pilot, and they're supposed to assassinate the political figure. So for you, your Queen of Diamonds is p equals 2. When you see p equals 2, you go, phase separation. Lever rule. All right? No guns, just this. I just want the formula. You go, phase separation, lever rule. Whenever p equals two. So what happens in here? Phase separation. Boom, boom. Right there. Oh, I'm not supposed to say boom, boom. Phase separation. Ta-da! You know, something. PC, whatever. OK. So let's keep going. Now I want to look at a second type of phase diagram. So the first type of phase diagram was complete solubility. Now the second type of phase diagram has partial solubility. So let's look at that one. So i call this type two. And you don't predict this stuff. We would give you the phase diagram and simply ask you, you know, you're the specialist. Tell me happens if I cool this to such and such a temperature? You can tell me you get phase separation, and the composition of the two end members, and so on. So the characteristics, the bonding characteristics here, are partial or limited solubility of A and B. We're doing all of this for a binary system A and B, and no change of state. So that means either always solid, or always liquid. That's not to say that the A and B never become solid. I'm just saying that a type two diagram is limited to a single state. So let's take a look at the diagram. The diagram looks like this. We're going to plot temperature versus composition. So pure B on the right, pure A on the left. Concentration is the abscissa, and then the ordinate, of course, is temperature. And so the shape of the diagram is this. There's the coexistence curve. It's called a synclinal coexistence curve. So above, we have all solid. And in here, we have two solids. So using the metallurgical terms, this is alpha plus beta. Two phase regime, OK? So let's get those labels up right off the bat, so we know who's where. So outside the coexistence curve, p equals 1. Inside the coexistence curve, p equals 2. And wherever I'm saying all solid, alpha plus beta, I could write all liquid, and then it goes to l1 plus l2. So maybe we should do that, to show that we're multilingual. So it's either all solid, or it could be all liquid. And then this would be liquid 1 plus liquid 2. And so why do we call it synclinal? What's this? This is an incline. If I put two inclines and I synchronize them, I get a syncline. And if I don't synchronize them-- these are two ladders. I could prop them up. That's called a syncline. And if I put two ladders like this, if I'm really stupid and I fail physics, if I do this, they fall down. So this is called an anticline, OK? That's an anticline. This is a syncline. It's not a U-shape, or a hump, or something like that. This is 3091. This is a syncline. Synclinal coexistence curve. All right. So what's on here? What's this equilibrium, then? Well, the equilibrium must be this equals this. So that would be, in the case of the solid, it's the solid solution goes to alpha plus beta, or it could be the liquid goes to liquid 1 plus liquid 2. That's the equilibrium. You can think about it almost as a solubility. So let's start over here on the left. Let's pick a temperature. Let's call this T1. So I can put B into A, and I continue to get an all-solid solution, up to this concentration here. What happens at this concentration? I've hit a solubility limit. And if I try to put any more B into A, I get a tie line-- lever rule. In here is lever rule time, isn't it? Lever rule. p equals 2, OK? So it's solubility limits. Solubility limits. All right. So let's look at some examples. Oh, here's this one, actually. That's why I had a-- you know, if you go to lower temperature, gold, nickel-- look! They actually phase separate. That's shocking. I always find this one shocking, because you think gold, nickel, nice FCC metals, they should substitute for one another. Look what happens. If you start putting nickel into pure gold at 700 degrees, you get the 10 weight percent, you put any more in, boom. Right across here. So this is going to be two phase. So if you look at the solid, what's that going to look like? If I'm at, say, 30%, now I look underneath and I've got a polygrain system, and I'm going to have an alpha and a beta, and an alpha and a beta. I'm going to have two different grains. And now, what's the relative amount of alpha and the relative amount of beta? It's given by the lever rule. OK. Let's look at a few other examples. Here's hexane nitrobenzene. And so they've even put l-alpha, l-beta. They put the-- actually, that's probably a-- it's hard to read. But I think it's really an F, but it didn't come through very well, so it's the beta fraction, the alpha fraction. So you can see how you use the lever rule. So up here it's all p. That's the not pressure, that's my circle p. Single phase, two phase. Drop down to 290 degrees, it separates into two liquids. With the hexane-- with two liquids, you'll actually have them floating on top of one another, right? Well, maybe. Depending if the density difference is tiny, you might get a dispersion. I'll show you that in a second. All right. So p equals 2, lever rule. OK? Hexane nitrobenzene. Look at this one. This is surprising to me. Potassium chloride, sodium chloride. It's almost lenticular with a little bit of depression, but down here, it actually separates. This is polymer. It's a polystyrene, polybutadiene, depending on what the-- at low index, polymerization index, they mix. At high polymerization index, the two are mixing. You get two phase, single phase. Changes the mechanical properties, too, doesn't it? Now this one-- some systems actually have a lower critical point. Oh, I meant to tell you. You see, there's this temperature here. Above this temperature, they mix in all proportions. You can ever have phase separation. This is the maximum temperature at which you have no phase separation. So this is called the consolute temperature. And it's usually an upper consolute. The higher you go in temperature-- it's like saying, do you dissolve more sugar in cold water or warm water? You dissolve more sugar in warm water. And eventually, you get to a temperature high enough that you can mix in all proportions, right? That's the consolute. Some systems, for entropic reasons-- and again, if you take 560 or some of the other thermal classes, you'll understand this, but for entropic reasons, you actually have homogeneous solutions at low temperature, and at elevated temperature, you get phase separation. That's funny, isn't it. So this is water triethylamine, and they mix in all proportions at this zone here. And then above a certain-- it's called lower consolute temperature. It actually phase separates. Doesn't matter. All you care about is p equals 2, phase separation, lever rule. The rest is details. This is a real hoot. This is water nicotine. It has a lower consolute temperature and an upper consolute temperature. So it's got a solubility bubble. You see? Out here, things are soluble. In here, they're insoluble. So they are emissible. So we call this zone, this region here under the dome, so to speak, this it's called the miscibility gap, because things in there are not soluble in one another. But with nicotine, you have a lower consolute and an upper consolute, so you have a miscibility gap that's a complete ring. That's cool! I like this. All right. So now I'm going to do a little experiment. We have a few minutes, so I'm going to show you ouzo water. I'm going to mix it, and I'm going to actually mix ouzo and water, and go into the two phase regime, and then out. So this is what it is, all right? It's oily stuff, you know. If you're of Greek ancestry, you know what this stuff is. All right? But it's got licorice, it's got a fair bit of oil in it. So what I'm going to do, is I'm going to come in here, and I'm going to start adding ouzo to water. And they're both clear, colorless liquids. OK, Dave. Let's go to the-- All right. So what we're going to do-- this is distilled water. So I'm going to put some distilled water in here. A little bit of distilled water. And this is ouzo. It's clear and colorless. OK. Can we show you this? Let's do this. OK. Ouzo, it comes from Greece. All right. So it's also clear and colorless. That was the point. Not to show you the label. See? It's clear and colorless. So this is clear and colorless, and the water, as you know, is clear and colorless. So now what I'm going to do-- see, I don't have to wear a smock, or goggles, or anything. This is great. So what I'm going to do, is I'm going to add some of this stuff. Clear and colorless. OK. It's turned milky. Why has it turned milky? Because we've now crossed into here, and the second phase is coming out. But the density difference and surface tension differences are so slight, that instead of having the second phase float on the first phase, we have a dispersion, G. If you go to the dairy case, and you look at something called milk, you'll have the same thing. Why is milk milky? Because the fatty phase is a fine dispersion, and the particle sizes is dadadada, versus the wavelength of visible light, et cetera. Now if my phase diagram is correct, and I'm in here, if I keep adding ouzo, I should eventually emerge on this side, and instead of having a milky dispersion, I eventually should come here, and now I'll have a homogeneous, single-phase solution. But now it's going to be ouzo-rich, instead of water-rich. You never thought phase diagrams were interesting. You don't know. There we go. Je vous presente. There it is. So what we've done is, we've come across the thing. Yeah, that's good. Just put that there. Now, David, please, back to the slides. May I have the next slide, please? That's a joke. When you have these terrible speakers at conferences, they get up there, really nervous, and they stand up there, and the first thing they say is, may I have the first slide, please? That's sort of a gag among scientists. What's your opening statement? May I have the first slide, please? May I have the next slide, please? OK. So we've done this. All right. Now-- yeah, OK, this is just more. All right. So now I'm going to show you absinthe. Absinthe is the same thing. Absinthe also comes from the same family. And there's a little culture here. It contains wormwood. The wormwood was up at around 200, 250 parts per million. And what wormwood does, it's got a hormone they call thujone. And thujone antagonizes the gamma aminobutyric acid, which moderates firing of the neural synapses. Basically, we've got this GABA that regulates how-- our brains could work much faster than they do, but the GABA regulates. If GABA doesn't work, you can start moving so fast that you get muscle thing, and you get epileptic, is one example. But anyways, what it does, is if you drink this stuff, instead of being a stupid drunk, you become a very high-functioning, very alert drunk. And I'm going to show you what the consequence is in art. Anyways, the cultural piece here was that there was a destruction of the French wine industry in the 1800s due to a blight called Phylloxera. It's like a beetle, and it ate the vines. And in fact, the French wine industry was saved by the American wine industry. Vines from New York state were brought over, grafts were made, and virtually all French wine today is on American stock with the exception, occasionally you'll see something says vieille vin. Old vines. There was some areas that were spared. So what's that have to do with anything? It has to do with the fact that when the Phylloxera hit, the price of wine went very high, and the people at the bottom of the socioeconomic ladder couldn't afford wine. So they turn to absinthe. Absinthe was easily produced. Thirty years later, when the vines come back, the French wine industry wants to recapture the market. So there's a Faustian bargain between the French wine industry and the Women's Temperance Union to try to disparage absinthe. And there were a few major show trials that involved vicious murders in which it was alleged that the killer was deranged on absinthe. And with this, they slandered the name of absinthe so badly that it was eventually banned and literally taken off the market. I mean-- you know, we had the Rosalind Franklin commentary a few lectures ago. I hope I never read about you doing something like this in order to help your start-up company take over market share. That's not the way you're supposed to do it. Anyway, so what-- now, how did they drink it? They drank it by mixing-- so it's now back, but it's got very low levels of thujone, so it's-- David, may we go to this? So we're going to make a mix called the louche. It's a-- OK, here it is. And this stuff here is beautiful. Because it's a green color, a soft green color. And the French even have a name for it. They call it the green fairy, la fee verte. So this is what absinthe looks like. Beautiful. Very nice. And it's got the same kind of licorice smell to it. And so you make the louche by one part absinthe and five parts water. So this, at no upcharge, you get this glass. It's a beautiful piece of, you know, late Belle Epoque glassware. And the interesting thing is, these people were absolutely crazy, scientifically. And so what they did is, the markings on this glass are such that if you put absinthe up to this ring here, and then you put water the rest of the way, you get exactly the five to one ratio for louche. So you don't even have to measure. You just put like so. OK. Let's see. I'm going to get this right! It's science. Here's our distilled water. You see how it's milky? So there's the louche. And if you wanted to be a real hipster, you had this special Belle Epoque spoon. And what you'd put on top of this would be a sugar cube. You know, this is how it was done. You have to know some culture. You don't know. You've led sheltered lives. You pour it through here, like this. So now I'm going to show you the art. So David, may we cut to the slides again? So there's the louche. All right. So here's the poster. Here's the man. He's pouring the water through the slotted spoon with the thing, and there's a very well-to-do lady, you can tell, she's got a beautiful hat, and she's well-dressed. And he's inviting her to join him for a glass of absinthe. L'absinthe oxygenee. That's the name of the company. The oxygenated absinthe. C'est ma sante. This is my health. Van Gogh painted this. This is Picasso. The absinthe drinker. There's the absinthe again. This is Picasso after many absinthes. You can see the slotted spoon and the sugar cube. What else is there? That's up to you. All right. When you saw Moulin Rouge, you may not have known all of this cultural history. So now let's take a look at what's going on here. There's the absinthe, and I think we've got a little-- [VIDEO PLAYBACK] -I don't even know if I am a true Bohemian revolutionary. -Do you believe in beauty? -Yes. -Freedom? -Yes, of course. -Truth? -Yes. -Love? -Love? Love. Above all things, I believe in love. Love is like oxygen. Love is a many-splendored thing-- [END VIDEO PLAYBACK] Love is like oxygen. Chemistry is everywhere! All right, so this is what happens eventually. This is the poster. This is from Switzerland. October 7, 1910. Gentleman, This Is the Hour! And there's the clergyman with the Bible, and there's the green fairy. And she has a wand. She's been stabbed, but she's lying here with a wand. The wand is an opalescent wand, because this stuff is opalescent. Last thing I'll show you is some really, really cool chemistry that you must remember. When Toulouse-Lautrec drank-- let's go back, David, to the thing. So when Toulouse-Lautrec drank absinthe, and he drank lots of it, he didn't like the milky color. So he wanted to make the milky color disappear. So I'm in the two-phase regime, and I've got an oily phase. And so how do I get rid of the oily phase? What he did is he added cognac And why did he add cognac? Not because he was a hard-drinking alcoholic. It's because if you've got a fat phase here, and you've got an aqueous phase here, and if you add alcohol, you've got CH3CH2OH. This can bond to the water by a hydrogen bond, and this aliphatic tail can stab the fat and bring them into solution. That's why you have these recipes. Do you ever wonder why the recipe says, add brandy, or add this, and then two steps later, it says flame? You say, geez, I just put the brandy in, now it's vaporizing away. Isn't that kind of stupid? No, it's not! This is what you're doing. You're cosolvating. If you're ever making a cream sauce or something, and all of a sudden, everything just curdles? First you scream. You go, ahhh! Phase separation. The second thing you do, is you get some of this, and you cosolvate it. All right. So let's see what Lautrec did. Lautrec-- I might have to dilute the volume here. We've got quite a lot in here. So just to make the point. So here's the louche, ad now we're going to add cognac. He called this drink, Le Tremblement de Terre. The earthquake. He had a walking stick a meter long that was hollow, and it always had this in it. Look! Look at that. Isn't that something? So now it's this beautiful blonde, golden color. It's clear. So this is all phase separation and so on. At the end, it's all about chemistry. All right. We'll see you on Wednesday for our wrap-up. Good.
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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: Let's do a case that this mostly solvable and illustrates all these things. It's a very entertaining case. It's called Landau-Zener transitions. For these two people, Lev Landau, who you've probably heard from Landau and Lifshitz. He's the first person that tried to do this. And Zener did it more carefully. In fact, apparently found that Landau made a factor of two error. And the paper of Zener, it's actually quite nice, and it's a very nice example that illustrates the physics of this transition. So we'll devote the rest of the lecture to that Landau-Zener thing. OK. So it will give us a little bit into the spirit of the adiabatic approximation in the language that Berry used. So Landau-Zener transitions. OK, Zener and Landau were interested in molecules, and some way of thinking of molecules is to think of nuclei as fixed, separated by some distance R, and then you assume they are fixed, and they're separated by some distance R. And then you calculate what is the electronic configuration. So Zener imagined that you would have psi 1, one electronic configuration. It's a wave function that depends on some x's for the electrons, but it represents the situation where the two protons, say, for a simple molecule, maybe they're more distances, but in particular, they are separated by a distance, R. So that's an electronic configuration. Electronic configuration, protons a distance R away. And suppose there's another configuration psi 2 of R. It's another configuration, so two configurations. Two different states. Maybe in the first state, the electrons are in some ground state. In the second state, they're in some kind of excited state, two different configurations. Now, we could plot. So we'll have [INAUDIBLE] here, distance R, and here's the cloud of electrons. We could plot a graph as a function of the separation, what are the values of the energies. And here is one possibility for the states. And here's another one. And that's the plot of the style that Zener drew in his paper. And this represents E1 of R and this E2 of R. That is the energy of the first state, the energy of the second state as a function of R. So we are having here-- oops-- two energy eigenstates. So we have H of R. The Hamiltonian depends on the R. And basically you're putting the two protons, the distance R, and calculating the electrons, how they move. Psi i of x R equals E i of R psi i of xR. This is for i equal 1 and 2. The case that the people were interested in was the case where this molecule here, for example, in the state 2, for this value of R, there is a critical R 0, where things, the levels get very close. For some value of R, this molecule, for example, could be a polar molecule. A polar molecule is a permanent dipole moment. It has plus charges and minus charges, not evenly distributed. So you get a dipole. And maybe here, the molecule is non-polar. And here, it's non-polar. Here polar. So if you would follow one of the energy eigenstates, there's a critical value of R, where the electronic configuration is such that it goes from non-polar to polar and in the other energy eigenstate, it goes from polar to non-polar. So the question is well, OK, what-- first of all, what does all this have to do with instantaneous energy eigenstates and time dependence? Why are we thinking about this? The issue is that sometimes, you can think of this molecules as forming or being subjected to extra interactions in which you will have a process or a reaction in which the radius changes in time. So it's possible under some configuration that R becomes R of t. And then, this Hamiltonian is a Hamiltonian that depends on R of t. This wave functions psi i are R of t E i's become R of t, psi i's become x of R of t. This is an important point. It's simple, but important. The most important points in physics are simple. But you have to stop and recognize that something slightly new is happening here. If you have solved this equation for all values of R, if you know those energy eigenstates for all separations of the molecule, you now have found instantaneous energy eigenstates if it so happens that R is a function of t, because if this is true for any value of this [INAUDIBLE] R, well, then this is true for all times. Because for any specific time, this is the R, the same R is here, the same R is here, and the same R is here. And that equation holds for all R. So if this can be solved for all R, this holds for all times. And you have your instantaneous energy eigenstates. You have found those instantaneous energy eigenstates. And therefore, the instantaneous energy eigenstate are these ones. And the instantaneous energies are this ones. So many times in quantum mechanics, you do that. You solve for the energy eigenstates for a whole range of some parameters. And then it so happens that those parameters may change in time. But then you have found the instantaneous energy eigenstates for all times. So in that picture, we have the following situation, in which the energies now could be thought if R is some function alpha of t, then the same picture would basically hold true for time here and the energies as a function of time, because as time changes, R changes, and as R changes, you already know how the figure looks. So this is a figure of the energy levels as a function of time. And now the physical question is do we get a transition or not? So the adiabatic theorem would say, OK, you should state in your instantaneous energy eigenstate, but we're going to get precisely to this situation where these things could be so small, so little, that there is a possibility of a non adiabatic transition, in which you jump to the other one, because the gap is small. So this goes to the real physics of the adiabatic theorem, can we get an estimate or a calculation that tells us how much probability you have of jumping the gap and going to the other branch? That's what we're going to try to do. So for that, we'll do a particular example. So let's do that. It's an easy one to begin with. I'll erase this. So baby example, toy example. So take a Hamiltonian, H of t, which is going to be of this form, time dependent one, but relatively simple. Elements just along the diagonal. OK, that's your Hamiltonian, two by two matrix, elements on the diagonal, but just simple things, the same thing. So let's calculate the instantaneous energy eigenstates. OK, sounds like a task. It's actually pretty simple. The instantaneous energy eigenstates are 1, 0 and 0, 1. They don't depend on time, because essentially this Hamiltonian is just alpha t over 2, 1 minus 1, 0, 0 is-- it's a constant matrix times an overall factor. The eigenstates of this matrix are 1, 0 and 0, 1. And they are the eigenstates of this matrix for any time, because the time goes in front. The matrix doesn't quite change shape. So these are the instantaneous energy eigenstates. They are good forever. To plot this, I will assume from now on that alpha is positive. The energy of the first state is-- well, what do you get when you add with the Hamiltonian on this state? The matrix [INAUDIBLE] and this is just alpha t over 2. And the energy of the second state is going to be minus alpha t over 2. We can plot those energies, and here is the energy of the first state is alpha t over 2 with alpha positive. This is like this, thick output of here. This is the state 1, 0 is here, 1, 0 is here, alpha t over [? 6. ?] Here is the energy E1 of t. The energy time dependent. Here is time. And here are energies. This is E1. And then we have the E2 is the other one that goes like this. It's state 0, 1. That's the state 2, 0, 1. And the energy is E2 of t, which is minus alpha t over 2. So it's negative for large positive time and positive for the other one. So these are your instantaneous energy eigenstates. OK, and this is not quite what we wanted here. We wanted things to avoid themselves. But this is going to illustrate an important effect. I claim, actually, that the true solutions of the Schrodinger equation are in this case dressed up versions of the instantaneous energy eigenstates. So what I claim is the kind of-- you do the adiabatic state corresponding to this, the adiabatic state corresponding to that, and those are exact solutions. So there is no coupling between the states, 1 and 2. So this is plausible. So let's write those solutions. I claim here is psi 1 of t I claim is the exponential of minus i over h bar integral up to t of E1 of t prime dt prime times the state 1. I claim this solves the Schrodinger equation. i h bar dt of this psi should be equal to H psi. Is it clear? Yes, I think it's clear. It solves it, because if you take the time derivative of this thing, it multiplies by E1. The i h bar cancels that factor of minus i over h bar. The time derivatives brings out an E1 of t. But this state, despite the phase when h [INAUDIBLE],, it goes through the phase, hits the state 1, and produces the E1 energy. So this is solved by that equation, and you can do the integral. It looks OK. It's exponential of minus i alpha t squared over 4 h bar 1, and the state psi 2 of t is the same exponential with E2 with 2 here, and it's the exponential of plus i alpha t squared over 4 h bar 2. OK. Let's appreciate the lesson again. We got a very simple system, two levels, crossing-- they cross. The energy levels cross. That generally doesn't happen. You have to have a very special Hamiltonian for the energy levels to cross. We found the instantaneous energy eigenstate, and we found two exact solutions of this Schrodinger equation, two perfect complete exact solutions of the Schrodinger equation that represent the system doing just zoom, like that, or doing like that. Totally oblivious that there's a state they're crossing, the Schrodinger equation doesn't couple them in this case.
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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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JOANNE STUBBE: The way we teach the course is we show them that all of these chemical transformations can be described by 10 pretty simple steps. And if you understand the basic chemistry of these simple steps, you can really understand almost all of the kinds of interconversions you see and basic metabolism. And what you'll see is while this looks overwhelming, really, with a few central pathways, which is what we focus on in this course-- glycolysis, fatty acid oxidation, biosynthesis, sugar biosynthesis as well as degradation, the Krebs cycle, which feeds into the respiratory chain-- knowing those central reactions-- almost everything in metabolism feeds in and out of these pathways. And so then, it really is a question of what is the environment like and how do you enhance breakdown of sugar under the appropriate environment versus synthesize sugar under a different environment. So then it's a question of regulation. So what you're going to learn in this course is really focused on central metabolism. And it doesn't matter whether you study a bacteria or a human, the central metabolism is pretty much the same. The thing that's different is the detailed regulation and the complexity of the regulation. And we don't really talk that much about regulation in 5.07. What we do is introduce you to five or six basic regulatory mechanisms that are used over and over again. But then, regulation is really distinct, even between organisms. And so then that becomes much more complicated. And when you go off and become a biochemist, you've got to study your own system and figure out what the environment is.
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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 [email protected]. PROFESSOR: All right, so let's get started. So today we're going to review local alignment, we talked about last time, and introduce global alignment, also talking about issues related to protein sequences, which include some more interesting scoring matrices. So just some info on topic one, which we are still in. So I will have an overview slide. It'll have a blue background, and there will be a review slide with a purple background in every lecture. So last time we talked about local alignment and some of the statistics associated with that, and also a little bit about sequencing, technologies, both conventional Sanger DNA sequencing as well as second generation sequencing. And at the beginning of the local alignment section, we introduced a simple BLAST-like algorithm, and then we talked about statistics, target frequencies, mismatched penalties, that sort of thing. So there were a couple questions at the end which I just wanted to briefly answer. So I believe it was Joe asked about how the dye is attached to the DNTP in dye terminator sequencing. And it appears that it's attached to the base, sort of the backside of the base, not the Watson-Crick face, obviously. That seems to be the common way that it's done. And then there was another question from somebody in the back. I don't remember who asked about when you're making libraries, how do you make sure that each of your insert sequences has the two different adapters, one adaptor on one side and the other adapter on the other side? And there are at least three ways to do this. So simplest is in RNA ligation, when you take advantage of the different chemistry at the five prime and three prime ends of the small RNA that you're trying to clone. So you just use the phosphate and NLH to ligate two different adapters. Another more complicated way occurs in ribosome footprint profiling, which is a method for mapping the precise locations of ribosomes along mRNAs, and involves polyA tailing, and then introducing the adapters together, the two adapters, with a polyT primer that primes off the polyA tail. And then you circularize, and then you PCR off the circles. And it's a little bit complicated, but you can look it up in the reference that's up here on the slide. It's working now. And then, finally, the way that's actually most commonly used for protocols like RNA seq and genomic DNA sequencing is that after you make your double strand DNA, there's an enzyme that adds a single A to the three prime end of each strand. So now you have a symmetrical molecule. But then you add these funny Y-shaped adapters that have and overhanging T on, say, the red guy here. And so what will happen is that each of these Y's can be ligated here. But each of the inserts, independent of which strand it is, will have a red adapter at the five prime end and a blue adaptor at the three prime end. Any questions about this or about sequencing technologies before we go to local alignments? OK, good. It was a good question, and that's the answer. So we motivated our discussion of local alignments last time by talking about this example, where you have a non-coding RNA that you found, inhuman. You BLAST it against mouse, and you get this alignment. Is this significant? So is this really likely to be a homologous sequence? And how do you find the alignments? And so we said that, well, there's this theory that's exact, at least exact in the asymptotic sense for large query and database sizes that tells us the statistical significance of the highest scoring ungapped local alignment. And it's given by this formula here, which is the extreme value or Gumbel distribution. And then we talked about the constraints or the expected score has to be negative, but positive scores have to be possible for this theory to work. And we also talked about an algorithm. But if you remember, the algorithms was very simple. It involved-- this is zero-- keeping track of the cumulative score. So we have a mismatch and a match, mismatch, mismatch, mismatch, match, match, match. That is a high scoring segment, et cetera. So you keep track of the lowest point you've ever been to as well as the current score. And when the current score exceeds that lowest point you've ever been to by more than we've ever seen before, more than this, then that's your high scoring segment. Now it turns out, if this is not intuitive to you, there's another algorithm which I find, personally, more intuitive. So I just want to tell you about that one as well. And it's basically the same thing, except whenever you go negative, you reset to zero. So here, we were going to go negative, so we just reset to zero. That was on this mismatch here. Then we have a match. Now we're at plus 1. That's fine. Now we have a mismatch. Now we're down to zero. We don't need to do anything. Now we have another mismatch. Here, we're still at zero. Remember, we just stay at zero. We were going to go negative, but we stayed at zero. Another mismatch, we still stay at zero. And now we have these three matches in a row. My line is not staying very flat. But this should've been here flat at zero. The point is that now the highest scoring segment is the highest point you ever reach. So it's very simple. So this is actually slightly easier to implement. And that's sort of a little trick. So for local alignments, you can often reset to zero. Any questions about that? So well, we talked about computational efficiency, this big O notation, where you consider the number of individual computations that are required to run an algorithm as a function of the size of the input, basically the number of units in the problem base pairs, amino acid residues, whatever. So computer scientists look at the asymptotic worst case running time. That's either because they're pessimistic, or perhaps because they want to guarantee things. They want to say, it's not going to be worse than this. Maybe it'll be faster, and then you'll be happy. But I can guarantee you, it's not going to be worse than this. And so in this case, the algorithm we talked about was order n times n, where that's the lengths of the two sequences. So toward the end last time, we get we talked about this lambda parameter and said that lambda is the unique positive solution to this equation here, where sij are the scores and pi and rj are the nucleotide frequencies. And then there's this target frequency formula that comes up that says that if you use a scoring system sij to apply to sequences, and then you pull out just the high scoring segments, the ones that are unusually high scoring, they will have a frequency of matching nucleotides qij that's given by the product of the frequencies in the two sequences basically weighted by e to the lambda sij. So matches will occur more strongly, because that has a positive work, and mismatches less strongly. And that then gives rise to this notion that there's an optimal mismatch penalty, if you just consider scoring systems that have plus 1 for a match and m for a mismatch, some negative number, that's given by this equation here, and here I've worked out a couple of values. So the theory says that to find matches that are 99% identical, you should use a mismatched score of minus 3, but for 75% identical, you should use minus 1. And I asked you to think about does that make sense, or how is that true? So y is minus 3 better than minus 1 for finding nearly identical matches. Anyone have an idea or thought on this? There's some thoughts on the slide. But can anyone intuitively explain why this is true? Yeah, what's your name? AUDIENCE: Eric. PROFESSOR: Yeah, go ahead. AUDIENCE: With a mismatch penalty of minus 3, you actually need more steps climbing back up to get back to some local maximum. And therefore, you do require a longer stretch [INAUDIBLE] matches in order to get a significant hit. That's my guess as to why a score of m equals minus 3, a penalty, [INAUDIBLE] why it would be better at finding the higher identity matches. PROFESSOR: OK, because the minus 3 makes you go down faster, so it takes longer to recover, so you can only find nearly identical things with that kind of scoring system. Is that your point? OK, that's a good point. So yeah, when would you want to use a mismatched penalty of minus 1? AUDIENCE: When you're trying to look for things that are [INAUDIBLE], but maybe not so close. Well, when you're looking for a [INAUDIBLE], you're looking for [INAUDIBLE]. That kind of situation. PROFESSOR: And so let's say I'm using a mismatch penalty of minus 2. Can I find regions that are 66% identical? AUDIENCE: Probably. PROFESSOR: Probably. AUDIENCE: But not guaranteed. PROFESSOR: Anyone else have a comment on that? Match is plus 1. Mismatch is minus 2 regions of 66% identity. Yeah, with Levi, yeah. AUDIENCE: No, since your score will just be zero. PROFESSOR: Yeah. That's correct. So Levi's comment is your score will be zero. Well, I'll just say plus for match, plus, plus, minus, plus, plus, minus. I mean, it'll be interspersed. It doesn't have to be like this for every triplet. But but on average, you'll have two matches for every match. That's what 66% identity means. And so these will score a total of plus 2. And this will score minus 2. And so you basically will never rise much above zero. And so you you can't really use that mismatch penalty. There's a limit. 66% is sort of at a point where you can no longer see. You could potentially see things that are 75% identical if they were incredibly long with that kind of mismatch penalty. But you just can't see anything below 2/3% identity with minus 2. So to find those low things, you have to use the lower. You have to go down to minus 1 if you want to find the really weak matches. But they will have to be correspondingly very long in order to achieve statistical significance. So correspondingly, the reason why it's better to use a harsher mismatch penalty of minus 3 to find the nearly identical regions is that, in this equation, when you go from having a plus 1, minus 1 scoring system to plus 1, minus 3, lambda will change. This equation will no longer satisfied so that a new value of lambda will be relevant. And that value will be larger. That's not totally obvious from this equation because you sort of have one term, which is either the minus lambda in one term or either plus lambda. But it turns out that that making the mismatch penalty more negative will lead to a solution that's a bigger value of lambda. So that means that the same score, x, will lead to a larger negative exponent here. and? How will that affect the p-value? Anyone take us through this? It's a little bit convoluted with all these negative exponentials and stuff, but can someone explain to us how that affects the p-value? Same x. We're going to increase lambda. What happens to the p-value? This gets bigger, more negative. That means this e to the minus thing gets closer to 0. That means that this is inside an exponential. As that thing gets closer to 0, this whole term here gets closer to 1. Therefore you're subtracting it from 1. Therefore the p-value gets smaller, closer to 0, more significant. Does that made sense? So it's good to just work through how this equation works. So that's all I wanted to say about mismatch penalties for DNA. Any questions about that? So how do you actually use this in practice? So if you just Google "BLAST end," you'll get to this website. It's been set up at NCBI for about 20 years or so. And of course, it's gone through various iterations and improvements over the years. And if you look down at the bottom, there is a place where you can click and set the algorithm parameters. And there are a number of parameters you can set. Some of them affect the speed. But we're focused here mostly on parameters that will affect the quality, the nature of the alignments that you find. And so here, you can set not arbitrary penalties, but you can set within a range of standard mismatch penalties. You can do 1 minus 1, 1 minus 2, et cetera. So what about sequences that code for protein? So exons, for example. So you can search them with a nucleotide search, like BLAST. But it can often be the case that you'll do better if you first translate your exon into the corresponding amino acid sequence using a genetic code and then search that peptide. Now you may or may not know the reading frame of your exon a priori, or even know that it is an exon, so BLAST automatically will do this translation for you. So for example, with this DNA sequence, it'll translate in all three of the reading frames, leading to essentially this bag of peptides here, where sometimes you'll hit a stop code on, like right here. And then it just treats it as, OK, there's a little PR dipeptide there. And then there's a longer peptide here, [INAUDIBLE], and so forth. So it just makes these bags of peptides for each reading frame and searches all those peptides against some target, which can be approaching database or a DNA database, again, translated in all the reading frames. So the folks at NCBI have made all these different flavors of BLAST available. So BLASTP is for proteins. N is for nucleotides. And then the translating ones are called things like BLASTX for a nucleotide query against a protein database. TBLASTN for a protein query against a nucleotide database, which gets translated in all frames, or TBLASTX, where you translate both the nucleotide sequences in all frames. And then there's a number of other versions of BLAST which we probably won't discuss but that are well described in the textbook and other accessible online sources. Let me ask you this. So remember ESTs. So ESTs are segments of cDNAs that typically correspond to one ABI 3700 Sanger sequenc off of that cDNA, so one read, like 600 bases or so. So let's say you have some ESTs from chimp. And you don't have the chimp genome yet. So you're going to search them against human. What would you do? Would you use a translating search? Or would you use a BLASTN search? Or does it matter? Chimp is a 98% identical human, very high. Any ideas? Yeah, Tim. AUDIENCE: You could use a translating search, because you know that the cDNAs are at least coding for RNAs. And so if you just use a nucleotide search, then you're not going to have functional significance in terms of the alignment. But if it's going for a protein, then-- PROFESSOR: You mean you won't know whether it is the protein coding part of the cDNA or not? AUDIENCE: So I just mean that if you're looking between chimp and human, then you're expecting some sort of mismatch. But it's possible that it could be a functional mismatch. Then you know that the cDNA is maybe coding for a protein. Therefore, if the mismatch is between two similar amino acids, then that would be picked up by a translating search, but it would be skewed against it in a nucleotide search. PROFESSOR: OK, fair enough. But if you assume that the two genomes are, let's say, 97% identical, even in a non-coding region, which they're very high. I don't remember the exact percent, but very high. Then if you're searching 600 nucleotides against the genome, even if it's 95% identical, you'll easily find that under either. So either answer is correct, BLASTN or BLASTX. And the UTRs could only be found by-- if it happened that this was a sequence from a three prime UTR, you could only find that by BLASTN typically. What if it's a human EST against the mouse genome? So mouse exons are about 80% identical to human exons at the nucleotide level, typically. Any ideas? What kind of search would you do? BLASTN, BLASTX, or something else? TBLASTX. Yeah, go ahead. AUDIENCE: I have another question. What exactly is the kind of question we're trying to answer by doing this BLAST search? PROFESSOR: Oh, well, I was assuming you're just trying to find the closest homologous cDNA or exons in the genome-- exons, I guess, yeah, the exons. of the homologous gene. Yeah, that's a good question. Exons of a homologous gene. We've got a human EST going against the mouse genome. When do we do? AUDIENCE: I suggest BLASTP because-- PROFESSOR: Well, BLASTP, that's protein. This is a nucleotide sequence against nucleotide. So we can do BLASTN or TBLASTX, let's say. AUDIENCE: TBLASTX. PROFESSOR: TBLASTX. You translate your EST, translate the genome, search those peptides. TBLASTX, why? AUDIENCE: The nucleotide sequences may be only about 80% similarity, but the protein sequences functionally, due to the functional constraints, you might actually get higher similarities there. PROFESSOR: Yeah. It's exactly right. So they are about, on average, about 80% identical. It varies by gene. But a lot of those variations that occur are at the third side of the codon that don't effect the amino acid, because there's a lot of constraint on protein sequence. And so you'll do better, in general, with a translating search than with a nucleotide search. Although, they both may work. But you may find a more complete match with a translating search. That's good. Everyone got that? Sally, yeah. AUDIENCE: Is there a reason why you wouldn't use BLASTX and instead you use TBLASTX? PROFESSOR: Yeah, I just gave the example of searching against the genome. But you could search against the mouse proteome as well. You might or might not. It depends how well annotated that genome is. Mouse is pretty well annotated. Almost all the proteins are probably known. So you probably get it. But if you were searching against some more obscure organism, the chameleon genome or something, and it wasn't well annotated, then you might do better with searching against the genome, because you could find a new x on there. OK, good. Question yeah, go ahead. AUDIENCE: So when we do these translations, these nucleotide, amino acid things, do we get all frames? Do the algorithms to all frames? PROFESSOR: Yeah, all six frames. So three frames on the plus strand, and three frames on the reverse strand. Yeah. All right, great. So that's the end of local alignment, for the moment. And we're going to now move on to global alignment using two algorithms. For global alignment, Needleman-Wunch-Sellers, and then for gapped local alignment, Smith-Waterman. And toward the end, we're going to introduce the concept of amino acid substitution matrices. So the background for today, the textbook does a pretty good job on these topics, especially the pages indicated are good for introducing the PAM series of matrices. We'll talked a little bit today and a little bit next time. So why would we align protein sequences? So the most obvious reason is to find homologues that we might, then, want to investigate, or we might, for example, if you have a human protein and you find homologous mouse protein, and that mouse protein has known function from a knockout or from some biochemical studies, for example, then you can guess that the human protein will have similar function. So we often use this type of inference that sequence similarity implies similarity in function and/or structure. So how true is this? So it turns out, from a wide body of literature, that this inference that sequence similarity implies functional and structural similarity is almost always true when the sequence similarity is more than about 30% identity over the whole length of a protein, over 300, 400 amino acids. That's a good inference. Below that, sort of in the 20% to 30% sequence similarity, that's often referred to as the Twilight Zone, where sometimes it's a good inference, and sometimes it's not. So you need to be a little bit careful. And below that, it's deeper into the Twilight Zone, where most of the time you probably shouldn't trust it. But occasionally, you can see these very remote homologies. You might want to have additional information to support that kind of inference. And I want to just point out that the converse is just not true in biology. So structural similarity does not imply sequence similarity or even derivation from a common ancestor. So you may think, well, every protein has a really complex, elaborate three dimensional structure, and there's no way that could ever evolve twice. And it's true that probably that exact structure can never evolve twice. But a very similar structure, a similar fold even, in terms of the topology of alpha helices and beta strands, which Professor Frank will talk about later in the course, the identical fold can involve more than once. It's not that hard to evolve a pattern of alpha helices and beta strands. And so this point about structural similarity not implying sequence similarity, the way I think about it is like this, like here are two organisms. This is a hummingbird, you've all seen. And some of you may have seen this. This is a hawk moth, which is an insect that is roughly two inches long, beats its wings very fast, has a long tongue that sips nectar from flowers. So it basically occupies the same ecological niche as a hummingbird, and looks very, very similar to a hummingbird at a distance. From 10 or more feet, you often can't tell. This is an insect, and that's a bird. The last common ancestor was something that probably lived 500 million years ago, and certainly didn't have wings, and may not have had legs or eyes. And yet, they've independently evolved eyes and wings and all these things. So when there's selective pressure to evolve something, either a morphology or a protein structure, for example, evolution is flexible enough that it can evolve it many, many times. So here's an example from the protein structure world. This is homophilous iron binding protein. This is just the iron coordination center. And this is now a eukaryotic protein called lactoferrin. Turns out these guys are homologous. But eukaryotes and bacteria diverged 2 million years ago or so, so their ancestry is very, very ancient. And yet, you can see that in this iron coordination center, you have a tyrosine pointing into the iron here. And you have a histidine up here, and so forth. So the geometry has been highly conserved. It's not perfectly conserved. Like, here you have a carboxylate. And here you have a phosphate. So there's been a little bit of change. But overall, this way of coordinating iron has basically evolved independently. So although these are homologous, the last common ancestor bound anions-- that's known from [INAUDIBLE] construction. So they independently evolved the ability to bind cations, like iron. And here is actually my favorite example. So here's a protein called ribosome recycling factor. And that's its shape. So it's a very unusual shaped protein that's kind of shaped like an L. Does this remind anyone of anything, this particular shape? Have you seen this in another biomolecule at some point? AUDIENCE: [INAUDIBLE]. PROFESSOR: Something like [INAUDIBLE]. OK, could be. Any other guesses? How about this? That's a tRNA. So the 3D structure of tRNA is almost identical, both in terms of the overall shape and in terms of the geometry. Sorry, I'm having issues with my animations here. The geometry of these, they're both about 70 angstroms long. So why is that? Why would this protein evolve to have the same three dimensional shape as a tRNA? Any ideas? AUDIENCE: [INAUDIBLE]. PROFESSOR: [INAUDIBLE]. Right, exactly. It fits into the ribosome, and it's involved, when the ribosome is stalled, and basically releasing the ribosome. So it's mimicking a tRNA in terms of structure. And so the point about this is that, if you were to take a bunch of biomolecules and match them up using a structure comparison algorithm to find similar ones-- these two are clearly similar. And yet, they probably never had a common ancestor right, because one's an RNA in one's a protein. OK. So now what we're going to talk about a few different types of alignments. So we talked about local alignments, where you don't try to align the entire sequence of your query or your database. You just find smaller regions of high similarity. Global alignment, where you try to align the two proteins from end to end, you assume that these two proteins are homologous, and actually that they haven't had major insertions or rearrangements of their sequence. And then semi-global, which is sort of a little twist on global. And we'll talk about a few different scoring systems-- so ungapped, which we've been talking about until now, and then we'll introduce gaps of two types that are called linear and affine. And the nomenclature is a little bit confusing, as you'll see. They're both linear, in a sense. So a common way to represent sequence alignments, especially in the protein alignment-- you can do it for protein or DNA-- is what's called a dot matrix. Now we've got two proteins. They might be 500 amino acids long each, let's say. You write sequence one along the x-axis, sequence two along the y-axis. And then you make a dot in this matrix whenever they have identical residues, although probably there would be a lot more dots in this. So let's say, whenever you have three residues in a row that are identical-- OK, that's going to occur fairly rarely, since there's 20 amino acids. And you make that dot. And for these two proteins, you don't get any off diagonal dots. You just get these three diagonal lines here. So what does that tell you about the history of these two proteins? What's that right there? Sally. AUDIENCE: An insertion or deletion. PROFESSOR: An insertion or deletion. An insertion in which protein? AUDIENCE: Sequence two. PROFESSOR: Or a deletion in? AUDIENCE: One. PROFESSOR: OK. Everyone got that? OK, good. There's extra sequence in sequence two here that's not in sequence one. You don't know whether it's an insertion or deletion. It could be either one, based on this information. Sometimes you can figure that out from other information. So sometimes you call that an indel-- insertion or deletion. And then, what is this down here? Someone else? Insertion, I heard, insertion in sequence one or deletion in sequence two. OK, good. All right, so what type of alignment would be most appropriate for this pair sequences, a local or a global? AUDIENCE: I would do global, because they're very, very similar. [INAUDIBLE]. PROFESSOR: Yeah. They are quite similar across their entire lengths, just with these two major indels. So that's sort of the classical case where you want to do the global alignment. All right. So what about these two proteins? Based on this dot matrix, what can you say about the relation between these two, and what type of alignment would you want to use when comparing these two proteins? Yeah, what's your name? AUDIENCE: Sonia. PROFESSOR: Go ahead, Sonia. AUDIENCE: It looks like they've got similar domains, maybe. So local alignment might be better. PROFESSOR: And why wouldn't you do a global alignment? AUDIENCE: Local, because the local alignment might actually find those domains and tell you what they are. PROFESSOR: So a local alignment should at least find these two guys here. And why do these two parallel diagonal lines, what does that tell you? AUDIENCE: That the two different proteins have similar sequences, just in different parts of the protein, different areas relative to the start. PROFESSOR: Right. Yeah, go ahead. AUDIENCE: Doesn't it just basically mean that there's a section in sequence two that's in sequence one twice? PROFESSOR: Yeah, exactly. So this segment of sequence two, here-- sorry, having trouble, there we go, that apart-- is present twice in sequence one. It's present once from about here over to here, and then it's present once from here over to here. So it's repeated. So repeats and things like that will confuse your global alignment. The global alignment needs to align each residue-- or trying to align each residue in protein one to each residue in protein two. And here, it's ambiguous. It's not clear which part of sequence one to align to that part of sequence two. So it'll get confused. It'll choose one or the other. But that may be wrong, and that really doesn't capture what actually happens. So yeah, so here a local alignment would be more suitable. Good. So let's talk now about gaps, again, which can be called indels. In protein sequence alignments, or DNA, which many of you have probably seen, you often use maybe just a dash to represent a gap. So in this alignment here, you can see that's kind of a reasonable alignment, right? You've got pretty good matching on both sides. But there's nothing in the second sequence that matches the RG in the first sequence. So that would be a reasonable alignment of those two. And so what's often used is what's called a linear gap penalty. So if you have end gaps, like in this case two, you assign a gap penalty A, let's say. And A is a negative number. And then you can just run the same kinds of algorithms, where you add up matches, penalize mismatches, but then you have an additional penalty you apply when you introduce a gap. And typically, the gap penalty is more severe than your average mismatch. But there's really no theory that says exactly how the gap penalty should be chosen. But empirically, in cases where you should know the answer, where you have, for example, a structural alignment, you can often find that a gap penalty that's bigger than your average mismatch penalty is usually the right thing to do. So why would that be? Why would a gap penalty-- why would you want to set it larger than a typical mismatch? Any ideas? Yeah, what's your name? AUDIENCE: I'm Chris. PROFESSOR: Chris. AUDIENCE: Because having mutations that shift the frame or that one insert would have insertions or deletions is far more uncommon than just having changing [INAUDIBLE]. PROFESSOR: Mutations that create insertions and deletions are less common than those that introduce substitutions of residues. Everyone got that? That's true. And do you know by what factor? AUDIENCE: Oh, I couldn't give you a number. PROFESSOR: So I mean, this varies by organism, and It varies by what type of insertion you're looking at. But even at the single nucleotide level, having insertions is about an order of magnitude less common than having a substitution in those lineages. And here, in order to get an amino acid insertion, you actually have to have a triplet insertion, three or six or some multiple of three into the exon. And that's quite a bit less common. So they occur less commonly. A mutation occurs less commonly, and therefore the mutation is actually accepted by evolution even less commonly. And an alternative is a so-called affine gap penalty, which is defined as G plus n lambda. So n is the number of gaps, and then G is what's called a gap opening penalty. So the idea here is that basically the gaps tend to cluster. So having an insertion is a rare thing. You penalize that with G. But then, if you're going to have an insertion, sometimes you'll have a big insertion of two or three or four codons. A four codon insertion should not be penalized twice as much as a two codon insertion, because only one gap actually occurred. And when you have this insertion event, it can any variety of sizes. You still penalize more for a bigger gap than for a smaller gap, but it's no longer linear. I mean, it's still a linear function, just with this constant thing added. So these are the two common types of gap penalties that you'll see in the literature. The affine works a little bit better, but it is a little bit more complicated to implement. So sometimes you'll see both of them used in practice. And then, of course, by changing your definition of gamma, you could have a G plus n minus 1. So that first gap would be G, and then all the subsequent gaps would gamma. So you're not going to have to double score something. All right. OK. You've got two proteins. How do you actually find the optimal global alignment? Any ideas on how to do this? So we can write one sequence down one axis, one down the other axis. We can make this dot plot. The dot plot can give us some ideas about what's going on. But how do we actually find the optimal one where we want to start from the beginning? In the end, we're going to write the two sequences one above the other. And if the first residue or the first sequence is n, and maybe we'll align it to here, then we have to write the entire sequence here all the way down to the end. And below it has to be either a residue in sequence two or a gap. And again, we can have gaps up here. So you have to do something. You have to make it all the way from the beginning to the end. And we're just going to sum the scores of all the matching residues, of all the mismatching residues, and of all the gaps. How do we find that alignment? Chris. AUDIENCE: Well, since we're using dynamic programming, I'm guessing that you're going to have to fill out a matrix of some sort and backtrack. PROFESSOR: And so when you see the term dynamic programming, what does that mean to you? AUDIENCE: You're going to find solutions to sub problems until you find a smaller solution. Then you'd backtrack through what you've solved so far to find the global sequence. PROFESSOR: Good. That's a good way of describing it. So what smaller problems are you going to break this large problem into? AUDIENCE: The smaller sub-sequences. PROFESSOR: Which smaller sub-sequences? Anyone else? You are definitely on the right track here. Go ahead. AUDIENCE: I mean, it says at the top there, one sequence across the top and one down the side. You could start with just the gap versus the sequence and say your gap will increase as you move across. Basically, each cell there could be filled out with information from some of its neighbors. So you want to make sure that you fill out old cells in some order so that we can proceed to the next level with what we've [? written down. ?] PROFESSOR: So if you had precisely to find a sub problem where you could see what the answer is, and then a slightly larger sub problem whose solution would build on the solution that first one, where would you start? What would be your smallest sub problem? AUDIENCE: I'd start with the top row, because you could just the gap versus gap, and then move in the row, because you don't need anything above that. PROFESSOR: And then what's the smallest actual problem where you actually have parts of the protein aligned? AUDIENCE: One row in column two, basically. If it's a match, you have some score. And if it's a mismatch, you have some other score. And you want the best possible one in each block. PROFESSOR: Yeah, OK. Yeah. That's good. So just to generalize this-- hopefully this is blank-- in general, you could think about we've got, let's say, 1 to n here, and a sequence 1 to n here. You could think about a position i here and a position j here. And we could say finding the optimal global alignment, that's a big problem. That's complicated. But finding an alignment of just the sequence from 1 to i in the first protein against the sequence from 1 to j in the second protein, that could be pretty easy. If i is 2 and j is 2, you've got a dipeptide against a dipeptide. You could actually try all combinations and get the optimal alignment there. And so the idea, then, is if you can record those optimal scores here in this matrix, then you could build out, for example, like this, and find the optimal alignments of increasingly bigger sub problems where you add another residue in each direction, for example. Does that makes sense to you? The idea of a dynamic programming algorithm is it's a form of recursive optimization. So you first optimize something small, and then you optimize something bigger using the solution you got from that smaller piece. And the way that that's done for protein sequences in Neeleman-Wunsch is to, as we were saying, first consider that there might be a gap in one aligning to a residue in the other. So we need to put these gaps down across the top and down the side. This is a linear gap penalty, for example. And so here would be how you start. And this is a gap penalty, obviously, of minus 8. So if you're the optimal solution that begins with this V in the top sequence aligned to this gap in the vertical sequence, there's one gap there, so it's minus 8. And then if you want to start with two gaps against this V and D, then it's minus 16. So that's how you would start it. So you start with these problems where there's no options. If you have two gaps against two residues, that's minus 16. By our scoring system, it's unambiguous. So you just can fill those in. And then you can start thinking about, what do we put right here? What score should we put right there? Remember, we're defining the entries in this matrix as the optimal score of the sub-sequence of the top protein up to position i against the vertical protein up to position j. So that would be the top protein position one up to the vertical protein position one. What score would that be? What's the optical alignment there that ends position one both sequences? It'll depend on your scoring system. But for a reasonable scoring system, that's a match. That's going to get some positive score. That's going to be better than anything involving a gap in one against a gap in the other or something crazy like that. So that's going to get whatever your VV match score is. This is your Sij from your scoring matrix for your different amino acids. And then, basically the way that this is done is to consider that when you're matching that position one against position one, you might have come from a gap before in one sequence or a gap in the other sequence, or from a match position in the other sequence. And that leads to these three arrows. I think it gets clear if I write up the whole algorithm here. So Sij is the score of the optimal alignment ending at position i in sequence one and position j in sequence two. Requires that we know what's above, to the left, and diagonally above. And you solve it from the top and left down to the bottom and right, which is often called dynamic programming. And let's just look at what the recursion is. So Needleman and Wunsch basically observed that you could find this optimal global alignment score by filling in the matrix by at each point taking the maximum of these three scores here. So you take the maximum of the score that you had above and to the left, so diagonally above, plus sigma of xi yj. Sigma, in this case, is the scoring matrix that you're using that's 20 by 20 that scores each amino acid against each other amino acid residue. You add that score if you're going to move diagonally to whatever the optimal score was there, or if you're moving to the right or down, you're adding a gap in one sequence or the other. So you have to add A, which is this gap penalty, which is a negative number, to whatever the optimal alignment was before. I think it's maybe easier if we do an example here. So here is the PAM250 scoring matrix. So this was actually developed by Dayhoff back in the '70s. This might be an updated version, but it's more or less the same as the original. Notice, it's a triangular matrix. Why is that? AUDIENCE: It's symmetric. PROFESSOR: It's symmetric, right. So it has a diagonal. But then everything below the diagonal, it would be mirrored above the diagonal, because it's symmetric. Because you don't know when you see a valine matched to a leucine, it's the same as a leucine matched to a valine, because it's a symmetrical definition of scoring. And here are two relevant scores. So notice that VV has a score of plus 4 in this matrix. And over here, VD has a score of minus 2. So I'll just write those down. Anyone notice anything else interesting about this matrix? We haven't said exactly where it comes from, but we're going to. Yeah, what's your name? AUDIENCE: Michael. PROFESSOR: Go ahead. AUDIENCE: Not all the diagonal values are the same. PROFESSOR: Not all diagonals are the same. In fact, there's a pretty big range, from 2 up to 17, so a big range. And anything else? OK, I'm sorry, go ahead. What's your name? AUDIENCE: Tagius. PROFESSOR: Tagius, yeah. Go ahead. AUDIENCE: There are positive values for things that are not the same? PROFESSOR: Yeah. So all the diagonal terms are positive. So a match of any particular residue type to its identical residue is always scored positively, but with varying scores. And there are also some positive scores in the off diagonal. And where are those positive scores occurring? Notice they tend to be to nearby residues. And notice the order of residues is not alphabetical. So someone who knows a lot about amino acids, what can you see about these scores? Yeah, go ahead. AUDIENCE: I think these amino acids [INAUDIBLE] based on their [INAUDIBLE]. PROFESSOR: So the comment was that the residues have been grouped by similar chemistry of their side chains. And that's exactly right. So the basic residues, histidine, arginine, and lysine, are all together. The acidic residues, aspartate and glutamate, are here, along with asparagine and glutamine. And notice that D to E has a positive score here. It's 3. It's almost as good as D to D or E to E, which are plus 4. So recognizing that you can often substitute in evolution an aspartate for a glutamate. So yeah, so it basically, to some extent, is scoring for similar chemistry. But that doesn't explain why, on the diagonal, you have such a large range of values. Why is a tryptophan more like a tryptophan than a serine is like a serine. Tim, you want to comment? AUDIENCE: Perhaps it's because tryptophans occur very rarely in all proteins [INAUDIBLE]. So if you've got two [INAUDIBLE], that's a lot rarer of an occurrence and [INAUDIBLE]. PROFESSOR: So Tim's point was that tryptophans occur rarely, so when you see two tryptophans aligned, you should take note of it. It can anchor your alignment. You can be more confident in that. Sally. AUDIENCE: Well, tryptophans are also incredibly bulky, and also have the ability to make electric interactions, electro-static interactions. PROFESSOR: Not really electro-static, you would say, more-- [INTERPOSING VOICES] AUDIENCE: Yes. But they do have a lot of abilities to interact with other side chains. And cysteines contribute very, very strongly to the three dimensional structure of the protein. PROFESSOR: Why is that? AUDIENCE: Well, because they can form [INAUDIBLE]. PROFESSOR: OK. Yeah. So maybe you don't put your tryptophans and your cysteines into your protein by chance, or you only put them when you want them, when there's enough space for a tryptophan. And when you substitute something smaller, it leaves a gap. It leaves a 3D spatial gap. And so you don't want that. You don't get good packing. When you have cysteines, they form disulfide bonds. If you change it to something that's non-cysteine, it can form that anymore. That could be disruptive to the overall fold. So those ones tend to be more conserved in protein sequence alignments, absolutely. Whereas, for example, if you look at these hydrophobics, the MILV group down here, they all have positive scores relative to each other. And that says that, basically, most the time when those are used-- I mean, there are sometimes when it went really matters. But a lot of time, if you just want a transmembrane segment, you can often substitute any one of those at several positions and it'll work equally well as a transmembrane segment. So these are not random at all. There's some patterns here. So let's go back to this algorithm. So now, if we're going to implement this recursion, so we fill on the top row and the left column, and then we need to fill in this first. I would argue the first interesting place in the matrix is right here. And we consider adding a gap here. When you move vertically or horizontally, you're not adding a match or adding a match. So from this position, this is sort of the beginning point. It doesn't actually correspond to a particular position in the protein. We're going to add now the score for VV. And we said that VV, you look it up in that PAM matrix, and it's plus 4. So we're going to add 4 there to 0. And so that's clearly bigger than minus 16, which is what you get from coming above or coming from the left. So you put in the 4. And then you also, in addition to putting that 4 there, you also keep the arrow. So there's that red arrow. We remember where we came from in this algorithm. Because someone said something about backtracking-- I think Chris-- so that's going to be relevant later. So we basically get rid of those two dotted arrows and just keep that red arrow as well as the score. And then we fill in the next position here. And so to fill in this, now we're considering going to the second position in sequence one, but we're still only at the first position in sequence two. So if we match V to V, then we'd have to add, basically, a gap in one of the sequences. Basically it would be a gap in sequence two. And that's going to be minus 8. So you take 4, and then plus minus 8, so it's negative 4. Or you could do minus 8 and then plus negative 2, if you want to start from a gap and then add a DV mismatch there, because minus 2 was the score for a DV mismatch. Or again, you can start from a gap and then add another gap. OK, does that make sense? So what is the maximum going to be? Negative 4. And the arrow is going to be horizontal, because we got some bonus points for that VV match, and now it's carrying over. We're negative, but that's OK. We're going to just keep the maximum, whatever it is. All right, so it's minus 4, and the horizontal arrow. And then here's the entire matrix filled out. And you'll have a chance to do this for yourself on problem set one. And I've also filled in arrows. I haven't filled in all the arrows, because it gets kind of cluttered. But all the relevant arrows here are filled in, as well as some irrelevant arrows. And so then, once I fill this in, what do I do with this information? How do I get an actual alignment out of this matrix? Any ideas? Yeah, what's your name? AUDIENCE: [INAUDIBLE]. PROFESSOR: Yeah, so what he said is start at the bottom right corner and go backwards following the red arrows in reverse. Is that right? So why the bottom right corner? What's special about that? AUDIENCE: [INAUDIBLE]. PROFESSOR: Yeah. It's a score of the optimal alignment of the entire sequence one against the entire sequence two. So that's the answer. That's what we define as the optimal global alignment. And then you want to know how you got there. And so how did we get there? So the fact that there's this red arrow here, what does that red arrow correspond to specifically? AUDIENCE: [INAUDIBLE]. PROFESSOR: Right. In this particular case, for this particular red arrow, remember the diagonals are matches. So what match is that? AUDIENCE: [INAUDIBLE]. PROFESSOR: Yeah, that's a Y to Y match, right? Can everyone see that? We added Y to Y, which was plus 10, to whatever this 13 was and got 23. OK, so now we go back to here. And then how do we get here? We came from up here by going this diagonal arrow. What is that? What match was that? That's a cysteine-cysteine match. And then how do we get to this 1? We came vertically. And so what does that mean? AUDIENCE: [INAUDIBLE]. PROFESSOR: We inserted a gap, in which sequence? The first one. The second one? What do people think? Moving down. AUDIENCE: [INAUDIBLE]. PROFESSOR: Yeah, the top one. And so that got us to here. Here's a match, plus 2 for having a serine-serine match. Here's a plus 3 for having a D to E mismatch. But remember, that's those are chemically similar, so they get a positive score. And then this is the V to V. So can you see? I think I have the optimal alignment written somewhere here, hopefully, there. That's called the trace back. And then that is the alignment. OK, we align the Y to the Y, the C to the C. Then we have basically a gap in this top sequence-- that's that purple dash there-- that's corresponding to that L. And you can see why we wanted to put that gap in there, because we want these S's to match, and we want the C's to match. And the only way to connect those is to have a gap in the purple. And the purple was shorter than the green sequence anyway, so we kind of knew that there was going to be a gap somewhere. And good. And that's the optimal alignment. So that's just some philosophy on Needlemen-Wunsch alignments. So what is semi-global alignment? You don't see that that commonly. It's not that big a deal. I don't want to spend too much time on it. But it is actually reasonable a lot of times that, let's say you have a protein that has a particular enzymatic activity, and you may find that the whole, the bulk of the protein is well conserved across species. But then at the N and C termini, there's a little bit of flutter. You can add a few residues or delete a few residues, and not much matters at the N and C termini. Or it may matter not for the structure but for, you know, you're adding a single peptide so it'll be secreted, or you're adding some localization signal. You're adding some little thing that isn't necessarily conserved. And so a semi-global alignment, where you use the same algorithm, except that you initialize the edges of the dynamic programming matrix to 0, instead of the minus 8, minus 16 whole gap, and go to 0. So we're not going to penalize for gaps of the edges. And then, instead of requiring the trace back to begin at the bottom right, Smn, you allow it to begin at the highest score in the bottom row or the rightmost column. And when you do the trace back as before, these two changes basically find the optimal global alignment but allowing arbitrary numbers of gaps at the ends and just finding the core match. It has to go basically to the end of one or the other sequences, but then you can have other residues hanging off the end on the other sequence, if you want, with no penalty. And this sometimes will give a better answer, so it's worth knowing about. And it's quite easy to implement. Now what about gapped local alignments? So what if you have two proteins? Do you remember those two proteins where we had the two diagonal? I guess they were diagonal lines. How where they? Something like that. Anyway, diagonal lines like that. So where in this protein on the vertical, there is a sequence here that matches two segments of the horizontal protein. So for those two, you don't want to do this global alignment. It'll get confused. It doesn't know whether to match this guy to this or this other one to the sequence. So you want to use a local alignment. So how do we modify this Needleman-Wunsch algorithm to do local alignment? Any ideas? It's not super hard. Yeah, go ahead. AUDIENCE: If the score is going to go negative, instead of putting a negative score, you just put 0, and you start from where you get the highest total score, rather than the last column or last row. Start your traceback from the highest score. PROFESSOR: So whenever you're going negative, you reset to 0. Now what does that remind you of? That's the same trick we did write previously with ungapped local alignment. So you reset to 0. And that's as a no penalty, because if you're going negative, it's better just to throw that stuff away and start over. We can do that because we're doing local alignment. We don't have to align the whole thing. So that's allowed. And then, rather than going to the bottom right corner, you can be anywhere in the matrix. You look for that highest score and then do the traceback. That's exactly right. So it's not that different. There are a few constraints, though, now on the scoring system. So if you think about the Needleman-Wunsch algorithm, we could actually use a matrix that had all positive scores. You could take the PAM250 matrix. And let's say the most negative score there is, I don't know, like minus 10 or something, and you could just add 10, or even add 20 to all those score. So they're all positive now. And you could still run that algorithm. And it would still produce more or less sensible results. I mean, they t wouldn't be as good as the real PAM250, but you would still get a coherent alignment out of the other end. But that is no longer true when you talk about the Smith-Waterman algorithm, for the same reason that an ungapped local alignment, we had to require that the expected score be negative, because you have to have this negative drift to find small regions that go in the positive. So if you have this rule, this kind of permissive that says, whenever we go negative we can just reset to 0, then you have to have this negative drift in order for positive scoring stuff to be unusual. All right, so that's another constraint there. You have to have negative values for mismatches-- I mean, not all mismatches. But if you took two random residues in alignment, the average score has to be negative. I should probably rephrase that, but more or less. And here's an example of Smith-Waterman. So you right zeroes down the left side and across the top. And that's because, remember, if you go negative, you reset to 0. So we're doing that. And then you take the maximum of four things. So coming from the diagonal and adding the score of the match, that's the same as before. Coming from the left and adding a gap in one sequence, coming from above and adding a gap in the other sequence, or 0. This "or 0" business allows us to reset to 0 if we ever go negative. And when you have a 0, you still keep track of these arrows. But when you have a 0, there's no arrow. You're starting it. You're starting the alignment right there. So that's Smith-Waterman. It's helpful. I think on problem set one, you'll have some experience thinking about both Needleman-Wunsch and Smith-Waterman. They sort of behave a little bit differently, but they're highly related. So it's important to understand how they're similar, how they're different. And what I want to focus on for the remainder of this lecture is just introducing the concept of amino acid similarity matrices. We saw that PAM matrix, but where does it come from? And what does it mean? And does it work well or not, and are there alternatives? So we could use this identity matrix. But as we've heard, there are a number of reasons why that may not be optimal. For example, the cysteines, we should surely score them more, because they're often involved in disulfide bonds, and those have major structural effects on the protein and are likely to be conserved more than your average leucine or alanine or whatever. So clearly, scoring system should favor matching identical or related amino acids, penalize poor matches and for gaps. And there's also an argument that can be made that it should have to do with how often one residue is substituted for another during evolution. So that commonly substituted thing should have either positive scores or less negative scores than rarely substituted things. And perhaps not totally obvious, but it is if you think about it for a while, is that any scoring system that you dream up carries with it an implicit model of molecular evolution for how often things are going to be substituted for each other. So it's going to turn out that the score is roughly proportional to a [INAUDIBLE] score for the occurrence of that pair of residues, divided by how often it would occur by chance, something like that. And so that if you assign positive scores to things, to certain pairs of residues, you're basically implying that those things will commonly interchange during evolution. And so if you want to have realistic, useful scores, it helps to think about what the implicit evolutionary model is and whether that is a realistic model for how proteins evolve. So I'm going to come to Dayhoff. And so unlike later matrices, she had an explicit evolutionary model, like an actual mathematical model, for how proteins evolve. And the idea was that there are going to be alignments of some proteins. And keep in mind, this was in 1978. So the protein database probably had like 1,000 proteins in it, or something. It was very, very small. But there were some alignments that were obvious. If you see two protein segments of 50 residues long that are 85 identical, there's no way that occurred by chance. You don't even need to do statistics on that. So you're sure. So she took these very high confidence protein sequence alignments, and she calculated the actual residue residue substitution frequencies, how often we have a valine in one sequence as a substitute for a leucine. And it's actually assumed it's symmetrical. Again, you don't know the direction. And calculated these substitution frequencies. Basically estimated what she called a PAM1 one matrix, which is a matrix that implies 1% divergence between proteins. So there's, on average, only a 1% chance any given residue will change. And the real alignments had greater divergence than that. They had something like 15% divergence. But you can look at those frequencies and reduce them by a factor of 15, and you'll get not exactly 15 but something like 15. And you'll get something where there's a 1% chance of substitution. And then once you have that model for what 1% sequence substitution looks like, turns out you can just represent that as a matrix and multiply it up to get a matrix that describes what 5% sequence substitution looks like, or 10% or 50% or 250%. So that PAM250 matrix that we talked about before, that's a model for what 250% amino acid substitution looks like. How does that even make sense? How can you have more than 100%? Is anyone with me on this? Tim, yeah. AUDIENCE: Because it can go back. So it's more likely, in some cases, that you revert rather than [INAUDIBLE]. PROFESSOR: Right. So a PAM10 matrix means, on average, 10% of the residues have changed. But a few of those residues might have actually-- so maybe about 90% won't have changed at all. Some will have changed once, but some might have even changed twice, even at 10%. And when you get to 250%, on average, every residue has changed 2 and 1/2 times. But again, a few residues might have remained the same. And some residues that change-- for example, if you had an isoleucine that mutated to a valine, it might have actually changed back already in that time. So it basically accounts for all those sorts of things. And if you have commonly substituted residues, you get that type of evolution happening. All right. So she took these protein sequence alignments-- it looks something like this-- and calculated these statistics. Again, I don't want to go through this in detail during lecture, because it's very well described in the text. But what I do want to do is introduce this concept of a Markov chain, because it's sort of what is underlying these Dayhoff matrices. So let's think about it. We'll do more on this next time. But imagine that you were able to sequence the genomes of cartoon characters with some newly developed technology and you chose to analyze the complicated genetics of the Simpson lineage. I'm assuming you all know these people. This is Grandpa and Homer eating doughnut and his son, Bart. So imagine this is Grandpa's genome at the apolipoprotein A locus. And a mutation occurred that he then passed on to Homer. So this mutation occurred in the germ line, passed on to Homer. And then when Homer passed on his genes to Bart, another mutation occurred here, changing this AT pair to a GC pair in Bart. So this, I would argue, is a type of Markov chain. So what is a Markov chain? So it's just a stochastic process. So a stochastic process is a random process, is sort of the general meaning. But here we're going to be dealing with discrete stochastic processes, which is just a sequence of random variables. So X1 here is a random variable that represents, for example, the genome of an individual, or it could represent the genotype, in this case, at a particular position, maybe whether it's an A, C, G, or T at one particular position in the genome. And now the index here-- one, two, three, and so forth-- is going to represent time. So X1 might be the genotype in Grandpa Simpson at a particular position. And X2 might be the genotype of Homer Simpson. And X3 would be the genotype in the next generation, which would be Bart Simpson. And what a Markov chain is is it's a particular type of stochastic process that arises commonly in natural sciences, really, and other places all over the place. So it's a good one to know that has what's called the Markov property. And that says that the probability that the next random variable, or the genotype at the next generation, if you will-- so Xn plus 1 equals some value, j, which could be any of the possible values, say any of the four bases, conditional on the values of X1 through Xn, that is the entire history of the process up to that time, is equal to the conditional probability that Xn plus 1 equals j given only that little xn equals some particular value. So basically what it says that if I tell you what Homer's genotype was at this locus, and I tell you what Grandpa Simpson's genotype was at that locus, you can just ignore Grandpa Simpson's. That's irrelevant. It only matters what Homer's genotype was for the purpose of predicting Bart's genotype. Does that make sense? So it really doesn't matter whether that base in Homer's genome was the same as it was in Grandpa Simpson's genome, or whether it was a mutation that's specific to Homer, because Homer is the one who passes on DNA to Bart. Does that make sense? So you only look back one generation. It's a type of memoryless process, that you only remember the last generation. That's the only thing that's relevant. And so to understand Markov chains, it's very important that you all review your conditional probability. So we're going to do a little bit more on Markov chains next time. P A given B, what does that mean? If you don't remember, look it up in the Probability and Statistics, because that's sort of essential to Markov chains. So next time we're going to talk about comparative genomics, which will involve some applications of some of the alignment methods that we've been talking about. And I may post some examples of interesting comparative genomic research papers, which are going to be optional reading. You may get a little more out of the lecture if you read them, but it's not essential.
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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 look at a typical application of Newton's second law for a system of objects. So what I want to consider is a system of pulleys and masses. So I'll have a fixed surface here, a ceiling. And from the ceiling, we'll hang a pulley, which I'm going to call pulley A. And this pulley will have a rope attached to it, wrapped around it. And here we have object 1. And the rope goes around the pulley. And now it's going to go around another pulley B and fixed to the ceiling. So that's fixed. And hanging from pulley B is another mass 2. And our goal in applying Newton's second law is to find the accelerations of objects 1 and 2. Now, how do we approach this? Well, the first thing we have to do is decide if we're going to apply Newton's second law, what is the system that we'll apply it to? And there's many different ways to choose a system. When we look at this problem, we'll have several different systems. So let's consider the ones that we're going to look at. And the first one is very simple. It will be block 1. And the second system that we look at, we'll call that AB is pulley A. Now that brings us to an interesting question about pulley B and block 2, because we could separately look at pulley B, and we could separately consider block 2, or we consider them together. And I want to first consider separately pulley B and block 2. Now in some ways when you're looking at a compound system, and it has four objects, it makes sense to apply Newton's second law to each object separately. But we'll pay careful attention to the fact that object 2 is connected to pulley B. And eventually, we'll see that we can combine these two things. So the next step is once we have identified our object is to draw a free body force diagrams for each of the objects. So in order to do that, let's start with object 1. And we want to consider the forces on object 1. Now that brings us to our first issue about what types of assumptions we're making in our system. For instance, we have a rope that's wrapped around this pulley. And we have two pulleys that in principle could be rotating. But what we'd like to do to simplify our analysis-- so let's keep track of some assumptions here. Our first assumption will be that the mass, MP, of pulley A and the mass of pulley B are approximately zero. Now the reason for that is that we're not going to consider any of the fact that these objects have to be put into rotational motion. Later on in the course, we'll see that this will give us a more complicated analysis. We're also going to assume that our rope is not slipping. So the rope is actually is just slipping on the pulleys. So what that means is it's just the rope is sliding as the objects move. Now again, what this is going to imply is that the tension in the rope-- this rope is also slipping. And the rope is massless as well. It's very light rope. And all of these assumptions we've seen when we analyze ropes tell us that the tension T is uniform in the rope. So that's our first assumption. And we need to think about this before we even begin to think about the forces on the object. And now we can draw our forces. What do we have? We have the gravitational force on object 1. And now we can identify the tension pulling in the string, pulling object 1 up. Now for every time we introduce a free body diagram, recall that we have to choose what we mean by positive directions. And in this case, I'm going to pick a unit vector down, j hat 1 down. So that's my positive direction for force. Now before I write down all of Newton's laws, I'll just write down our various force diagrams. So for pulley A, I have two strings that are pulling it downwards. So I have tension and tension. And this string, I'm going to call that T2, is holding that pulley up. So we have the force diagram. Now I could write MAG, but we've assumed that the pulley is massless. And again, I'll call j hat A down. For object 2, let's do pulley B first. Now what are the forces on pulley B? I have strings on both sides, T. Pulley B is massless, so I'm not putting gravitational force. And this string is pulling B downwards, so that's T3. And again, we'll write j hat B downwards. And finally, I have block 2. So I'll draw that over here. I'll write block 2. In fact, let's say a little space here. We'll have j hat B downwards. Now block 2, what do we have there? We have the string pulling up block 2, which we've identified as T3. And we have the gravitational force on block 2 downward, M2 g. And there we have j had 2. So I've now drawn the free body diagram of the various objects. And that enables me to apply Newton's second law for each of these objects. So let's begin. We'll start with object 1. We have-- remember in all cases, we're going to apply F equals m a. So for object 1, we have m1g positive downward minus T is equal to m1 a1. And that's our F equals m a on object 1. So sometimes we'll distinguish that the forces we're getting from our free body diagram. And A is a mathematical description of the motion. For block 2, we have m2g minus T3 is equal to m2 a2. And now for pulley A, we have 2T pointing downwards minus T2 going upwards. And because pulley A is massless, this is zero even though pulley A may be-- it's actually fixed too. So it's not even accelerating. And what we see here is this equation-- I'm going to quickly note that it tells us that the string holding pulley 2 up, T2, is equal to 2T. So we can think of if, we want to know what T2 is, we need to calculate T. And finally we have B. And what is the forces on B? We have T3 minus 2T. And again pulley B is 0. And so we see that T3 is equal to 2T. Now if you think about what I said before about combining systems, if we combine pulley B in block 2, visually what we're doing is we're just adding these to free body diagram together. When we have a system B and block 2. Let's call this j hat downwards. And when we add these free body diagram together, you see that the T3 is now internal force to the system. It cancels in pair by Newton's second law. And all we have is the two strings going up, so we have T and T. And we have the gravitational force downward. And separately, when we saw that T3 equals 2T and we apply it there, then if we consider a system B2, and look at our free body diagram, we have m2g minus 2T-- and notice we have the same result their 2T equals m2 a2. So in principle now-- and I'll outline our equations. We have equation 1. We have equation 2. And in these two equations, we have three unknowns, T, A1, and A2, but only two equations. And so you might think, what about this missing third equation here? However, in this equation, we have a fourth unknown, T3. And this equation is just relating to T and T3. So in principle, we would have four unknowns and three equations. Or if we restrict our attention to these two equations, we have three unknowns and two equations. Are unknowns T, A1 and A2. These are our unknowns. And now our next step is to try to figure out what is the missing condition that's relating the sum of these unknowns. And that will be a constraint condition that we'll analyze next.
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https://ocw.mit.edu/courses/7-013-introductory-biology-spring-2013/7.013-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. HAZEL SIVE: All right, let's look at some of your questions. A bunch of them are-- you know what, I'm probably going to use that screen most of the time. Because this one's not fitting. But let's look on this screen. Your questions really focused around IPS cells and the kind of magic of IPS cells. And there were a couple of major questions that came up. What's the problem with using IPS cells therapeutically? Well, there are a number. One, they're so new that really we don't know what kinds of cell types these IPS cells can make, so it's not clear how to use them. Another problem actually, which my colleague, Professor Jaenisch works on is the question of how to actually grow these cells. I have a voice issue this morning. So you know, the quieter you are, the more you'll be able to hear my words. IPS cells, human cells, grow very slowly in the laboratory. And it's very difficult to grow them. So there are some basic questions in biology, as to how to grow these cells to enough numbers that would actually be useful. But here's another one that's very important. The transcription factors that we use to convert adult cells into IPS cells included an oncogene called c-Myc. And Myc is the kind of gene you don't want floating around your body nice and active in your cells, because it will give you cancer. So the challenge is how to actually use Myc and other potent transcription factors to turn adult cells into stem cells but not have the stem cells give the recipient cancer. And there's some very clever ways that people are trying to get around this. It is really impossible to teach you, if there are groups of you talking. I just cannot do it. So those of you talking, please don't. Thank you. All right, so what's the big deal about IPS cells? Well, the big deal is they can be your own. In theory-- and maybe in practice, a decade, five years maybe from now-- your own cells, your skin cells could be removed from you, could be dealt with in the laboratory, turned into your own stem cells, and put back into your own body, and they'd be your own. So of course, they wouldn't be rejected by you. And that is an enormous deal. They're called autologous cells. There's an ethical issue, in that you don't have to harvest any embryos to get stem cell lines. And that's a big deal. And here's a conceptual deal. When we talked about development, we talked about this directional pathway, where uncommitted cells became committed became differentiated. It was really thought that that was a one-way pathway. But what IPS cell technology has shown-- let's look at this-- all right, we can get the idea from one of these two screens. What IPS cell technology has shown is that you can reverse differentiation. You can take differentiated adult cells and with the right regulatory factors, you can turn them into stem cells. And so conceptually, that's a big deal. Very good, great questions. My next office hours are on Monday, 12-1. Come along or email me. But now we're going to turn to a new module, and that is the nervous system. And this is where we are in the course. You've done all the foundational material. We've talked about formation. And now we're going to talk about systems and the nervous system in particular. Let's start with the question of what a system is. Building up, the complexity of what we're talking about in life. A system refers to many organs that work together with one common function. Many organs working together with one overall function. And you will talk about two in the course. You'll talk about the nervous system and the immune system with Professor Jacks. The nervous system, which is the topic of the next three lectures, has to do with communication-- has to do with communication from the outside to the inside of an animal's body, has to do with communication within the body, and has to do with communication to get the organism-- the animal, because plants don't have nervous system-- the animal to do something. So the nervous system is about communication to or from or within the body of an animal. We've had a number of electrical analogies in the course. You will remember the famous signaling analogy of turning on the light switch over there on the wall. But now we move on to another electrical analogy, which is actually a truer one. And I'm going to use the analogy for the nervous system of talking about wires that transmit the signal, by which cells communicates. I'm going to talk about connectors between the wires. And then I'm going to talk about circuits. And those are going to be the topics of the next three lectures. So wires, connectors, and circuits. And those will be nervous system 1 to 3, the lectures in this module. Today, we're going to talk about the wires. And there will be three topics. The first is the cell type that's got something to do with the wiring for this communication. The second is something called the action potential, which is the signal by which cells communicate within themselves. And the third has to do with the ion channels and pumps. But let's start by phrasing the problem in a kind of a cool way. If you look in the human-- let's look on this screen. If you look in the human, and you outline the nerves. And if you went to see the living, plasticized human exhibits, you would have seen the plasticized nerves. They're really extraordinary. The network of nerves, which are the unit of communication throughout the body. Is enormous. And it embraces every single part of the body, almost. The cell that's involved in communication-- we'll draw this out in a moment-- is the neuron. And one of the things about neurons is that they've got very long processes, called axons, that we will deal with in depth today. But let's phrase things more intuitively. This is what our brains look like. They in fact do not have nerves innovating them. And so if you actually have brain surgery, you can really be awake during surgery. Because there are no nerves within the brain, or at least no pain receptors within the brain. Of course, the brain has nerves, but there are no pain receptors in the brain. Human brain. Billions and billions and trillions of neurons. We'll talk about numbers in future lectures. But let's look a little more deeply in the brain, so you can see how packed it is with neurons and how the connections between the neurons are so unbelievably complex that thinking about how you use circuitry to construct the human nervous system-- or indeed that of most animals-- is an enormous problem. Professor Sebastian Seung here-- whose name is off the screen, and I apologize to him, it's the screen snafu, it will be on your power points when you download these from the web-- is working on the task of putting together all the connections in the human brain. And starting with a little cube of brain-- this is about 100 microns, it's not quite a cube, but it's approximately a 100-micron cube-- he's worked to try to figure out what all the cell connections are and what all the cells are, within this little tiny cube of brain, which is just a tiny, tiny fraction of your brain. So look at this. This is an electron micrograph that he's put a bunch of electron micrographs together, to build this 3D structure. And now his students and those of Professor Lichtman at Harvard go and outline a particular cell in serial sections, through this cube. These are very tiny sections. These are about five nanometer sections. And then they put these sections together. And you can get the three dimensional structure of the neuron, as it's going through the cube. And you can start to map how this neuron lies next to other neurons. OK, there it's going backwards. And there are two cells lying next to one another that you can get in 3D rendering, with this very painstaking process of first, getting sections of the brain, putting them all together into a chunk, and then deconstructing the individual cells shapes within that chunk. If you do that for the whole chunk of tissue, this is what you get. There's that red and that green neuron. Here is a blue one, there's a yellow one, lime, purple, red, dark blue, yellow, orange. It's daunting. There you go. In that tiny little chunk of brain, that is how packed the cells are. And the connections between them are enormous. And that is just less than a millionth of your brain. So to figure out all the connections, all the circuitry in the nervous system, is an enormous task. And we don't know it. We'll talk about what we do know later on. But I wanted to frame the problem for you, so that you have a sense of where we're trying to go. Let's go back to the neuron. And let's talk about cell type and how cell type is important for thinking about signaling in the nervous system. So like everything else in the body, communication uses cells as its currency. And the cell type is the special kind of cell type, which is the neuron. So neurons are the connecting/wiring cells. There's a second type of cell in the nervous system that is really pivotal for nervous system function that are called glia. And these are cells which are referred to as support cells. But that's not really fair. They guide neurons, and as we'll mention later, they also insulate neurons, so that the wires don't short circuit. So they guide and insulate neurons. The structure of the neuron is important to understand its function. Like all cells, it's got a nucleus and cytoplasm. And here it is, the nucleus, the cytoplasm. And this region of the neuron is called the cell body. But unlike other cells, coming out from this cell body there are processes. And they are very substantial processes. On one side of the cell body are usually fairly short processes. They can be branched, and there can be bunches of them. And these are called dendrites. Dendrites are processes that receive a signal. So they are the place where there is a signaling-- let me get rid of that dendrite-- there is a signaling input. The signal that dendrites receive moves through the cell body and into another process, which is called the axon and is very long. And the fact that it is very long is actually what this lecture is all about. So the axon is the wire. Axons can be up to a meter long. Here's the axon. And there are axons that start in your spinal cord and move all the way down your leg, from a single cell. We'll talk about why that is in a moment. These axons branch at their ends. And they connect to another neuron or something else, but we'll draw another neuron. Here's another neuron with its dendrites and another axon. The connection-- here is neuron 1 and neuron 2. And the connection between axons and dendrites-- or as you'll discover, axons in the cell body-- is called a synapse. Some people say synapse-- that's the connection-- either is OK. And the thing is that this input that is way over on the left hand side of the board is transmitted, along the axon, into the next neuron, and then along the next neuron. This is the signal. And we have to think about why cells might want to have these very long processes to do this, and then how the signal is transmitted. The reason that cells have got these long processes rather than-- well, let's actually step back a moment. Let's think about how cells might communicate with one another. You could imagine a whole bunch of little round cells, all lined up, so that there a meter of them that go from your spinal cord down to your leg. And that would give you a chain of communication from your spinal cord to your leg. And you could have one back to your brain and so on. And that in theory would work OK. But it turns out that cell-cell communication is very slow, and that cells have figured out a way to transmit a signal along their own length that is very rapid. You know that there is a finite time between getting a stimulus and a response. You know, you touch something hot, you can tell it actually takes a moment before you figure out it's hot. That's the speed of transmission of the signal, up into your brain. And you say, wow, that's hot, move my finger. If you had cells that were connecting rather than long processes of one cell, it would take you that much longer to actually make that-- maybe 10 times longer-- to make that connection. And you'd get a bad burned finger. So the axon is the thing that allows rapid transmission of the signal. So the axon is long. It leads to an intracellular signal. And this is very rapid, relative to an intercellular signal. But how do you transmit a signal along a cell? Well, axons do this by using movement of ions. So the signal along the axon is due to movement of ions. And this is called an action potential, as we will discuss. And this process draws on a property of all cells that neurons have capitalized on. So almost all cells have got a potential difference across their membrane, because there is a charge difference across the plasma membrane. So almost all cells have what is called a membrane potential, which is a membrane potential difference. And in general, cells are more positive outside than they are inside. So outside the cell-- and this is worth your remembering-- it is more positive. And it's more positive because there's a lot of sodium ion. You'll see the why this is important. There's low potassium ion, and there's some high chloride ion. But really, the thing that's important is that there's very high sodium concentration outside the cell. Conversely, inside the cell is obviously more negative. Sodium is low. Potassium is higher but still not very high. And there's a bunch of ions that are kind of trapped in the cell. Why is this important? Let me see what I have next here. OK, most cells show a potential difference. Here it is. Here is written the potential difference. Neurons are somewhere between minus 70 and minus 60 millivolts, where you are talking about the relative potential difference inside to outside, that's why it's negative. And you can see tumor cells actually have got a very low membrane potential, which may or may not be significant. What is the nature of the signal that neurons use to transmit from the input, along the axon, to the next cell? Particularly, what is the signal that's transmitted along the axon? And the answer is something called an action potential. It's the signal transmitted along the axon, the wire that I referred to. And an accident potential, which I'm going to abbreviate as AP, will define-- and you'll understand this in a moment-- as a local, transient depolarization. Local, transient-- just lasts a moment-- depolarization, change in membrane potential. And I'm going to do most of this on the board. You have a hand-out, but I'm going to do most of this on the board because it works better as a conversation than a demonstration. Let's draw a a bit of an axon. Here is the axon. Two plasma membranes-- outside, inside, outside. So this is the axon. Plasma membrane, PM. And what we are going to do-- and here is a cytoplasm-- what we are going to do is take a chunk of the axon and blow it up and focus on just one plasma membrane, on one side of the axon, and look and see in detail what is going on there. So let's take this chunk and blow it up, so that we now have the plasma membrane--and you remember it's a lipid bilayer. But I'm drawing it as a single line, because it's really a pain control to draw it as a lipid bilayer. But you know it's a lipid bilayer. On one side-- and here is outside the cell and inside the cell. On one side, there are lots of positive charges. And on the other side, there are fewer and relatively more negative charges. When the axon looks like this, with this balance of positive and negative charges, it's said to be at resting potential. And resting potential is about minus 60 millivolts. OK, what's our goal? Our goal is to start here, at this asterisk, and to transmit a signal along the length of this piece of axon and to transmit the signal in a directional way. So our goal is to transmit a directional signal. Let's draw three time points, each of which have a plasma membrane of this particular segment of axon. And let us-- I'm going to move over a bit, to this side of the board-- let's have them here, here, and here. And so we're going to have a time vector going diagonally across the board. And we started off with something that looked like it did on the board above at resting potential. And now, we're going to-- over a very short segment of membrane, we're going to reverse the membrane charge, or the cell's going to do it. So that on the outside now, there's a little part of the membrane that's negative outside and positive inside. And the rest is positive outside and negative inside. Over time, this is called a depolarization, a reversal of the membrane potential. Over time, that depolarization, that initial depolarization is going to rectify. It's going to go back to how it was. So you'll get positive charges outside again. But the segment of membrane next door is going to depolarize, so it now becomes negative outside and positive inside. And the rest is positive outside and negative inside. That little piece of membrane, the second-- so that's depolarization 1, here's depolarization 2. Again, over time, you're going to get rectification of that second depolarization. And you've got the idea now, it's going to move further down the axon. So here is depolarization 3. And you recall that this is outside and inside the axon. So if you look at my diagram-- I haven't given you any mechanism-- but you can see here we have got a signal that is moving in this direction, along the axon. Each of these depolarizations that I've drawn is called an action potential. And I'll give you, in a moment, some more properties, so that you'll know an action potential when you see one. But there are a couple of questions that arise from this easy-to-draw diagram. Firstly, how does this really happen? How do charges reverse across the membrane? Secondly, why is the signal unidirectional? Why doesn't it go backwards? And thirdly, how do you reset the depolarization once it's happened? And all of these things you will see are connected, but let's raise these questions. So how does this happen? Why is it unidirectional? And what does it mean-- or what is the mechanism of resetting the membrane potential, after a depolarization has happened? As for example here, you have reset the membrane potential. So the answer to all of this is complex. And we'll answer it in chunks, as we tend to do in this class. And the first thing we'll answer with respect to is changing membrane potential. Let's start off again with our axon chunk, with outside and inside, and the charge distribution that is at resting potential. And along comes some kind of input. It might be touch, it might be another neuron touching a second neuron. It might be vision, light, that comes along, some kind of input. Here it is. And this input acts on a very local part of the membrane. And it changes the membrane potential just a tiny bit, such that the membrane potential might reach something called threshold potential. So let's look, let's draw it out. Here it is. Just over a one little tiny bit of the membrane, there's some kind of charge reversal. The positive charges come from outside, and they move inside. And we'll call this potential difference threshold potential. And if you want a number, it's about minus 55 millivolts. What happens after threshold? Well threshold, you understand what threshold is. It means something happens because you have reached a point of no return. And what happens is that there is now an action potential, and there is a massive movement of the positive sodium ions into the cell. So from threshold potential there is a massive movement, again out and in. So now you've got instead of this little tiny region of depolarization, you've got a large region of depolarization. So this is a small, depol-- for depolarization-- leading to a very large depolarization of the kind that I drew on the board before. This large depolarization is called the action potential. And it has several properties. The action potential reverses the membrane potential almost completely. So now in this region of the membrane, it's at about plus 60 millivolts. So it's a massive depolarization. You completely reverse the ion distribution over a small region of the membrane. It's very local, however. An action potential occurs over-- or this massive depolarization occurs over about a micron of membrane. It takes one to two milliseconds to set up. It involves the movement of about 10 to the 5th ions, from the outside in. And I've told you the potential. The other thing that is really critical that you have to understand is something called all or none. The depolarization you get with an action potential is either complete, or it doesn't happen. If you reach threshold, you reverse polarity, and you get this complete depolarization to plus 60 millivolts from minus 60. You do not get a partial depolarization to plus 10 or 20 or 30 sometimes, or 35. For a given neuron, you get a specific action potential. And it either happens, or it doesn't happen. So all on none, very important. No little or big action potentials. And now we've got a depolarization, but we haven't answered two questions. We haven't answered the question of unidirection, and we haven't answered the question of resetting. So let's do that on the next board. And let's start off, actually, with an action potential. And I'm actually going to draw an action potential, kind of in the middle of this axon. You'll see why. When you get an action potential, what's happening is that sodium ions are moving from the outside inside. And those sodium ions-- because they're just ions-- will start to diffuse in the cytoplasm. And as they diffuse next door, they'll change the membrane potential, which will reach threshold, which will trigger an action potential in the membrane chunk next door. And then those ions will diffuse, to make the chunk of membrane next door reach threshold potential. And you'll get an action potential triggered, and so on. But the ions, of course, because they're just ions, can move in either direction. So the ions can move back. If this is where your action potential took place, the ions could move in that direction and trigger an action potential going back, up the axon, towards the cell body. Why doesn't that happen? It doesn't happen because once you've triggered an action potential, that membrane becomes refractory, unable to trigger another action potential for a while. And during that time where the membrane is unable to respond and make another action potential, the ions have diffused away and gone on down the axon. And so you get a unidirectional propagation. Let's try to draw that out. So here's an action potential. And the ions that are moving in will diffuse. And they will take the membrane next door to threshold. And so they'll trigger an action potential next door. Those ions fusing backwards can't do anything. Because once the membrane has had an action potential, it can't have another one for a while. So this membrane here, next door to the action potential, is refractory to depolarization-- that is a really horrendous spelling job there, depolarization-- for some period of time. Let's say for about a second or a little less than that, but somewhere around there. And so that means that the action potential is unidirectional. The ions can diffuse in both directions, but the action potential can only go in one. So that gives you a direction of your signal. And also, I have cavalierly drawn on there that the membrane potential reverses and resets itself, where the action potential previously occurred. And we'll talk about that more in a moment. So here, the membrane potential has reset. So this is a theoretical walk through action potentials. And I gave you a bunch of handouts. But I'm not going to go through them, because you can use them as a test or as an exercise after class, to see how much you understood. One of the things about conductance along an axon is that it's very quick. It takes a very short time from touching that hot thing to realizing you've touched it. But one of the reasons it's so short is because you're not sending an action potential all the way along an axon like we're drawing. You don't really get successive parts of the membrane depolarizing. Because that actually, although it's faster than intercellular connections, is still quite slow. So there's a way that a cell insulates itself, an axon insulates itself, to give you action potentials just to particular places. And that really speeds up the rate of transmission of an action potential. And I've got that on the slide here, and we're right on the board. So that during depolarization and after and all the time, ions leak from the axon. And this decreases the frequency of action potential formation. And so what cells do, it's kind of like a short-- no, it's not quite short circuiting, but it's a bad electrical wire. And so what the cell has done is to insulate itself with layers of fatty cells. And these cells are really kind of amazing. Most of the cells that insulates the neurons in the nervous system wrap around the neurons, as in this diagram. You can see here is a cell. And these lines are because a single cell has wrapped itself around the neuron. You know that the plasma membrane is lipid. It doesn't conduct ions, and so you've got really a fatty layer of insulation. And the thing that insulates the cells is something called a myelin sheath, which is lipid plus some protein. But it's a really hydrophobic layer that wraps around the neurons and insulates them. Along the neurons that are insulated, there are specific places where there is no insulation. And that's where action potentials take place. So action potentials take place at nods without myelin. And this is one way that neurons really speed up their conductance rate. And so I put here action potential frequency, but actually that's not correct. I'm going to talk about rate of transmission. So how does this all go together? Let's look at a movie, where here's the neuron, and here is the axon, transmitting an action potential along its length. And here's a different way of depicting the action potential, as a graph of voltage against time. And that's something that you'll practice in section. But what I want you to see is that there's an action potential moving along the axon. And the axon can transmit many, many action potentials, one after the other, with a short recovery period in between. We still have not answered quite the question of how action potentials work. And the answer to that is to consider ion channels and pumps, because all of this charge distribution doesn't just happen. It's set up by the cell, and it's set up by ion channels and pumps, which we can write Regulate Membrane Potential. Let's review very briefly what ion channels and pumps are. We've talked about them a bunch. But you need to know some essences for this particular module. Ion channels allow ions across the membrane by diffusion. So here is an ion channel. And I'm drawing a channel which is open. And the ions move by diffusion, across the channel. But there are other classes of ion channels, which are not always open. They are called gated. And we talked about them, when we talked about protein secretion, protein localization. So gated channels under a particular stimulus can be closed and then change to the open confirmation, after they've been given the appropriate stimulus. So here there is some kind of stimulus. And a channel that is gated will open. A third kind of way of getting ions across the membrane is to use a pump, where a pump is localized in the plasma membrane as well. But instead of a diffusion-governed process to get ions across the membrane, the pump is actively moving ions across the membrane. So ion pumps actively transport ions. And they generally require energy, ATP, in order to do so. And all of these things are essential to set the membrane potential and to change it during action potential formation. If we consider the resting potential-- actually, let me see what I have on the slides here. This is a really cool thing. All right, let me get through our board work, and then we'll do what is really cool after or on Monday. In order to set up the resting potential, there are several kinds of channels and pumps that you need to be aware of. One of them-- which is a biggie and for which a Nobel Prize was given some decades ago-- is called the sodium potassium pump. And this is really a big thing. It's also called the sodium potassium ATPase. And what it does is to pump three sodium ions out of the cell and put two potassium ions into the cell. And this is an enormously important pump for life. And you can see what it does is to increase the sodium concentration outside the cell and increase the potassium concentration inside. There is also something called an open potassium channel that will allow all this potassium that's being pumped in by the sodium potassium pump to start diffusing out of the cell. But in actual fact, it doesn't all diffuse out. Because it hits the positive charges of the sodium ions on the outside, and there's an electrostatic repulsion. And so that limits how much potassium-- by Monday, I will either be able to speak by Monday or I will have completely lost my voice. You'll have the option. So the open potassium channel allows potassium out by diffusion, until it is repelled or stopped by electrostatic forces coming from the sodium ions. And then a third ion that's open, an ion channel that's open is the chloride channel, which we won't discuss right now. During the action potential, there is an enormously important ion channel that is the last thing I'll mention today. And that's called the voltage gated sodium channel. This is an ion channel that, like many ion channels, consists of a complex of proteins. We'll make a note of that, I'll make a note of that next time. But the voltage gated sodium channel is sensitive to membrane potential. And when threshold potential is reached, there is a change in the confirmation of this channel, which is closed normally at resting content but becomes open at threshold potential, to lead to the action potential. And it becomes open through actually the sliding of one of the alpha helices that make up the proteins of the channel. And the sliding alpha helices slide because their charges change, the charges of the amino acids change. And that opens up the channel. So I'm going to show you one picture of the last of the sodium channel, the voltage gated sodium channel. Here it is. And then we'll finish. Take a look at this quickly. Here is the voltage gated sodium channel closed. Amino acids blocking up the pore. And there it opens up, to let the ions in. And we'll finish this 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: So in quantum mechanics, you see this i appearing here, and it's a complex number-- the square root of minus 1. And that shows that somehow complex numbers are very important. Well it's difficult to overemphasize the importance of i-- is the square root of minus 1 was invented by people in order to solve equations. Equations like x squared equals minus 1. And it so happens that once you invent i you need to invent more numbers, and you can solve every polynomial equation with just i. And square root of i-- well square root of i can be written in terms of i and other numbers. So if you have a complex number z-- we sometimes write it this way, and we say it belongs to the complex numbers, and with a and b belonging to the real numbers. And we say that the real part of z is a, the imaginary part of z is b. We also define the complex conjugate of z, which is a minus i b and we picture the complex number z by putting a on the x-axis b on the y-axis, and we think of the complex number z here-- kind of like putting the real numbers here and the imaginary parts here. So you can think of this as ib or b, but this is the complex number-- maybe ib would be a better way to write it here. So with complex numbers, there is one more useful identity. You define the norm of the complex number to be square root of a squared plus b squared and then this results in the norm squared being a squared plus b squared. And it's actually equal to z times z star. A very fundamental equation-- z times z star-- if you multiply z times z star, you get a squared plus b squared. So the norm squared-- the norm of this thing is a real number. And that's pretty important. So there is one other identity that is very useful and I might well mention it here as we're going to be working with complex numbers. And for more practice on complex numbers, you'll see the homework. So suppose I have in the complex plane an angle theta, and I want to figure out what is this complex number z here at unit radius. So I would know that it's real part would be cosine theta. And its imaginary part would be sine theta. It's a circle of radius 1. So that must be the complex number. z must be equal to cosine theta plus i sine theta. Because the real part of it is cosine theta. It's in that horizontal part's projection. And the imaginary part is the vertical projection. Well the thing that is very amazing is that this is equal to e to the i theta. And that is very non-trivial. To prove it, you have to work a bit, but it's a very famous result and we'll use it. So that is complex numbers. So complex numbers you use them in electromagnetism. You sometimes use them in classical mechanics, but you always use it in an auxiliary way. It was not directly relevant because the electric field is real, the position is, real the velocity is real-- everything is real and the equations are real. On the other hand, in quantum mechanics, the equation already has an i. So in quantum mechanics, psi is a complex number necessarily. It has to be. In fact, if it would be real, you would have a contradiction because if psi is real, turns out for all physical systems we're interested in, H on psi real gives you a real thing. And here, if psi is real then the relative is real, and this is imaginary and you have a contradiction. So there are no solutions that are real. So you need complex numbers. They're not auxiliary. On the other hand, you can never measure a complex number. You measure real numbers-- ammeter, position, weight, anything that you really measure at the end of the day is a real number. So if the wave function was a complex number, it was the issue of what is the physical interpretation. And Max Born had the idea that you have to calculate the real number called the norm of this square, and this is proportional to probabilities. So that was a great discovery and had a lot to do with the development of quantum mechanics. Many people hated this. In fact, Schrodinger himself hated it, and his invention of the Schrodinger cat was an attempt to show how ridiculous was the idea of thinking of these things as probabilities. But he was wrong, and Einstein was wrong in that way. But when very good physicists are wrong, they are not wrong for silly reasons, they are wrong for good reasons, and we can learn a lot from their thinking. And this EPR are things that we will discuss at some moment in your quantum sequence at MIT. Einstein-Podolski-Rosen was an attempt to show that quantum mechanics was wrong and led to amazing discoveries. It was the EPR paper itself was wrong, but it brought up ideas that turned out to be very important.
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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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Good morning. Good morning. So, what I would like to do today is pick up on our basic theme of molecular biology. We've talked about DNA replication. The transcription of DNA into RNA, and the translation of RNA into protein. We discussed last time some of the variations between different types of organisms: viruses, prokaryotes, eukaryotes, with respect to the details of how they do that in general that bacteria have circular DNA chromosomes typically that eukaryotes have linear chromosomes, etc. What I'd like to talk about today is variation, but variation not between organisms but within an organism from time to time and place to place, namely, how it is that some genes or gene activities are turned on, on some occasions, and turned off on other occasions. This is, obviously, a very important problem to an organism, particularly to somebody like you who's a multi-cellular organism, and has the same DNA instruction set in all of your cells. It's obviously quite important to make sure that the same basic code is doing different things in different cells. It's important, also, to a bacterium to make sure that it's doing different things at different times, depending on its environment. So, I'm going to talk about a very particular system today as an illustration of how genes are regulated, but before we do that, let's Ask, where are the different places in this picture? DNA goes to DNA goes to RNA goes to protein, in which you might, in principle, regulate the activity of a gene. Could you regulate the activity of a gene by actually changing the DNA encoded in the genome? So, why not? Because what? It becomes a different gene. Yeah, that's just a definition. Why couldn't the cell just decide that I want this gene now to change in some way? Oh, I don't know, I'll alter the DNA sequence in some way. And, that'll make the gene work. Could that happen? Is that allowed? Yeah, it turns out to happen. It's not the most common thing, and it's not the thing they'll talk about in the textbooks a lot but you can actually do regulation. So, the levels of regulation are many, and one is actually at the level of DNA rearrangement. As we'll come to later in the course, for example, your immune system creates new, functional genes by rearranging locally some pieces of DNA, some bacteria, particularly infectious organisms control whether genes are turned on or off by actually going in there, and flipping around a piece of DNA in their chromosome. And, that's how they turn the gene on or off is they actually go in and change the genome. There's some protein that actually flips the orientation of a segment of DNA. Now, these are a little funky, and we're not going to talk a lot about them, but you should know, almost anything that can happen does happen and gets exploited in different ways by organisms. So, DNA rearrangement certainly happens. It's rare, but it's always cool when it happens. So, it's fun to look at. And, something like the immune system can't be dismissed as simply an oddity. That's an incredibly important thing. The most common form is at the level of transcriptional regulation, where whether or not a transcript gets made is how it's processed can be different. First off, the initiation of transcription that RNA polymerase should happen to sit down at this gene on this occasion and start transcribing it is a potentially regulatable (sic) step that maybe you're only going to turn on the gene for beta-globin and alpha-globin that together make the two components of hemoglobin, and you're only going to turn them on in red blood cells, or red blood cell precursors, and that could be done at the level of whether or not you make the message in the first place. That's one place it can be done. Another place is the splicing choices that you make. With respect to your message, you get this thing with a number of different potential exons, and you can regulate how this gene is used by deciding to splice it this way, and skip over that exon perhaps, or not skip over that exon. That alternative spicing is a powerful way to regulate. And then finally, you can also regulate at the level of mRNA stability. Stability means the persistence of the message, the degradation of the message. It could be that in certain cells, the message is protected so that it hangs around longer. And, in other cells, perhaps, it's unprotected and it's degraded very rapidly. If it's degraded very rapidly, it doesn't get a chance to make a protein or maybe it doesn't get to make too many copies of the protein. If it's persistent for a long time, it can make a lot of copies of protein. All of those things can and do occur. Then, of course, there is the regulation at the level of translation. Translation, if I give you an mRNA, is it automatically going to be translated? Maybe the cell has a way to sequester the RNA to ramp it up in some way so that it doesn't get to the ribosome under some conditions, and under other conditions it does get to the ribosome, or some ways to block in other manners than just sequestering it, but to physically block whether or not this message gets translated, what turns out that there's a tremendous amount of that. It's, again, not the most common, but we're learning, particularly over the last couple of years, that regulation of the translation of an mRNA is important. There are, although I won't talk about them at length, an exciting new set of genes called micro RNA's, teeny little RNAs that encode 21-22 base pair segments that are able to pair with a messenger RNA and interfere in some ways partially with its translatability. And so, by the number and the kinds of little micro RNAs that are there, organisms can tweak up or down how actively a particular message is being translated. So, the ability to regulate translation in a number of different ways is important. And then, of course, there's post-translational control. Once a protein is made, there's post-translational regulation that could happen. It could be that the protein is modified in some way. The proteins say completely inactive unless you put a phosphate group on it, and some enzyme comes along and puts a phosphate group on it. Or, it's inactive until you take off the phosphate group. All sorts of post-translational modifications can occur to proteins after the amino acid chain is made that can affect whether or not the protein is active. Every one of these is potentially a step by which an organism can regulate whether or not you have a certain biochemical activity present in a certain amount at a certain time. And, every one of these gets used. This is the thing about coming to a system that has been in the process of evolution for three and a half billion years is that even little differences can be fought over as competitive advantages, and can be fixed by an organism. So, if a tiny little thing began to help the organism slightly, it could reach fixation. And, you're coming along to this system, which has had about three and a half billion years of patches to the software code, and it's just got all sorts of layers and regulation piled on top of it. All of these things happen. But, what we think is the most important out of this whole collection is this guy. The fundamental place at which you're going to regulate whether or not you have the product of a gene is whether you bother to transcribe its RNA. But I do want to say because, yes? And, which exons you used and which aren't? Yeah, well, there are tissue-specific factors that are gene-specific that can influence that. And, surprisingly little is known about the details. There are a couple of cases where people know, but as you'd imagine, you actually need a regulatory system in that tissue to be able to decide to skip over that exon. And, the mechanics of that surprisingly are understood in very few cases. And, you might think that evolution wouldn't like to use that as the most common thing because you really do have to make a specialized thing to do that. So, that's what happens on these. That's one in particular where I think a tremendous amount of more work has to happen. mRNA stability, we understand some of it but not all the factors in this business. I was telling you about translation with these little micro-RNAs is stuff that's really only a few years old that people have come to understand. So, there's a lot to be understood about these things. I'm going to tell you about initiation of mRNAs, because it's the area where we know the most, and I think it'll give you a good idea of the general paradigm. But, any of you who want to go into this will find that there's a tremendous amount more to still be discovered about these things. So, the amount of protein that a cell might make varies wildly. Your red blood cells, 80% of your red blood cells, protein, is alpha or beta-globin. It's a huge amount. That's not true in any other cell in your body. So, we were talking about pretty significant ranges of difference as to how much protein is made. How do things like that happen? Well, I'm going to describe the simplest and classic case of gene regulation and bacteria, and in particular, the famous lack operon of E coli. So, this was the first case in which regulation was ever really worked out, and it stands today as a very good paradigm of how regulation works. E coli, in order to grow, needs a carbon source. In particular, E coli is fond of sugar. It would like to have a sugar to grow on. Given a choice, what's E coli's favorite sugar? It's glucose, right, because we have the whole cycle of glucose. The whole pathway of glucose goes to pyruvate, which we've talked about, and glucose is the preferred sugar to go into that pathway, OK, of glycolysis. Glycolysis: the breakdown of glucose. But, suppose there's no glucose available. Is E coli willing to have a different sugar? Sure, because E coli's not stupid. If it were to refuse another sugar, it wouldn't be able to grow. So, it has a variety of pathways that will shunt other sugars to glucose, which will then allow you to go through glycolysis, etc. Now, given a choice, it would prefer to use the glucose. But if not, suppose you gave it lactose. Lactose is a disaccharide. It's milk sugar, and I'll just briefly sketch, so lactose is a disaccharide where you've got a glucose and a galactose. Glucose plus galactose equals lactose. So, if E coli is given galactose, it is able to break it down into glucose plus galactose. And it does that by a particular enzyme called beta galactosidase, which breaks down glactosides. And, it'll give you galactose plus glucose. How much beta-galactosidase does an E coli cell have around? Sorry? None? But how does it do this? When it needs it, it'll synthesize it. When it needs it, like, there's no glucose and there's a lot of galactose around, how much of it will there be? A lot. It turns out that in circumstances where E coli is dependent on galactose as its fuel, something like 10% of total protein can be beta-gal under the circumstances when you have galactose but no glucose. Sorry? Sorry, when you have lactose but no glucose. Thank you. So, when you have lactose but no glucose, E coli has 10% of its protein weight as beta-galactosidase. Wow. But when you have glucose around or you don't have lactose around, you have very little. It could be almost none, trace amounts. So, why do this? Why not, for example, just have a far more reasonable some compromise? Like, let's always just have 1% of beta-galactosidase. Why do we need the 0-10%? 10%'s actually extremely high. So what. It's a good insurance policy. So, if I only have galactose, I need more. Well, I mean, 1% will still digest it. I'll still do it. What's the problem? Sorry? So what, I do it at a slower rate. Life's long. Why not? Ah, it has to compete. So, if the cell to the left had a mutation that got it to produce four times as much, then it would soak up the lactose in the environment, grow faster, etc. etc., and we could have competed. So, these little tuning mutations have a huge effect amongst this competing population of bacteria. And so, if E coli currently thinks that it's really good to have almost non at sometimes and 10% at other times, you can bet that it's worked that out through the product of pretty rigorous competition, that it doesn't want to waste the energy making this when you don't need it, and that when you do need it, you really have to compete hard by growing as fast as you can when you have that lactose around. OK. So, how does it actually get the lactose, sorry, keep me honest on lactose versus galactose, into the cell? It turns out that it also has another gene product, another protein, which is a lactose permease. And, any guesses as to what a lactose permease does? It makes the cell permeable to lactose, right, good. So, the lactose can get into the cell, and then beta-gal can break it down into galactose plus glucose. These two things, in fact, both get regulated, beta-gal and this lactose permease. So, how does it work? Let's take a look now at the structure of the lack operon. So, I mentioned briefly last time, what's an operon? Remember we said that in bacteria, you often made a transcript that had multiple proteins that were encoded on it. A single mRNA could get made, and multiple starts for translation could occur, and you could make multiple proteins. And, this would be a good thing if you wanted to make a bunch of proteins that were a part of the same biochemical pathway. Such an object, a regulated piece of DNA that makes a transcript encoding multiple polypeptides is called an operon because they're operated together. So, let's take a look here at the lack operon. I said there was a promoter. Here is a promoter for the operon, and we'll call it P lack, promoter for the lack operon. Here is the first gene that is encoded. So, the message will start here, actually about here, and start going off. And, the first gene is given the name lack Z. It happens to encode beta-galactosidase enzyme. Remember, they did a mutant hunt, and when they did the mutant hunt, they didn't know what each gene was as they isolated mutants. So, they just gave them names of letters. And so, it's called lack Z. And, everybody in molecular biology knows this is the lack Z gene, although Z has nothing to do with beta-galactosidase. It was just the letter given to it. But, it's stuck. Next is lack Y. And, that encodes the permease. And, there is also lack A, which encodes a transacetylase, and as far as I'm concerned you can forget about it. OK, but I just mentioned that it is there, and it actually does make three polypeptides. We won't worry about it, OK, but it does make a transacetylase, OK? But it won't figure in what we're going to talk about, and actually remarkably little is known about the transacetylase. There's also one other gene I need to talk about, and that's over here, and that's called lack I. And, it too has a promoter, which we can call PI, for the promoter for lack I. And, this encodes a very interesting protein. So, we get here one message encoding one polypeptide here. This mRNA encodes one polypeptide. It is monocystronic. This guy here is a polycystronic message. It has multiple cystrons, which is the dusty old name for these regions that were translated into distinct proteins. And so, that's that mRNA. So, lack I, this encodes a very interesting protein, which is called the lack repressor. The lack repressor, actually I'll bring this down a moment, is not an enzyme. It's not a self-surface channel for putting in galactose. It is a DNA binding protein. It binds to DNA. But, it's not a nonspecific DNA binding protein that binds to any old DNA. It has a sequence-specific preference. It's a protein that has a particular confirmation, a particular shape, a particular set of amino acids sticking out, that it combined into the major groove of DNA in a sequence-specific fashion such that it particularly likes to recognize a certain sequence of nucleotides and binds there. Where is the specific sequence of nucleotides where this guy likes to bind? It so happens that it's there. And this is called the operator sequence or the operator site. So, this protein likes to go and bind there. Now, I've drawn this, by the way, so that this operator site is actually right overlapping the promoter site. Who likes to bind at the promoter site? RNA polymerase. What's going to happen if the lack repressor protein is sitting there? RNA polymerase can't bind. It's just physically, blocked from binding. So, let's examine some cases here. Let's suppose that we look at here at our gene. We've got our promoter, P lack. We've got the operator site here. We've got the lack Z gene here, and we've got the lack repressor, lack I, the repressor sitting there. Polymerase tries to come along to this, and it's blocked. So, what will happen in terms of the transcription of the lack operon: no mRNA. So, that's great. So, we've solved one problem right off the bat. We want to be sure that sometimes there's going to be no mRNA made. This way, we're not going to waste any metabolic energy, making beta-galactosidase. Are we done? No? Why not. We've got to sometimes make beta-galactosidase. So, we've got to get that repressor off there. Well, how is the repressor going to come off there? When do we want the repressor off there: when there's lactose present. So, somehow we need to build some kind of an elaborate sensory mechanism that is able to tell when lactose is present, and send a signal to the repressor protein saying, hey, lactose is around. The signal gets transmitted all the way to the repressor protein, and the repressor protein comes off. What kind of an elaborate sensory mechanism might be built? Use lactose as what? So, this is actually pretty simple. You're saying just take lactose, and you want lactose to be its own signal? So, if lactose were to just bind to the repressor, the repressor might then know that there was lactose around. Well, what would it do if lactose bound to it? Sorry? Why would it fall off? Yep. More interested in the lactose. So, if you're suggestion, this is good. I like the design work going on here. The suggestion is that if lactose binds to this here, binds to our repressor, it's going to fall off because it's more interested in lactose than in the DNA. Now, how is the interest actually conveyed into something material? Because the actual level of cognitive like or dislike for DNA on the part of this polypeptide is unclear, you may be anthropomorphizing slightly with regard to this polypeptide chain. So, mechanistically, what's going to happen? Shape. Yes, shape? Change confirmation, the binding act, the act of binding lactose creates some energy, may change the shape of the protein, and that shape of the protein may, in the process of wiggling around to bind lactose may de-wiggle some other part of it that now no longer binds so well to DNA. That is exactly what happens. Good job. So, you guys have designed, in fact, what really happens. What happens is what's called an allosteric change. It just means other shape. So, it just changes its shape, that it changes shape on binding of lactose. And it falls off because it's less suitable for binding this particular DNA sequence when it's bound to lactose there. So, in this case, in the presence of lactose, lack I does not bind. And, the lack operon is transcribed. Yes? Uh-oh. OK, all right designers, here we've got a problem. You have such a cool system, right? You were going to sense lactose. Lactose was going to bind to the lack repressor, change its confirmation falloff: uh-oh. But, as you point out, how's it going to get any lactose, because there's not a lactose permease because the lactose permease is made by the same operon. So, what if, in fact, instead of getting one of these DOD mill speck kind of things of some repressor that is absolutely so tight that it never falls off under any circumstances, what if we build a slightly sloppy repressor that occasionally falls off, and occasionally allows transcription of the lack operon? Then, we'll have some trace quantities of permease around. With a little bit of permease around, a little lactose will get in. And, as long as even a little lactose gets in, it'll now shift the equilibrium so that the repressor is off more, and of course that will make more permease, and shift, and shift, and shift, and shift. So, as long as it's not so perfectly engineered as to have nothing being transcribed, so no mRNA is really very little mRNA. See, this is what's so good, I think, about having MIT students learn this stuff because there are all sorts of wonderful design principles here about how you build systems. And, I think this is just a very good example of how you build a system like this. Now, all right, so we now have the ability to have lack on and lack off, and that is lack off, mostly off because of your permease problem: very good. Now, let's take a little digression about, how do we know this? This kind of reasoning, I've now told you the answer. But let's actually take a look at understanding the evidence that lets you conclude this. So, in order to do this, and this is the famous work in molecular biology of Jacobin Manoux in the late '50s for which they won a Nobel Prize, they wanted to collect some mutants. Remember, this is before the time of DNA sequence or anything like that, and wanted to collect mutants that affected this process. So, in order to collect mutants that screwed up the regulation, they knew that beta-galactosidase was produced in much higher quantity if lactose was around. The difficulty with that was that wild type E coli, when you had no lactose would produce very little beta-gal, one unit of beta-gal, and in the presence of lactose, would produce a lot, let's call it 1, 00 units of beta-gal. But, the problem with playing around with this is lactose is serving two different roles. Lactose is both the inducer of the expression of the gene by virtue of binding to the repressor, etc., etc. But, it's also the substrate for the enzyme because as beta-galactosidase gets made, it breaks down the lactose. So, there's less lactose in binding, and if you wanted to really study the regulatory controls, you have the problem that the thing that's inducing the gene by binding to the repressor is the thing that's getting destroyed by the product of the gene. So, it's going to make the kinetics of studying such a process really messy. It would be very nice if you could make a form of lactose that could induce beta-galactosidase by binding to the repressor, but wasn't itself digested. Chemically, in fact, you can do that. Chemically, it's possible to make a molecule called IPTG, which is a galactoside analog. And, what it does is this molecule here which I'll just sketch very quickly here, it's a sulfur there, and you can see vaguely similar, this is able to be an inducer. It'll induce beta-gal, but not a substrate. It won't get digested. So, it'll stick around as long as you want. It's also very convenient to use a molecule that was developed called ex-gal. Ex-gal again has a sugar moiety, and then it also has this kind of a funny double ring here, which is a chlorine, and a bromine, and etc. And, this guy here is not an inducer. It's not capable of being induced, of inducing beta-galactosidase expression. But, it is a substrate. It will be broken down by the enzyme, and rather neatly when it's broken down it turns blue. These two chemicals turned out to be very handy in trying to work out the regulation of the lack operon. So, if I, instead of adding lactose, if I think about adding IPTG, my inducer, when I add IPTG I'm going to get beta-gal produced. When I don't have IPTG, I won't produce beta-gal. But then I don't have a problem of this getting used up. So now, what kind of a mutant might I look for? I might look for a mutant that even in the absence of the inducer, IPTG, still produces a lot of beta-gal. Now, I can also look for mutants that no matter what never produce beta-gal, right? But, what would they likely be? They'd likely be structural mutations affecting the coding sequence of beta-gal, right? Those will happen. I can collect mutations that cause the E coli never to produce beta-gal. But that's not as interesting as collecting mutations that block the repression that cause beta-gal to be produced all of the time. So, how would I find such a mutant? I want to find a mutant that's producing a lot of beta-gal even when there's no IPTG. So, let's place some E coli on a plate. Should we put IPTG on a plate? No, so no IPTG. What do I look for? How do I tell whether or not any of these guys here is producing a lot of beta-gal? Yep? So, no IPTG, but put on ex-gal, and if anybody's producing a lot of beta-gal, what happens? They turn blue: very easy to go through lots of E coli like that looking for something blue. And so, lots of mutants were collected that were blue. And, these chemicals are still used today. They're routinely used in labs, ex-gal and stuff like that, making bugs turn blue because this has turned out to be such a well-studied system that we use it for a lot of things. So, mutants were found that were constituative. So, mutants were found that were constituative mutants. Constituative mutants: meaning expressing all the time, no longer regulated, so, characterizing these constituative mutants. It turns out that they fell into two different classes of constituative mutants. If we had enough time, and you could read the papers and all, what I would do is give you the descriptions that Jacobin Maneaux had of these funny mutants which they'd isolated and were trying to characterize, and how to puzzle out what was going on. But, it's complicated and hard, and makes your head hurt if you don't know what the answer is. So, I'm going to first tell you the answer of what's going on, and then sort of see how you would know that this was the case. But, imagine that you didn't know this answer, and had to puzzle this out from the data. So, suppose we had, so if there were going to be two kinds of mutants: mutant number one are operator constituents. They have a defective operator sequence. Mutations have occurred at the operator site. Mutant number two have a defective repressor protein, the gene for the repressor protein. How can I tell the difference? So, I could have a problem in my operator site. What would be the problem with the operator site? Some mutation to the sequence causes the repressor not to bind there anymore, OK? So, a defective operator site doesn't bind repressors. Defective repressor, the operator site is just fine, but I don't have a repressor to bind at it. So how do I tell the difference? One way to tell the difference is to begin crossing the mutants together to wild type, and asking, are they dominant or recessive, or things like that? Now, here's a little problem. E Coli is not a diploid, so you can't cross together two E colis and make a diploid E coli, right? It's a prokaryote. It only has one genome. But, it turns out that you can make temporary diploids, partial diploids out of E coli because it turns out you can mate bacteria. Bacteria, which have a bacterial chromosome here also engage in sex and in the course of bacterial sex, plasmids can be transferred called, for example, an F factor, is able to be transferred from another bacteria. And, through the wonders of partial merodiploid, you can temporarily get E colis, or you can permanently get E colis, that are partially diploid. So, you can do what I'm about to say. But, in case you were worried about my writing diploid genotypes for E coli, you can actually do this. You can make partial diploids. So, let's try out a genotype here. Suppose the repressor is a wild type, the operator is wild type, and the lack Z gene is wild type. And, suppose I have no IPTG, I'm un-induced. I have one unit of beta-gal. When I add my inducer, what happens? I get 1,000 units of beta-gal. Now, suppose I would have an operator constituative mutation. Then, the operator site is defective. It doesn't bind the repressor. Beta-gal is going to be expressed all the time, even in the absence. All right, well that was, of course, what we selected for. Now, suppose I made the following diploid. I plus, O plus, Z plus, over I plus, O constituative, Z plus. So, here's my diploid. What would be the phenotype? So, in other words, one of the chromosomes has an operator problem. Well, that means that this chromosome here is always going to be constituatively expressing beta-gal. But, what about this chromosome here? It won't. So, this would be about 1, 01, give or take, because it's got one chromosome doing that and one chromosome doing this, and this one would be about 2, 00. Now, that quantitative difference doesn't matter a lot. What you really saw when you did the molecular biology was that when you had one copy of the operator constituative mutation, you still got a lot of beta-gal here even in the absence of IPTG. So, that operator constituative site looked like it was dominant to this plus site here. But now, let's try this one here. I plus, O plus, Z plus, over I plus, operator constituative, Z minus. What happens then? This operator constituative site allows constant transcription of this particular copy. But, can this particular copy make a working, functional beta-gal? No. So, this looks, when you do your genetic crosses, you find that the operator constituative, now, if I reverse these here, suppose I reverse these, I plus, O plus, Z minus, I plus, O constituative, Z plus, same genotypes, right, except that I flipped which chromosome these are on. Now, what happens? This chromosome here: always making beta-gal and it works. This chromosome here: not making beta-gal. Even though it's regulated, it's a mutant. So, in other words, from this very experiment, you can tell that the operator site is only affecting the chromosome that it's physically on, that it doesn't make a protein that floats around. What it does is it's said to work in cys. In cys means on the same chromosome. It physically works on the same chromosome. Now, let's take a look, by contrast, of the properties of the lack repressor mutants. If I give you a lack repressor mutant, I plus, O plus, Z plus is the wild type. I constituative, O plus, Z plus: what happens here? This wild type is one in 1, 00. This guy here: 1,000 and 1, 00, and then here let's look at a diploid: I plus, O plus, Z plus, I constituative, O plus, Z plus. What's the effect? The I constituative doesn't make a functioning repressor. But, I plus makes a functioning repressor. So, will this show regulation? Yeah, this will be regulated just fine. This works out just fine, and in fact it'll make 2,000, and it'll make two copies there. But again, the units don't matter too much. And, by contrast, if I give you I plus, O plus, Z minus, and I constituative, O plus, Z plus, what will happen? Here, I have my mutation on this chromosome. But, it doesn't matter because I've got my mutation on this chromosome in the repressor. I've got a mutation on lack Z here, but as long as I have a functional copy, one functional copy of the lack repressor, it works on both chromosomes. It will work on both chromosomes, and so in other words this lack repressor, one copy works on both chromosomes. In other words, it makes a product that diffuses around, and can work on either chromosome, and it's said to work in trans, that is, across. So, the operator is working in cys. It's operating on its own chromosome only. A mutation in the operator only affects the chromosome it lives on, whereas a functional copy of the lack repressor will float around because it's a protein, and that's how Jacobin Maneaux knew the difference. They proved their model by showing that these two kinds of mutations had very different properties. Operator mutations affected only the physical chromosome on which they occurred, which of course they had to infer from the genetics they did, whereas repressor, a functional copy repressor, could act on any chromosome in the cell. So, OK, we've got that. Now, last point, what about glucose? I haven't said a word about glucose. See, this was a big deal to people. This model, the repressor model, we have this repressor. What about glucose? What's glucose doing in this picture? So, glucose control: so here's my gene. Here's my promoter, P lack. Here's my operator, beta-gal. It's encoded by lack Z. You've got all that. When this guy is present, sorry, when lactose is present, the repressor comes off. Polymerase sits down. Wait a second, polymerase isn't supposed to sit down unless there's no glucose. We need another sensor to tell if there's glucose, or if there's low glucose. So, we're going to need us a sensor that tells that. Any ideas? Yep? Yeah, if you work that one through, I don't think it quite works. But, you've got the basic idea. You're going to want another something, and it turns out there's another site over here, OK? There's a second site on which a completely different protein binds. And, this protein is the cyclic AMP regulatory protein, and it so happens that in the cell, when there's low amounts of glucose, let me make sure I've got this right, when there's low amounts of glucose, what we have is high amounts of cyclic AMP. Cyclic AMP turns out, whereas lactose is used directly as the signal, cyclic AMP is used as the signal here. When the cell has low amounts of glucose, it has high amounts of cyclic AMP. Now, what do you want your cyclic AMP to do? How are we going to design this? It's going to bind to a protein, cyclic AMP regulatory protein, it's going to sit down, and now what's it going to do? Is it going to block RNA polymerase? What do we want to do? If there's low glucose, high cyclic AMP, we sit down at the site, we want to turn on transcription now, right? So, what it's got to do is not block RNA polymerase, but help RNA polymerase. So, what it actually does is instead of being a repressor, it's an activator. And what it does is it makes it more attractive for RNA polymerase to bind, and it actually does that by, actually it does it slightly by bending the DNA. But, what it does is it makes it easier for RNA polymerase to bind. It turns out that the promoter is kind of a crummy promoter. It's actually just like, remember the repressor wasn't perfect; the promoter's not perfect either. The promoter's kind of crummy. And, unless RNA polymerase gets a little help from this other regulatory protein, it doesn't work. We have two controls: a negative regulator responding to an environmental cue, a positive activator responding to an environmental cue, helping polymerase decide whether to transcribe or not, and basically that's how a human egg goes to a complete adult and lives its entire life, minus a few other details. There are some details left out, but that's a sketch of how you turn genes on and off.
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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 we continue our discussion now of electron and proton scattering. And we dive deep into the structure using deep inelastic scattering. Inelastic here means that we are destroying the structure of the proton in the scattering process. But we have a way to look at the remnant of the proton and also of the scattered electron, and then compare our theoretic expectation for the cross-sections with the finding in experiments. Let me just talk about this in more general terms. The energy of the probing electron or the photon in the scattering process allows us to look at the proton with varying resolution. So at very low energies, we basically see a point-like particle. And then the scattering process looks very much like the scattering of an electron with a muon. If we increase the energy of the electron, we can see that there's an extended charge distribution in the proton. Further increase allows us to resolve the fact that the proton is made out of three quarks. And if you increase the energy further, we see a lot of new particles appearing, quarks and antiquarks and gluons, which make up the structure of the proton. The picture here is-- I like this very much, I drew this myself some years ago-- is what I would like to have you remember. So in deep inelastic scattering experiments, we basically use the photon scattered-- radiated off the electron as a magnifying glass for the proton. So we can look into the structure of the proton here, and we see the distribution of the charged particles in the proton-- only the electrically-charged particles. We don't have scattering between-- direct scattering between photons and gluons. So in measurements, what we can do is we can test scattered electron, and we can look for the remnant of the proton in our measurement, and then we do differential cross-section measurements, compare with our theory, and can infer information about the structure of the proton. To do this, we have special kinematic variables which turn out to be very useful. The most important one is probably this x year, which is called the Bjorken scaling x or Bjorken x, which you can think about-- so this is q squared, the momentum transfer of the photon, p is the momentum of the proton, q is the momentum transfer, this q here, factor of 2. And what this is basically the fraction of the momentum carried by the parton here in the scattering process. There's a few other useful variables, but I don't want to go into any of the details yet. So there's a number of very important scattering experiments. The first one I mentioned before is SLAC-MIT experiment, which led to the discovery that the proton is made out of quarks, and the Nobel Prize in Physics 1990 to Jerry Friedman, Henry Kendall, and Taylor. And what they did is they basically had a beam at SLAC of electrons of 5 to 20 GeV. And they scattered this beam off of hydrogen target protons. And they used this spectrometer here in order to then make a differential measurement of the scattered electrons. So that's very cool. Even higher energies were available at HERA, the electron-positron collider, where the energies of the electrons were in the order of 30 GeV of the protons up to 830 GeV. And so what we find then in those collisions, the differential cross-section measurements, is shown here. And it's very-- not an easy plot to read. So you see our structure functions here in the logarithmic plot. Remember, this is the a log 10 plot here. And you see here q squared, the momentum transfer of the photon, so the energy used in the scattering process. When we try to read this here, we can look at a fixed q squared, for example. And at a fixed q squared, you see that if you probe-- if you are testing for a fixed fraction of the partons taking away partons, a fraction of the parton's momentum of the proton, you see that the lower the fractions, the more particles you see. So at a fixed energy, you see many, many more particles the lower you go in the fraction. So there seem to be like an increase of particles the lower the fraction is. If you then check for a fixed fraction, let's say 0.4-- it means that the parton carries 40% of the proton's energy-- you see that it's almost flat as a function of q squared. So it seems like there's 40%-- the number of particles you see at 40% of momentum fraction is constant, this q squared. However, if you look at smaller energies-- sorry, smaller momentum fractions, you see that higher energies seem to show even more particles at this momentum fraction. And the ways to understand this is two diagrams. So the first one is this one here, where you see a quark radiating a gluon. And so what you see is that the deeper you look, you are able to then resolve this part here. And so you see more quarks and gluons which carry even smaller momentum fractions than the initial quark here. You also see diagrams like this, which is called gluon splitting, as a gluon splits into a quark/antiquark here. And you start resolving those. And those also carry lower momentum. The evolution of our-- of this parton distribution function, or the structure functions, can be calculated and described in the so-called DGLAP equations. And all you do here is calculate the contributions from the so-called quark and gluon splitting. So you calculate the splitting functions, these higher-order corrections to a very simple quark model. And you find that you can actually very nicely describe those curves here. So you see this yellowing-- yellow here is the QCD fit, which basically uses those splitting functions as input. All right. So we learned quite a bit about this proton already. If we want to now calculate a cross-section of a proton scattering with a proton, we are actually interested in the energy distributions or the momentum distributions of the partons in the protons. And so we'll come back to how we use this later. But I want to introduce parton distribution functions which do exactly that. They're defined as the probability to find a parton in the proton that carries energy between x and x plus dx. You can write them using the structure functions before. But they literally describe this probability. And so what you find inside the protons are now the valence quark, the down quark and the up quarks, c quarks and antiquarks, and gluons. So you want to describe those momentum distributions or energy distributions of those particles. There's a number of sum rules. If you integrate momentum fractions from 0 to 1, 1 being the momentum of the proton, if you integrate them all together, you have to find 1, because that is the momentum of the proton you start with. If you integrate the down quarks and the up quarks, you find 1 or 2, meaning that those are the number of valence quarks we have available. If you integrate the distributions of strange and antistrange and charm and anticharm, you get 0, because there needs to be the same number of strange and antistrange and charm and antistrange. And because of energy conservation, the sum needs to-- this sum, those sums need to be 0. And then we can look at those. So what's shown here is x times the Parton Distribution Functions, the PDFs, as a function of f. And what you see here for our valence quarks, you see a distribution which is kind of what you expect-- it almost peaks at 0.3, a surge, and has the distributions because there's kinematics involved, and also interactions-- because of the interactions with the gluons. And then you see the c quarks and antiquarks. And you see them increasing in numbers quite significantly here as you go to small fraction of the momentum carried. It's exactly what we just discussed in the previous plot already. An interesting way to look at this very same distribution function is if you plot them proportional to the momentum fractions, or the area proportional to the momentum fractions. And what you see there is that a very significant part of the momentum of the proton is carried by the gluon. So you see here again, our valence quarks, our c quarks, and the gluons itself. So that's all I wanted to say on the structure of the proton. We'll later in a lecture see how we can use those PDFs, those Parton Distribution Functions, in order to calculate a cross-section in proton scattering.
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https://ocw.mit.edu/courses/5-08j-biological-chemistry-ii-spring-2016/5.08j-spring-2016.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. JOANNE STUBBE: So the key question is, do these-- and I think this is a general question you can ask, metabolically, inside any cell is do these enzymes that are on different polypeptides cluster. And is there an advantage, kinetically or whatever, is there some kind of an advantage to have clustering inside the cell. And where have you seen something like this before? Do you remember? Do you remember the section. Where have you see multi enzyme complexes and clustering before? Yeah? So or the classic one, PKS has been around. But it's completely analogous to fatty acid synthesis, right? And so in bacteria, they're all single polypeptides. In humans, they're all activities that are on single chains. OK, so that's sort of what's going on here with the purine pathway. We'll see there are ten. This just sort of helps us focus if we get to the data at the end, which I think we will from what we did the last time. Is that you have six enzymes for 10 activities. So that just means you have more than one enzyme per polypeptide, OK? And so I guess the key thing that I wanted to focus on is do you think it's important to cluster? Here's a pathway. These are the names. We're not going to go through the names. The names really aren't important for what we're doing. There'll be two names that we'll be looking at over and over again. These are the papers that you guys did, in fact, read. One is the original paper, which got a lot of press. And I just want to show you that there have been there's actually been four papers published in the last six months on this topic. And one of which was published. Yeah. One of which is published with Science, where they are now claiming that this complex is localized to the mitochondria. OK, so you take pictures. And it's that this is looking at super resolution fluorescence methods. And you can clearly see clumps of blobs focused on the mitochondria. Why would you want it to be at the mitochondria? So then you have to ask your question. You might need purines, because that's where you make, through a proto mode of force in respiration. Remember, when you convert oxygen to water, you get a huge amount of energy released. You make ATP. But it's going to be made from something. So maybe you would want. That's the way they rationalize it. And they do. And then they connect it to the other latest hot topic, which is EM torque, which is the major signaling switch for fatty acids and for amino acids. And now in the last two years, purines and pyrimidines, I decided-- I've done a lot of reading about it, decided and believe. I mean, I believe it. But I don't believe the connections yet. So again, this is what you're going to see in the next decade is connecting signaling to primary metabolic pathways, like the purine pathway. That's going to be a big thing and how do you connect them is going to be the key question. So anybody that wants to do some more reading, this is an updated version. I kept updating this three or four times. And so I think these are the key questions we want to focus on. And so what I'm going to do. Well, define the questions a little bit and whether the things we need to think about just to determine whether this is really important, biologically. Then we'll define fluorescence and what you can do with fluorescence. And then we'll come back and look at the data. In the paper, we're also going to look. We probably won't get all the way through all of the data. But we will look at some of that data again in either the next lecture or Wednesday's lecture. So you will see it again if we don't get through the data. So they claim they have a multi enzyme complex. Did you believe that from the data? I mean, they didn't look at all 10 enzymes simultaneously, right? Or six enzymes. AUDIENCE: Whenever they show an image of the cells, and then they fluorescent, trying to show local sections, it's always so hard for me to figure out-- JOANNE STUBBE: What you see. AUDIENCE: Yeah. JOANNE STUBBE: OK, so that was said. We'll look at some of those pictures. But I completely agree with that, that you can't see anything from fluorescence pictures. So everybody, all chemists or chemical biologists, now have huge numbers of these pictures in their papers. And with Alice's group, I'm always on their case that I can't tell a damn thing. This is on thesis. I can't see anything. And Alice says she can't see anything either. So it's very hard to see things in these pictures. The contrast isn't very good. And what her lab now does is it goes to EM, where you can see things much more clearly. The fluorescence things are tough. So you're not the only one. And if somebody says it's obvious that this. And you don't see it. Raise your hand and say, I don't see it. Show me what I should be looking at, OK? So that's a good take home message, because everybody and his brother is doing this. And this goes back to knowing how to do it correctly. We're not going to talk about any of that stuff. I mean every one of the methods I'll sort of show you that's out there. You have to really study it to make sure you're handling it correctly. So I mean I think, to me, this has been a problem that I've been interested in. And I started working on this a long time ago in the purine pathway is not are things sticking together important. Actually, I don't think those are important. You immunoprecipitate all these things. OK, so you say obviously these are talking to each other. But the key thing is the kinetic competence. And lots of times when you mess around. You get it in a state. And you post-translationally modify it. So it's sitting in this state there probably isn't on the pathway. You need to then show it's on the pathway. So I think a lot of protein, protein interactions, especially now that we know that proteins move around. And they're in this complex. And they're in that complex. And they're in that complex. The key, I think, is to transient interactions. So why? So this is just my personal take on this. I'm letting you think about this. But is it easy to look at transient interactions? No. OK, so anyhow, I think people need to start doing a lot more thinking about how to look at that. And one way you could look at transient interactions is if you can fluorescent label something. And they come together on a certain timescale and then move apart. And can you do that inside the cell with the right spatial and time resolution. You might be able to start looking at that. So that the methods that are being developed and continue to be developed are incredibly powerful. And I think will allow us to ask this question happens inside the cell, which you've all seen pictures in your introductory courses of, man, how complicated the inside of the cell is. That's part of the issue. So the issue is that you might have a purinosome somewhere in the cell, depending on the growth conditions. But those enzymes might be involved in other things. And so you have only a tiny amount of it, as opposed to trying to make the cell by growth conditions, putting it into all one state. So you can see it. So the question is, how do you see it? And so that's the key issue. And if you perturb it enough. And you do see it. Then you have to ask the question. And this is a question that you might want to think about in terms of these two papers you were reading. That's what. If you do this, that's what the Marcotte paper said, that the cells were incredibly sick when you take out all the purines. And in fact, Alice is-- because of this mitochondria connection between the purinosome and Alice's interest in the mitochondria, she's had people trying to repeat this. And Vicki Hung worked on this and couldn't repeat it. So she didn't spend that much time on it. But all I'm saying is it's not a slam dunk to be able to do this. But that being said, I think this has been an issue that people have been thinking about for decades. And it's just really hard to test experimentally inside the cell. This is where we need chemical biologists to figure out new ways of being able to look at this, so that you can actually make a measurement that's interesting. So I guess the question I want to start with, before we researched looking at fluorescence, is why do you think it would be important to do this. Or do you think it would be important to have a complex. What's the advantage of doing that? Yeah. AUDIENCE: You were saying in lecture that you want to increase the effective molarity. And so by having all these things right next to each other, there's-- obviously you're going to have more interactions per second. JOANNE STUBBE: Well, you may or may-- you may not. It depends. So I think this is the key question. Is diffusion fast inside the cell? AUDIENCE: Yeah. JOANNE STUBBE: Yeah. It's still very fast. For small molecules, it's incredibly fast. Even for proteins, it's incredibly fast. So even if this guy is over here. If you're turning over here at a much slower rate, and you have enough of them so you can interact at diffusion control, do you need this organization? There are a lot of smart people who think you don't need that. There are a lot of smart people who think you do need that. But this is the question I want to raise. However, so catalytic efficiency is absolutely it. But where might you really need catalytic efficiency. And so that goes back. There are places where you really need this. So if you look at the first intermediate in the pathway, this guy, what do you think about that guy? Do you think he's stable? So if you look at the first intermediate in the pathway, which we'll talk about next time. So this is amino phosphoribosine-- phospho-- I'm drawing a complete-- I think I'm tired. Anyhow it's the amino analogue of ribose 5-phosphate. Phosphoribosylamine, that's what it's called, PRA. OK, do you think that's stable, as chemists? So what do you think that could do? AUDIENCE: Could you release the amine? JOANNE STUBBE: Yeah, so how would you do that? AUDIENCE: So if it's proteinated, and then the ring opens till-- JOANNE STUBBE: OK, so that would be one way. You want to release it that way. OK, so it would have to be under conditions. We could do that under neutral conditions. What else can happen to this ring. That doesn't happen. There are lots of ways this molecule can break down. OK, it depends on the details of the environment. How else could this molecule ring open? You wouldn't need to ring open here. You just go through an oxocarbenium ion and have water attack. So what if it opens that way. So that's the way you form aldehydes. All sugars are in equilibrium with aldehydes. These things are in equilibrium. So you have a ring open species. But then what happens if a ring closes? It can ring close from the top face or the bottom face. You have an imine. What can happen to the imine? It can hydrolyze. This molecule, and this is a molecule my lab worked on decades ago, has a half life in solution of 10 seconds. So is 10 seconds short or long, biologically? What do you think? AUDIENCE: I'm going to say short. But I don't know JOANNE STUBBE: Yeah. I think it's amazingly long inside the cell. So I think as a chemist, nobody could ever. Nobody ever saw this intermediate. My lab was the first one that figured out how to look at it. And I won't go through that. But the fact is that 10 seconds is a long time inside the cell, if you think about how small the cell is and how fast diffusion is. OK, so one place, though, where you might want to have organization is if you have something chemically really unstable. OK, because then when you generate it, it could potentially be passed off, or as you say in the immediate vicinity, it's a competition. But if it's right there, you're effective molarity, that would get into that first question, the effective molarity. It would be high enough to get passed on. It would get high enough to get passed on to the next guy in the pathway. So that would be one thing is instability. And in the purine pathway. We will go through this a little bit. But really, that's one of the things that's most amazing about Buchanan's elucidations of the pathway is only intermediates are unstable. Nobody, still, if you're looking at omics, looking at nucleotides, nobody knows how to deal with these molecules. They're all chemically unstable. And they don't get that they're chemically unstable. They don't ever see them. The reason they don't see them is because they don't know how to handle them to keep them alive during the analysis part of the project OK, so you have this instability problem. And in the purine pathway, the instability problem is a real problem for not just this guy. This guy is obvious. But for other guys. OK, so then the next question is where else. And if you're thinking about metabolism in general, where else might you want to have organization of your enzymes? You might want to have it if you generate an intermediate in the pathway. And then there. It's a branch point for other metabolic pathways. OK, so there's an intermediate in this pathway that can go to thiamine biosynthesis to histidine to tryptophan in biosynthesis. I'm not going to go through that. But that would be another place that I think it's obvious that you could sequester, under a different set of conditions, and prevent the other pathways from happening. So if you have an intermediate branch point, you can prevent other pathways. So those two things I think are important. One of the questions is, do you increase the flux through the pathway? OK, so there's been a lot of engineering people. People really care about this in terms of engineering. If you want to engineer a metabolic pathway, should you be linking all your proteins together? And there have been a lot of papers published. If you look at bioengineering papers, where they link all the pathways, all of the enzymes together in a way, because they want them to cluster, because they think they're increasing the flux through the pathway. And so there are some people that do calculations that show you increase the flux. Other people do calculations so you don't increase the flux. So I think, again, this is an area that I think is very active. And it's pertinent, because everybody and his brother is trying to make biofuels. You'd need to do a lot of engineering from a lot of enzymes from different places, putting them together. How do you make them efficient? OK, so we asked the question about flux. And I think, mathematically, people are looking at that. You need to know a lot about the kinetics of your system. These systems, there's a lot known about the kinetics. So and then this goes to the question of how, what is unstable. And you need to think about diffusion. I think this is not so easy to think about this. But we do need to think about flux through the pathway. And then the other thing that's interesting in terms of regulation is it turns out in eukaryotes, where things are much more regulated than in prokaryotes, because of the increased complexity of everything. Almost all of these pathways are organized on multiple activities on one polypeptide. That's telling us something, I think, since we see this over and over and over again. So there must be some reason to do that. So for all of these reasons in terms of the purine pathway this has been sort of a target for people for a long time. That's one of the reasons I decided to talk about it, because this was one of the first papers where people were excited that they thought they had evidence for this kind of organization in the cell. Not in the animal. But in the cell. OK. Let's see what I want to say next. I'm trying to keep this on some kind of a schedule. OK, so this is the hypothesis. The hypothesis is that these things are organized in some way. And this was taken out of-- probably it was a review paper. It wasn't taken out of a paper you had to read. Here's the cell. That's the nucleus of the cell. And what do you see. I think you can see this right. You see these little dots which they call punctate staining. So what else do you need to know that they don't have in this picture that's really sort of key to thinking about this model. So here they've just have a bunch of enzymes stuck together and all in a little ball. OK, so if you read the paper there was a couple of things. AUDIENCE: How you're getting the fluorescence? JOANNE STUBBE: How you're getting them? AUDIENCE: If it's by effusion or fluorescence. JOANNE STUBBE: Yeah, so how you're getting the fluorescence becomes key. OK, so we're going to talk about that. How did what was a major way they got the data. We'll talk about this in a minute in more detail but. But whenever you're going to use fluorescence, you have to figure out how to get a probe onto your protein. So that's like a major focus. And this again is where chemical biology needs to play a role. We still need better ways to be able to do this. You've seen over the course of the semester. I think in a lot of ways you could potentially do this. We'll come back to that in a minute. But if you look at this, what's missing? And this is something that drove me crazy when I reviewed the original paper. AUDIENCE: I just noticed, so you're getting that. But they didn't stain the membranes really. There's not a good-- I mean, you can kind of see the shape of the cell. But it would be nice to have a clear sort of-- JOANNE STUBBE: OK, so they might have done that. Did you look at the supplementary material? They might have stained the membrane. OK, so I think everybody would believe you see little blobs. OK, so what do you need to think about in terms of the little blob. AUDIENCE: The size. JOANNE STUBBE: The size. Right. Yeah, so that's one thing. They don't ever they don't ever talk about this. They might in some of the very later papers. But if we know this, we have structures of all the enzymes in the pathway. So you could make a guesstimate about how big these blobs should be, if you had one of each of these. And these things are huge. So this would tell you that you would have many, many of these. This is one thing that I think they need to do some more thinking about that they could have many, many of these things. And then the question is, why would you want many, many of these things. And how were they organize? Are they just sort of randomly organized or are they really organized in something like that with this big huge protein in the middle. That's one of the ones they look at. FGAM synthase, that has a molecular weight of 150,000, which is huge for an enzyme. And so for a long time-- and the catalytic activity-- my lab has studied that-- is way over here. And so you have a lot. Could it be a scaffold. OK, so that's where that idea actually came from. So but the hypothesis is that these guys are organized. And they're under certain growth conditions. That's the key. And we'll look at those pictures that come together if they do this when you need to make purines. And then they can go apart. OK, so the key thing, I think, is. And I wanted to just remind you why we're spending this time looking at fluorescence. And we probably should have spent two or three recitations on fluorescence methods. But we didn't. Is that we've seen this many times before. We've seen stopped-flow fluorescence in the Rodnina paper, where we were looking at the kinetics of fidelity of EF-Tu. And somehow they put a fluorescent probe onto the piece of tRNA. That was not trivial. How you got the probe there. And that probe could-- and we'll talk about this in a minute. But it could change. It changes when it's in different environments. And so you can use it as a way to monitor changes. So reactive oxygen species, we just looked at this. And I decided to put this up, since we didn't have the structures up last time. Fluorescein is one of the dyes that. This is fluorescein that people use. This is a version of fluorescein. But we talked about how do you know that epidermal growth factor is generating hydrogen peroxide? OK, so what we need is a sensor of hydrogen peroxide. So we talked about that last time. And this is the sensor that people use. Why did they use it. We talked about it. But we didn't have the structure. So they use the dye acetate of this molecule. This one they use, the triacetate. The one that they use in paper was the diacetate. Anyhow, you need to get the fluorescent probe into the cell. So that's something you're going to have to deal with. And so if you acetylate it, you don't have phenols or phenolates which might not get through the membrane, which apparently they don't. So then when they get in the cell, what do they do? They hydrolyze, OK. So what happens is when they hydrolyze, they are now-- you have these hydroxylated compounds that are able to be oxidized by an oxidant. And one of the oxidants that can do this. And there are others that can do it as well, is hydrogen peroxide. So people use this as an indicator of hydrogen peroxide. But it's not specific. Yeah? AUDIENCE: So are they also trapped after that esterase, like from diffusing that out to the-- JOANNE STUBBE: No. I mean, I don't think they diffuse back out, because I think they're the phenolates. So I think the diffusion out, like with many of these things, like if you use-- lots of times you esterify phosphates to get them into the cells. Once they hydrolyze, they charge. They don't get back out. So I don't really know. But that's what I would guess. So I guess the key thing and the basis for some comments that I made in class was that we don't really have. We don't know that this is specific for one reactive oxygen species. And so there are lots of people in the chemistry, biology interface trying to make specific sensors. OK, that's not easy to do. The hydrogen peroxide, they're getting better. In fact, Ting's APEX, which is a peroxidase, sort of similar to what we talked about with peroxireductions in the myeloperoxidase can actually function as a hydrogen peroxide sensor. So anyhow, what happens is that when it gets oxidized, it becomes fluorescent. So it's not fluorescent. It becomes fluorescent. So it just gets turned on. And you can see something, OK? So that's something we talked about. In Liz's part of the course, we talked about the fact that we can watch protein unfolding in the e. Coli proteasome. OK. And what did you look at in the proteasome, clip X clip P? You looked at titin. That had a little tryptophan on it. And tryptophan can absorb. It's not a very good thing, because it absorbs in the UV. But tryptophan fluorescence is used a lot. There are lots of tryptophans, so it's also really hard to use. But titin was this little tiny protein. And it was the only tryptophan. And they also did experiments with green fluorescent protein, which is what we're using in this paper. We remember. They pull on it. And you pull and you pull when you pull and then all of a sudden it unfolds. And you lose your chromophore. So you go from the on state to the off state. So all of these things. Binding measurements. You talked about. You had one problem set. I don't know whether you guys did that problem set or not. But there was a-- what was the calcium sensor? Does anybody remember? Anyhow, there was the calcium sensor, where you were asked in the problem set for a something or other that you asked to measure the KD for. And you can do binding assays. So fluorescence is an incredibly powerful tool as is the take home message. And we've seen it throughout the course. We just haven't talked about it. So now the key thing. And we're going to talk a little bit about fluorescence at probably a freshman level. Many of you guys, who were the undergraduates. You guys, have you done fluorescence experiments? You haven't done in the lab? I thought we had two Eureka labs that were fluorescence oriented. AUDIENCE: [INAUDIBLE]? JOANNE STUBBE: No? AUDIENCE: Yeah, yeah. so we-- [INTERPOSING VOICES] JOANNE STUBBE: So doesn't Tim's? He does sensors to sniff. I don't know what to sniff, but to sniff something, TNT or-- AUDIENCE: Right. but we didn't use fluorescence with that. JOANNE STUBBE: You didn't use fluorescence for that. OK, or the Tokmakoff lab? AUDIENCE: The one experiment we did in lab is we labeled a protein, the green absorbing dye. And it used laser anisotropy to measure KD rotations. And so the-- JOANNE STUBBE: OK, so you guys are experts, then, on fluorescence. Well, hopefully you-- anyhow, so one of the questions is we need to ultimately the key thing for any of this is we're going to have to have a fluorophore. So that's it. So we need the key starting point is a fluorophore And what are fluorophores. So you want something that's usually aromatic and large. It could be-- it could have a lot of nitrogens in it. Oh, I knew I forgot something. So there's a book called, Molecular Probes. OK so I gave you a handout on fluorescence. I forgot to bring the book, if anybody wants to see it. This book is worth its weight in gold if you're a chemical biologist. Because this has everything in the world you need to know about fluorescence. It's described in a thoughtful way. They sell all the probes. If you want to do something to tweak something, they'll help you do all of that. So this book, this is molecular probes book. I think it's online now. I have a copy that's five years old. I use it a lot. It's a really important book. And I got this out of the book. And it just shows you in the book, they have all these pictures of these fluorophores. So they're just big, huge, greasy molecules. You have to worry about solubility a lot of the time. So you have to stick sulfates, or something that ends up making it soluble. So that's going to be a key thing. So we need to have a fluorophore. And we have many options that we can buy these things. OK, so what's this? OK, so what we want to think about is this. So in your, the latest version of your handouts, I've written down what I'm going to say. Butt it's pretty simple for-- I'm talking about this in a pretty simplified viewpoint. But what we're going to see is these fluorophores are going to allow us to. They allow us to do assays. I'm going to show you a quick example of that. That is you can have something that is. You can have a molecule that is quenched, so you have a quencher on one side. I'll show you. And I'll show you the way the quenching comes from, something fluorescent on the other side. You can't see anything. You cut it in half. It could be a protease. It could be a nuclease. The quencher goes away. And you see fluorescence. You could have a sensor for metal binding, which Liz talked about. So you have two fluorophores. OK, you've got to figure out what the right fluorophores are. Something binds. They change confirmation. And they change confirmation in some way that you can actually detect a shift in the wavelength. And then you're looking. In our case, we're just sticking something on the end to see something. You were making a protein fluorescent. That's all we're doing. So you can use it for assays. You can use it for FRET. And in the current-- so you can measure distances. We're not going to go into that. But any of you that are interested in the current version of the handout, I have sort of short tutorial on what FRET is and where you should go to look this up. And then we just basically have a fluorescent tag. OK, and we'll come back and talk about the tag. We already talked about the fact that we have green fluorescent protein, red fluorescent protein tags. But we'll come back and talk about the other tags. So we have a fluorophore And so what does that mean in terms of what's going on. So you have your molecule. And your molecule has a ground state, which we'll call this S0. This is the ground state. And you have many vibrational modes. And you have this big huge fluorophore that can absorb your electron. And your fluorophore can't absorb a photon. And so what happens is. So we're going to have excitation with a photon in a certain way, in a wavelength that can be absorbed by the electron in your molecule to the excited state, which they call S1. And so you can have your electron going to an excited state. And we have a wavelength of light when that happens. And that depends on the structure of your molecule. So you don't want to be in the UV region. You want to be out in the region where you have less interference. And so that's the key game you have to play to get into that region in the visible. You really have to put a lot of stuff on here. You just can't make a small little molecule that absorbs at 600 nanometers. So that's part of the problem. So you're making big things of necessity, so you can actually see something happen. And so then what happens under those conditions. So we're going to have the excitation wavelength of light at a certain lambda max. You absorb. It's just like absorption. You have a certain wavelength that it absorbs more frequently. Then what happens in the excited state on a very fast timescale, you lose energy. OK, so under these conditions, you're doing a relaxation. And then we'll see in a minute. I'll talk about what are the mechanisms of relaxation. But that can tell you. You can use those relaxation mechanisms in a different way to design your fluorescent experiments. So what you see in this cartoon is that you're relaxing on a very fast timescale. And physical chemistry has told us that to see fluorescence, it needs to go down. So these are the vibrational modes. So you're exciting your electron electronically and vibrationally. And then you need to go down in vibrations. You're losing energy somehow. What happens to that energy? OK, we can talk about what can happen to that energy. And when it gets to the lowest level of the excited state, you have fluorescence. OK, and so that also happens on a pretty fast timescale. So the key thing here. So when you get to the lowest. So this is the lowest level, it fluoresces. And so this is where the photon emits. OK, so the photon wavelength for emission or h nu emission. And the key thing that you've probably heard about, again, when you were introduced to fluorescence is because you're losing energy here, what happens to the energy? You're going to a longer wavelength. OK, so the excitation and the emission wavelengths are distinct. And that's called the stoke shift. So it's the wavelength of excitation vs. the wavelength of emissions. So you have a stokes shift, which is the wavelength of excitation minus the wavelength of emission. And so you need to look at molecules. People have spent a lot of time. You saw those 25 lists of things where people have designed things that actually work quite effectively. OK, and so then the question is you losing energy. You are always going to be at longer wavelengths. OK, so that's good, that makes it easier to see, because there aren't that many things inside the cell that give you a background, which is what you need to worry about in all of the experiments you're doing inside the cell. The brightness, we'll come back to that in a minute. So what kinds of models can give you. What kinds of mechanisms are there for relaxation of the excited state. And so there are a number of mechanisms that can be involved. And one is, again, non-radiative relaxation. And how does that happen? So you're changing vibrational modes. And when you're in the excited state, if you're in solution, you have interactions with solvent or other molecules, all of which can affect this kind of transition. If you're in the active site, there can be other things. So the key here is the environment. And again, it could be solvent. It could be protein. And the only way you can tell is by actually looking at the fluorophore on your molecule to end up seeing what you end up seeing. OK, so a second way that you can see. And you probably saw this in your introductory. Yeah? AUDIENCE: So what would be an example? Like, if a unit of the energy being released is a photon in one case for non-radiative, what's the unit of energy? JOANNE STUBBE: What is the unit of energy? So energy, heat is one way that you lose all of this. So it's vibrational energy. I would say, it's mostly heat. So you're changing excitation levels somehow. And the beauty of fluorescence. And this is the key to the sensitivity is you're not doing anything to your molecule. So your electrons got excited. They give off a little heat or whatever. They somehow change a little bit. And then they go back down to the ground state again. So what can you do? You can excite them again. So this can happen over and over and over again, unless the molecule in the excited state becomes destroyed. So that's called photo bleaching. So the key thing here. And this is, I think, this ability to recycle is the key to sensitivity. But again, I haven't used fluorescence inside the cell. I've never done this myself, experimentally. So I don't really know. But you hear about photo bleaching all the time. So I think this is not a trivial thing that you can just blow off. It would be nice. But what you're doing is you're using the same excitation. And then loss and excitation and loss over and over and over again. And so it provides a much more sensitive assay than what you normally see for something like absorption. OK, so let's see. There was one other thing. Oh, so we talked about this mechanism, non radiative relaxation. How else could you relax? You can go from a singlet state to a triplet state. OK, I'm not going to talk. But intersystem crossing, yeah. So you can go from the singlet excited state to the triplet state. I'm not going to talk about this. But the triplet state then can phosphoresce. We're not going to be discussing that at all. But that's one possibility. We just talked about the fact that you can have something in there that quenches the fluorescence. It interacts with something in a distance dependent fashion. And that, again, affects the intensity of your fluorescence. So you also have reaction with the second molecule. And that can become. And it could be good or bad. If it reacts with oxygen, what happens is oxygen, the energy is immediately transferred to the oxygen. So that's why in many fluorescence experiments, you remove oxygen from all of your samples. It acts as a quencher. So you have. And it could be oxygen, which acts as a quencher. Or it could be another fluorophore. In which case. and if everything is set up correctly, you can get the energy to shift the energy of emission can get shifted to longer wavelengths. So that's what FRET is all about. OK, so it could not. A second molecule could be another fluorophore. OK, so those are sort of ways that you can relax. And then you can set up different kinds of experiments, depending upon what the objective is of using fluorescence. So I've written this out in more detail. And for those of you who want to look at FRET, I've defined FRET. I've given you the equations. And people use this quite a bit inside the cell. You need to study this. There are a lot of issues associated with it that you need to think about. And I'll come back. You need to think about. It's not. There are a lot of constants that determine the rate constant for your FRET, OK. And so you just you need to think about all these constants to be able to interpret the data in a thoughtful way. And I've given you a tutorial that I felt was pretty good that I get off the web that just shows what FRET is. And that we have many, many dyes that we can measure distances from 10 to 100 Angstroms using FRET. That's not in this paper. So I didn't. And this just sort of is a cartoon of what I was just telling you. So here, you might have an interaction. But if you cut it, the interaction could be gone. Here, you might have no interaction. But when some small molecule binds, you see an interaction. And you can pick this up using fluorescence changes. OK, so people do these kinds of experiments all the time. And this kind of an assay is extremely-- there are two kinds of assays that one does. So if you work in a pharmaceutical company, people do this all the time. They want a very sensitive assay. Everybody uses fluorescence. They might use an assay like this, where you go from nothing to something. OK, so you have high sensitivity. And the other thing they use is, which I gave you in your handout, is fluorescence polarization, which I'm not going to be talking about. But those are the two major methods that people develop assays around in the pharmaceutical industry. So fluorescence is here to stay. We still need better tools. It can be quantitative. You can measure a quantum efficiency of the electron, light, that's involved in the excitation and the photon that's involved in the emission. If it's 100 percent efficient, then you're quantum efficiency is 1 anyhow. So you have a whole range of quantum efficiencies. OK, so now what I want to do is we're late. But we'll at least get to the other sources telling you what I just told you. OK, so I want to just introduce to you some of the issues that we're going to be facing. And we are going to talk about this in class, probably Monday or on Wednesday morning, OK. So I'll extend this in class. But they've attached green fluorescent protein to all of these things. So this is issue number one. What should they have done in these papers that they didn't do? If you read the paper carefully. I mean, it's hard to read a science paper, because all the key pieces of data are in supplementary information. So they made a few. In all of these, I can't remember what they made. But they made fusion proteins, right? So here, you have a purine enzyme. And here we have some kind of fluorescent protein. So that's the probe they're using. OK, so what's wrong with that, with the way they did their experiments? Can anybody look at the details of what's going on? So what, if you made this fusion, what would be the first thing you would do with a fusion protein? AUDIENCE: My first thought would be, if it changes the activity of the original protein. GFP's a very large [INAUDIBLE]. JOANNE STUBBE: Right. Exactly. GFP I'm going to show you in a second. I think I can show you this in a second. These are just the ways they were looking. But you have all these probes. GFP, it's over here. These are the organic dyes. Here's an antibody. We'll come back to that. So GFP is big. So does it change activity? They didn't assay that. To me, that's mind boggling. OK, because I've dealt with these. I know these proteins, that one protein there. So two of them they're dealing with. One of them is a trifuctional protein. The other one's 150 kilodaltons. These proteins are not easy to deal with, OK. So to me, this is a key thing. So this goes back to the Marcotte paper, where he's saying, well, I mean, maybe these things don't express very well. And they aggregate. They don't fold. We saw how complicated the folding process is. What is the second thing? How did they get the proteins into the cell? How did they get? They don't get proteins into the cell. How did they get? Yeah, how did they get GFP constructs into the cell? AUDIENCE: Transient transfection? JOANNE STUBBE: Yeah, transient transfection, what is the issue there? Without going into details, but what's the issue? AUDIENCE: Like, when the cell's normal mechanism, like the cell's own enzymes maybe-- JOANNE STUBBE: So you do have a normal-- you do have the normal enzyme. They didn't make any effort to knock out the purine enzymes. OK, but I think the key thing with transient transfection is the levels. First of all, a lot of cells don't have anything. But then you don't care about that, because you don't look at them, because they're not fluorescent. OK, but do you think the levels are important. I think the levels are incredibly important. So the question is 100-fold, 1,000 fold over the endogenous levels. And so to me, the first experiments I would have done before I did any of these other experiments is I would have looked at. You might have chosen the trifunctional protein, which they did, because it has activities 2, 3, and 5. And this other big huge protein, which-- so these are the proteins they focus on is activity 4. So 4 is huge. You might think it could function as a scaffolding protein to interact with activities 2, 3, 5. All of that's totally reasonable. OK, but they didn't deal with those issues. So you need to figure out how to attach something that's fluorescent. So one way is genetically. OK, and we've seen this. So we're just fusing GFP onto the protein of interest. Another way in this paper, also, and you mentioned that, is they were using endogenous antibodies. OK, so antibodies can't get into cells. So how do you assay this? So these are also tough experiments. So somehow you fix the cells. So they aren't falling apart when you're trying to perturb the cell to allow the antibodies to get in. And then you permeabilize the cells. Have any of you ever done that? I've done it in yeast. In yeast, it's brutal. I mean, it works. But it's the conditions are like it's a witch's brew. Anyhow, so then you get the antibody in. And that's what you're looking at. And if you look in the-- I have. We're not going to get that far. But I have pictures of-- So when they compared the transient transfection with the endogenous levels, that might give them some feeling for what levels, the levels of expression actually are. And of course, the way that people really want to attach things is using small things, whatever these lists of dyes are that we have. And what are the methods that you guys have learned about to attach these fluorophores. So instead of using a genetic fusion, which is probably. That's a really good way, except the protein, the green fluorescent protein is big. Green fluorescent protein is also a dimer. So people have spent a lot of time engineering green fluorescent protein to be a monomer. So the ones you buy commercially now are all monomers. That would add complexity to everything on top of this. How would you attach some of these things? So we know what the structures of these things are. AUDIENCE: You can do like a halo tag. JOANNE STUBBE: So you could do a halo tag. Have you talked-- we haven't talked about that. So give me another method. Give me a method we've talked about. AUDIENCE: A handle, [INAUDIBLE] handle to attach? JOANNE STUBBE: Yeah, but how would you do that? How do you attach these handles? You want to attach a fluorophore. OK, so it turns out that all of these things here, which you can't see. But these little aromatic things have been synthesized. So click it on. If they could have a settling there. But then it needs to be clicked to something. AUDIENCE: [INAUDIBLE]. JOANNE STUBBE: So you can't just. So how do you click it? AUDIENCE: [INAUDIBLE]. JOANNE STUBBE: So but is that easy to do inside the cell? No. And in mammalian cells, it's impossible. OK, so you can't use unnatural amino acids inside the cell. The technology is not there at this stage. So the question of how you attach this. You could make your. If you could make your protein outside the cell. You might be able to do that. But then you have the problem of getting your protein inside the cell. So getting a probe that's fluorescently, you're labeling the protein of interest is not easy. And Alice Ting's lab, again, has spent a lot of time, not that successfully. But using ligases that you can then incorporate into the cells that can then react with things you put onto your protein to attach fluorophores. But this is an area that's really important, because in my opinion, looking at regulation inside the cell, we don't really want to perturb. We don't want to be at very high levels. And we want to be able to see something to understand regulation. So I think. So anyhow, the issue is that we want to be as small as possible. We don't want to be Brad's lab. What is Brad's lab? Does he use these nanobodies that are antibodies? AUDIENCE: Like an [INAUDIBLE]? JOANNE STUBBE: No. They have all these. They have things called the nanobodies now. So I think they are like the little guys you make on your solid phase peptide synthesizer. But they are specific. They specifically bind to proteins. So there are only five examples that I've seen in the literature. So they act like antibodies. But they're-- huh? AUDIENCE: Like [INAUDIBLE],, like little-- JOANNE STUBBE: They're little tiny proteins that are maybe. I don't know. 50 amino acids that somehow, some guy at the University of Chicago-- not Kent-- developed these things. And they specifically. They act like an antibody. They can specifically interact with a protein of interest. And then you attach a green fluorescent protein onto it. So again, what you have something smaller. So because with these antibodies. What you see is the non-specific, right? I mean, we've seen that. And with fluorescence, that means you have fluorescence background in everything you do. So anyhow, I think we're not that. So that's just you're using fluorescence microscopy. This tells you why you're interested in fluorescence microscopy. And we'll just close here. And we're going to come back and talk about this in class. But this is sort of the example of the data that you need to think about. So what we hear is in the presence of purines, you don't see any of these little dots. You remove the purines. OK, so this is not so easy either, because the way we grow cells, we don't have defined media, right? I mean we're using. I don't know what you guys use now, but fetal calf serum or something. It's got all this stuff in it that we don't really know what it is. We don't use defined media. And apparently, when they-- the Marcotte paper-- when they were describing this, said it was not so easy to remove the purines. And the method they used to remove the purines also removed other stuff, OK. So you're stressing the cell. That was the take home message. So under those conditions, you see something different. OK, and so then they did another experiment, because they were worried about levels. Here, they are they have an antibody to the trifuctional protein. And so this is what they see under low purine conditions. Does this look like this? I don't know. So you can't tell by looking at one picture. OK, so you've got to do statistical analysis of all these things. So I think this sort of-- we'll come back and talk about this in class. But I think this is the first example where people are trying to look at this. The data is interesting. But we've already raised issues of what some of the problems are. And hopefully, you can think about more of the problems.
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https://ocw.mit.edu/courses/7-91j-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: All right. So today, we're going to briefly review classical sequencing and next-gen or second-gen sequencing, which sort of provides a lot of the data that the analytical methods we'll be talking about work on. And we'll then introduce local alignment a la BLAST and some of the statistics associated with that. So just a few brief items on topic one. All right. So today, we're going to talk about sequencing first. Conventional-- or Sanger sequencing-- then next-gen or second-gen sequencing briefly. And then talk about local alignments. So background for the sequencing part, the Metzger review covers everything you'll need. And for the alignment-- we'll talk about local alignment today, global alignment on Tuesday-- then chapters four and five of the text cover it pretty well. So here's the text. If you haven't decided whether to get it or not, I'll have it up here. You can come flip through it after class. Sequencing is mostly done at the level of DNA. Whether the original material was RNA or not, usually convert to DNA and sequence at the DNA level. So we'll often think about DNA as sort of a string. But it's important to remember that it actually has a three dimensional structure as shown here. And often, it's helpful to think of it in sort of a two dimensional representation where you think about the bases and their hydrogen bonding and so forth as shown in the middle. My mouse is not working today for some reason, but hopefully, we won't need it. So the chemistry of sequencing is very closely related to the chemistry of the individual bases. And there are really three main types that are going to be relevant here. Ribonucleotides, deoxyribonucleotides, then for Sanger sequencing, dideoxyribonucleotides. So who can tell me which of these structures corresponds to which of those names? And also, please let me know your name and I'll attempt to remember some of your names toward the end of the semester probably. So, which are which? Yes, what's your name? AUDIENCE: I'm Simona. So the ribonucleotide is the top right. The deoxy is the one below it. And the dideoxy is the one to the left. PROFESSOR: OK, so that is correct. So one way to keep these things in mind is the numbering of the bases. So the carbons in the ribo sugar are numbered one, so carbon 1 is the one where the base is attached. Two is here, which has an OH in RNA and just an H in DNA. And then three is very important. Four, and then five. So five connects to the phosphates, which then will connect the base to the sugar phosphate backbone. And three is where you extend. That's where you're going to add the next base in a growing chain. And so what will happen if you give DNA polymerase a template and some dideoxy nucleotides? It won't be able to extend because there's no 3-prime OH. And all the chemistry requires the OH. And so that's the basis of classical or Sanger sequencing, which Fred Sanger got the Nobel Prize for in the 1980s-- I think it was developed in the '70s-- and it's really the basis of most of the sequencing, or pretty much all the DNA sequencing up until the early 2000s before some newer technologies came about. And it takes advantage of this special property of dideoxy nucleotides that they terminate the growing chain. So imagine we have a template DNA. So this is the molecule whose sequence we want to determine shown there in black. We then have a primer. And notice the primer's written in 5-prime to 3-prime direction. The ends would be primer sequences and then primer complimentary sequences in the template. So you typically will have your template cloned-- this is in conventional sequencing-- cloned into some vector like a phage vector for sequencing so you know the flanking sequences. And then you do four sequencing reactions in conventional Sanger sequencing. And I know some of you have probably had this before. So let's take the first chemical reaction. The one here with a DDGTP. So what would you put in that reaction? What are all the components of that reaction if you wanted to do conventional sequencing on, say, an acrylonitrile? Anyone? What do you need and what does it accomplish? Yeah, what's your name? AUDIENCE: I'm Tim. PROFESSOR: Tim? Oh yeah, I know you, Tim. OK, go ahead. AUDIENCE: So you need the four nucleotides-- the deoxynucleotides. You will need the dideoxy P nucleotides. In addition, you need all the other [INAUDIBLE]. You need polymerase. Generally, you need a buffer of some sort, [INAUDIBLE], to [INAUDIBLE]. PROFESSOR: Yeah, primary template. Yeah. Great. That's good. It sounds like Tim could actually do this experiment. And what ratio would you put in? So you said you're going to put in all four conventional deoxynucleotides and then one dideoxynucleotide. So let's say dideoxy G just for simplicity here. So in what ratio would you put the dideoxynucleotide compared to the conventional nucleotides? AUDIENCE: To lower the concentration. PROFESSOR: Lower? Like how much lower? AUDIENCE: Like, a lot lower. PROFESSOR: Like maybe 1%? AUDIENCE: Yeah. PROFESSOR: Something like that. You want to put it a lot lower. And why is that so important? AUDIENCE: Because you want the thing to be able to progress. Because you need enough of the ribonucleotide concentration so that [INAUDIBLE] every [INAUDIBLE] equivalent or excess and you're going to terminate [INAUDIBLE]. PROFESSOR: Right. So if you put equamolar deoxy G and dideoxy G, then it's going to be a 50% chance of terminating every time you hit a C in the template. So you're going to have half as much of the material at the second G, and a quarter as much as the third, and you're going to have vanishingly small amounts. So you're only going to be able to sequence the first few C's in the template. Exactly. So that's a very good point. So now let's imagine you do these four separate reactions. You typically would have radiolabeled primer so you can see your DNA. And then you would run it on some sort of gel. This is obviously not a real gel, but an idealized version. And then in the lane where you put dideoxy G, you would see the smallest products. So you read these guys from the bottom up. And in this lane there is a very small product that's just one base longer than the primer here. And that's because there was a C there and it terminated there. And then the next C appears several bases later. So you have sort of a gap here. And so you can see that the first base in the template would be a complement of T, or C. And the second base would be, you can see, the next smallest product in this dideoxy T lane, therefore it would be A. And you just sort of snake your way up through the gel and read out the sequence. And this works well. So what does it actually look like in practice? Here are some actual sequencing gels. So you run four lanes. And on big polyacrylamide gels like this. Torbin, you ever run one of these? AUDIENCE: Yes. PROFESSOR: Yes? They're a big pain to cast. Run for several hours, I think. And you get these banding patterns. And what limits the sequence read length? So we normally call the sequence generated from one run of a sequencer as a read. So that one attempt to sequence the template is called a read. And you can see it's relatively easy to read the sequence toward the bottom, and then it gets harder as you go up. And so that's really what fundamentally limits the read length, is that the bands get closer and closer together. So they'll run inversely proportional to size with the small ones running faster. But then the difference between a 20 base product and a 21 might be significant. But the difference between a 500 base product and a 501 base product is going to be very small. And so you basically can't order the lanes anymore. And therefore, that's sort of what fundamentally limits it. All right. So here we had to run four lanes of a gel. Can anyone think of a more efficient way of doing Sanger sequencing? Is there any way to do it in one lane? Yeah, what's your name? AUDIENCE: Adrian. You can use four different types of the entities. Maybe like four different colors. AUDIENCE: Four different colors. OK, so instead of using radio labeling on the primary, you use fluorophore on your dideoxy entities, for example. And then you can run them. Depending where that strand terminated, it'll be a different color. And you can run them all in one lane. OK, so that looks like that. And so this was an important development called terminator sequencing in the '90s. That was the basis of the ABI 3700 machine, which was really the workhorse of genome sequencing in the late '90s and early 2000s. Really what enabled the human genome to be sequenced. And so one of the other innovations in this technology was that instead of having a big gel, they shrunk the gel. And then they just had a reader at the bottom. So the gel was shrunk to as thin as these little capillaries. I don't know if you can see these guys. But basically it's like a little thread here. And so each one of these is effectively-- oops! Oh no. No worries, this is not valuable. Ancient technology that I got for free from somebody. So the DNA would be loaded at the top. There would be a little gel in each of these-- it's called capillary sequencing. And then it would run out the bottom and there would be a detector which would detect the four different flours and read out the sequence. So this basically condensed the volume needed for sequencing. Any questions about conventional sequencing? Yes? AUDIENCE: Where are the [INAUDIBLE] where you'd put the fluorescent flags? Like the topic from the [INAUDIBLE]? PROFESSOR: Yeah, that's a good question. I don't actually remember. I think there are different options available. And sometimes with some of these reactions, you need to use modified polymerases that can tolerate these modified nucleotides. Yeah, so I don't remember that. It's a good question. I can look that up. So how long can a conventional sequencer go? What's the read length? Anyone know? It's about, say, 600 or so. And so that's reasonably long. How long is a typical mammalian mRNA? Maybe two, three kb? So you have in a typical exon, maybe 150 bases or so. So you have a chunk. You don't generally get full length cDNA. But you get a chunk of a cDNA that's say, three, four exons in length. And that is actually generally sufficient to uniquely identify the gene locus that that read came from. And so that was the basis of EST sequencing-- so-called Expressed Sequence Tag sequencing. And millions of these 600 base chunks of cDNA were generated and they have been quite useful over the years. All right. So what is next-gen sequencing? So in next-gen sequencing, you only read one base at a time. So it's often a little bit slower. But it's really massively parallel. And that's the big advantage. And it's orders of magnitude cheaper per base than conventional sequencing. Like when it first came out it, it was maybe two orders of magnitude cheaper. And now it's probably another four orders of magnitude. So it really blows away conventional sequencing if the output that you care about is mostly proportional to number of bases sequence. If the output is proportional to the quality of the assembly or something, then there are applications where conventional sequencing still is very useful Because the next-gen sequencing tends to be shorter. But in terms of just volume, it generates much, much more bases in one reaction. And so the basic ideas are that you have your template DNA molecules. Now typically, tens of thousands for technologies like PacBio or hundreds of millions for technologies like Illumina that are immobilized on some sort of surface-- typically a flow cell-- and there are either single molecule methods where you have a single molecule of your template or there are methods that locally amplify your template and produce, say, hundreds of identical copies in little clusters. And then you use modified nucleotides, often with fluorophores attached, to interrogate the next base at each of your template molecules for hundreds and hundreds of millions of them. And so there are several different technologies. We won't talk about all of them. We'll just talk about two or three that are interesting and widely used. And they differ depending on the DNA template, what types of modified nucleotides are used, and to some extent, in the imaging and the image analysis, which differs for single molecule methods, for example, compared to the ones that sequence a cluster. So there's a table in the Metzger review. And so I've just told you that next-gen sequencing is so cheap. But then you see how much these machines cost and you could buy lots of other interesting things with that kind of money. And I also want to emphasize that that's not even the full cost. So if you were to buy an Illumina GA2-- this would be like a couple years ago when the GA2 was the state of the art-- for half a million dollars, the reagents to run that thing, if you're going to run it continuously throughout the year, the reagents to run it would be over a million. So this actually underestimates the cost. However, the cost per base is super, super low. Because they generate so much data at once. All right, So we'll talk about a couple of these. The first next-gen sequencing technology to be published and still used today was from 454-- now Roche-- and it was based on what's called emulsion PCR. So they have these little beads, the little beads have adapter DNA molecules covalently attached. You incubate the beads with DNA, and you actually make an emulsion. So it's an oil water emulsion. So each bead, which is hydrophilic, is in the little bubble of water inside oil. And the reason for that is so that you do it at a template concentration that's low enough that only a single molecule of template is associated with each bead. So the oil then provides a barrier so that the DNA can't get transferred from one bead to another. So each bead will have a unique template molecule. You do sort of a local PCR-like reaction to amplify that DNA molecule on the bead, and then you do sequencing one base at a time using a luciferase based method that I'll show you on the next slide. So Illumina technology differs in that instead of an emulsion, you're doing it on the surface of a flow cell. Again, you start with a single molecule of template. Your flow cell has these two types of adapters covalently attached. The template anneals to one of these adapters. You extend the adapter molecule with dNTPs and polymerase. Now you have the complement of your template, your denature. Now you have the inverse complement of your template molecule covalently attached to the cell surface. And then at the other end there's the other adapter. And so what you could do is what's called bridge amplification where that now complement of the template molecule will bridge over hybridized to the other adapter, and then you can extend that adapter. And now you've regenerated your original template. And so now you have the complementary strand, and the original strand, your denature. And then each of those molecules can undergo subsequent rounds of bridge amplification to make clusters of typically several hundred thousand molecules. Is that clear? Question. Yeah, what's your name? AUDIENCE: Stephanie. How do they get the adapters onto the template molecules? PROFESSOR: How do you get the adapters onto the template molecules? So that's typically by DNA ligation. So we may cover that in later steps. It depends. There's a few different protocol. So for example, if you're sequencing microRNAs, you typically would isolate the small RNAs and use RNA litigation to get the adapters on. And then you would do an RT step to get DNA. With most other applications like RNA-seq or genome sequencing-- so with RNA-seq, you're starting from mRNA, you typically will isolate total RNA, do poly(A) selection, you fragment your RNA to reduce the effects of secondary structure, you random prime with, like, random hexamers RT enzyme. So that'll make little bits of cDNA 200 bases long. You use second strand synthesis. Now you have double stranded cDNA fragments. And then you do, like, blunt end ligation to add the adapters. And then you denature so you have single strand. AUDIENCE: I guess my question is how do you make sure that the two ends sandwiching the DNA are different as opposed to-- PROFESSOR: That the two ends are different. Yeah, that's a good question. I'll post some stuff about-- It's a good question. I don't want to sweep it under the rug. But I kind of want to move on. And I'll post a little bit about that. All right so we did 454 Illumina. Helicos is sort of like Illumina sequencing except single molecule. So you have your template covalently attached to your substrate. You just anneal primer and just start sequencing it And there's major pros and cons of single molecule sequencing, which we can talk about. And then the PacBio technology is fundamentally different in that the template is not actually covalently attached to the surface. The DNA polymerase is covalently attached to the surface and the template is sort of threaded into the polymerase. And this is a phage polymerase that's highly processive and strand displacing. And the template is often a circular molecule. And so you can actually read around the template multiple times, which turns out to be really useful in PacBio because the error rate is quite high for the sequencing. So in the top, in the 454, you're measuring luciferase activity-- light. In Illumina, you're measuring fluorescence. Four different fluorescent tags, sort of like the four different tags we saw in Sanger sequencing. Helicose, it's single tag one base at a time. And in PacBio, you actually have a fluorescently labeled dNTP that has the label on-- it's actually hexaphosphate-- it's got the label on the sixth phosphate. So the dNTP is labeled. It enters the active site of the DNA polymerase. And the residence time is much longer if the base is actually going to get incorporated into that growing chain. And so you measure how much time you have a fluorescent signal. And if it's long, that means that that base must have incorporated into the DNA. But then, the extension reaction itself will cleave off the last five phosphates and the fluorophore tag. And so you'll regenerate native DNA. So that's another difference. Whereas in Illumina sequencing, as we'll see, there's this reversible terminator chemistry. So the DNA is not native that you're synthesizing. So this is just a little bit more on 454. Just some pretty pictures. I think I described that before. The key chemistry here is that you add one dNTP at a time. So only a subset of the wells-- perhaps a quarter of them-- that have that next base, the complementary base free-- as the next one after the primer-- will undergo synthesis. And when they undergo synthesis, you release pyrophosphate. And they have these enzymes attached to these little micro beads-- the orange beads-- sulfurylase and luciferase, that use pyrophosphate to basically generate light. And so then you have one of these beads in each well. You look at which wells lit up when we added dCTP. And they must have had G as the next base and so forth. And there's no termination here. The only termination is because you're only adding one base at a time. So if you have a single gene in the template, you'll add one base. But if you have two Gs in the template, you'll add two Cs. And in principle, you'll get twice as much light. But then you have to sort of do some analysis after the fact to say, OK how much light do we have? And was that one G, two G, and so forth. And the amount of light is supposed to be linear up to about five or six Gs. But that's still a more error-prone step. And the most common type of error in 454 is actually insertions and deletions. Whereas in Illumina sequencing, it's substitutions. David actually encouraged me to talk more about sequencing errors and quality scores. And I need to do a little bit more background. But I may add that a little bit later in the semester. OK, so in Illumina sequencing, you add all four dNTPs at the same time. But they're non-native. They have two major modifications. So one is that they're three prime blocked. That means that the OH is not free, I'll show the chemical structure in a moment. So you can't extend more than one base. You incorporate that one base, and the polymerase can't do anything more. And they're also tagged with four different fluors. So you add all four dNTPs at once. You let the polymerase incorporate them. And then you image the whole flow cell using two lasers and two filters. So basically, to image the four fluors. So you have to sort of take four different pictures of each portion of the flow cell and then the camera moves and you scan the whole cell. And so then, those clusters that incorporated a C, let's say, they will show up in the green channel as spots. And those incorporated in A, and so forth. So you basically have these clusters, each of them represents a distinct template, and you read one base at a time. So, first you read the first base after the primer. So it's sequencing downwards into the template. And you read the first base so you know what the first base of all your clusters is. And then you reverse the termination. You cleave off that chemical group that was blocking the 3-prime OH so now it can extend again. And then you add the four dNTPs again, do another round of extension, and then image again, and so forth. And so it takes a little while. Each round of imaging takes about an hour. So if you want to do 100 base single and Illumina sequencing, it'll be running on the machine for about four days or so. Plus the time you have to build the clusters, which might be several hours on the day before. So what is this? So actually the whole idea of blocking termination-- basically Sanger's idea-- is carried over here in Illumina sequencing with a little twist. And that's that you can reverse the termination. So if you look down here at the bottom, these are two different 3-prime terminators. Remember your base counting. Base one, two, three. So this was the 3-prime OH, now it's got this methyl [INAUDIBLE], or whatever that is. I'm not much of a chemist, so you can look that one up. And then here's another version. And this is sort of chemistry that can cleave this off when you're done. And then this whole thing here, hanging off the base, is the fluor. And you cleave that off as well. So you add this big complicated thing, you image it, and then you cleave off the fluor and cleave off the 3-prime block. These are some actual sequencing images you would image in the four channels. They're actually black and white. These are pseudocode. And then you can merge those and you can see then all the clusters on the flow cell. So this is from a GA2 with the recommended cluster density back in the day, like a few years ago. And nowadays, the image now since the software has gotten a lot better, so you can actually load the clusters more densely and therefore get more sequence out of the same area. But imagine just millions and millions of these little clusters like this. Notice the clusters are not all the same size. Basically, you're doing PCR in situ, and so some molecules are easier to amplify by PCR than others. And that probably accounts for these variations in size. So what is the current throughput? These data are accurate as of about, maybe, last year. So the HiSeq 2000 instrument is the most high performance, widely used instrument. Now there's a 2500, but I think it's roughly similar. You have one flow cell. So a flow cell looks sort of like a glass slide, except that it has these tunnels carved in it like eight little tubes inside the glass slide. And on the surfaces of those tubes is where the adapters are covalently attached. And so you have eight lanes and so you can sequence eight different things in those eight lanes. You could do yeast genome in one and fly RNA-seq in another, and so forth. And these days, a single lane will produce something like 200 million reads. And this is typically routine to get 200 million reads from a lane. Sometimes you can get more. You can do up to 100 bases. You can do 150 these days on a MiSeq, which is a miniature version. You can do maybe 300 or more. And so that's a whole lot of sequence. So that's 160 billion bases of sequence from a single lane. And that will cost you-- that single lane-- maybe $2,000 to $3,000, depending where you're doing it. And the cost doesn't include the capital cost, that's just the reagent cost for running that. So 160 billion-- the human genome is 3 billion, so you've now sequenced the human genome over many times there. You can do more. So you can do paired-end sequencing, where you sequence both ends of your template. And that'll basically double the amount of sequence you get. And you can also, on this machine, do two flow cells at once. So you can actually double it beyond that. And so for many applications, 160 billion bases is overkill. It's more than you need. Imagine you're doing bacterial genome sequencing. Bacterial genome might be five megabases or so. This is complete overkill. So you can do bar coding where you add little six base tags to different libraries, and then mix them together, introduce them to the machine, sequence the tags first or second, and then sequence the templates. And then you effectively sort them out later. And then do many samples in one lane. And that's what people most commonly do. So, questions about next-gen sequencing? There's a lot more to learn. I'm happy to talk about it more. It's very relevant to this class. But I'm sure it'll come up later in David's sections, so I don't want to take too much time on it. So, now once you generate reads from an Illumina instrument or some other instrument, you'll want to align them to the genome to determine, for example, if you're doing RNA-seq mapping reads that come from mRNA, you'll want to know what genes they came from. So you need to map those reads back to the genome. What are some other reasons you might want to align sequences? Just in general, why is aligning sequences-- meaning, matching them up and finding individual bases or amino acid residues that match-- why is that useful? Diego? AUDIENCE: You can assemble them if you want. PROFESSOR: You can assemble them? Yes. So if you're doing genome sequencing, if you align them to each other and you find a whole stack that sort of align this way, you can then assemble and infer the existence of a longer sequence. That's a good point. Yes, your name? AUDIENCE: Julianne. Looking at homologs. PROFESSOR: Looking at homologs. Right. So if you, for example, are doing disease gene mapping, you've identified a human gene of unknown function that's associated with a disease. Then you might want to search it against, say, the mouse database and find a homolog in mouse and then that might be what you would want to study further. You might want to then knock it out in mouse or mutate it or something. So those are some good reasons. There's others. So we're going to first talk about local alignment, which is a type of alignment where you want to find shorter stretches of high similarity. You don't require alignment of the entire sequence. So there are certain situations where you might want to do that. So here's an example. You are studying a recently discovered human non-coding RNA. As you can see, it's 45 bases. You want to see if there's a mouse homolog. You run it through NCBI BLAST, which as we said is sort of the Google search engine of mathematics-- and you're going get a chance to do it on pump set one, and you get a hit that looks like this. So notice, this is sort of BLAST notation. It says Q at the top. Q is for "query," that's the sequence you put in. S is "subject," that's the database you were searching against. You have coordinates, so 1 to 45. And then, in the subject, it happened to be base 403 to 447 in some mouse chromosome or something. And you can see that it's got some matching. But it also has some mismatches. So in all, there are 40 matches and five mismatches in the alignment. So is that significant? Remember, the mouse genome is 2.7 billion bases long. It's big. So would you get a match this good by chance? So the question is really, should you trust this? Is this something you can confidently say, yes mouse is a homolog, and that's it? Or should you just be like, well, that's not better than I get by chance so I have no evidence of anything? Or is it sort of somewhere in between? And how would you tell? Yeah, what's your name? AUDIENCE: Chris. You would want to figure out a scoring function for the alignment. And then, with that scoring function, you would find whether or not you have a significant match. PROFESSOR: OK. So Chris says you want to define a scoring system and then use the scoring system to define statistical significance. Do want to suggest a scoring system? What's the simplest one you can think of? AUDIENCE: Just if there's a match, you add a certain score. If it's a mismatch, you subtract a certain score. PROFESSOR: So let's do that scoring system. So the notation that's often used is Sii. So that would be a match between nucleotide i and then another copy of nucleotide i. We'll call that 1, plus 1 for a match. And sij, where i and j are different, we'll give that a negative score. Minus 1. So this is i not equal to j. So that's a scoring matrix. It's a four by four matrix with 1 on the diagonal and minus 1 everywhere else. And this is commonly used for DNA. And then there's a few other variations on this that are also used. So good, a scoring system. So then, how are we going to do the statistics? Any ideas? How do we know what's significant? AUDIENCE: The higher score would probably be a little more significant than a lower score. But the scale, I'm not sure-- PROFESSOR: The scale is not so obvious. Yes, question? AUDIENCE: My name is Andrea. So if you shuffled the RNA, like permute the sequence, then we'll get the [INAUDIBLE] genome you get with that shuffled sequence. And the score is about the same as you'd get with the non-shuffled sequence [INAUDIBLE] about very significant scores. PROFESSOR: Yeah, so that's a good idea. BLAST, as it turns out-- is pretty fast. So you could shuffle your RNA molecule, randomly permute the nucleotides many times, maybe even like 1,000 times, search each one against the mouse genome, and get a distribution of what's the best score-- the top score-- that you get against a genome, look at that distribution and say whether the score of the actual one is significantly higher than that distribution or just falls in the middle of that somewhere. And that's reasonable. You can certainly do that, and it's not a bad thing to do. But it turns out there is an analytical theory here that you can use. And so that you can determine significance more quickly without doing so much computation. And that's what we'll talk about. But another issue, before we get to the statistics, is how do you actually find that alignment? How do you find the top scoring match in a mouse genome? So let's suppose this guy is your RNA. OK, of course, we're using T's, but that's just because you usually sequences it at the DNA level. But imagine this is your RNA. It's very short. This is like 10 or so, I think. And this is your database. But it goes on a few billion more. Several more blackboards. And I want to come up with an algorithm that will find the highest scoring segment of this query sequence against this database. Any ideas? So this would be like our first algorithm. And it's not terribly hard, so that's why it's a good one to start with. Not totally obvious either. Who can think of an algorithm or something, some operation that we can do on this sequence compared to this sequence-- in some way-- that will help us find the highest scoring match? I'm sorry. Yeah? AUDIENCE: You have to consider insertion and deletion. PROFESSOR: Yeah, OK. So we're going to keep it simple. That's true, in general. But we're going to keep it simple and just say no insertions and deletions. So we're going to look for an ungapped local alignment. So that's the algorithm that I want. First, no gaps. And then we'll do gaps on Tuesday. Tim? AUDIENCE: You could just compare your [INAUDIBLE] to [INAUDIBLE] all across the database and turn off all the [INAUDIBLE] on that [INAUDIBLE], and then figure out [INAUDIBLE]. PROFESSOR: Yeah, OK. Pretty much. I mean, that's pretty much right. Although it's not quite as much of a description as you would need if you want to actually code that. Like, how would you actually do that? So, I want a description that is sort of more at the level of pseudocode. Like, here's how you would actually organize your code. So, let's say you entertain the hypothesis that the alignment can be in different registers. The alignment can correspond to base one of the query and base one of the subject. Or it could be shifted. It could be an alignment where base 1 of the query matches these two, and so forth. So there's sort of different registers. So let's just consider one register first. The one where base 1 matches. So let's just look at the matches between corresponding bases. I'm just going to make these little angle bracket guys here. Hopefully I won't make any mistakes. I'm going to take this. This is sort of implementing Tim's idea here. And then I'm going to look for each of these-- so consider it going down here. Now we're sort of looking at an alignment here. Is this a match or a mismatch? That's a mismatch. That's a match. That's a mismatch. That's a mismatch. That's a match. Match Match. Mismatch. Mismatch. Mismatch. So where is the top scoring match between the query and the subject? Tim? Anyone? AUDIENCE: 6, 7, 8. PROFESSOR: 6, 7, 8. Good. Oh-- AUDIENCE: 5, 6, 7. PROFESSOR: 5, 6, 7. Right. Right here. You can see there's three in a row. Well, what about this? Why can't we add this to the match? What's the reason why it's not 2, 3, 4, 5, 6, 7? AUDIENCE: Because the score for that is lower. PROFESSOR: Because the score for that is lower. Right. We defined top scoring segment. You sum up the scores across the map. So you can have mismatches in there, but this will have a score of 3. And if you wanted to add these three bases, you would be adding negative 2 and plus 1, so it would reduce your score. So that would be worse. Any ideas on how to do this in an automatic, algorithmic way? Yeah? What's your name? AUDIENCE: Simon. So if you keep shifting the entire database, [INAUDIBLE]. PROFESSOR: OK so you keep shifting it over, and you generate one of these lines. But imagine my query was like 1,000 or something. And my database is like a billion. How do I look along here? And here it was obvious what the top scoring match is. But if I had two matches here, then we would've actually had a longer match here. So in general, how do I find that that top match? For each of those registers, if you will, you'll have a thousand long diagonal here with 1's and minus 1's on it. How do I process those scores to find the top scoring segment? What's an algorithm to do that? It's kind of intuitively obvious, but I want to do something with, you define a variable and you update it, and you add to it, and subtract. Something like that. But, like a computer could actually handle. Yeah? What was your name? Julianne? AUDIENCE: Could you keep track of what the highest total score is, and then you keep going down the diagonal, And then you update it? PROFESSOR: OK. You keep track of what the highest total score was? AUDIENCE: Yeah. The highest test score. PROFESSOR: The highest segment score? OK. I'm going to put this up here. And we'll define max s. That's the highest segment score we've achieved to date. And we'll initialize that to zero, let's say. Because if you had all mismatches, zero would be the correct answer. If your query was A's and your subject was T's. And then what do you do? AUDIENCE: As you go down the diagonal, you keep track of-- PROFESSOR: Keep track of what? AUDIENCE: So you look at 1 in 1 first. And then you go 1 in 2, and you find a score of zero. But that's higher than negative 1. PROFESSOR: But the score of the maximum segment at that point, after base 2, is not zero. It's actually 1. Because you could have a segment of one base alignment. The cumulative score is zero. I think you're onto something here that may be also something useful to keep track of. Let's do the cumulative score and then you tell me more. We'll define cumulative score variable. We'll initialize that to zero. And then we'll have some for loops that, as some of you have said, you want to loop through the subject. All the possible registers of the subject. So that would be maybe j equals 1 to subject length minus query length. Something like that. Don't worry too much about this. Again, this is not real code, obviously. It's pseudocode. So then this will be, say, 1 to query language. And so this will be going along our diagonal. And we're going to plot the cumulative score. So here you would you have an update where cumulative score plus equals the score of query position i matched against subject position j. And update that. So that's just cumulative score. So what will it look like? So in this case, I'll just use this down here. So you have zero, 1, 2, minus 1, minus 2. So you'll start at position zero in the sequence. At position 1 you're down here at minus 1 because it was a mismatch. Then at position 2, as you said, we're back up to zero. And then what happens? Go down to minus 1, down to minus 2. Then we go up three times in a row until we're up here to 1. And then we go down after that. So where is your highest scoring match in this cumulative score plot? People said it was from 5 to 7. Yeah, question? AUDIENCE: So would it be from like a local minimum to a local maximum? PROFESSOR: Yeah. Exactly. So, what do you want to keep track of? AUDIENCE: You want to keep track of the minimum and the maximum. And look for the range which you maximize to different-- PROFESSOR: Yeah, so this is now sort of more what I was looking for in terms of-- so this was the local minimum, and that's the local maximum. This is the score. That's your mass s there. And you also want to keep track of where that happened in both the query and the subject. Does that make sense? So you would keep track of this running cumulative score variable. You keep track of the last minimum. The minimum that you've achieved so far. And so that would then be down here to minus 2. And then when your cumulative score got up to plus 1, you always take that cumulative score, minus the last minimum cumulative score. That gives you a potential candidate for a high scoring segment. And if that is bigger than your current max high scoring segment, then you update it and you would update this. And then you would also have variables that would store where you are. And also, where did that last minimum occur. So I'm not spelling it all out. I'm not going to give you all the variables. But this is an algorithm that would find the maximum score. Yeah, question? AUDIENCE: So you're keeping track of the global maximum, local minimum, so that you can accept the most recent local minimum following the global maximum? PROFESSOR: I'm not sure I got all that. But you're keeping track of the cumulative score. The minimum that that cumulative score ever got to. And the maximum difference, the maximum that you ever in the past have gone up. Where you've had a net increment upwards. Like here. So this variable here, this max s, it would be initialized to zero. When you got to here, your last minimum score would be minus 1. Your cumulative score would be zero. You would take the difference of those, and you'd be like, oh I've got a high scoring segment of score one. So I'm going to update that. So now, that variable is now 1 at this point. Then you're going down, so you're not getting anything. You're just lowering this minimum cumulative score down to minus 2 here. And then when you get to here, now you check the cumulative score minus the last minimum. It's 1. That's a tie. We won't keep track of ties. Now at here, that difference is 2. So now we've got a new record. So now we update this maximum score to 2 in the locations. And then we get here, now it's 3, and we update that. Does that make sense? AUDIENCE: Imagine the first dip-- instead of going down to negative 1, it went down to negative 3. PROFESSOR: Negative 3? AUDIENCE: That first dip. PROFESSOR: Right here? So we started back a little bit. So back here, like this? AUDIENCE: No. PROFESSOR: Down to negative 3? AUDIENCE: No. PROFESSOR: But how do we get to negative 3? Because our scoring is this way. You want this dip to minus 3? AUDIENCE: No PROFESSOR: This one minus 3? Imagine we are at minus 3 here? AUDIENCE: Yeah. Imagine it dipped to minus 3. And then the next one dipped to higher than that, to minus 2. And then it went up to 1. And so, would the difference you look at be negative 2 to 1, or negative 3 to 1? PROFESSOR: Like that, right? So, minus 3, let's say, minus 2, 1. Something like that. What do people think? Anyone want to--? AUDIENCE: Minus 3 to 1. PROFESSOR: Minus 3 to 1. It's the minimum that you ever got to. This might be a stronger match, but this is a higher scoring match. And we said we want higher scoring. So you would count that. AUDIENCE: So you keep track of both the global minimum and the global maximum, and you take the difference between them. PROFESSOR: You keep track of the global minimum and the current cumulative score, and you take the difference. AUDIENCE: The global maximum-- PROFESSOR: It's not necessarily global maximum because we could be well below zero here. We could do like this. From here to here. So this is not the global maximum. This just happens to be, we went up a lot since our last minimum. So that's your high scoring segment. Does that make sense? I haven't completely spelled it out. But I think you guys have given enough ideas here that there;s sort of the core of an algorithm. And I encourage you to think this through afterwards and let me know if there are questions. And we could add an optional homework where I ask you to do this, that we've sometimes had in the past. It is a useful thing to look at. This is not exactly how the BLAST algorithm works. It uses some tricks for faster speed. But this is sort of morally equivalent to BLAST in the sense that it has the same order of magnitude running time. So this algorithm-- what is the running time in Big-O notation? So just for those who are non-CS people, when you use this Big-O notation, then you're asking, how does the running time increase in the size of the input? And so what is the input? So we have two inputs. We have a query of length. And let's say subject of length n. So clearly, if those are bigger, it'll take longer to run. But when you compare different algorithms, you want to know how the run time depends on those lengths. Yes. What's your name? AUDIENCE: Sally. m times n. PROFESSOR: So with this, this is what you would call an order mn algorithm. And why is that? How can you see that? AUDIENCE: You have two for loops And for each length, essentially, you're going through everything in the query. And then, for everything that you go through in the query, you would [INAUDIBLE]. PROFESSOR: Right. In this second for loop here, you're going through the query. And you're doing that nested inside of a for loop that's basically the length of the subject. And eventually you're going to have to compare every base in the query to every base in the subject. There's no way around that. And that takes some unit of time. And so the actual time will be proportional to that. So the bigger n gets and m gets, it's just proportional to the product. Does that make sense? Or another way to think about it is, you're clearly going to have to do something on this diagonal. And then you're going to have to do something on this diagonal, and this one, and this one. And actually, you have to also check these ones here. And in the end, the total number of computations there is going to be this times that. You're basically doing a rectangle's worth of computations. Does that makes sense? So that's not bad, right? It could be worse. It could be, like, mn squared or something like that. So that's basically why BLAST is fast. So what do these things look like, in general? And what is the condition on our score for this algorithm to work? What if I gave a score of plus 1 for a match, and zero for a mismatch? Could we do this? Joe, you're shaking your head. AUDIENCE: It would just be going up. PROFESSOR: Yeah. The problem is, it might be flat for a while, but eventually it would go up. And it would just go up and up and up. And so your highest scoring segment would, most of the time, be something that started very near the beginning and ended very near the end. So that doesn't work. So you have to have a net negative drift. And the way that's formalized is the expected score has to be negative. So why is the expected score negative in this scoring system that has plus 1 for a match, and minus 1 for a mismatch? Why does that work? AUDIENCE: It should be wrong three quarters of the time. PROFESSOR: Yeah. You'll have a mismatch three quarters of the time. So on average, you tend to drift down. And then you have these little excursions upwards, and those are your high scoring segments. Any questions about that? AUDIENCE: Question. Is there something better than m times n? PROFESSOR: We've got some computer scientists here. David? Better than m times n? I don't think so, because you have to do all those comparisons. And so there's no way around that, so I don't think so. All right. But the constant-- you can do better on the constant than this algorithm thing. AUDIENCE: With multiple queries-- PROFESSOR: With multiple queries, yeah. Then you can maybe do some hashing or find some-- to speed it up. OK, so what about the statistics of this? So it turns out that Karlin and Altschul developed some theory for just exactly this problem. For searching a query sequence. It can be nucleotide or protein as long as you have integer scores and the average-- or the expected-- score is negative, then this theory tells you how often the highest score of all-- across the entire query database comparison-- exceeds a cut off x using a local alignment algorithm such as BLAST. And it turns out that these scores follow what's called an extreme value or Gumbel distribution. And it has this kind of double exponential form here. So x is some cut off. So usually x would be the score that you actually observed when you searched your query against the database. That's the one you care about. And then you want to know, what's the probability we would've seen something higher than that? Or you might do x is one less than the score you observed. So what's the chance we observed something the same, as good as this, or better? Does that make sense? And so this is going to be your P value then. So the probability of S. The score of the highest segment under a model where you have a random query against a random database of the same length is 1 minus e to the minus KMN e to the minus lambda x. Where M and N are the lengths of the query and the database. x is the score. And then K and lambda are two positive parameters that depend actually on the details of your score matrix and the composition of your sequences. And it turns out that lambda is really the one that matters. And you can see that because lambda is up there in that exponent multiplying x. So if you double lambda, that'll have a big effect on the answer. And K, it turns out, you can mostly ignore it for most purposes. So as a formula, what does this thing look like? It looks like that. Kind of a funny shape. It sort of looks like an umlauf a little bit, but then has a different shape on the right than the left. And how do you calculate this lambda? So I said that lambda is sort of the key to all this because of its uniquely important place in that formula, multiplying the score. So it turns out that lambda is the unique positive solution to this equation here. So now it actually depends on the scoring matrix. So you see there's sij there. It depends on the composition of your query. That's the pi's. The composition of your subject, that's the rj's. You sum over the i and j equal to each of the four nucleotides. And that sum has to be 1. So there's a unique positive solution to this equation. So how would we solve an equation like this? First of all, what kind of equation is this, given that we're going to set the sij, and we're going to just measure the pi and the rj? So those are all known constants, and lambda is what we're trying to solve for here. So what kind of an equation is this in lambda? Linear? Quadratic? Hyperbolic? Anybody know what this is? So this is called a transcendental equation because you have different powers. That sounds kind of unpleasant. You don't take a class in transcendental equations probably. So in general, they're not possible to solve analytically when they get complicated. But in simple cases, you can solve them analytically. And in fact, let's just do one. So let's take the simplest case, which would be that all the pi's are a quarter. All the ri's are a quarter. And we'll use the scoring system that we came up with before, where sii is 1, and sij is minus 1. If i does not equal j. And so when we plug those in to that sum there, what do we get? We'll get four terms that are one quarter, times one quarter, times e to the lambda. There's four possible types of matches, right? They have probability one quarter times a quarter. That's pi and rj. And the e to the lambda sii is just e to the lambda because sii is 1. And then there's 12 terms that are one quarter, one quarter, e to the minus lambda. Because there's the minus 1 score. And that has to equal 1. So cancel this, we'll multiply through by 4, maybe. So now we get e to the lambda plus 3. e to the minus lambda equals 1. It's still a transcendental equation, but it's looking a little simpler. Any ideas how to solve this for lambda? Sally? AUDIENCE: Wouldn't the 1 be 4? PROFESSOR: I'm sorry. 4. Thank you. Yeah, what's your name? AUDIENCE: [INAUDIBLE] I think [INAUDIBLE] quadratic equation. If you multiply both sides by [INAUDIBLE] then [INAUDIBLE]. PROFESSOR: OK, so the claim is this is basically a quadratic equation. So you multiply both sides by e to the lambda. So then you get e to the 2 lambda plus 3. And then it's going to move this over and do minus 4 e to the lambda equals zero. Is that good? So how is it quadratic? What do you actually do to solve this? AUDIENCE: Well, [INAUDIBLE]. PROFESSOR: Change the variable, x equals e to the lambda. Then it's quadratic in x. Solve for x. We all know how to solve quadratic equations. And then substitute that for lambda. OK, everyone got that? If you use 16 different scores to represent all the different types of matches and mismatches, this will be very unpleasant. It's not unsolvable, it's just that you have to use computational numerical methods to solve it. But in simple cases where you just have a couple different types of scores, it will often be a quadratic equation. All right. So let's suppose that we have a particular scoring system-- particular pi's, rj's-- and we have a value of lambda that satisfies those. So we've solved this quadratic equation for lambda. I think we get lambda equals natural log 3, something like that. Remember, it's a unique positive solution. Quadratic equations are two solutions, but there's going to be just one positive one. And then we have that value. It satisfies this equation. So then, what if we double the scores? Instead of plus 1 minus 1, we use plus 2 minus 2? What would then happen? You can see that the original version of lambda wouldn't necessarily still satisfy this equation. But if you think about it a little bit, you can figure out what new value of lambda would satisfy this equation. We've solved for the lambda that solves with these scores. Now we're going to have new scores. sii prime equals 2. sij prime equals minus 2. What is lambda prime? The lambda that goes with these scores? Yeah, go ahead. AUDIENCE: Half of the original? PROFESSOR: Half of the original? Right. So you're saying that lambda prime equals lambda over 2. And why is that? Can you explain? AUDIENCE: Because of the [INAUDIBLE]. PROFESSOR: Yeah, if you think about these terms in the sum, the s part is all doubling. So if you cut the lambda apart, and the product will equal what it did before. And we haven't changed the pi's and rj's, so all those terms will be the same. So therefore, it will still satisfy that equation. So that's another way of thinking about it. Yes, you're correct. So if you double the scores, lambda will be reduced by a factor of 2. So what does that tell us about lambda? What is it? What is its meaning? Yeah, go ahead, Jeff. AUDIENCE: Scale of the distribution to the expectant score? Or the range score? PROFESSOR: Yeah. It basically scales the scores. So we can have the same equation here used with arbitrary scoring. It just scales it. You can see the way it appears as a multiplicative factor in front of the score. So if you double all the scores, will that change what the highest scoring segment is? No, it won't change it because you'll have this cumulative thing. It just changes how you label the y-axis. It'll make it bigger, but it won't change what that is. And if you look at this equation, it won't change the statistical significance. The x will double in value, because all the matches are now worth twice as much as what they were before. But lambda will be half as big, and so the product will be the same and therefore, the final probability will be the same. So it's just a scaling factor for using different scoring systems. Everyone got that? All right. So what scoring matrix should we use for DNA? How about this one? So this is now a slight generalization. So we're going to keep 1 for the matches. You don't lose any generality by choosing 1 here for matches, because if you use 2, then lambda is just going to be reduced to compensate. So 1 for matches. And then we're going to use m for mismatches. And m must be negative in order to satisfy this condition for this theory to work, that the average score has to be negative. Clearly, you have to have some negative scores. And the question then is, should we use minus 1 like we used before? Or should we use like minus 2 or minus 5, or something else? Any thoughts on this? Or does it matter? Maybe it doesn't matter. Yeah, what's your name? AUDIENCE: [INAUDIBLE]. Would it make sense to not use [INAUDIBLE], because [INAUDIBLE]. PROFESSOR: Yeah, OK. So you want to use a more complicated scoring system. What particular mismatches would you want to penalize more and less? AUDIENCE: [INAUDIBLE] I think [INAUDIBLE] needs to be [INAUDIBLE]. PROFESSOR: Yeah, you are correct in your intuition. Maybe one of the biologists wants to offer a suggestion here. Yeah, go ahead. AUDIENCE: So it's a mismatch between purine and pyrimidine [INAUDIBLE]. PROFESSOR: OK so now we've got purines and pyrimidines. So everyone remember, the purines are A and G. The pyrimidines are C and T. And the idea is that this should be penalized, or this should be penalized less than changing a purine to a pyrimidine. And why does that makes sense? AUDIENCE: Well, structurally they're-- PROFESSOR: Structurally, purines are more similar to each other than they are to pyrimidines. And? More importantly, I think. In evolution? AUDIENCE: [INAUDIBLE]. PROFESSOR: I'm sorry, can you speak up? AUDIENCE: C to C mutations happen spontaneously in [INAUDIBLE] chemistry. PROFESSOR: Yes. So C to C mutations happen spontaneously. So basically, it's easier because they look more similar structurally. The DNA polymerase is more likely to make a mistake and substitute another purine. The rate of purine, purine or pyrimidine, pyrimidine to transversions which switch the type is about three to one, or two to one in different systems. So yeah, that's a good idea. But for simplicity, just to keep the math simple, we're just going to go with one mismatch penalty. But that is a good point. In practice, you might want to do that. So now, I'm saying I'm going to limit you to one mismatch penalty. But I'm going to let you choose any value you want. So what value should you choose? Or does it matter? Or maybe different applications? Tim, yeah? AUDIENCE: I've just got a question. Does it depend on pi and ri? For example, we could use all these numbers. But if the overall wants to be negative, then you couldn't use negative .1. PROFESSOR: Right, that's a good point. You can't make it too weak. It may depend on what your expected fraction of matches is, which actually depends on pi and ri. So if you have very biased sequences, like very AT rich, your expected fraction of matches is actually higher. When you're researching an AT rich sequence against another AT rich sequence, it's actually higher than a quarter. So even minus one might not be sufficient there. You might need to go down more negative. So you may need to use a higher negative value just to make sure that the expected value is negative. That's true. And yeah, you may want to adjust it based on the composition. So let's just do a bit more. So it turns out that the Karlin and Altschul theory, in addition to telling you what the p value is of your match-- the statistical significance-- it also tells you what the matches will look like in terms of what fraction of identity they will have. And this is the so-called target frequency equation. The theory says that if I search a query with one particular composition, p, subject meta-composition r-- here, I've just assumed they're the same, both p just for simplicity-- with a scoring matrix sij, which has a corresponding of lambda. Then, when I take those very high scoring matches-- the ones that are statistically significant-- and I look at those alignments of those matches, I will get values qij, given by this formula. So look at the formula. So it's qij. So pipj e to the lambda sij. So it's basically the expected chance that you would have base i matching based j just by chance. That's pipj. But then weighted by e to the lambda sij. So we notice for a match, s will be positive, so e to the lambda will be positive. So that will be bigger than 1. And you'll have more matches and you'll have correspondingly less mismatches because the mismatch has a negative. So get the target value score. And that also tells you that the so-called natural scores are actually determined by the fraction of matches that you want in your high scoring segments. If we want 90% matches, we just set qii to be 0.9, and use this equation here. Solve for sij. For example, if you want to find regions with R% identities. Little r is just the r as a proportion. qii is going to be r over 4. This assumes unbiased base composition. A quarter of the matches are acgt. Qij, then, is 1 minus r over 12. 1 minus r is a fraction of non-matching positions. They're 12 different types. Set sii equal to 1, that's what we said we normally do. And then you do a little bit of algebra here. m is sij. And you sort of plug in this equation twice here. And you get this equation. So it says that m equals log of 4 1 minus r over 3 over log 4 r. And for this to be true, this assumes that both the query and the database have uniform composition of a quarter, and that r is between a quarter and 1. The proportion of matches in your high scoring segment-- you want it to be bigger than a quarter. A quarter is what you would see by chance. There's something wrong with your scoring system if you're considering those to be significant. So it's something above 25%. And so it's just simple algebra-- you can check my work at home-- to solve for m here. And then this equation then tells you that if I want to find 75% identical matches in a nucleotide search, I should use a mismatch penalty of minus 1. And if I want 99% identical matches, I should use a penalty of minus 3. Not minus 5, but minus 3. And I want you to think about, does that make sense? Does that not make sense? Because I'm going to ask you at the beginning of class on Tuesday to explain and comment on this particular phenomenon of how when you want higher percent identities, you want a more negative mismatch score. Any last questions? Comments?
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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: All right, spin orbit. We have it still there, delta H spin orbit, orbit. Well, we know what V is. So that derivative, dv dr can be taken care of, the 1 over r. That gives you e squared over 2m c squared 1 over r cubed S, L. And you remember that S, L, from 805, J squared minus S squared minus L squared. The reason that's why you tend to do addition of angular momentum, because S and L, calculating the matrix elements of this thing is very easy, in that base is because every state there has a fixed value of S squared, a fixed value of L squared. And j squared plus a couple of possibilities. So we must work with the couple basis. Basis. And therefore, we can attempt to find E1 of a N, L, J, MJ, spin orbit, equal e squared over 2m squared c squared and LJ, MJ, S, L, R cubed. The R cube has to stay inside the expectation value, because the expectation value includes integration over space. So this is a very important. N, L, J, MJ. And again the useful question, the couple basis will have have degeneracies. All the states are degenerate there. So this time, we fixed n, because the degeneracies happen only when you fix n. So do we have the right to do this, to use the formula form perturbation, the non-degenerate perturbation theory to do this calculation? And the answer is yes, because the perturbation S.L over R cubed commutes with with L squared, with J squared, and with Jz. you need all that because you can have degeneracies by having different L values. And that would be taken care by this operator that has different eigenvalues when L is different. You can have the degeneracies involving different J values. This would be taken by this operator. And you can have degeneracies when m has different value so that involves the Jz operator. So you really need a perturbation that commutes with all of them. And why does it commute with all of them? You can see it in several ways. Let's do L squared. L squared is Casimir, it commutes with any LI. It doesn't even think about S because it doesn't know anything about S. So it commutes with S. So L squared with any LI and commutes with S. And L squared is an invariant, it commute with r squared because r squared is rotational invariant. So everything commute with that. In order to do the other ones, you can also think in terms of this matrix J squared over here and S squared and do all of them. You should do it and convince yourself that they all commute. So we can do this. If we can do it, it's good because then we can evaluate these quantities. So let's do a little of the the evaluation. So this E1nljmj is equal to. Let's evaluate the S dot L part by using 1/2 j squared s squared minus L squared. So that gives you a factor of h bar squared over 2, with this 2 over there. So you get e squared h squared over 4m squared c squared, J times J plus 1 minus l times l plus 1 minus 3/4, times nljmj, 1 over r cubed, mljmj. OK. That should be clear from the fact that you have a j squared minus s squared. That's a 3/4 spin is always 1/2, and l squared. This is known. It's equal, in fact, to nlml 1 over cubed nlml. Which is equal, let me discuss that again. It's a little-- a0l, l plus 1, l plus 1/2. OK. So this is a known result. It's one of those expectation values that you can get from Feynman, Hellman, or for other recursion relations. And this is always computed in the original uncoupled basis. But we seem to need it in the coupled basis. So again, are we in trouble? No. This is actually the same. And it is the same only because this answer doesn't depend on ml. Because this states involve various combinations of ml and ms. But doesn't depend on ml, so the answer is really the same. So these things are really the same. Happily, that simplifies our work. And now we have E1 nljm is equal to En 0-- this is yet another notation, this is the ground state energy here-- mc squared n, j, j plus 1 minus l, l plus 1 minus 3/4, over l, l plus 1/2, l plus 1. OK this is spin orbit.
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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 ocw.mit.edu. PROFESSOR: All right. The quiz on Thursday will cover up through piezoelectricity. You've had a set of notes in your hands that cover just about everything that I wanted to say about piezoelectricity. There are other modulae that one could talk about. But these follow quite directly from the one or two that we will do. So we will not have anything to say about elasticity until the last lecture of the term, which is a nice outcome because you certainly don't want to take a quiz on forthright tensors. You'll spend the entire hour just writing out all these cumbersome equations. All right. So the quiz will cover up through piezoelectricity, including a few of the representation surfaces that we'll examine today. The thing that we'll be looking at is-- I'm sorry. You have a question? AUDIENCE: Yes. A question. Well, you said [INAUDIBLE] example. PROFESSOR: I'm sorry. Third-rank tensors. AUDIENCE: OK. PROFESSOR: Also, we had just a little bit of that going into the second quiz. And we didn't really ask anything about that on the second quiz. So it'd be third-rank tensors. But you have to use second-rank tensors to define third-rank tensors. So it really will be not the emphasis, but certainly you should be familiar with the early part of what we did with second-rank tensors. All right. So today we'll look at some representation surfaces to your wonder and delight at how incredibly anisotropic the variation of these third-rank properties are with direction. One of the things that I love to do for problems when we get to these different piezoelectric effects is make up hypothetical devices just for fun. So among those that I've invented, the first one is a earthquake sensing device because it's well known that California is going to split in half, and half will fall into the sea any day now. So if this is the San Andreas fault, I have developed large, prismatic monoclinic crystals that I embed into the San Andreas fault at regular intervals. The bottom of these crystals is grounded. There's an electrode at the top. And if the San Andreas fault starts to move, and there is shear on these crystals, there will be a charge developed on the top. And I ask you to relate the charge on the top in terms of the piezoelectric modulae to the shear along the San Andreas fault. So there's a very clever little device that clearly is going to be lucrative because these crystals will have to be huge in size. And they'll cost more even than silicon. Another device that I've invented is used down at the Boston fish pier. This is a crystal on which I hang a pan, and you put fish in it. That creates a tensile stress in this direction. We measure the charge on this face. And we put a little meter that measures the charge that's accumulated. So this is how the fishermen can weigh their fish after they haul them in off the boat. So you see this is really practical material we're dealing with. This has applications in all realms of life. So today I'd like to show you, also, another problem that sets up. And I pass this out just so you can, again, have a look at some of the questions that you should be equipped to answer. So here-- not fully expecting anybody to do it, but you can see the sorts of problems one might ask. Here is problem set number 16, which asks you, should you be so inclined, to think about manipulations of third-rank tensors. But then the second question is an idea for a device which the Center for Material Science and Engineering is considering employing here in building 13. And you can read all about that.. This is the famous soup cell. I think probably you will see some sort of for fun modulus for a particular device on the quiz. I haven't made one up yet. But I think you'll have a look at something like that. And when we do the longitudinal piezoelectric modulus for quartz, which we'll do momentarily, this will give you an idea of how to set up these expressions. The problem, basically, is that the direct piezoelectric effect measures a polarization in terms of all of the elements of applied stress. So there is a modulus dijk times all of the elements of stress, sigma jk. So this, until you use the condensation of subscripts, contains nine terms going this way and three equations going this way. As we discussed earlier, since only six of the nine elements of stress are independent, you can condense this down into a 3 by 6 array of terms. The problem in doing that, though, even though one has only 18 different modulae to work with, instead of 27-- that's a considerable economy in notation, and an elimination of a great deal of redundancy-- the matrix form of this relation-- and I emphasize that it is a matrix, and it's no longer a tensor-- the matrix form cannot be transformed. Again, there are only six terms in the matrix elements of stress. And there are three equations, again, one for each component of polarization. So instead of having 27, one has only 18. But if you are considering changing the reference axes-- and that is one of the things that it's interesting to do for these various modulae that one can define-- change the orientation of a particular rod-shaped specimen that you cut out of a crystal to different crystallographic orientations and then ask how the scalar modulus changes as you change the direction in which you've sliced out the wafer or the rod of material. In order to do that, you want to transform the piezoelectric tensor to a new set of reference axes. And you cannot transform the dij's because they are a matrix and not a tensor. And no law of transformation is defined. So what you have to do in any generic problem of this sort is, for a particular single crystal, look up the form of the dij matrix that will have the equalities between tensor elements written in and the 0's, those modulae which are identically 0, entered into the array. And then you have to work your way backwards to get to the full tensor notation. So we'll see this when we look at the modulae for symmetry three two. You'll have to write in the exact matrix subscripts without absorbing the equalities in the notation. And then you'll have to expand the matrix terms into full three-subscript tensor terms. Then if you want to transform the axes, which is to say you want to cut out your specimen, be it a plate or a rod, in a different orientation, you have to transform the full three subscript tensor elements to a new setting. And then if you want to continue to work in that setting in the compact matrix form, collapse it back down to matrix form, insert the equalities, and then you're back to where you started from. So this problem of seeing how modulae that describe different phenomena vary with crystal symmetry, you have to go through this problem of expanding the compact form and then collapsing back down when you've got it as a function of some orientational angle or in terms of a coordinate system that you want to work in. The problem is exactly the same for elasticity. And we'll look at some of these modulae next term. You've heard these names before, I'm sure-- Young's modulus, shear modulus, and so on. We'll take a look on next Tuesday, a week from today, at Young's modulus, which is one of the more important ones. And probably are used to seeing this in the form of information for a polycrystalline material, which is essentially isotropic. The modulus is much more interesting for single crystal materials. And then the surfaces that are defined particularly for the lower symmetries are absolutely wild things with lumps and wiggles and lobes and things of that sort, nothing like the dumb old, uninteresting ellipsoids that we encountered for second-rank properties. So let's, then, take a look at how we had set up and defined the direct piezoelectric effect. We set this up as a proper tensor relation, saying that P1 is equal to d1. V1 And then, you'll recall, we have nine elements of strain-- sigma 11, sigma 12, sigma 13, sigma 21, sigma 22, sigma 23, sigm a 31, sigma 32, and sigma 33. But the tensor is symmetric, so we really only need to enter into our relation six of these nine terms explicitly. And what we did was to replace the two subscripts on the elements of strain, which, again, you need if you want to refer to those elements of stress through a different coordinate system. You have to know the elements of stress in tensor form. But we convert it to six terms by going and replacing the pairs of subscripts with a single one, two, and three, marching down the diagonal of the tensor this way and then marching up the right-hand side, calling two, three, four, and calling one, three, five, and finally ending up in this slot here. And we call that six. So that was the notation we used to get to a matrix representation. But the place where all this started is-- and I'll write just one line of this to be merciful-- d 111 times sigma 11, so this pair of subscripts goes with this pair, plus d 122. I'm putting that one in next because we're going to number the terms for stress in this order, one through six. Times sigma 22 plus d 133 times sigma 33 plus d 123 times sigma 23 plus d 132 times sigma 32 plus d113 times sigma 23 plus d 132 times sigma 32. And finally we end up in slot number six. And we have a d 112 times sigma 12 plus a d 121 times sigma 21. Look at that. There's an equation that covers two whole blackboards. So if we now condense this down to matrix form we would say that P1 is d 11 times sigma 1 plus d 12 times sigma 2 times d 13 times sigma 3 plus-- and now we have this messy problem with the 2's-- we have a d 14 times sigma 4 plus, again, a d 14 times a sigma -- 2 3 is equal to 32, so I can call this 4-- and then these terms become 15 sigma 5 and, again, a 15 times sigma 5 plus a 16 times sigma 6 plus d 16 times sigma 6. Now we have to make a choice. Either we are going to have, in a general matrix relation, that P sub i is equal to dij times sigma j, if j is equal to 1, 2, or 3. But it's equal to 2 dij times sigma j if j is equal to 4, 5, or 6. And that is something we like to avoid, if possible. That's ugly. That's ugly. And it's going to be a hell of a matrix if we have factors of two in front of some of the matrix elements but not in terms of others. So this is something we could do. Hey, it's our ballgame. We make up the rules. But that's going to be an ugly thing to have to deal with. So instead, as we mentioned last time, what we will do is to lump together these terms and define d 14 not as these individual tensor elements, but define those matrix elements as the sum of these two elements. And then we saw before that-- we saw last time in our earlier meeting-- that from the converse piezoelectric effect, which expresses strain in terms of an applied field, where the elements of strain 1, 23, and 32 appear in separate equations. And knowing that the same array of piezoelectric coefficients amazingly describes the converse piezoelectric effect as well as the direct piezoelectric effect, we know that d 123 equals d 132. And that is from the converse effect. So equivalent to saying this is to define d 14 as twice d 123 because these two elements are equal. So we're eating the factor of two here so that we can write a nice matrix relation that doesn't involve a factor of two. So making this combination of terms for the shear stresses, we would have simply d 14 sigma 4 plus d 15 times sigma 5 plus d 16 times sigma 6. And we can say, in general, for the other two equations, by analogy to this one, that P sub i is dij times sigma j-- a nice, neat matrix relation but one for which, unfortunately, there's no law of transformation for the matrix modulae dij. If you want to change to another coordinate system, we have to be prepared to resurrect this full three subscript notation on the piezoelectric modulae. OK. Comments or questions at this point? All right. If not, let me remind you that in the notes which I distributed last time, there are summarized all of the constraints imposed on the piezoelectric modulae for single crystals. And, again, these constraints, these requirements that the tensors remain invariant for the change of axes produced by a symmetry element that the crystal possesses, these transformations show that no third-rank tensor property can exist in a crystal that has inversion. So the 11 [INAUDIBLE] group, so-called, that possess inversion have absolutely no property and can be not considered further for third-rank properties. And then one must consider all of the 32 minus 11 21 point groups that lack conversion separately. There's no reason why they should behave the same way. Remember that for second-rank tensor properties we pulled the argument that inversion imposes no restrictions or constraints whatsoever on second-rank properties so, therefore, two different symmetries that differ only by the presence or absence of an inversion center. That is to say, you change 2 to 2 over m if you add inversion. But the argument was inversion requires nothing, so the constraints or symmetry, too, look exactly like those for symmetry. And here you've got to plod through every single one of the non-centrosymmetric point groups separately. And they all have tensors that have different forms. So what I'd like to do is look at one specific one. And that is the matrix for symmetry 32. And that is a point group that you'll recall has a threefold axis and twofold axes at intervals of 60 degrees. And in your list of the qualities and absences, 32, where the 3 is parallel to the axis x3, and the twofold axis is parallel to x1. So we're defining this as x1, this as x3, and x2 comes out halfway between the two. So let me draw this looking down along the threefold axis. These are all twofold axes. And we'll take x1 in this direction. x2 pokes out in between two twofold axes. And x3 comes straight up along the threefold axis. With that coordinate system, the form of the piezoelectric modulus matrix has this form-- d 11 minus d 11 0 d 14 0 0 0 0 0 d 15 minus d 14 0 and in the bottom row d 31 d 31 d 33 0 0 0. So this matrix with the equalities put in is obtained by looking at the full three-subscript tensor and requiring that it look the same before and after any of the rotations involved by these three distinct axes-- the threefold and the pair of twofolds. OK. Now, there are many different scalar modulae that one could define, some of them serious and worth the consideration because of the application in devices or other practical situations. The modulus that I'd like to examine is something called the longitudinal piezoelectric effect. And let's emphasize, again, that it is impossible to come up with one representation surface that fits every need because, again, the direct piezoelectric effect relates the components of a vector to a tensor, sigma ij. There are six independent tensor elements. So how can you describe how this vector is going to change as you change orientation of a crystal whose behavior is described by all of these modulae? But I would point out, however, that there are only two distinct-- oops, I'm sorry. This is d 14. I don't know how I made that d 15. There are only, in this array-- and I slipped a notch. Excuse me. Last couple of days are such that I am not able to even read from my notes. So these bottom lines, my apologies, are all 0. And there are two modulae, d 11 and d 14. So there are two independent numbers, but they appear as different matrix elements and, therefore, different tensor elements, as well. And this should be minus 2 d 14. I'm sorry. I slipped down a notch, and I got some for 32 and some for 6. And I'm glad I found it at this point, or I'd really be in deep trouble. OK. So two numbers and that is, in fact, the form of the matrix for symmetry 32. So the longitudinal piezoelectric effect is one of the representation surfaces that gives you the way in which the polarization will change for one very, very specific type of stress. In particular, what we'll do is set this up as a coordinate system. And we will look at a coordinate system where this is the reference axis, x1. We'll come down with a compressive stress, sigma 11, along that axis. And, therefore, we're applying a uni-axial stress, which has just one component of stress. So that's very specialized. In general, there would be six different components of stress. But we're looking at one specific stimulus applied to this crystal plate. In response to that sigma 11 there are charges induced on all of these surfaces. And these charges, this charge per unit area, is proportional to the component of polarization P1, the component of polarization P2 along the surface out of which x 2 comes, and the charge per unit area or the polarization along x3. So this would be P3. And this would be the direction of x3. Now I deliberately tried to show this sample as a thin wafer, which has a much larger surface area here than it does on the other two surfaces-- a much smaller area there. Therefore, since polarization is charge per unit area, if this area normal to x1 is a very large area, there's a lot of charge accumulated there. It's going to be easy to measure. If we make the wafer vanishingly thin, then the charge per unit area is high, but the total area is small. So there's going to be a negligible accumulation of charge on these two side surfaces. So the longitudinal piezoelectric effect and the longitudinal piezoelectric electric modulus is an effect, where we look at the component of polarization, P1, in response to an applied stress, sigma 11. So it's that simple. Look at all the terms we've thrown out. We've thrown out a whole bunch of elements of stress, which we could impose if we wanted to. And we've thrown away two of the three components of polarization by designing a specialized sample. So all that's left then is that P1 equals sigma 11. And the relation between those two parameters is the 111. So all this is going to hinge on one single piezoelectric electric modulus, d 111, and how that changes with direction. So this is for one orientation of a plate. And I had not specified how the orientation of this plate is related to the symmetry axes. So let's do that now. What I'm going to assume is that this is a crystal. It doesn't look like it's hexagonal. But imagine that this is a crystal of quartz. And we could look at an x1 that's in this direction. And imagine that we have cut out of this crystal a wafer that has a normal along x1. And then relative to this coordinate system, if this is x1, the modulus d 111 would tell us what charges accumulated on these two surfaces. But now, what we could do if we wanted to know how this modulus changed with direction would be to cut out a plate, a thin plate, in another orientation, where this is x1 prime. And this has changed relative to the orientation of the cell edges in the crystal. The crystal is fixed. We're just cutting a wafer out in a different orientation. So this is x1 prime. We're going to, again, squeeze it with a tensile stress sigma 11 prime. And we'll ask how the polarization P1 prime is related to sigma 11 prime. And the answer is that P1 prime will be a tensor element d 11 prime times sigma 11 prime. So what we are asking, essentially, is how does d 11 transform when we take the direction of x1 in a different orientation and, thus, change the value of d 11 prime? It's going to change all of the piezoelectric matrix elements. But we're looking at an effect in a sample that is deliberately prepared such that we will measure only the surface charge given by P1 prime. And, therefore, the way in which the properties of this plate change as we vary the way in which we've sliced it out of the single crystal is going to be simply the variation of d 11 prime with direction. So this is the general nature of what we will do when we define any of the scalar modulae related to the piezoelectric modulus tensor. We can change our notation a little bit in that we have a modulus which I'll define as a scalar modules d. And that d is going to be d 111 prime. And I know how to evaluate that. d 111 priime will be C 1l, C1m, C 1n, where these are direction cosines, times all of the elements in the original tensors, dlmn, in the tensor referred to the original coordinate system. So even though this looks simple-- it's just one modulus-- when we transform it, we've got a product of three direction cosines out in front at every single one of the 27 tensor elements in the original tensor. So it's not as trivial as it seems. So this is how this modulus that relates compressive stress to induced surface charge will change with orientation. But what are these direction cosines? These are the direction cosines-- not the full direction cosine matrix. These are the direction cosines for x1 prime. OK? So we can get rid of this two-subscript notation if it's understood that these are the direction cosines of x1 and simply call these l l, l m, and l n, just as we did for the direction cosines of a vector because we're only concerned about the orientation of one of the axes, namely x1 prime. We don't care diddly-bop about x2 prime or x3 prime because these don't enter into the modulus that we have defined. And this will be times dlmn. OK. Is what we're doing clear? So we have defined this particular effect in terms of those of the 27 piezoelectric modulae which are necessary to describe it. And then we've established how they will change with a change of the direction of one particular direction. And we don't care anything about x2 or x3. All right. Now we go through this process of inserting for matrix notation with the equalities built in. The proper matrix notation in the first term is d 11 in matrix notation. That's this term up here in the upper left-hand corner. The second term, the term that we've written as minus d 11, that's not d 11 at all. This is, by definition, d 12. And we need the true subscripts, if we're going to transform this. So this really is not even a matrix because the subscripts have lost meaning, and we're just using them to identify equalities. Then comes a 0. And next comes something that we've labeled d 14. And the subscripts there are correct. That is, indeed, the fourth term in the first row. But we're going to want to convert d 14 into a tensor element momentarily. Now let's get the rest of the terms that are non-zero. This is really d 25. So the next term that is non-zero is d 25, which just happens, because of symmetry, to be equal to minus d 14. But this is the true matrix subscripts, and this is the true matrix subscript here. The next term over to the right is the fifth and final non-zero term. This is minus 2 d 11. And this is really d 26. Those are the true matrix elements. So we put in the proper matrix subscripts. And now the next, final, step in the expansion is to convert these terms into actual tensor elements. So this is d 111. And these are tensor subscripts, so this is something we can transform. This is d 12. In tensor notation this is d 122. And that's something that has a law of transformation. d 14 is really d 123 plus d 132. We lumped two tensor elements together to define this modulus. Minus d 14 that appears in the next to the last non-zero spot is d 25. d 25 is really d 231 plus d 213. Then, finally, d 26 is d 121 plus d 112. AUDIENCE: Shouldn't that be d221? PROFESSOR: Sorry. d 26, you're right. That's down in the second row. d 221 and d 212. Now we've got something we can transform. The law for transformation is l sub l, l sub m, l sub n, dlmn. So And these are the direction cosines of x1. So this term will transform as l 1, l 1, l1 times d 111 . 1 The next term will transform as l 1, l 2, l 2 times d 122. And that will be d 122 prime for different orientation of x1. This will be two terms. This will be l 123. And I can write them in any order. So this is l 123 times d 123 plus d 132. And these terms prime, when we change axes, are going to be equal to l 2, l 1, l 3 times d 231 plus d 213. And this last term will be l 1, l 2 squared times d 221 plus d 212. All right. So this now is our new tensor element, d 11 prime. And that's given by this sum of terms. So we'll have a first term l 1 cubed times d 111. And if I look through these other terms, that's the only term in l 1 cubed that I'll have. The next term will involve the product of three cosines-- l 1 and l 2 squared times d 122. And if I go down here, here's an l 1, l 2 squared again. So I have plus d 221 plus d 212. And then, finally, the other coefficient that I have is plus l 1, l 2, l 3-- which is what this should be. And that will be times the sum of terms d 123 plus d 132. Up here, you've got the same thing again-- plus d 231 plus d 213. And I have a total of 1, 2, 3, 4, 5-- 1, 2, 3, 4, 5 terms. OK, that is how the longitudinal piezoelectric modulus will change as we change the direction of the normal to the plate that we have cut out of the crystal. So these are direction cosines relative to the crystallographic axes. l 3 is the angle between the normal to the plate and the threefold axis. l 1 is the angle cosine to the angle between the normal to the plate and one of the twofold axes. And l 2 is the direction cosine for the normal to the threefold axis and the twofold axis. OK. So we've got it now in terms of tensor elements. And now-- yeah? AUDIENCE: Is it at all reasonable to assume instead of taking those sums in d 123, d 122, just saying 2 d 123? Is that OK in assuming? Or not necessarily? PROFESSOR: Well, we could do that. But I did it the long way to not obscure what we're doing. OK? This is a well-defined summation over subscripts. And we're going to collapse immediately down to the sums. And we're going to replace the equalities. So let's see what comes out of this, if we now, having reached the zenith, having transformed the tensor elements, go down and replace this with a consolidation of terms and an insertion of the equalities between the matrix elements. OK The first term is d 11. So I will have l 1 cubed times d 11. Notice I'm getting third powers of direction cosines, which is going to be what causes the exotic nature of these anisotropies. And then I have a product of l 1 and l 2 squared. And this is d 12. And this second term is-- where did it go? This is d 21 plus d 212. And that is what we call d 26. And then, finally, this product of three different direction cosines-- l 1, l 2, l 3. And we have d 231 plus d 213. And this is d 25. And, again, an l 1, l 2, l 3 times d 14. And the second term here is d 25, if I've done it correctly. d 25 -- this is d 24. And this one is d 24. 2/4 OK. Let's now insert the equalities-- back to where we came from. d 11 is d 11. d 12, however, for symmetry 32-- I'm going to my handy-dandy chart of symmetry restrictions. I don't want to do that. That's fourth rank. AUDIENCE: It's still on the board. PROFESSOR: It's still on the board? Yes. Thank you. When your nose is in it, it's hard to see. d 12 is minus d 11. 1 And d 26 is minus 2 d 11. So these two terms can be consolidated. l 1 cubed plus l 1, l 2 squared times d 11 minus d 11. So these two terms die. And I have a minus 2 d 11 that's left. If I insert the equalities here, I'll have l 1, l 2, l 3. d 25 is minus d 14. And here's a d 14 itself. So these two terms die. And then I had d 25. And that is-- AUDIENCE: You don't have d25. PROFESSOR: I don't have d 25. Where did I get the extra one? AUDIENCE: [INAUDIBLE]. PROFESSOR: OK. I'll take your word for it. And I know how it has to turn out. OK. So these two terms kill each other. And I'm left with, then, an l 1, l 2, l 3. Or have I left something out? This is d 1-- this is d 15 and d 231 and d 23 -- uh, this is d 2 -- and the combination of 13 and 31 is d 25. Right? And if I look at my equalities, this is l 1, l 2, l 3 minus d 14 plus d 14 plus d 25. And d 25 is minus d 14. AUDIENCE: Why would you add this d25? PROFESSOR: Let me check my notes and see what I've got here. AUDIENCE: [INAUDIBLE]. PROFESSOR: OK. See what I -- d 25 is minus d 14. And then I have just a d 14. I don't know -- I see what I did. I put it in the wrong slot. This is d 14. And d 25 is the one that's minus d 14. so this term dies, which is nice because that cross term is messy. So what I'm left with, then, if I check against my notes, is d l 1 cubed plus l 1, l 2 squared times d 11. And then I have minus d 1 minus-- this doesn't belong in here. This wants to end up being l 1 cubed minus 3 l 1, l 2 squared times d 11. The first term is l 1 cubed. That's correct. The second term is l 1, l 2 squared. We've got a minus d 1 plus minus 2 d 11. So I have minus 3. It should be a minus. Yeah, that carries down to a minus. So I have minus 3 l 1, l 2 squared all time d 11. OK. So what we have ended up with is an expression for the longitudinal piezoelectric modulus as a function of orientation. The surprising thing is that l 3 does not appear here at all. It doesn't depend on the angle between the normal to the plate and the threefold axis. It depends only on one modulus, and that is a remarkable thing. This says that the shape of this surface is independent, essentially, of the property, any property that relates the one one prime to a uni-axial stimulus, sigma 11. And you measure a vectory component in the same direction is always going to have this universal surface. And it involves just a geometric term and then one modulus that changes the magnitude of the longitudinal piezoelectric modulus but does not change the asymmetry. So let's see what this function looks like as a function of direction. Maybe we better wait for that until we come back because that's going to take a few minutes. So this is what we found. That is correct. And we have to now decide what this looks like, which will take a few more minutes. But let's stop here rather than run late. All right. Let's take our 10-minute break as usual.
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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: I'd like to begin by reviewing the last lecture, where we introduced the idea of non-Euclidean spaces. The important idea here is that general relativity describes gravity as a distortion of space and time. And that idea becomes crucial in cosmology, so we want to understand what that's about. As I explained last time, there are really two aspects to this general subject. There's the question of, how do we treat curved spacetimes and how do particles, for example, behave when they move through curved spacetimes. That we will be doing. There's also the important question of how does matter distort the spacetime, which is the Einstein field equations. That we will not do. That we will leave for the general relativity course that you may or may not want to take at some point, or maybe you already have. The idea of non Euclidean geometry really goes back to Euclid himself, and the fifth postulate. The distinction between Euclidean geometry and what is generally called non Euclidean geometry is entirely in the fifth postulate. We never bothered changing the first four postulates. This fifth postulate is explained better in the next slide, where we have a diagram. The fifth postulate says that if a straight line falls on to other straight lines, such that the sum of the two included angles is less than pi, then these two lines will meet on the same side where the sum of the angles is less than pi, and will not meet on the other side. This postulate was something that attracted attention, really from the very beginning, because it seems like a much more complicated postulate than the other four. And for many years, mathematicians tried to derive this postulate from the others, thinking that it couldn't really be a fundamental postulate. But that never worked, and eventually in the 17 and 1800s, mathematicians realized that the postulate was independent. And you could either assume it's true or assume it's false. And you get different versions of geometry in the two cases. We learned last time that there are four other ways, at least, of stating the things that are equivalent to the fifth postulate, and they're diagrammed here. If a straight line intersects one of two parallel lines, it will also intersect the other. If one has a straight line and another point, there is one and only one line through that point, parallel to the original line. If one has a figure, one can construct a figure which is similar to it, of any size. And finally, the famous statement about sum of the angles that make up the vertices of a triangle. If there exists just one triangle for which it's 180 degrees, then that's equivalent to the fifth postulate, and you can prove that every triangle has 180 degrees. The fifth postulate was questioned in a serious way by Giovanni Geralamo Saccheri in the 16, 1700s, who wrote a detailed study of what geometry would be like if the fifth postulate were false. He wrote this believing that the fifth postulate must be true. And he was looking for a contradiction, which he never found. Things went further in the later 1700s, with Gauss and Bolyai and Lobachevski, who independently developed the geometry that we call Gauss Bolyai Lobachevski geometry, which is a two dimensional geometry, non Euclidean. It corresponds to what we now call an open universe, that we'll be learning about in more detail later today. The Gauss Bolyai Lobachevski geometry was treated purely axiomatically by the three authors that I just mentioned. But it was given a coordinate representation by Felix Klein in 1870, which was really the first demonstration that it really existed. When one treats it axiomatically, one still always has the possibility that some contradiction could be found someplace. But by the time you put it into algebraic equations, then it becomes as consistent as our understanding of the real numbers, which we have a lot of confidence in, even though I don't think mathematicians really know how to prove the consistency of anything. But we have a lot of confidence in this kind of mathematics. So by 1870, it was absolutely clear that this open geometry, this non-Euclidean geometry, was a perfectly consistent, is a perfectly consistent, formulation of geometry. An important development coming out of Kline's work is that the idea about how one describes geometry changed dramatically. Prior to Kline, essentially all of geometry was done in the same way that Euclid did it, by writing down axioms and then proving theorems. Kline realized that you could gain a lot of mileage by taking advantage of our understanding of algebra and calculus by describing things in terms of functions. And in particular, geometry is described by giving a distance function between points. This was further developed by Gauss, who realized that distances are additive. So if distances mean anything like what we think distances mean, it would be sufficient to describe the distance between any two arbitrarily close points. And then if you want to know the distance between two distant points, you draw a line between them, and measure the length of that line by adding up an infinite number of infinitesimal segments. So the idea that distances need only be defined infinitesimally was very crucial to our current understanding of geometry. Gauss also introduced another important idea, which is a restriction on what that infinitesimal distance function should look like. Gauss proposed that it should always have the same quadratic form that it has for Euclidean distances. For Euclidean distances, the Pythagorean theorem tells us that the distance between any two points is the sum of the squares of the coordinate distances. And for non-Euclidean geometry, we generalize that by allowing each term in this quadratic expansion to have its own prefactor, and those prefactors could be functions of position. So g sub xx of xy is just a function of x and y. And g sub xy is another function of x and y. And g sub yy is another function of x and y. And the distance function is taken as the sum of those three terms. The important feature that that quadratic form corresponds to, which was noticed by Gauss, is that if the distance function has that form, it means that even though the space is not Euclidean and will not obey, in general, the axioms of Euclidean geometry-- and in particular the fifth postulate-- it is still true that in a very tiny neighborhood, it will resemble Euclidean geometry, where the resemblance will become more and more exact as you confine yourself to tinier and tinier neighborhoods. And we're kind of aware of this in everyday life. The surface of the earth is approximately spherical-- we'll ignore little things like mountains and roads and bumps, and pretend the surface of the Earth is spherical. Nonetheless, the surface of the Earth looks flat to us. And the reason it looks flat is that we see only a tiny little bit of it. And a tiny piece of a curved surface always looks flat. And mathematically speaking, the way to introduce enough assumptions to validate that conclusion is to assume that the local distance function is a quadratic function of this form. And what Gauss originally proved is that if the distance function is of that form, it is always true that in a tiny neighborhood, to an arbitrary accuracy, you could define new coordinates-- x prime and y prime in the notation of this diagram-- where in terms of the new coordinates in the tiny neighborhood, the distances are just the Euclidean distances. ds squared equals dx prime squared plus dy prime squared. And that's a very crucial fact that we will be making use of, Einstein made use of, in the context of general relativity, which we'll be getting to. OK, that finishes the review. Any questions about anything that we talked about last time? OK, great. Now what we want to do is to go on to apply these ideas in detail. In one of them in particular, build up a full description of closed and open universes today. And we are going to begin by giving a mathematical description of the simplest non-Euclidean geometry that we have available, which is just the surface of a sphere-- a two dimensional sphere embedded in three dimensions. That is, a two dimensional surface embedded in a three dimensional space, as is intended to be shown in that diagram. The sphere is described simply by x squared plus y squared equals plus z squared equals r squared, where x, y, and z are just Euclidean coordinates. So in this case, our curved space, which is the surface of the sphere, can be embedded in a Euclidean space of one higher dimension. That's not always the case. We should not pretend that that will always be the case. But when it is the case, it allows us to study that curved surface in a very straightforward way, because everything is really determined by the Euclidean geometry of the space in which this sphere is embedded. Nonetheless, when we're done formalizing our description of the surface of the sphere, the goal will be to concentrate on what Gauss called the "inner properties," namely the properties of the surface itself. And we will try to pretend that the three dimensional space never even existed. It won't be required for anything that we'll be left with, once we have a solid description of the surface itself. And this will be very important for what we'll be doing later. So it is important to get in touch with the idea that we're going to study this sphere, making use of the fact that it could be embedded in three Euclidean dimensions. But in the end, we want to think of it as a two dimensional geometry, which is non Euclidean. OK, so our goal will be to write down the distance function for some coordinization of the surface of the sphere. I should say at the beginning, when this picture makes it most obvious, that one of the reasons we might be interested in the surface of a sphere, if we're interested in cosmology, is that we know that we're trying to build cosmological models that are consistent with homogeneity and isotropy. Because we discussed earlier those are, to a very good approximation, valid features of the universe that we're living in. So the surface of a sphere has those properties. It's certainly homogeneous, in the sense that any point on the surface of a sphere will look exactly like any other point. If you were living on that sphere, and you didn't have any other landmarks, you'd have no way of knowing where on the sphere you were. Furthermore, it's isotropic-- same in all directions. And when I say that, it's important that I really mean it in the context of the two dimensional surface, not the three dimensional geometry. So the three dimensional geometry is isotropic. If you sat at the center of that sphere and looked any direction in three dimensions, everything would look the same. But that's not the isotropy that's important for us. We want to imagine ourselves as two dimensional creatures living on the surface. And then you can imagine that if you were a two dimensional creature living on the surface-- so you happen to be at the North Pole, because that's the easiest to describe-- you could imagine looking around in a circle, 360% available, and the world that you'd be living in would look exactly the same in all directions on the surface. And that's the isotropy that's important to us, because the two dimensional surface here is what we're soon going to generalize to be our three dimensional world. And it's isotropy within that world that we're talking about. OK, so, first thing we wanted to do is to put coordinates on our two dimensional surface. If we want to ultimately forget the third dimension and live in the surface, we want to have coordinates to use in the surface. It's a two dimensional surface, so it should have two coordinates. And when we use the usual coordinization of a sphere, polar coordinates, well, it will be two angles, theta and phi. And there are some different dimensions that are used in different books, but I think almost all physics book use these conventions. Theta is an angle measured from the z-axis, and phi is an angle measured by taking the point that you're trying to describe, which is that dot there, projecting it down into the xy plane, and in the xy plane, measuring the angle from the x-axis. So that's phi. And theta and phi are the polar coordinates describing a point on the surface of the sphere. And what we want to do is describe the distance function in terms of those polar coordinates-- that's our goal. OK, to describe the distance function, what we want to imagine is two infinitesimally nearby points-- one described by coordinates theta and phi, and one described by theta plus d theta and phi plus d phi. So we have one point described by theta and phi, and another point described by theta plus d theta, phi plus d phi. So the coordinate changes are just d theta and d phi. What we want to know is how much distance is undergone by moving from the first point to the second point. And the easiest way to see it is to make the changes one at a time. So first, we can just vary theta. And if we just vary theta, we see that the point described by theta and phi moves along a great circle, which goes through the z-axis. And the distance that the point goes is just an arc length as a piece of that great circle. And since this attended angle is d theta, and the radius is r, the distance of that arc length really just follows from the definition of an angle in radians. The arc length is r times d theta, and that's really the definition of d theta in radians. So if we vary theta only, the distance ds is just equal to r times d theta. Everybody happy with that? OK. Now if we vary phi, it's slightly more complicated, but not much. If we vary phi, the point being described would-- if you vary phi all the way around-- make a circle around the z-axis. That circle does not have radius r. That's the one thing that may be a little bit surprising, until you look at the picture and see that it's true. The radius of that circle is r times sine theta. So in particular, if theta were 0, if you're up around the North Pole, going around that circle would just be going around the point. 0 radius. And you have maximum radius when you're at the equator, going all the way around. So again, we're going in a circle through an angle-- in this case d phi. So the arc length is just the angle times the radius. But the radius is r times sine theta, not r itself. So when we vary phi only, ds is equal to r times sine theta times d phi. Any questions? OK, now, the next important thing to notice is that these two variations that we made are orthogonal to each other. When we varied phi, we moved in the horizontal plane. There's only motion in the x and y directions when we vary phi. When we vary theta, we move in the vertical direction. And those two vectors are orthogonal, as you can see from the diagram. And because we have two orthogonal distances that we're adding up, and because we're in underlying Euclidean space here-- we can think of those distances as being distances in the three dimensional Euclidean space that we're embedded in-- we get to use the Pythagorean theorem. So putting together these two variations, we get ds squared is equal to an overall factor of r squared, times d theta squared plus sine squared theta, d phi squared. And that formula then describes the metric on the surface of a sphere. And it describes a non-Euclidean geometry. And once we have that metric, we can forget the three dimensional picture that we've been drawing, and just think of a world in which there are two coordinates-- theta and phi-- with that distance function. And that's the way we want to be able to think about it. OK, everybody happy? OK, I want to mention, because it is perhaps useful in other cases, depending on your taste of how you like to solve problems-- the description I just gave of deriving this formula was geometric, that is, we drew pictures and wrote down the answer based on visualizing the pictures. But it can also be done purely algebraically. To do it purely algebraically, one would first write down formulas that relayed the angular coordinates-- I should go back to slides. What's going on here? Is my computer frozen? [INAUDIBLE] Yes? AUDIENCE: For that formula, are you assuming that d theta and d phi are really small so that you can [INAUDIBLE] the triangle? PROFESSOR: Yes, these are infinitesimal separations only. That's the key idea of Gauss. And yes, we're making use of that. This formula will not hold if d theta and d phi were large angles. Holds only when they're infinitesimal. Yes? AUDIENCE: And then similiarly, we can use the line integral for calculating the distances based on this metric? PROFESSOR: Yes, that's right. If we wanted to know the distances between two finitely separated points, we would construct a line between them, and then integrate along that line. And by line what we mean is the path of shortest distance, which we'll be learning about more next time, probably. Those are not necessarily easy to calculate. In this case, they're calculable, but not really easy. Any other questions? OK, so what I wanted to do was to look at the definition of the coordinate system, as now shown on the screen. And from that, we can write down the relationship between x, y, and z, and theta and phi. And those relationships are that x is equal to r times sine theta times cosine phi. y is equal to r times sine theta times sine phi. And z is equal to r times cosine theta. And once one writes those formulas, then one can just use straightforward calculus to get the metric, without needing to draw any pictures at all. You may have wanted to draw pictures to get these formulas, but once you have these formulas, you can get that by straightforward calculus. I'll sketch the calculation without writing it out in full. But given this formula for x, we can write down what dx is by calculus, by chain rule. So dx-- it's a function of two variables. So it would be the partial of x, with respect to theta, times d theta, plus the partial derivative of x, with respect to phi, times d phi. And we'll go work out what these partial derivatives are if I differentiate this with respect to theta. The sine theta turns into cosine theta. So the first term becomes r cosine theta, cosine phi, d theta. And then plus, from here we have the partial of x with respect to phi. We just differentiate this expression with respect to phi. The derivative of cosine phi is minus sine phi. So the plus sign becomes a minus sign-- r sine theta, sine phi d phi. And then I won't continue, but we could do the same thing for dy and dz. And once we have expressions for dx, dy, and dz, we can calculate ds squared, using the fact again that all of this is embedded in Euclidean space. That's where we're starting, although in the end, we want to forget that Euclidean space. But we could still make use of it here, and write ds squared is equal to dx squared, plus dy squared, plus dz squared. One can then plug in the expression that we have for dx, in terms of d theta and d phi, and the analogous expressions that I'm not writing down for dy and dz. And when one puts them in here, one makes lots of use of the identity that cosine squared plus sine squared equals 1 And after using that identity a number of times, what you get when you just put together this algebra is exactly what we had before-- r squared times d theta squared, plus sine squared theta, d phi squared. So the important point is that once you have the identities that relate these two different coordinate systems, and if you know the distance function in the xyz coordinate system, you're home free, as far as geometry is concerned. One could just use calculus from there on if one wants to. Although usually the geometric pictures make things easier. OK, any questions about that? OK in that case, we are now ready to move on, having discussed the two dimensional surface of a sphere embedded in three Euclidean dimensions. The next thing I'd like to do-- and this really will be our closed universe cosmology-- we're going to discuss the three dimensional surface of a sphere in four Euclidean dimensions by analogy. The previous exercise was a warm- up. Now we want to introduce a four dimensional Euclidean space. So our sphere now will obey the equation x squared plus y squared plus z squared plus-- we need a new letter for our new dimension in our four dimensional space, and I'm using w. x, y, z, and w. So that equation describes a three dimensional sphere in a four dimensional Euclidean space. And our goal is to do the same thing to that equation that we just did to the two dimensional sphere embedded in a three dimensional space. Now of course it becomes much harder to visualize anything. If you ask how do we do visualize things in four dimensions, I think probably the best answer is that we usually don't. And what we're going to take advantage of mainly is that once you know how to write things in terms of equations, you don't usually have to visualize things. Or if you do, you can usually get by, by visualizing subspaces of the full space. If you want to visualize the full sphere in some rational way, I think the crutch that I usually use, and that most people use, is if you have just one extra dimension, in this case w, try to think of w as a time coordinate. Even though it's not really a time coordinate, it still gives you a way of visualizing things. So if we think of w as a time coordinate here, the smallest possible value of w would be minus r. The maximum possible value of w would be plus r, consistent with this constraint, consistent with being on the sphere. So when w is equal to minus r, the other coordinates have to be 0 to be on the surface of the sphere. So you can think of that as a sphere that's just appearing at time minus 1, with initially 0 radius. Then as w increases, x squared plus y squared plus z squared increases. So you could think of it sphere that starts at 0 size, gets bigger, gets as big as r in radius when w equals 0 and then gets smaller again and disappears. So that's one way of thinking of this four dimensional sphere. Yes? AUDIENCE: Is that kind of like looking at the cross sections of the sphere? PROFESSOR: Is that kind of like looking at the cross sections of the sphere? Exactly, yes. One is looking at cross sections of the sphere-- successive cross sections at successive values of w. And if you do it in succession, it makes w act like a time coordinate. But you're right. The fact that w is a time coordinate is kind of irrelevant. You could imagine just drawing them on a piece of paper in any order. And for any fixed value of w, you're seeing a cross section of what this sphere looks like. That is how the xyz coordinates behave for a fixed value of w. Now to coordinatize the surface of our sphere. Last time we used two coordinates, because we had a two dimensional surface. This time we're going to want to use three coordinates, because this is a three dimensional surface that we're describing. And that means that we need at least one new coordinate. And the new coordinate that I'm going to introduce will be another angle, which I'm going to call psi. And the angle I'm going to define, as in this diagram, as the angle from the new axis, the w axis. So psi is the angle of any arbitrary point to the w axis. And therefore the w coordinate itself is going to be r times cosine psi, just by projecting. And the square root of the sum of x squared plus y squared plus z squared is then the other component of that vector. And beyond the sphere, it's easy to see that the square root of x squared plus y squared plus z squared has to be r times sine psi. OK, now we still need two more coordinates. This is only one coordinate-- we want to have three. But the two other coordinates will just be our old friends, theta and phi We're going to keep theta and phi. And in order to keep them, what we'll imagine doing is that for any point on the surface of this three dimensional sphere of the four dimensional space, we could imagine just ignoring the w coordinate. And then we have x, y, and z-coordinates, and we can just ask what are the values of theta and phi that would go with those x, y, and z-coordinates. So theta and phi are just defined by quote, "projecting" the original point into the three dimensional xyz space, which just means ignore the w coordinates, look at xyz, and ask what would be the angular coordinates, theta and phi, for those values of x,y, and z. And it's easy to take those words I just said and turn them into equations. We like to write down the analog of these equations. But we want to have four equations now that will specify x, y, z, and w as a function of our three angles, psi, theta, and phi. But that's not hard. I'll show you at the bottom with w. w we already said is just r times the cosine of psi. Just coming from the fact that psi is defined as the angle from the point to the w-axis, and that's enough as you see from the picture to imply that w is equal to r times cosine psi. The other point, x,y, and z, really just follow by induction from what we already know. Each of these formulas will hold in the three dimensional subspace, except that r, the radius of a sphere in the three dimensional subspace, is not r anymore but is r times sine psi. So x is equal to r times sine psi times what it was already-- sine theta cosine phi. y is equal to 4 times sine psi times sine theta sine phi. And z is equal to r times sine psi times cosine theta. So I just take each of these equations and multiply them by sine psi to get the new equations. And you can straightforwardly check-- if we take x squared plus y squared plus z squared plus w squared here, it makes successive use of the identity that sine squared plus cosine squared equals 1. We'll be able to show that x squared plus y squared plus z squared plus w squared equals r squared, like it's supposed to. OK, so is everybody happy with this coordinatization? OK, I should mention, by the way, that if you ever have the need to describe a sphere in 26 dimensions or whatever, this process easily iterates, once you've known how to do it once. That is, every time you add a new dimension, you invent a new letter for the new axis. You define a new angle, which is the angle from that access. And then the new coordinatization is just to set the new coordinate equal to r times the cosine of the new angle. And then take all the old equations and put in an extra factor of the sine of the new angle. And you got it. So you could do that as many times as you want, if you want to describe a very high dimensional sphere. We should say something about the range of these angles. The original angle phi-- maybe I should go back a few slides now. The original angle phi, as you can see from the slide, goes around the xy plane, so it has a range of 0 to 2 pi. The original angle theta is an angle from the z-axis, and the furthest you could never be away from pointing towards an axis is pointing away from it. And that's pi, not 2 pi. So theta has a range of 0 to pi. And similarly, psi is also defined as an angle from an axis. So again, the furthest you could ever be away from pointing towards an axis is pointing away from it. So psi, like theta, will have a range of 0 to pi. And if you ever need to coordinatize a 26 dimensional sphere, as I just mentioned, you keep adding new angles. Each of the new angles goes from 0 to pi. Thus each new angle was introduced as an angle between the point that you're trying to describe and the new axis. So, they're all angles like theta and psi. So 0 is less than phi, is less than 2 pi. But 0 is-- these should be less than or equal to's-- 0 is less than or equal to theta, is less than or equal to pi. And 0 is less than or equal to psi, is less than or equal to pi. OK, next we want to get the metric of our three dimensional spherical surface embedded in our four dimensional Euclidean space. And I'm going to do it by the geometric sort of way. I'll try to just motivate the pieces. There'll be some cross here between actually algebra and geometry. Let's see, first I should mention that once you have this, you can get the answer by the same brute force process that we describe, but didn't really carry out here. That's tedious, but it's pretty well guaranteed to get you the right answer if you're careful enough, and does not require drawing any pictures or having any the visualization of the geometry, so it does have some advantages. But I will not do it that way. I will do it in a geometric sort of way, because I think it's easier to understand the geometric sort of way. So here goes. As we did over there, we will vary our coordinates one at a time, and then see how we can combine the different variations. Again we'll find that they're orthogonal to each other. So I'll be able to combine them just by adding the squares. But we don't necessarily know that the beginning. So let's start by varying psi, the new coordinate. And there, things are very simple because our new coordinate is just defined as the angle from an axis. So if we vary psi, the point in question just makes a circle around the origin, and the circle has radius, capital R. So the variation if I vary psi is just r times d psi. So ds is equal to r times d psi. What could be simpler? OK, now we'll imagine varying either theta or phi or both. And since we already pretty well understand this three dimensional subspace-- this is the previous problem-- I'm going to talk about varying them simultaneously. So vary theta and phi. Well then we know that ds squared is really given by this formula. We are just varying theta and phi in a three dimensional space. The fourth direction doesn't change in this case. But the radius involved is not what we originally called r. But the radius in the three dimensional space is r times sine psi. That's the square root of x squared plus y squared plus z squared. So what we get is ds squared is equal to R squared times sine squared psi times d theta squared plus sine squared theta, d phi squared. Everybody happy with that? OK, now I'm going to first jump ahead and then come back and justify what we're doing. But if these variations are orthogonal-- which I will argue shortly that they are, so I'm not doing this for nothing-- if the variations are orthogonal, then we just add the sum of squares using the generalized Pythagorean theorem. In this case, Pythagorean theorem in four dimensions-- that's four Euclidean dimensions, so we should be able to use it. So what we get for our final answer is ds squared is an overall factor of r squared, times d psi squared plus sine squared psi, times d theta squared, plus sine squared theta, d phi squared. I just added the sum of the squares. Now I need to justify this orthogonality. OK, to do that, let me introduce a vector notation in the four dimensional space of x,y z, and w. OK, we're justifying this in the Euclidean embedding space. And the Euclidean embedding space of these transformations are orthogonal. So let me imagine varying psi. And then I could construct a four dimensional vector dr-- I'll call it sub psi because it arises from varying psi. So this is the four dimensional vector that describes the motion of this point r as psi is varied. And first let me just give these components names. I'll call it-- rather than repeat the r, I'm just going to call this d psi sub x, d psi sub y, d psi sub z, and d psi sub w. This is just by definition. I'm just naming the components of that vector. And since the vector already has a subscript, I don't want to have two subscripts. So I've changed the name of the vector for writing as components. And similarly, here I will just vary one of these two angles, theta and phi I'll just vary theta, and let you know that varying psi is no different, and you can easily see that. So if I vary theta, the variation of r when I vary theta will be dr sub theta. And its components I will just call d theta x, d theta y, d theta z, and d theta w. So these are just definitions. I haven't said any actual facts yet. But I've defined these two vectors, and given names to their components. And now we want to look at them and take their dot product. Their dot product is just a four dimensional Euclidean dot product. So the dot product is just dx times d theta d psi y d theta y plus d psi z times d theta z plus d-- blah. I think we should write it. So this actually is now a fact about Euclidean geometry in four dimensions. The dot product of these two vectors is just equal to the product of the x components, plus the product of their y components, plus the product of their z components, plus the product of their w components. And now what we want to do is to look at this sum and argue they're 0, because if they're orthogonal, the dot product of two vectors should be 0. OK, so let me first look at the dr sub theta vector. OK, what do we know about it? Well, we know that when we vary theta, from these formulas, w does not change-- and we can easily see that from the picture as well. So d theta sub w equals 0. And since these are all products, that means that this last term vanishes, no matter what d psi w is. So we know we don't need to worry about that term. We only need to worry about these three terms. Now what do we know about those three terms? If you look at dr sub theta, and look at its three spacial components-- x,y, and z-- from here, we could see that varying theta does the same thing to x, y, and z as it did over here, except a different overall factor out front. So in particular, what I want to point out is that varying theta does not change x squared plus y squared plus z squared. It leaves it constant. So if we think of the xyz space, we could imagine a sphere in the xyz space, and varying theta always causes a variation that's tangential to that sphere. It never moves in the radial direction. So the three vector, defined by d theta x, d theta y, d theta z, is tangential in the three dimensional subspace xyz. Just as it was when we didn't have a w coordinate. The w coordinate doesn't change anything here. So that's a little bit [? soluable. ?] Are people happy with that? Do you know what I'm talking about? OK. Now we want to look at dr sub psi, and it will have a w component-- d psi sub w-- but we don't care about that. We know we don't care about it because that piece already dropped out of our expression. So we want to know what this vector looks like in the xyz space. We don't care about what it looks like in the w space. So in the xyz space, we can look at these formulas here. As we vary psi, x, y, and z could change. But they all change by the same factor-- whatever factor psi changes by. The same sign psi appears in all three lines. So changing psi can only multiply x, y, and z all by the same factor. And what that means is that if you think of this geometrically in the xyz space, varying psi moves the point only in the radial direction. If you multiply all of the coordinates by a constant, you are just moving in the radial direction. So dr psi has the property that when we look at only its x, y, and z components-- d psi x d psi y, d psi z, it is radial in the three dimensional subspace. So, the sum of these three terms-- this is what we're trying to evaluate-- is the dot product of a radial vector and a tangential vector. And the dot product of a radial vector and the tangential vector will always be 0, because there are orthogonal to each other. Sorry for the overlap here, but equal 0, and that's because radial is perpendicular to tangential. OK, everybody happy with that? OK if so, we have important result now. We have derived the metric for the three dimensional surface of a four dimensional sphere embedded in four Euclidean dimensions. And that, in fact, is precisely the closed universe of cosmology. It's the homogeneous isotropic description of a closed universe. OK, next thing I want to point out is just a definition. An important feature of non Euclidean geometry and general relativity-- because they're connected to each other-- is that there never is a unique, useful coordinate system. And Euclidean spaces, there is a unique, useful coordinate system. It's the Cartesian system. Sometimes it's also useful to use polar coordinates or something else, but by and large the Cartesian coordinate system is the natural description of Euclidean spaces. And the coordinates of a Cartesian coordinate system really are distances. Once, however, you go from Euclidean geometry to non Euclidean geometry-- from flat spaces to curved spaces-- you're usually in a situation where there just is no natural coordinate system. When we invented this psi, theta, phi, we really made a number of arbitrary choices there. We could have defined things quite differently if we wanted to. So in general, one has to deal with the fact that the coordinates no longer represent distances, and therefore there's a lot of arbitrariness in the way you choose the coordinates in the first place. So in particular, we could think of psi equals zero as the center of our new coordinate system, with coordinates psi, theta, and phi. psi equals 0 corresponds to being along the w axis, so it's a unique point. When you say that psi is equal to 0, it no longer matters what theta and phi are. You're at the point w equals r and x, y, and z equals 0. So we can think of that as the origin of our new coordinate system. And we can think of then the value of psi as measuring how far we are from that origin. So psi will become our radial coordinate. So thinking of it as our sphere, psi equals 0 we might think of as the North Pole this sphere. But we're also going to think of it as the origin of psi, theta, phi space. And we will sometimes use other radial variables-- other variables for the distance from the origin. So in particular, another coordinate that's very commonly used is u, which is just defined to be the sine of the angle psi. And notice that the sine of the angle psi shows up in a lot of equations. So taking that as our natural variable is a reasonable thing to do on occasion. Both are useful. If we do use this, then we can rewrite the metric in terms of u, instead of psi. And in order to do that, we just have to know how du relates to d psi, because the metric is written in terms of the differentials of the coordinates. So that's easy to calculate. du would be equal to cosine of psi d psi. But if we're trying to rewrite the metric solely in terms of u, we don't want to have to divide or multiply by cosine psi, because that's written in terms of psi. But we could express cosine psi in terms of u, because if u equals sine psi, then cosine psi is the square root of 1 minus u squared. So I can rewrite this as the square root of 1 minus u squared times d psi. And then d psi squared, which is what appears in our metric, can be rewritten just using that as du squared divided by 1 minus u squared. And the full metric now, in terms of the u theta and psi coordinates, can be written as r squared times du squared over 1 minus u squared, plus u squared times d theta squared plus sine squared theta d phi squared. So this is another way of writing the metric for this three dimensional sphere embedded in four Euclidean dimensions. Any questions? Yes? AUDIENCE: The value of u doesn't uniquely determine a point on the sphere, right? Because [INAUDIBLE]. PROFESSOR: Very good point. In case you didn't hear the question, it was pointed out that the value of u, unlike the value of psi, is not uniquely indicate a point, because on the entire sphere, there are two points u for every value of sine psi-- one in the northern hemisphere and one in the southern hemisphere, if we think of hemispheres as dividing whether w is positive or negative. So in fact, that's right. If we use the u coordinate, we should remember that if we want to talk about the whole sphere, we should remember that the whole sphere is twice as big as what we see if we just let u vary between 0 and its maximum value, 1. u equals 1 corresponds to the equator of this sphere. Another point which I'd like to make now that we've written it this way is that writing it this way is the easiest way to see-- although we can see it in other ways as well-- that if u is very small, if we look right in the vicinity of the origin of our new coordinate system, if u is very small, 1 minus u squared is very close to 1. The square of a small quantity is extra small, so this denominator is extra close to 1. And that means that for very small values of u, what we have is du squared plus u squared times this quantity. And this is just polar coordinates in Euclidean space. So for u very small, we do see that we have a local Euclidean space. And that, if you might remember, was one of the key points about writing the metric as the sum of squares in the first place. And it's true about every point, although the coordinate system here only makes it obvious about the origin. But we know that the space actually is homogeneous, from the way we constructed it. So what's true about the origin is true about any point. So the coordinate system, the metric around any point, if you look close enough in the vicinity of the point, looks like a Euclidean space, which is what we expected from the very beginning, but we can see it very explicitly here. OK. So far, this is just geometry. But this will be a model for a homogeneous, isotropic universe. We know it's homogeneous, we know it's isotropic from the way we discussed it. When we write the metric this way, it's obvious that it's isotropic about the origin, because this construction we know is just polar coordinates, and we know polar coordinates don't really single out any direction, even though manifestly they look like they do, because you're measuring angles from the z-axis. But we know that really describes an isotropic sphere. It's not obvious that this formula describes a homogeneous space, because it makes it look like u equals 0 is a special point. But we do know that it did correspond to a homogeneous space from the way we constructed it. It really is just a three dimensional sphere embedded in four Euclidean dimensions. And if you think of it as the sphere, there clearly is no special point on the sphere. So the homogeneity is a feature of this metric, but a hidden feature. It's hard to see how you would transform coordinates to, for example, put a different point at the origin. But it is possible, and we know it's possible, because of the way we originally constructed. And if we had to do it, we go back to the original construction and actually do it. That is, if I told you I wanted some other point in that system to be the origin, you could trace it all way back to the four dimensional Euclidean space, and figure out how you have to do a rotation in that four dimensional Euclidean space to make the point that I told you I wanted to be the origin to actually be the origin. So you'd be able to do that. It would be some work, but you would in fact know how to do that. We really do understand that the space is homogeneous, which is guaranteed by our original construction. OK. Finish with the basic geometry-- one more chance to ask any questions about it. OK, next I want to go on now to talk about how this fits into general relativity. And here we are going to be confronting-- actually the only place we ever will confront-- the issue of how matter causes space to curve, which is the aspect of general relativity that we're not really going to do it all. So I'll basically just be giving you the answer. Although we will in fact know enough to narrow down the range of possible answers, pretty much. But there will be a fudge factor that I'll have to just tell you the right answer for. So what I want to do now is to make a connection between this formalism and the model that we already discussed of the expanding universe whose dynamics we derived using Newtonian mechanics. So using internet Newtonian mechanics, we introduced a scale factor a of t. And convinced ourselves that a dot over a squared is equal to 8 pi over 3 g rho, minus kc squared over a squared. And furthermore, that this a of t describes the relationship between physical distances and coordinate distances. Namely, if we have objects that are at rest in this expanding universe, comoving objects is the phrase usually used to describe that. If we have comoving objects, the comoving objects will sit at fixed coordinates in our coordinate system. And the distance between any two of them will be some fixed distance-- l sub c-- a coordinate distance. But the physical distance will vary with time, proportional to this scale factor. So the physical distance between any two points will be a function of time, which is the scale factor times the time-independent coordinate distance. OK, if we look at our metric for this sphere, and say we're going to assume that this is going to be the metric that describes the space that we're describing over here, then clearly this r that sits out front rescales all of the distances. All the distances are proportional to r, just as all the distances here are proportional to a. So that could only work if r is proportional to a. So that's our key conclusion here-- that r is going to have to vary with time, and be proportional to a. But we can even say a little bit more than that, because we can look at dimensionality. And here comes in handy that I insisted from the beginning in introducing this idea of a notch. The notch helps us here to get this formula right. The units of r-- r is just distance-- so the units of r are just distance units-- and I'll pretend that we're using meters. It doesn't really matter what actual units we're using, so I'll call this m for meters. On the other hand, a of t comes from this formula, where physical distances are measured in meters, but coordinate distances are measured in notches, so a is meters per notch, as we've said many times before. So a of t is meters per notch. So that tells us something about this constant of proportionality-- it has to have the right units to turn meters per notch into meters. That is, it has to have units of notches. Where did we get notches from? The other thing that we know is that the little k that appears in the Friedman equation, which we know is a constant-- we also know it's a constant that has units of 1 over notches squared, as we worked out some time ago. Units of k is 1 over notches squared. And that's the only thing around that we can find that has units of notches, so we're going to use that to make the units turn out right in this equation. So to get the units right, we can write it as r of t is equal to some constant-- actually, it's more convenient to square this-- r squared of t is equal to some constant times a squared of t, divided by k. And now that constant is dimensionless. The units are all built into the a's and the k's and everything else. AUDIENCE: Is the constant multiplied by that, or-- PROFESSOR: Yes. That equals sign was a big mistake. Constant times a squared over k. And that constant is now dimensionless. Because this is meters squared, this is meters squared per notch squared, and this is just per notch squared, so the notches cancel over here. OK, now what this constant is really is the statement of how curved is our space-- r is really a measure the curvature of our space-- how curved is our space for a given description of what the matter is doing? a of t is directly related to the [INAUDIBLE] calculations you've already done. So this clearly is a formula of exactly the type that I told you we weren't going to learn how to deal with. We're not capable in this course of describing the Einstein field equations, which determine how matter causes a space to curve. So I'll just tell you the answer. The answer is that this constant is equal to 1. So it's certainly a simple answer, but we won't be able to derive it. So we end up with just r squared of t is equal to a squared of t, divided by k. Now it may be useful at this point to remind ourselves what little k meant in the first place. We introduced little k in the context of describing a purely Newtonian model of an expanding universe, where we imagined just a finite sphere of matter expanding. And in doing that, we defined k to be equal to minus 2 e divided by c squared, where e itself was not a quantity that we proved was conserved. And it's related to an energy, but as we discussed, there are various ways that you could relate it to the energies of different pieces of the system. But e was given by that expression. And if I put this into that, just to see more clearly how our discussion relates to our Newtonian discussion, we get r squared is equal to a squared of t times c squared over 2 e. And I wrote it this way mainly to illustrate, or demonstrate, an important point, which is that our calculation was non-relativistic, but there's a c squared appearing in this formula. This c squared really just arose from our definitions. And if we had some other quantity here whose units were-- we have to sub units of meters per second for this formula to come out right. If we put some other velocity here, then this constant would not be 1 anymore, but something else. So saying that this constant is 1 is saying that this formula is meaningful. Putting the c squared there simplifies things. And that in turn means that the curvature really is a relativistic effect. OK, we think of relativistic effects as effects that disappear as the speed of light goes to infinity. So this formula tells us that as the speed of light goes to infinity-- for fixed values of things like the mass density, which are buried in a squared of t-- as the speed of light goes to infinity for fixed values of the mass density, r squared goes to infinity. Now infinity may sound like it's backwards, but it's the right way. 1 over r is really the curvature. r is the radius of curvature of the space. As r goes to infinity, our curved space looks more and more flat. So we're saying that if you could imagine varying the speed of light, as you made the speed of light larger and larger, this space would become flatter and flatter. So this curvature of the space really is a relativistic effect, which is related to the fact that the speed of light is finite and not infinite. Yes? AUDIENCE: Sorry, when we replace a with that, are we missing a minus sign? PROFESSOR: Oh we might be, yeah. Minus sign now fixed. The point is that for the case we're talking about, e would be negative and k would be positive. So this formula needs an absolute value sign in it. Thank you. OK, it may also be useful to relate r more directly to astronomical observables, which we can do, because we have the Friedman equation up there, which relates a to rho. And a dot over a is also the Hubble expansion rate, so that's h squared. So this formula tells exactly how to write a in terms of rho and h squared. And in fact, it tells us how to write a over the square root of k in terms of rho and H squared. And that's exactly what r is-- it's a squared over k. So putting those equations together, we could write r is equal to c times the inverse of the Hubble expansion rate, over the square root of omega minus 1. Where omega you would call is equal to rho divided by rho sub c. And rho sub c is 3 h squared over 8 pi-- Newton's constant. So this formula says two things. It says that the radius of curvature becomes infinite if c were infinite. And that says what I already said, that's a relativistic effect. It also tells you the radius of curvature goes to 0 as omega goes to 1. So omega approaching 1 is the flat universe case, which is what we've already mumbled about, but this formula shows it very directly. As omega approaches 1, for a fixed value of h and a fixed value of the speed of light, the radius of curvature goes to infinity. The space becomes more and more flat. Look, I'm just going to write one more formula, which is really just a redefinition, but an important redefinition as far as where we're going to be going next. And then we'll finish today's lecture, and continue next week. What I wanted to do is to put these definitions back into the metric itself. So we can write ds squared is equal to a squared of t divided by k-- which is what we previously called r squared-- times du squared over 1 minus u squared, plus u squared times d theta squared plus sine squared theta d phi squared. And now what I want to do is make one further redefinition of this radial variable, which, remember, initially was psi. Then we let u be equal to the sine of psi. Now I'm going to make one further substitution. I'm going to let little r be equal to u divided by the square root of k, to bring this k inside. And that then is also equal to sine of psi, divided by the square root of k. And when we do that, the metric takes a slightly simpler form. ds squared is equal to a squared of t all by itself on the outside now. And then this, after we factor in the k, becomes dr squared over 1 minus k r squared, plus r squared times d theta squared plus sine squared theta, d phi squared. And this is the form that the metric is usually written in. It's called the Robertson-Walker metric. So we've only discussed closed universes. I had hoped to discuss closed and open, but open will in fact follow very quickly from what we already have. So we'll begin next time by discussing open universes.
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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 material from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: All right. So today we're going to start the last section. We're going to do three lectures on phase diagrams. And I've given this the label here of Stability, Sustaining the Solid State. We've talked a lot about the solid state in 3.091 as the vehicle for teaching the rudiments of chemistry. One of the things we have not talked about, is what are the conditions that sustain the solid state? How do I know whether something's going to be a solid, or a liquid, or a gas? This is very important. For example, if you're in a foundry, you're running places, making auto parts, you have to know what the solidification temperature is of the alloys, so you can get the parts out quickly and keep productivity high. It's even important in failure analysis. You know, when they were looking at things like the rubble from the World Trade Center by understanding phase stability, the thermal history was imprinted in the metal. It's possible to determine what temperatures were achieved, and therefore, what the chain of events was that led to the collapse of those buildings. Now you might say, well, isn't this straightforward? I mean, for example, you just look up the melting point or boiling point. Water, 0 degrees C. Melting point, boiling point, 100 degrees C. Well, suppose you decide to realize your life's ambition. You're going to go and scale Mount Everest. So you pay the $10,000, you get a license from the Nepalese Government, and you're at base camp at 20,000 feet. You're sitting there, you've got your campfire going, you've got a hankering for a hard-boiled egg. And you start, you put the eggs into boiling water and you cook them 10 minutes, you take them out, and they're still runny. You say, I must've lost track of time. So you cook them 20 minutes, and they're still runny. And you cook them 30 minutes, and they're still runny. And you eventually come to the realization that at 20,000 feet, the boiling point of water is below the denaturing temperature of egg yolk. So now the boiling point is a function of pressure. So you know, we could have one of these Iron Chef cookoffs, only the dish is souffle, only we'll have the contest at Flagstaff, Arizona, where the altitude is so high that the classical recipes won't work, because the boiling point, the atmospheric pressure, and everything conspire so that you're not going to be able to support the souffle. So you're going to have to understand phase diagrams. In fact, if you understand phase diagrams, that's your ticket to being a four-star chef in the kitchen. Well, here's another place we can look at, where pressure has an important role. Let's look under the hood of a car. You're running an internal combustion engine. So this is the engine block, and there's combustion going on inside. We've got to keep this thing from overeating. It could damage the metal. In extreme, you could melt the metal. So we've got cooling channels in here with water flow. But then the water's going to heat up, and the water's going to boil. So we've got, over here, the radiator. So it's really a double cooling system. The water cools the block, and then the radiator cools the water. So how's that going to work? Well, here, the temperature is on the order of about 90 degrees C, and here the temperature inside the engine block is much greater than 90 degrees C. It's several hundred degrees C inside that engine block, and we've got water going like this. So we've got cooling water flow in this manner. And now cold water goes in and hot water comes out, and then we get into the radiator, and we've got channels here, and we've either got a fan or some kind of air flow. And so now, what's the gambit here? I want to see what's going on inside this radiator. So if this is the wall of the radiator, and in here I've got water, and over here I've got air, the idea is to have a high heat flux to get the energy out of the water and cool it down. Now, if things get really, really hot in here-- let's say you're zooming along down the highway at about 90, I mean, about 65 miles an hour, and all of a sudden you come upon a collision, and you have to go to a dead stop, all that energy in the engine is now dumped, and the water could overheat. And in the extreme, if it overheats, it can actually go into a boil. And this is bad. Because heat transfer is really good between a liquid and a solid, and it's really poor between a gas and a solid. This is a low heat flux, and this is bad. High heat flux, that's good. This is the bubbles. This is boiling. And why? Because what transfers energy? It's the atoms. And what's the atom density in a liquid versus the atom density in a gas? You just have really crummy heat transfer. That's why insulation involves dead airspace. Because you have very poor atom density. So what can we do in order to try to repress the bubble formation? Well, we said if we went up Mount Everest, we went to high altitude, the boiling point went down. And why did it go down? Because the atmospheric pressure is lower. So why don't we go the inverse here, and we'll put a pressure cap on the radiator, and make the pressure go up, and when the pressure goes up, it exerts a back pressure and represses bubble formation? So by increasing pressure, we can increase the temperature, and get the temperature up. So let's say p equals-- if you look on the top of the pressure cap, it will say 15 psi. And 15 psi is almost 14.7 psi, which is exactly 1 atmosphere, which in SI units is, and I know this to six significant figures, 101325 Pascals. Or, you know, from Torricelli, 760 millimeters of mercury, that's the column that's supported. So we're at 2 atmospheres pressure in there, and with two atmospheres of pressure, we can raise the boiling point to 106 degrees C. And then what we can do, is now change the composition. So we change the pressure, and I change the boiling point. I change the composition. add 50% ethylene glycol. So now I've got a one-to-one mix, ethylene glycol and water. And that takes the boiling point up to 130 degrees C, or 265 degrees Fahrenheit. So another example of how we can manage systems. So when I'm showing you by these several examples, is that the boiling point is a function of pressure and composition. So we can tune the boiling point. See we were tuning materials properties before. Now we're tuning physical chemical properties, because that's what chemistry is all about. It's about control. It's about management. Management to advantage. So how do we know what to do here? Can we predict this from first principles? No. Systems are still too complicated. Our models aren't robust enough. So instead, we turn to an archive. So we have phase diagrams. And the phase diagrams guide us in our search for better management of material systems. And these are stability maps. They tell us which phases are stable under which conditions of pressure, temperature, composition. You can consider them as archives. So that's what I want to talk about, because it's really important to know this stuff. OK. So what are we going to do? So first of all, let's-- I don't want to write on that one. This is interesting. This could be a work of art. I'd hate to desecrate it. You like? This is really good! So I'm just going to leave that. That's very good. That's excellent. OK. So I've been talking about phase. What is a phase? Let's get the right definition here. What is a phase? It's a region uniform chemical composition. Just to get it down. it's a region in a material, region of uniform chemical composition, and it has these other properties. Uniform chemical composition. It's physically distinct. And I'll show you by examples, but let's just get the definition down. It's physically distinct, and in the extreme case-- so that means it's bounded. It's physically distinct. That's a fancy way of saying that you can put a boundary around it. It's bounded. And then the other property that it has, it's mechanically separable in the extreme. And so that's in the case where you have a multi-phase system. You can point to the different p. So what I'm going to do, is I'm going to give you some examples. And so I'm going to let circle p equal number of phases. I'll give you some examples. Why am I using circle p? Because just plain old p is already pressure. So pressure, so this is kind of a Western motif, you know? Like a Circle Bar Ranch or something. So it's a p with a circle. That's my font. That's how I say number of phases. So let's take a look at some examples. So here's some examples of p equals 1. So simple one is pure water. Pure H2O liquid. It has all of this. Wherever I have the water, all right, here's some water, it's all of the same chemical composition. It is physically distinct. I can put a boundary around it. And in this case, there's nothing to separate, because it's just a one-phase system. White gold. White gold is one phase. Because we have gold. Obviously we have gold. If you market something as white gold, it has no gold, you'll go to prison. And how do we make white gold go to its white color from the normal yellow? We add silver and we add nickel. And these are all FCC metals, and they substitute for one another on the lattice, and if you sample anywhere in the white gold, you get the same chemical composition, and so on. Air is an example of a one-phase system. It's a gas, and in decreasing order, we have nitrogen, oxygen. So it's a solution. We're breathing a solution. Nitrogen, oxygen, argon are the main constituents. There's some CO2, some sulfur dioxide if you live near a power plant, NOx if you're near a tailpipe, and so on and so forth. And then the last one I'll give as an example is calcia-zirconia. Or cubic zirconia. Let's just call it cubic zirconia. The holidays are coming, so we better get cubic zirconia up here. That's the poor man's diamond. Cubic zirconia, which consists of a solid solution of calcium oxide in zirconium oxide. This is a solid solution, so it's continuous and chemical composition identical. Now, contrast that with two-phase system. So p equals 2 looks like this. So let's look at, instead of simple water liquid, if I have ice cubes in water. So now I have two different phases, and they're separated by state of matter. State of matter is the issue here. Because I have a solid-- this is solid-- and I'm going to put the boundary around them. The composition of the solid is different from the composition of liquid. They're both the same water, but you can tell it's a different crystal structure and so on. And I can put boundaries around, and it's mechanically separable. You can pull the ice cubes out of the mix. Milk is two-phase. Milk has a fatty phase and it has an aqueous phase. The aqueous phase contains the minerals. That's why you can drink skim milk and still get your calcium, because calcium, even the general public knows calcium is a mineral. It's not elemental calcium, it's a calcium compound. But it's ionic, it's soluble in water, and the fat contains the other parts. You know from last day, this is one of the lipids. And so you have fat globules that are mechanically separable and distinct from aqueous. So they actually have a different chemical composition, different type of bonding. Right? This is largely nonpolar, and nonpolar doesn't like to dissolve in something that is polar and hydrogen-bonded. So they phase separate. I've been saying phase separate all along. Now you know what we mean by phase separate. Here's a third one. This is really cool, because you won't get this in any other chemistry class, because only 3.091 talks about such things. David, can we cut to the document camera? I want to show you a piece of snowflake obsidian. Obsidian is a volcanic glass. So here's the piece of obsidian. And what you're looking at is, the black area is the glassy part, and the white is actually some of the obsidian that has decided to devitrify. And it is turning crystalline. And so those are islands of obsidian crystal sitting in a matrix of obsidian glass. If we go back to the-- I picked that up in Yellowstone Park. It's fantastic. There's a wall, just a wall of obsidian that comes down. And I think I've got an image here. There it is. This one I pulled off the internet. So this is crystalline phase, and this is the amorphous phase. And the overall composition, it's a glass, it's a silicate. And why isn't it transparent to visible light? Because it's got iron in it. The iron gives it the absorption and capabilities. The band gap is invisible, so it's black. But then when this crystallizes, it's got a different index, and it's mainly SiO2. It's really the cristobalite phase of the obsidian. So that's cool. So this is actually differentiated on the basis of amorphous and crystalline of something that's virtually the same chemical composition. So this could give you an idea of what happens in the area of phase separation. OK. Now the second thing that I have to bring to your attention, is that these phase diagrams treat systems at equilibrium. They're stability maps, and they treat systems at equilibrium. So what is equilibrium? Let's just get one time up here a definition of equilibrium. Equilibrium is a condition which there is no net reaction taking place. The system could be reacting. If you're looking at the ice cubes in water, the ice could be dissolving, and some of the water could be freezing, but overall, with time, there's no net reaction. It's a dynamic system with no net reaction. And it's characterized as the lowest energy state. Systems strive towards equilibrium. If it's the lowest energy state, no net reaction then. If a system is truly at equilibrium, the properties should be invariant over time, because there's no net reaction. You can't prove equilibrium. You can only demonstrate the absence of equilibrium. Because at equilibrium, there's nothing, there's no net reaction. And it could be attainable by multiple paths. So in other words, if I have ice as the equilibrium phase, I can make it by freezing water, and get to the same ice at the given temperature and pressure. Or I could sublime. But if I specify the pressure and the composition, I should always end up with the same system. So attainable by multiple paths. Because it's a stable state, so it doesn't matter how you got there. It's a state function. And you know from your physics, a state function has the same value, regardless of how you arrived there. And then the last thing is, I want to differentiate. See, these are all single phase, but some of them are just one element, some of them are just one compound, and some of them have a whole mix. So we have a way of quantifying the degree of chemical complexity by using the term called component. It's a measure of chemical complexity. It's the number of chemically distinguishable constituents. And again, these are definitions. I'll give you some examples so then the definition comes to life. I like to think of it as the number of bottles you've got to pull off the shelf to make the final product. The building blocks. So for example, water is just water. You could say, but it's hydrogen and oxygen. But if you take 560 or some other thermal class, you'll learn that they're bound by mole ratio. You don't have freedom of hydrogen and oxygen. So it's just one bottle, whereas to make white gold, you need three bottles. You need a bottle of gold, a bottle of silver, and a bottle of nickel. So that's a three component system. And I'm going to designate the components by c with a circle around it. Because c without a circle is already composition, concentration. OK. So now we can put all of this together, and we'll show the difference between multicomponent and multiphase. So we'll make a little table here, and here I'm going to talk about one-phase systems, and distinguish on the basis of components. So if it's single phase single component, that's just water. One phase, one component. Simple liquid. If it's two components one phase, that could be calcia-zirconia. I need a bottle of calcia and a bottle of zirconia in order to make this two component single-phase system. And then to have three components, that's the white gold. It's still single phase, but I need the gold, silver, and the nickel, whereas if I come over here, I've got two phases. That could be slush. Slush, that's ice water. It's just one component. I just start with water, and drop it down to 0 Centigrade, and I'm going to end up with ice crystals in the water. So that's one component. I just needed one bottle. Literally, one bottle. All right. Now what about two components, two phases? Well, we saw earlier when we started thinking about solubility, we did the bilayer with carbon tetrachloride and water. Remember, we dropped potassium permanganate and iodine, and the potassium permanganate dissolved in the water, and the iodine dissolved in the carbon tetrachloride? Two different components. I need a bottle of carbon tet and a bottle of water. And they'll phase separate. They don't dissolve in one another, because this is a nonpolar liquid, and this is polar hydrogen bonded. That's the origin of phase separation. It's all about bonding, and bonding is all about electronic structure, and that sounds like a good topic for a class in chemistry. And then the last one here is obsidian. The snowflake obsidian, in which the components are SiO2-- so I'm going to have three components and end up with two phases. SiO2, there's some magnesia, and then the black comes from the magnetite. Fe304. So I've got three bottles off the shelf, and I end up with something that is crystalline and amorphous. So that's where I get the two phases. Three components, two phases. It's great. So now I want to look at some stability maps. And I'm going to start with the simplest of all, which is the one we know best from human experience, and that's water. So we're going to look at the one-- so it's a one component. So for water, we start with c equals 1, and the example is going to be water. One component phase diagram. So we don't have a composition axis, because the only thing we can vary is pressure and temperature. If we change composition, we've lost water. So it has to be axiomatically, no variation composition. All we have is that the coordinates of temperature and pressure. So there it is out of the book, and I'm going to draw it a little bit differently, just to emphasize three of my favorite words. Not to scale. Otherwise you can't see some of the subtle features. So overall, it's a y-shaped diagram. And this is pressure. Let's make it pressure in atmospheres. Then our human experiences is at 1 atmosphere. So at the highest temperatures, we expect to have vapor, and at the lowest temperatures, we expect solid. And in between, we expect liquid. And then if we look at the 1 atmosphere isobar, what's happening on this line? On the one side of line it's liquid, and the other side of line, it's vapor. At this temperature, liquid and vapor coexist. I call that the boiling point. So I'm going to put 100 here. And we already learned from our little example of Mount Everest that this locust here, this line, is the liquid equals vapor two phase equilibrium. And you see how it changes? If I go to the top of Mount Everest, the pressure is less than 1 atmosphere. which means the boiling point of water is below the denaturing point. So this actually gives you the map. Or if I put the pressure cap on, I go up here. And you can use this in the kitchen again. If you love wild rice, and you can't wait 45 minutes when you get home, use a pressure cooker. What the pressure cooker does, is it allows you to have liquid up to 130 degrees Celsius. See, you can't steam it. You've got to soak it in water. And you know the Arrhenius Law. If you can get the temperature up by 30 degrees, you'll cut the cooking time in half. So how do you get water, liquid, at 130 degrees? Then you can cook quicker! Pressure. Seal it. And then you've got liquid water way, way up here. Now over here, this is solid goes to liquid. So this is the freezing point line. At 1 atmosphere, that's 0. 0 Celsius. And you know, if you go skating-- so let's say the ice is minus 5 Celsius. And you're here, but you put your weight on the blades, and the blades have small area. So pressure is force per area. So that puts you up here, and now you cross, and now you glide on water. Film of water. You ever watch a hockey game? I grew up in Canada. We used to say, well, the ice is fast. Why is the ice fast? Because you get down to temperatures where you just get the right slickness of water. If the ice is slow, the temperature is so close that, you know, you get these big, bruising hockey players, and they're moving up into here, and it's like they're dragging. And if you ever come from Minnesota or North Dakota, if you've ever skated when the temperature gets down about minus 20, minus 30? It's a different experience, because you never get across this line. Your blades are just, they can't get a good grip. See? Everything you need to know is here. This is four-star chef, this is prize-winning hockey player. We haven't got off the 1 atmosphere line yet. OK. So now I'm going to say, this is single phase, right? It's just vapor. This is single phase, and this is single phase. It's all solid. And along this line, it's two phase, isn't it? Solid-liquid. This is ice cubes, isn't it. Anywhere along this line is ice cubes in water. Look at this one. This is liquid vapor. This is solid, liquid-- all three of them coexist here. This is p equals 3, which means, ice cubes floating in boiling water. And this happens at only one point. It's called the triple point, and its coordinates are-- this is not to scale. So this is 0.01 degrees C. It's a 1/100 of a degree higher. And the pressure here is 4.58 millimeters of mercury, which then you can convert using the 760 as 1 atmosphere. Oh, and this is solid-vapor. You can go directly from solid to vapor down here. If you hang up your clothes on a line and it's minus 20 degrees outside, first thing that happens, the clothes freeze. You come back five hours later, and the clothes are dry. Well, what happened? Did somebody take them in? Run them around the clothes dryer, then hang them up for you? No. You're going directly from solid to vapor, down in here. So here's the whole story of the phase diagram. One last thing I want to point out. Here, you're at a given temperature and it's vapor. And then you squeeze, squeeze, squeeze, and now things get closer together, and you go from paper to liquid at constant temperature. That makes sense, right? I start from vapor, constant temperature. I squeeze it, I get liquid, I keep squeezing, I should make solid. Here's vapor. I squeeze, I get solid, and I squeeze harder, and I get liquid. That makes no sense at all, but that's what happens. What does this tell me? It tells me that the number of nearest neighbors in the solid is greater than the vapor. Yeah, I get that. But the number of nearest neighbors in the liquid must be greater than the number of nearest neighbors in the solid. This is an exception. When I see this negative slope, it means that ice cubes are going to float on water. That's what I learn from this negative slope. So let's put that down. dp by dt when dp by dt is less than zero. This is for solid equals liquid. That means that the density of the solid is less than the density of the liquid. All from here. OK. Good. So let's look at a few other phase diagrams. I think I've got a few things up here. Here's silicon. Pressure versus temperature. Its normal melting point is about-- see, now I said normal melting point. After today's lecture, if somebody says, what's the boiling point of water? You don't go, 100 degrees. You say, at what pressure? And they go, oh, I hate you. I hated you before. I'm going to use an adverb here. Now I really hate you. OK. So this is the normal melting point of silicon. It's about 1430 degrees Centigrade, or a 1700 something Kelvin. And it also has this negative slope. And what does that tell you? That liquid silicon is denser than solid silicon, and it's a good thing it is, because we wouldn't have the microelectronics age if it weren't for that. We couldn't have Teukolsky crystal growth and everything else that makes silicon dirt cheap. You know why it's dirt cheap? Because it's made from dirt. That's why it's dirt cheap. It's true. Here's aluminum. This is normal. This is the normal solid-liquid equilibrium. This is FCC metal, which is the closest packed. Axiomatically, the liquid must be less well-packed. All right? So this is dog bites man, this is man bites dog. This is the unusual one. OK. What else do we have? Nitrogen! Here's nitrogen. I drew this. So there's the 1 atmosphere line. Nitrogen boils at 77 Kelvin, and it freezes at about 63 Kelvin. And it's got the positive slope that you'd expect, and so solid nitrogen is denser then liquid nitrogen. And you can remember 77 Kelvin. You want to remember a few of these things. I wanted to point out to you that at atmospheric temperature, ambient temperature, you can remember 20-20. 20-20 vision. So at 20 degrees C, it's roughly 20 millimeters of mercury, is the vapor pressure of water. It's not exactly true, but it's close enough. I think it's really 24. But 20-20 is nice to know. And what's the street address for MIT? 77 Mass Ave. That's the boiling point of liquid nitrogen, 77 Kelvin. So I thought we'd do a few things here. And I'm going to show you a little bit of fun and games with liquid nitrogen. But first, I've been instructed I have to practice safe laboratory-- So I'm going to put on my lab coat. Yes. It's a nice lab coat. Mm-hm! Yeah, look. You want to know-- let's go back to this. They have to know where the-- this is not just any old lab coat. This is a nice lab coat. This is from France. It's from MAISON DUTECH. See? Notice the collar. It's not that typical, you know. MAISON DUTECH. OK. Let's go back to the phase diagram. So Dave Broderick is going to help us here. So we're going to have some fun. I'm going to put on a face shield so I don't blind myself. All right. Here we go. All right. So let's get some liquid nitrogen. Sounds different, huh? Sounds terrible for me. All right. So we've got some liquid nitrogen here, and I'll pour it into a Dewar. So what happens when you wet a sheet of paper with ink with liquid nitrogen? What kind of bonding is there in liquid nitrogen? Look. No running, no running. If it's liquid nitrogen. I tell you, everything you learn here is so valuable compared all the other stuff you learn here-- All right. So there's a couple of things. So let's look at glass transition temperature. So this is a latex glove. You can see it's above its glass transition temperature, and the cross-links are snapping it back. What I'm going to do, is take it down below its glass transition temperature, and turn it glassy and brittle. So that it's below the TG. Here's some paraffin. Paraffin, which is also a polymer here. Take the paper off. OK. So here's paraffin. Look. Look at the van der Waals bond. All right? This is van der Waals bond. Now we're going to go below the TG. It's liquid nitrogen! It doesn't matter. OK, you can hear it already. So it's now below the glass transition temperature. So what else have we got? Oh, you've heard of the glass flowers at Harvard? Well, we've got glass flowers here, so. Here's a rose. See, in the proteins, you know, it's above the-- you know, it's soft. But now, we're going to put the-- mm-hm. Here we go. You ready? Oh, come on. It's not a kitten. What's the matter? It's still burbling here. We have to wait until it's dead. This is for science. OK. Now it's below its glass transition temperature. Oh, I'm sorry, David. I should put this over on the other side. But maybe it's the red, or maybe it's because it's a rose. So here's the chrysanthemum. Today is your day. Let's go. Get this thing moving! Yeah. So there you have it. There you have it. I think that's-- oh, there's one, may we go back to the slide? Better put the gloves on for this. I'll have to write apology letters. OK. So we're going to do next, is we're going to look at phase equilibriums. There's the melting points and boiling point of oxygen, argon, and nitrogen. And what you notice is that nitrogen boils at a lower temperature then oxygen and argon. So what I'm going to do, is I'm going to pour some liquid nitrogen into this beaker. And then inside the graduated cylinder, I'm going to let it sit here, and we're going to condense air. And the result will be that we're going to start condensing liquid oxygen, and they'll be little bits of argon ice floating in liquid oxygen. And liquid oxygen is faintly blue, and it's paramagnetic, as you know, because it's got the unpaired electrons, which means that it will levitate in a magnetic field. I couldn't bring a big magnet in here, but we'll see if we can make some blue stuff. And it's foaming, because this would be like pouring water into something that's-- I don't know. 300 degrees Centigrade. But eventually, the heat transfer gets low enough that you have stability here, and-- yeah. OK. Oh, that's terrific. OK. What else have we got? Let's go back to the slide. God, I hate this thing. OK. So you've been looking at that nitrogen. Let's look at what's next. Ah. Now if you solidify nitrogen here, you get a whole bunch of different polymorphs. Different crystal structures. So there are about five or six different crystallographic forms of solid nitrogen. For that, we'd have to bring liquid helium. 4.2 Kelvin. Very cool stuff. Here's carbon dioxide. Now, it also has the positive slope, so solid will sink in liquid. But look, the interesting point here. The triple point is 5 atmospheres. At atmospheric pressure, you go directly from solid to gas. It sublimes. And that's advantageous. When we use carbon dioxide as a coolant, when it gives up its enthalpy to cool, it doesn't then turn into liquid. If you chill something with ice, with water ice, when it gives up its enthalpy, the object is now swimming in water. And that's no good. So hence we have the term dry ice, because we go directly from solid to vapor. And we want to prove this. So what we've got here is a big block of dry ice. So this is at minus 78 degrees Celsius, and it's still stable here. Well, it's subliming before your eyes. It's smaller than it was when we started the lecture. So what I'm going to do now-- oh, I'm not wearing that stuff. I guess I'd better. They're going to write me up on some log, teaching you bad practice. But anyways. All right. So what we're going to do, is we're going to test the proposal that this actually goes-- so I've got some water. Here, David, let's go out to the video. So let's take some water. I think we got water from France. Goes with the lab coat. This is Evian, OK? So I pour the Evian into the Florence flask. This is Florence. Florence is round, and the Erlenmeyer is the flat one. So let's move this guy. You're going in the backstage for a while. You just keep condensing All right. So now, this is here. Now what we're going to do, is we're going to put some dry ice in there, and see if we go directly from solid to a vapor. Take a chip off the old block, here. OK. So now you can see that it's going directly from solid to vapor. So that's carbon dioxide bubbles. And we assume that the phase diagrams is correct. I mean, another thing we could do, is-- I mean, I know what club soda tastes like. I could just test it, right? Yeah. It's carbon dioxide. Now! Here's the other thing. You know, Schweppes likes to pride itself on tiny bubbles. They have a tagline, Tiny Bubbles. Right? So here's Schweppes. And you can see Schweppes, with their wimpy little tiny bubbles. Can you get that, David? But we have, we have big bubbles. I mean, this is pure. You can't get any better than that. I got Evian water, and I got a block of carbon dioxide. I like making my own carbonated water. I've taken it to a new level. OK. What else do we have? Well, let's see if we made any liquid oxygen here. Try this. I don't know if we got anything here. Oh yeah! Oh, wow. Look. OK. So that liquid in there-- oh, I see. You've got a separate thing. I don't know if you can see. It's condensing. That liquid in there, that's liquid oxygen. I won't drink this. But I can inhale it. Pure oxygen. I could have a cigarette, and just put the cigarette in there. What happens if I put the lit cigarette into the oxygen? All that happens is that it burns quickly, that's all. There's no explosion. It burns quickly. If you walk into a room full of oxygen, all that happens is you'll singe your nose, that's all. There's no explosion. It's just oxygen. There's no fuel. The fuel is just in tobacco, and there's not much. It's not a problem. A room full of oxygen? Pfft! OK. Let's go back to the slides. What have we got here? We've got some examples of other phase diagrams. All right. Yeah. [MUSIC PLAYBACK: FROM 'PHANTOM OF THE OPERA'] PROFESSOR: Actually, it's quite tasty. All right. So. Now, here's zirconia. Now, here's the thing to know about zirconia. Look at all the different phases. And cubic zirconia is the one that has a high index of refraction, but it's stable only at temperatures exceeding 2000 degrees C. So that's no good. So what I'm going to show you next day is that by changing the composition, and adding calcium oxide, we open up this zone here that's labeled cubic, and we get it stable all the way down to room temperature. It's an example of changing composition to get the stability of a phase that we want. This is carbon. So down here, the normal form, as we've learned earlier, the normal form is graphite. Diamond is formed at elevated temperatures and pressures. And this is liquid carbon. All right? Now, to make artificial diamond, you can try putting graphite in an anvil, and squeezing, and squeezing, and squeezing, and coming up into here. And then it becomes metastable, and then maybe, because the activation energy to jump from that sp3 hybridized state to sp2 is so high, the diamond is stable. So you don't have to go running home to see if your diamond studs are now turning into graphite. They won't. Once you get into this zone, they'll stay. The activation energy is too high. You need more than 1/40 eV to do it, is necessary. But in the 1950s, General Electric, its Schenectady research labs, reasoned that because carbon is so soluble in iron, they could dissolve carbon in liquid iron, and then raise the pressure and temperature, and then change to exsolve and cause carbon to precipitate out of molten iron. That's the birth of artificial diamond. That's how they did it. Using this concept, but then dissolving it in molten iron. [MUSIC PLAYBACK: 'DIAMONDS ARE A GIRL'S BEST FRIEND'] PROFESSOR: I don't know how that got in there. All right. Here's one for mercury. I looked it up, using the tools that you've been taught here in 3.091. So this is a paper phase diagram for mercury. And physicists drive me nuts, because they plot temperature versus pressure. I want pressure versus temperature. So I just turned it around. And so here's the liquid line. So you know that liquid mercury is going to float on solid mercury, and there's a plurality of solid phases. And these are all different crystal structures. Now I've got a treat for you. I was in Barcelona several years ago, and I saw the mercury fountain that was made by Joan Miro in collaboration with Calder. Calder has the big sail outside of East Campus. So Calder collaborated with Miro, back in the '30s and '40s. And this is in honor of the miners of Almaden, where there are cinnabar mines. And the miners fought bravely against the Fascist forces of Franco, and Miro wanted to honor them. So here's their museum. It's a beautiful place up on top of an escarpment over Barcelona. And what I'm going to show you is a mercury fountain. [MUSIC PLAYBACK] This is liquid mercury. The speed has not been altered. This is a liquid at room temperature, density 13.5, metallic bonds. Look at it. Very high surface tension. Look at the way it shimmers. Look at this shot. This has not been altered. That's the way it pours into gravity field. It's really fantastic. And of course, mercury vapor is toxic, so the whole thing is in a-- [END MUSIC PLAYBACK] PROFESSOR: Oh, that's the-- I cut that snippet out of the documentary. OK, we don't want to see that. So they have the whole thing inside of a polymethylmethacrylate cage, to keep the mercury vapor from affecting the people that work in the museum. Here are some phase diagrams from hell. This is bismuth. And look at bismuth. Bismuth is like water, and like silicon. So the solid is less dense than the liquid. But look at all of these different phases. Here's sulfur. Sulfur is crazy. Look, phases in the solid phase, there's even phases in the liquid! Different phases. They're allotropes. Here's sulfur. Elemental sulfur, the disulfide, hexasulfide. It makes rings, it makes long chains. And these are called allotropes. They're different forms of the same element. So oxygen is O. There's the diatomic molecule. And the triatomic molecule, which is ozone, these are all allotropes of oxygen. What else do we have? OK. This is polymorphs. So this is alpha goes to beta. See, different crystal structures. You could see with your naked eye what's going on here. Different crystal structures. And then this goes to lambda, and then finally, they're pouring here. And if you get a very high cooling rate, and you've got those long chains, what happens? You don't form the crystal. You form the amorphous form. So they'll call this forming plastic sulfur. We know better. It's amorphous sulfur. It's not plastic sulfur. Here's water, a complete diagram. Now, this is in kilobars. So there's all the different phases of water. There's the negative line. That's the thing here. But at very, very high. Look at all the different phases of water. And they have numbers on them. If you read Kurt Vonnegut's Cat's Cradle, there's ice nine. You get that going, it freezes the whole world, right? But here's, I draw your attention on this one. You see this point here? This is ice seven. At 100 degrees C, it's solid. You need 25 kilobars. So we could go over to the lab, take some water, put 25 kilobars on it, make solid ice at 100 degrees C. And it'll be stable. Look! This is minus 78. It's been here for a long time. So now I come into a cocktail party with this, I drop this ice cube into a glass of aqueous beverage, it sinks to the bottom, and the beverage heats up to the boiling point. That's what this is telling you! That'll get you popularity instantly. How did you do that? How did you do that? And up here is the supercritical fluid, the last thing we haven't said anything about. Look. If you take a gas, and you keep squeezing it and squeezing it, eventually the gas molecules are close enough together that they're indistinguishable from the liquid phase. Or if you start here, with the liquid, and you keep raising the temperature, the density goes lower and lower until the molecules are far enough apart that you can't tell this zone, whether it's a rarefied liquid or a highly compressed gas. The properties criss-cross. This is supercritical. It has liquid-like transport properties, and vapor-like chemical properties. So you can end up doing things that are very, very different. Now, I'm going to give you an example of decaffeinating coffee by taking the coffee, going up into the supercritical regime, which allows you to extract the caffeine, and not brew the coffee. So you want to just get the-- well, if you just heat it up, you'll make the coffee. So here's an example. It's called solvent extraction. Historically, this solvent was methylene chloride, which is also used as paint stripper. We don't do that anymore to coffee. They did in the '60s! They did. They also used trichloroethylene. It works really well to leech out the caffeine. How much is left in there for you to taste in your coffee? I don't know. Maybe that's why people were so excited in the '60s to drink their decaffeinated Sanka. I don't know. But anyway, so you take the green beans in carbon dioxide as supercritical, and drop the caffeine level, and then you play with temperature and pressure to go over the KSP. The caffeine salts out, and then you recycle the CO2. OK. I hope I've given you some introduction to phase diagrams. And we'll see you on Monday.
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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 the last two videos, we looked at the Dirac equation and we looked at solutions Dirac equations. And in the last lecture we found that, along with positive energy states, we had those negative energy states. Since we cannot simply drop them or disregard them, we do have to find physical interpretation for these negative energy solutions. The first one which was put forward, is the one where you think about negative energy states all being populated-- and that is the vacuum. The vacuum is basically a sea full of negative energy states which are all populated. So if you have a positive energy state, and there are electrons sitting in this energy state here, the electron, because of the Pauli exclusion principle cannot fall down into the negative energy state. But you are able to kick them out, for example, to excite them with a photon. Very excited. The negative energy state, you get an electron out. This process is then will lend to the creation of a positron and an electron pair with a photon. [INAUDIBLE] pair production. It can also explain undulation where there's an empty and a negative energy stage where the electron just folds into creating a photon. So while the interpretation is useful and it explains pair production and undulation processes, they fail to explain what this vacuum, the sea of negative energy state even is. So a more useful interpretation is one part forward that Feynman and Stückelberg, which came out of the discussion of quantum field theory. And we already discussed this interpretation when we looked at Feynman [INAUDIBLE].. So have a look at this Feynman diagram here where you have an electron with a positive energy and an electron with a negative energy, building a photon, which is twice the energy in the symmetric configuration of the electrons before. And you're interpreting the negative energy solution here of the electron as the electron moving backward in time. This is an equivalent to a positron with a positive energy and an electron with a positive energy where the positron and the electron move forward in time. Again, in both cases, you see the energy of the photon is two times the energy of those two particles. All right. So this is a very short discussion. And we will see later on how we use the spinodes for antiparticles together with spinodes for particles order to make relations that could have matrix element. And so we move forward with our discussion of Feynman rules, this time now, with spin-1/2 particles included.
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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: Let us consider the anharmonic oscillator, which means that you're taking the unperturbed Hamiltonian to be the harmonic oscillator. And now, you want to add an extra term that will make this anharmonic. Anharmonic reflects the fact that the perturbations are oscillations of the system are not exactly harmonic. And in the harmonic oscillator, the energy difference between levels is always the same. That's a beautiful property of the harmonic oscillator. That stops happening in an anharmonic oscillator. The energy differences can vary. So the things, if you have a transition from one level, first level to the ground state, or second level to the ground state, one is not the harmonic of the other because they're not exactly twice as big as each other. So let's try to add an x to the 4th perturbation, which is intuitively very clear. You have a potential. And you're adding now an extra piece that behaves like x to the 4th. And it's going to make this potential blow up faster. Now, in order to get the units right, we need the length scale. There's a length scale d in the harmonic oscillator. We're going to use it. One way to derive that length scale is to recall from units that p, a momentum, has units of h divided by length. So b squared has units of h squared over d squared. There's an m. So that's an energy. And this is an energy too. So we can set it equal to m omega squared d squared. From where we get, for example, as 1 over d to the 4th is equal to m squared omega squared over h squared. And d squared is equal to h over m omega. So that's a length scale, something with units of length in the harmonic oscillator. So if we want to add the perturbation that is x to the 4th, we'll add a lambda delta H that is going to be a lambda, which is unit freedom, something that has units of energy. So you can put h bar mega has units of energy. And then you can put the operator x to the 4th divided by d to the 4th. It's a perturbation with units of energy. This is good. So there's a couple of ways of thinking of it. You may remember that in the harmonic oscillator x, the operator x, was given by the square root of h over 2m omega, a plus a dagger. So this is d times a plus a dagger over square root of 2. So this perturbation lambda delta H is equal to lambda. H bar omega would be here. H bar omega. Now x/d is this a plus a dagger divided by square root of 2. So you have a 4a plus a dagger to the 4th. So that's lambda delta H. That's your perturbation. So what do we want to do with this perturbation? It's a nice perturbation. We want to compute the corrections to the energy, corrections to the states, and see how they go. We can do that easily. Let's compute first that corrections to the ground state energy. These are calculations that are not difficult, but they take some care. As any calculation, it's based on just getting a lot of numbers right. So what is the ground state energy correction? So we had a series of formulas that I just erased. But this index k0 for our state, we can think of k as being the eigenvalue of the number operator. So we'll get n. And remember in the harmonic oscillator, we label states by 0, 1, 2, 3. And that's the eigenvalue of the number operator in the harmonic oscillator. So the ground state energy would correspond to 0, has a first order correction that would be given by the unperturbed ground state and delta H 0. So what is that? Delta H is h omega over 4. And we have 0 a plus a dagger to the 4th 0. Easy enough. That's what we have to do. Now, the evaluation of that matrix element is a simple exercise, the kind of things you've been doing before. You have to expand and just use the aa dagger computation relation that this aa dagger is equal to 1. You said repeatedly a kills the vacuum. a dagger kills the other vacuum on the left. Please do it. If you feel you're out of practice, the answer is the number 3 here. So e01 is equal to 3/4 h bar omega. So how does that tell you anything? Well, better use the full formula. So e0 of lambda is supposed to be equal to the ground state energy unperturbed, which was h omega over 2 plus lambda times the first order correction, which is 3/4 h bar omega plus order lambda squared. So e0 of lambda is h omega over 2, 1 plus 3/2 lambda plus order lambda squared. That's what we got. And that makes nice sense. Your ground state energy you knew used to be h omega over 2. If you introduce this perturbation, you don't expect it to jump abruptly. It's going to grow up with the lambda to the 4th term with the x to the 4th term into 3/2 lambda here, a small correction. So far, so good. What takes a bit more work is doing a higher order correction. So let's do the next one as well. So for this second order correction, we have a formula. And let's see what it tells us. It tells us that the second order correction to the energy is minus the sum over all states that are not the ground states. So it's k different from 0. That's good enough. Delta H 0k squared over Ek0 minus e00. So potentially, it's an infinite sum. And that's the kind of thing that is sometimes difficult. In many examples, you have infinite sums. Sometimes you can do an approximation, say a few terms is all you need to do. But in this case, happily, it's not going to be an infinite sum. So what do we have? Delta H0k is what we need to calculate. The energy differences, we know. These are the unperturbed harmonic oscillator energy differences. What we need is this object, which is h bar omega over 4. Remember, what is delta H there in that box formula there. And we have 0 a plus a dagger to the 4th K, where K is a state with number K. If you wish, you could say K is a dagger to the K over square root of K factorial acting on the ground state. But sometimes you don't need that. You actually don't need to put the value of K. So our challenge here is to see which values of K exist. For that, you can think of this part of the term, the a plus a dagger to the 4th acting on the vacuum. This is like, in a sense, this term is nothing else but in saying what happens to the wave function when you multiply it by x to the 4, to the ground state wave function. That's physically what you're doing here. You're multiplying an x to the 4th times the ground state wave function. So it should give you an even wave function. So you would expect that x to the 4th multiplied by 5 0 should be proportional to 5 0, but maybe 5 2 and 5 4th. Should not have the odd ones because this function is not odd. It's even. So as you express x to the 4th 5 0 in terms of the other ones, it should be that. So indeed, again, I ask you to do a little computation. If you do 0 a plus a dagger to the 4th, this is equal to-- you can show. It gives you 4 factorial square root the state 4 plus 6 square root of 2 times the state 2 plus 3 times times the ground state. OK. So we have that look let's continue therefore. Given that [INAUDIBLE] has given us three states-- one proportional to the vacuum, one proportional to the state with occupation number 2, and one with a state of occupation number four-- the values of k that are relevant, since k is different from 0, are only 2 and 4. 0 doesn't matter. And only 2 and 4 can couple here. So what do we get? Delta H0 4 would be, from this formula, h omega over 4 times this inner product when k is equal to 4. So you get here square root of 4 factorial, which is h omega over 2 times square root of 6. Delta H02 is equal to h omega over 4. And the overlap of this state with 2, which is 6 square root of 2. So this is h omega 3 square root of 2/2. OK. So that's your first step. When you have to compute a perturbation, you have to compute all the relevant matrix elements. And you have to think what you can get. And only k equal 2 and 4 happen, so this is the sum of two terms. So the second order correction to the energy is minus delta H0 2 squared divided by E2 0 minus E0 0 minus delta H0 4 squared over E4 0 minus E0 0. So what is this? It's minus delta H0 2 squared is h bar omega squared. 9 times 2/4. 9 times 2/4. And you divide by this difference. And the difference between E2 and E0 is 2 h bar omega. There's an h bar omega difference every state. So you go from E0 to E1 to E2, two of them. Minus delta H0 4 squared, which is H omega squared times 6/4 over 4 h bar omega. Well, this gives you minus 21/8 h bar omega. And therefore, the energy, this is the E0 2, so you can go back here and write E0 lambda is equal to H omega over 2. 1 plus 3/2 lambda. And then the next term that we now calculated is 21/4 lambda squared. Very good. We work hard. We have lambda squared correction to this energy. Now, this is not really trivial. There's no analytic way to solve this problem. Nobody knows an analytic way to solve this problem. So this is a nice result. It's so famous the problem that people have written hundreds of papers on this. The first people that did a very detailed analysis was Bender and Wu. And they computed probably 100 terms in the series or so. The next term turns out to be 333/8 lambda cube. Followed by 30,855/64 lambda to the 4th plus ordered lambda to 5th. Funnily, these people also proved that this series has no radius of convergence. It never converges. It just doesn't converge at all. You would say, oh, what a waste of time. No, it's an asymptotic expansion, this series. So it can be used. So it's a very funny thing. Series sometimes don't converge so well. And this one is a peculiarly non-convergent series. So what does it mean only that it's an asymptotic expansion? Basically, the expansion for the factorial that you use in statistical mechanics all the time, Sterling's expansion, is an asymptotic expansion. Quantum electro dynamics with Feynman rules is an asymptotic expansion. Very few things actually converge in physics. That's the nature of things. What it means is that when you take lambda, for example, to be 1 in 1,000, a very small number, then this is 1. This will be very small. This will still be small. This will be small. And eventually, the terms go smaller, smaller, smaller, smaller, smaller, smaller. And eventually, start going up again. And if you have an asymptotic expansion, you're supposed to stop adding terms until they become the smallest. And that will be a good approximation to your function. So this can be used. You can use it. You can do numerical work with the states and compare with asymptotic expansion. And it's a very nice thing. Another thing is that you can compute the first correction to the state. I will leave that as an exercise. And in the notes, you will find the expression of the first correction to this state. It's not much work because, after all, it involves just this kind of matrix elements. So it's basically using the matrix elements to write something else.
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https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/8.06-spring-2018.zip
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PROFESSOR: Today, we begin with our study of scattering. And so scattering is the next chapter in our study of approximations. What is the purpose of scattering? Physicists want to study the interactions between particles, the forces. We know, for example, in a hydrogen atom there is an electromagnetic force within the proton and the electron. We understand it rather well. But there are other forces in nature that we don't understand very well, are harder to understand. Those at higher energies, the interactions between particles can be rather complicated. Also, it may happen that you have a situation where there's a potential affecting particles, maybe a trap or something that holds particles. And there's a potential. And you really cannot measure directly this potential. There's no simple way of doing so. And then what you do is you send particles in. And you see how they get affected. And by studying the reflection or the scattering of the particles off of this potential, you'll learn a lot about what this potential can be. So this subject today is scattering. And in general, you have a beam of particles, a target, and a number of detectors. That's what happens in a big accelerator. And in general, you have such situations. OK? So a beam, some sort of target, and particles get scattered. And you have detectors. And in general, the detectors are all over in all directions because particles may back scatter, may go in different ways. So collisions, in general, are rather intricate. Particles can change identity, can do all kinds of different things. For example, you can have a proton colliding with a proton. This could be the case when the beam is a beam of protons. And here, maybe there's a target in which you study what happens when a proton encounters a proton. Or it may be that in some cases there's another beam that is coming also with protons. And they collide. But when protons collide with protons, funny things can happen. For example, you can get the two protons plus a pion, a neutral pion, 0 denoting zero charge. That's a hadron, a strongly interacting particle, like the proton is. It's made of quarks. And this can happen. There's no such thing as a conservation of the number of particles. That is one of the reasons you need quantum field theory to do this kind of computation. And this reaction is allowed because it, at least the first thing you can check is that it conserves charges. Proton and proton go to proton and proton, so a charge is conserved. And pi 0 is neutral. So that, for example, can happen. Or proton plus proton can go to proton plus neutron. Now, charge is not conserved. But this time, a positively charged pion called pi plus. The particles can completely change identity. You can have an electron and a positron colliding. So that's electron. This is the positron, the anti-particle of the electron, equal mass, oppositely charged. And then they can go into a mu plus plus a mu minus. These are charged laptons. Again, just like the electrical laptons, as opposed to these particles here. They're called hydrons. These particles are, as far as we know, are elementary. These particles are made out of quarks. And in this case, you have a complete change of identity. The original particles have disappeared. And new particles have been created. And you can look forward to study these processes as you continue your studies of quantum mechanics. But these are like reactions. Now, in our name, in our nomenclature for scattering, we will call the process a scattering process where the identity of the particles is unchanged. So we will have scattering, strictly speaking, when there is no change of identity in the initial and final states. No change of identity of the particles in the process of going from initial to final state. So a scattering process looks like a plus b goes to a plus b. We don't change identity. Those processes where you change identity are harder to discuss. And we won't discuss them here. Even here, there is still a lot of things that can happen. It's very intricate. So we will demand even more. We will call or look for elastic scattering. And that means that if these particles or objects have internal states, those internal states are not changed. So the internal states of the particles do not change. Of course, that can happen essentially when the particles are complex. They're bound states of other particles. So you may have a bound state of one form. And then when you have bound states, you know you have all kinds of energy levels internal states. And then when we say we have an elastic process, it means that those internal states are not changed. A classic example of an inelastic scattering experiment, the most famous historically of those experiments, is an experiment by Frank and Hertz in 1914. They were having a chamber with mercury gas. And they shot electrons from one side to the other. And the electrons and the mercury atoms collided. And they found that the electrons were slowed down by some quantized amount of energy. And that was the first evidence that atoms had energy levels. Bohr's theory of atoms had been proposed one year before that, 1913. And then, the process in which these electrons hit the atoms and lost energy corresponding to producing transitions in the atoms. And that would be an inelastic scattering because the atom changed its internal state. This is a collision between an electron and an atom, but the atom has changed internal states. So we will want to consider cases when we don't change the internal state. And that will be when we have elastic scattering. So a few more things that we're going to assume as we do elastic scattering. We will work without spin. All we will do in the next lectures, one or two more lectures in this, will be particles without spin. As we're doing in our course, we also work in the nonrelativistic approximation. So these are the first things that we will assume. One, no spin. This doesn't complicate matters as far as the scattering is concerned. It just complicates the algebra and the calculations you have to do because you have more degrees of freedom. We will be working nonrelativistically. Moreover, we will assume there are interactions. There are interactions between the particles that produce the scattering. And those interactions are simple enough that they just depend on the difference of position. So the interaction potentials are of the form v of R1 minus R2. So processes in which will have two incoming particles and they interact, and they scatter elastically. And the potential depends on just the difference of the two positions. Eventually, we will even assume it's only on the magnitude of these differences for particular cases. But if you have a potential like this, like when we were studying with hydrogen atom for the first time, a potential that just depends on the differences of positions can be treated in the center of mass frame. It's a nice frame where you can work on it. And then you can think of it as a single particle scattering of a potential. So you translate the Schrodinger equation into a center of mass degrees of freedom that are generally simple. And we don't worry about and a relative degree of freedom. So this can be done here. So this makes a process equivalent to scattering of a particle of a potential V of r. And that particle has the reduced mass, just like happened with a hydrogen atom. The real mass of the equation we solve for showing the Hamiltonian for a hydrogen atom, the mass that shows up there is the reduced mass, which is approximately equal to the electron mass. But in general, in this scattering process, it may be between two identical particles in which case a reduced mass would be half the mass of the particle. So again, we will work scattering with energy eigenstates. You may have studied already in 804 a little bit of problems of scattering off of rectangular barriers, tunneling, all these things. And we work with energy eigenstates in those cases. And we will work with energy eigenstates as well here. Now, in some versions of 804, you discuss a lot wave packets. I used to discuss a lot wave packets. We would build wave packets and send them in. And use a transmission on reflection coefficients to figure out what the wave packets do. And in general, a wave packet is a somewhat more physical way of thinking of the processes. Because the energy states, eigenstates, are unnormalizable and don't have a direct interpretation in terms of particles. Rigorously speaking, you should always do things with wave packets. But the fact is that we all work with energy eigenstates. And most of the times, what happens with wave packets can be more or less gleaned from what is happening with energy eigenstates. So we will not new wave packets here. We will not bother to construct wave packets. They would not teach us too much at this moment. So we will work with energy eigenstates.
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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 explore an example of a force in which the work done it is not path independent. And the classic example is the friction force. So let's consider the following setup. Suppose we have a horizontal surface with friction and we have an object. And we're moving this object. So let's choose an origin. We'll call this plus x, our i hat direction, it's all going to be one-dimensional motion. And we're going to move this object from an initial to a final state. And were going to move it directly in a straight line from the initial to the final state. And this will be our path 1. And in our second case, what we'd like to contrast with that, is that we'd like to move the object out to a point xa and then back to the final point. So this is our path 2. Has two legs. And we'd like to compare the work done on these two paths. So for path 1 we'll begin by calculating-- our force here is the kinetic friction force. And the kinetic friction force remember is, in this case, it's going to oppose the motion. So we have force kinetic for path 1, and that is minus mu k mg in the i-hat direction. And so when we do the integral for the work from x initial to x final, this is path 1, then we have minus mu k mg i-hat dotted into-- Now what is the ds for this path? It's simply dx i-hat, so dx i-hat. Notice we're not putting any sign into dx. The sign will show up in terms of our end points of our integral. So we do the dot product here, we have i-hat dot i-hat, that's 1. And so this interval, we can pull out all the constants, mu k mg. We're just integrating dx from x initial to x final. And so we get mu k mg times x final minus x initial. Now for path 2 we have two separate integrals. So for path 2 we'll just show the first part where we're going from x initial to xa. Then the friction force is opposing the motion. And we always just write dx in terms of the coordinate system, dx i-hat because you'll see that the signs show up in the end points of the integral. And then when we're coming back-- I'll put that in a different color and I'll put it below it. So when we come back, notice the friction force is going to change direction. ds will still be written that way but pay close attention to the end points of the integral. So now what we have is two integrals. So W is the integral from x initial to xa. And now I'm going to take the dot products here directly. It's the same friction force, we still have this same integral, which is minus mu k mg dx. Now here's where it's a little bit tricky. Notice on this path fk is plus mu k mg i-hat. And so when we dotted into dx we have a plus sign, we'll just continue that integration here, of mu k mg dx from xa to x final. Both of these integrals are straightforward integrals to do. This is minus mu k mg xa minus x initial. And over here, we have a plus mu k mg x final minus x initial. Notice x final minus xa, rather, is negative. And so both of these integrals are negative, as we expect. And so what we see here is that we have two pieces, so minus 2 mu k mg xa. And then we have that other piece, mu k mg x final minus x initial. Hang on, this is actually a plus sign. So our answer is very different because the displaced the amount that we've traveled is different. So what we see here is an example of a force which is the work done is not path independent but depends on the path taken from the initial to the final states.
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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, welcome everyone. It's 8:05, well, 9:05. So today's lecture and discussion is about how to make good problems, whether it's on homework problems, exams, even problems in class. Its general principles that you can use constructing questions that you ask, so problems broadly conceived. Now why problems? So problems are, I think, one of the central ways of thinking about teaching. And I'll show you how problems fit into the grand scheme of this course, of rethinking all the aspects of teaching. So one way that I like to think about teaching is called backward design. So this is a way of designing a whole course, could even be for a curriculum. And doing problems is one aspect of that. So why is it called backward design? Well, it's backward from the usual way. So I'll show you what backward is and then show you what the usual way is. So in backward design the highest-level thing you figure out-- not necessarily the first thing you figure out, but the highest level thing-- is your course goals. And that then feeds to OK, if those are your course goals how do you know whether students have reached them? And that's problems. Problems, tasks, projects, whatever it may be, things that students are going to do. And so generally speaking, things to do. And then given that you've operationalized the goals in these problems what are you going to do in class, so that students actually are able to do these kind of things? OK, so that is backward design because the usual way is the following. You say, OK, well, you have a bunch of lecture notes. Where do you have those? Well, either the last person who taught the course gives you their lecture notes. Or you just take the standard book in the subject and you just do the lectures one per chapter or one week per chapter, something like that. And you just go through in the order. So in the usual way you start with this. And then you think oh, whoops, oh yeah, I've got to make some problem sets. And oh, some exams. Well, what have we done this last couple weeks? OK, that's what we're going to do. So the usual way is you start with here. And then you get here. And you somehow never get here. The course goals are, basically, to get through the book. Right? That's the implicit course goal. But it's never explicitly planned. So that's a terrible way, but the normal way of designing a course. So this is backward design. So this is due to Wiggins and McTighe. OK, so here we're going to be discussing this today. So we're doing this first. And then the next session is this one. And I think the session after that is, what do you do in class, interactive teaching and things like that. So why am I doing it in this order? Well, in the levels of abstraction these are the most abstract. And this is the most concrete, OK, what question I'm going to ask students in class now. And this is things like goals. I want them to really understand conservation. They're really high level. So I think if you start here, by trying to figure out course goals, I haven't found that works very well for me because it's too high level. I can't think about course goals in absence of actual problems and things I want students to be able to do. So this has the right level between concrete and abstract. So it's actually where I start my thinking. Now, you may actually find you want to start somewhere else on the continuum. But this is what I found works quite well. You start here, you operationalize what your goals are in the problems that you think are interesting. And then you look at the problems. And that sort of gives you an idea of the course goals. And then you think about the course goals. And that helps you think of problems. But I start here. And then, once I have things like that, then class time is actually much easier to plan. Because class often is going to be problems like this, related to this, maybe shorter. But it's easier to plan here after that. Whereas we start here, you don't know, you're back in the old way. So I don't want to start here. This is too abstract, so I start here. So that's why we're doing this now. And then next week we're going to do this one. Question? AUDIENCE: So when you say that you're setting goals for your class-- PROFESSOR: Pardon? AUDIENCE: So if you say that, basically, setting the goals for your class, or for your own course. So when you design your course, basically you write out some kind of goals you want to reach? So how do you then do the problem? Do you go in detail before you've even started the whole class? Or do you do it that on a week-to-week basis? PROFESSOR: Oh, what do I tell the students about the course goals? AUDIENCE: Yeah, I mean, if you say, OK, basically, you have your goals and then you design problems so you can reach the goals? But then do you also design these problems before you even start the whole lecture? The whole semester? PROFESSOR: Oh, well, it's pretty rough. I mean, usually the first time you teach the thing it's too abstract to design-- I find-- everything in advance. Maybe that's my extrovert nature, which is that I find it hard to design things for people I don't know. If I don't know the audience I don't even know what to say. So I want to actually teach the course one time at least. So then the audience becomes concrete for me. So really, the first time through the course nothing is ever ideal. And you're doing everything on the fly. But the second time through the course, you can actually, basically, make all the problems ahead of time. So the second time through the course you actually refine your goals and your problems together. So that shows you the structure of where we are and where we're going to go and then later. So this is two weeks from now. Today, here. So how do you make problems? Well, the fundamental principle is, again, deduced by induction from what happens with regular problems. So if you remember, I gave you an example earlier of the results of standard problem solving. And I'll just remind you that we're putting up the percentages again. OK, so after doing lots and lots of standard multiplication problems, students were-- if you remember-- unable to do the following problem. So these 13-year-Olds and 17-year-Olds were asked to estimate 3.04 times 5.3. So here were the answers. So they were given four possible answers. And they could also not answer. So this is no answer. So that's the age 13. So of the four possible answers-- or let's say no answer is possible as well-- so five, they're 1% over 1/5 will get the correct answer. So that's a bit depressing. So you think, oh, OK, the 17-year-olds are going to be better. And they are, but they only have 37% correct. OK, so here is a serious problem, which is rote learning. And that's the fundamental thing to avoid when doing problems. So these students-- especially the 17-year-olds-- were actually perfectly capable of doing exactly multiplication. So on the same test they were asked questions like multiply 2.7 by 8.32 and given just empty space on the page to multiply it out. About 80% did it right. So it's not that they don't know how to multiply. They just don't know what multiplication means, which is a far more serious problem. I'd rather they knew what multiplication meant and didn't really understand, didn't really know how to multiply. Because the algorithm you can teach them later if they understand what it means. But if they don't even understand what it means that basically it's 3 times 5, so it's got to be somewhere around 16, they don't understand what multiplication means or the number system means. That's a far more serious problem. And that's produced by the traditional kind of problems. The traditional kinds of problems are, generally speaking, too low level and can be solved, to easily, by rote. Even the complicated traditional problems can eventually be turned into rote learning, and that's the danger. So the main goal in making good problems is-- I would say-- how to make problems that fight rote learning. OK? That produce long-lasting actual conceptual understanding along with whatever else you're trying to produce and teach. But really, fight the rote learning because that is the fundamental problem. So a way of thinking about this is to categorize the levels of thinking that you're expecting of students when you're asking them to do a problem. And for that there is work from 50 years ago now-- which is still useful-- which is Bloom's Taxonomy. So Bloom's Taxonomy-- let's see, when was it? I think it was 1956. I think it was '56. I'll put a handout on the website for you which summarizes in one page. But I'll summarize it even shorter here. So what is it a taxonomy of? Well, when you read the title the book, it's Bloom's Taxonomy of Educational Objectives. And it's not really clear if that's going to be useful for anything at all. Because you think, what the hell is that? So the shortest way of understanding it is that it's six levels. And each level is higher than the one previous to it in terms of the level of thought required of a student. So it's not really a strict order. You can have really hard questions that seem to be low level just because they address misconceptions. But generally speaking, it's a rough order. So the order is knowledge questions. Those are things like define, name, state, list, You know, it's basically almost rote. I mean, these are the kind of problems that everyone would agree are rote, state the definition of Newton's Second Law. The second is comprehension. So that's sort of like knowledge but it's a higher level. So that's being able to grasp the meaning of the material, not just recite the facts. But generally, that doesn't include the idea of being able to transfer it to new situations. That's where application comes. So can you apply the material to something? For example, often compute, demonstrate, show that, those are typical verbs that are used when you have application questions. The next level is analysis. So that's really trying to understand the structure of the material. So things like diagram, differentiate, infer, outline, point out, identify, distinguish, discriminate, all those kinds of verbs com into analysis kinds of questions. Synthesis-- so this is breaking things apart. This is the ability to put things back together in new ways. OK? So design, explain, categorize, create, revise, rewrite, reorganize, devise. And the highest level is evaluation. So judging the value of material, for what purposes is this good versus that good? Explain, contrast, summarize, support, recommend, evaluate, criticize, appraise, so generally quite high-level skills. So this is the taxonomy and this is a higher level. OK, so a useful exercise-- whenever you're thinking of problems-- is to try to place them on this. Now it's not that you want no-knowledge kind of questions and you want only to ask these kind of questions. Because if you just ask these kind of questions people will not be ready for them. And they're not prepared. It doesn't actually teach anything. You want a mix of them. But generally speaking, what happens to often, is that most of the questions that happen are usually in this zone, maybe a few of those. And then only later in people's careers do they get over here. For example, in engineering curricula you learn a ton and ton of application, maybe comprehension, Maxwell's equations, differentiation, integration. Then we start to synthesize stuff. You know, in your senior year you do design and project classes. But you don't have to wait all the way until then. You can actually start putting stuff in much earlier. And it doesn't have to be an entire class. You can have tasks that are like that in a short time, on a problem set in the middle of lecture time. OK? So questions about Bloom's Taxonomy? OK, so what I'll do, is I'll give you an example of questions that in freshman physics mechanics-- on the topic of Newton's laws-- that fit into each of those levels. OK, so just to give you an example, to make it concrete, to see what those kinds of questions are. And some of the categorizations I'm going to give you are debatable. But the rough idea is that they go from high to low. So this is on the subject of Newton's second and third law. So a knowledge question, state Newton's second and third law. Newton's second law, state Newton's third law, so you can see, just because people can state it doesn't mean they can actually do anything with it. That's why this is way up just at the knowledge end. Computers are actually really good at this. Right? You can look it up on Google. But your computer can tell you what Newton's second and third law are but you can't actually do anything with them. So comprehension-- so give an example of Newton's Second Law, using Newton's Second Law. OK, this next one is an application. It turns out to be, actually, quite a hard application. But it's still an application. So it's to apply Newton's second and third law to show the following, which the following is suppose I have a piece of chalk in my hand and I stand on a weighing scale. I want the students to show that the weight that the scale reads is equal to the sum of the two individual weights. So I weigh myself. And then I put the chalk on the scale. And I add up those two weights. And that should be the same as when I hold the chalk in my hand. OK? So I'll diagram that as-- OK, so that's me plus chalk equals me plus chalk. So this is a scale. OK? So there is the piece of chalk. So now this is an example of where that's quite a hard application question, even though it's kind of high up on the Bloom's list, so towards the lower end. Because this is a serious misconception that students have. Newton's second and Newton's third law, people are completely confused about, you'll find. They don't know when to use which. And they think any time two forces are equal it's sort of 50-50, whether it's because of Newton's third law or Newton's second law. So this-- if you're addressing misconceptions-- makes the problem, actually, all the much harder. So at analysis-- so the analysis comes out-- oh, question sorry. AUDIENCE: What are Newton's second and third laws? PROFESSOR: Ah, fair question. So Newton's Second Law says force equals mass times acceleration. Third law is-- the way it's usually stated, which it's terrible-- action equals reaction. Or for every action there's an equal and opposite reaction. Oh, which reminds me, one of the questions from before, which is what name would I give? Because I said I slagged off the reaction force as the name and the normal force as a name. People ask me, what should you call it? And I actually thought about that. And I think the answer is contact force. Because that tells you what the thing is. It's a force of contact. So Newton's third law says action equals reaction. So now, the problem is that students have, is that whenever they see two forces are equal they assume it's because the action equals reaction. Or they think Newton's third law is the reason for it. So for example, the bouncing-ball question that we talked about last time. So in the bouncing ball the question is, when it's stationary on the ground what are the forces on it? And they'll often say, well, there's a weight downwards. And there's the force upwards. And they'll often say, well, the force upwards equals the weight downwards. These are both forces on the ball because action equals reaction. And they won't realize that action and reaction have to be on different objects. OK? So they're fundamentally confused about the meaning of action and reaction, which you can bring out with the following analysis question-- When two forces are equal in magnitude how do you tell if it's Newton's two or Newton three? So how do you know if it's Newton two or Newton three? So just ask them that. Yes, question? AUDIENCE: This might be a little bit off subject, but should we use Newton's third law ever? It seems like it's a natural consequence from saying a equals 0. PROFESSOR: Ah, oh, OK, interesting. So OK, let me actually explain that one. I'll put it over there because I didn't do the boards right over here. So that is exactly the issue brought out by this question. And let me show you the difference. And then there was another question. That reminds me of another whole set of questions from the sheets, which was how do you develop your physics intuition? So let me answer that in a moment. After this and then we'll have a break. OK, so the bouncing-ball question-- actually let's not do the bouncing ball. Let's just do an object sitting on the ground. Oh, great, can you pass them around? Oh, Leeann's just coming around with a handout on Bloom's Taxonomy. And let's see, I'll do these after the break. OK, suppose you have this object sitting on the ground and it has Mass m. OK, let's look at what the forces on it are. Well, there's mg down-- and actually, in this case, the contact force, our normal force is equal to mg. OK, so now the question is, are these two equal because of Newton's Second Law or Newton's third law? That's what this question asks. So it's actually worth understanding the difference here. So these guys are equal because a equals 0. So this thing's just sitting on the ground, a equals 0. So the net force is 0, so these two forces have to cancel. OK, so that's Newton's Second Law. So then the question is, where does Newton's Third Law show up in this? Ah, Newton's Third Law shows up here. Here is the Earth. And here is your object. Let's just separate them just so you can see them separate, although this is sitting on there. So this is mg. Well, this is the mg because of gravity, right? The earth gravity is attracted to mass, m. Well, the mass is also attracted to the earth. So what's the force on the earth from the mass? AUDIENCE: mg. PROFESSOR: mg, it has to be. Now this mg equals that mg. Call this f2 and this is f1. So f1 equals f2 by Newton's third law. So that's a fundamentally different set of forces. So are equal and opposite. They're an action/reaction pair. So what Newton's third law really says-- when you understand it-- is that all forces in the world come in pairs. So if you ever see one force you have to hunt around for his other half. And the other half here is this. So actually, there's another question you could use for Newton's third law. So I don't have it on the list here. But I'll ask it right now, which is suppose someone comes to you and proposes the following gravitational law-- f gravity equals g m1 squared m2 over r squared when the g is the proper units. So m1 squared times m2, is that a legal force law for gravity? No, because when you switch m1 and m2 as you're doing here-- you're switching the mass and the earth-- you actually change the force. So then this would violate Newton's third law. OK? So Newton's third law, it actually is a constraint on the laws of physics. It's actually conservation of momentum. And for momentum to be conserved you can't have any arbitrary law of physics that you want. Does that help clarify that? AUDIENCE: Come up with a better set of taxonomy for describing it that doesn't lead people-- it leads people into thinking about that. PROFESSOR: I know. I agree. So I think the terms are very important. So that's why don't like this form, action equals reaction. I actually would prefer it were stated all forces come in pairs that are equal and opposite. Right, and language does have a powerful effect. Yes? AUDIENCE: The technical [? route's ?] obvious, just in calling the novel force the contact force, but that's not refined enough. Because both friction and novel are contact force? PROFESSOR: Yeah. AUDIENCE: [INAUDIBLE] constraint force because it could constrain into the [INAUDIBLE]. PROFESSOR: Yeah, that's probably a better name. Yeah, I agree with you. So the comment was that contact force isn't quite right because there's friction force as well. And that's also a contact force. And maybe a better name is constraint force. I think that is a better name. OK, so let me finish this list of two more examples of Newton's Second and third law. And then we'll have a break. OK, synthesis. So creating a problem using Newton's second and third law. So for example, this is a problem like that. But here I'm asking the students to make one up. And here is a very hard evaluate question, which is what, if any, are the limits of validity of Newton's Second and third law? So that's, actually, a very interesting historical question. Interesting historically because Einstein thought about that question. He wanted to know whether Newtonian gravitation and special relativity were compatible. And basically, they can't be because of Newton's third law. Because gravity should propagate with some speed, not the speed of light. So you can't actually conserve momentum right away. Suppose the sun does something strange. Well, it takes a while for the earth to actually respond to that. And it's in that intervening time Newton's third law seems to be violated. So Einstein thought very carefully about that question and came up with his theory of gravitation. So this is quite a high-level question. But it's to show the whole range of things you can ask about the second and Newton's third law as you climb Bloom's Taxonomy. OK, so what we're going to do in the second part is we're going to actually practice rewriting a couple of questions. And I'll show you a couple more examples. And seeing how we can make questions worse and better, because I find that's actually quite a good way to get practice with designing questions is taking good questions, making them bad, taking bad questions and making them good. Both ways are useful. OK, so we'll do that. Let's say it's 9:04, 9:10 let's say. So it's 904 on that clock. So see everyone, sorry it's 10:04 on that adjusted clock. So see everyone at 10:10 or 10:11. And meanwhile, here are the feedback sheets. So I'll just put a bunch in the two back things. So when you come back just grab those. OK, so I'll just answer a couple questions which I promised to answer. So the comment made by several people about last week's examples, say, of the rock and the rolling was that oh no, that showed me that my physics intuition is quite off, and bummer. And so then if you're going to teach it, how do you make sure, well, A, how do you learn to think intuitively about these things? And B, if you're going to teach it how do you make sure-- related to that-- you actually have the intuition before you teach it. There's two answers to that. One is that it's very rare that anyone ever achieved a goal that they didn't set out to do. And then just happened upon it. So I would say that's true, also, with developing intuition in these areas. You have to make it your goal. So it's not going to just happen by chance. So then questions, OK, if it is your goal how do you go about it? One way is anytime you do a calculation, at the end of the calculation ask yourself this question. So John Wheeler-- who was Feynman's adviser-- he recommended the following question, which I think is a fantastic question. So the question is if you could time travel back and talk to your earlier self what one or two sentences would you tell your earlier self before they started solving the problem that would make the problem just sort of flow smoothly? OK? So what sentence would you tell your earlier, pre-problem solved self? So this is after you solved the problem. So that forces you to give a insightful summary of what you learned from the problem. So it forces you towards the intuitive way of reasoning. And you'll find it's quite hard at first. Because you don't have the intuitive library. But at least it points you in the direction to be going. OK, now what else? Well, in chemistry there's various books that help develop intuition. They have all kinds of intuitive reasoning questions in them. And some are quite good. Some of the books are quite good. So the one in chemistry that I like is called Voyages in Conceptual Chemistry. And in physics one of the ones I like is called Thinking Physics by Epstein. I think Epstein's like that. So what that is, is that's basically, one question per page. And it's multiple-choice intuitive reasoning question. And they're about all kinds of interesting physics, like why do tea kettles whistle? If you shake bottle of soda up, why does it explode? And so it's filled with really, really fascinating puzzles and ways of thinking. And then with explanations. So just by thinking about them, reading the explanation, trying to understand, OK, then asking yourself Wheeler's question at the end of reading the explanation, even if you didn't solve it, you can develop a lot of intuition. I don't know a good one for math or biology. And if anyone does, I definitely appreciate any suggestions. OK? But there are resources in some fields. And the other is just whenever you find paradoxes go after them and share them with everyone. And ask everyone you know,do you know any paradoxes I can work on? That'll also developed intuition. OK, and then related to that, one comment was I've noticed that teaching is the best way of learning. How can you incorporate that into a course? So yeah, that's true. So sometimes you just have to accept that the first time you teach the thing you're going to be learning a lot. So that's fine. And the second time you teach it, it'll be much better. So you'll actually develop a lot of intuition the first time you teach a course. So for example, I found when I was a graduate student, I was a teaching assistant for Physics 1A at Caltech, which was like 801. And I found I had a huge number of misconceptions about tension. I just had no idea what tension was. For example, I thought tension was a force. And now that seems ridiculous to me. And it doesn't seem ridiculous to you that's because you're in exactly the same state I was in before I was taught physics 1A. And I've got myself all in a twist when I was trying to explain it. And I finally tracked it down to actually, that particular misconception. So then I developed a bunch of questions for myself, which I also asked students about tension. So teaching is learning. And then how do you incorporate that into a class for the students? Well, you can ask them to teach each other and argue with each other. So we'll talk about that in the session on interactive teaching. But yeah, it is the best way of learning. And so if you can get the students to do that as well you'll be increasing their learning. And then the last question for the moment was what's my response to the following article-- so it was in the globe on February 15, 2009. And it said, don't open with a joke. So it was summarizing a paper called "Increased Interestingness of Extraneous Details in a Multimedia Science Presentation Leads to Decreased Learning." Journal of Experimental Psychology, December, 2008. So they summarized that. So basically, what they found is that if you start with a joke that's extraneous to the material it actually decreases the learning. And that, I think, is quite plausible. So I recommend that you start with something interesting in the beginning. But it should be related to the material. So the problem wasn't the joke. The problem was that the joke wasn't related to the material. So the thing should be interesting. Because you want to draw people in. You don't want to conclude from that article that you shouldn't start with some interesting. But you should start with something interesting that's related to the material. OK, so redesigning questions. So here is an actual question that we will redesign. And I've left up the Bloom's Taxonomy there. So actually, I have a bunch of questions. We'll do a couple of them. And we'll review, probably, one of the questions we did earlier, which is about the cones. OK, so the question to start with-- and we're going to go that way and that way-- is find the eigenvalues of this matrix. So for those whose linear algebra is rusty, the eigenvalues of a matrix so a matrix, m, times some eigenvector e is equal to lambda times e. So if there's some vector that the matrix just brings back to itself with some constant, that's the eigenvalue. And this is the eigenvector. So eigen meaning own in German. So somehow this vector belongs to the matrix. It's said to be an eigenvector of the matrix. It's preserved when you hit it with the matrix. You get back the thing itself. And in that way it belongs to the matrix. OK, so there are recipes for computing it. And so here the question is, I'm asking students to use the recipe for computing the eigenvalue. OK, so now I'm going to ask you the following question, which is to design-- well, first of all, where would you say that guy goes? It's sort of comprehension. You know, it's basically computing. Maybe it's a bit of application. But it's basically comprehension. OK, so can we go lower on the taxonomy towards knowledge? Can we go higher towards the higher-level reasoning? Let's try to construct questions on this end. OK, so everyone understand the thing to work out? I'm going to ask you to try to think of stuff that could be put here and here. So this is lower level and this is higher. Does anyone need me to explain-- or want me to explain-- more about eigenvalues or eigenvectors? OK, so find a neighbor or two. And try to think of stuff that can go in this box or this box. And generally, the higher-level questions requires more knowledge about the material, which is one aspect of their being higher level. You need a bigger margin of safety when you're trying to design them. But don't let that bother you. See what you can construct. And you can use the levels on Bloom's Taxonomy to try to create some kind of questions around it. By the way, the answer to this question, by the way, is i and minus i, just in case you are curious. OK, so let's see what examples you've come up with, so related to this question. Obviously, you'll change it around some. Let's say either side, this side or this side. Someone I haven't heard from? Yes, can you tell me your name? AUDIENCE: Mike. PROFESSOR: Mike, yeah? AUDIENCE: What is an eigenvalue? PROFESSOR: OK, so what is an eigenvalue? Where would you put that? AUDIENCE: Lower. [INAUDIBLE]? PROFESSOR: Oh, yeah, good question. It depends where you put this one. This one is computing, which I roughly put that somewhere between comprehension and application. So right now I would say, we're somewhere around there. So your question is to define eigenvalue? OK. I'm just short-handing it as defined lambda. OK, so that's one question. OK, another question. Can you tell me your name? AUDIENCE: Wendy. PROFESSOR: Wendy, yes? AUDIENCE: So related to that, is that you could put that question higher. If you asked them, what does an eigenvalue mean? If you word it differently instead of [INAUDIBLE]. PROFESSOR: OK, which may have been what you meant as well and I may have defined lambda. OK, so you're saying, well, you can actually ask them to say, what does eigenvalue mean? OK, so how would you if they know what it means? So what would you look for in something like that? AUDIENCE: If they could describe it to you, or maybe tell you a problem in which it'd be useful, in using it? PROFESSOR: OK, so yeah, let me write that too. Create a problem that would benefit from lambda, that would need lambda. OK, so for example here, create. That's gone pretty far down towards synthesis. I really should have flipped the list around. But synthesis is quite high on Bloom's taxonomy. So you're actually going to create a problem. It's like the Newton's law example. Yes? AUDIENCE: [INAUDIBLE] instead of saying, again, what does lambda mean? You could say what do the results from finding lambda? What are the results of that solution? PROFESSOR: OK, so how does it help you to know the eigenvalues? Is that the question? AUDIENCE: No, I'm just saying instead of the general asking for the definition, what does lambda mean? What is an eigenvalue? What does it mean? But after having solved the problem saying, what does your answer mean? PROFESSOR: Ah, OK, so even if you ask them this first you can say, OK, interpret your answer. Yeah, and that's a-- generally-- very good policy. So interpret. and where would I put that? Eh, interpret is somewhere like analysis, maybe a bit of synthesis. But yeah, it's quite high on the list. So interpret, that's another really key word. In fact, it's often neglected, right? We just ask students to solve the problem. And wander, wander, wander. They get the thing that looks like an answer. They put a box around it and they're done. But actually, you can massively increase the level of a problem just by adding to the problem, interpret your answer. What does it mean, that that's an eigenvalue? And maybe they would say something like this. But even that would be better than just having done the computational procedure and got an answer out of it. Yes, can you tell me your name? AUDIENCE: Ben. PROFESSOR: Ben, yes? AUDIENCE: I might ask, does that matrix have any eigenvalues? PROFESSOR: OK, so yeah, so let's put it over here. Right, so there what you're doing is you're not presuming. You're not key-wording what they should do. When you do this-- especially if it's a linear algebra course-- they've probably seen a bunch of things like this. Like in a calculus course, integrate these things by parts. They just know I have to integrate by parts, whereas here you're at least one step higher. Because they don't know whether there's an answer to the problem or not. And they have to think, well, how do I know if they're eigenvalues? So there's actually one intervening level of sophistication. Other suggestions? Yes, Adrian? AUDIENCE: Just give them two numbers and say write a matrix that has these as the eigenvalues? PROFESSOR: OK, so let me continue the list over here. OK, so create a matrix which has the following eigenvalues. So say lambda equals 2 comma 3. So create a matrix, create a meaning. We're already high up in the taxonomy toward synthesis. And actually, I would even use a related one, which is-- so create a real matrix with eigenvalues i comma minus i. So that one, is quite related to this. And this is the answer. So it's quite a hard question. Even if people can do this and find i comma minus i, this one is very hard because you're asking to apply a constraint that the thing be real. So it's not just any matrix. But it's that the thing be real. And then you have to really understand quite a lot about matrices to do that. Yes, can you tell me your name? AUDIENCE: Scott. PROFESSOR: Scott. AUDIENCE: So, I was thinking of a synthesis question which would be is after multiplying the item the eigenvector by the matrix how is it rotated? PROFESSOR: OK, so what is the geometric effect on the eigenvector of the matrix on the eigenvector? AUDIENCE: Right, but the question is if they understand what an eigenvalue is. PROFESSOR: They should be able to answer that right away. Right, but it really tests for understanding of what the meaning is. So it's another way of approaching the question of do they understand eigenvalue means? What is the geometric effect of the matrix on the eigenvector? And then hopefully, they'll be able to say well, actually, it doesn't rotate it. It just scales it. And the eigenvalue is the amount of scaling. OK, and that demonstrates quite a high level of understanding too. And This is one reason that oral exams, generally, are quite good ways of evaluating. Because you can ask these kind of questions. And as soon as you see that they understand you move on to something else. Which of the flip side is that's why oral exams are very disconcerting for the students. Because you spend no time on the things they understand. Right? So as soon as you know they understand it, you move on to something else. So the student just feels like, what the hell happened to me? I was doing fine. And they asked me about something else. Yeah, that's exactly right. That's why they asked you about something else. They knew there was nothing more here to find out. You understood it. But that's why oral exams are such a good way-- so efficient-- for evaluating. So in Cambridge we used to do oral exam interviews for admission for undergraduate. So people would be admitted to the major right away. So we had two 20-minute interviews. And we found those were much more reliable than all the exams that they did in high school, the equivalent of AP exams. Because you could really ask questions that really probed understanding and right away see what people could do and couldn't do. So geometric effective of m on e creating a matrix, so you're asking, so what's the geometric effect? It's not just comprehension. You're asking them to apply it and maybe, in a bit in a new area. They hadn't really thought of eigenvalues and eigenvectors as geometric at all. And now, all of sudden, you're asking them that. You're asking them to transfer their knowledge, which is one of the fundamental goals. Yes? AUDIENCE: Once you give them two matrices that has the same eigenvalues you can ask them to see if they are related in some ways. PROFESSOR: Right, so for example, let's do this one. So how are those two matrices related? So they both have eigenvalues i and minus i. This one obviously does because those are the diagonal values and there's nothing else. And this one, you have to do it by the magic procedure. You don't have to. And so how are they related? So that is some kind of analysis question. And it's quite a difficult analysis question. It tests for really, do they understand what matrices mean? So that's why I've underlined this word. The questions all in this column are all somehow testing that they understand what things mean. So that ties back to the point I made earlier, which is that the whole fundamental point of asking good questions is to avoid rote learning. Is that by asking good questions you're fighting rote learning. And rote learning is the bane of most education. Right? Rote learning is too often the result of education. And it's the-- in my view-- fundamental problem with education. So by asking questions that are different you're actually starting to fight that and break the habit. So for example, how does rote learning show up? Well, in the bouncing-ball question, so the force is-- in the instant that it's stationary, while it's bouncing-- people say mg. While this mg is right. The gravity, but the contact-- let's say the constraint force-- in this case it is a very full-contact force is mg. One of the reasons they say that is the misconception that they think a equals 0 because v equals 0. So that's one reason. But the other reason is that they've done so many problems with books resting on tables, balls resting on tables, where the force is mg. Right? So they've now induced a new rule of physics, let's call it Newton's fourth law. That anytime you see a contact force it is mg. So contact force equals mg maybe times cosine of theta. [LAUGHTER] PROFESSOR: Right? or sine theta. I guess it's mg-- well, it depends how you measure theta. But theta is the tilt of the plane. So if the plane is tilted 0 degrees just to ground, well it's then mg. So that kind of learning is the result of rote learning because they've solved problems without understanding what they mean. So instead of figuring out that oh, actually, you could understand it from Newton's Second Law and understanding acceleration, they think oh, there's a new pattern here. Right? Oh, a new pattern is whenever I see contact forces it's mg times maybe cos theta or sine theta. Yes? AUDIENCE: What's the exact question with that system and what's the normal [INAUDIBLE]? PROFESSOR: Yeah, so the question was so you drop the ball, say from here. And when it bounces off the steel table there's one instant when it's stationary on steel table. And what are the forces on it at that instant? AUDIENCE: I remember from last week. PROFESSOR: Yeah, no, fair enough. I should have stated the question out again. So this is stationary for an instant. All right, so actually the ball is quite compressed here. Yes? AUDIENCE: Isn't it impossible to give an exact answer for the force with the [INAUDIBLE] if you don't have the duration? PROFESSOR: It is, right. So it is impossible to give the exact answer. But what you ask them is, well, how big is it roughly? So I always protect myself by saying roughly. And then you can ask anything you want. And because, really, all I'm interested in is do they think it's 1 times mg or do they realize it's 10 to the 4 times mg? And those are so far apart that I don't care about factors of two or three or even 10. Now, how can you know that it's 10 to the 4 times mg? Well, that is quite a hard question. That's way down there. Even though you'll see some lists of Bloom's Taxonomy say estimate is a low-level skill, not that high in Bloom's Taxonomy. It's almost like computing somewhere in comprehension. I don't agree with that. Because as you saw from the example where students couldn't estimate 3.04 times 5.3, they're terrible at estimation. So estimation actually hits one of the big misconceptions and reveals a lot of the misconceptions. So I put estimation almost at the level of synthesis because it forces you to really synthesize your understanding. You can't hide behind formulas. So the next thing I'll do is say, OK, let's estimate this and see if you can do that. And if they really understand the system they can estimate it. But it's quite hard. And I find almost no students can do it on their own. But the method of estimation is, as you say, you do need to know the contact time. So the acceleration is not 0 as you say. But it's delta v over delta t, where delta t is the contact time. Well, it's going v and then it's pretty much going v up, so that's about 2v. But I don't care about factors of 2. And then, what's the contact time? Well, there's a couple methods for doing the contact time. One is you listen for the pitch. And whatever the highest frequency in the pitch is-- suppose it's 10 to the 4 hertz-- well, then 1 over that time, 1 over the frequency is roughly a time. So it's probably the contact time. So that's one approach. The other approach is to say well, actually, physically what's happening? The ball is hitting the bottom. And this end compressing. But this end keeps going. Because it has no knowledge that the bottom of the ball hit the ground. How's it going to know about that? Well, the bottom of the ball has to tell it. How's it going to tell it? Well, it has to send some kind of signal. How does it send signal? Sound. So it sends a sound wave upward. And the sound wave travels in steel about 5 or 10 kilometers per second. So you suppose it's a 5-centimeter ball, then it's about 10 to the minus 5 seconds is your contact time. So this is really short. So suppose v is something like a meter per second, and divide it by 10 to the 5 seconds. You're talking about something like 10 to the 5 meters per second squared, which is 10 to the 4 times the acceleration due to gravity. OK? So I went through because I think it's interesting but also to show that estimating actually requires generally, very high-level skills. And it's quite hard for the students. Therefore you should do it. So there are some people who say, well, it's very hard for the students. Don't do it. But again, there's another rule of thumb in teaching, which is nobody ever learned anything that you didn't actually try to help them with. Well, that's not quite true. But if you never teach something they're not very likely to learn it. So if it's something hard but valuable you have to start teaching at some point, which is why I put up that graph earlier. Oh, it's gone now. Where in the earlier stages of teaching you do things more approximately. And then in the later stages you do it more exactly. So what I've told you now is a slight lie. So this brings up the question from last time, what about lying? Is it useful? Yes, it's essential because this argument is simple enough that once you understand it you can just keep it in your head. And it's a bit of a lie, the reason being that the ball doesn't have a uniform cross section. So as the ball compresses the contact area changes, whereas if the ball were cube or a rod the contact area would stay the same as it compressed. So there's a correction factor because the contact area changes. Now, that correction factor is not very big. It turns out to be the fifth root of some dimensionless thing. And so this sound-wave argument you put in to the 4/5 power and then you put in the other thing to the 1/5 power. So it's almost correct for this spherical ball. So in a later of course it actually adjusts for that factor. And then in a much later course, probably, what I would do is calculate the exact 2 pi over whatever factor that shows up as well, by way of increasing the symbolic complexity. But even this estimate, doing that is conceptually hard. So I would say start that early. AUDIENCE: I think the point I was trying to make was that this-- I suspect that the students answered that wrong because of real problem solving, and that they expect there to be a solution given the data that you [INAUDIBLE]. PROFESSOR: Well, if you've worked with them a bit on estimating they know that they're going to have to do some thinking. If you say compute the force I agree with you for sure. They're going to think they have to be able to do it from the data there. And they'll just say it's got to be mg because that's the only other force I know. But if you say estimate then there is a bit of a queue that they have to include some other information which might not put in the problem. And so that mitigates the rote learning. But what I find, even then, is they usually still can't do it because their misconceptions about acceleration are so deep and about equilibrium and force. They think f equals mv, basically. OK, so let's just look at what we've done here. So I would say you can define lambda. You can define eigenvector. You can even say state the computational procedure. I'm not sure where I would put that. Depending on how people learn it, this could be purely a knowledge question. Because people could state the procedure but not really understand it, not be able to do anything. Or actually, you could say program the procedure. Well, that would be something over here probably. Because to program it you probably have to understand what you're doing pretty well. So stating the procedure, it depends how you ask them to state it, whether it's a low-level or a high-level question. So I'll put program over here. OK, so we've gone down and up. And again, I want to remind you, I'm not saying all your questions need to be here. But generally, the questions are too much over here and not enough over here. So this gives you a recipe for thinking of new questions. Yes? AUDIENCE: So throughout the course do you basically, maybe start on the first three in the beginning and then you move it more through until you reach the end of the course? PROFESSOR: More fractal. The question was as the course moves on do you move more towards synthesis? I actually try to make it fractal. So within each unit and each day even, I try to maybe have something, knowledge or comprehension just to make sure everyone's tracking. But try to have something interesting as well, all the time. So something synthesis, analysis, maybe some evaluate. And so each problem set-- for example-- I'll put, generally, warm-up questions. So basically they're marked warm up so that people know OK, those are going to be knowledge and comprehension. And you need that. Because otherwise there's no point trying to do synthesis if you don't have the base knowledge. But just make sure you understand those, just in your sleep. And then regular problems, which tend to be somewhere in the middle, say application or analysis. And then bonus problems which are just optional. But they're for people who either find all the other problems easy or are really curious. For them I put either synthesis or evaluation. For example, in the approximation class the last problem set had design an email indexing system. So that was the bonus problem. And there was a bunch of warm-up and regular problems which were involving Unix text processing. And then for the people who are really curious actually, put it all together and make a really fast email-indexing system so you can just find any email that you want just by doing keyword search. Sort of like a web search but just on your email inbox. Because I actually use that myself. Because I find folders are just hopeless. It just takes all my time. I have to figure out what folder things are in. And that could be many. So I just keep it all in one giant inbox with, I think, 150,000 messages in it. And I just search through that. And I have an index that's rebuilt every week or so. And so I can search through it really fast. So it's actually very practical problem and it gives you an example of the levels of things you can ask. So try to do it on a fractal scale. Each class should have something, mix the levels each homework, the exam. So the general rule is to mix them up all the way. OK? So hopefully, from that, you feel confident that you have tools that you can make interesting questions. And this does have lots of payoffs, which one of them is that questions that are enjoyable and educational, students love them. So when I practiced what I preached and made these kind of problems, mix of problems for my classes, often students have said, oh, can you please have more problems? Which is very rare. They said, oh, these are some of the most interesting problems I'm likely to get at MIT. And I used to read them out in my living of my independent-living group just so that other people could hear the interesting problems and we'd try to solve them together. And then I've had several emails after class finished, saying, oh, you promised to post the optional problem set. Where is it? So it's because the problems, actually, somehow connect to things that people find real and interesting. So that's a mix of the perceptual. Don't make it just purely symbol manipulation. Try to connect to ways people look at the world. But also, try to connect to higher-level reasoning skills. Question? AUDIENCE: Do you always use bonus problems? PROFESSOR: I do. I find that works really well. Because it's like a pressure-escape valve. It caters for the diversity of backgrounds in the class. The people who are feeling very worried by the material, they can just skip them, no problem. And they don't feel like they have to do them. And they don't feel like, oh my god, I can't do them. I'm going to get points off. And the people who find the other stuff easy, they find something for them too. So yeah, I try to always use bonus problems. And I learned that from Knuth's book, called Concrete Mathematics. So actually, I think they have five levels of problems at the end of every chapter. There's warm ups. There's homework problems. There's exam problems, bonus problems, and research problems. So the research problems were the ones that are not yet solved. So at least you know they're not yet solved. So you're going to be spending a while on them. And then in the second edition I think some of them were actually solved. OK, so if everyone could just take a one minute and fill out the feedback sheet? Oh yeah, and let me just say, why I'm doing this now at 10:50, of the questions on the sheets before was the feedback sheet, the person said, I've been using them in my class and they've been really useful. But I find that the response rate has been dropping. What do you do about that? And I have to say, that I found the same thing. This year I found the same thing. Not in this class but in my other class, whereas I didn't find that last year. And I was trying to figure out why. And theory I have is that I was trying to put in too much material in the class and not leaving the students time before the five-minute mark to fill out the sheet. So one thing, I've been trying to mend my ways and make sure I end class at 9:50 or 10 minutes to the hour so the students have at least one or two minutes to fill out the sheet without feeling pressured to get to their new class. So I'll let you know how that goes over the next week. So take a minute now and then that'll maybe make you think of some questions which we can answer in the next two minutes. And then while you're doing that, just a couple announcements. So sorry I didn't say anything before about what to do with the equation treatments that you shared with each other. So just discuss them with each other. You have a collaborator. And then revise your treatment based on what that person told you. You don't have to turn in your revised treatment. You guys are all graduate students, or almost all of you. You're all taking the class because you're interested. I'm not trying to police everyone. But the goal here, and what you should do, is take the benefit of the collaboration, revise your equation treatment, and share it with each other. And then just bring in your original one next week. And then I'll also have a couple of readings for you to do for next week. And also a short problem to work on, things to think about to bring in next week as well. OK, so sheets and then questions. Question? AUDIENCE: So I tried doing these sheets for a class that I was teaching yesterday. In fact, I found that I got a response rate of 15%. If your response rate from the beginning's very low and [INAUDIBLE] to increase that rate? PROFESSOR: Right, so 15%, yeah, I find I'm getting that around now too. And I think what's happened is that I've got people in the habit of not filling them out because I wasn't giving them enough time. So I may be hosed because now they're in the habit. So I'm trying to mend it. And I'll let you know if that fixes it by giving people actual time before they feel rushed. But it may be too late. And I worry that it is too late. Because last year I was actually more diligent about making sure they had time. And I found the response rate stayed about 60%, 70%. But even I find the 15% very useful. And I found that-- people even in the past, when the response rate was 30% in another class people said, I didn't have a question all the time. But I knew when I did I could ask it. So even the 15% isn't bad. It's still useful. But yeah, I would like it to be higher. And I'll tell you how that goes. Question? AUDIENCE: [INAUDIBLE]? Wiggins and-- PROFESSOR: McTighe, yeah. AUDIENCE: Oh, McTighe. Where are they based? PROFESSOR: So where are they? They're in New Jersey, and they actually consult on educational course design and run workshops and courses. And we actually invited Wiggins last year to come and speak. So I'll put a short reading selection from their book called Understanding By Design for optional reading for people so then you can learn more about that. I find the basic form that they say very useful. I find applying it the way they describe, sometimes I just can't square it. But it may resonate with you quite well. And the overall approach resonates very strongly with me. So hopefully you will like it too. So I'll put that on the website and some readings for next time about course design. Yes? AUDIENCE: Just a comment, actually on your web page, where the response sheet [INAUDIBLE] and situation where only a few people were very dislike the current [INAUDIBLE]. They don't like it this way and so they want to change it. But the 80% who don't feel [INAUDIBLE] is actually OK with it. [INAUDIBLE] the change. It makes most people unhappy. PROFESSOR: Unhappy, right, so that can happen. And so that's the silent majority problem. So what I do when I get comments like people saying, well, I wasn't happy about this. Sometimes what I do is-- especially if it doesn't resonate with my intuition about how things were going, and I thought they were going well-- I'll say, well, there were a few comments about this issue. What do people think? And just take an informal straw pole before I make big changes. AUDIENCE: I think it helps to have a positive question on [INAUDIBLE], like, what did you take away from this lecture? Or what did you really understand? PROFESSOR: Right, I think that's right. It helps to have a positive question. So that's why I say, either what you take away, or what helped or hurt? So there's an opportunity to say something helpful. And any other comments is also interpreted in a positive way. Answers from Lecture 5 to questions generated in Lecture 4. PROFESSOR: OK, and questions from last time. So they grouped into several ones. One is how should you choose your levels of your problems-- say, in class or in homeworks-- among the different levels of Bloom's Taxonomy? Should it be, for example all towards the evaluation? So your side would be here, sort of the high level. A Mix? A flat mix? Should it be peaked towards the middle with some tails? So that depends, partly, on your course design. So to answer that question, you can't really answer that question until you figure out what your course goals are. So in some ways there's an argument for doing course design first and then working on what problems to do and how to design problems. But actually, I like the other way around. I like doing course design second. Because actually, doing problems is concrete enough to do something. And then it leads to questions about course design, which we'll answer how to mix the levels. OK, another one was several people wanted to know, why is tension not a force? Or in fact, maybe is a force. So let me explain that. Because that actually gives you another example of sorting out misconceptions. So it's always useful. So I'll show you the sequence of questions I used with myself when I finally convinced that tension wasn't a force. And here's how I did it. And then I used the same sequence of questions with students to fight the misconception that's been induced by many pictures of-- for example-- a pulley. And here's a rope and there's tension. Sorry, this is actually done the other way. So here's the rope. And there's a t labeled there. And then on the mass there's a force labeled as t upward. So when you see that you think, oh, tension's a force. So you see that a whole bunch of times and you're pretty sure of it. So the sequence of questions is as follows-- say OK, here's a tree. And I'm going to tie a rope to the tree, rope or a string, it's a mass-less thing. And I put a force on it. OK, so I'm going to pull that tree. The tree's not going to go anywhere. We're going to pull on the rope with 100 newtons of force. OK, so you can try this yourself too. What's the tension in the string? OK, so take 30 seconds and I'll take a tension. OK, a tension, anyone? AUDIENCE: 100 newtons. PROFESSOR: OK, so in fact that is 100 newtons. And most people will say 100 newtons. OK, so now the next question is this-- here is a river and here's a rope. And me and a friend each pull on the rope with 100 newtons. OK? OK, so discuss with your neighbor-- a, b, or c, what's the tension in the rope? And we'll take a vote. OK, so take 10 seconds, get your vote ready. We'll take a quick straw pole. Who votes for 0 newtons? Who votes for 100 newtons? Who votes for 200 newtons? OK, so now here, how do you know what it's going to be? Well, you make a force diagram. And there's a way to convince yourself that there's no doubt about what it should be. And we just draw the forces on the rope here. Oh, we've already done that. Let's draw the forces on the rope up there. Well, what are the forces on the rope here? Well, it's 100 newtons from you. But there must also be 100 newtons from the tree. All right? So in fact, these two situations are identical. So yeah, 100 newtons seems pretty clear here. So it has to be 100 newtons up there too. OK, now that-- so far-- isn't the hard part. But now is when you actually create the contradiction with the idea of it being a force. It can't be a force because if it were a force you would just add up vectors. Right? So there's a couple answers you get if it was a force. If you're a bit sloppy you'd just add up the magnitudes and get 200 newtons. If it's really a vector-- you're putting two vectors on it, two forces-- you should get 0 because they're two opposite forces. But actually you get 100. The only way to get 100 is that the tension has to be something completely different from a force. So that's how you know it's not a force. But how do what it is? Well, it turns out to be a tensor. So it's not a vector. Forces are vectors. It adds differently than vectors do. It adds like a tensor. So tension is more like a pressure. It doesn't have a particular direction. Pressure is in all directions. So things like that, they can't be described by vectors. So you need something else. And that's why we introduced tensors. So now, in freshman physics I wouldn't give them a whole story about tensors. But I do actually teach this much of it so that they know that is a different object. And then what do I do about this? Well, I won't write t unless it's for myself. But for the students I think it's sloppy and it increases misconception. It's much better to write the force sub t, which is the force due to tension. OK? So even a slight change like that can mitigate the rote learning, where people think oh, tension's a force. Because you always see it drawn like every other force. There's mg down, There's t up, must be a force too. AUDIENCE: [INAUDIBLE]? PROFESSOR: What are the two objects you hit it with to make a scalar? It's the two directions that you're interested in, basically. So it would be the x direction or the direction along the string and the direction along the string. AUDIENCE: So it's the two direction then? PROFESSOR: You can use the same direction twice. So it's the direction along the string and along the string. There's no cross direction. So it has no shear. So there's no cross direction. So it's some component, tension is one component of the stress tensor. So now, again, freshman don't need to know all the details. This is back to the subject of lying they do need to not be told fundamental lies. Like for example, tension is a force. OK, so now, how do you deal with misconceptions that show up on problem sets? This is a good question. So the point was made well, in class you can sort it out right then. Suppose you ask a question, say, about the Ideal Gas Law. And you find that people have confused the Ideal Gas Law and the Adiabatic Gas Law. Well, you can talk about it right then, which is a good reason to do those things in class. What happens on the problem set if you find there's a bunch of misconceptions? Well, one thing is, the first time you do the problem set, yeah it's a bit of a problem. Because you don't know what the misconceptions are unless you've researched ahead of time. But the second time you do the class you know what they are. So you can address the misconceptions in the solution set. So when I'm not totally sleep deprived I always try to get the solution set out the exact same day that the problem set is turned in. So the students can right away see. They can turn in their problem set, pick up a solution set or look at it online. So they can right away get quick feedback. And they can sort out the misconceptions that way. So that's a good question. So it is harder. Yes, question? AUDIENCE: What do you do about the fact that a lot of people will just ignore the solution sets? Or do you just consider that their problem? PROFESSOR: So the question is, what do I do about people who just ignore the solution sets? Because a lot of students do that. Yeah, partly I consider it their problem. Because look, that's there for them to learn. But partly, to the extent that they're ignoring the solution set because they only cared about the grade. So one cause of that is they really only care about the grade. So they just want to wait for the grader to give them back their solutions, their problem set with a mark on it. And they don't care what's right and wrong. That, I think, partly, is our responsibility that we've chosen boring problems. So to fix that, I think, one of the key areas is to use more of the Bloom's Taxonomy and make interesting problems. And the flip side is to deemphasize the grading. So actually, I like the PDF scale on homework. I think we are too micromanaging, generally, about homework, you know, grading it on a percentage scale, on A, B, C, D. P, you did a reasonable. And D, you blew it off. You did something but it was a joke. And F, you didn't turn it in. I think that scale puts the right emphasis on the grading. Look, we just want you to do the thing and try it. And if the problems are interesting that allows people to actually get involved in it. So there's lots of studies that show that if you reward people for things they already enjoy-- so you start paying them more for doing things they really like-- they actually start liking them less. So I'll talk about that as one of the political barriers to educational change. Because that's completely against the conventional psychology of this society, where everything has to be rewarded, merit-based pay and this and that. So how do you go against that? It's quite difficult. But one way, in general, in those problem sets is to deemphasize the grading. Related method-- which I haven't tried yet but I'm planning to-- is reading memos on the solution set. So I talked about reading memos before. And maybe one of you will try it before me and can tell me how it works. A reading memo is-- I talked about-- where you have the students take a reading and mark-- either online or on a piece of paper-- things that were confusing or things that were interesting and things they noticed. Well, you can do that with the solution set too. You say, OK, your assignment, homework 3A is to look at the solution set for homework two and mark anything that's confusing. So that forces people to read the solution set. And often what you'll find is that they'll find mistakes in the solution set, which is good to know because you can correct them right away. But you'll see what parts are confusing to students or not. Did that help answer your question? Great. Do I ever asked Wheeler's question explicitly? Yes, sometimes. I should do it more in class in lecture. But I do it a lot to myself. Whenever I solve a problem I think oh, what was the key idea? And then once I see the key idea happen many times I think oh, that's a very key idea. That should be taught explicitly in class. So I do that. And I'm going to talk about that principle of course design today, basically organizing courses around large themes and principles, which is one of the main [? cures. ?] OK, so what's wrong with tension as a force? How far should I go down the taxonomy? Basically, where in the course should I be? Really low in the taxonomy early and higher later? Well, I talked about that a bit last time. The fractal picture is not bad. So yeah, you have a few low-level examples and higher-level examples. But you do that throughout the course. So you don't wait until the end of the course to get the reward of the really interesting problems. And then related to that, what about exams and homework? Should they be the same or different? My view is they should be the same given the constraint that maybe the exam is done in a shorter time. So if anything, the exam should be easier than the homework. Generally, homework people will spend six or seven hours on. When I was an undergraduate we had take-home exams. And we would spend 24 hours on them. You pick them up, you signed it out. You'd turned it in in 24 hours. So that was misery. And it was interesting. You learnt a lot. But it was kind of miserable because it just wiped out a whole day from your quarter. And you had that for four classes. So four days from your quarter were just gone. So generally, exams are three hours, maybe two hours. And so that's much less time than a homework set. So if anything, exams should be easier, definitely not harder. Student loath when the exam does different things than the problem sets. And that is a general principle of course design, that if you want to prepare them for being able to do stuff-- for example on the exam-- you have to prepare them with things. You have to "teach to the test." That's actually good. You want your homeworks to be like the exam. And that's fine if the test is good. If the test tests real-world, interesting skills sure, then it's fine to teach the test because they'll come out being able to do that, no problem. So the exam and homework-- as a rough rule of function-- will be pretty much the same. Or if anything, the exam should be easier. Oh yeah, the first time you teach a class. So all this Bloom's Taxonomy and designing questions, it takes a long time. And it's really hard to make good questions. It's easy to make define XYZ questions. But what about the higher-level questions? What do you do to mitigate that problem of a huge amount of time? Well, one of the cures is to cooperate with other teachers. So you steal their good problems. So I gave you the T.S. Eliot saying before, that talent invents and genius steals. So be a genius. Whenever you see a good problem just use it. And whenever someone wants a good problem, if you have one, share it. And now, should you guard your problems? I think those days of guarding problems are gone. They've gone by because of things like OpenCourseWare, putting problem sets online. It's just not possible anymore to guard problems. I'm not sure it was ever a good idea. But now it's not even practical. So everything is online, all from the previous year. So the problems can't be guarded. And it's not practical to make up a whole set of new good problems for the next year. You'll never sleep. So I just trust the students not to cheat by looking at the old solution sets. And if you're grading, that's yet another reason to minimize how strict the grading is. So if the grading is really, just that they make an effort and try to learn something, then there's no benefit to cheating and looking at the old solution sets. So you actually make the incentive structure so that they don't need to do all that. Question? AUDIENCE: Do you have to change your problem to a certain extent in terms of plagiarism? PROFESSOR: So the question is, do I have to change the problems to a certain extent to avoid plagiarism? I don't even care. I don't even change them. I think, again, it's sort of what Adrian has said, is it their problem if they do it? And to some extent it is their problem. If they're going to cheat and look at the old solution set and not acknowledge it. There's only so much you can do. And you then get in an arms race with them. You change the problem to some extent. But then someone else makes it. So they can always subvert that. One person can solve the new problem and give it to everybody else. So you have no hope of actually winning that war, even if it were a moral war to try to win. So I don't even change them. And I say look, just acknowledge, just like you would in science. I think it's Larry Lessig, the copyright scholar, professor of law. He used to be at Harvard. Now it's Stanford. He founded Creative Commons and various fantastic copyright projects. So he said, well, there's a couple ways to get-- not compliance-- people to do things. One is you can do it through rules. For example, this is illegal. You just can't do this. We'll punish you if you do this. The other one is through norm. So this is just not done. In our community we just don't do things like that. And generally, this kind of-- not exhortation-- but push works much better than the punishment kind. So I just tell them, look, in science when you grow up and become a scientist of course you acknowledge your peers and people you learn from. Because that's just polite. It's just how we are. We say, please. We don't say gimme, gimme, gimme. I'm working on teaching my daughter that right now. She says, daddy read. Daddy read. Mommy read. Elsa read. I'm like OK, I'm fine with the Elsa read. But daddy read, mommy read, that's nice. And we love reading. But how do you ask? And now she says mommy read please, daddy read please. And so it's the start. So people already learnt that from young. So you can tap into that. Look, how do you act as a member of the community? We're social animals. So you don't have to put too much pressure in that direction. Whereas if you try to do it through a punishment you have to work much harder. The punishment one-- underneath-- has the idea that we don't trust the students. Right? And in that way it's common with changing the numbers on the problems set. It's that we don't trust you so we're changing the numbers. So I try to avoid those things as much as possible and try to go more towards this. Yes? AUDIENCE: Has anyone tried to have problem sets without solutions? So then there's no grading. And then the problem sets are just for practicing. PROFESSOR: Yeah, have problem sets for practice with no solutions or no grading on the problem sets? Have people tried that? So there's actually whole educational systems based on that. Not so much in America, but in England, the Oxford and Cambridge tutorial system. You're given problems to work on. So I was on both sides of that system. You're given problems to work with your tutor. And they're not graded. I mean, your tutor helps you make sure that you can do the problems. But everything that's graded happens at the end. Well, when I was an undergraduate it happened at the end of your entire degree. There was eight three-hour exams that counted for everything. And nothing else counted. So it had some craziness. That time at the end was pretty stressful. But before that your tutor was actually your friend. And your tutor wasn't the person doing the exams. So you would do problems because you wanted to learn and your tutor would help you so that you would do well on the exam that other people were setting. So you had a nice alliance with your tutor. AUDIENCE: Does that work well? PROFESSOR: I would say it works well. At MIT it would work great. Given [INAUDIBLE] that people are too overloaded. And they would say, oh well, if there's not going to be any grading at all, I don't have to do anything, while I'd really like to do this problem set. But I have four other classes that are making me work 20 hours a week. And I don't have any time. So then because of the other pressures they might fall away. But they would like to do it. So that's why I compromise by just saying, you have to do the thing and make an effort. And then I trust that once they start trying it they'll enjoy doing it. So I tried to make it as light as I think you can get away with at MIT, given the pressures from all the other teachers. But in the ideal world I think that's what we would do here. There wouldn't be that pressure. AUDIENCE: [INAUDIBLE]? PROFESSOR: Yeah, in Oxford and Cambridge it works very well, provided you have a good tutor. And there's many students who are there just because it's a thing to do, to go to Oxford and Cambridge. But if you factor out those students who are just there for the stamp of the place, for the other students it works fantastically well. I mean, the tutorials are a great discussion time. You can really kindle people's curiosity because it's really not based on grading, it's based on helping. AUDIENCE: [INAUDIBLE]? PROFESSOR: It's so hard to do. I mean, I don't know of any comparative studies within one culture. And I don't think you could really do them well between, say, the English universities and the American just because there's so many other social variables that are so different like the population that goes to university, which universities are talking about the exam system itself, so I don't know of any studies that have been done across cultures and that I would trust. And within I don't know of any. But it's a good question. I know studies in general about rewards, which is that the more you reward people for things that they like the less they like doing them. So I might give you some readings for that for very last penultimate session. And for the reading that I'm going to give you for the next session is going to be about tutoring, actually. So I'll put that on the website today. OK, questions? Yes? AUDIENCE: [INAUDIBLE] of a question. So I was wondering, if you do exams or homework problems throughout the semester, you can, maybe, also better monitor how the students develop compared to having one big exam on the end. And then maybe using [INAUDIBLE] lectures. PROFESSOR: Right, so having continuous assessment is one phrase for it. It allows you to monitor, make sure things are going OK. So all of a sudden you don't find, oh my god, the person didn't know anything. They got a third, which is like straight D's in your whole undergraduate. So in Oxford and Cambridge things are graded on first, second, third. And all of a sudden you had had no idea the student was going to get a third, or they were sick or something like that. So I think, yeah. So homework problems and exams have a really good purpose in assessment and helping students get feedback on how they're doing as well. So for that purpose, grading is useful. And one solution to that dilemma-- because I don't like to grade exactly because it stresses them out with the grade-- is to give them the solution set right away. So they can get feedback from the solution set whether they understood things, not whether they had a 90 or and 80, but whether it made sense to them. And that's, I think, a much more important thing. But then the other part of grading, which is that it's used publicly-- as part of your record and written to other people-- that part of grading I think is not right. I don't think it's morally right to, basically, reveal information about students to other people without their permission. I don't think that should be the purpose of the university. But the other purpose of grading, helping students know where they are. And helping us know where they are so we can help them more, I think, is a completely valid purpose. OK, so there was another suggestion which was to use as many non-physics and math examples as possible. So I'll try to reform my ways for the non physicists and mathematicians. It's just those come most naturally for me. OK, so I think that was the main-- yes, question? AUDIENCE: [INAUDIBLE] for evaluating the students or for teaching the students. PROFESSOR: Pardon? AUDIENCE: Homeworks? PROFESSOR: Yes, for evaluating versus for teaching? AUDIENCE: Yes, because that's one thing to evaluate, right? [INAUDIBLE]? PROFESSOR: It's both. So there's two, yeah. So the other purpose is that they learn and they do things. That's for sure. So you want them to do that. And you can do that in class and homework. And the take-home exams that we had when I was an undergraduate had that purpose too. You spent 24 hours really learning stuff. So that purpose is very valid too. But it's also very valid to know where the students are. So for example, that basically closes the feedback loop. You're doing all this stuff. And now you're seeing the result of it. And you want to know, well, did any of it take? Did any of it stick? And if none of it did, well, maybe you want to change what you're doing. AUDIENCE: Oh, but you cannot evaluate a person every week, right? And that's what drives students to copy solutions because they care about the grade. [INAUDIBLE]. PROFESSOR: That's why distinguish two senses of evaluate. There's evaluate for the purpose of helping the students. And there's evaluate for the purpose of publishing their grades elsewhere. And it's the publishing their grades elsewhere that worries them. So you want to minimize that part. And you want to do as much evaluation towards helping them as possible. So a solution set-- for example, one thing you can do is say OK, here's the solution set. Ask me about anything that's confusing. And then you can put some of the responsibility for this evaluation onto them so it's self evaluation. But you're still using the homework for evaluation. It's just you're not trying to say look, that's the main thing. I'm going to use that to tell everyone else about it. That's what worries them and produces copying. So you want to minimize that as much as possible. And it's hard. And that's the subject of the political obstacles to educational change session that'll be in the second-to-last lecture. OK, so yeah, I don't pretend any of these have easy answers, unfortunately. Oh, there was another suggestion, which is that I make a web forum where people can ask questions afterwards. Because it's hard sometimes think of a question on the fly. And that's true. And the problem is it's not easy to make web forms on the MIT.edu site. So I'll try to figure out a way to do that. But for security reasons you can't really easily make web forms that just append to a file or do something like that, although I'm going to look into it and see what workarounds I can find. Yes? AUDIENCE: [INAUDIBLE] scripts? PROFESSOR: Yeah, you can use scripts at MIT.edu. So I have to figure out how to use that. But yeah, I had that in the back of my mind. But thanks for pointing that out. Yeah? AUDIENCE: What about forms? PROFESSOR: Pardon? AUDIENCE: Forms? PROFESSOR: Forms? AUDIENCE: [INAUDIBLE]? PROFESSOR: Yeah, so then you get email. So the only problem with that is that you get a ton of spam. So there are all these spam bots that float around and just fill out forms and send junk emails all the time. So maybe that can be filtered out and it's not so bad. So but yeah, that's one of the few things you can do on MIT.edu, directly on the web.MIT.edu. So there's a very solutions to it. I'll try to think of something around that because it is a good suggestion. 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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For one-dimensional collisions, let's talk about two objects, 1 and object 2, moving with velocity V1 and another object V2 moving with velocity V2. Let's say they're moving on the ground. Now I'd like to introduce the concept of relative velocity, a concept that we experience all the time in our lives. But let's see what it actually means. So V relative, I'm going to define this to be the velocity of V1 minus the velocity of V2. Now because of this minus sign that seems a little bit about abstract. But one typical example where we see this all the time is for people traveling on highways. You might have two cars, one car overtaking the other car. But if you're sitting in car 1, it looks like car 2 is going quite slow. So let's just take typical highway example. So you might have V1. And we'll give it some speed. So we'll make it 60 miles per hour. And we'll just call this one-dimensional problem i hat. And V2-- notice we're not speeding on a highway-- V2 is going at 50 miles per hour, i hat, very slow. And the relative velocity, V1 minus V2-- so that's what we're calling V relative-- that's 60 miles per hour minus 50 miles per hour i hat. And that's just 10 miles per hour. And that's what people experience when one car is approaching another car. If you're in car 2, car 1 seems like it's coming at you at 10 miles per hour. This is what we mean by relative velocity. There's another important example-- so that's example 1-- the other important example to look at, example 2, is when two objects are moving in opposite directions. So let's just see write them in terms of components this time. So we have V2 x1. And we have V2. And let's make V1 x positive. So object 1 is moving in that direction. And let's write this one as V2 x. We don't have to call this initial. We'll just call it V2 x i hat. And here V2 x is equal to minus V1 x. So its component is negative. And even though we drew an arrow in this picture, the picture is still fine, because if the component is negative, it means it's moving in the opposite direction. The key arrow is the unit vector when we are writing components. And now V relative in this case is V1 x i hat minus V2 x i hat. That's the V1 x i hat minus minus. So there's another V1 x i hat. So the relative velocity in this case has a component that's twice the speed of V1. If two objects are moving together at the same speed, the relative velocity, the way we've defined it, has twice the magnitude of either velocity. And this is an important example to consider in collisions. Now this relative velocity concept we'll see we'll add a new way of thinking about elastic collisions with no external forces in one dimension.
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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. PROFESSOR: In this session, we're going to be looking at carbohydrate biosynthesis. Please look at Storyboard 29. Looking at Panel A, the next set of pathways on which we're going to focus concerns carbohydrate biosynthesis. First, we're going to look at the pathway by which glycogen is made. As you know, glycogen is the polymeric form of glucose that's very readily available for energy production. The second pathway is gluconeogenesis. The organs that utilize glucose as their metabolic fuel prefer to have glucose at a concentration of about 100 milligrams per deciliter in the blood. The challenge comes from the fact that we eat only sporadically, and thus, levels of glucose will go up and down depending upon the time that has passed since our last meal. When glucose levels drop, gluconeogenesis is the pathway that's activated. It takes non-carbohydrate precursors, converts them to glucose, and then secretes the glucose into the blood, ultimately to help maintain a constant glucose concentration. Looking at Panel B, our first topic is the synthesis of glycogen. If you look back at my first two lectures, I describe the structure of glycogen, which is also shown here. The piece of glycogen that I've shown consists of a linear chain of glucose molecules connected together through the 1 and 4 carbons. Chemically, we call this an alpha 1, 4 linkage. To the left of the glycogen molecule is the non-reducing end and to the right is the reducing end, which is connected by a tyrosine residue to a protein called glycogenin. Glycogenin is a variant of the glycogen synthase enzyme we'll talk about later. Glycogenin has the property that it can synthesize a polymeric glucose chain, such as the one shown, to boot up the synthesis of glycogen. In effect, glycogenin forms a primer molecule such as the one shown that provides a non-reducing end that can be extended by its sister enzyme, glycogen synthase. While I've drawn a linear glucose, I want you to keep in mind, the branches off of the six carbon of the glucoses in the chain are possible. As I mentioned earlier, glycogen is designed for fast breakdown. Its glucose units will quickly enter the pathway of glycolysis and generate ATPs in a manner of seconds. Let's now look at Panel C. It's useful to review the biochemical steps by which glycogen is broken down because the exact same intermediates appear in the reverse order in the synthesis of glycogen. The key enzyme for glycogen breakdown, or glycogenolysis, is glycogen phosphorylase. Glycogen phosphorylase progressively will nip off the non-reducing end-- that is the left-most sugar as shown-- producing oxonium ion intermediate. And as we saw in my first lecture, glycogen phosphorylase stereospecifically adds inorganic phosphate to the bottom face-- that is the alpha face of the oxonium ion-- to give glucose 1-phosphate with this stereochemistry shown. Once again, these same intermediates are going to appear in the synthesis of glycogen. Now let's take a look at the way that glucose 1-phosphate interfaces with the other biosynthetic and catabolic pathways. Let's look at Panel D. This metabolic map shows glycogen in the lower right-hand corner. We can see how glycogenolosis results in glycogen breakdown to glucose 1-phosphate, and in the reverse direction, we can see how glucose 1-phosphate can be used to make glycogen by way of two enzymes-- UDP Glucose Pyrophosphorylase, UGP, and Glycogen Synthase, GS. Let's look a little deeper at other pathways with which glucose 1 phosphate interfaces. This schematic shows that glucose 1-phosphate can be converted to glucose 6-phosphate by phosphoglucomutase, PGM. In this case, the phosphate is moved from the 1 carbon to the 6 carbon of the glucose moiety. The opposite reaction also occurs. That is, if you have glucose 6-phosphate, phosphoglucomutase will convert it to glucose 1-phosphate. Glucose 6-phosphate is an intermediate in glycolysis, gluconeogenesis-- which is the next pathway we're going to look at-- and another future pathway, the pentose phosphate pathway. All of this shows us that glucose 6-phosphate is a crossroads and one of the branches that leads from it and it is by way of glucose 1-phosphate. Let's imagine a scenario in which we eat a meal. Glucose appears in the blood as shown. It is taken into the cell. The enzymes hexokinase or glucokinase will phosphorylate the glucose into glucose 6-phosphate. If the glucose is not needed for glycolysis or the other pathways that I mentioned, phosphoglucomutase will convert the glucose 6-phosphate to glucose 1-phosphate, and then in the pathway of glycogen synthesis, glucose 1-phosphate will be polymerized into glycogen for energy storage. At this point, let's turn to Storyboard 30 and look at Panel A. Now let's take a look at the detail pathway by which glucose 1-phosphate is converted to glycogen. At the left is the structure of glucose 1-phosphate. The first enzyme involved is UDP Glucose Pyrophosphorylase, or UGP. The second substrate in this reaction is UTP, uridine triphosphate. The UDP Glucose Pyrophosphorylase catalyzes attack by the phosphate on the 1 carbon of glucose 1-phosphate on the alpha phosphorus of UTP. The two products are pyrophosphate and UDP glucose. The reaction is made thermodynamically irreversible by hydrolysis of pyrophosphate by inorganic pyrophosphatase into two molecules of inorganic phosphate. The UDP glucose is the substrate for the next enzyme in the sequenced glycogen synthase. I'm going to divide the glycogen synthase reaction into two parts. In the first, we see cleavage of the bond between the 1 carbon of the glucose and the beta phosphate of the UDP. This reaction liberates the UDP and generates the oxonium ions shown in the brackets. Structurally, this oxonium ion is the same intermediate we saw in glycogen breakdown, but here, we're using it as a biosynthetic reagent. Let's look at Panel B. In this panel, we continue the glycogen synthase reaction. In Panel A, we generated the oxonium ion of glucose, which is shown here, in Panel B in the lower left. Glycogen synthase activates the hydroxyl group on the 4 carbon of the terminal sugar residue. That is the sugar on the non-reducing end. The oxygen on the non-reducing sugar residue of glycogen then attacks the bottom face of the oxonium ion to give rise to a glycogen unit that has been extended by one glucose residue. I have put an asterisk on the 6 carbon of the glucose residue that's been added to the growing glycogen chain.
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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. All right. Last time, what we had done is that we had looked at the first evidence for the particle-like nature of radiation. And that evidence was a photoelectric effect. The evidence was that what you had to have was a photon or a particle of energy, a quantum of energy, a packet of energy, in order to get an electron out. And that energy had to be at least the energy of the work function of the metal. And so for every packet you put in there, you got one electron out. That is an example of the particle-like nature of radiation. But Einstein went on to show an even more convincing property of the particle likeness of radiation or a photon. And that is that what he did was showed that a photon has momentum. It has momentum, even though a photon does not have mass, although a photon does not have rest mass, for those of you in the know in this area. And having momentum is very much a particle-like property, right? Because you know how to write down momentum. Momentum is mass times velocity. You've got a mass in here. That is a particle-like property. And, yes, I am starting out with the lecture notes from number four, which I didn't finish last time. That is a particle-like property. But what Einstein showed was, from the relativistic equations of motion, what drops out from the relativistic equations of motion is the fact that a photon, at a frequency nu, has a momentum h nu over c. And because we know the relationship between nu and c, nu times lambda equals c, I can write the momentum of a photon as h over lambda. If you have some radiation, this is the photon momentum here. If you have some radiation, at a wavelength lambda, that radiation or those photons have this momentum p given by h over that wavelength. Now, that was a prediction from the relativistic equations of motion. And it took another eight, ten years or so before there actually was an experiment that demonstrated the momentum of a photon. And that experiment was called the Compton experiment. What went on in that experiment is that an X-ray beam came into some material or some molecule, some atoms, and they could actually see the transfer in the momentum from the photon to the atom. Kind of like in this website from the University of Colorado, here. This is just a cartoon of what is happening, but I got this photon done and I got this atom coming at me. And I cannot move this fast enough. I am going to get clobbered. You have a different computer than I have. Oh, I have to push down. Okay. Well, if I get aimed here, these photons are coming at this atom, coming at me. And, boy, if I do it fast enough, I can turn it around. Hey, I did it. But now, of course, if I go and lower the power. Come on. Come on. Aah! I got killed. [LAUGHTER] Christine, I don't like your computer. Oh, wait. I have got to get it. Get it. Get it. Get it. Please. Please. Please. Aah. All right. Well, you guys are going to be a lot better at this than I am. You can go and play with this. Christine is going to try now. Oh, look at that. She is going to get it. She is going to get it. She is going to get it. Yeah! [APPLAUSE] Three cheers for Christine. Oh, now it something else. This is going to keep going here. You need more power there. [LAUGHTER] Hey, not that guy. Fantastic. All right. And it is actually just this effect that was used by Steve Chu at Stanford and Bill Phillips at NIST and Cohen and Tanugi who provided some of the theory behind it. It is just that effect that they used to literally trap an atom in space. How do you do that? Well, you take an unsuspecting atom, and you bring in a high power laser beam coming out in this direction. And those photons transfer momentum, and they push that atom this way. But you are smarter than that, so you bring in a laser beam this way. And so you have momentum transfer this way and this way. You just trapped the atom in this dimension. And then you bring in a laser beam this way. Bring in a laser beam that way. You have trapped the atom now in this dimension. What is that? Your dog. That is not part of my lecture. [LAUGHTER] And then you bring in the laser beam this way. And so now you have constrained in the three dimensions. And so the atom is trapped in space by this photon pressure, by this momentum transfer. And this is called laser trapping. And these three gentlemen, whose names I gave you just a moment ago, are laser atom trapping and received a Nobel Prize in 1997 for this demonstration. But the other reason why this laser atom trapping was really so important is because it is actually the first step in another experiment. It is the first step in producing a Bose-Einstein condensate. What this laser trapping does is literally to slow the atom down or to cool the atom, because temperature and the velocity of the atom, the speed of the atom are related. The slower the speed, the lower the temperature. And to produce a Bose-Einstein condensate, you have to have bosons, which you lower in temperature. And ultimately, they condense. And the temperatures have to be on the order of micro-Kelvin. And so this is the first step in producing that Bose-Einstein condensate. This will bring you down to temperatures of, say, a Kelvin or so. And then there are lots of other techniques, a couple of other steps that bring you down to the microKelvin range. And then, finally, you can get the bosons to condense. And one of my colleagues in the Physics Department, Wolfgang Ketterle, also received the Nobel Prize for the formation of the Bose-Einstein condensate. I actually think he is teaching a recitation section in 8.01. Maybe some of you have him. You do? No, you don't have him. Okay. But you will be able to meet him and talk to him. Question? I'm sorry. Fantastic. Has he told you about this yet? Oh, he went to a conference. Okay. Well, you can imagine he is in demand. But you will see him, right? I hope. Very important effect here. We have radiation that is exhibiting both wave-like properties and particle-like properties. And, just in general, experiments where the radiation produces a change in the state of the matter such as the photoelectron effect. In photoelectron effect, the matter changes in the sense that an electron is pulled off of it. In those experiments, the radiation usually exhibits the particle-like behavior. In experiments where there is a change in the spatial distribution of the radiation, or where the radiation interaction results in a change in the spatial distribution of the radiation, that is when the radiation exhibits its wave-like behavior. And so it really is not appropriate to ask, is light or radiation a particle or a wave? The appropriate question to ask is, how does light behave? Does it behave like a particle or does it behave like a wave under particular experimental circumstances? And having both behaviors, this wave-particle duality of radiation is not a contradiction. It just is the fundamental nature of radiation, of light. You may think it is a contradiction because in your everyday experience, you either see a wave or you see a particle. But that is your everyday experience. And there are parts of nature that you cannot see every single day. And those deeper parts of nature have different rules. And you have to be accepting of those different rules. And so it is not a contradiction in terms. It just seems strange to you just because that isn't your everyday experience. It is the fundamental nature of radiation. Well, not only is that the fundamental nature of radiation, but the wave-particle duality of matter is also the fundamental nature of matter. And that is what we are going to talk about right now. We are going to move to matter, particles. The particle-like nature of matter is within your everyday experience, but it is the wave-like nature of matter that is not within your everyday experience. And so let's take a look at that. Suppose we did this experiment. That is, we had a nickel crystal. And these two atoms here are just two of the atoms on the surface of a nickel crystal. And we know the spacings between these two atoms in the crystal because we know the crystal structure of the nickel. That spacing is about 2x10^-10 meters. Naively, if you brought in a beam of electrons, particles, and we know they have mass. J.J. Thompson taught us they were particles, they had mass. But, naively, if you brought them in, you might expect these electrons to scatter isotropically. That is that they would scatter equally in all directions so that when they ultimately hit this screen here, this curved phosphor screen, and I changed the geometry here to a curved screen just so that it will be a little bit easier to analyze the geometry of this problem, which we are going to do in a moment, you might expect this screen to be lit up uniformly at all angles. Well, this is exactly the experiment that Davidson and Germer did in 1927, along with this gentleman, G. Thompson, George Thompson, son of J.J. Thompson. And J.J. Thompson actually did an experiment a little different than Davidson and Germer. I am going to show you the Davidson and Germer experiment here. But here is the same diagram that I had before, except that I made the nickel atoms a little bit smaller just so that this diagram would be a little bit easier to understand. I cleaned up the diagram, but kept the spacing between the two nickel atoms the same. And so Davidson, Germer and Thompson came in, scattered these electrons and looked how they scattered back. And, lo and behold, what they saw is that these electrons seemed to scatter back at a preferential angle. The angular distribution was not isotropic. Instead, it looked like the electrons scattered back at a pretty well-defined angle here, 50.7 degrees. And not only did they scatter back at that angle, they also scattered right back at themselves. Backscattered this way, so this scattering angle is zero degrees. And under some particular conditions, the electrons also scattered at a larger angle, here. But the bottom line is that the scattering pattern was not isotropic. There was a bright spot, lots of electrons scattered at this angle, a dark spot, no electrons scattered at this angle. A bright spot, dark spot, bright spot. This looks like interference phenomena, just like the two slit experiment. Bright spot, dark spot, bright spot, constructive, destructive, constructive interference, back and forth. That was their observation. How do we understand that? Well, it is looking like these electrons are behaving like waves. Suppose these electrons are coming in, so we have this constant stream of electrons impinging on our nickel crystal. Well, what is happening here is that when these electrons are reflecting back from the individual nickel atoms, these individual nickel atoms are kind of functioning like those little slits we saw in the two slit experiment. That is, they are scattering back as a wave. These electrons seem to be scattering as a wave, so isotropically in all directions. This semicircle around each one of the atoms, and I only show you two atoms here, each semicircle is the maximum of the wave front. It is the crest of the wave front. And then, as time goes by, of course, these waves propagate out. And then another wave front, another wave maximum appears and a distance between these two maxima is, of course, the wavelength. And as time goes on, they scatter further. And as time goes on, they still scatter. And they keep the propagating out until they reach the screen. And, lo and behold, on the screen you see a bright spot, dark spot, bright spot, dark spot. Interference pattern. Let's analyze this. Here is the diagram again. I just moved it over and cleaned it up again. I want you to look at this spot right in there. That is where we have the maximum of a wave scattered from atom one at the same point in space as the maximum of waves scattered from atom two. Constructive interference. Here is another point of constructive interference. Here is another point of constructive interference. Everywhere along this line, we have constructive interference, which results in a large intensity right at this scattering angle here, a bright spot. And we already know the condition for constructive interference. That is, in order to get this constructive interference, the difference in the distance traveled by the two waves that are interfering has to be an integral multiple of the wavelength lambda. Now, I use the term d instead of r, but it is the same thing for the condition for constructive interference, here. And if you went and analyzed what the difference in the distance was for this constructive interference along this line, you would find it was n equals 1. The difference in the distance traveled is one lambda. And, if you looked at the points of constructive interference along this line that led to this bright spot, the difference in the distance traveled will be two lambda. This is our second-order interference feature, our second-order diffraction spot. And, if you look at it along the center here, normal to the crystal, that would be the zero-order spot. d2 minus d1 is equal to zero lambda. All right. That is what looks like is happening. Now, what we are going to do is we are going to actually analyze this geometry a bit more. We didn't do so in the two slit experiment. We could have. We didn't. We are going to do it here. And what we are going to be after is if these electrons are acting like waves, then they have a wavelength. And we want to know what the wavelength is. Davidson and Germer wanted to know what the wavelength was. And we are going to use this scattering angle here, theta. This angle from the normal to where the electrons are scattering, that angle theta, we are going to use that information, theta equals 50.7 degrees, to back out the wavelength. And we know what the condition is for constructive interference. We just talked about it. Here it is, d2 minus d1. That is the wavelength we are after in this analysis. Now, what is d2 here? Well, the length of this line, d2, is the distance that the wave that scatters from electron two travels from electron two to the screen. d1 is the distance that the wave that scatters from atom one travels to the screen. That is what d2 and d1 are, here. Now, I am going to draw a perpendicular from atom one to this line d2. There is my right angle. Now, you can see, then, that this leg of the triangle is d2 minus d1, the difference in the distance traveled. That is this quantity here. That is going to be important. Now, you have got to convince yourself that this angle right here in the triangle is the same as this scattering angle. You can convince yourself of that pretty easily. Now we have a well-defined triangle. We know one length of it, we have measured the angle theta, and d2 minus d1 is something that we would like to know. Let's do a little geometry. The sine of theta, the sign of this angle is equal to the opposite length, which is d2 minus d1, divided by A, this distance between the two atoms in the nickel crystal. We have two equations. This is the equation that in theory should obtain for constructive interference. This is the equation that we set up given the particular physical geometry of our problem. These two (d2 minus d1)'s better be equal to each other. We have n lambda is A sine theta. Let's solve for lambda. We can do that. That is A sine theta over n. We already know what theta is, we know what A is, so what is n? Well, n is going to be one because this is the bright spot that is closest to the zero-order spot, which is always present at the normal angle there if you are coming in at normal orientation. So n is equal 1 so I can plug things in. And, when I do that, I find that the wavelength that I predict is 1.66x10^-10 meters. We've got this wavelength. Now, before I go on, I just want to point out that this geometry of the problem that I set up here is identical to the geometry in a technique known as X-ray diffraction. X-ray diffraction does not use electrons coming in, but uses X-rays, photons. And it is a technique that is going to be important to you if you do any kind of science involving materials or biological systems. And it is important because X-ray diffraction allows you to determine the structure, in particular of proteins, crystal proteins. You crystallize the protein, and you use this X-ray diffraction to get out the structure. And the reason why you want the structure of the proteins is because the structure gives you a hint as to what the function of the proteins are. And so in the use of X-ray diffraction, we don't go and calculate what lambda is. We already know what lambda is in X-ray diffraction. We know the wavelength of the incident photons, the X-rays. What we don't know in the technique of X-ray diffraction is the distance between the atoms in an unknown structure. And so in X-ray diffraction, we know the wavelength, and we can figure out what order it is and we measure the scattering angle. And we use that to determine the distance between the atoms. And, in that way, we back out the structure of the sample. Yes. We are just going to get there. We are going to do that. All right. Same geometry here. Now, this experiment of Davidson and Germer was really an important one because just three years before this, there was a prediction for what the wavelength of particles ought to be. And that prediction was made by this gentleman, Louis de Broglie. In his Ph.D. thesis, no less, what Mr. de Broglie did was that he looked at the relativistic equations of motion that Einstein wrote down and used to propose that a photon or radiation with a wavelength lambda had momentum p. Well, he took those same equations and said, well, these relativistic equations of motion apply to matter just as well as they apply to radiation. Therefore, if you have radiation with a wavelength lambda, you then have this momentum p for the radiation. This is what Einstein said. But he turned it around and said, if you have matter with a momentum p, well, that matter ought to have a wavelength lambda. He turned around Einstein's equations of motion and proposed that the wavelength of a particle be given by h over p where, of course, p here is the mass times the velocity. Fantastic. What a great Ph.D. thesis. I'm impressed. Now, let's see how well, as you can imagine, that predicts the wavelength that Davidson and Germer actually measured. We know we have 54 electrons coming into this nickel crystal. That is their kinetic energy, one-half m v squared. Kinetic energy can also be written in terms of the momentum. The momentum is p square over 2m. You can convince yourself of this. This is a good thing to know for doing these problems, that the kinetic energy is p squared over 2 times the mass of the electron. And so if you solve that, what you get for the momentum of the electrons is 4.0x10^-24 kilograms meters per second. And now I can take this momentum and plug it into the expression for de Broglie's wavelength, 6.6x10^-34 joule seconds, over the momentum, 4x10^-24. What do I get? 1.7x10^-10 meters. Absolutely the same as the experiment. De Broglie made a prediction. A few years after that, experiments demonstrated that de Broglie was absolutely correct in his prediction. What do we have here? We have matter, particles, exhibiting wave-like behavior. And, those particles can be measured to have a wavelength that actually agrees with a prediction, some theory, the de Broglie wavelength. And we also have another phenomena here, which I really enjoy, and that is Davidson and Germer and George Thompson. They demonstrated that electrons behave like waves. And what did J.J. Thompson do, father of George Thompson, well, he demonstrated that an electron was a particle. Here, we have both the father and the son talking about seemingly opposite behavior, but they are both right. How often does that happen? That, I think, is really amazing. But if matter is wave-like and if electrons can be represented by a wavelength, then what about your wavelengths and my wavelengths? We should have a wavelength. And we do. And just briefly, here, let's talk about what the wavelength is of a baseball pitched by Curt Shilling at 90 mph. What is that wavelength? Well, a baseball is five ounces. 90 mph. You can calculate the momentum. It is in your notes there. We will calculate, here, the wavelength. And what you are going to find is that it is 1.2x10^-34 meters. That is pretty small. What is the diameter of a nucleus? 10^-14, right. That is a good number to know. Here, we have a wavelength that is 10^-34 meters. Is that wavelength of a macroscopic size object going to have any consequence in our world? No, absolutely not. Why? Because in order to see any effects from this small wavelength, we are going to have to have slits or atoms that are going to be on the order of this close together. But there is no way that we are going to have two nickel atoms that are this close together or two slits in a two slit experiment this close together. And so for macroscopic objects, the wavelike properties have no consequence in this world. And that is simply because the mass is too large. It makes the wavelength too small to have any effects in our everyday lives. And it is actually -- Yes? Okay. Well, they are actually coming in as waves. They are behaving as waves. Remember my beach picture? I drew them coming in like a circle. But remember my picture of this barrier here on the beach, and I am laying here on the sand, and then the waves are coming in? Here is blue. These electrons, as they are coming in, really need to be thought of as these kind of plane waves. I drew them as kind of just particles. But you have to really think of them as plane waves and that they are reflecting off of these two atoms in the way that I just explained. Another question? The thing is that you cannot get that velocity slow enough to make the wavelength large enough to be of consequence. If you could then you would, right? You would see the wave-like behavior. But you cannot, practically speaking, get it to that extent. No. In this particular case, anything that is so massive, any smaller effects, like what you are talking about, exactly the point of observation is not going to have an effect on anything that is so massive. Pardon? Yes, your point of observation will have an effect on your interpretation of the experiment, if you are talking about something that has a much larger wavelength. Absolutely. In high energy physics experiments, for example. Good question. It is just this observation, here, of the wave-like behavior of electrons, of particles, that led to the interpretation, then, or led to the realization that maybe, you have to treat electrons as waves. Or maybe you have to treat the behavior of electrons as wave-like behavior. And that is what this gentleman Schrˆdinger did. He said, well, you know what? This gave him an idea. Maybe what is wrong is that an electron in an atom, I cannot treat that electron as a particle. Instead, what I have to do is treat is as a wave. I have to treat its wave-like properties. And it was that impetus that led him to write down a wave equation of motion. An equation of motion for waves. He realized, or he was guessing at the moment, well, maybe in the case when a microscopic particle has a wavelength that is on the order of the size of its environment, in that case, maybe the wavelength has an effect, makes a difference. For example, in the case of the electrons, we had a wavelength calculated, there, of 1.7x10^-10 meters. That is a wavelength that is on the order of the size of the environment of the electron, which is on the order of the size of an atom. Maybe in that case I have to pay attention to the wavelength. The reason you and I don't have to pay any attention to our wavelength is because our wavelength is 10^-30 meters or so. And that is much, much larger than the size of the environment. In this case, we don't have to pay any attention to our wavelength. But, for an electron in an atom, we have a problem, here. What did Schrˆdinger do? Schrˆdinger said, I have to write down a wave equation. An equation of motion for matter waves. And what is that equation of motion? Well, that equation of motion is H hat operating on Psi, and it gives us back an energy, E, times a Psi. What is this? Well, Psi, here, is a wave. I am somehow going to let my electron in an atom be represented by Psi. This is going to be a wave form. This is going to be a wave function. I am going to let my electron be represented by the wave function. Exactly how it is going to be represented by the wave function is something that I am not going to tell you quite yet. But it is going to represent the electron. This energy, here, that energy is going to turn out to be the binding energy of the electron in the atom. This thing, here, is called the Hamiltonian Operator. That Hamiltonian Operator is specific to a particular problem, and we will look at the Hamiltonian Operator for a hydrogen atom. But this operator is operating on Psi, and it gives you back a Psi, the same function, multiplied by a constant. That constant is the binding energy. Now, you think, well, let me just cancel this and this. But you cannot do that because this is an operator. This has some derivatives in it. Its operating on Psi gives you back the same function times the constant. Now, how did Schrˆdinger actually derive this equation, so to speak? Well, what he did was to just guess at a wave function. We are going to use a one-dimensional wave function. He is going to say, let me represent my electron by Psi of x equal to 2 a times cosine 2 pi x over lambda. That is going to be my wave function. Why not? And then he said, well, what I really need here is an equation of motion. I need to know how Psi changes with x. If you wanted an equation of motion, if you wanted to know how Psi changes with x, what would you do to Psi? Take the derivative. Let's take a derivative of Psi of x, with respect to x. That is going to be equal to minus 2 a times 2 pi over lambda times sine of 2pi x over lambda. That is an equation of motion. Now, this is actually kind of an equation of position. But it is telling us how Psi changes with x. But now, since that gave us some information about how Psi changes with x, how would we get the rate of change of Psi with x? Second derivative. Let's take the second derivative of that. Second derivative of Psi of x with respect to x, that is minus 2a times 4 pi squared over lambda squared times cosine 2pi x over lambda. That's pretty good, but now, what do you see in this equation? Psi. You see what I started with. It is recursive, right. You see the Psi of x here. Let me write that equation in terms of the original function. Second derivative of Psi of x with respect to x, minus 2pi over lambda squared times Psi of x. lambda)^2 Psi(x)** That is pretty good. You know where I am trying to go? I am trying to derive, so to speak, Schrˆdinger's equation. See, it is not very hard. Yes. Now, this equation right here, this equation is just a classical wave equation. The only thing I have done so far is take derivatives. I have done nothing else. I just took derivatives. It could represent any kind of wave. There is nothing quantum mechanical about it. But here comes the big leap that Schrˆdinger made. He substituted in here for lambda the momentum of the particle. In other words, if this is a wave equation and that wave has some wavelength, here, lambda. He said, well, if this is a wave equation for a matter wave, well, then I better get the momentum of that particle in this wave. And he knew how to do that because de Broglie told him how to do that. De Broglie said, lambda is equal to h over p. That is pretty good. What can I do here? Well, what I can do is I can rearrange this and write this in terms of the momentum of the particle. Second derivative of Psi of x with respect to x is going to be minus p squared over h bar squared times Psi of x. Now, let me explain what h bar here is h bar is a shorthand way of writing h over 2pi. If you don't know it already, you should know it. You are going to need it. That is h over 2pi, so this is h bar squared. Now, what do I have? Now I have a matter wave. I have an equation of motion. I have something that tells me how Psi moves with respect to X. The rate of change of Psi with X. And I had the momentum of the particle buried in here. From this form, I am going to get to that. And I guess I am going to get to that next Wednesday.
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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: Time to learn. Time to get back to business here in 3.091. So let's do the announcements first. Tomorrow quiz one, based on homework one. Ten minutes. Bring your Periodic Table, your table of constants, something to write with and a calculator. There's about a dozen or so copies of the text on reserve at the Hayden Library. I want you to be aware of that. Now some of you approached me last day and said is it true that we have to memorize the Periodic Table? Yes. But not the whole Periodic Table. I know some of you are going to say, well that's so 19th century, rote learning and so on. To which I say, you're wrong. Every educated person should know that potassium lies under sodium, which lies under lithium. And you are going to be educated whether you like it or not. But this is a lot to learn and so we're just going to look at the main block. We're not going to ask you to memorize the lanthanides, actinides and the super heavies. We'll leave that out. So it's quite straightforward. We've been doing this for many years. This is from 2004. And it's not that hard. I give you the blank here. And I even put the numbers on for you. And all you have to do is add the two-letter or one-letter symbol. You don't have to give me the full name of the element. You don't have to get me its density or its electron affinity. You've just got to put argon, in there. Ar. That's all you've got to put in there. OK? And every year people grouse when I announce this. But you'll find out that you'll breeze through it. You'll have 10 minutes. This will be on September 24. So next week there will be two minor celebrations. One on Tuesday and one on Thursday. So it's going to be a very festive week. And people blast through it. It's one mark off for every error down to zero. We don't give negative scores. And people just go right through it. And then they urge me to make sure that in future years I victimi-- I mean I ask students to continue this. People are very proud of the fact they know the S, P, and D blocks of the Periodic Table by memory. But I don't want to neglect, disrespect the lanthanides and actinides. So we're going to have two contests. Mnemonics to help remember the names of the lanthanides and actinides. That's tricky to remember. So these are mnemonics contests. So here are some examples. Here's one, see lanthanum, cerium, praseodymium, neodymium and so on, all right? Obviously this one came from industry because of the slur against the Academy. Obviously, somebody in industry has no idea how hard we work here. And plus, look at this: "to dramatically help". This gigantic split infinitive. Clearly, clearly someone from industry. Here's another one that would help you memorize, I think. Now, this is obviously not referring to 3.091. It must be referring to some other chemistry professor here. I'll let that go unchallenged. Now, occasionally people ratchet up. And this was about six, seven years ago, a young man by the name of Blake Stacey, who was aware of the fact that there are 14 lanthanide elements because there are fourteen F electrons. And an Elizabethan sonnet has 14 lines. And so he took Sonnet 57 by Shakespeare in reference to the fact that lanthanum is element 57. And he wrote this in reference to Sonnet 57. And, of course, he knew he was going to win when he made samarium the word smelting, and being the chemical metallurgist I am, he had me at the fifth line. Actually, I showed this to Professor Sonenberg in Humanities, and she said, actually, this is pretty good. That's one of those times you look at and say, you wouldn't see this at Harvard. So now I'm going to show you one. Sometimes people ratchet up. A couple of years ago, there was a video. So let's take a look at that one. [VIDEO PLAYBACK] PROFESSOR: And she goes through various poses as a superhero of different elemental value. [END OF VIDEO PLAYBACK] PROFESSOR: But we don't have that much time. So anyways, September 25, 5:00 p.m. Eastern Daylight Time. Submit your entry, whether it's the big canisters of the Hollywood footage or whether it's just a 14-line mnemonic device. And the prizes-- it's a contest, there are prizes-- I have ties and scarves from the American Chemical Society. [LAUGHTER] PROFESSOR: What are you laughing at? These things! Now look at me! This is going to be stylish. They're black. They've got elements from the Periodic Table on them. Very hot. Very hot. They're black. They look great. And we're going to present them on the Friday of the parents' weekend because the parents eat this stuff up. So now I think it's time to get back to learning. So last day we ended with the stoichiometry, and I showed you the Kroll process, how to make titanium. So the question that we never got to was how do we tell that magnesium has a higher affinity for chlorine than titanium does? How do we know that magnesium is going to reduce titanium tetrachloride to a sponge? So for that we have to look inside the atom. We have to look inside the atom. So that's what we're going to start doing today. And we're going to begin, as often we will, with a history lesson. So let's go back for a status report around the end of the 19th century. So what did we know towards the end of the 19th century? The atom is electrically neutral. The negative charge is carried by electrons. We knew that there was some carrier of negative charge. We knew further that this electron is very, very tiny. Small mass. And so if it has a very small mass, then by difference, then the bulk of the mass of atom must be contained in the positive charge. So now the question is, what is the spatial distribution of charge inside an atom? You might say, well why do we want to know that? Well, we're trying to get a physical model of the atom so that then we can start to address the question of chemical reactivity. So we have got to know where all the players are before we can attribute action to them. So let's now take a look at models of the atom. For the first modern model of the atom we go to J. J. Thompson. J. J. Thompson, he was a professor of physics and director of the Cavendish lab at Cambridge University. The other Cambridge. Cambridge University. And this Cavendish Lab was named after Henry Cavendish. The same one who is credited with the discovery of hydrogen in 1766. He never married, and when he died he bequeathed this fortune to Cambridge University, and they took that money and established a physics laboratory, which is still in operation to this day. So what did J. J. Thompson tell us? He said the electrons were distributed throughout a uniformly charged positive sphere of atomic dimensions. So that's the text. Now let's make this cartoon, because that makes more sense. Visualize it a little bit better. So it looks a little bit like this. So here's a sphere of atomic dimension. Electrons are these tiny little things distributed throughout the positive sphere. And this is massive because the electrons have very, very low mass. And just to be clear, there are no protons. No protons. So this isn't positive. It's a positive chargeness. It's just a big positive blur. And then the little electrons inside. So the electrons are negative, and they're tiny and low-mass, and they're mobile. He didn't say much about how they move, but you can imagine that they move, and maybe they even spin. And so on. And this was known as the Plum Pudding Model. There's a big cultural bias here. So even though I grew up in British Canada, I never had this stuff. But my understanding is plum pudding is this dish that's served at holidays and Christmas-time. So the custard is the charge, the positive charge. And these little bits of fruit are the electrons. And they're throughout. At Christmas-time, they might even put little charms in here. Like a little wishbone or something for good luck. And you pull it out and all that kind of stuff. Yeah that's merry, old England. So this is the Plum Pudding Model. And that's what was in place. By the way, we've been using the term electron, but J. J. did not like the term electron. Actually, he won his Nobel Prize for the-- quote, unquote-- the discovery of the electron. Just as Cavendish didn't discover hydrogen, he characterized it. J. J. Thompson characterized the electron. What did he in particular characterize? He characterized the charge-to-mass ratio. He made the first quantitative measure of the charge-to-mass ratio of the electron. But he never used the term, electron. He called it the corpuscle. He called it the corpuscle of electrical charge, whereas this is really sort of an elementary charge. The corpuscle. And fortunately there was another British chemist. In fact, he was an electrochemist. His name was John Stoney, who, while I think that Thompson was a brilliant man, I wasn't crazy about his literary choices. And it was Stoney who said, let's call the term electron, coming from the Greek term, elektra, which is the Greek word for amber. And you know if you rub amber, you'll build up static charge, and so on. That's where we got the electron from. So that looks pretty good. So that's our theory. And what happens next? Well, what happens next is that we've got to put this theory to the test. And the person who puts the theory to the test is a man by the name of Ernest Rutherford. Rutherford was an interesting character. He came from New Zealand and he was born on a farm. It was a family with 12 children. You notice Mendeleyev came from a family of 14, this guy comes from a family of 12. So you learn how to survive. He worked on a farm so he was very handy, and that came in to play as his career progressed. He was a brilliant experimentalist. He was able to do experiments where others failed. And he did this experiment where he was a professor of physics at Victoria University in Manchester in the UK. Now, he came out of New Zealand, got a scholarship to study at Cambridge University. And he was actually a research student for J. J. Thompson. And he conducted his thesis research under J. J. and during that time his thesis research was about the properties of charged particles. And they were very interested in gases and radioactive elements, ionizing radiation, the whole thing. The Cavendish lab was a beehive of activity. And what he did under his PhD was he understood both the alpha particle and the beta particle. First, he identified what they are. The alpha particle was determined to be the helium nucleus. So this is helium that's lost both of its electrons. So it's just protons and neutrons in the nucleus. Completely naked. No electrons. And then the beta particle, which other people in physics had been referring to this mysterious beta particle, he made the connection between the beta particle and the electron. So the same electron that's over here and attributed to the negative charge within the nucleus is also a free particle that can go through space. And these are taken from your reading. And you can see that if you send the particle beam through a charged plate, the negative plate here being the lower plate, the positive particle, which in this case is the alpha particle, will bend down towards the negative plate. And the negative particle, being the free electron, is bent in the opposite direction. So these are the kinds of experiments they would do. And they also looked at their ability to penetrate different media. So when it comes to their ability to penetrate solids, they differentiate themselves. So they alpha particle is quite poor at penetrating solids. So you see in the cartoon there, the alpha particle is stopped even by that little sheet of paper. So it doesn't do so well here, whereas the beta particle, the electron, can go zooming right through the paper and is stopped by the lead. And then over here they were also studying their ability to ionize gases. So if you take a gas and you bombard it with alpha particles, will you get a lot of ionization? Yes or no, et cetera. And what they found was that it was a complementary, that the alpha particle is pretty good at ionizing gases, whereas the beta particle is poor at ionizing gases, and in fact, is stopped. This is why, if you have an electron beam you need a vacuum. Otherwise, the electron will be dissipated in very, very short order. So after doing this work with J. J., he got a teaching job in Canada at McGill University up north. And while he was there he continued to do work on the origin of alpha particles and won the Nobel Prize for his work while he was at McGill. And so when he determined that the alpha particle origin in the decomposition, the disintegration of elements. And believe it or not, in this work back in 1906, 1907, he was already anticipating nuclear fission because he was saying that if you put thorium or polonium into a tube, you can expect that it will emit alpha particles. Well, conservation of mass says they have got to be coming from somewhere. And if they're not coming from out here, they have got to be coming from inside. This was brilliant work. This is Rutherford, young man in Montreal, gets the Nobel Prize in 1908. And then he goes to Victoria University, and he sets up this experiment. Critical experiment. I mean, imagine this guy. I mean, he's already got the Nobel Prize. He could just put his feet up. No, he keeps going. So this is the experiment. It's called the Rutherford-Geiger-Marsden experiment. Because Rutherford was the brains behind it and Geiger and Marsden were the guys in the lab. And so here's the experiment. So what he's going to do is he's going to put this model to the test. We're going to see if this model makes sense or not. So he's got a gold foil-- now they used different elements, but ultimately they loved gold because gold, first of all, is noble and it doesn't oxide. So it's pure, even in atmospheric oxygen. And secondly, we've known since antiquity we can physically deform gold and make it into something sub-paper thin. So that's what they did. They beat this thing down to 600 nanometers thick, 0.6 micron. And they used alpha particles here. Radium, polonium, thorium as the source in the lead container and so on. And this stream of alpha particles comes zooming out at energies of 7.68 million electron volts per particle. So this is like a cannon with a pumpkin in front of it. And they're going to shoot these at the foil. So what's the purpose here? They want to see what's going on. And if you imagine that we are shooting alpha particles, which are positive, and I've got a wall here of all of these nuclei that are essentially positive chargeness, uniformly distributed with these tiny, tiny little electrons of almost no mass, you'd expect that most of these particles are just going to go zooming on through. Beyond a certain thickness, I might start seeing some attenuation. But if I get the gold film really thin, I should just go blasting these all through. Well, what they find is, two things. First of all, most of the particles do go through. When they go through, they're deflected a little bit. They don't just go through, they are deflected. And most of them are deflected through low angles. But a small number of them are deflected through high angles. They almost go straight back at the source. So if you've got that set of data, that doesn't support this model. This model cannot be sustained in the light of those data. It's a very important experiment. By the way, how do they count these things? Well, Geiger is the same guy who give us the Geiger counter. Geiger was a technician that worked for Rutherford. And he developed this screen, they called it a scintillation screen. It was coated with zinc sulfide. And so when alpha particles bounced and they hit the screen, they would excite electrons in the screen and then cause a scintilla, a spark. And then what you do is you just get a graduate student or some other source of abundant, cheap labor-- and you have a whole bunch of graduate students sitting here-- and they count. That's how it was done in good old days, 1909. And so they counted what the distribution was. So that's the experiment. So this is taken from your book. So this what you'd expect. And this is what they saw. So this is small angles and this is large angles. Theaters of the angle here. Now Marsden, who's the third? There's a third person. Marsden. Marsden's not too different from you. Marsden was about 20 years old at the time. He dropped out of school and then he came to Rutherford, and he said, I'm looking for a job. In those days, you couldn't return to college. If you dropped out, finished. You had to stay all the way through. Once you dropped out, they slam the door behind you. But Rutherford, he was different. He looked at people on the basis of their intrinsic value. He didn't care about title. And so he said to Marsden, OK, you want to do something for me? Here's your question, I like you to imagine an alternative model of the atom. And here's the alternative model of the atom. What Rutherford said was, suppose instead, if this is the dimension of the atom, instead of having positive charge uniformly distributed, he said, what if we have positive charge concentrated at one point and then out here we have the negative charge? So here's the positive charge. And here's a whole bunch of negative charge. And now here's the next one. And here's the next one. And so on. And now we're going to shoot these helium nuclei. See what's in between here? This is a void, taking from Democritus. And so now if I shoot the positive helium nuclei here, if they get close to this positive charge they'll be deflected. If they don't get close to the positive charge, they just go right through. And once in awhile they get almost on axis, in which case they fling back. This model would be consistent with the data set I just showed you. And then to go further, he says to Marsden, I'm going to tell you what the distribution of high-angle scattering was, you tell me what is the ratio between the radius of the nucleus, where the positive charge is to the-- excuse me, I'm showing you the diameter, so this is the diameter of the nucleus versus the diameter of the atom. What would be the ratio of the two? I mean, it's gold, so I know I got 79 plus here. And I've got 2 plus. And I've got a 79 plus. What would be the size of the sphere of 79 plus that would deflect something of 2 plus through this angle? So that's Marsden. It's not bad for a kid that's been out of school for a couple of years. That's his problem. So he did. He did the calculation. He did the calculation and Marsden figured out that the radius of the nucleus to the radius of the atom is on the order of about 1 to 10,000. That means most of the gold foil is void. Most of the gold foil is nothing. And you've got all this mass, highly dense, concentrated here. So this is the Rutherford model, which was called the nuclear model. This was proposed by Rutherford because it has such a thing is a nucleus. He coined the term, nucleus. So that's a big step forward from this thing here. So you'd think that the physics community would be ecstatic. That now we've got a model that squares with the data. Well, what do you think the reaction to Rutherford's announcement was? Condemnation. Derision. People didn't like it. They laughed at him. They said, well, this is crazy. There's a couple of problems with this. So strong negative reaction to Rutherford's model. And here the two major objections that people posed. First of all, they talked about nuclear collapse. They said if you've got a system of negative particles revolving around a positive nucleus, electrostatic forces are going to cause them to be drawn to one another. What's going to keep the electrons from collapsing into the nucleus? That was the first objection. Coulombic attraction. Coulombic, electrostatic, same thing. If you want to use a man's name, Coulombic. If you don't want to use a man's name, electrostatic. Coulombic attraction between plus and minus. And the second one was, OK, suppose for some reason we don't understand that the electron doesn't collapse, it stays in orbit, it's still not going to work. Because you run out of energy. If you take a charged particle, a charged particle that accelerates consumes energy. Acceleration, as you know, can mean either change in speed or change in direction. So if the electron is revolving at a constant speed, it has to change direction; otherwise, it will fly out of the room. So changing direction means acceleration, which means what keeps the energy in the electron? Why doesn't the electron just run out of energy and just stop? Even if you can prevent it from collapsing, it will just stop. So this is the problem of energy deficit. Energy deficit. Well, what powers the electron? What powers the accelerating electron? So that's a problem. So now the story continues. The next chapter is 1912. Because this is 1911. In 1912, a young Danish businessman by the name of Niels Bohr. Niels Bohr just finished his PhD in Copenhagen and he got a fellowship from the Carlsburg Brewery Foundation. Carlsburg Brewery put up money for science. So he won a Carlsburg fellowship. And he was a young physicist. And he knew that in the UK there was a controversy brewing between the model of Rutherford and the model of J. J. Thompson. So he says, you know what I'm going to do? I've got a year, I got money. I'm going to spend six months with J. J. Thompson, I'm going to spend six months with Ernest Rutherford. And that's what he did. So he goes and he spends six months at the Cavendish lab with J. J. and then he goes up to Manchester and he spends three months at Manchester with Rutherford. Why didn't he spend six months? Because he got a job offer and he went back to Copenhagen to start his teaching duties. And what he did is he developed a quantitative model for us. He developed a quantitative model. And that's all part of what I showed you the other day. We recognize patterns. That's what the Rutherford, Geiger, Marsden experiment gave us. The pattern of the distribution of alpha particle deflections. So now we need to develop a quantitative model that will explain the observations, and if it's a really good model, it will make predictions that, again, can be tested by experiment. So we ratchet back and forth between experiment, model. Experiment, model. So he proposes the model, not with J. J., because J. J., I don't know what he did there. This is where he wants to propose the model. Quantitative model. To Rutherford's atom. Let's take a look at what we've got there. Now, big lesson here. How do you announce a model? You don't put an ad in the newspaper. This is the scientific literature. This is not Google. This is not Wikipedia. This is a primary source. This is how science is communicated. So this is a journal. It's called Philosophical Magazine and Journal of Science. It goes all the way back to the 1600s. In fact, when I was a freshman at the University of Toronto we had copies of Phil Mag going back all the way to the 1600s. And these were not facsimiles, these were real copies. And I was in the stacks and I was reading something by-- I don't know-- Lord Kelvin, you know late 1800s, and I said gee, well, if I've got Lord Kelvin, then maybe I can go over here to the early 1800s and read Humphry Davy and, gee, if can keep going, and going and I pull out a leather-bound edition of Phil Mag from the 1660s, and I open the book and there's a paper about gravity by Isaac Newton. And that's how science is communicated. So I urge you to go to the library and read primary sources. Most of this stuff is available online. You can just sit at your desk and you can zoom in to history. So this is Phil Mag and this is London, Edinburgh and dublin. See this is 1913, so Ireland was not a free state yet. So Dublin was a British city. London, Edinburgh, Dublin. There's a close-up of it. July 1913, On the Constitution of Atoms and Molecules by Niels Bohr, Doctor of Philosophy, Copenhagen. And it was communicated by Professor Ernest Rutherford, FRS, Fellow of the Royal Society. So, let's read. It's in English. In order to explain results of experiments on scattering of alpha rays by matter, Professor Rutherford has given a theory of the structure of atoms. You see the dagger here? Down here, Ernest Rutherford, Phil Mag, duh-duh-duh-duh. There's a citation. See, attribution. According to this theory, the atoms consist of a positively charged nucleus surrounded by a system of electrons kept together by attractive forces from the nucleus. That's what keeps them from just fleeing anywhere, and they don't go in, but they vwomp. The total negative charge of the electrons is equal to the positive charge of the nucleus. Further, the nucleus is assumed to be the seat of the essential part of the mass of the atom and have linear dimensions exceedingly small compared with the linear dimensions of the whole atom. This is beautifully written. Clear. It's textbook quality and it's written by a man whose native language isn't even English. Read the literature. The number of electrons in an atom is deduced to be approximately equal to half the atomic weight. Great interest is to be attributed to this atom model. You have to know a little bit about Bohr. Bohr used this phrase, great interest. That was his way of saying, embroiled in controversy. If someone said something that he didn't agree with, instead of saying, I think that's nonsense, he say, very interesting. Very interesting. So great interest is to be attributed to this atom model. That means people are in pitched battle on both sides. For as Rutherford has shown, the assumption of the existence of nuclei as those in question seems to be necessary in order to account for the results of experiments on large-angle scattering of the alpha rays. Another footnote: Geiger and Marsden. Geiger and Marsden published their results in this journal. In an attempt to explain some of the properties of matter on the basis of this atom model, we meet, however, with difficulties of a serious nature arising from the apparent instability of the system of electrons. Difficulties purposely avoided in atom models previously considered. For instance, in the one proposed by Sir J. J. Thompson, which is to say, mmm, you know, it's curious. You condemn the Rutherford model because he's got electrons in motion, which means they're accelerating in energy deficit. But in the Plum Pudding Model, the electrons are in motion, and J. J. Thompson is silent about it. So this is a very aimed barb right at J. J. Thompson done elegantly. With this elegant slap in the face. You know this is how scientists do it. They don't go, oh you're ugly , and your mother dresses you funny. Instead they write things like this, difficulties purposely avoided in atom models previously considered for instance. So this is sort of yeah, how about you, huh? But anyway, we can go on and on. But down here, the result of the discussion of these questions seems to be a general acknowledgment of the inadequacy of classical electrodynamics in describing the behavior of systems of atomic size. So he's saying that you can't use classical electrodynamics down to subatomic dimensions. He's saying that those models, any more than you can use planetary models down to human dimensions, this, in many respects, is foretelling nanotechnology. He's saying the properties we know, of how things behave in the Newtonian world are not necessarily valid at atomic dimensions. This is very important stuff. Anyway, so what I did is I went through the paper and I've reduced the content of the paper to these things called postulates. So Rutherford's atom is correct. That's the first thing. Classical electromagnetic theory not applicable to the orbiting electron. And you're going to get-- the PDF and all this is going to be posted, so don't feel like you have to be a super stenographer here. Newtonian mechanics is applicable to the orbiting electron. So you can see, wait a minute it's sort of like cafeteria physics here. Electrodynamics doesn't apply, Newtonian mechanics does apply. And the answer is, yeah. That's what we're going to do here. We'll build a model and we're going to try these ideas out. And if they're crazy, how are we going to know they're crazy? Because there's going to be data that refute. But if we get data that supports all this, then there's only one conclusion. This thing makes sense up to a point. The energy of the electron is conserved in the system. It's bond is kinetic plus potential. The quantization through angular momentum. And lastly, the Planck-Einstein relation applying to electron transition. So now I'm going to take all this and I'm going to write out the model for you. Now, I don't want you to say, oh, he's going to derive something. Do we have to know derivations? This is the only time I'm going to derive something for you. The reason is, it teaches the model. It teaches how the model is based on assumptions and so on. But I don't expect you to memorize things because that's not the way things work. So let's go starting with this point here. This point here is Bohr model atom. So it's Bohr's representation of the Rutherford model. We can call it a nuclear model or a planetary model. Because you've got orbiting electrons. Now, first important point. The Bohr model is the simplest model. It works for a one electron system. So he has a single electron. Planetary model, single electron. And this orbits the positive nucleus. So you might say, one electron, well, what's that mean? Well, obviously it means atomic hydrogen. But it could mean helium plus. That's a one electron system. It could mean lithium 2 plus. It could mean roentgenium 110 plus. That's a one electron system. And this is all gas. Gas phase. No solids here. Gas phase. So it's an isolated nucleus with one electron around it. That's how we're going to begin. And here's what it looks like. I'm going to put the nucleus, first of all not to scale, because we're not going to draw 10,000 to 1. Over here is the nucleus with z positive charge. And then out here is the electron in orbit. This is the lone electron in orbit. And this distance is r. The distance from the nucleus. You're going to say, well, is that from the outside or the inside? It's 10,000 to 1; it doesn't matter. Forget about it. So now let's use this concept. We're going to say it's a conservative system and so the energy of the system is simply going to be the energy of the electron. You might say, well, why don't you consider the energy of the nucleus? Because the nucleus is, relatively speaking, stationary. To use Rutherford's colorful language, he says when an elephant has fleas, it's the fleas that do the jumping. So you don't care about that energy of the nucleus in this case. So the energy of the electron is the kinetic plus the potential. So the kinetic energy, we're saying it's 1/2 mv squared because it's Newtonian mechanics apply. See number 3. So that's 1/2 mv squared, where v is the velocity. So that's the kinetic, and then what's the potential energy? The potential energy here is electrostatic. Because that's the energy that's stored here is due to the Coulombic forces between them. So that's going to be q1, q2 over 4 pi epsilon 0 divided by r. So what I'm going to do here, we have to just for convention I'm going to always choose that q1 in my derivation, q1 will be the charge on the nucleus. so q1 will equal z, which is the proton number here, times the elementary charge. Remember e is the elementary charge. It's not the charge on the electron. So sometimes I use the word e with a minus sign, meaning the electron. But this is e meaning the elementary charge. z times e and then q2, which is the charge on the electron, is minus the elementary charge e. So we're always going to have minus e out here. What varies is z. And so we're going to make some substitutions. So this is going to be what? It's going to be 1/2 mass of the electron times velocity of the electron squared. And now q1 is ze and q2 is minus z, so that's minus ze squared. And then that'll be over 4 pi epsilon 0 r. And I'm going to call this thing equation 1. And by the way you can find all of this information by looking at your table of constants. So see, epsilon 0 sitting here. Permittivity of vacuum. There it is, 8.85 times 10 to the minus 12 farads per meter. Elementary charge 1.6 times 10 to the minus 19 Coulombs. Electron mass, 9.1 times 10 to the minus 31 kilograms. And what's this 4 pi epsilon 0 doing here? Well we know this is going to give us joules. And this is going to give us joules. And the epsilon 0 is the factor that renders electrostatic units on to the same plane as mechanical units. If I plug in-- this is going to be Coulombs, it's Coulombs squared, and this is going to be meters, and now this things is farads per meter. And I got farads per meter times meters divided by Coulombs squared. I don't care. This, with impunity is joules. Because this is an energy unit. And this is Systeme Internationale. And this is kilograms times meters per second, meters per second and how many apples in a bushel and-- forget it! Put this in kilograms. Put this in in meters per second. With impunity, it's in joules. So that's what that thing does there. Alright. Good. So let's keep going. Postulate 3. Newtonian mechanics applies to orbiting electrons. So orbiting electron, it's in a stationary orbit. So that's the thing. So postulate 3 implies that the sum of the forces acting on the electron must be 0. This is a force balance, right? If the forces are out of balance, the electrons are either going to move closer to the nucleus or move farther from the nucleus. So that's equal to, again-- it's a sum of a dynamic force and a Coulombic force, or electrostatic. Here I just noticed a nice parity here. So these are technical terms, but we can give them human terms. So the dynamic force, there's a Newtonian force, this is a Newtonian force and this is a Coulombic force, after the two people that are associated with the ideas. Charges interaction are Coulombic, blah, blah, blah, blah. So now lets substitute in. So we know that if you've got something orbiting, something tethered, then the dynamic force on that, that pulls the object away on a tether is mv squared over r. And the electrostatic is q1 q2 over 4 pi epsilon 0 r squared. Force goes as 1 over r squared. Energy goes as 1 over r. How do you know? Well, one way to think about it is, you know energy is the result of a force moving through a distance. You say, I'm working really hard, and I say I don't see any force moving through a distance. That's a joke. So here's energy. So force. So if this is 1 over r squared, the integral of 1 over r squared is minus 1 over r. If you get these backwards, you integrate 1 over r, you're going to get natural log of r, which makes no sense. That's one way. Or the other way is, you can just remember. So anyways we're going to plug into those values. And what do we get? What we get there is this. We get mv squared over r. q1 is z times e. q2 is minus z. So it's minus ze squared over 4 pi epsilon 0 r squared. And I'm going to call this equation 2. So now we keep going. Next one, energy quantized through its angular momentum. This is really critical. This is the major breakthrough by Bohr. Now what do we mean by quantization? I told you that in 1909, Millikan had figured out that electrical charge is quantized. Now quantization had already been annunciated by Max Planck in 1909. You see, the classical theory of radiation gave this prediction for intensity as a function of wavelength. They called this the explosion, the collapse of the theory. As you go to low wavelength, the intensity goes without abatement. But these are the spectra that were measured. Depending on the temperature of the gas, you get this distribution. So how to get the distribution to turn around at lower values of wavelength? And in order to do that, Planck suggested that light is composed of energy packets, or quanta. That light is a stream of individual energy bits. And the elementary unit of electromagnetic radiation is the photon. And you can say, well I don't understand, where goes he get that from? That's not the way to think about it. Say, if you make this assumption you get these curves. And then figure out what it means. You can't make everything anthropomorphic. If this were the case, and furthermore if the energy of one of these quanta was related to its frequency-- and wavelength and frequency are inversely related-- then the proportionality constant is h, which we now have designated the Planck constant in honor of Planck, then you get this. People grew to accept it. 1900, he gets a Nobel Prize for it. La de da! But who knows what light is anyways. It has no mass, it's kind of mysterious. 1900, the whole concept of action at a distance was strange to people. The notion that something could affect you, from me, without physical contact, this was a new idea. We take it for granted, but this was very new. So it's kind of mysterious, they said, OK, he's a physicist, who knows? And it worked! But it's light. Now what Bohr is going to say is the electron moving in orbit, he says, I'm going to apply the same concept of quantization to the electron moving in orbit. Well, that's different. And using this value he gives us mvr, which is the angular momentum is quantized n h over 2 pi. h is the Planck Constant and n is an integer. 1, 2 3 4, et cetera. So this is the quantum condition and this will be equation 3. And I have three equations and three unknowns, and what I'm going to do next day is come back and show you the solution of the equations and how they give us a set of equations that allow us to compare with data. So let's jump to the thing. And by the way the Planck constant is there. Now Niels Bohr eventually won the Nobel Prize. He was widely revered. He achieved the stature on a par with Einstein. If you go to Denmark, it's still not part of the European Union, you can still get the 500 Kroner note if you change a 100-dollar bill there. And there's Niels Bohr with this pipe. And you can see the concept of things rotating around it an so on. There's a young Niels Bohr. This is Bohr with Werner Heisenberg. Heisenberg was a post-doc who worked with Bohr in 1925 and annunciated the Uncertainty Principle, which we will study a few lectures from now. This is a chemical conference sponsored by chemical company Solvay. Brussels 1930, there's Bohr and Einstein. They loved hats. This is the Hamburg, this is Borsalino. Here's Bohr mixing it up with royalty. This is a young Queen Elizabeth of England. There's Prince Phillip. This is the royalty Danish family. Bohr loved music and here is with Louis Armstrong undoubtedly talking about the resonant structures within the tube here and how to get the various harmonics and overtones. Now last thing, a little bit about hydrogen. We've been studying hydrogen here, the one electron atom. But there are isotopes of hydrogen. The original was discovered by Henry Cavendish. There is one form of hydrogen that has a neutron in addition to the proton. That's the deuterium, from which we can get heavy water, D2O. This was discovered here in New York at Columbia University by Harold Urey. And then lastly, there's a second isotope of hydrogen that consists of two neutrons and a proton. That's tritium and it was discovered at the Cavendish lab by Ernest Rutherford. He was active right up to his death. In fact, it is argued that Rutherford did his best work after he got the Nobel Prize. Your book is showing this. It's nice to follow. We started with Dalton and now we're here and then later, in another week or so, we'll be down here. And here's the sort of the intellectual road map that takes you to modern atomic theory. So with that I'll dismiss the class and we'll see you on Wednesday.
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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: All right. Let's just do 10 more seconds. All right. Does someone want to tell me up here-- hello everyone up here-- how they got the right answer? Over-- everyone over there, I guess. OK, I'm coming over there. So for the answer we have an MIT chemistry bag. That's quite special. AUDIENCE: Well, since your delta H value acts as the y-intercept, you know that it's negative because it's negatively. And then you also see that you're ascending positively, so that you know your slope is positive. And so delta S also has to be negative to make it positive. CATHERINE DRENNAN: Interesting. That was a different answer than I would have given, but that works really well, which is always great to ask people. So you could also think about this in terms of your going from a situation with a negative delta G to a positive value. So you know that temperature is going to make a difference there. And so that would also tell you how the temperature changes is also going to tell you the answer to that. So that's great. All right. So a couple of announcements. As you can see, see what I am wearing on my legs. So I was pleased with the performance on exam 2. The average was 84.7, not quite as celebratory as 87.4, which was the average on exam 1. Yeah, those are real numbers. That's a little weird. But anyway, I was still very happy with the performance on the exam. I would like for exam 3 for the average to be back at 87, though. I like it to be very close to 90 because I want everyone to have an average around 90 in the class to demonstrate excellent knowledge of chemistry. So I try to write an exam where I feel like if you have excellent knowledge of chemistry, you can get a 90% or above. So this is my goal. I all want you to have this fundamental knowledge of chemistry that you can go out and solve the energy problem, that you can go out and be ready to tackle the next great challenge in health. There's so many challenges facing us, and we don't even know what they are right now. So the Ebola situation, I think, is a real case in point, that we don't know what the next challenge facing us is going to be. So we need to be ready with our chemistry knowledge. The only thing I can tell you about the challenges facing us is that knowing chemistry is going to be really important in tackling those problems. So everyone needs to have an excellent understanding of chemistry to go on and do well and solve the problems of the world. So 84.7, really good. You're getting there. Awesome. So celebratory tights, or leggings. Not really sure the difference between tights and leggings. Anyway, I was very pleased. But next unit, next exam, has thermodynamics, chemical equilibrium, and acid-base. We're already done with thermodynamics and moving on. So it's just for the rest of the semester, we're about halfway there, the rate seems to accelerate. So problem set 5, thermodynamics. It's a little shorter because you have fewer days to do it. So we wanted to make sure that you would be able to get it done by Friday at noon. But that's when it's due. It's all thermodynamics. We've already covered all the material on the problem set. You can do it already right away. All right. So also in my wardrobe-- I'm commenting a lot on my wardrobe today-- you'll notice maybe that I've worn this shirt once before, and my goal was really to have a different shirt every time. Now, this shirt is appropriate for today's lecture. But I just wanted to say that I did try to get a different shirt. And I ordered new shirts on thermodynamics and chemical equilibrium this year, and one of the shirts had to do with entropy. And I just thought I would share with you briefly, because it's a good review of thermodynamics, what happens when you do things related to entropy. So just to sort of show you what happened. In September, I was super organized. I ordered my shirts. And on October 3, they shipped. They went through Indianapolis to Massachusetts and to Newton Highlands about somewhere around October 8 or 9. That's where I live, so that's really good. But the shirts were about entropy, and then look what happened. Wait-- rerouted for delivery to a new address. Jersey City, Cincinnati, Ohio, Warrendale, Pennsylvania. And then yesterday they were in Springfield, Massachusetts, to which I say, entropy! This does not happen when you order T-shirts about kinetics. That's all I'm saying. Those arrive record time. Thermodynamics is not always your friend, but you can embrace it. And I'm going to try to calm myself down because we should be at equilibrium today. Today, the topic is chemical equilibrium. It's the calm part of thermodynamics. We'll get back to entropy. There will be some mention of entropy. But I'm going to try to calm down. I'm going to remember my reaction quotients and my equilibrium constants and think about chemical equilibrium and embrace the T-shirt I am wearing. Because I feel powerful in this T-shirt, like I can control thermodynamics with the power of the Green Lantern, with chemistry knowledge at my fingertips, and I will move forward. So chemical equilibrium is calming, but it's also a dynamic process. We can't forget that the reactions have not stopped. It's just the rate of the forward reaction is equal to the rate of the reverse reaction. So there's no net change, but the dynamic process is still happening. So my goal in life is to have equilibrium in my life. The amount of work-- the rate at which work comes into my office equals the rate at which I complete the work and it leaves my office. Most of you should also have this goal. The rate at which problem sets come in equals the rate at which you complete the problem sets and the problem sets get turned in. You don't want to get yourself in a situation where that equilibrium is messed up, where the rate of things coming in is just nowhere at all equal to the rate at which things are completed and going out. So goal-- chemical equilibrium. We want to be at this calm place where we're still working hard. It's still dynamic. But the net change is good. It's in our favor. The amount of work we're doing equals the work coming in. So I like chemical equilibrium. I feel like it brings hope to thermodynamics. So let's look at an example of a reaction, and we're going to talk a lot about this reaction in chemical equilibrium. So we have N2, nitrogen, H2, hydrogen coming together to form ammonia, and it has a delta G0 of minus 32.9 kilojoules per mole. So let's think about what's happening in the beginning of this reaction, when we're just starting this reaction. So we can plot concentration on one axis, and we can have time on our other axis, so time increasing over here. And we can think about what happens when we're starting our reaction. We can have hydrogen starting out. We have some hydrogen in a certain concentration, and it's going to come down and level out. So that's our H2. We'll also have nitrogen as a reactant. It comes down, levels out. And we're starting with our product. We have no product. We're just starting the reaction with the reactants. So product increases and then levels out. So we're decreasing our reactants, increasing our product, but the lines go flat. The reaction is still happening. The reaction in the forward direction is happening, and then in the reverse direction is happening. But there's no net change at equilibrium because it's reached a state where the rate of the forward reaction equals the rate of the reverse reaction. And let me also just point out some arrows and some nomenclature here. So we have this double arrow, and you can see this a couple of different ways. But this indicates the reaction's going forward, and the reaction is going back. And that's necessary if you're at equilibrium. The definition of equilibrium-- rate forward equals rate back at equilibrium. So you'll be seeing those arrows a lot. All right. So let's think about delta G now in terms of what's happening at different times in this reaction. So if we're just starting out this reaction, and you don't have a lot of products, so you're in this case of pure reactants. At this point, you're going to be moving in the forward direction. You need to make products. And so at this point, the forward direction is negative. It's spontaneous. You're spontaneously making your products. You don't have any at this point. And so your delta G is going to be negative. So up here with pure reactants, your delta G is negative. Now, if you started with pure products-- you could do that as well-- the NH3 will dissociate and form the reactants. And so that would be over here. And so when you have excess products, now the forward direction is positive. It's not spontaneous because it's the reverse direction that is spontaneous. So over in this part of the graph over here, our delta G is positive. So if you have pure reactants, you spontaneously go to products. If you have pure products, you're going toward reactants. And when you have the right amount of reactants to products, the right ratio for equilibrium, then you get to equilibrium and delta G equals 0 at that point. So delta G is going to change depending on your ratio of products to reactants. So delta G changes as the proportion of the reactants and the products change, and there's an equation to describe that. So we're back now to having a few more equations in this unit. We noted a lot of people forgot to bring calculators to exam 2. There weren't so many calculations. We're back to more calculations again. So delta G equals delta G0 plus RT natural log of Q. So let's think about these terms. So delta G is Gibbs free energy at some particular point, at some ratio of product to reactant. Delta G0 is Gibb's free energy under standard conditions, in their standard state. Q is our reaction quotient, products over reactants. And R is the universal gas constant, and T is temperature. So this equation describes how delta G will compare to delta G0 at a particular condition, a particular ratio, of products and reactants. So let's talk about this reaction quotient Q. So let's look at a made-up reaction-- A plus B goes to C plus D. And think about that reaction in the gas phase. When we think about that reaction in the gas phase, we're going to be talking about partial pressures. And so Q over here is products over reactants, but here we're writing it out in a more complicated long way. So let's go through this. So we have this P sub C. Well, what is that? That is partial pressure. So partial pressure of gas X, and this is partial pressure of gas C, and C is a product and D is a product. So we have products over reactants. Partial pressure of gas C over P reference, partial pressure reference, which is 1 bar. The reference is 1 bar, so we're dividing that by 1. And this is raised to the little c, which has to do with the coefficients of the reaction. Then we have partial pressure of gas D over our reference raised to the small d-- those are our products, products are C and D-- over reactants-- partial pressure of gas A over reference raised to little a, partial pressure of gas B over reference raised to little b. But 1, the number 1, the reference is 1. So most of the time you'll see the following expression-- Q, our reaction quotient, partial pressure of gas C to the stoichiometry little c partial pressure of gas D raised to the stoichiometry d, over reactants, partial pressure of A little a partial pressure of B little b. And you will note that because we had 1 bar and these are in bars, our units are going to cancel here. So Q does not have units. I knew this should be very exciting for you. You won't be losing one point for lack of units with Q. There are no units. So this is very exciting. So there's the expression that you'll mostly see for Q. And now we can kind of forget that the reference is there. We can ignore the reference. You just don't freak out later if you have no units. It's OK. All right. So we also could be talking about solution, and here we're going to talk about concentrations. In a lot of these problems, they're talking about things in the gas phase, and they're giving you concentrations. Don't worry about it. It's all OK. So here our concentration reference, C for concentration, is 1 molar. And you'll see this term-- C in brackets. That means concentration of. And so if you see that like this, you would express this in words as the concentration of C-- again, that's a product-- over the reference-- we'll get rid of the reference in a minute, but we'll keep it for now-- raised to the little c, concentration of product D over this concentration reference of 1 molar to the little d, concentration of reactant A over this concentration reference raised to the little a, concentration of other reactant B over our concentration reference of 1 molar raised to the little b. And then, again, this is concentration. And we can get this expression, which is usually the one you see, of just products over reactants. But I will make one point here, that you need to know how to balance equations to be able to do these. So in the beginning of the book there was something about balancing, limiting reagents, stuff like that. If you feel like you did not master that, you want to go over it. You need to be able to balance. And in a lot of these problems you may be thinking about limiting reagents again. All right. So that's Q. You can see it as partial pressure-- and I'm going to do a little partial pressure gas review on Friday-- or concentrations with these brackets. So at equilibrium, delta G equals 0. And it's a dynamic process-- rate of the forward reaction equals rate of the reverse reaction. Q now is the equilibrium constant, because the ratio of products to reactants at equilibrium is the definition of the equilibrium constant. So when delta G equals 0, Q equals K. So we now can go back to this expression that we had and rewrite this for the situation at equilibrium, when delta G equals 0. So when delta G equals 0, this is 0 and Q is K. K is products over reactants at equilibrium. And we can rewrite this or rearrange it now-- delta G0 equals minus RT natural log of K. And that's a very important equation that you'll be using a lot in these units. So it relates delta G0 with our equilibrium constant K, and it depends on temperature. All right. So K-- same expression as Q, except for something very important, which is that the concentrations are the concentrations at equilibrium. So it has the same form as Q, but it's only the amount of products and reactants at equilibrium. So in the gas phase, we would write K in terms of our partial pressures. And again, this is this little symbol to remind you these are the concentrations at equilibrium. And in solution it would be expressed in molar or something else. So K is, again, products over reactants, but only those concentrations at equilibrium. Whereas Q is at any point, any concentrations of products over reactants for Q. For K, it's those concentrations at equilibrium. All right. So now let's think about Q and K together. So we can rewrite this expression again. We just derived a new expression for delta G0, and that was minus RT natural log of K. So we can say delta G equals, and now have this other expression, minus RT natural log of K plus RT again-- gas constant and temperature-- natural log of Q. And we can simplify it, bring out the RTs, and now we have delta G equals RT natural log of Q/K. And this is, again, a very important expression that you'll use a lot. Because it tells you about delta G, whether the reaction is going to be spontaneous in the forward direction or the reverse direction, depending on the ratio of Q and K. This is a very important equation for chemical equilibrium. So let's think about what this means, what comes out of this. So the relationship between Q and K, if Q is less than K, what is the sign of delta G? You can just yell it out. AUDIENCE: Negative. CATHERINE DRENNAN: Yep. So it's going to be negative. Just mathematically, you can look at that expression. So it'll be negative. And again, we give you all the equations on the equation sheet. That will be there. So you just need to know how to think about it. And so the forward reaction will occur. And so if we think about this, it's going to mean that at equilibrium there are more products than there are at this time for Q. There's less products in the Q expression. K is greater than Q. So we need to make more products. So delta G will be negative, and the forward reaction will occur. So when Q is greater than K, delta G is positive. And so when Q is greater than K, that means there's more products now in Q than there are at equilibrium, too many products. So we need to shift it in the reverse direction. So delta G will be positive. So again, thinking about the ratio of Q and K tells you about what direction is going to be spontaneous. Is it spontaneous in the forward direction or in the reverse? OK. So let's continue doing a little example. We can do a little calculation here on the board for the same reaction. We're given a value of K. And now we're told some partial pressures of these gases and asked which direction the reaction will go. So let's write the expression for Q. So Q, again, is going to be equal to products over reactants, and our product here is our NH3. So we're going to be talking about the partial pressure because it's a gas of NH3. And am I done with the top part of this expression? AUDIENCE: No. CATHERINE DRENNAN: No. What do I need? AUDIENCE: [INAUDIBLE]. CATHERINE DRENNAN: Yep. Again, you need to remember the stoichiometry of the reaction. So now we have the partial pressure of N2 on the bottom, and I'm good, and the partial pressure of H2. And again, I have to remember that there are three H2s in this balanced reaction. So we have the partial pressure of H2 to the third. I can put in my numbers. 1.1 squared over 5.5 over 2.2-- these numbers may be made up, that's OK-- equals 2.1 times 10 to the minus 2. And we're back to thinking about significant figures a lot again in this unit. So I have two there, so I'm going to have these two here, and I'm good. But now this is the value of Q. And I want you to tell me with a clicker question, what direction is this reaction going to go? All right. Let's just take 10 more seconds. All right? So recognized, for the most part, that yeah, most of these numbers, that Q is a bigger number than K. And so then you have to think about what that means. And so when Q is greater than K, then you're going to go toward reactants. And so you're going to dissociate the product until you achieve equilibrium again. And so this means when you have a big Q number, you have too many products compared to the equilibrium state, and you want to dissociate your products. So you're going to go spontaneously in the reverse direction. Now, I just want to mention one point. For the first part of the course, I only taught that material once before. But the second part I've taught a lot. So I have lots of years of experience of what people do wrong on exams on this part, and I will share one thing. A lot of students-- and faculty too, I'm this way-- right-left challenged. So they write left when they mean right. They write right when they mean left. And so for those of you who are like me and have this issue, draw an arrow. When people draw an arrow, they always draw it in the direction that they mean, or they say toward reactants or toward products. And so I've seen so many times on the test they'd explain the answer beautifully, and then write the wrong direction down. So if you are right-left challenged, try toward reactants or toward products or draw an arrow. So that is my suggestion to you, as someone who's also very bad with saying the right direction, left or right, when I mean that direction. OK. So what does K tell us? So K tells us about the ratio of products to reactants at equilibrium. And if you have a very big number for K, it's going to tell you something about the ratio of those products to the reactants, and there'll be a lot of products for reactants. So let's just think about this for a minute. So when K is greater than 1-- K is products over reactants at equilibrium-- that's going to mean high products. So you'll have a higher number there. But if you have a small value for K, then you're going to have low products at equilibrium. So this is a good thing just to think about and check yourself when you're doing these problems. Does my answer actually make sense? And if you see a big value for K, you're like, great. If I want products, I want a big value for the equilibrium constant, means that this reaction is going to give me a lot of what I want, a lot of my products. Because again, K is products over reactants at equilibrium. All right. So let's do an example here. So here is an example where K is greater than 1. It's actually 6.84 at room temperature. We have a delta G0 of minus 4.76 kilojoules per mole. Two NO2 molecules going to N2O4. So we're going to start with 1 bar of our reactants and no products. So we can look at what that's going to look like in our plot of concentration versus time. And here the concentration is indicated as partial pressure, which is one type of concentration. So we're starting only with our reactants. It will go down. We'll have no products in the beginning. At time 0, no products, and that will grow up. The curves will level off as you reach the equilibrium. Again, the reaction's still going in the forward and reverse directions, but the rates are equal so there's no net change in the concentrations then. So now we can actually do some math and figure out how this changes. And so this is one way that you can set up these kinds of problems. And we will be doing these types of problems in chemical equilibrium, acid-base equilibrium. There'll be many examples, so it's good to become familiar with them if you haven't seen it before. So we can calculate the partial pressures at equilibrium using this information. So when we started, we only had our reactant. We had 1 bar, 1.000 bar, to be specific, and we had no product. As the reaction goes, we will form product, so it's plus X. Some amount of product is going to be formed. What is the change in partial pressure as you go to equilibrium? What happens to this? What changes? What do I put in this area? AUDIENCE: [INAUDIBLE]. CATHERINE DRENNAN: What? AUDIENCE: [INAUDIBLE] CATHERINE DRENNAN: I heard minus-- AUDIENCE: [INAUDIBLE]. CATHERINE DRENNAN: So I heard minus X. I heard somebody say [INAUDIBLE]. It's minus 2X because you have to remember the stoichiometry of the reaction. And so we have two of these going to one of those, so it's minus 2X. And then at equilibrium, we have 1.000 minus 2X is our concentration of our reactant, and our concentration of product is plus X. So now we're going to solve for this, and we need to write an expression for the equilibrium constant. So why don't you write that for me. All right. 10 more seconds. I'd like 90%. Oh, so close. OK. So again, the trick here, you have products over reactants, and you have to remember the stoichiometry. So we can go back over here. So we have partial pressure of our product over partial pressure of our reactant. And we have the stoichiometry there, and now we can continue to plug in. So the concentration, the partial pressure, of our product is X at equilibrium. The partial pressure of our reactant is 1 minus 2X, and that whole term is raised to the 2. So it's all that whole term is squared. And if you solve for X, you will get 0.38 bar. So you definitely want to remember calculators and things on this exam. And that value, then, is our product. So that's the answer to the partial pressure of the product because that's what X is. Another thing that I've seen on exams is that people solve for X. They're happy to solve for X. But then they don't remember what X was. So keep track of what things belong to what. So then we want to also find the partial pressure of our reactant. That was 1 minus 2X. So here we have 1 minus 2 times X. And you can see the significant figure fun that one can have in this because we have multiplication, division, subtraction, and pretty soon we're going to have log significant figure rules. So there's going to be a lot of fun. And of course, then this is our reactant over here. So let's just go back to the diagram, which is up above in your notes, on the same page in your notes, and plug these in and think about what this means. So we've done the math. And we see that our reactant at equilibrium is 0.238 bar, and our product at equilibrium is 0.381 bar. So K is greater than 1, more products, and you see that that works out. So if you were asked just to explain what you expected, you could say K is greater than 1. I expect more products. But if you do the math, and you'll often do the math, you can calculate what the partial pressures are at equilibrium of the reactants and of the products. All right. So now let's think about the relationship, again, between delta G0 and K. So here is our expression we saw before. We can also rewrite this to solve for K. So sometimes you will be given information about delta G0 and asked to calculate a K at a particular temperature. But we can also think about what we would expect. So if K is large, what is going to be true about delta G0? And so why don't you tell me what you think. All right. Let's just do 10 more seconds. So it would be a large negative number. And we can think about this, that if you really are lying on the side of products in your reaction, then that would be consistent with a bigger value and a negative value. So you can think about-- again, we talked about in terms of formation, is the thing that's being formed more or less stable compared to its elements, the same sort of idea. You can think about the relative stability and whether you'd expect more products or more reactants at equilibrium based on these values. So let's do an example and prove that this is true. So let's consider baking soda again. Baking soda works really well for this unit on thermodynamics and chemical equilibrium. So again, we have our baking soda going to CO2, which is very important, the CO2 gas, which helps our bread rise. And we calculated last week that the delta G0 at room temperature for this process was plus 36 kilojoules per mole. And that would mean that it's not spontaneous in the forward direction, which is really bad for our bread, because we need the CO2 gas to cause it to rise. But the good news was that if we remember to turn on the oven in baking bread and put it at a normal temperature in the oven, that the delta G0 is minus 15 kilojoules per mole. So then it becomes a spontaneous reaction. Now we can think about this in terms of our equilibrium constant K. So if we do the math here, at room temperature K would be 4.9 times 10 to the minus 7. That is a small number. That means that there is very little product at equilibrium at this temperature. Very little CO2 gas to be able to be used to make our bread rise. But now if we calculate K at this value, at a negative number for delta G, K is now 55. So we have quite a lot of product then to be used to make the bread rise. So you can think about things in terms of delta G's and whether something's going to be spontaneous and give you the product you want. But the equilibrium constant gives you that information as well. If it's a very small number, you have very little product. And if you want product, if you're trying to industrially make something, that's a very bad thing. But if you have a big value for K and this negative delta G0 value, then that's good if you want a lot of the product at equilibrium. So equilibrium doesn't just matter for things like baking soda or forming ammonia from nitrogen and hydrogen. Chemical equilibrium applies to large molecules as well, such as enzymes in your body that are catalyzing reactions. So I'm going to share with you an In Their Own Words. And this is Nozomi Ando, who I will just mention is an MIT undergraduate-- was a graduate of MIT, was an undergraduate here in physics, majored in physics. And she is now a professor of-- do you want to guess? AUDIENCE: Chemistry. CATHERINE DRENNAN: Chemistry, that's right. She's a professor of chemistry at Princeton University. So here is a good example of what can happen. And she was very happy-- actually, I think 5.111 didn't exist then. I think it was 5.11 when she was here. But nonetheless, so here in her words about chemical equilibrium and the proteins she studies. [VIDEO PLAYBACK] -My name is Nozomi Ando, and I study a protein called Ribonucleotide Reductase, or RNR for short. It catalyzes the reaction of converting ribonucleotides, or the building blocks of RNA, into deoxyribonucleotides, or the building blocks of DNA. It's the only means of getting those letters for DNA, so it's important for DNA synthesis and repair and replication. It knows, for example, when there's an imbalance in the pools of the letters for DNA or if there's a lot of letters of the RNA. And this controls the sort of state that RNR is in. And RNR can be in an equilibrium of active and inactive states that are sort of regulated by the alphabet soup in the cell. When it's active, it's very compact, but then it has to make a really dramatic structural change to go into an inactive state. I have this imagery of Transformers because it's just so dramatic. So when it's active, it's compact, like when a Transformer's a car. And when it's inactive and it makes a sound, che, che, che, che, and then it expands into a robot. The letter A, or adenosine, pushes this equilibrium from the active to inactive state. And it tells RNR, OK, we have enough of the DNA letters, so stop. For humans, it's really important to study RNR because it's the protein that is essential for making letters of DNA. So it's essential for DNA replication, which is essential for cells to divide. And we want RNR to function normally for our health. But in cells that are dividing too quickly, such as tumor cells, we want to slow it down. So actually, RNR is a really important target for anti-cancer drugs. But also, because RNR exists in every organism, we can start looking at differences between different species. So for example, it could be anti-bacterial and not just anti-cancer. [END PLAYBACK] CATHERINE DRENNAN: So that's an example of how you have a shift between two states, an inactive and active state, that it's just a chemical equilibrium. And binding one thing shifts the equilibrium one way. Binding something else shifts it back. So chemical equilibrium-- a lot of nature works by just suddenly shifting the equilibrium between different states. So understanding chemical equilibrium is pretty important. All right. So now we're going to apply stress to our chemical equilibrium, and we're going to talk about the principle of Le Chatelier. So here, a system in equilibrium that's subject to stress will react in a way that tends to minimize that stress. And whenever I talk to MIT students about this, I feel like I need to really emphasize this point. MIT students experience a lot of stress, but often do not tend to react in a way to minimize that stress. They say, all right, I am just struggling with this double major. I don't know what to do. It's just so much work. So maybe I should triple major instead. No. [LAUGHTER] Ask yourself, what would Le Chatelier do? Minimize. Double major, single major. So this is a principle that can apply to your life. And I recently saw a chemistry major, and we had lunch. He was a former MIT student, and now he's a CEO of a company, small company, in Cambridge that's designing computer software platforms. He was a chemistry major, and he's doing software design and building like little computer tablet things for restaurants. So I said, would you use your chemistry at all? And he goes, oh, I use some things all the time, especially Le Chatelier's principle. I'm all about minimizing the stress. So that was one thing that he really grabbed onto in chemistry. So again, if you think about this, about minimizing the stress, you can predict the direction of the reaction. And in nature, this really works pretty well. So Le Chatelier's principle gives us a way to predict the direction the reaction will go if you ask, which direction will minimize the stress? So let's look at an example. We're back to N2 and H2 making ammonia. So here's a slightly different plot than I drew over here. Now, this is a reaction sensitive to temperature, so the equilibrium constant's going to change with temperature. So every plot of this may look a little different depending on what temperature it's at. But some things are the same. If you start with hydrogen and nitrogen, you'll have some of those to begin with. They will be above 0. But ammonia will start-- if you had no product, you'll have ammonia at 0 and it'll rise in. The other thing that should look similar is that as the reaction runs for long enough it'll reach equilibrium, and you'll have the lines level out. You'll reach an equilibrium state. Now what happens if you stress that equilibrium state? So say you add hydrogen. And so this line adds the hydrogen, then it goes down. Then you're also going to use up some nitrogen as you're using up the hydrogen. You're going to shift it to make more product. Now, say, you make product. You add product-- sorry, you're adding product. It's going to shift to minimize the stress, use up the product, have the product dissociate, and make more hydrogen and more nitrogen. So here's the plot. Now, let's just think about what's happening at each step. So if we're adding more reactant, you have more reactant. You have too much reactant. Shift to minimize the stress, and you will shift the reaction toward product. Get rid of the reactant, use up the reactant. Let's get back to the equilibrium condition, minimize the stress. Now, we can think about this in terms of Q and K again. So when you have reactants added, then Q is going to fall below K momentarily. And so that means-- if you think about our equation over here, recall this equation, this important equation. So with Q less than K, you get a negative delta G, and that's going to be spontaneous toward the right, toward products. So again, you respond by making more products. You shift to the right. And you saw that over here. It's shifting toward products. You're using up the hydrogen. You're using up the nitrogen. You're trying to return to an equilibrium condition. Now what happens if we add more product? And so when you're adding more product, Q is greater than K momentarily. You have too many products now compared to equilibrium. And when Q is greater than K, you get a positive delta G. And that means that it's spontaneous in the reverse direction, or non-spontaneous in the forward direction. So you shift toward reactants. You shift to the left. So again, you can use this equation to think about what direction is now going to be favorable. And so here you added product. The product gets used up. It's shifting toward reactants, and you're making more reactants until you reach equilibrium again. And we have one minute left for a last clicker question. What happens when you remove products? All right. So let's just take 10 more seconds. All right. So we'll just put it up over here. So that means that delta G is going to be negative in that case because you remove products. So Q is less than K, and so the reaction is going to be spontaneous in the forward direction. Delta G will be negative, and you'll move to make more products. Again, minimize the stress. You took away products. You need to make more. Minimize the stress. Do problem set 5 and minimize your stress.
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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: Superposition is very unusual and very interesting. Now we've said about superposition that in classical physics, when we talk about superposition we have electric fields, and you add the electric fields, and the total electric field is the sum of electric fields, and it's an electric field. And there's nothing strange about it. The nature of superposition in quantum mechanics is very strange. So nature of superposition-- I will illustrate it in a couple of different ways. One way is with a device that we will get accustomed to. It it's called the Mach-Zehnder interferometer, which is a device with a beam splitter in here. You send in a beam of light-- input- beam splitter and then the light-- indeed half of it gets reflected, half of it gets transmitted. Then you put the mirror here-- mirror 1, you put the mirror 2 here, and this gets recombined into another beam splitter. And then if there would be just a light going in, here there would be two things going out. There's another one coming from the bottom. There will be two. There will be interference. So you put a detector D0 here and a detector E1 here to detect the light. So that's the sketch of the Mach-Zehnder interferometer-- beam splitters and mirrors. Take a beam, spit the light, go down, up, and then recombine it and go into detectors. This was invented by these two people, independently, in the 1890s-- '91 to '92 apparently. And people did this with light-- beams of light before they realized they're photons. And what happens with a beam of light-- it's interesting-- comes a beam of light. The beam splitter sends half of the light one way half of the light the other way. You already know with quantum mechanics that's going to be probabilistic some photons will go up maybe some photons will go down or something more strange can happen. If you have a superposition, some photons may go both up and down. So that's what can happen in quantum mechanics. If you send the beam, classical physics, it divides half and half and then combines. And there's an interference effect here. And we will design this interferometer in such a way that sometimes we can produce an interference that everything goes to D0 or everything goes to D1 or we can produce suitable interferences that we can get fractions of the power going into D1 and D2-- D0 and D1. So we can do it in different ways, but we should think of this as a single photon. Single photos going one at a time. You see, whatever light you put in here, experimentally, the same frequency goes out here. So what is interference? You might think, intuitively, that interference is one photon interfering with another one, but it can't be. If two photos would interfere in a canceling, destructive interference, you will have a bunch of energy. It goes into nothing. It's impossible. If they would interfere constructively, you would add the electric fields and the amplitude would be four times as big because it's proportional to the square. But two photos are not going to go to four photons. It cannot conserve energy. So first of all, when you get light interference, each photon is interfering with itself. It sounds crazy, but it's the only possibility. They cannot interfere with each other. You can send the photons one at a time and, therefore, each photon will have to be in both beams at the same time. And then, each photon as it goes along, there will be an interference effect, and the photon may end up here or end up there in a probabilistic way. So you have an example of superposition. Superposition. A single photon state a single photon is equal to superposition of a photon in the upper beam and a photon in the lower beam. It's like two different states-- a little different from here, you had photons in two different polarizations states superposed. Here you have photons in two different beams-- a single photon is in both beams at the same time. And unless you have that, you cannot get a superposition and an interference that is consistent with experiment. So what does that mean for superpositions? Well, it means something that we can discuss, and I can say things that, at this moment, may not make too much sense, but it would be a good idea that you think about them a little bit. We associated states with vectors. States and vectors are the same thing. And it so happens that when you have vectors, you can write them as the sum of other vectors. So the sum of these two vectors may be this vector. But you can also write it as the sum of these two vectors-- these two vectors add to the state. And you can write any vector as a sum of different vectors, and that's, actually, quite relevant. You will be doing that during the semester-- writing a state a superposition of different things. And in that way you will understand the physics of those states. So for example, we can think of two states-- A and B. And you see, as I said, states wave functions, vectors-- we're all calling them the same thing. If you have a superposition of the states A and B, what can happen? All right, we'll do it the following way. Let's assume if you measure some property on A, you always get value A. So you measure something-- position, momentum, angular momentum, spin, energy, something-- on A, it states that you always get A. Suppose you measured the same property on B. You always get B as the value. And now suppose you have a quantum mechanical state, and the state is alpha A plus beta B-- it's a superposition. This is your state. You superimpose A and B. And now you measure that property. That same property you could measure here, you measure it in your state. The question is, what will you get? You've now superimpose those states. On the first state, you always get A; on the second state, you always get B. What do you get on the superimposed states, where alpha and beta are numbers-- complex numbers in general? Well the most, perhaps, immediate guess is that you would get something in-between maybe alpha A plus beta B or an average or something. But no, that's not what happens in quantum mechanics. In quantum mechanics, you always get A or you always get B. So you can do the experiment many times, and you will get A many times, and you may get B many times. But you never get something intermediate. So this is very different than in classical physics. If a wave has some amplitudes and you add another wave of different amplitudes, you measure the energy you get something intermediate. Here not! You make the superposition and as you measure you will either get the little a or the little b but with different probabilities. So roughly speaking, the probability to get little a is proportional to the number in front of here is alpha squared, and the probability to measure little b is proportional to beta squared. So in a quantum superposition, a single measurement doesn't yield an average result or an intermediate result. It leads one or the other. And this should connect with this. Think of the photon we were talking about before. If you think of the photon that was at an angle alpha in this way, you could say that the polarizer is measuring the polarization of the object. And therefore, what is the possible result it may measure the polarizations say oh, if it's in the x direction you get it right, and what is the probability that you get it to be in the x direction is proportional to cosine squared alpha-- the coefficient here squared. So the probability that you find the photon after measuring in the x direction is closer in squared alpha, and the probability that you'll find that here is sine squared alpha. And after you measure, you get this state which is to say the following thing. The probability to get the value A is alpha squared, but if you get A, the state becomes A because this whole state of the system becomes that. Because successive measurements will keep giving you the value A. If you get B, the state becomes B. So this is what is called the postulate of measurement and the nature of superposition. This is perhaps the most sophisticated idea we've discussed today, in which in a quantum superposition the results are not intermediate. So when you want to figure out what state you have, you have to prepare many copies of your state in this quantum system and do the experiment many times. Because sometimes you'll get A, sometimes you'll get B. After you've measured many times, you can assess the probabilities and reconstruct the state.
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https://ocw.mit.edu/courses/5-08j-biological-chemistry-ii-spring-2016/5.08j-spring-2016.zip
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The following content is provided under a Creative Commons license. Your support will help MIT Open Courseware continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT Open Courseware at ocw.mit.edu. ELIZABETH NOLAN: What we'll do today is have an overview looking at the ribosome structure, and also an overview of translation to get everyone on the same page for the discussions we'll start next week on the elongation cycle of translation. So I'll post, within lecture notes, reading as it applies to a given module and information about the problem sets, so you have that here. So before we get into some more molecular level details about ribosome structure, it's important to appreciate how we've gotten to where we are now in terms of our understanding, and so where we'll start is back in some early studies of electron microscopy. And this is in the '50s, and this researcher, Palade, obtained images, that looked like this, of rat pancreas tissue. And what was seen in these images were a lot of dark spheres. You can see them throughout, and they were called the particles of Palade. And one thing the scientists questioned is whether these black spheres or dots were something real or an artifact from his methods, so the perennial and arduous question of artifact versus reality. So this is something that we all question everyday when doing our experiments as well. And so he was quite a thorough scientist and experimentalist, and he repeated these experiments using different procedures to fix the tissues. And he observed these types of features in many different types of samples, and what was determined later on is that these black spheres are actually ribosomes. So one of the things we're going to look at today is how did we get from an image like this, just seeing some black dots, to the crystal structures we have today and the atomic resolution and understanding. And so he received the Nobel Prize back in 1974 for this contribution. So just to keep in mind the hypothesis of translation, which is easy for us to take for granted these days. Goes back into the '60s, so there were studies during the '60s that resulted in the discovery that the 50S subunit of E. coli ribosomes catalyzes peptide bond formation. And it was discovered that the anticodon of the tRNA interacts with the 30S subunit, and that was important for translation. So this decoding problem-- effectively, how do we get from mRNA to protein-- was also articulated in the early '60s, and this was a puzzle for basically four decades. If we think about this from the standpoint of structure analysis and crystal structure-- so we'll look at images and data from crystal structures of the ribosome subunits today. If you take a look, what's important to appreciate here is that there was huge amounts of effort over many, many years to get where we are now. So in 1980, first crystals of the ribosome were obtained, but these crystals weren't of suitable quality for analysis. If we look at 20 years later, in 2000, the first crystal structure of the 50S subunit was reported, and since then, there's been a flurry of activity. So in 2001, first crystal structure of the 30 subunit-- 30S subunit with mRNA bound, and in this time, too, single molecule spectroscopy was well on its way, and so there were studies beginning of ribosome dynamics. And later, 2011, we have a crystal structure of a eukaryotic 60S subunit here. And so we're going to focus our discussions on the prokaryotic ribosome, their similarities and differences between prokaryotic ribosomes and eukaryotic ribosomes, just to keep in mind if you've heard about eukaryotic ribosomes in other classes. So also to note, in 2009, the Nobel Prize in chemistry was awarded for structural studies of the ribosome to these three researchers here. And their contributions are shown ranging from basically the first low quality crystals of the 50S ribosomal subunit to understand how important that was, to the first crystal structures. And something Professor Stubbe and I like to remind everyone and keep in mind is, with these types of problems and areas, there's often many contributors, and they can't all be recognized by this prize because it's limited to three individuals at maximum. And so other folks like Harry Noller, Peter Moore, and Joaquim Frank made really seminal contributions to our understanding of this macromolecular machine. So what are the questions we're going to address in this module? And then we'll go over some of the basics in ribosome structure. So first, one is that we learn from structural studies of the ribosome, and really, what does ribosome structure at an atomic level tell us about its function? How does the ribosome recognize, bind, and decode mRNA? How are amino acids recognized and delivered, and how is the correct amino acid delivered? The genetic message needs to be read, and it needs to be read properly. And what happens if a wrong amino acid is delivered? So that's a possibility. How does the ribosome cope? So this brings up the notion of fidelity. How is fidelity of translation maintained? And we'll address that next week. How is translation initiated? How does the ribosome catalyze peptide bond formation? So we're interested in that mechanism within the context of this course. How does the polypeptide leave the ribosome, and what happens to that polypeptide after it exits? So that will be a transition for us into module 2 on protein folding. How is translation terminated, and what happens to the ribosome after? So a given polypeptide chain is made. What happens after that? And where we'll close this module is thinking about how our understanding of the ribosome, from all of these basic and fundamental studies, allows for the development of new technologies. And we'll specifically think about how it's possible to use the ribosome to incorporate unnatural amino acids into proteins. So where we're going to move forward today is really structure-- focusing on ribosome structure and a general overview of translation, basically to have everyone here up to speed for the discussions to come next week. So first of all, we'll do an overview of key players in translation, a brief look at the cycle, and then we'll go into structural studies of the ribosome. And within this set of lecture notes are several tables that have lists of the players and detailed overall cycle that I encourage you to use, just as a reference throughout this module for keeping everything straight. So first, of course, we have the ribosome. So the ribosome, as we all know, reads the genetic code via the mRNA, and it catalyzes peptide bond formation. So in addition to the ribosome, we have the mRNA. So this mRNA delivers the genetic code to the ribosome, and it provides a template-- [AUDIENCE MEMBER SNEEZES] Bless you-- for protein synthesis. So effectively, we can think about this process as a template-driven polymerization. So somehow, the amino acids need to get to the ribosome, and so we need the help of the tRNAs. So these transfer RNAs deliver the amino acid monomers, to the ribosome, and they transfer the amino acids during synthesis of the polypeptide. So in addition to the ribosome, the mRNA, and the tRNAs, the ribosome needs some help, so we have translation factors. And there's translation factors that are involved in each step of the translation cycle. So these are proteins that are required at specific points during the translation process. And so in terms of translation factors, we can break the process of translation into three or four steps-- I prefer three-- which are initiation, elongation, and termination. Some review articles and papers will divide this into four steps, because termination, you can think about peptide release and then ribosome recycling. But regardless to that detail, at each of these stages, there are translation factors that help. So we have initiation factors that help with the process of initiation, and in prokaryotes, we have initiation factors 1, 2, and 3, so 3 translation factors that help during elongation. So the process of making the peptide bond-- there are elongation factors. EF for Elongation Factor. IF for Initiation Factor. We have EF-Tu, EF-G, and others, and we'll spend quite a bit of time thinking about EF-Tu and EF-G over the course of the next week and in recitation and in problem sets, thinking about how these factors are really facilitating the elongation process here. And during termination, there are release factors, so we have release factors 1, 2, and 3. And we can also-- these are involved in release of the polypeptide that's been synthesized from the ribosome, and there's other players as well that I'll list here, including ribosome recycling factor, so the subunits get recycled, as we'll see. And we can also include a protein called trigger factor here. That is involved in folding of nascent polypeptide chains or the polypeptide chain as it's coming off the ribosome. And then just to summarize in terms of three stages of translation as I'll present them to you within this course, we have the initiation process; two, elongation; and three, termination. And where we'll be focusing the lectures next week, and really, this whole module, is here on elongation. And I'll just note that the elongation cycle is highly conserved. Termination and initiation vary quite a bit between prokaryotic and eukaryotes there in terms of the processes and involved players. So where are we going? Just as a brief overview of the cycle, and we'll come back to this later within today's lecture, or if not, on Monday. So we start with initiation, and we're going to have to ask ourselves, how is it that this 70S prokaryotic ribosome or initiation complex is assembled? And so there's a special tRNA involved, the initiator tRNA that we see binds here, and we'll talk more about these E, P, and A-sites in a moment. So we see the ribosome is assembled, the mRNA is bound, and there's an initiator tRNA bound. In order for the elongation cycle to be entered, an amino acid needs to be delivered, and that's delivered by an aminoacyl tRNA. That's in a ternary complex, so three components. We have the tRNA, the elongation factor Tu, and GTP. So this complex here somehow delivers an aminoacyl tRNA to the ribosome, and we're going to look at this process in detail next week. So this will be one of our case studies thinking about experiments and how experiments have supported a specific kinetic model here. So here, we have a complex where the tRNA is ready to occupy the A-site. What happens here-- we see that there's a GTP hydrolysis event. We'll talk about more as we go forward. Peptide transfer reaction-- so we have formation of a peptide bond, and then this elongation factor G comes in to facilitate the elongation cycle. And then this cycle will continue until some point that signals to stop synthesis, so a stop codon will enter the A-site. And there's a termination process, ribosome recycling, and you can imagine this whole cycle happening again. So how do we get to this cartoon to some more detailed understanding? That's where we're going. So come back to this cartoon at various stages throughout the course. So first, we'll do a cartoon overview of the prokaryotic 70S ribosome, and then we're going to look at some of the data from crystallography studies here. So as I think we all know the ribosome is comprised of RNA and proteins, and by mass, it's about 66% RNA and about 34% protein. And it's comprised of two subunits, and those are indicated in the cartoon by different colors. So in prokaryotes, we have the 50S, which is the large subunit. This is made up of 23S ribosomal RNA, a piece of 5S ribosomal RNA, and proteins. In terms of size this is huge, so it's approximately 1.5 megadaltons. And what we find within this subunit is the catalytic center, or peptidyl transferase center. This is sometimes abbreviated as PTC. And what we also find in the 50S subunit are three sites for tRNA binding. And so the other subunit in prokaryotes is the 30S. This is a small subunit. It's comprised of 16S rRNA and proteins, and it's also quite large. Just smaller than the 50S, so on the order of 0.8 megadaltons. And in terms of function, what we have in the 30S is the decoding center, so for decoding the mRNA, and the site of mRNA binding. So if we draw this in cartoon form-- and this is something I really encourage you all to do when thinking about the experiments and the problem sets because that's going to help you understand the experimental design and what actually happened. Here, what we have on top is the 50S subunit. On bottom, the 30S subunit. We have the mRNA, and note the directionality, so 5 prime end, 5 prime end of the ribose, 3 prime end here. And then within this 50S, we can think about these sites for tRNA binding ordered as such, so E, P, and A here, so this is the catalytic center or peptidyl transferase center here. So overall, this assembled ribosome is on the order of 2.3 megadaltons and is about 200 Angstroms in diameter. So just in terms of these names, 50S, 30S-- this is overall the 70S assembled ribosome. What do these numbers-- where do they come from? What does this 50S, 30S, 70S mean? So what is the S? AUDIENCE: It's [INAUDIBLE]. It has to do with the sedimentation [INAUDIBLE].. ELIZABETH NOLAN: Yeah, it has to do with the sedimentation. So there's a type of experiment called analytical ultra centrifugation, and effectively, you can use this to ask about the sedimentation of a biomolecule. So effectively, what is the rate at which a biomolecule moves in response to the centrifugal force in a centrifuge there? And so you can use optics to monitor the sedimentation and use mathematics to fit those data to come up with an S value. So typically, the larger the S value, the larger the size. It's not always directly proportional to the mass because things like shape play a role as well, but effectively, we see 50S, and that subunit is larger than the 30S. And note, when they come together, it's not additive. It's 70S there, if you're to look at the assembled ribosome in one of these experiments. So that's where those values come from here. So if we take a look from my cartoon depiction to actual image from cryoelectron microscopy-- so this is just rotated basically 90 degrees. What do we see? We have the 50S here, the catalytic center. We have the 30S. Here's the mRNA, and what we're seeing is that, in this particular structure, there's some tRNAs bound, and they've indicated also a ribosomal protein here. Just as a sense of complexity-- so in E. coli, the 50S subunit has over 30 proteins associated with it. That's a lot-- 30 different proteins. And the 30S has 21 ribosomal proteins associated with it, so we need to think about the proteins in addition to the RNA. So let's take a look at an image from the crystal structure reported in 2000, of the 50S ribosome from a particular prokaryote shown here. So this is what's described as the crown view, and in this particular depiction, what we're seeing is that the ribosomal RNA of the 50S is in gray or white, and the ribosomal proteins that are bound are in gold. So taking a look at this, what do we see? We can ask ourselves some questions from this structure. So the first question I'll ask is about the RNA. What does this RNA look like? So do we see any obvious domains? If anyone has some experience looking at structures. I don't see any. What I see is a compact mass of RNA here. There's not obvious domains or regions that are somehow different here. To me, in this structure, it looks like one big glob of RNA. But then the question is, is that truly the case, or is there an organization we're just not seeing at this level? The next question we can ask is where are the proteins? So if we look at the proteins and how they're arranged on this compact mass of RNA, what do we see? Where are they? AUDIENCE: The edges? ELIZABETH NOLAN: On the edges, yeah. There's many on the edges, like L1 here, this one on the outside here, over here. So it looks like these proteins, at least in this view, are mostly on the outside. Is there anything unusual or potentially unusual we can see in addition about these proteins? Maybe looking at this one here or here. What's going on? AUDIENCE: I can't see very well, but I think that there's not just [INAUDIBLE]. ELIZABETH NOLAN: It looks like there's some unfolded parts? AUDIENCE: Yeah. ELIZABETH NOLAN: Right, so look here. So it looks like there's some unfolded regions to these proteins. And why is that? And where are these unfolded regions going? So what we can do is look at the RNA separately and look at the protein separately now and see what we learn from these analyses. So effectively, if we consider the 23S rRNA, despite that structure we saw before that looked like a compact mass of RNA, it's structured, and it consists of six domains. And these domains have quite complicated shapes, and they fit together. And here is just a schematic diagram of this structure. So if we take a look, we can see that there's domain 1, domain 2, 3, 4, 5, and 6. And on the left here, it's indicated where, in that crown view we just looked at, right here, these domains are located. So there is organization, even though in that structure, it looks like one compact mass of RNA. So let's think about these proteins a bit more. And in addition to the crown view and the observations we had from this particular face of the ribosome, where it looks like many proteins are on the outside, and there's some unfolded regions, what happens if we look elsewhere? So here, we have rotation, so 180 degrees from here, effectively looking, we can say, on the backside. And here, we can look at the view from the bottom of this subunit. So what do these images suggest? Do they support what we were thinking from this one view here, that proteins are mostly on the outside? Yeah, I see some shaking heads "yes." It looks like the surface of this 50S is covered, effectively, by a protein lattice here. So what might a role be for these proteins, an important role? AUDIENCE: Structural? ELIZABETH NOLAN: Yeah, so some structural role. So these proteins can help with stabilizing this 3D structure of the RNA. And they have other functions as well, and some of those will come up as we discuss this elongation cycle. But one function is certainly structural. If we just think about the distribution of the proteins along the surface of this 50S, it looks more or less uniform. There aren't patches where there's no protein or patches where there's a lot of protein. They're pretty much evenly distributed here. So as it turns out, most of the segments of the 23S do interact with protein, and if we look at these proteins more closely, we're going to follow up on the observation that it looks like they have some unfolded regions. So what we're looking at here are just a selection of the 50S proteins in the absence of the RNA. So these structures have been taken out of that total structure. In terms of nomenclature, l means large and s means small, in terms of thinking about ribosomal proteins. And so what's found in the 50S is that we can categorize 17 of the proteins as globular or folded and 13 of the proteins as cases where there's extensions that are non-globular or have no clear structure. And that's color coded in these examples, where we have folded regions in green and then unfolded regions in red. So why is this, and where are these red extensions going? So what's seen is that these non-globular extensions work their way into the interior of the ribosome, so we can think about them kind of like tentacles, for instance, going into the interior. So how might they interact with the RNA? So I'll give you a hint. In these regions in red, there are quite a number of arginine and lysine residues compared to other regions. So what properties of arginine or lysine would be important? AUDIENCE: [INAUDIBLE] ELIZABETH NOLAN: Positive charge. Right, we have positively charged amino acids. What about PKAs? So who votes for arginine having a higher PKA than lysine? The opposite? So that's a point for review. Lysine around 10.5. arginine around 12.5. arginine's higher here. So if we have a bunch of positively charged residues in these extensions, how are they going to interact with the rRNA? What are the molecular features there that are important? AUDIENCE: [INAUDIBLE] ELIZABETH NOLAN: Pardon? AUDIENCE: Phosphates? ELIZABETH NOLAN: Yeah, the phosphate backbones. So we have the negatively charged phosphates, positively charged amino acids-- effectively formation of salt bridges here. AUDIENCE: So I know structure for a lot of these, well, non-globular regions. Does it mean that they're more disordered, or do they still have relatively similar B factor compared to the rest of the globular region? It's just that they don't fall under [INAUDIBLE]---- ELIZABETH NOLAN: I don't know what the B factors are for the different regions of these proteins, and for the case of discussion here, I would have it fall under a lack of secondary structure. And keep in mind, the ribosome is quite dynamic, and in isolation, are all the proteins there and in their native way or not is just something else to keep in mind. But these are certainly lacking a fold and going into the interior and working from salt bridges here. Here's just an example of the 50S with tRNAs bound. So we have the 50S. We see tRNA in the E-site, the P-site, and the A-site. And so what are these three sites? Effectively, their names indicate what they bind or what they do in terms of these letters. The A-site binds aminoacyl tRNAs with the exception of initiator tRNA, which cannot bind to the A-site. The P-site binds the initiator tRNA during the initiation process of translation, and then it also binds peptydil tRNAs, so effectively the tRNA that has the growing peptide chain attached. And then the E-site binds the DA slated tRNA, and this is called the E-site because it's the exit site. And eventually, this tRNA that has lost its amino acid needs to get kicked out of the ribosome. So one more point-- just going back about these proteins to highlight. We stated that these proteins are mostly on the exterior, and there's just these extensions that go in. One thing I didn't explicitly say is that this peptidyl transferase center is devoid of protein. So in this catalytic center that's responsible for peptide bond formation, there's no protein. So based on all of the structural evidence, the nearest protein is 18 Angstroms away. That's quite far when thinking about making a peptide bond in a catalytic center. And also we'll learn that magnesium ions are important for ribosome assembly. I'll just point out that the closest magnesium ion is 8 Angstroms away. So if there's no protein in this catalytic center that's responsible for formation of peptide bonds in this growing polypeptide chain, what does that tell us right off the bat about the ribosome and catalysis? AUDIENCE: [INAUDIBLE] ELIZABETH NOLAN: Pardon? AUDIENCE: You have many functional component of [INAUDIBLE] ribozymes. ELIZABETH NOLAN: Yeah, so the ribo-- the ribosome is a ribozyme, yes. So there's many functional components, but in terms of peptide bond formation, it's the RNA that's catalyzing that reaction. So it's a ribozyme, or an RNA based catalyst. And so this is something many of us may take for granted right now, but it was a big surprise to see this here. And to the best of my knowledge, the ribosome is the only natural ribozyme that has a polymerase activity. So many of these natural ribozymes are involved in RNA maturation here, so for those of you interested in evolution and hypotheses about RNA world, this observation that there's no protein in the catalytic center of the ribosome supports an RNA world hypothesis, the idea that the RNA, which stores genetic information, can perform chemical catalysis predates DNA and proteins. One thing I'll just, though, point out is that, prior to this structural study, roughly two years before, there was some experimental work done just looking at isolated 50S rRNA with no proteins. And it was found that isolated 50S rRNA could catalyze peptide bond formation, and that, specifically, domain 5 was important for that reaction here. So if you're curious about that, I can point you in the direction of a paper. One last observation about the 50S subunit involves a peptide exit tunnel. And so somehow, the growing polypeptide chain needs to get out of this macromolecular machine, and in order for that to happen, there's an exit tunnel in the 50S subunit. So here, if we go back to that cryo-EM image, what's shown in this particular depiction is a polypeptide chain emerging from the 50S here. If we look at this view, a top or bottom view, what we see is that there's a hole here, and that hole is this exit tunnel. This is just another view of the same thing rotated, and a macrolide is a type of antibiotic that can bind in the region and is thought to block exit of the polypeptide. So there's some features about this exit tunnel that are interesting and that we need to consider. First of all, it's long, so approximately 100 Angstroms. And the diameter is relatively small, so the diameter is on the order of 15 Angstroms. So what we need to think about, from the perspective of this diameter, is what can fit, and so this week in recitation, you're looking at using PyMOL and ubiquitin as an example. If you just ask yourself, would ubiquitin, folded ubiquitin, fit in this exit tunnel based on its size? And so where does protein folding occur? We think about this as primarily and predominantly happening after the polypeptide comes out of the ribosome because there just isn't room in this exit tunnel for some folded structure to exist here. Also, the exit tunnel not shown in these images is lined with hydrophobic residues, just as another feature. So it's narrow, and it cannot accommodate folded proteins. So briefly on the 30S, similar to the 50S as said before, this 30S is comprised of RNA and proteins. It has the sites of mRNA binding and decoding. Here's just a structural overview of the 30S with different regions named, and similar to what we saw for the 23S rRNA of the 50S subunit, the 16S rRNA also has structure. And I just show you the domain organization here, so we see that there are four domains, and they're color coded in green, yellow, blue, and red here. And so another point just to make in passing about 16S-- 16S rRNA is highly conserved amongst species, so sequencing the 16S is commonly done in studies of, say, the microbiome to figure out something about the distribution of different types of prokaryotic organisms there for that. So why spend so much time on the individual subunits? What we find is that the structures are very similar when the ribosome is assembled. So we can think of the 30S and the 50S as coming together to give the 70S, and these subunits basically look the same as they do in isolation. And that's depicted here, in just another example. So if we're looking at this structure based on the cartoon and our discussions, you should be able to identify the different components. So here, what do we have? AUDIENCE: 50S. ELIZABETH NOLAN: Yeah, and here? AUDIENCE: 30S. ELIZABETH NOLAN: 30S, right. What's this? AUDIENCE: [INAUDIBLE] AUDIENCE: [INAUDIBLE] ELIZABETH NOLAN: Yeah, we have a tRNA bound here. Here, a protein. So bring yourself back to this cartoon and its simplicity as we work through problems next week. So another point to make, just to think about, is how is it that these subunits actually come together, and what mediates that interaction there? And so these subunits basically come into contact at about 12 positions, and magnesium ions are really important for mediating the interaction between the 30S and the 50S. So there's bound magnesium ions that mediate interactions between these subunits here. And so in week four recitation, we're going to think about how to purify ribosomes. And if you're interested in purifying ribosomes, how do you get an assembled 70S prokaryotic ribosome? And based on the need for magnesium ions here, we'll see how that's an important variable in these procedures. So we'll close just with some overview points about the translation process as a whole. So during translation, mRNA is read from the five prime to the three prime end. Polypeptides are synthesized from the N terminus to the C terminus, so there's directionality. As I said earlier, translation factors are required at each stage-- initiation, elongation, and termination. Something that I haven't highlighted yet is the importance of GTP. So in that initial overview of the cycle, we saw that there were some instances of GTP hydrolysis by certain translation factors, and in this translation process, GTP hydrolysis provides a means to convert chemical energy into mechanical energy. And so we're going to think a lot about how GTP hydrolysis plays a role next week. And although we're going to look at many structures, keep in mind that conformational changes are essential for catalysis by the ribosome, and that this is a very dynamic system here. So just some additional facts-- so ribosomes will synthesize six to 20 peptide bonds per second. The error rate is less than 1 in 1,000, which brings up fidelity again. How does the ribosome maintain this? And the rate accelerations are on the order of 10 to the 7-fold, so less than many enzymes, but quite good. And in all living organisms, these ribosomes carry out protein synthesis, so all ribosomes contain two subunits that reversibly associate during the translation cycle. Protein synthesis occurs through the binding of the aminoacyl tRNAs to the 70S ribosome in an order dictated by the mRNA. And next week, we're going to dissect how this actually occurs, and we think this will be quite new for all of you, even if you've learned about the ribosome in other courses. These tRNAs move sequentially through these three ribosome binding sites, as we saw before here. So we can return to our overview cycle here, that we saw before. And so we'll briefly address how initiation occurs. So how is this 70S ribosome assembled? We'll have a detailed case study of EF-Tu and then look through this elongation cycle in more detail. In terms of the players and the outcomes-- so this is a reference slide for all of you, where the stages are listed, that all of the players are listed, so some more detail than what's up here, and then the outcome. So what you can see from this overview, and go back and study it outside of lecture, is that, in each case, we see GTP, which means GTP hydrolysis occurs at each step. In initiation, our outcome is assembly of the 70S with mRNA bound and with an initiator tRNA in the P-site. The outcome of elongation is synthesis of this nascent, or new, polypeptide chain, and termination is the hydrolytic release of the peptide, release of the tRNAs and mRNAs and dissociation of the 70S and, ultimately, recycling. So there's many factors that need to be taken into account and dealt with it at every stage here. This is just another reference table. It has some additional players, like EF-Ts, and this is a nucleotide exchange factor for EF-Tu. So EF-Tu is a GTP-ase that we'll hear more about in lecture next week, and in recitation next week. Briefly, some topics for review-- if you need to review the genetic code, please do. We're not going to spend much time on it here. But in brief, I think we all know this genetic code is based on codons. They're read sequentially from a fixed starting point, and the code, which is a triplet code, is degenerate and non-overlapping. So why do we have a triplet code? We have four bases. AUDIENCE: We need enough combinations to [INAUDIBLE].. ELIZABETH NOLAN: Exactly, there needs to be enough combinations for all the amino acids. So we have 20 proteinogenic amino acids, and what else do we have? We have selenocysteine. We have pyrrolysine. So a triplet code with four bases covers everything we need here. We have start codons and stop codons we have to keep in mind, listed here. And as a reminder, in translation, the amino acids are delivered by the aminoacyl tRNAs. So the mRNA does not recognize these amino acids directly. We need the tRNAs that allows this reading to occur. Throughout this course, we're going to refer-- well, throughout this section with the ribosome, we'll be referring to nucleotides, et cetera, by the letter abbreviations. There are structures, chemical structures, associated with these abbreviations, and it's important to know those and be thinking about those as you work problems. So just as review, we have the DNA bases, C, G, A, and T. In RNA, we have uracil instead of thymine. The purines, A and G, have two rings, and the Pyrimidines, one ring. For nomenclature, nucleoside versus nucleotide-- so the nucleoside is a base plus a sugar, so there's this glycosidic bond here between the base and the carbon here of the ribose. And then the nucleotide is this nucleoside with one or more phosphate groups attached at the five prime carbon. So we go one prime, two prime, three prime, four prime, five prime for the numbering of the ribose. And keep in mind, from 5.07 or 7.05-- I think this should be known, but these phosphates, we have alpha, beta, and gamma phosphates. And depending on whether your ATP or some other nucleotide is being hydrolyzed to, say, an AMP or and ADP, you're going to have attack at different positions. So if you need to review, visit your basic biochemistry textbook for these details. Also just to keep in mind, the Watson-Crick based pairing, so G and C. We have three hydrogen bonds here, A and T, two hydrogen bonds. And after spring break, Professor Stubbe will be presenting a module on nucleotide metabolism, where we'll be thinking about these things in some more detail. So where we'll begin on Monday is briefly looking at an overview of initiation, and then we're going to begin to ask how did these amino acids get attached to tRNAs, and how did those aminoacyl tRNAs get to the A-site of the ribosome. So we'll see you then.
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https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/8.01sc-fall-2016.zip
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Newton's third law states that forces always come in equal and opposite pairs. One way we can write that is if you imagine two objects, object one and object two, the force exerted by object one on object two is equal and opposite to the force exerted by object two on object one. This makes explicit that real forces always arise from a physical interaction. For any force in a problem, you should always be able to identify the other member of the interaction pair. Newton's third law is the most subtle and sometimes the most confusing of his three laws of motion so I'd like to do an example that will help clarify how to think about it. So I'm going to pick a very extreme example. Let's imagine the collision between a marble moving this way and a train moving that way. And so we'll say that the mass of the train-- I'll write that as "m sub train"-- and the mass of the marble is m sub marble. And obviously, the mass of the train is much larger-- I'm going to write that as several greater-than signs-- much, much larger than the mass of the marble. And so the question I want to consider is, at the instant that these two objects collide, which experiences the greater force? So think about that yourself for a moment. In terms of outcomes, clearly, the marble will be smashed by the train whereas the train will not be noticeably affected by the marble. So your intuition might therefore suggest that it's the marble that feels the greater force. But that's incorrect because Newton's third law tells us that forces come in equal and opposite pairs. And what that tells us is that each object will exert an equal but oppositely directed force on the other. Now I chose a very extreme example to capture your attention. But that might seem like a surprising result but Newton's third law tells us that the forces on each object are going to be equal and opposite. But just because the forces are equal doesn't mean that the motion will be equal. The accelerations of the two objects are vastly different because of their different masses. And the accelerations and the forces are related by Newton's second law, F equals m a. So if I write that in terms of the acceleration, the acceleration of the marble, "a marble," is equal to the force divided by the mass of the marble whereas the acceleration of the train is equal to-- so since I wrote "F" for the force acting on a marble, I'm going to write minus F for the force acting on the train. So that's minus F divided by the mass of the train. And so if I want to look at the ratio, how big is the acceleration of the marble divided by the acceleration of the train? And let's take the absolute value so we're just talking about magnitudes here. That's going to be equal to the mass of the train divided by the mass of the marble. But the mass of the train is much, much, much larger than the mass of the marble so the right-hand side here is a very, very big number. And that tells us that relative to the acceleration of the train, the acceleration of the marble is going to be enormous. Even though the force experienced by each object is identical, because of their different masses, their accelerations will be very different. So that gives us an example of what we mean by Newton's third law, in terms of the interaction pair and equal and opposite forces acting. I want to reiterate that for any force in a problem, you should always be able to identify the other member of the interaction pair. So forces always come in pairs. It's important to keep in mind that these force pairs don't both act on the same object. They never act on the same object. The interaction pair always involves a pair of objects, two different objects. And let me just make that explicit with an example. The example I'll consider is, imagine a person standing on the ground. What are the forces acting on this person? I'll draw the force diagram here, say. There's gravity, mg, acting downwards and there's a normal force upward exerted by the ground. And those two balance to give a net force of zero, which is why the person is standing on the ground and not sinking down into the ground, for example. Now, you might look at gravity and the normal force and wonder if those are an interaction pair. And they're not because notice that these two forces are both acting on the same object, the person. The interaction pair always comes from realizing what is exerting the force on the object. So let's look at each of these in turn. Gravity is exerted by the Earth. So if the Earth-- and this means really the entire Earth-- exerts a force mg on the person, Newton's third law tells us that the person exerts a gravitational force on the entire Earth of mg upwards. Now, that might seem remarkable to you if you're just standing around on the floor that you are exerting a gravitational force on the planet but you are. However, this is like the example we just talked about a moment ago. The masses are extremely different even though the forces are the same. So the acceleration of the Earth due to this person's mass is negligible because the Earth is so much more massive than the person. But Newton's third law tells us that there is a tiny acceleration on the Earth due to the person. That is the third law pair for gravity. mg downward exerted by the Earth on the person is paired with mg upward on the Earth exerted by the person. Now, for the normal force acting on the person, that force is exerted by the ground. So the ground exerts an upward force N on the person. Newton's third law tells us that that means that the person must exert a downward force N on the ground. That is the interaction pair for Newton's third law for the normal force.
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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: Hi, I'm Sal. Today we will be doing problem number 4 of fall 2009, exam 1. Now before you attempt the problems there's certain background information that you should know before attempting it. One is knowledge in ionic bonding. Because the problem deals with ions. Two, how to draw an energy-level diagram. That's very important. And three, what Madelung's Constant is. Now before attempting the problem, it's a good thing to read the whole problem in detail and make sure you don't skip anything out of it. So that you get all the information that's given to you. Because that's very important. So the problem reads as follows. For a given cation, c and anion a, show that the following four energy states on the same energy-level diagram. 1, Ions in infinite separation. 2, 9-pair c and a, c stands for cation, a stands anion. 3, An ion line, c a c a c a, just a repetitive line of anions and cations. And 4, crystalline solid of c a. So a three-dimensional crystal structure with t and a. And it says also that, assume that the comparison is based upon identical numbers of ions in all four states. The diagram need not to be drawn to scale, however you must convey relative values of different energy states. So that's the problem reads. Now if you recall from lecture there's an important equation that actually describes the energy between two ions, between a cation and an anion. And this energy is directly proportional to the product of the two charges. And it's inversely proportional to the separation distance between them. And it's multiplied by some constants. So what does this mean? Well the fact that it's directly proportional to the product of your two charges means that your energy for any ion or anything that ionically bonds, it's going to be negative. Because your z minus will be negative and that's the whole value of your energy will be negative. And that's what dictates stability. How negative the energy is. So with that in mind, we can go ahead and start the problem. So we have a cation a and anion c. Now I like to draw diagrams, or draw little pictures because that helps me when I'm solving the problem. So I'm just going to draw pictures of a cation and an anion. So I have a cation, c and then I have an anion, a. So c and a, and the value of r naught from our equation is pretty much the separation distance between the two. So project this up, this is r naught. And if you look from the picture, r naught is just simply the radius of your cation plus the radius of your anion. Now this is already describing the energy between just a cation and an anion with these magnitude of charges. So all that tells me is that, this is already answering number 2, pretty much from the problem that we're asked for, which is the ion pair. So the problem asked you to draw an energy-level diagram. That's what I'm going to do over here. So I'm going to go ahead and start drawing the energy-level diagram. Now I'm going to go ahead and label my energy axis to be positive to be up. And I'm going to draw a baseline of 0 here. So I know that to be able to incorporate my ion pair-- like I said the energy is negative, so if the energy goes up to here and let's say the value of the energy here is 0, then that ion pair should lie somewhere below it. So I'm going to go ahead and draw this line. And I'll label this, s 2 for our energy level diagram. So this is the ion pair. Now number 1 says ions at infinite separation. What is the energy of two charged species that are separated by infinity? Well if you look at your equation, your equation tells you that it's inversely proportional to the separation distance. So all that tells you is that if you divide by infinity, your energy should be essentially 0. Which makes physical sense because those two charged species can't feel each other when they're separated. So this is actually already my number 1, which is 0. Separated away. So now we need to find what the energy is of an ion pair. Which is the line of c a. So we want to draw a little picture. It's pretty much going to be a repetition of that. And if I draw a line-- that's not very straight-- I'm going to draw a bunch of cations and anions. Essentially I can just imagine that it extends to infinity in both ways. So with my energy between these two ions right here is given by my equation, then I should be able to figure out what the energy is of the whole system. Because we're talking about electrostatic interaction, which you can just add linearly. So I can make one assumption. But look at my cation and anion pair, I can go ahead and simplify my life by letting z plus be 1. So I'll write that down. I'll let z plus equal to plus 1, which is the charge on our cation. And I'll let z minus equal minus 1, which is the charge on the anion. So that's to help with the math. So if I come back over here and if I look at my equation line then I know that if I draw-- let's see-- a reference access here. Then the separation between the cation and anion is r naught. And between the cation and the other cation is going to be 2 r naught. And it's going to extend for that. So if I go ahead and relabel my energy equation for a line. I'll go ahead and call it e line of function of r naught. It's simply going to be the first one, which is going to be negative e squared over 4 pi epsilon naught, r naught, multiply the 1 over minus 1 over n. So that's the energy between these two ions. So now because the energy is electrostatic, this cation also feels the repulsion of this other cation. So therefore with that you have got to add what that interaction is. So the product of the two charges is plus 1, which gives it a plus in front of your equation. And I end up getting e squared over 4 pi epsilon naught. Now here's something that you should pay particular attention. The separation between these two cation is not r naught. It's actually 2 r naught. Because of where exactly where it sits on the line. So my separation distance now becomes 2 r naught and it's multiplied by the same 1 minus 1 over n factor. Then if you repeat it again, you're analyzing the interaction energy between the cation and the other anion that's next on the line. So as you can see you end up getting this nice-- 4 pi epsilon naught, 3 r naught, 1 minus 1 over n. And essentially it repeats until it ends. Now I'm just analyzing on the right side. Because it extends, it's an infinite line, because it extends from negative infinity to infinity, then you can also assume that there's an anion to the left. So all you really have to do is, you take your overall energy for the right side and you multiply it by 2. Because it's additive. So this whole thing gets multiplied by 2. And that's essentially what the energy is of a line. Now in order to be able to know where it lies on the energy-level diagram, it's important to try to get it into the same basic form as the energy of an ion pair, just something simple like over there. So just by looking at the equation I know there's a lot of factors here that are common. Like that e squared, 4 pi epsilon naught, even r naught and this factor can all come out of the equation. You can take it out. So if I do that, my equation of my line then simply becomes negative e squared over 4 pi epsilon naught. Now I'm going to go ahead and take out the r naught as well. And I'm going to multiply it by the factor. And essentially what stays from here is just your 1 over 1. From this one it's your 1/2 and then your 1/3 and then we're happy. And because I also took out the negative, I want to make sure that I take out the negative from all the other ones too. So this product then gets multiplied by 1 minus 1 over 2, plus 1/3 minus 1/4 plus... and it goes on forever. Now it so happens that this is actually a series that we have an answer or an answer to. And this product that's being multiplied by your fundamental base of your equation is just simply-- oh sorry I forgot the times 2, don't forget that, that's very important because you're summing from both sides-- this essentially right here it's simply arised because of your geometry. You're analyzing your systems. And this value happens to be greater than 1, which is a good thing. And the value of this is actually 1.386. So the fact that the whole equation is still negative, because we're talking about coulombic interaction between a negative charged species and a positive charged species. And we're multiplying that by a value that is greater than 1 this means that if this is 1 right here, then our energy level diagram should lie somewhere below it, like right here. So this is 3. And that's your ion line. Now in real life we know that crystals don't exist in lines really. Or they're not just an ion pair. it's literally something that is three-dimensional and has an ordered structure. Sometimes could be different geometries, but normally it's a cubic structure. So because that forms in nature, then that tells me that the energy of a three-dimensional crystal, which is the fourth one that we need to put on the energy-level diagram, the value should be greater than 1.386. And this value of 1.386 is actually Madelung's Constant, which you should know. It was a required preliminary to the problem. So I know that below this value of 1.386 I have my three-dimensional crystal. Now an energy-level diagram is pretty much quantized by different steps. Now what is quantizing our problem here? Well this is 0 because our energy is 0. So we can almost assume that we're multiplying-- or that the Madelung's Constant for ion pair with infinite separation is 0. And this is 1, so we can assume that the Madelung Constant is 1 or we know that it's 1 for the ion pair. So if I start writing this out on the side here-- give me some space-- I know that's one. I'm going to go ahead and label this by Madelung's Constant, which is 1.386. So I come over here. Relabel this. This value here's is 1.386. And this value here has to be greater than 1.386 for a three-dimensional crystal. And this little diagram here is the answer to your question from the exam. So when you take the exam you don't really have to go through all this math to answer the problem. If you know the material from lecture, then the solution should be very straight-forward very easily. And also if you keep in mind that there's a lot of material that you need to know before you take the exam, and knowing how to absorb that material and apply it will help you in solving these type of problems, which can be pretty difficult.
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https://ocw.mit.edu/courses/8-03sc-physics-iii-vibrations-and-waves-fall-2016/8.03sc-fall-2016.zip
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YEN-JIE LEE: So during the semester, we have a few recitation instructors. They help with the students during the recitation section. So during those sections, usually the instructor will solve a similar problem, like what is actually covered during the lecture. And that gives the students another chance to look at more example and to get familiar or get used to the calculation which we carried out for the first time during the lecture. So also the recitation section-- during the recitation section, the instructors help the students. So they can actually ask more questions during the recitation sections, including the questions they have in the p-set or any additional questions they thought about after the lecture. OK. So that they get another chance to clear some of the concepts which they don't really understand during the lecture. So we did not record the recitation sections during the Fall semester in 2016. On the other hand, we included problem-solving videos from Professor Wit Busza, which is similar to what we do during the recitation sections from previous years, to go with the lecture videos done during the semester.
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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: As you know, Professor Guth is away. I'm substituting for today, he didn't leave me with a particularly coherent game plan, so I'm going to begin with where he thinks we should start. Please jump in if I am just repeating something that he has already described to you guys, or if there's anything you like me go over a little bit more detail, I will do my best here. So, I'm working off of a fairly rough plan. But let me just quickly describe what-- based on what Alan has explained to me --what we're planning to talk about today, and if there's any adjustments you think I should be making that would be great. So, the game plan for today. What I want to do very quickly is hit on a couple of the key points which I believe you talked about last week, which is a quick review of the essential features of symmetries of the gauge fields the make up the standard model. Now, I believe you guys did in fact talk about this last week, at least briefly. And you talked about how you can take these things and embed them in a larger gauge group, the group SU(5). I'm not going to talk about that too much, but I want to just quickly hit on a few elements related to this before we get into that. From this we'll then talk about the Higgs mechanism-- really I'm going to talk about the Higgs field, I'm not going to talk about the Higgs mechanism quite so much as motivate why it is necessary-- and then talk about how the Higgs field behaves and why it's important for the next problem, which is what is called the cosmological monopole problem. To be more specific magnetic monopole problem. I confess I feel a little bit awkward talking about this problem on behalf of Alan. This would be kind of like if you were planning on studying Hamlet and there was this guy W. Shakespeare who was listed as the instructor and you walk in and discover there's this guy Warren Shackspeare, who's actually going to be teaching or something like that. I kind of feel like Warren here. This stuff really is Allen's thing, so it's sort of, I'm probably going to leave this at the denouement of all this when you actually get into inflation to him. I may have a little bit of time at the end to just motivate it a little bit, but the grand summary will come from him. OK, so, as discussed by Alan the standard model describes all the fundamental interactions between particles via gauge theories. OK, and these gauge theories all have a combined symmetry group that is traditionally written in a somewhat awkward form, SU(3) cross SU(2) cross U(1). U(1) could be an SU(1) for reasons which I'll elucidate a little bit more clearly in just a moment. There's really no point in putting the S on that one. So each of these things essentially labels the particular symmetry group. So, the "S" an element of SU(n) is a matrix that is n x n, that is unitary-- that's the U. Unitary just means that the inverse and the transpose of the matrix at the same, actually the Hermitian conjugate because they can be complex, in fact, they generally are. And it has determinant of 1. That's what the special refers to, special, the S in SU(n) stands for special unitary n. So, the S means that the determinant is one-- that's what's special about it-- unitary is this idea that the inverse Hermitian conjugate are the same, and then n refers to all these things. So, that tells us that the gauge degrees of freedom are related by a symmetry that looks like a 3 by 3 matrix with these properties, as listed there for the SU(3) piece of the symmetry. SU(2) means it's a 2 by 2 matrix. And U(1) means it's a one by one matrix, what's a 1 by 1 matrix? It's a number, its a complex number. And that's why we don't really need to put an "s" in front of it. If it's a complex number its determinant is 1 if it's just a complex number whose modules is one. That's why we don't bother with the S on the U(1). So, I think you've already hit on some of this but this is sort of useful to review because it's going to set up why we need to introduce a Higgs mechanism in a little bit. Let me just quickly hit on what the details structure of this looks like for you want to think is the easiest one understand, So, as I just said, a one by one matrix is just a complex number. So that means that any element of this group is a complex number, which we can write in the form z equals ei theta, where theta is a real number. Now, the thing which is I want to hit on in this, the reason I want to describe this a little bit is, this may not smell like the gauge symmetry that you're used to if you study classical E&M. Some of you here are in 807 with me right now, and we've gone over this quite a bit recently. How is this akin to the gauge group that we are normally used to when we talk about the gauge freedom of electricity and magnetism? Well, it turns out there's actually a very simple relationship between one and the other, rather between this view of it and the way we learn about it when we study classical E&M. It's simply that we use a somewhat different language, because when we talk about it in this group theoretic picture we're doing it in the way that is sort of tuned to a quantum field theory. So, the way we have learned about electromagnetic gauge symmetry in terms of the fields sort of goes as follows. We actually work with the potentials, and so what we do is we note that the potentials Amu, which you can write as a four vector, whose time-like component is the negative of the scalar potential, and whose spatial components are just the three components of the vector potential. So, this potential and this potential-- --okay this is possibly module of factor of c somewhere in here but I'm going to imagine the speed of light has been set equal to 1. Both of those potentials generate the same E&B fields. OK, again you still should be looking at this and thinking to yourself what the hell does this have to do with the U(1) as we presented it here. I've given you a bunch of operations that involve some kind of a scale or function of time and space. And I've added particular components of this four vector in this way, what does that to do with this multiplication by a complex number? Well, where it comes from is that when we study E&M, not as a classical field theory but as a quantum field theory, we have a field that describes the electron. So, where it comes from is that when you examine the Dirac field, which is the quantum field theory that governs the electron, when you change gauge the electron field acquires a local phase change. So in particular, what we find is that if we have a field 5x, which those of you who have taken a little bit of quantum field theory should know this is actually a spinner field, but for now, just think of it as some kind of a field that under the field equations of quantum electrodynamics-- the Dirac equation or high order ones that have been developed by Feynman, Schwinger, and others-- under a change of gauge this goes over to si prime of x, which equals e to the-- terrible notation I realized-- 1e is obviously the root of natural logs, e sub 0 is the fundamental electric charge. OK, can everyone read that? I didn't block it too badly here I'm not used to this classroom. So, here's the thing to note, is that this field lambda, which we learned about in classical E&M directly connects to the phase function of the Dirac field in quantum electrodynamics. So, our gauge symmetry is simply expressed in the quantum version of electrodynamics by a function of the form e to the i real number, where that real number is the fundamental electric charge times the classical gauge generator. So, this is what is meant when people say that electrodynamics is a U(1) gauge theory. Now, I'm not going to go into this level of detail for the other two gauge symmetry that are built into the standard model. But, what I want you to understand is that the root idea is very, very similar. It's just now, instead of my gauge functions looking like e to the i, some kind of a local gauge phase of x multiplying my functions, my quantities which generate the gauge transformation are going to become complex value matrices. So that makes them a lot more complicated, and it's responsible for the fact that the weak and the strong interactions are non-abelian paid which order you perform the gauge transformation in matters. Question. AUDIENCE: What's the physical significance of them being non-abelian? PROFESSOR: Yes. So, what is a physical significance of them being non-abelian? I'm trying think of a really simple way to put this, it's-- Alan would have an answer to this right off the top of his head, so I apologize for this-- this isn't the kind of thing that I work on every day so I don't have an answer right at the very top of my head, unfortunately. Let me get back to you on that one, OK, that's something I can't give you a quick answer to. It's an excellent question and it's an important question. Any other questions? OK, so, here's a basic picture that we have. So, we find is that the strong interactions have a similar structure where my need to e to the i factor goes over to a 3 by 3 matrix, and the weak interactions in a similar structure with my e to the i factor going over to a 2 by 2 complex matrix. OK, what does this have to do with cosmology? In fact, as an enormous amount to do with cosmology, as we'll see over the course of the rest of course. Part of the thing which is interesting about all this is that we have strong experimental reasons, and theoretical reasons to believe, that the different symmetries that these interactions participate in, the different symmetries that we see them having, that isn't the way things have always been. So, in particular when the universe was a lot hotter and denser these different symmetries actually all began to look the same. In particular the one which is particularly important, and you guys have surely heard of this, is that the SU(2)-- if we just focus on electric and the weak piece of this-- SU(2) cross U(1). So, this is associated with the gauge boson that carry the weak force, OK, the z boson, the w plus, and the w minus. And your U(1) ends up being associated with the photon. In many ways, when you actually look at the equations that govern these things, they seem very, very similar to one another except that the-- here's partly an answer to your question I just realized-- the gauge generators of these things have a mass associated with them. That mass ends up being connected to the non-abelian nature of these things. That's not the whole answer, but it has a connection to that. That's one thing which I do remember, like I said I feel this is really Alan's perfect framework here and I'm just a posture in bad shoes. So if we look at this thing, what we see is that these symmetry groups, what's particularly interesting is that U(1) can be regarded as a piece of SU(2). And we would expect that in a perfect world they would actually be SU(2) governing both the electric and the weak interactions. Whereby perfect I mean everything is a nice balmy 10 to the 16th GeV throughout all of space time, and all the different vector bosons happily exchange with one another, not caring with who is who. It's actually not very perfect if you want to teach a physics class and have a nice conversation, but if you are interested in perfect symmetry among gauge interactions it's very, very nice. So, the fact that these are separate is now-- I was about to use the word believed but it's stronger and that, we now know this for sure thanks to all the exciting work that happened at the LHC over the past year or two-- the fact that these symmetries are separate is due to what is called spontaneous symmetry breaking. So, let's talk very briefly about what goes into this spontaneous symmetry breaking. So SU(2) turns out to actually be isomorphic to the group of rotations on a sphere. So, when you think about something that has perfect SU(2) symmetry it's as though you have perfect symmetry when you move around through a whole host of different angles. OK, so you move through all of your different angles and everyone looks exactly identical to all the others. If you break that symmetry it may mean you're picking out one angle as being special, and then you only retain a symmetry with respect to the other angle. And essentially, that is what happens when SU(2) breaks off in a U(1) piece of it. Something has occurred that picked out one of these directions. And by the way, you have to think very abstractly here. This is not necessarily a direction in physical space we're talking about here but it's a direction in the space of gauge fields. So, if we imagine that all of these, my gauge fields in some sense the different components of them defined in some abstract space direction, initially these things are completely symmetric with respect to rotations in some kind of an abstract notion of a sphere. And then something happens to freeze one of the directions and only symmetries with respect to one of the angles remains the same. Let's just write that out, when SU(2)'s symmetry is broken so one of the directions in the space of gauge fields is picked out as special. That direction then ends up being associated with your U(1) symmetry. So, what is the mechanism that actually breaks the symmetry and causes this to happen. Well, this is what the Higgs field is all about. The idea is there is some field that fills all of space time. It has the property that at very high energies it is extremely symmetric, with to respect all these gauge fields, all directions and sort of gauge field space look exactly the same. And then as things cool, as the energy density goes down by the temperature of the expanding universe, cooling everything off, the Higgs field moves to a particular place that picks out some direction in the space of gauge fields as being special. So let's make this a little bit more concrete. OK. You guys have probably heard quite a lot about the Higgs field over the past couple years, months-- what actually is it? Well, the field itself is described by a complex doublet. So, if you actually see someone write down a Higgs field what they will actually write down is h, being a two components spinner, whose components are h1 of x, h2 of x-- where x really stands for space time coordinates, so that's time and all of your spatial coordinates-- and both h1 and h2 are complex fields. The thing which is particularly key to understanding the importance of this thing is that h transforms, under gauge transformations, with elements of SU(2). So, if you want to change gauge the way you're going to do it is you're going to have some new Higgs field. So remember, if U(2) is an element of SU(2) we call it the two by two matrix. This is what they look like in a new gauge OK-- pardon me a second I don't see a clock in this room, I just want to make sure I know the time, thank you. OK, so, what are we going to do with this? Well, there's a couple features which it must have, so the Higgs field fills all of space time and it has an energy density associated with it, which we will call just the potential energy. It's really an energy density, but, whatever. The energy density that is associated with this thing must be gauge invariant. OK, even when you're working with strong fields and weak fields, the lesson of gauge invariance from E&M still holds. OK, one of the key points was that the gauge fields affect potentials, they allow us to manipulate our equations to put things into a form where the calculation may be easier. But at the end of the day, there are certain things it actually exert forces that cause things to happen, those must be invariant to the gauge transformation. Energy density is of those things. If you were to get into your spaceship and go back to the early universe and actually take a little scoop of early universe out and measure the energy density, A, that would be cool, but B it would be something that couldn't actually depend on what gauge you were using to make your measurements. That is something that is a complete artifice of how you want to set up the convenience of your calculation. So, in order for the energy density to be gauge invariant we have to find a gauge invariant quantity that is constructed from this, which is the only thing the energy density can depend on. This means, let's call our energy density V, it's the potential energy density. So, it can only depend on the following combination of the fundamental fields Pretty much just what you'd expect. This is sort of the equivalent to saying that if you're working in spherical symmetry the electrostatic potential can only depend on the distance from a point charge. This is a very similar kind of construct here, where I'm taking the only quantity that follows in a fully symmetric way, of calling the fact that this is a special unitary matrix that I can construct from these things. So then, where all the magic comes in is in how the Higgs field potential energy density varies as a function of this h, this magnitude of h. So, as I plot v as a function of h, in order to get your spontaneous symmetry breaking to happen what you want is for the minimum of V, the minimum potential energy, to occur somewhere out at a non-zero value of the Higgs field H. Now, why is that so special? The thing that is so special about that is that when I constructed this magnitude of h, I actually lost a lot of information about the Higgs field. OK, let's just say for the sake of argument that this minimum occurs at a place where the Higgs field in some system of units has a value of 1. So, all I need to do is as my universe cools what I'm going to want is energetically, my potential is going to want to go down to its minimum. So, that just means that as the universe is cooling, maybe at very, very early times when everything is extremely hot and dense, I'm up here where the potential energy is very high. As the universe expands, as everything cools, it moves over to here, it just moves to someplace where the Higgs field takes on a value of 1. And that's exactly correct, that is what ends up happening. But remember, the minimum occurs at some value in which the magnitude of this field does not equal zero, but given that value-- where again let's just say for this for sake of specificity that we set it equal to the magnitude of this thing equal to 1 in some units-- there's actually an infinite number of configurations that correspond to that because this is a complex number, this is a complex number. I could put it all into little h1, and I could set into the value where that thing is completely real, or I could put it all into little h2 being completely imaginary or all on to h1 being all imaginary, halfway into h1, halfway into h2. There are literally an infinite number of combinations that I can choose which are consistent with this value of the magnitude of H. So, yeah-- AUDIENCE: So, I don't know if I'm putting too much physical significance on the gauge, but with the other cases of spontaneous symmetry, briefly, that we discussed you can always measure. OK, I've broken my symmetry, and now it's lined up this way, or there's something measurable. Now, the field has to be physical because the fact that you have gauge symmetry gives you some concerned quantity, right? But, how can I measure what direction in gauge space that I picked out? PROFESSOR: So, that is, let me talk about this just a little bit more. I think answering your question completely is not really possible, but there is a residue of that is in fact very interesting, and let me just lay out a couple more facts about what actually happens with this gauge symmetry, and it's not going to answer your question but it's going to give you something to think about. OK, so that's an excellent and very deep question, and there are really interesting consequences. And this is a case where my failure to answer the previous one is because there's details I can't remember, in this case, I think it's because there's details we actually don't understand fully. Research into the mechanism of electroweak symmetry breaking, which is what this is all about, is one of the hot topics in particle physics right now. AUDIENCE: I was just wondering if gravity has any gauge symmetry associated with it. PROFESSOR: It does, but it fits in a very, very different way, and with the exception of the fairly speculative framework of string theory-- which I think is very, very promising, but it's just sufficiently removed from experimental verification that I'm going to have to label it speculative-- it doesn't quite tie in in the same way. And that's the best I can say right now. The gauge symmetries of general relativity are, at the classical level, they correspond to coordinate transformations, at a quantum level, there's not such a simple way to put it. All right, where was I, OK, sorry I didn't get to your question. So, the point we made here is that we have spontaneously, when we actually choose which one of these infinite number of values we're going to have, we just randomly break the symmetry. OK, and you guys apparently have already talked a little bit about spontaneous symmetry breaking. The analogy that people often make is to the freezing of water, OK, prior to the water entering its solid phase its completely rotationally symmetric, then at a certain point crystalline planes start to form, the water forms, all the molecules get set into a particular orientation, you lose that rotational symmetry. In this case, we started out with a theory, with a set of interactions that were completely symmetric in sort of gauge field space. And now by settling down and picking a particular special value of h1 and h2 we have at least nailed down one direction. It's like we've defined a crystalline plane, and so now things, suddenly, aren't as symmetric. And we start to pick out preferred directions in our gauge fields. What we can do with this is really a topic for a whole other course, and that course is called quantum field theory, but I will sketch a couple of the consequences and this gets directly to the answer your questions. So, one of the consequences of this is that once we have picked out a particular direction, electrons and neutrinos are different. When the Higgs field is equal to zero there is no difference between an electron and a neutrino. They obey exactly the same equation, there's literally no difference between them. Once we have actually settled on an h1 and an h2 some combination of the fundamental underlying fields comes together, acquires a mass, acquires an electric charge, and we say A-HA thou beist an electron. It wasn't like that in the original unbroken symmetry. AUDIENCE: Also, [INAUDIBLE]? PROFESSOR: Presumably, but I'm going to stick with just these for now, but I've I'm pretty sure that's the case, yeah. That gets into even more complications of course because the additional generations are actually consequence presumably of some broken higher level symmetry, which is even poorly, more poorly understood. But you raise a good point. So, that's one partial answer your question. How one can actually walk that backwards to understand this thing about the initial state? That's hard to say. I actually think this particular one is one of the profound and interesting aspects of this, in part because we now know the neutrino has a mass. We have no idea what that is, and in fact we only really have bounds on the mass, such that we know it is non-zero, and we have upper limits that are set by very indirect measurements. But the actual values of the mass are very, very poorly constrained. Within the standard model you just take the electroweak interaction, introduce a Higgs coupling and allow the symmetry to be spontaneously broken, the neutrino mass is zero. Full stop zero. So something's not right, we're actually missing something here. People have kind of jury rigged the standard model to put in the masses by hand, and it works OK, but it's not completely satisfying. And a lot of experiments going on right now to explore the neutrino sector are hopefully going to open us up to a deeper understanding of this and may say a lot about all this physics, which is at present, pretty poorly understood. The consequence, which has received the most popular press, and what you guys have certainly seen about in newspapers, given the results that came out from the LHC over the past year is that quarks and leptons have mass, or put more specifically, rest mass. To understand what this actually means I think you really need to ask yourself what is mass meant to be. Well, the idea is you calculate the spectrum of oscillations associated with the fields of your theory, and then if your theory predicts a discrete spectrum of oscillations, it doesn't even have to be discrete but predict some spectrum of oscillations, then for every oscillation frequency omega there's an associated mass that is just H bar omega over c squared. If your omega has some lower bound that is greater than zero, then your theory has particles with nonzero rest mass. Without going into the details-- and this again is something which those of you who are going to go on to study this in more detail in a higher level course, which is fairly standard stuff is done in probably the first or maybe late in the first or early in the second semester of a typical quantum field theory course-- what you'll find is that when the Higgs field is zero then quarks and leptons have, the field that describes quarks and leptons-- and yes including mu and tau, so including all the leptons, this one I'm very confident on-- the spectrum goes all the way to zero if the Higgs field is zero. But when the Higgs field becomes non-zero, roughly speaking, it shifts the spectrum over for these particles. There's an interaction between the things like the electron field in the Higgs field or the up quark field and the Higgs field, which shifts the spectrum over just enough so that the frequency is never allowed to go below some minimum. AUDIENCE: Going back a bit, I'm confused about how picking a specific value to the Higgs field is breaking SU(2) symmetry and not U(1), because it seems like we're fixed on a circle, right? PROFESSOR: That's right what U(1) is a symmetry on a circle, SU(2) is kind of like symmetry on a sphere, essentially. AUDIENCE: Right, so how are we not picking a specific value [INAUDIBLE] circle [INAUDIBLE]? PROFESSOR: Well, what we're doing is, think of it this way, imagine SU(2) is a symmetry on a sphere, and then when we break the SU(2) symmetry it's like we're picking some circle on that sphere. So, we've broken one circle, we've picked one circle, but now we're allowed to go anywhere on that remaining circle, which is a U(1) symmetry. Does that help? Yeah, OK good. And it comes down to the fact if you sort of count up your degrees of freedom, it has to do with the fact you you've got four, you have two complex numbers, so there's four real parameters associated with this thing, and they are isomorphic to sort of rotations in a three space and you're adding one constraint. OK, so let me just finish making this point here again. So, when h does not equal zero, spectrum get shifted for the quarks and leptons, so everything picks up a little bit of a mass. And the final one, final consequence which we're going to talk about today, is that the universe is filled with magnetic monopoles. We all remember studying Maxwell's equations learning that del dot b is equal to 4 pi times the density of magnetic charge-- this all makes perfect sense, right? Well, this is actually something that when it first sort of came out and people begin to appreciate this thing with sort of a "Um, well everything else works so well, maybe we're just not looking hard enough. " So, it was a bit of a surprise. So, where do these magnetic monopoles come from? And essentially, the magnetic monopoles are going to turn out to be a consequence of the fact that when spontaneous symmetry breaking happens it doesn't happen everywhere simultaneously. So, think again about-- yeah? AUDIENCE: Doesn't that bring up possibility that the symmetry could break in different ways in different places? PROFESSOR: That is in fact exactly what this is going to be. Magnetic monopoles are in fact exactly a consequence of this, yes. Give me a few moments to step ahead to fill in a couple of the gaps, but you're basically already there. So, think about crystalline crystal formation again. Imagine you have, we could do ice if you like or choose something that's got a little bit more of an interesting crystalline structure. Imagine you have a big bucket full of molten quarts, OK. So, if you have a big thing of quartz that you want to sort of freeze into a single gigantic crystal, what you typically do if you'd like to do this is you actually seed it with a little bit of a starter crystal. So, you put a little bit of crystal into this thing, and what that does is it sort of defines a preferred orientation of the crystal axes, so that as things start to cool in the vicinity of that they have a preferred orientation to grab on to. And that seed then gradually gets bigger and bigger and bigger, and all the little crystals as they form near it tend to latch onto the preexisting crystalline structure, and that allows you to grow actually extremely large crystals. I don't know if anyone here is doing a year off with the LIGO project but these guys have to make these sort of 100 kilogram mirrors of very pure either Sapphire or silicon dioxide, and when you make 100 kilograms of crystal you need to build it really, really carefully. It's extremely important for the optical purposes that all the axes associated with the crystal will be pointing in the right direction. Otherwise you spend $100,000 on this thing and it ends up being the world's prettiest paperweight. So, similar things happen when the Higgs field cools. Let's imagine that we've got our universe, time going forward like this, and at some point over here the universe cools enough that's the Higgs field condenses into some particular direction. And symmetry is spontaneously broken right at this one point over here. So, I'm going to draw my diagram over there and put some words over here. I shouldn't say Higgs field cools enough, the universe cools enough so that the Higgs field breaks the symmetry. So, just to be concrete, let's imagine that at 0.1 over here it takes on a field of the value one for h1 and I for h2. So just for concreteness imagine it looks something like this at this point. And so what happens is as the university continues to expand other areas are going to cool off. The bits that are closest to it are going to see that there is already a preferred orientation defined by the Higgs field. And so it's energetically favorable for those regions of the universe to fall into the same alignment and so there'll be a region in space times that grows here as the universe cools, in which the Higgs field all falls into this configuration, which I will call h1. But suppose somewhere over here at 0.2, and the key thing is that initially 0.2 is going to be so far away from 0.1 that these points are out of causal contact with one another. I can not send a message from event one to event two. The Higgs field also reaches a point that the universe cools enough that at 0.2, just you know, it's a system that's not in thermal equilibrium. So, some places are going to be a little bit hotter than others, some are going to be a little bit cooler. And so, at these two points it just so happened that the Higgs field got to the point where it could spontaneously break the symmetry. So at 0.2 the Higgs field also got to the point where it could spontaneously break its symmetry. And the only thing that's got to happen is, remember the only constraint we have is that the magnitude of the Higgs field be equals to some value-- I should normalize that to root 2 in units I want to use but whatever. Let's say on this one my h1 is equal to y, and h2 is equal to minus 1. So, it's basically the same thing but all the fields are multiplied by i. It's the same magnitude, so it's going to have the same potential energy. So that's cool. Clearly this is allowed, and now all the regions in the universe that are close to the this are going to sort of smell this particular arrangement of the Higgs field and say OK, that's preferred arrangement I want to go into. So, we have two separate values of the Higgs field that are happily swooping out space time here. This gets to the excellent question I was just asked a moment ago-- what happens when they collide? As the universe expands and gets cooler, all of it is going to end up getting swooped into either the field that was seeded at event one, or the field that was seeded in to event two, but at a certain point we're going to get the bits where they're smashing into one another. So what happens when these different domains come into contact with one another? The absolutely full and probably correct answer is we don't know. The reason is that we don't really, to be perfectly blunt, fully understand every little detail about the symmetry breaking, or about the structure of whatever grand unified theory brings all these things together at the temperatures at which this is happening. Because this is happening when the universe has a temperature of like 10 to the 16th GeV. And so it's way beyond the domain of where we can push things. But we can, as physicists are fond of doing, we can paramaterize our ignorance, and we can ask ourselves, well what happens if these various parameters that characterize my grand unified theory take on the following plausible kinds of parameters. And what we find is that generically, when you have two different domains where the Higgs field takes on different values like this, when these domains come into contact you get what are called topological defects. The topological defects come in three different flavors. To understand something about those flavors you have to know a little bit about what happens in general when you have phase transitions, and different regions of your medium go through a phase transition with different values of the parameters. So, it's a general case that whenever you have some kind of a phase transition and you have domains of different phase that come into contact with one another, your field will attempt to smoothly match itself across the boundary. But that can be very difficult. So if you imagine these particular two cases that I have here, that's essentially saying that when these two domains coming to contact with one another there's going to be sort of a transition zone where the field is attempting to rotate from one value of the Higgs to the other. And it's going to pick some value that is in some sense intermediate to those two things. So that, let's say we continue these up here, so that the collision is occurring right in this place here, in this little locus of events in space time. I have Higgs field 2 over here, Higgs field 1 over here, and I've got some crazy intermediate field that goes between the two of them, which is trying to sort of force itself to smoothly transition from one to the other. In so doing, I might end up pushing my field away from the minimum, in which case there will then be some energy trapped in that layer. And there's a reason we do this level the class in a bit of a hand wavy way, I mean it's very, very complicated to get the details right. But the key thing we see is that in doing this match, the field has to do some pretty silly shenanigans order to make everything kind of match up and we can be left with odd observable consequences from the energy associated with the Higgs field getting pinned down at that boundary here. Now, the details of the forms of this boundary vary a lot depending upon to the specific assumptions you make about your underlying grand unified theory. OK, so I should back up for a bit. I'm sort of assuming here when I discuss all this that there is some underlying SU(5) theory which describes the strong weak and electromagnetic interactions are very, very high temperatures as one gigantic thing. And we're getting to the point now where all the different interactions are beginning to just sort of crystallize out of it. There's a lot of different ways you can pack your underlying, fundamental, what we now think of as our standard model, into SU(5) grand unified theories. And so the ways in which we can get different topological defects depend upon how we choose to do that. So defect flavor one is you get something called a domain wall. When we do this the fields attempts to make itself smoothly match from one region of Higgs field, say from Higgs 1 to Higgs 2. It succeeds, but you end up with kind of a two dimensional structure-- a wall-- in which there's some kind of anomalous field that is just pinned down there. And so we end up with a big sheet. So in a theory like this, it would predict that somewhere out in the universe if there were regions in which the Higgs field had taken on a different value than the one that we encounter around us right now, it could be somewhere out gigaparsecs away, essentially a giant sheet of some kind. And there would be weird, anomalous behavior associated with it. People have really looked long and hard to try to find things like this and in fact it would be expected to leave interesting residuals in the cause of microwave background. My understanding of the literature is that there are actually now very strong bounds on the possibility of having a grand unified theory that leads to domain walls. And so this kind of a topological defect is observationally disfavored. So this, I should mention, only occurs in some grand unified theories. Basically, As we move on to the other flavors of defects we end up just going down a step in dimensionality associated with the little kinks that are left over when the different domains come into contact with one another. Flavor two, we would get what's called a cosmic string. Some of you may have heard of this. This is essentially, at its core, just a one dimensional, it could be gigparsecs long, but one dimensional, truly one dimensional-- essentially just a point in the other two dimensions-- string of mismatch Higgs field with some kind of an energy density associated with it when the different domains get in contact. AUDIENCE: Do we have any estimate of how close in actual space these different regions would have started? PROFESSOR: We do and I'm actually going to get to that. So, let me give you two answers to that. One of them is you are going to estimate that apparently on PSET 10, according to the notes that Alan left for me. But I'm going to spell out for you the arguments that go into it in the last 10 minutes of a class. But yeah, so let me just quickly finish up this one because this again-- so a cosmic string is sort of like a one dimensional analog of a domain wall. And because it would be this sort of long one dimensional structure, that has actually up a lot of energy sort of pinned down to it by the fact it has a Higgs anomaly associated with it, it would be strongly gravitating and so it would leave really interesting signatures. It was thought for a while that cosmic strings might have been the sort of original gravitational anomalies that seeded some of the structures we see in the universe today. Again, it's now pretty highly disfavored. If cosmic strings exist, they don't appear to contribute very much to the budget of mass in our universe. I should also mention that this is only predicted by some grand unifying theories. If you guys are curious about this I suggest when Alice back you ask him what the difference between these sums, why some predict a domain wall, some predict the cosmic strings. Flavor three is where you end up with the Higgs field essentially being able to smoothly transition without leaving any defect anywhere except at a zero dimensional point. So you end up with just a little knot in the Higgs field. And for reasons that I will outline very soon, it turns out that this little not must carry magnetic charge, and so it must be a magnetic monopole. The domain walls and the cosmic strings are, as I've emphasized, only predicted by certain specific grand unified theories. Magnetic monopoles are actually predicted by all of them. Question. AUDIENCE: What does it mean to have a one dimensional domain wall, because there's no different region separated by one [INAUDIBLE]. PROFESSOR: That's right. So what ends up happening, and this is where I think you're going to have to ask Alan to sort of follow up on this a little bit. So, as the domains come into contact with one another. The fields do their best to smoothly transition from one to the other. And grand unified theories that predict a cosmic string, they succeed pretty much everywhere. They're able to actually smoothly make it all go away so you don't end up with feel being pinned down anywhere, except in a little one dimensional singularity that is somewhere along where the two dimensional services originally met. And that is-- there's details there that I'm not even pretending to explain. And as I say, those are only predicted by certain kinds of grand unified theories. All of them will then predict that even if you don't have that, that cosmic string will then shrink itself down and it'll just be left with a little knot of Higgs field, where there's a little bit of residual mismatch between the two regions. AUDIENCE: Do all three types of defects carry a magnetic charge, or only the knots? PROFESSOR: I think only the knots. They do carry other kinds of fields, though, in particular the other ones gravitate, in fact all them gravitate, and so that's one of the ways in which people have tried to set observational limits on these things. In particular there have recently been a fair amount of work of people trying to set limits on cosmic strings from gravitational lensing, and there was really a lot of excitement because people thought they discovered want a couple years ago. And they saw basically two quasars that looked absolutely identical, that were separated or scale that was just right to be a cosmic string. And then people actually looked at with better telescopes, and saw they had absolutely nothing to do with one another. They were not cosmic, they were not lenses, it's just every now and then God is screwing with you. OK, so without going into some of the details what you have, these little point like defects-- and I'm short on time so I'm going to kind of go through this a little bit in a sketchy way enough so that I can pay for you how to do some calculations you're going to need to do. So the point like defects end up being regions, where at that point the Higgs field actually takes the value zero. So remember I was describing how when you have two regions where the Higgs fields are both taking on values such as there at the minimum of the Higgs potential energy, and they come in to match one another, and what we have a boundary condition that very far away the Higgs field has values such as the energy is minimized. And there is a theorem, which in his notes Alan-- the way he describes it is he gives you a figure and outlines the various things that are necessary for the theorem to be true, and invites you to think deeply for a moment and until insight comes to you, I guess. And when you put this ingredient that the Higgs field has this asymptotic, very far away value that drives you to the minimum of the field, and yet it must change value somewhere in the middle, the theorem requires that there be one point at which H equals 0. And apparently, this is a consequence in all grand unified theories. So, recall, H equals 0. This is a point at which the potential energy density can be huge. So, when you have a little point like defect like this, it looks like a massive nugget, little massive particle. You can in fact calculate the total amount of energy associated with this particle. If you do so just including the influence of the Higgs field, the calculation basically goes like this. It's very similar to the way we calculate the energy associated with electric and magnetic fields in electrodynamics. Ask yourself, how much energy is contained in a sphere of radius, capital R, centered on this little knot of Higgs field. Well, it's going to look like 4pi times an integral of the gradient of the Higgs field squared r squared dr It turns out, when you calculate the [INAUDIBLE] of the Higgs field around one of these little defects, it's actually very complicated close to the defect, but as you get far away it has a very simple form. The gradient goes as 1 over r, it tells you the field itself actually goes something like log. That means your energy looks something like, R squared, 1 over R squared dr which goes as R, which diverges as you make the sphere bigger and bigger and bigger. So, what's the mistake we made? Well, the Higgs field doesn't always just sit there and operate on itself. The Higgs field actually couples pretty strongly to all of our vector bosons. Particularly, it couples pretty strongly to electric and magnetic fields. So, we have to repeat this calculation including the interaction of the Higgs field with the E&D field. And in Alan's notes he gives you some references on this because this is not the kind of calculation you can really sketch out very easily in an undergraduate class. To make this integral convergent, the only way it can be done is if that little nugget of Higgs field is endowed with magnetic charge. You need to have a monopolar magnetic field that ends up putting in interaction terms, that make the divergence of this integral go away. So, I at last get to the punchline of all this, we are left inevitably, if we accept the whole foundation story of particle physics that the different interactions were unified in some high energy scale and then froze out. We are driven inevitably to the story that defects in the Higgs field create magnetic monopoles. Now, I realize I'm out of time, so let me just quickly sketch a few interesting facts about this and there's a few exercises that you guys are apparently going to look at in your homework assignment. When we do this calculation, one which is I believe just referenced in the notes that Alan has for the class, we learn a couple of things about this magnetic charge. One of them is that if you work in the fundamental unit, say CGS units, the value of the magnetic charge, we'll call that g, is exactly 1 over 2 alpha where alpha is a fine structure constant, times the electric charge. So if you have two magnetic monopoles they attract each other with a force that is-- so 1 over 2 alpha is approximately 68.5 I think-- and so it would be 68.5 squared times the force of two electric charges at that same distance. We also end up learning the mass. It turns out to be 1 over alpha times the scale of GUT symmetry breaking. Anyone recall what the scale of GUT symmetry breaking is? 10 to the 16 GeV. So, this is a particle, 1 over alpha is approximately 10 to the two, so this is a particle that has a mass of about 10 to the 18th GeV, in other words it's a single particle with a mass of 10 to the 18th protons. This is approximately one microgram. If you put one of these things on a scale it could measure it, that's bloody big. So getting to the last bit of the class, which I am just going to very basically quote the answer. The question becomes how often do these things get created and here I'm going to refer to Alan's notes. What you'll find is that, remember when we sketched our original picture of this thing we looked at regions of the universe where the Higgs field was initially seeded with different values. In order for the Higgs field to take on different values, initially, these regions had to be out of causal contact with one another. So we are going to require that the initial seed areas be separated by a distance, which is the correlation length, which has to be less than or of order the horizon distance. You can get a lower bound on this thing by imagining that it's-- sorry let me say one other thing. If you do that, then you can estimate that the number density associated with these things, the number density of these monopoles will be 1 over the correlation length cubed. To get a lower bound on the number density of these things, set the correlation length exactly to the horizon distance, and then do the following exercise. So first, let's set up the correlation length equal to the horizon distance. Set the density in monopoles equal to the mass of a monopole over rH cubed, normalized to the critical density. If you do this, you will find that just due to magnetic monopoles alone, the density of the universe. PROFESSOR 2: Excuse me, professor PROFESSOR: Yes, I'm wrapping up right this second. PROFESSOR 2: It's seven minutes. You were supposed to end at 10:55. PROFESSOR: I'm substitute teaching, I'm sorry. OK, so this tells us that we are at 10 to the 20 of the critical density. And a consequence is that the universe is approximately two years old. I will let Alan pick it up from there.
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https://ocw.mit.edu/courses/5-08j-biological-chemistry-ii-spring-2016/5.08j-spring-2016.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. JOANNE STUBBE: So what I want to do today is continue where we left off to try to get further in into this module on regulation of iron right now in terms of humans. And we're talking about the fact that regulation occurs at the translational level. And so I'm going to introduce to you the model. And I introduced you, last time, to two key players that we'll look at in a little more detail-- proteins and little pieces of RNA. And what happens is the proteins bind to the little pieces of RNA and prevent the translation of the messenger RNA into protein, or prevent degradation of the messenger RNA, allowing the translation to proceed. So that's really the take-home message. And I'll show you what the model is. So the last time, we were talking about one of the protein players, IRP1 and IRP2. And I told you that IRP1 was a cytosolic aconitase and that you had seen the aconitase reaction, which I drew in the board last time, which is conversion of citrate into isocitrate. If you look at the model up there, citrate to isocitrate, you're simply losing a molecule of water. And then over here, you're generating isocitrate. And the chemistry is facilitated by the presence of a single unique iron in the form of iron 4 sulfur cluster, which was the first example of these kinds of clusters doing chemistry in addition to electron transfer reactions that you've been exposed to before. OK. So the question then is, what is the signal? And so we're going to see that the signal is going to be related to-- let me just get myself organized here. So the signal-- the question is, what's recognized under low iron and high iron conditions? So that's what we'll be talking about, and how does this switch work to allow translation or not allow translation to occur. And the other player that we need to be introduced to before we look at how this signal works is the iron responsive element. And this is a piece of RNA-- and I'll show you that on the next slide-- within the messenger RNA. OK. So you have a structure like this, and there are so many base pairs, in this little stem loop, that's part of the messenger RNA. And you have a three nucleotide sequence, and this is a bulge sequence. And what we're going to see is that there are many of these structures. People have now done a much more extensive-- this was the model that came forth a long time ago when it was discovered. It was discovered a long time ago, but people have since done a lot of bioinformatics analysis to try to define really what do we know about this little sequence here. Is it three nucleotides? Is it more? People now think it's a little bit more, but it-- there's variability in that you need a bulge. And so what's going to happen is our iron responsive protein is going to interact with this bulge, and that's going to be what's related to depending on the location of this bulge within the message. So we'll see this bulge can be at the 3 prime end or the 5 prime end. And this location and its interaction with this protein is going to regulate the translational process. So that's what I'm going to be presenting to you. So what do I want to say? What I want to say then is for the iron response protein 1, it has, as we saw with the mitochondrial aconitase of 4 iron 4 sulfur clusters. So that could be a switch. We're doing iron sensing. What's going to cause us-- what is the sensor of iron that allows us to translate or not translate all of the proteins-- transfer and receptor, transfer and ferroportin, all of the things we were introduced to in the previous lecture. So this protein has a 4 iron 4 sulfur cluster. And when it loads the cluster, as with mitochondrial aconitase, the protein is active. So it's found in the cytosol as opposed to the mitochondria. And it can convert citrate to isocitrate. OK. So the question is, what is the switch that allows this IRP1 to interact with this little piece of RNA-- the stem loop piece of RNA? And so the switch is that you have to lose this cluster. And what you generate then is apoIRP1. And apoIRP1-- so somehow the cluster magically disappears. And when it disappears it can bind to the IRE. So in the apo form-- that means no metal-- it binds to IRE. Whereas in 4 iron 4 sulfur loaded form, it does not bind. So what that would imply-- if you think about it sort of superficially-- if you have low iron and there's no iron sulfur cluster, the apo form is going to bind. OK. I'm going to show you the model in a minute. OK. So the switch really is related to-- in this case, the sensor is related to-- the fact that we have a 4 iron 4 sulfur cluster. So we also have-- I told you before-- in addition to an IRP1, we have an IRP2. And IRP2 also looks, structurally, like a cytosolic aconitase, but it has no aconitase activity. OK. So we have the second protein, IRP2. It's also a cytosolic aconitase lookalike, but it has no activity. And why does it have no activity? You've seen, over here, the iron sulfur cluster is required to do the dehydration reaction. So it's required for activity in the mitochondrial enzyme. This has no iron sulfur cluster. So it has no iron sulfur cluster. And what it has in addition, even though it looks like IRP2-- the structurally homologous-- it has a 73 amino acid insert. So this is a distinction between the two. OK. But now, this raises the question here-- at least superficially, you can understand that you might be able to sense iron, because you have a cluster, and you can go to no cluster. OK. And you can go back and forth. And so, remember, in a couple of lectures ago I told you about biosynthetic pathways, and I showed you a picture of iron sulfur cluster assembly-- very complicated. At the end of the notes in this part of the lecture, you'll see what the model is. I'm not going to go through that. But how you assemble and disassemble, even though this model has been around for a long time, is only recently beginning to be understood. It's not trivial, because there are 10 steps to assemble a 4 iron 4 sulfur cluster. OK. But here, we don't even have any iron. So how is the IRP2, which binds to the same IREs-- and again, in vivo we don't really know all of this. People are trying to sort that out as what the what the functions of the different proteins actually are. But how does it sense? And so I just told you that the apo form of the IRP1 binds. That's also true of the IRP2. And in fact, it can only be apo, because it can't bind a cluster. So the active form, the binding form, the apoIRP2 binds the IRE. And then the question is, what is the switch? And so what we'll see is that the switch relates to the fact that IRP2 gets degraded. So when IRP2 is degraded, it can't bind. And that's how you turn the thing off. So then that takes you back a step further-- how do you target IRP2 for degradation? And this goes back to one of the reasons that I'm going to spend some time talking about degradation in mammalian systems. And so it turns out-- how does this relate then to iron sensing? And what I'm going to show you is that you have an E3 ligase. I'll show you this in cartoon form. And I'll just say, see PowerPoint. We're not going to-- it's not really completely understood, so I'm not going to talk about it in detail. But what it has attached to it is an FXBL5 domain that looks like a protein we've seen before earlier that has an iron in it. So many of you probably don't remember hemerythrin, but that's the little enzyme in worms that reversibly binds oxygen. So that was incredible. It's structurally homologous to that little protein. So this is-- again, the details are not known. But it can bind iron and it can sense oxygen. So if you're at low iron, there's no iron bound to this little domain. And so there's a consequence. I'll show you what that is. But if it's high iron, it has a different consequence. So the sensing is back a step. Its back a step into-- remember, I told you E3 ubiquitin ligases are multienzyme complexes. So this is part-- you'll see in a minute-- of the multienzyme complex. And so under certain sets of conditions when you have high iron, what happens is this is targeted for degradation. I'll show you what the model is for how this works. So the models are the same. That is, apo in both cases bind. In one case, the iron sensor is directly related to the IRP itself, because it has an iron sulfur cluster. And in the second case, it's indirectly related to an iron cluster that's associated with the E3 ubiquitin ligase. So another point I think I want to raise-- and this will get us into the next module, which I'm not going to spend very much time on-- the reason that we have iron module juxtaposed to the reactive oxygen species is they're really intimately linked. We've talked about how iron 2 can generate hydroxide radical or hydrogen peroxide. These are iron sulfur clusters are also oxygen-sensitive. This is oxygen-dependent. So again, what you're seeing is not only do we have iron sensors, but we will see that iron sensing and oxygen sensing are linked. And I would say-- I was trying to make up your exam, and I was trying to put in a linkage so you would all of a sudden see this, and the more I read, the more confused I got. So the fact is, there are many, many papers published on this now, and the linkages-- the proteins involved-- do many things. And so sorting this out into a very simple model is really still tough. But what I what I believe right now is both iron and oxygen sensing are linked through this type of a model. So let me now just show you a little bit about IRP1. We know a lot about IRP1, because we have structures. So this is the structure actually of the cytosolic aconitase, and this is with the 4 iron 4 sulfur cluster bound. So what happens when you get to the apo form? What happens in the apo form, you now have a little piece of RNA bound. And this little piece of RNA always has a bulge with a cysteine in it. And it always has some kind of a loop. And we'll see in a second that the sequence of that loop can be variable. But here you can see that. So these little balls here that are iron sulfur clusters are the cytidine bulge in this loop. So you see the thing changes confirmation and as binding to an IRE. So that is the switch. And then the question is, how does that work at level of controlling translation, which I'm going to show you in a second. OK. So where do we see these iron-responsive elements in our messenger RNA? So messenger RNA-- go back and look at all of the players I introduced you to the last time. We have a transferrin receptor that's involved in uptake. We have DMT1. That's a dimetal transporter involved in iron uptake. So intuitively, you should ask the question, if you are at low iron, do you want to take up more iron? So you want to turn on the transferrin receptor. You want to turn on the DMT1 protein. So I think most of it makes intuitive sense. The linkage to oxygen, I think, is less intuitive. If you have a lot of iron, what do you want to do? You want to store the iron. So you in some way want to make more of the ferritin. And then the other thing, this HIF 2 alpha is a transcription factor hypoxia-- inducible transcription factor-- that's linked to many, many things-- a huge number of people are working on this now-- one of which is this linkage to iron. But it senses anaerobiasis. And so you can see, it's also linked by one of these little elements. And the next slide just shows you a more recent one where people started doing a lot of bioinformatics on this. The previous slide was from a few years ago. And again, the details, what you need to see is you, in all cases, have stem loops, a little bulge of a cytosine, and then you have some kind of a loop at the top of the stem loop. And if you look down here and you go through-- so these little stem loops are going to be either at the 3 prime or the 5 prime end of your messenger RNA. And so, for example, one of the things you see is aminolevulinic acid. Does anybody know what pathway that's involved in? AUDIENCE: Heme synthesis. JOANNE STUBBE: Yeah. So it's a rate-limiting step in heme synthesis. So there would be a place that would make sense. Remember, I told you all the iron is in heme and hemoglobin. OK. So almost all of these stem loops that you'll see, if you go back and you look through your notes, will make sense in terms of the big picture of how you want to control the levels of these proteins to deal with high iron or to deal with low iron. OK. So that's iron 2. And so here's the picture of IRP2. And this is the model for how IRP2 works. And so here's the case when you have high iron. And when you have iron, this part here, the Fbox, Skp1, and Cul are all part of the SCF E3 ubiquitin ligase. And I don't expect you to remember the names, but remember I told you the E3 ubiquitin ligase is the one that does what? It attaches ubiquitin on to the proteins, targeting it for degradation. So this little part is the ubiquitin ligase. Here's your E2. Remember, you always need an E2 and an E3. And somehow the E2 is attaching this onto the IRP2, which is targeting it for degradation by the proteasome. OK. So this is exactly like the model we put forth a couple of lectures ago. So again, this is the part that's most interesting. If you go back and you look at hemerythrin, which irreversibly binds oxygen with two irons, you have a diirons site. And earlier in your notes, I showed you what that site looks like. This site is intact because the protein is folded. Under conditions of very low iron, what happens is this becomes unfolded. And then this part of the protein gets targeted for degradation by another E3 ligase-- not this one. And then you've lost your sensor. So the IRP2 remains stable. So you might think this is complicated-- and maybe I didn't spend enough time going through this-- but you should go back, and you should look at the explanation again. So the two key switches are here and here. This one is the more complicated switch. Everybody thought everything was understood once they found this little iron binding protein that models hemerythrin, but nothing could be farther from the truth. We still really don't understand, overall, how this fits into the big picture. But it's not an accident that iron and oxygen are required to fold this into this little bundle that looks like hemerythrin. So those are the switches. And so now, what I want to do is put forth a model. So let me see. So what is the model for how you want to turn these things off and on? OK. So we have two things-- we have IREs-- Iron-Responsive Elements-- that can bind in front of the message to be transcribed or at the end. So what we're going to look at two sets of conditions-- one is under conditions we have low iron, and one is under conditions where you have high iron. How do you sense those conditions? So that's the question. So again, we're sensing low versus high iron. So let's look at low iron first. So what we're going to see is we're going to have a stem loop. So here's my messenger RNA, and here's the 3 prime end, and here's the 5 prime end. And here, you initiate translation. And so if it binds-- if this protein IRP2 or IRP1, both in the apo form-- IRE-- and they do both bind. People are trying to sort all of this out. So this is IRP1 or IRP2. What happens to the translation? What happens to the translation is it's inhibited. So if you have this little stem loop in the front of your message, you inhibit translation. And so what I'm showing you now-- and then I will give you some examples-- is the key to thinking about this. And most of it actually is intuitive once you remember what all the factors are that are involved in iron homeostasis that we've already gone over. So binding inhibits translation. OK. So then we have the second case at low iron. And again, we have a 5 prime end, and then we have a 3 prime end, again, of our messenger RNA. And here is the initiation of translation. And in this case as well, you have the same sort of structures of stem loops. They are similar but distinct. And what can happen with the apo form of IRP1 or IRP2? Again, it can bind to these stem loops. So you can have-- this chalk is not working-- your proteins, they're all bound-- they may or may not be all bound. I don't think we know that much. But what you see is the number of stem loops at the 3 prime end is variable. It depends on the message. So number of stem loops is variable. So what does this binding do? What this binding does is it prevents the messenger RNA from being degraded. So it basically stabilizes the messenger RNA. So this model binding prevents messenger RNA degradation. So it stabilizes the messenger RNA. So that's the model. OK. So now, let's just look at a couple of examples. And then what you can do later on is go back and think about this more of what's going on. So again, this is the model. And let me just make sure I go through the ones I want to do. So we're still at low iron. And we'll do two at low iron. And then we'll look at the consequences of what happens at high iron, and does it make intuitive sense based on what we think the function of these proteins are that are going to be translated? So let's look at low iron. So we're at low iron. And let's look at ferritin. OK. So what is ferritin? It's the iron storage protein. So under low iron, do we want to store iron? No. So if you have the choice of these two modes of regulation, what would you choose? Where would you put your iron-responsive element? At the 5 prime end the 3 prime end? So we're at low iron. We don't want to store iron. So we don't want storage. So what would we do? If these are the two choices-- and these are the two choices from experimental data. There are many other variations on this you could have imagined, but this is the model that everybody agrees on at this stage. So where would you put your stem loop? Yeah. You'd put it at the 5 prime end. And why would you put it at the 5 prime end? Because it prevents conversion of your message from ferritin into the protein. So you have less of the ferritin. So what you see now is you have-- again, so this is a stem loop at the 5 prime end prevents translation and have lower concentration of ferritin. So that's exactly what you would expect. Some of the others are less intuitive, but we've seen ferroportin. Remember, ferroportin is the iron 2 transporter in many cells which allows the iron to come from the inside to, get picked up by transferrin, and redistributed to the tissue. So it, in conjunction with the hepcidin peptide hormone we briefly talked about plays a really important role actually in controlling where the iron ends up going. And in fact, what you would like to be able to do-- say you had not very much iron, where would you want to put your iron? Would you want to put your iron in some metabolic pathway that's not so important, or would you want to put your iron in a metabolic pathway that's very important? You would want to put it into the pathway where you really need it to survive. And so this is a subtle tuning on all of this. And so an example of how this can be tuned if you look at an iron-responsive element binding protein is succinate dehydrogenase. Any of you ever heard of succinate dehydrogenase? And where have you heard of it? You have heard of it, you just probably don't remember it. [INTERPOSING VOICES] JOANNE STUBBE: Yeah. So it's in the TCA cycle. So it converts succinate, which is a hydrocarbon, into an olefin, an alpha beta unsat-- into fumarate. So remember the TCA cycle, you can tune it down or you can tune it up. So if you really were desperate for iron, you would probably tune down the TCA cycle. So in fact, if you look, you'll see a stem loop in front of succinate dehydrogenase which prevents its translation and tunes down the pathway. So there's a subtle example of how nature has-- at least is the way we rationalize the experimental observations of what nature has done. Now ferroportin, which is the way I started on this, sets priorities. And it does this in conjunction with hepcidin, which we already talked about. Remember, hepcidin can target ferroportins for degradation. And this allows the iron to be distributed in defined ways within the cell. And in fact, what you want to do, in this case, is have the stem loop at the 5 prime end so that you don't export the iron inside the cell to the outside. So that's what it does. And some of these, as I'm saying, are easier to rationalize than others. The ferritin one is really easy to rationalize. The ferroportin is easy to rationalize based on what I just told you. But what you see also is that in some of these systems-- I don't know how much you guys thought about RNA, but you know messenger RNA can be spliced. In different cells it's spliced differently. You've also seen that cell types, in terms of iron homeostasis, the enterocyte, the macrophage system in the spleen, red blood cells are much more important, it might be, if you're in some other tissue, the splicing site is different, and you don't have a stem loop. So you can alter the regulation by alternate splicing systems. So these are these two are at the 5 prime end. What about the transferrin receptor? So let me put this down here. What about the transferrin receptor? What would you expect at low iron-- we're still at low iron-- the regulation to be from the transferrin receptor? What do you want to do at low iron? So this is another example, low iron. Let's look at the transferrin receptor. What does the transferrin receptor do? Hopefully you know this. Yeah. AUDIENCE: [INAUDIBLE] JOANNE STUBBE: You need to speak louder, I can't hear anything you said. You just went like this. That didn't mean anything to me. AUDIENCE: It helps to intake iron. JOANNE STUBBE: Yeah. So it helps to intake iron. So if you have low iron, what do you want to do? AUDIENCE: You want to increase-- JOANNE STUBBE: Yeah. So you want to increase that. So where would you put the stem loop? AUDIENCE: 3 prime end. JOANNE STUBBE: Yeah. So you put it at the 3 prime end, because that stabilizes the messenger RNA of the transferrin. So here, at low iron, you want to increase iron uptake. And that means that if you have the 3 prime end, you're going to stabilize the message. So you can go through each one of the proteins that we discussed in the last lecture. And before you look at it, try to rationalize under different sets of conditions. This is low iron. What would you expect to happen at high iron? Here, let's just look at this one so I don't have to draw this again. But what would happen to the transfer receptor at high iron? Do you want to take more iron into the cell? No, you don't. So what you want to do is get rid of the transferrin receptor. So now what do you do? At high iron, if you're IRP1, you switch to pick up the iron sulfur cluster. It no longer binds. And so now what happens? So this is all bound. So it's stabilized and bound. In this case now, messenger RNA is degraded. So the big players in iron homeostasis, I think, are easy to rationalize. If once you know-- this might not be so rational why you would stabilize messenger RNA or whatever, but this is the way nature designed this. Once you remember this-- and remember, the switches are just apo binding, and somehow they sense iron and they no longer bind, whatever the details are-- you should be able to understand in different kinds of cell types how you might regulate the iron at the translational level. So I think that is all I wanted to say. This is just a summary of what we've done in the human part. And we're thinking about-- I gave you a big picture of what happens in humans. This is the summary of that big picture with all these factors that are regulated at the translational level by the iron-responsive binding proteins. And so you can go back and look at this cartoon. Whether you want to store it, whether you want to distribute it, whether you want to put some in the mitochondria-- all of that kind of stuff is regulated at the translational level. So in this module, the second lecture, which was longer than I wanted it to be-- but that's life-- was focused on the big picture for human and how iron is transported. And uptake, which we talked about by divalent metal irons, transporters, and by transferrin in the plus 3 state, and this question of regulation at the translational level. Now everything-- the hepcidin, we didn't touch on very much. Very complicated, but it plays a major role systemically. Whereas these others-- what we were just talking about is more specific for each cell type. And different cell types want to have regulation in different ways. So the bigger picture is the hepcidin. And it was discovered a while back, but I still would say we don't understand a lot about what's going on in terms of that hormonal regulation. So now what I want to do, as advertised, is move into bacteria, and how do bacteria do the same thing. They have the same problem. We talked about metal homeostasis-- exact same problem in human and bacteria. But in the end, the bacteria want to survive and we want to survive, so we have the battle between the bacteria and us for iron. So what I want to do is introduce you to the bacteria-- generically, how they take up metal to use for the same things that we use it for-- a little less complicated, maybe, than humans. And then what we're going to do is I'll introduce you to this war between bacteria and humans. And then we're going to focus on one bacteria that's a major issue nowadays-- Staphylococcus aureus-- because of resistance problems. This is a problem that Liz's lab has worked on. And specifically, I'm going to give you one example of how Staphylococcus aureus gets iron out of our hemoglobin. That's going to be the example. And the system that you'll see, it's amazingly cool. But you'll see, there's still many things we don't really understand in a lot of detail. So what I want to do now is introduce you-- how am I doing timewise? OK. So what I really would like to do is draw this out on the board, because it's complicated. And I know what happens if you use PowerPoint, you go through it at 100 miles an hour. But I'm going to be using more PowerPoint to get through something. So anyhow, this is an overview of where we're going if you forget. So what I wanted to do, at least a little bit, we're going to be focusing on gram-positive and gram-negative systems. And I want to tell you what is the difference. You all know or have heard about gram-positive and gram-negative bacteria. They use different strategies. They use the same strategies, but they use distinct strategies because of their structures. And so what I want to do is give you an overview, and then we'll focus specifically on Staph aureus. So in gram-positive, here we have our plasma membrane. So this is the plasma membrane. And this big guy here is PG-- the peptidoglycan. And we'll see, in Staph aureus, the peptidoglycan is going to play a key role. So you need to understand the structure of the peptidoglycan. So I am going to spend a little bit of time describing to you the structure. It's also the major target of many antibacterial agents that are currently used. Why? Because it's unique to bacteria. So you have only this plasma membrane. There's no outer membrane. That's going to be distinct from gram-negative. And so the question is, how do you get iron from the outside to the inside? And so one of the ways you can take in iron is-- you've already seen this, and you've talked about it in detail in the first half of the course-- siderophores. And we've already talked about the fact that we have many, many different kinds of siderophores. And somehow these siderophores-- and we'll look at a few structures-- can get to the outside of a cell. They pick up iron in the plus 3 state, and then they need to bring it back to the plasma membrane. And then somehow it needs to get transported across the plasma membrane. This is a transporter. Most of them are called ABC transporters and they require ATP. And when they get across, they take the siderophore with the iron into the cell. So that looks simple enough. We'll see that the strategy of gram-negative bacteria is going to be distinct, because it has an outer membrane. So how would you get the iron out of the siderophore? And so I'm going to push this up, and I'll come back down again. So how do you get the iron out of the siderophore? And of course, what that depends on is the reduction potential of the iron. So we will see with enterobactin, in which you already looked at, the reduction potential under neutral conditions is minus 750 millivolts. Minus means that it's really hard to reduce. It wants to be oxidized. It's outside the realm of what you can do inside the cell. So And we want to reduce iron 3 to iron 2. Why? Because we increase the exchangeability of our ligands. That's why that was introduced before. So you could reduce this, potentially. And so I'll just put a question mark there. And so then what you have is a siderophore. So let me just write this down so you don't forget. So this is the siderophore. And then you have your iron. So what happens to the siderophore with no iron? It can now get recycled back to pick up more iron. So let me just put this here. This is recycled. And we're going to be focusing on here taking up iron from siderophores, but we'll see that you can take up iron from hemes. And you have the same issue. You're going to use the same strategy. You'll bring it into the cell. You've got to get the iron out of the heme, and you have to recycle it. So if you can't reduce it, what do you do? Does anybody remember what you do with enterobactin? Anybody remember the KD? We talked about this last time, but it bonds like a son of a gun. It's hard to reduce. You have ester linkages in enterobactin. If you go back and look at the structure, there are proteins that can hydrolize the ester linkages. So ring opens-- makes it bind less tightly, and so it can be released. So in the case of enterobactin, you have an esterase. So let me just show you that, and then we'll come back again to the gram-negative. But if you look at the siderophores, there are 500 siderophores. Here is enterobactin. here are the esters. You can hydrolyze them to release. You can't reduce, because, again, the more negative, the more it wants to be oxidized. And the range of reduction inside the cell is maybe minus 500. You can't get that much above that. But if you look down here at citrate-- remember, we were talking about citrate-- unusual in that citrate is part of this aconitase IRP1 and IRP2 system. But what's the reduction potentially are completely different. So if you had iron citrate, you could easily reduce it under physiological conditions. So the strategies you need to be able to release the iron to then use the iron for what you want to do is distinct depending on the siderophore. So if we go back, now let's just look over here and I'll draw that in parallel. So what's the difference between gram-positive and gram-negative? So let's draw that out. And then what I'm going to show you, rapidly, is, again, the strategies with heme are subtly different, but very, very similar. We have different sets of proteins. So with gram-negative we have an outer membrane. So this is gram-negative. And what we have in the outer membrane are proteins. It has a lot of proteins. And it has a big protein with a ball in it. And these proteins-- it has 27 beta strands, and these are beta barrels. So there are many, many of these proteins. In fact if any of you heard Dan Cohn's talk this past semester, he's figured out how do these things get made down here and get inserted in the outer membrane. It's an interesting problem. So these are beta barrel proteins, and they have 27 strands. We then have a peptidoglycan. But the peptidoglycan is distinct. It's much smaller. It doesn't take up anywhere near as much space. And then you have your plasma membrane. And then in the plasma membrane-- so this is a plasma membrane-- you still need to do the same thing. You need to get your siderophore into the cell. So what do you have here? You still have transporters. And those transporters are going to allow your siderophore to go into the cell, just like we saw with the gram-positive. So over here then, we have a siderophore-- again, the same types of siderophores. So somehow it needs to get inside the cell. And we have many of these beta barrels, and a lot of them are specific for a given siderophores. There are many, many of these things. We'll look at E. coli. There are 10 different ways to get iron from the environment into the cell. That tells you how important all of this is. And it turns out that you also have, inside the cell, a periplasmic binding protein that can pick up the iron when it gets transferred across here. So you have a periplasmic binding protein. And one of the questions is, how does the siderophore get transferred? And to do that in gram-negative bacteria, you need a machine. And that machine is composed of three proteins. It's called the tan protein. If you look over there in pink, you have tanB. It's exbB. And this should be not C, but exbD. So there are three proteins required. And they somehow can use the proton motor force from the inner plasma membrane to allow transport across the outer membrane. So in all of these, one has tanB. So this is tanB. And tanB can recognize part of the beta barrel. So it interacts with the beta barrel protein. And this is exbD and exbB. And again, you generate a proton motor force which allows the siderophore to get into the cell. It then gets transferred to a periplasmic binding protein. And then what does it have to do? So from here, it has to go through our transporter. So let me put this up here, just like we just did before. So your siderophore-- so you can't see the bottom of my transporter-- comes through So this is the plasma membrane. And what do you have? You have the same problem. You have to get the iron out of the siderophore. And so the problem is exactly the same and gram-negative and gram-positive. So you somehow have to get it in. It's more complicated to get it in with gram-negative because of the different constructions of the peptidoglycans and the outer membrane. The other thing I wanted to say about the other outer membrane is-- which I don't know if you guys know, but I think it's incredibly important and is a major issue in a human disease-- the fact that, in addition to this outer membrane in these beta barrels, the whole outer surface is covered with sort of amazing molecules called lipopolysaccharides. So the whole outer surface is covered with LPS-- I'm not going to write it out-- lipopolysaccharides. Which, actually, one of my best friends elucidated the whole pathway for how that works. It's a beautiful, beautiful set of biochemical studies to figure out how this thing is made. It's got lipids. It's got all these sugars. It's got all this stuff hanging off of it. And this thing is really important in human health. If you read about infections, they're always talking about lipopolysaccharides. So what I'm going to do next time, just by way of showing you to introduce you to this-- you can see here in the next cartoon we have the same problem when we want to take up hemes as opposed to siderophores. And we're going to focus on hemes. So this is a cartoon, very similar to the one you just saw. And there are a couple of proteins on the outside that you need to think about. How are you going to get the heme across the peptidoglycan or into the cells? So the model is very similar. You should look at that. And then we'll see this is what your problem set is going to be on. This is for Staph aureus. And we'll see that if you get a heme, there's going to be bucket brigade that can transfer the heme through proteins covalently bound to the peptidoglycan into the cell. It's sort an amazing system, and that's what we're going to talk about for probably the first half of the next lecture. So you're going to have to read on that on your own to solve the. problem.
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https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/8.04-spring-2016.zip
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BARTON ZWIEBACH: The next thing I want to talk about for a few minutes is about the node theorem. Theorem. And it's something we've seen before. We've heard that if you have a one-dimensional potential and you have bound states, the ground state has no nodes. The first excited state has 1 node. Second, 2, 3, 4. All I want to do is give you a little intuition as to why this happens. So this will be an argument that is not mathematically very rigorous, but it's fairly intuitive and it captures the physics of the problem. So it begins by making two observations. So in the node theorem, if you have psi 1, psi 2, psi 3, all energy-- energy-- eigenstates of a one-dimensional potential-- bound states. Bound states. With energy E1 less than E2, less than E3 and E4, psi n has n minus 1 nodes. Those are points where the wave function vanishes inside the range of x. So for this square well, you've proven this by calculating all the energy eigenstates. The first state is the ground state. It has no nodes. The next state is the first excited state. It has one node. And you can write all of them, and we saw that each one has one more node than the next. Now I want to argue that in an arbitrary potential that has bound states, this is also true. So why would that be true for an arbitrary potential? The argument we're going to make is based on continuity. Suppose you have a potential like this-- V of x-- and I want to argue that this potential will have bound states and will have no node, 1 node, 2 nodes, 3 nodes. How could I argue that? Well, I would do the following. Here is the argument. Identify the minimum here. Call this x equals 0. Oh, I want to say one more thing and remind you of another fact that I'm going to use. So this is the first thing, that the square well realizes this theorem, and the second is that psi of x0 being equal to psi prime at x0 being equal to 0 is not possible. The wave function and its derivative cannot vanish at the same point. Please see the notes about this. There is an explanation in last lecture's notes. It is fact that for a second order differential equation, psi and psi prime tell you how to start the solution, and if both psi and psi prime are equal to 0, the general solution of the differential equation is always 0 everywhere. So this kind of thing doesn't happen to a wave function-- the point where it's 0 and the derivative is 0. That never happens. This happens-- 0 wave function with the derivative. But this, no. Never happens. So those two facts. And now let's do the following. Let's invent a new potential. Not this potential, but a new one that I'll mark the point minus a here and the point a here and invent a new potential that is infinite here, infinite there, and has this part I'll write there. So this will be called the screened potential. Screened potential. Va of x in which Va of x is equal to V of x for x less than a, and it's infinity for x greater than a. So that's a potential in which you turn your potential into an infinite square well whose bottom follows the potential. It's not flat. And now, we intuitively argue that as I take a to infinity, the bound states of the screened potentials become the bound states of your original potential. Because when the screen is very, very, very far away, up to infinity, you've got all your potential, and by the time you have bound states that are decaying, so the screen is not going to do much at infinity. And anyway, you can move it even further away. If you move it one light year away or two light years away, shouldn't matter. So the idea is that the bound states-- bound states-- of Va of x as a goes to infinity are the bound states of V of x. And moreover, as you slowly increase the width of the screen, the bound states evolve, but they evolve continuously. At no point a bound state blows up and reappears or does something like that. It just goes continuously. These are physically reasonable, but a mathematician would demand a better explanation. But that's OK. We'll stick to this. So let's continue there. So here is the idea, simply stated. If a is going to 0, if the width of the screen is extremely narrow, you're sitting at the bottom of the potential at x equals 0. And the screened potential is basically a very, very narrow thing, and here, there's the bottom of the potential. And for sufficiently small a-- since you picked the bottom of the potential there-- it's basically flat. And then I can use the states of the infinite square well potential. As a goes to 0, yes, you have a ground state with no nodes, a first excited state with one node, and all the states have the right number of nodes because they are the states of the infinite square well, however narrow it is. So the only thing we have to now show is that if you have a wave function-- say, let's begin with one with no nodes-- as you increase the width of the screen, you cannot get more nodes. It's impossible to change the number of nodes continuously. So here it is. I'm going to do a little diagram. So for example, let's assume the screen is this big at this moment, that you have some ground state like this. You've been growing this, and then as the screen grows bigger, you somehow have maybe a node. Could this have happened? As you increase this screen, you get a node. Now I made it on this point. I didn't intend to do that, so let me do it again somewhere. Do you get a node? Well, here was the original screen, and here the derivative psi prime is negative. Psi prime is negative. On the other hand, psi prime here is already positive. So as you grew this screen, this PSI prime that was here must have turned from negative to positive, the way it looks here. But for that, there must have been a point somewhere here when it was horizontal if it's continuous. And therefore, there must have been some point at which psi and psi prime were both 0 at the endpoint x equals a, whatever the value of a was, because psi prime here is positive, and here is negative. So at some point it was 0, but since it's at the point where you have the infinite square well, psi is also 0. And you would have both psi and psi prime equal 0, which is impossible. So basically, you can't quite flip this and produce a node because you would have to flip here, and you can't do it. One could try to make a very precise, rigorous argument, but if you have another possibility that you might think, well, you have this wave function maybe. And then suddenly it starts doing this, and at some stage, it's going to try to do this. But before it does that, at some point, it will have to be just like this and cross, but at this point, psi and psi prime would be 0. So you can intuitively convince yourself that this thing doesn't allow you to produce a node. So if you start with whatever wave function that has no nodes, as you increase the screen, you just can't produce a node. So the ground state of the whole big potential will have no nodes. And if you start with the first excited state that has one node, as you increase the screen, you still keep one node. So the next state of the full potential will have one node as well. And that way, you argue that all your bound states of the complete potential will just have the right number of nodes, which is 0, 1, 2, 3, 4. And it all came, essentially, from the infinite square well and continued.
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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 let me go back to what we were doing. The plan for today is as follows. We're going to look at this unitary time evolution and calculate this operator u, given the Hamiltonian. That will be the first order of business today. Then we will look at the Heisenberg picture of quantum mechanics. And the Heisenberg picture of quantum mechanics is one where the operators, the Schrodinger operators, acquire time dependence. And it's a pretty useful way of seeing things, a pretty useful way of calculating things as well, and makes the relation between classical mechanics and quantum mechanics more obvious. So it's a very important tool. So we'll discuss that. We'll find the Heisenberg equations of motion and solve them for a particular case today. All this material is not so to be covered in the test. The only part-- of course, the first few things I will say today about solving for the unitary operator you've done in other ways, and I will do it again this time. So going back to what we were saying last time, we postulated unitary time evolution. We said that psi at t was given by some operator U of t t0 psi t0. And then we found that this equation implied the Schrodinger equation with a Hamiltonian given by the following expression. ih dU dt of t t0 u dagger of t t0. So that was our derivation of the Schrodinger equation. We start with the time evolution. We found that, whenever we declare that states evolve in time in that way, they satisfy a first order time differential equation of the Schrodinger form in which the Hamiltonian is given in terms of U by this equation. And we talked about this operator. First we showed that it doesn't depend really on t0. Then we showed that it's Hermitian. It has units of energy. And as you may have seen already in the notes, there is a very clear correspondence between this operator and the way the dynamics follows with the ideas of Poisson brackets that are the precursors of commutators from classical mechanics. So that's in the notes. I will not go in detail in this. Many of you may have not heard of Poisson brackets. It's an interesting thing, and really that will be good enough. So our goal today is to find U given H, because as we mentioned last time, for physics it is typically more easy to invent a quantum system by postulating a Hamiltonian and then solving it than postulating a time evolution operator. So our goal in general is to find U of t t0 given H of t. That's what we're supposed to do. So the first thing I'm going to do is multiply this equation by u. By multiplying this equation by a u from the right, I will write first this term. ih dU dt of t t0 is equal to H of t U of t t0. So I multiplied this equation by u from the right. This operator is unitary, so u dagger u is one. That's why this equation cleaned up to this. Now there's no confusion really here with derivatives, so I might this well write them with normal derivatives. So I'll write this equation as d dt of U t t0 is equal to H of t U of t t0. You should be able to look at that equation and say I see the Schrodinger equation there. How? Imagine that you have a psi of t0 here, and you put it in. Then the right hand side becomes h and t acting on psi of t. And on the left hand side, this psi of t0 can be put inside the derivative because it doesn't depend on t. Therefore this becomes ih bar d dt of psi of t. So the Schrodinger equation is there. OK so now let's solve this. We'll go through three cases. Case one, h is time independent. So we're doing this sort of quickly. So H of t is really H like that. No explicit time dependence there. So what do we have? ih bar. Let's write dU dt is equal H times U. And we tried to write a solution of the form U use equal to e to the minus iHt over h bar times U0. Does that work? Well, we can think du dt and ih. So we get ih. When I take dU dt, I have to differentiate this exponential. And now in this exponential, this full operator H is there. But we are differentiating with respect to time. And H doesn't depend on time, so this is not a very difficult situation. You could imagine the power series expansion. And H, as far as this derivative goes, is like if it would be even a number. It wouldn't make any difference if it's an operator. So the derivative with respect to time of this thing is minus iH over h times the same exponential. Moreover, the position of this h could be here, or it could be to the right. It cannot be to the right of U0 though, because this is a matrix, a constant matrix that we've put in here as a possible thing for boundary condition. So so far we've taken this derivative, and then i's cancel, the h bar cancels, and you get H. But this whole thing is, again, U. So the equation has been solved. So try this. And it works. So having this solution we can write, for example, that U of t t0 is going to be e to the minus iHt over h bar, some constant matrix. When t is equal to t0, this matrix becomes the unit matrix. So this is e to the minus iHt0 over h bar times U0. And therefore from here, U0 is the inverse of this matrix, which is nothing else but e to the iHt0 over h bar. So I can substitute back here what U0 is and finally obtain U of t t0 is e to the minus iH over h bar t minus t0. And this is for h time independent. And that's our solution. There's very little to add to this. We discussed that in recitation on Thursday. This unitary operator you've been seeing that from the beginning of the course in some sense, that you evolve energy eigenstate. If this acts on any energy eigenstate, h is an energy-- if you act here on an energy eigenstate, the energy eigenstate is an eigenstate precisely for H, you can put just the number here. That is e to the, say, alpha h on a state psi n is equal to e to the alpha en psi n if h on psi n is equal to en on psi n. So the function of an operator acting on an eigenstate is just the function evaluated at the eigenvalue. So this is a rule that you've been using a really long time. OK, so when h is time independent, that's what it is. How about when h has a little time dependence? What do I call a little time dependence? A little time dependence is an idea, the sign to make it possible for you to solve the equation, even though it has some time dependence. So you could have Hamiltonians that are time dependent, but still have a simplifying virtue. So H of t is time dependent. But assume that H at t1 and h at t2 commute for all t1 and t2. So what could that be? For example, you know that the particle in a magnetic field, the spin in a magnetic field is minus gamma B dot the spin. And you could have a time dependent magnetic field, B of t times the spin. I'm not sure this is the constant gamma that they usually call gamma, but it may be. Now then if the magnetic field is time dependent, but imagine its direction is not time dependent. So if its direction is not time dependent, then, for example, you would have here minus gamma Bz of t times Sz. And the Hamiltonian at different times commute because Sz commutes with itself, and the fact that it's time independent doesn't make it fail to commute. So if you have a magnetic field that is fixed in one direction but change in time, you can have a situation where your Hamiltonian is time dependent, but still at different times it commutes. And you will discuss such case because it's interesting. But later on as we do nuclear magnetic resonance, we will have the more interesting case in which a magnetic field rotates and therefore it's not that simple. So what happens if you have a time dependent Hamiltonian that actually commutes? Well, the claim is that U of t t0 is given by a natural extension of what we had before. You would want to put exponential of minus iHt, but the reason this worked was because the derivative with respect to time brought down an iH over h bar. So one way to fix this is to put t t0 H of t prime dt prime. So this is an answer to try this. Look at this. If the Hamiltonian were to be time independent, you could take it out. And then you would get t minus t0. That brings you back to this case, so this looks reasonable. So let me call this quantity R of t. And then you notice that R dot of t, the derivative of this quantity with respect to time. Well, when you differentiate an integral the upper argument, you get just the integrand evaluated at the time represented by the upper argument of the upper limit of integration. So this is H of t. And now here comes a crucial point. You're trying to differentiate. This U is really e to the R. And you're trying to differentiate to see if the equation holds dU dt. So what is the dU dt? Would be d dt of 1 plus R plus RR plus 1 3 factor RRR. And now what happens? You differentiate here, and the first term is R dot. Here You, would have one half R dot R plus R R dot. And then 1 over 3 factorial, but three factors. R dot RR plus R R dot R plus RR R dot. But here is the claim R dot commutes with R. Claim R dot and R commute. Why is that? Well, R dot depends on H. And R is an integral of H as well, but the H at different times commute anyway, so this must be true. There's no place where you can get a contribution, because R dot is like an H, and here's an integral of H. So since the Hamiltonians are assumed to commute, R dot commutes with R. And this becomes like a normal derivative of an exponential in which you can move the R dot to the left everywhere. And you're differentiating the usual thing. So this is R dot and times the exponential of R. So actually that means that we've got pretty much our answer, because R dot is minus i over h bar H of t. And e to the R is U, so we got dU dt equals this, which is the same as this equation. The only reason a derivative with respect to time will not give the usual thing is if R and R dot fail to commute, and they don't. So you could put the R dot here. You can put R dot on the other side, because it commutes with R, but it's better here. And therefore you've got this very nice solution. So the solution is not that bad. Now finally, I want to discuss for a second the general case. So that's case-- there was a 1, a 2, a 3 H of t general. What can you do? Well, if H of t is general, there's not too much you can do. You can write something that will get you started doing things, but it's not obviously terribly useful. But it's interesting anyway that there's a way to write something that makes sense. So here it is. U of t and t0. I'll write the answer and explain how it looks, and then you will see that it's OK. It's interesting. But it probably is not the most practical way you can solve this problem. So here it is. There's an acronym for this thing. T it's called the time ordered exponential. This operator does something to the exponential function. So it's a definition. So I have to say what this time ordered exponential is, and it's the following. You take the exponential and just begin to expand. So 1 minus i over h bar-- or I'll put like this, plus minus i over h bar integral from t0 to t of dt1 H of t1. So far, so good. I've just expanded this. Now if I would continue expanding, I would get something that doesn't provide the solution. You see, this thing is the solution when the Hamiltonian at different times commute. So it's unlikely to be the solution when they don't commute. In fact, it's not the solution. So what is the next term here? The next term is you think of the exponential as you would expand as usual. So you will have here plus one half of this thing squared. So I will put something and then erase it, so maybe don't copy. One half minus i over h bar squared. And you would say, well, t0 to t dt prime H of t prime. t0 to t dt double prime H of double prime. Well, that would be just an exponential. So what is a time ordered exponential? You erase the one half. And then for notation call this t1 and t1. And then the next integral do it only up to time t1, and call this t2. So t1 will always be greater than t2, because t2 is integrated from t0 to t1. And as you integrate here over the various t1's, you just integrate up to that value. So you're doing less of the full integral then you should be doing, and that's why the factor of one half has disappeared. This can be continued. I can write the next one would be minus i over h bar cubed integral t0 to t H of t1 integral t0 to t1 dt2 H of t2. And then they next integral goes up to t2. So t0 to t2 dt3 H of t3. Anyway, that's a time ordered exponential. And I leave it to you to take the time derivative, at least to see that the first few terms are working exactly the way they should. That is, if you take a time derivative of this, you will get H times that thing. So since it's a power series, you will differentiate the first term, and you will get the right thing. Then the second term and you will start getting everything that you need. So it's a funny object. It's reassuring that something like this success, but in general, you would want to be able to do all these integrals and to sum them up. And in general, it's not that easy. So it's of limited usefulness. It's a nice thing that you can write it, and you can prove things about it and manipulate it. But when you have a practical problem, generally that's not the way you solve it. In fact, when we will discuss the rotating magnetic fields for magnetic resonance, we will not solve it in this way. We will try to figure out the solution some other way. But in terms of completeness, it's kind of pretty in that you go from the exponential to the time ordered exponential. And I think you'll see more of this in 806. So that's basically our solution for H and for the unitary operator U in terms of H. And what we're going to do now is turn to the Heisenberg picture of quantum mechanics. Yes, questions? AUDIENCE: Why does R dot [INAUDIBLE]? PROFESSOR: Because that's really a property of integrals. d dx integral up to x from x0 g of x prime dx prime is just equal to g of x. This is a constant here, so you're not varying the integral over in this limit. So if this limit would also be x dependent, you would get another contribution, but we only get the contribution from here. What's really happening is you're integrating up to x, then up to x plus epsilon subtracting, so you pick up the value of the function of the upper limit. Yes? AUDIENCE: So what happens to the T that was pre factor? PROFESSOR: What happens to this T? AUDIENCE: Yeah, what happens? PROFESSOR: That's just a symbol. It says time order the following exponential. So at this stage, this is a definition of what t on an exponential means. AUDIENCE: OK. PROFESSOR: It's not-- let me say T is not an operator in the usual sense of quantum mechanics or anything like that. It's an instruction. Whenever you have an exponential of this form, the time ordered exponential is this series that we've written down. It's just a definition. Yes? AUDIENCE: So when we have operators in differential equations, do we still get [INAUDIBLE]? PROFESSOR: If we have what? AUDIENCE: If we have operators in differential equations do we still get unique [INAUDIBLE] solutions? PROFESSOR: Yes, pretty much. Because at the end of the day, this is a first order matrix differential equation. So it's a collection of first order differential equations for every element of a matrix. It's pretty much the same as you have before. If you know the operator at any time, initial time, with the differential equation you know the operator at a little bit time later. So the operator is completely determined if you know it initially and the differential equation. So I think it's completely analogous. It's just that it's harder to solve. Nothing else. One last question. AUDIENCE: So let's say that we can somehow fly in this unitary operator, and then we have a differential equation, and we somehow, let's say, get a wave function out of it. What is the interpretation of that wave function? PROFESSOR: Well, it's not that we get the wave function out of this. What really is happening is that you have learned how to calculate this operator given H. And therefore now you're able to evolve any wave function. So you have solved the dynamical system. If somebody tells you a time equals 0, your system is here, you can now calculate where it's going to be at the later time. So that's really all you have achieved. You now know the solution. When you're doing mechanics and they ask you for an orbit problem, they say at this time the planet is here. What are you supposed to find? x is a function of time. You now know how it's going to develop. You've solved equations of motion. Here it's the same. You know the wave function of time equals. If you know it at any time, you've solved problem. OK, so Heisenberg picture of quantum mechanics. Heisenberg picture. So basically the Heisenberg picture exists thanks to the existence of the Schrodinger picture. Heisenberg picture of quantum mechanics is not something that you necessarily invent from the beginning. The way we think of it is we assume there is a Schrodinger picture that we've developed in which we have operators like x, p, spin, Hamiltonians, and wave functions. And then we are going to define a new way of thinking about this, which is called the Heisenberg picture of the quantum mechanics. So it all begins by considering a Schrodinger operator As hat, which is s is for Schrodinger. And the motivation comes from expectation values. Suppose you have time dependent states, in fact, matrix elements. One time dependent state alpha of t, one time dependent state beta of t. Two independent time dependent states. So you could ask what is the matrix element of A between these two time dependent states, a matrix element. But then, armed with our unitary operator, we know that As is here, and this state beta at time t is equal to U of t comma 0 beta at time 0. And alpha t is equal to alpha at 0 U dagger of t0. So the states have time dependence. But the time dependence has already been found, say, in principle, if you know U dagger. And then you can speak about the time dependent matrix elements of the operator As or the matrix element of this time dependent operator between the time equals 0 states. And this operator is sufficiently important that this operator is called the Heisenberg version of the operator s. Has time dependence, and it's defined by this equation. So whenever you have Schrodinger operator, whether it be time dependent or time independent, whatever the Schrodinger operator is, I have now a definition of what I will call the Heisenberg operator. And it is obtained by acting with a unitary operator, U. And operators always act on operators from the left and from the right. That's something that operators act on states from the left. They act on the state. But operators act on operator from the left and from the right, as you see them here, is the natural, ideal thing to happen. If you have an operator that's on another from the right only or from the left only, I think you have grounds to be suspicious that maybe you're not doing things right. So this is the Heisenberg operator. And as you can imagine, there's a lot of things to be said about this operator. So let's begin with a remark. Are there questions about this Heisenberg operator. Yes? AUDIENCE: Do we know anything about the Schrodinger operator? PROFESSOR: You have to speak louder. AUDIENCE: Is the Schrodinger operator related to the Hamiltonian [INAUDIBLE]? PROFESSOR: Any Schrodinger operator, this could be the Hamiltonian, this could be x hat, it could be Sz, could be any of the operators you know. All the operators you know are Schrodinger operators. So remarks, comments. OK, comments. One, at t equals 0 A Heisenberg becomes identical to A Schrodinger at t equals 0. So look why. Because when t is equal to 0, U of t-- of 0 0 is the operator propagates no state, so it's equal to the identity. So this is a wonderful relation that tell us you that time equals 0 the two operators are really the same. And another simple remark. If you have the unit operator in the Schrodinger picture, what is the unit operator in the Heisenberg picture? Well, it would be U t 0 dagger 1 U t 0. But 1 doesn't matter. U dagger with U is 1. This is a 1 Schrodinger, and therefore it's the same operator. So the unit operator is the same. It just doesn't change whatsoever. OK, so that's good. But now this is something interesting also happens. Suppose you have Schrodinger operator C that is equal to the product of A with B, two Schrodingers. If I try to figure out what is CH, I would put U dagger-- avoid all the letters, the t 0. It's supposed to be t 0. Cs U. But that's equal U dagger As Bs U. But now, in between the two operators, you can put a U U dagger, which is equal to 1. So As U U dagger Bs U. And then you see why this is really nice. Because what do you get is that C Heisenberg is just A Heisenberg times B Heisenberg. So if you have C Schrodinger equals A Schrodinger, B Schrodinger, C Heisenberg is A Heisenberg B Heisenberg. So there's a nice correspondence between those operators. Also you can do is for commutators. So you don't have to worry about this thing. So for example, if A Schrodinger with B Schrodinger is equal to C Schrodinger, then by doing exactly the same things, you see that A Heisenberg with B Heisenberg would be the commutator equal to C Heisenberg. Yes? AUDIENCE: That argument for the identity operators being the same in both pictures. If the Hamiltonian is time independent, does that work for any operator that commutes with the Hamiltonian? PROFESSOR: Hamiltonian is [INAUDIBLE]. AUDIENCE: Because then you can push the operator just through the exponential of the Hamiltonian. PROFESSOR: Yeah, we'll see things like that. We could discuss that maybe a little later. But there are some cases, as we will see immediately, in which some operators are the same in the two pictures. So basically operators that commute with the Hamiltonian as you say, since U involves the Hamiltonian, and this is the Hamiltonian, if the operator commutes with the Hamiltonian and you can move them across, then they are the same. So I think it's definitely true. So we will have an interesting question, in fact, whether the Heisenberg Hamiltonian is equal to the Schrodinger Hamiltonian, and we'll answer that very soon. So the one example that here I think you should keep in mind is this one. You know this is true. So what do you knowing the Heisenberg picture? That X Heisenberg of t times P Heisenberg of t commutator is equal to the Heisenberg version of this. But here was the unit operator. And therefore this is just ih bar times the unit operator again, because the units operator is the same in all pictures. So these commutation relation is true for any Heisenberg operator. Whatever commutation relation you have of Schrodinger, it's true for Heisenberg as well. OK, so then let's talk about Hamiltonians. Three, Hamiltonians. So Heisenberg Hamiltonian by definition would be equal to U dagger t 0 Schrodinger Hamiltonian times U of t 0. So if the Schrodinger Hamiltonian-- actually, if Hs at t1 commutes units with Hs at t2, the Schrodinger Hamiltonian is such that for all t1 and t2 they commute with each other. Remember, if that is the case, the unitary operator is any way built by an exponential. It's this one. And the Schrodinger Hamiltonians commute. So as was asked in the question before, this thing commutes with that, and you get that they are the same. So if this is happening, the two Hamiltonians are identical. And we'll have the chance to check this today in a nice example. So I will write in this as saying the Heisenberg Hamiltonian as a function of time then is equal to the Schrodinger Hamiltonian as a function of time. And this goes Hs of t1 and Hs of t2 commute. OK, now I want you to notice this thing. Suppose the Hs of t is some Hs of x,p, and t, for example. OK, now you come and turn it into Heisenberg by putting a U dagger from the left and a U from the right. What will that do? It will put U dagger from the left, U dagger on the right. And then it will start working it's way inside, and any x that it will find will turn into a Heisenberg x. Any p will turn into Heisenberg p. Imagine, for example, any Hamiltonian is some function of x. It has an x squared. Well the U dagger and U come and turn this into x Heisenberg squared. So what I claim here happens is that H Heisenberg of t is equal to U dagger H Schrodinger of x, p, t, U. And therefore this becomes H Schrodinger of x Heisenberg of t, P Heisenberg of t, and t. So here is what the Heisenberg Hamiltonian is. It's the Schrodinger Hamiltonian where X's, and P's, or spins and everything has become Heisenberg. So the equality of the two Hamiltonians is a very funny condition on the Schrodinger Hamiltonian, because this is supposed to be equal to the Schrodinger Hamiltonian, which is of x, p, and t. So you have a function of x, p, and t. And you put X Heisenberg P Heisenberg, and somehow the whole thing is the same. So this is something very useful and we'll need it. One more comment, expectation values. So this is three. Comment number four on expectation values, which is something you've already-- it's sort of the way we began the discussion and wanted to make sure it's clear. So four, expectation values. So we started with this with alpha and beta, two arbitrary states, matrix elements. Take them equal and to be equal to psi of t. So you would have psi t As psi t is, in fact, equal to psi 0 A Heisenberg psi 0. Now that is a key equation. You know you're doing expectation value at any given time of a Schrodinger operator, turn it into Heisenberg and work at time equals 0. It simplifies life tremendously. Now this is the key identity. It's the way we motivated everything in a way. And it's written in a way that maybe it's a little too schematic, but we write it this way. We just say the expectation value of As is equal to the expectation value of AH. And this, well, we save time like that, but you have to know what you mean. When you're computing the expectation value for a Schrodinger operator, you're using time dependent states. When you're computing the expectation value of the Heisenberg operator, you're using the time equals 0 version of the states, but they are the same. So we say that the Schrodinger expectation value is equal to the Heisenberg expectation value. We right it in the bottom, but we mean the top equation. And we use it that way. So the Heisenberg operators, at this moment, are a little mysterious. They're supposed to be given by this formula, but we've seen that calculating U can be difficult. So calculating the Heisenberg operator can be difficult sometimes. So what we try to do in order to simplify that is find an equation that is satisfied by the Heisenberg operator, a time derivative equation. So let's try to find an equation that is satisfied by the Heisenberg operator rather than a formula. You'll say, well, this is better. But the fact is that seldom you know U. And even if you know U, you have to do this simplification, which is hard. So finding a differential equation for the operator is useful. So differential equation for Heisenberg operators. So what do we want to do? We want to calculate ih bar d dt of the Heisenberg operator. And so what do we get? Well, we have several things. Remember, the Schrodinger operator can have a bit of time dependence. The time dependence would be an explicit time dependence. So let's take the time derivative of all this. So you would have three terms. ih bar dU dagger dt As U plus U dagger As dU dt plus-- with an ih bar-- U dagger ih bar dAs minus dt. dAs dt and U. Well, you have these equations. Those were the Schrodinger equations we started with today. The derivatives of U, or the derivatives of U dagger. so what did we have? Well, we have that ih bar dU dt was HU-- H Schrodinger times U. And therefore ih bar dU dagger dt. I take the dagger of this. I would get a minus sign. I would put it on the other side. Is equal to U dagger Hs with a minus here. And all the U's are U's of t and t0. I ran out of this thick chalk. So we'll continue with thin chalk. All right, so we are here. We wrote the time derivative, and we have three terms to work out. So what are they? Well we have this thing, ih bar this. So let's write it. ih bar d d dt of As-- of A Heisenberg, I'm sorry-- Is equal to that term is minus U dagger Hs A Schrodinger U. The next term plus ih bar dU dt on the right. So we have plus U dagger As Hs dU dt, so U. Well, that's not bad. It's actually quite nice. And then the last term, which I have very little to say, because in general, this is a derivative of a time dependent operator. Partial with respect to time, it would be 0 if As depends, just say, on X, on P, on Sx, or any of those things, has to have a particular t. So I will just leave this as plus ih bar dAs dt Heisenberg. The Heisenberg version of this operator using the definition that anything, any operator that we have a U dagger in front, a U to the right, is the Heisenberg version of the operator. So I think I'm doing all right with this equation. So what did we have? Here it is. ih bar d dt of A Heisenberg of t. And now comes the nice thing, of course. This thing, look at it. U dagger U. This turns everything here into Heisenberg. H Heisenberg, A Heisenberg. Here you have A Heisenberg H Heisenberg, and what you got is the commutator between them. So this thing is A Heisenberg commutator with H Heisenberg. That whole thing. And then you have plus ih bar dAs dt Heisenberg. So that is the Heisenberg equation of motion. That is how you can calculate a Heisenberg operator if you want. You tried to solve this differential equation, and many times that's the simplest way to calculate the Heisenberg operator. So there you go. It's a pretty important equation. So let's consider particular cases immediately to just get some intuition. So remarks. Suppose As has no explicit time dependence. So basically, there's no explicit t, and therefore this derivative goes away. So the equation becomes ih bar dAh, of course, dt is equal to Ah Heisenberg sub h of t. And you know the Heisenberg operator is supposed to be simpler. Simple. If the Schrodinger operator is time independent, it's identical to the Schrodinger Hamiltonian. Even if the Schrodinger operator has time dependence, but they commute, this will become the Schrodinger Hamiltonian. But we can leave it like that. It's a nice thing anyway. Time dependence of expectation value. So let me do a little remark on time dependence of expectation values. So suppose you have the usual thing that you want to compute. How does the expectation value of a Schrodinger operator depend on time? You're faced with that expectation value of As, and it changes in time, and you want to know how you can compute that. Well, you first say, OK, ih bar d dt. But this thing is nothing but psi 0 A Heisenberg of t psi 0. Now I can let the derivative go in. So this becomes psi 0 ih bar dAh dt psi 0. And using this, assuming that A is still no time dependence, A has no explicit time dependence, then you can use just this equation, which give you psi 0 Ah Hh psi 0. So all in all, what have you gotten? You've gotten a rather simple thing, the time derivative of the expectation values. So ih bar d dt. And now I write the left hand side as just expectation value of H Heisenberg of t. And on the left hand side has to the A Schrodinger expectation value, but we call those expectation values the same thing as a Heisenberg expectation value. So this thing becomes the right hand side is the expectation value of A Heisenberg H Heisenberg like that. And just the way we say that Heisenberg expectation values are the same as Schrodinger expectation values, you could as well write, if you prefer, as d dt of A Schrodinger is equal to the expectation value of A Schrodinger with H Schrodinger. It's really the same equation. This equation we derived a couple of lectures ago. And now we know that the expectation values of Schrodinger operators are the same as the expectation value of their Heisenberg counterparts, except that the states are taking up time equals 0. So you can use either form of this equation. The bottom one is one that you've already seen. The top one now looks almost obvious from the bottom one, but it really took quite a bit to get it. One last comment on these operators. How about conserved operators? What are those things? A time independent As is set to be conserved if it commutes with a Schrodinger Hamiltonian. If As commutes with As equals 0. Now you know that if As with Hs is 0, Ah with Hh is 0, because the map between Heisenberg and Schrodinger pictures is a commutator that is valued at the Schrodinger picture is valued in the Heisenberg picture by putting H's. So what you realize from this is that this thing, this implies Ah commutes with Hh. And therefore by point 1, by 1, you have to dAh dt is equal to 0. And this is nice, actually. The Heisenberg operator is actually time independent. It just doesn't depend on time. So a Schrodinger operator, it's a funny operator. It doesn't have time in there. It has X's, P's, spins, and you don't know in general, if it's time independent in the sense of conserve of expectation values. But whenever As commutes with Hs, well, the expectation values don't change in time. But as you know, this d dt can be brought in, because the states are not time dependent. So the fact that this is 0 means the operator, Heisenberg operator, is really time independent. Whenever you have a Schrodinger operator, has no t, the Heisenberg one can have a lot of t. But if the operator is conserved, then the Heisenberg operator will have no t's after all. It will really be conserved. So let's use our last 10 minutes to do an example and illustrate much of this. In the notes, there will be three examples. I will do just one in lecture. You can do the other ones in recitation next week. There's no need really that you study these things at this moment. Just try to get whatever you can now from the lecture, and next week you'll go back to this. So the example is the harmonic oscillator. And it will illustrate the ideas very nicely, I think. The Schrodinger Hamiltonian is p squared over 2m plus 1/2 m omega squared x hat squared. OK, I could put x Schrodinger and p Schrodinger, but that would be just far too much. x and p are the operators you've always known. They are Schrodinger operators. So we leave them like that. Now I have to write the Heisenberg Hamiltonian. Well, what is the Heisenberg Hamiltonian? yes? AUDIENCE: It's identical. PROFESSOR: Sorry? AUDIENCE: It's identical. PROFESSOR: Identical, yes. But I will leave that for a little later. I will just assume, well, I'm supposed to do U dagger U. As you said, this is a time independent Hamiltonian. It better be the same, but it will be clearer if we now write what it should be in general. Have a U dagger and a U from the right. They come here, and they turn this into P Heisenberg over 2m plus 1/2 m omega squared x Heisenberg. OK, that's your Heisenberg Hamiltonian. And we will check, in fact, that it's time independent. So how about the operators X Heisenberg and P Heisenberg. What are they? Well, I don't know how to get them, unless I do this sort of U thing. That doesn't look too bad but certainly would be messy. You would have to do an exponential of e to the minus iHt over t with the x operator and another exponential. Sounds a little complicated. So let's do it the way the equations of the Heisenberg operators tell you. Well, X and P are time independent Schrodinger operators, so that equation that I boxed holds. So ih dx Heisenberg dt is nothing else than X Heisenberg commuted with H Heisenberg. OK, can we do that commutator? Yes. Because X Heisenberg, as you remember, just commutes with P Heisenberg. So instead of the Hamiltonian, you can put this. This is X Heisenberg P Heisenberg squared over 2m. OK well, X Heisenberg P Heisenberg is like you had X and P. So what is this commutator? You probably know it by now. You act with this and these two p. So it acts on one, acts on the other, gives the same on each. So you get P Heisenberg times the commutator of X and P, which is ih bar times a factor of 2. So we could put hats. Better maybe. And then what do we get? The ih there and ih cancels. And we get some nice equation that says dX Heisenberg dt is 1 over m P Heisenberg. Well, it actually looks like an equation in classical mechanics. dx dt is P over m. So that's a good thing about the Heisenberg equations of motion. They look like ordinary equations for dynamical variables. Well, we've got this one. Let's get P. Well, we didn't get the operator still, but we got an equation. So how about P dP dt. So ih dP Heisenberg dt would be P Heisenberg with H Heisenberg. And this time only the potential term in here matters. So it's P Heisenberg with 1/2 m omega squared X Heisenberg squared. So what do we? We get 1/2 m omega squared. Then we get again a factor of 2. Then we get one left over Xh. And then a P with Xh, which is a minus ih bar. So the ih bars cancel, and we get dPh dt is equal to-- the h bar cancelled-- m omega squared Xh. Minus m. All right, so these are our Heisenberg equations of motion. So how do we solve for them now? Well, you sort of have to try the kind of things that you would do classically. Take a second derivative of this equation. d second Xh dt squared would be 1 over m dPh dt. And the dPh dt would be [INAUDIBLE] 1 over m times minus m omega squared Xh. So d second Xh dt squared is equal to minus omega squared Xh, exactly the equation of motion of a harmonic oscillator. It's really absolutely nice that you recover those equations that you had before, except that now you're talking operators. And it's going to simplify your life quite dramatically when you try to use these operators, because, in a sense, solving for the time dependent Heisenberg operators is the same as finding the time evolution of all states. This time the operators change, and you will know what they change like. So you have this. And then you write Xh is equal to A cosine omega t plus B sine omega t where A and B are some time independent operators. So Xh of t, well, that's a solution. How about what is P? Ph of t would be m dX m dx dt. So you get minus m omega sine omega tA plus m omega cosine omega tB. OK, so that's it. That is the most general solution. But it still doesn't look like what you would want, does it? No, because you haven't used the time equals 0 conditions. At time equals 0, the Heisenberg operators are identical they to the Schrodinger operators. So at t equals 0, Xh of t becomes A, but that must be X hat, the Schrodinger operator. And at t equals 0, Ph of t becomes equal to this is 0 m omega B. And that must be equal to the P hat operator. So actually we have already now A and B. So B from here is P hat over m omega. And therefore Xh of t is equal to A, which is X hat cosine omega t plus B, which is P hat over m omega sine omega t. An Ph of t is here. A is-- Ph of t is m omega B, which is [INAUDIBLE] P. So it's P hat cosine omega t minus m omega X hat sine omega t. So let's see. I hope I didn't make mistakes. P hat minus m omega X hat sine omega t. Yep, this is correct. This is your whole solution for the Heisenberg operators. So any expectation value of any power of X and P that you will want to find its time dependence, just put those Heisenberg operators, and you will calculate things with states at time equals 0. It will become very easy. So the last thing I want to do is complete the promise that we had about what is the Heisenberg Hamiltonian. Well, we had the Heisenberg Hamiltonian there. And now we know the Heisenberg operators in terms of the Schrodinger one. So Hh of t is equal to Ph-- 1/2m Ph squared. So I have P hat cosine omega t minus m omega X hat sine omega t squared plus 1/2 m omega squared Xh squared. So X hat cosine omega t plus P hat over m omega sine omega t squared. So that's what the Heisenberg Hamiltonian is. So let's simplify this. Well, let's square these things. You have 1/2m cosine squared omega t P hat squared. Let's do the square of this one. You would have plus 1/2m m squared omega squared sine squared omega t X squared. And then we have the cross product, which would be plus-- or actually minus 1/2m. The product of these two things. m omega sine omega t cosine omega t. And you have Px plus XP. OK, I squared the first terms. Now the second one. Well, let's square the P squared here. What do we have? 1/2 m omega squared over m squared omega squared sine squared of omega t P squared. The x plus 1/2 m omega squared cosine squared omega t X squared. And the cross term. Plus 1/2 m omega squared over m omega times cosine omega t sine omega t XP plus PX. A little bit of work, but what do we get? Well, 1/2 m. And here we must have 1/2 m, correct. 1/2 m. Sine squared omega t P squared. So this is equal 1/2 m P squared. These one's, hall here you have 1/2 m omega squared. So it's 1/2 m omega squared cosine and sine squared is X hat squared. And then here we have all being over 2. And here omega over 2, same factors, same factors, opposite signs. Very good. Schrodinger Hamiltonian. So you confirm that this theoretical expectation is absolutely correct. And what's the meaning? You have the Heisenberg Hamiltonian written in terms of the Heisenberg variables. But by the time you substitute these Heisenberg variables in, it just becomes identical to the Schrodinger Hamiltonian. All right, so that's all for today. I hope to see in office hours in the coming days. Be here Wednesday 12:30, maybe 12:25 would be better, and we'll see you then. [APPLAUSE]
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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, well, it's time to start. So last time we were talking about e plus e minus to jets, and-- I should have written that-- in particular, e plus e minus to dijets. And we were talking about-- we talked about factorization and how to derive it. And I've schematically written the result here. Last time we had a more definite formula with all the arguments made explicit. So there were hard functions, jet functions, and the soft function, and we could either think about measuring invariant masses in two hemispheres and constraining them to be small to know we have dijets, or measuring kind of the sum of the two, which we called-- which is effectively the thrust, and then we get a slightly simpler formula because we can project these guys down onto guys that'd only have-- or functions of one variable. But either way, in either case, we have a factorization theorem that separates the degrees of freedom and kind of the picture is that we have these kind of scales in the problem. So this is new jet, new hard, and new soft, and we can use this factorization theorem to separate those scales into these functions. And then we can use-- I started talking about using renormalization group equations in order to sum the logs between these scales. And I wanted to finish up that discussion. So for the Wilson coefficient of the operator or, for that matter, for the hard function, which is the square of the Wilson coefficient, you have a very simple renormalization group equation. And that's because momentum conservation is enough to fix all the variables that label collinear fields. So you really just have this overall Q, which is the center of mass energy, and that's the only variable that it's fixed by kinematics. There's no convolutions. In the case of these things called jet functions, which we talked a little bit about last time, you end up with something like the Altarelli-Parisi where there's an integral. It's a little bit different in the sense that we actually know its structure-- tolerance and perturbation theory-- it has this structure. So it only has a very particular dependence on S, has to scale like 1 over S, so one thing that scales like 1 over S is a delta function or a plus function, and basically, there's only this single type of plus function that can show up, and that's the analog actually of something that happened to this guy where we said there was only a possibility of a single logarithm showing up. OK. So this is how far we got. Whenever you have something-- whenever you have equations where you have convolutions like this, what you should do is Fourier transform because if you Fourier transform something that's a convolution, it becomes a product. So we'll Fourier transform, and in order to be careful about convergence of our Fourier transform, we give it a small imaginary part. And we define a position space anomalous dimension, as the anomalous dimension up there that depend on S, but now Fourier transformed and likewise, we define a position space jet function the same way, and then this formula here, which is a convolution, if you use these in the inverse transforms, you arrive at a very simple formula for the position space. Gamma J. So the Fourier transform of a delta function is just the identity so there's this term here, becomes that term there. The Fourier transform of a plus function is actually a logarithm, and in order to get that result, you do need to have this convergence. This little i0 in there. So in general, if you have there's a general kind of relation between logs and plus functions like that, and in Fourier space, the kind of highest logarithm you get is L to the K plus 1. OK, so these logs here, you're basically-- if you think about counting them, you should count this one over S as an extra log, and in Fourier space, that becomes explicit. You really just get a log to the K plus 1. All right, but this formula here, is something that, again, is of this kind of multiplicative form. So what happens to the anomalous dimension in the Fourier space is that it has a multiplicative form. All right so we have mu debugging mu of J Y comma mu is simply gamma J Y comma mu J Y comma mu. So it's no more difficult than this formula up here. Just have to find the right space, which is Fourier space. So then we can make an all order solution of that formula. Given that we know this, we can just plug-in this and integrate, and we've done this a few times now. Let me just do it formally by not doing the integrals, but writing them out in a way that you could do them to show you what the all order solution would look like. So you can write the solution using this fact and using this fact. So we have to use that fact to convert this logarithm here into something that is just in terms of alphas. But using those two facts, we can do the same kind of things that we've done before and write the all order solution in the following way. The integrals are contained in this W and this K. Here's the noncusp piece, and then the cusp piece comes with two integrals because it had a logarithm, and we turned one of the logarithms into an integral as well. So hopefully I've gotten all my twos right. So we saw something similar when we were talking about the running of the Wilson coefficient of the C, because it also had this kind of form with just a logarithm and a constant term. And it's a similar thing here. So this is an all order solution for the running of the jet function. Of course, we don't know these functions to all orders. We only know them up to some given order, and so you plug those-- whatever order you want to work-- you plug it into this formula, and then you can do this integral, and then you figure out these factors here. And those factors are summing the logs. In the case of the jet function, what you would-- the way that you can think about this picture is that you want to-- you have to put the guys in the factorization theorem at a common scale, and so one way of thinking about it is as follows. Let's take our hard function and do perturbation theory at mu equals mu age, and then we'll sum logs down to say the soft scale. That's one way of doing it, and we'll do perturbation theory for the jet function, at a scale mu equals mu J, and then we'll use this renormalization group here, which will just say gamma J to run this guy down to the soft scale. In that kind of scenario, you wouldn't have to run the soft function. This is like a top down kind of picture where you're just running the objects at the higher scales down to the lowest scale. That's one way of doing it. You could also do it in an equivalent way where you run the soft function, for example, up to the jet scale, and you don't run the jet function. And that'll give you actually the same result. But in the way-- with the information I've presented you, this is the way that you would think about doing the renormalization group, and that would sum up logarithms in the cross section. And it's very easy if you write it in position space to see what type of logarithms you're summing out. So in position space, if we go back to a formula like the one at the top, basically, what happens is that in position space, you, again, have a product. So these convolutions here, which were also this form, if you Fourier transform, then you end up with a product. D sigma D, so schematically, D sigma Dy for, say, where y is the Fourier transform for thrust, would just be h times j in position space times the soft function in position space as well. And there's Q somewhere, but it would be a very simple formula which just has a product. And so you can figure out what logs you're summing. The logs that you're summing, again, just as we talked about before, are simplest to describe in when you take the log because you're really summing logs in an exponent. So if we talk about the log of D sigma Dy, then we can enumerate the types of logs that one sums as follows. So these are the leading logs. This is next leading log, and this is next to next leading log, just like before. When we were talking about enumerating the logs, we talked about before an example, we were enumerating logs in the Wilson coefficient C. Now we're doing it for a full cross section, but because in position space, the cross section is simply a product of objects, you can think about just-- if you take the logarithm, they kind of split apart. And it's very simple to enumerate what corresponds to the things that you would get by putting in anomalous dimensions at a given order. So if you use the leading log anomalous dimension, just the one loop cusp, then you get these terms. Get the higher terms from the running of the coupling. If you put in the next leading log terms, then you get those terms. OK. And so you supplement that resummation which comes from this K and this omega, in general, you supplement that with sort of fixed order calculations of the H and the J and the S. And that gives you a complete cross-section at some order and resum perturbation theory. So you could do that also in momentum space. You could write out the formula in momentum space just because after all that's what you want in the end. Position space is just really a nice way of deriving the results that then you eventually put back into momentum space in some way or another because that's where the data is. And let me just show you what the formula would look like, just so you get a kind of picture with all the arguments. So if I were to Fourier transform the resum formula, over here I didn't write the resummation in. I didn't write in the evolution factors that correspond to this. So let me do that, but let me do it in momentum space. So I guess I made a slightly different choice here. So I'm doing it for the thrust case. These convolutions just being integrals over the variables that are like this S prime, for example. So this is S prime integral. And this is an L prime integral. OK so there's three integrals in this formula. I guess this is-- it's kind of a funny notation, but I could have written integral instead of writing these tensor signs. That would have been probably more clear. Do that. OK so there's the formula in momentum space. And it's doing what I said. This factor here is running the hard function down to the soft scale, and this other factor here is running from the jet scale to the soft scale, and I didn't have to run the soft function. I should have written in here that I evaluate this guy at the soft scale. And this guy here remember is the non-perturbative guy which we also talked about a little last time. OK so that's kind of the basic structure these use as usual are summing the logarithms, and then you put fixed order results in for this guy, for that guy, and for this guy here. And in doing so you get the [INAUDIBLE] result. All right, so this fact that I could have done the running differently, that I could have run the soft function is a kind of consistency. So there are other ways of doing the r g, and they all lead to the same answer. And when I say the same answer, I really mean the same precisely exactly the same number. So I really mean the same, not just sort of approximately the same but exactly the same. And the other ways of doing the RGE are the basic idea of why there are more than one way is again this fact that if I do coefficient renormalization, that that's the inverse of doing operator renormalization. So there's two ways of thinking about doing the run. You run the operators, you run the coefficients. OK in this picture it's a little bit more complicated because we have three scales, but when I'm running the hard function, I quivalently could have run both the jet and the soft function and not run the hard function. So let me maybe give one other way of doing this just to show you. So I could have run the hard function say, just down to the jet scale. And then instead of running the jet function, I could have run the soft function up to the jet scale. That would be another way of doing the RGE, and that would lead to an equivalent picture. And there's more than two, because there's three scales in this. There's more than two natural things. There's three possible natural things, where you would run to either this scale, this scale, or this scale, but at the base of it, it all has to do with this equivalence that we talked about in simpler scenarios earlier. And you can write this also as a relation between anomalous dimensions. So it implies a non-trivial things about the formula that have to be trivial in order for this to work. So for example, it implies that the coefficients of the cusp anomalous dimension in the various renormalization groups would be in a certain way that they would be related, and even in the non-cusp anomalous dimensions, which is this formula, there's a relation between the jet, hard, and soft anomalous dimensions. OK? And that's needed-- this is an expression of this fact that you can do the renormalization group by running different objects. So you can derive this formula by just saying mu D divided by the cross section is 0. And going through it, if you were to allow, then you would find formulas like this one. OK? So questions? I went quick, because most of the concepts here are things that we've seen before in simpler guises. The new complication is really that we have this dependence on a variable which ended up being y, but by going to Fourier space, it looked just like a simple product again, and y behaved like a simple kinematic variable. The end of the day, we want to Fourier transform back, but we have a space where things look familiar. OK. So that's e plus minus the dijets, and that's the last SCETI example I'm going to do for a while. I'm now going to turn to doing SCETII, and we'll spend some time talking about SCETII. So in SCETII, instead of having ultrasoft interactions with collinear particles, we have soft interactions with collinear particles. So we have to call how that's going to work. So if we take a soft particle, consider a soft particle interacting with some collinear particle, and ask what do we get out? So if we just ask about momentum conservation, and we call this q, then q is equal to q soft plus q collinear. And since we know the scaling of these two, we know the scaling of q. It's just given by whatever the larger of the scalings is. So soft and collinear both have the same-- so let me remind you of the scaling. So this guy was q, lambda, lambda, lambda, and this guy was q lambda squared 1 lambda. So the difference between ultrasoft and collinear and soft and collinear is that soft and collinear have the same size of perp momenta. So the perp momenta are just of order lambda. Obviously, in the minus momenta, the 1 wins, so the minus momentum of order 1. And then the plus momentum, it's the soft that wins, because it's bigger than the collinear. So this is order lambda. OK. And that's actually much different than what we found when we added ultrasoft and collinear. Because when we added ultrasoft and collinear we got back collinear. Here, we're not getting back collinear. This is actually off-shell, from the perspective of our low-energy modes. Because q squared, the biggest part of q squared would be 1 times lambda. q squared's of order lambda which is much bigger than lambda squared. So adding a soft and a collinear particle in SCETII immediately gives you something off-shell. That's going to make some things easier and some things more complicated. Mostly, it's going to make things a little more complicated. At least at the start, it'll looking easier. So you could think about this from our mode picture, where we had softs that live here and collinears that live there. In order for them to interact, they have to go up to a place where we can have a common momentum, and that's a higher hyperbola, like this. It's not all the way up to the hard scale which is up here. That was hard, but there's an intermediate scale that comes in which is the scale of this q squared. So this is q squared. This is the order of the q squared there, and sometimes this is called a hard collinear scale. This would be called a hard collinear mode, hc. And you could even write down, if you wanted, an on-shell degree of freedom for this hard collinear mode. So an on-shell version of it, you could think of it as a mode, in theory. An on-shell version of it would have a scaling that's q 1 lambda root lambda. So you can think of one way of approaching it which we'll take this attitude in a minute. One way of approaching this problem would be to first think about doing some matching onto a theory with this hard collinear mode and then trying to match down onto a theory that just has the collinear and the soft mode, and we'll exploit that a little later. But first, let me do a different thing and just ignore this dashed line and just think about matching directly from the QCD onto the SCETII which just has these two degrees of freedom. So what's going to happen is that we're going to have to integrate out more things, because any type of interactions that we write down are basically giving us something off-shell. So let's do an example of this. So we'll do an example of a heavy-to-light current, but now, it won't be an SCETII current but rather an SCETI. So I just want to have one soft particle and one collinear particle. Simplest possible example, but now this is soft, and this is collinear, not ultrasoft and collinear. All right. So we have some current. I'm just going to label the lines with c's and s's. So imagine that you attached a soft gluon here. That would give you something that's off-shelf. So this line here is off-shelf, and likewise, on this side, if this is soft coming in but you attach something collinear here, then this is off-shelf. Before, when we had this picture, and we were doing it for SCETI, one set of attachments which still need to on-shell, in particular on this side. And on this side, we got something off-shell. We integrated it out, and we got a Wilson line. Right? Here, it's a little more complicated, because you touch alternate modes on either side. And you get something off-shell, and you just have to put out this pink line all at once, and that's going to give you another type of Wilson line. In this calculation, we're going to get W. It's both of the collinear modes, and another type of Wilson line from the soft modes. Let's call it Sn. So how do you build up a Wilson line? Well, first of all, think about attaching more gluons onto this line. Right? That's kind of the analog of what we did before. So just attach more soft gluons here, more collinear gluons here. This might be the first thing you would try, just adding those guys up, and that means you have a whole line of things that is off-shell, and you just calculate these diagrams. And if you do that calculation, it will give you Wilson, lines. And what it'll give is something that looks like this, where Sn dagger is a function of the n dot As component, and W as a function of said n bar dot Ac component. So there are Wilson lines along some direction, and it looks like this. This is c. This is S. Now, there's something wrong with this. It's not quite the right answer. The reason that the Wilson lines are the way they are is because I got the collinear ones from integrating them out from the next to the heavy quark which is the soft particle. So that's why W sits next to h. S dagger sits next to c because of the same thing on the collinear side. But if I wanted to make them into a gauge invariant thing, I want the W to sit next to the c and the S dagger to sit next to the h. That would be the analog of what we had in SCETI, and that's not what we got. If this was QED, then that would be fine, because this and this, I could just commute them. Well, I can always do that. I can just push this guy through that guy. But in QCD I can't, because these guys have color matrices, and they don't commute with each other. So that means that this isn't quite the whole story here. There's some diagrams that we missed. So there must be some diagrams that are non-abelian that we missed, and what are those diagrams? These are diagrams that involve triple gluon and four gluon vertices. And what these guys do, it turns out, is they do one thing in this calculation. They really just flip the order of the W and the S. So why do we have to consider those diagrams? So think about, instead of attaching gluons to quarks, attach gluons to gluons. So say we have soft collinear, and we attach them to each other but through a three-gluon vortex. Then, this guy's off-shell. If that off-shell guy attaches to this guy, then it's off-shell. And if you integrate out diagrams like this, that's going to change what our result over there look like. If you go to the same order on the other side, exactly analogous thing. OK. So these are also things that you have to integrate out. You have to integrate all these pink things out. And if you do that, then it does what I said. If you do that to all orders, what you can do with a axillary Lagrangian-type approach, obviously, you're not going to start calculating diagrams to all this, but there are some tricks to doing it. Then, you get them in the right order. And it's easy to check, for example, that at the order of two gluons that this guy does exactly what you want. So this is then a collinear gauge invariant object, and this is then something that's soft gauge invariant, and that's nice. That's what you're hoping. That's what you're expecting. AUDIENCE: What about diagrams like collinear soft all from the collinear line or something? IAIN STEWART: You want to add one more gluon where? AUDIENCE: I want to stick like a collinear right in between some softs. IAIN STEWART: Yeah. You mean like on here, like that? AUDIENCE: Well, yeah. Essentially, it'd be over on that diagram over there, like not even [INAUDIBLE]. IAIN STEWART: Yeah. It's the same thing, I think. So if you add this guy here, then I think it's power suppressed. Because it ends up being a power suppressed term that you don't need, if I remember correctly. Yeah. This guy's power suppressed. AUDIENCE: You have to define like a different lambda for each one? It seems like it would be-- IAIN STEWART: No. You can think of this as a full theory diagram. Right? Where this guy is off-shell, and you haven't changed how he's off-shell. You just change the value of the momentum, but one thing you've done is you've doubled the number of hard propagators. So if I remember correctly, this guy just gives you something off-shell. I'm pretty confident, something that's power suppressed in lambda. All right. So this is how soft collinear factorization works. So soft collinear factorization, if you think about it from the point of view of going from QCD to SCETII, it's just integrate out these pink lines, as usual, and you end up with something that's where things are splitting apart. And the reason that it's happening is because these lines are off-shell that are where you would try to interact, have interactions, and so your theory is forced apart by the fact that these things can't interact in an on-shell way. So it's different than SCETI that in that sense. So this is kind of cumbersome, as you can see, if you wanted to think about doing it arbitrarily. Because it seems like you just have to calculate diagrams, and who wants to calculate infinite classes of diagrams for arbitrarily complicated scenarios? Right? Maybe in this case, we can do it, but actually even in this case, we have to resort to some tricks to do it to all orders. And in more complicated scenarios, it would just get even more and more cumbersome. So we'd like to have a trick that's generic, where we could get the same answer. And the way that we can do that is by using this dashed line up in the picture. Think about formulating rather than directly this SCETII, formulate first an SCETI, and then we'll match that SCETI onto QCD. So another, in some ways, better method is to do QCD to SCETI and then SCETI to SCETII, so three steps. So we're going to first use a SCETI that doesn't have the collinear, just has the soft mode and the hard collinear mode. So it has what I call the hc in the picture and s. OK? So that would be a mode that has P squared in this picture of order, say let's call it lambda squared. So if I say that this is for some lambda squared, then you'd have the soft mode that has P squared of order lambda squared, and this hard collinear mode that has P squared of order q lambda. And that's exactly an SCETI-type situation, where we were calling these c and us. So that's step one. Step two is to factorize that theory, which we know how to do. In particular, the ultrasoft which is soft can be factorized with a field redefinition. So this has the advantage that at this first stage we still have a locality that protects us and helps us to understand the theory. We then factorize, and then in the third step, we match SCETI onto SCETII. And then, we're getting rid of these hard collinear modes, matching them onto some collinear modes, and then we still have our soft modes. So we think that the hard collinear modes contain the collinear modes in the first stage. But they also contain some other stuff which is unwanted baggage that we have to get rid of, and that's why we have this second stage of matching. But you can really think of it in the picture as doing first the matching, where you integrate out the hard scale, but you contain in your effective theory this scale associated to the dashed line. And then in a second stage, you integrate out the dashed line, and that gets you to the final thing you want which is just low energy modes on this line here. OK. So one thing this does is give a simple procedure of constructing SCETII operators, even though there's more non-locality in SCETII than there was in SCETI. Because there's non-locality, because you don't just have this one scale that could cause you, even have this smaller scale that's causing non-locality. And that's explicit in the soft Wilson line. The thing that's giving the soft Wilson line-- well, the things that are producing the Wilson lines are really modes of this scale. OK so it's more non-local. One way of saying it's more non-local is simply that there'll be 1 over l pluses, as well as 1 over l minuses. That's another way of saying why it's more non-local. OK so that's just an example of something that makes it look trivial. So let's say we wanted to do this calculation, this way. So then in step one, we would simply write down the current in SCETI. We would integrate out the off-shell pink lines, but there'd only be lines on one side, and we get this which is our SCETI result for heavy-to-light current. Then, in step two, we would do the field redefinition in order to factorize this theory, and we get that result. And in step three, in order to match this result onto a current in SCETII, it's really simply a renaming here. So why is this step so easy? It looks completely trivial. The reason that this step is so easy is that any T-product, any time-order product that you write down here, will have a correspondence in this other case. OK? So you can really-- it really is that simple. So when so if it's really that simple, then you can see the advantages of this procedure. It's not always that simple, and let me present as a kind of theorem or just a bit of a statement when it will be this simple and when it will not be this simple. So basically, you have to know when will the T-products between these two theories match up? And the T-products will match up under the following circumstances, or they won't match up under the following circumstances. So if in the SCETI one theory you had time-order products with greater than or equal to two operators that had soft and collinear fields, then you can generate some non-trivial matching. So I think that's best illustrated by an example again. So in this particular example, up here, we only had one operator that had both soft and collinear fields, this external current. And then we had Lagrangians, but they were totally decoupled after we made this field redefinition, the L0 Lagrangian. So they don't count as something that's mixing soft and collinear fields. We just had this operator. But if you had two of these operators, and you wanted to go through the same procedure, then you could get something non-trivial. So let's imagine that scenario. We have two operators, so let's think of having in the SCETI, two interactions that are like this, then we could string them together as follows. So this is a T-product between two different interactions that both had soft and collinear. I'm just taking two of these guys and putting them together. And if you look at the off-shell-ness of this line here, then P squared is of order, in our counting, it's like a k plus times a q minus which is of q times a lambda times a q. So it's really something that lives at that hard collinear scale. OK? But you need two terms in the time-order product that, in order for there really to be a propagator like this, that you're integrating out. So this guy here will match onto something when you do this matching, where you have two soft fields and two collinear fields, like that. Because this guy here really is something that you would want to integrate out at that dashed line. Really, it's an off-shell, hard, collinear mode. But if you didn't have two T-products, then effectively what happens is, when you change the external kinematics in order to do the matching, all what were called hard collinear lines become collinear lines, and then the matching is trivial, as it was up here. But if you're in a situation like this, then there is a line that, by changing the scaling of these external guys, it doesn't change the scaling of that internal line, and then you get some non-trivial Wilson coefficient. AUDIENCE: [INAUDIBLE] IAIN STEWART: So let's call these hc. Yeah, and so I start by calling them hc. Right? And then, what I want to do, when I calculate this diagram to do the matching, is I want to assign them a different scale. I want to define them to be c instead of hc. So I start out thinking of them as hc. That's what I do in step one. I can do a field redefinition. It doesn't matter. But now, I want to match from SCETI onto II. So therefore, what I do is I take my full theory, which is SCETI, and I evaluate it with fields that don't have the right power counting for that theory. And then I do a Taylor-series expansion of all the diagrams, so that's this. I change my external fields, and I call them c instead of hc. AUDIENCE: Oh, because you're doing the same thing with ultrasoft. So ultrasoft, it really means-- IAIN STEWART: And now I make these soft, but soft and ultrasoft, that's just really a renaming. AUDIENCE: Ultrasoft in the SCETI [INAUDIBLE].. IAIN STEWART: Ultrasoft is equal soft. Sorry. Maybe this makes it clearer. Ultrasoft and soft just two different names for the same thing. But hc and c aren't, because hc really lived on the upper hyperbola. I put the external particles onto the lower hyperbola, but I'm frozen in here with one particle that stays in the upper hyperbola and therefore has to be integrated out. OK. So the kind of thing that you would get by doing that is that you would get convolutions with some Wilson coefficient. In this case, they were-- well, let me just say, it depends on some P minuses. This is P1, this is P2, and then in my example over there, I guess I had some heavy quarks. OK, and there's some Dirac structures. So it would be something like this, where you'd get some function, whatever it is, that's coming exactly from integrating all the purple stuff. So this is the Wilson coefficient, and usually, you would call these things jet functions, because they look like jet functions. So this comes from the SCETI modes that got integrated out. So the difference between this and the SCETI matching, where you're integrating out the hard scale is you're already getting sensitivity here to the plus momentum of the soft. That's one difference. OK. So this is a general procedure for constructing SCETII. Let's see how much I want to say. So let's say a few words about power counting. So if you take some T-product of terms in SCETI, that will scale like some power of lambda to the 2K OK? So in SCETI, you have to be a little bit careful about power counting when you do this procedure. So in SCETI, you could just assign some power. Let's call it lambda to the 2K. And when you do a matching onto an operator in SCETII, what you're generically going to find is that there'll be a relation between the power counting here and the power counting there which is good, but there's also one thing to be aware of. So if I define eta to be the power counting parameter for SCETII-- just to give it a different name, so I can talk about a correspondence-- then basically, eta is like lambda squared. And that, well, that just follows from this formula, that if you want to identify lambda squared is lambda over q for the collinear modes, then this would be the right way of doing it. Little lambda would be square root of lambda over q. But in SCETI for the collinear modes, you'd say there's some parameter eta which is just lambda over q. OK? So basically, it looks like you just take 2K and go over to k, because you just changed the definition of what you're calling the parameter, but there's also this plus E. That means that you can get additional suppression, and that plus E comes about from the fact that you also change the power counting for your external fields, when you do this procedure. So something that was scaling like lambda in the hard collinear theory might become lambda squared-- i.e. eta-- in the SCETII theory. So this E which is greater than 0 comes from changing the power counting of external fields in the matching. One example is just having an external collinear quark or external collinear perp gluon, for example, where you would have c of order lambda in SCETII, and it becomes of order eta in SCETII which is lambda squared. Those are extra factors of the power counting parameter. OK. So what do we learn, if we put all these things together? Well, one thing since soft and the collinear fields don't talk to each other, when you write down Lagrangians for them, they're totally decoupled right from the start. And so there's actually no mixed soft collinear Lagrangian at leading order. And so all the meat of this theory is coming about when you integrate out these off-shell modes, and you construct the external operators. OK. I'll say little bit more about power counting. So there's no mixed soft collinear Lagrangian at leading order, and you can get mixed things at sub-leading order. So in that way, it's like SCETI after the field redefinition. But here, there is no analog. If you just want to go directly to SCETII, there wouldn't be an analog of the field redefinition. All the Wilson lines are coming from integrating off-shell particles. OK. So we're going to speed up a little. I'm going to skip over one thing, which I don't think I need. So you can write down power counting formulas, like we did in chiral perturbation theory. Maybe I'll just write those formulas down. So in general, in chiral perturbation theory, we had a formula that said, if we want to figure out the power counting of any given diagram, we can have a counting for that diagram, and we know what size it is. And there's analogous formulas like that in both SCETI and SCETII. So in SCETI it's pretty simple. You say you have something that's order lambda to the delta. Then delta, the formula for it follows. This is very analogous to what we talked about for chiral perturbation theory. So these guys here are vertices that are purely ultrasoft, and if you have purely ultrasoft vertices, you're subtracting 8 because the measure of the ultrasoft particles is 8. I'm not deriving this for you. I could, but I think it's fairly intuitive. So if you think about the lowest order Lagrangian, say psi bar ID slash psi for ultrasofts. This would already be something that's lambda to the 8, and this is then leading order, so you subtract 8. And this guy here is the rest, anything mixed or purely collinear. And then for that guy, you subtract 4, because as soon as you have at least one collinear particle, then you have to use the collinear measure. And so here, some operator like c bar, and dot dc was order lambda to the 4. So you subtract 4 and that's leading order. And so what this formula allows you to do is really just you do power counting entirely in terms of vertices. You never have to power count measures for loops or power count propagators. You just count vertices. So if you want to know how big a time-order product is, and it's a time-order product of an L3 with an L2 with this kind of setup, this will be basically you count this as lambda phi. And you could even get the absolute size right, taking into account the scaling of external particles, and that's what this constant 3 factor does. And U is equal to 1 only for pure ultrasoft. Otherwise, it's 0. OK? So there's formulas like this, and there's an analogous one in SCETII which is a little more complicated. But it allows you just to have a power counting where you can just power count Lagrangians or operators 01 with L1, and that's lambda squared without having to worry about what propagators do I have in this time-order product? What loops do I have? You never have to ask those questions. You just have power counting at vertices, and that makes things quite easy. AUDIENCE: When you write these expressions, one insertion of Lagrangian corresponds to a single vertex. [INAUDIBLE] that Lagrangian. IAIN STEWART: Yeah. That's right. OK. All right. So I want to do a couple of examples in SCETII. So let's do an example. Let's do an example which is in some ways simple, and it's an exclusive analog of our DIS example, and it's called the photon pion form factor. So it's exclusive. It's clearly exclusive. There's a pion, and really, all it's going to come in here is hard collinear factorization. So it's simple in the way that it actually is not really exploiting the full complications of SCETII. It's really just another example, where we have hard modes factoring from cleaner modes. But they're clearly SCETII modes, because they're going to be modes for this pion. So we're going to use again for this calculation a bright frame, which you'll see, I guess, when I write down some momenta. So what would we write down for this process in QCD, first of all? So you'd say, I have a pi 0. It has some momentum P pi. I have some current which I can take at 0, and I want to make a transition from the pi 0 to a single photon. So you could write that as follows. I can always replace the photon by current, and then I can parameterized that matrix on it like a form factor. And if we go through things like parity, charge conjugation, stuff like that, time reversal, we find out that there's one real form factor, and there's an epsilon symbol in this guy. It has to be linear in the polarization vector. That was already true here, and there's one way of getting the indices to work out which is with this epsilon. So what you're really after in this process would be some understanding of this form factor. That's the one piece of non-perturbative information. Everything else, Lorentz invariance is enough to tell you about. So we'd like to see if there's a factorization theorem for that form factor, if we take the limit where q squared is large. So we have an electromagnetic current. You can think about it is up and down quarks for the pion. So there's some matrix in the 2 by 2 space which is the charge, which you can write as 2/3 minus 1/3, like that. Or if you wanted to write it in terms of polymatrices, you could write it as identity over 6 plus a tau 3 for isospin polymatrix over 2. So there's some charge matrix that's going to show up. What happens if we expand q squared much greater than lambda QCD squared? In this formula over here, this form factor knows about lambda QCD, and it knows about q squared, and we haven't expanded. So what happens if we do expand? Can we simplify that form factor? And it does. So I'm going to do this in the bright frame again, and that effectively means I'm taking the momentum that corresponds to the off-shell photon as the same form as we did before for our calculation in DIS, purely in the z coordinate. The momentum of the photon, it's an on-shell thing, so I can just write it as E the photons say times a light-like vector. Then, P gamma squared will be 0, as we want it to be. And if I pick this kinematics, then P pi is just P plus P gamma, so that would be a P pi. OK? So in Feynman diagrams, what am I talking about? I'm talking about making a transition with an off-shell photon, through a diagram like this one, to a pi 0 or plus the cross graph. And with this setup with this kinematics, this intermediate line here, if we go through the scaling, it's going to be hard. So these guys with this kinematics, we make the pion collinear. In order to see that, you can impose P pi squared equals M pi squared. Right? This is never going to be made small, but E is going to have to be tuned to be basically q over 2. So this thing is going to become, if you impose this condition, something like M pi squared over 2 q. Then, you find out that the pion is collinear. So the pion is collinear. The photon is a photon. This line here is hard, and so we want to integrate out the hard line and match this guy in the effective theory onto some effective theory operators that would look like this. This guy is pink. OK. So we just have to write down what type of effective theory operators that could be, and again, it's an effective theory operator of two quark fields. So it's very much like what we did before. I'm not going to be going through all the indices that I have to go through in order to keep track of charge conjugation and all those things. I'll just write down the right answers for that part. So I could write like this. So after doing hard collinear factorization, there's again operators which is two quark's. Some hard Wilson coefficient which is this pink thing, integrating out the pink line that sits in the middle. This obey current conservation. Dimensional analysis fixes this 1 over q. Charge conjugation, actually, also provides constraints on the Wilson coefficients, just like it did in DIS. So charge conjugation tells us that there's a relation between flipping signs. But we just impose on this operator all the symmetries and things that we can think of and see what it says about the operator. You can think about just writing the operator down based on this picture without ever calculating any diagrams, just knowing what you're after. And then imposing on it all the symmetries that should be conserved. So one is dimensional analysis. That gave the q. Current conservation, that's partly responsible for the epsilon which is also like a parody. Go through flavor and spin, and you find that this has this structure, and there's some constant that I threw in here. And you actually know from the flavor that it's got two photons. Right? And I'm not ever adding any more photons, so that means there's two factors of q hat. And so there's a q hat squared, which I can stick inside this gamma. And it has to be a color singlet, because it's going to have to have a non-trivial matrix element with pi 0 which is a color singlet. And since it has to be-- so there's no TA inside the gamma that's what I mean. There'd be nothing for the index A to contract with. And since it's a color singlet, that means if you were to think about soft interactions here, they would just cancel. Right? If I were to think about putting in soft interactions and integrating them out, I'd get an S dagger S that would cancel out. So that's the sense in which this is a simple example. OK. So then, we would have a formula where we could equate the effective theory result with the full theory result. So we can equate matrix elements, and this is one way in which this example is different than dependent elastic scattering. We're really doing a matching at the amplitude level. Remember, the form factor was a parameterization of the amplitude, and if we do that-- If you like, if you think about the two currents that we had, and we've integrated them out. Now, think about just this operator here. The matrix element pi 0 to vacuum that we had to finding this thing which had two currents just becomes a matrix element of that operator. Again, like our DIS example, we form the sum and difference at the momenta. We can think of forming P bar plus which is P dagger plus or minus P. And one of these guys just gives the total momentum, just the minus with our sign conventions. So that gets fixed which, in this case, it just gets fixed to the pion momentum which is q. So if you like, you've inserted the operator here. You have some collinear lines, and then you have a matrix element of the pi 0, and then you have gluons. But you know that whatever momentum comes in has to be the momentum of the pion, and that's the one momentum constraint. So that means that the answer just involves one convolution again which is the one that's unconstrained. So there's a Wilson coefficient that depends on the other unconstrained guy which was P plus, and then there's a matrix element, where I can write in a delta function with that P plus, the usual kind of way that we've done. But here, it's a little different matrix element than the one we saw on DIS, because it's not forward. It's vacuum to pi 0. But it's actually the same operator that we were talking about in DIS. It's just a bilinear operator with two collinear quark fields, two collinear quarks in dress with Wilson lines. So one can go through a similar type of matrix element analysis for this operator, and it gives something that's called the light cone function. So let me define that for you. This matrix element can be written in terms of an object that has a dimensionless variable. It's an analog of the part-time distribution function but for this matrix element that we're dealing with here which is vacuum to pion. So this has some similarity to the formula we had in DIS. There's a delta function there. There's a delta function here. There's a dimensionless variable z, dimensionless variable z there, and this is a non-perturbative function. So this is what's called a light cone distribution function for the pion. So generically, when you have an exclusive process, and you're producing some hadron that's very energetic, like a pion, this is the type of thing that's going to show up, one of these light cone distribution functions. This example of photon to pion is in some sense a very, very simple, exclusive process, the simplest one in some sense. OK. So we could take this formula, plug it back in there, and then we'd have a factorization theorem. And I think that you can imagine what that would look like. I could write it, instead of an integral over W, as an integral over z, and that would be the factorization theorem involving this 5 pi. AUDIENCE: [INAUDIBLE] IAIN STEWART: Yeah. AUDIENCE: [INAUDIBLE] IAIN STEWART: That's right. So you should think of the z as like-- so the way to think about the z is as follows. So think about like when you initially produced these guys here, think about all the momentum going this way. When you initially produced them, after you integrated out the hard interactions, you had z and 1 minus z is the possible split of the quarks fields in the operator. So one of these guys carries z. Effectively, what this 2z minus 1 is doing is one of these guys is carrying z, and one of them is carrying 1 minus z. OK? The sum of these is 1, and that's the analog of the statement that the whole total momentum should be the pi 0 momentum. But you don't know how to split how much goes into each one of those. And what the wave function is, it's all the linear interactions that subsequently rearrange this thing before you annihilate it with the state. So all these things are dressing up the pion state. They're producing the pion pole. So this is like 5 pi. And so what you have is an operator that depends on z, a wave function that depends on z. So this is a Wilson coefficient that depends on z, and your final factorization theorem is exactly of that type, that you sort of-- of an integral of z of c of z Which also can depend on q and mu and then 5 pi of z and mu. AUDIENCE: Is there a sense of the 5 pi z as universal? IAIN STEWART: Yeah. It's universal. AUDIENCE: You can use it for other [INAUDIBLE]?? IAIN STEWART: Absolutely. Yeah. So as long as you can factor it so that it's these fields ans that pion, then you have this matrix element, you get this guy. Some people try to measure things about this guy, it's moments and stuff. All right. One thing that happens here has to do with this integral over z which is still in some sense an unsolved problem in SCET. So I have to mention it. So when you do this integral, you could ask, what does the c of z look like? And it turns out that c of z will have in it terms that go like 1 over z. And so you get integrals that look like dz over z of 5 pi of z. So at lowest order, this integral would show up. And it turns out that, for this matrix element here, you don't get anything worse than that. You never get 1 over z squared. And this integral here, because of properties of this 5 pi, is finite. There's no problems. But there are examples known in the literature, where that's not the case, where you actually get 1 over z squared, and people have some understanding of the physics that's happening in those cases. But there's not a complete understanding of how factorization works in those cases. OK? So that doesn't happen in this example, but there are other examples of exclusive processes that would lead to 1 over z squares, and then this integral is not well-defined. And people understand that there's a cut-off that's coming in, and they understand that that cut-off actually has to do with some rapidity. But how to explicitly write down an analog factorization theorem that involves those cut-offs and has renormalization group is an unsolved problem, unsolved SCETII problem. OK? But for the example we did, everything's kosher and beautiful. All right. So I think what I'll do next time, since we're out of time I'm not going to start my second example now. So next time, we'll do an example that does involve soft fields, both soft and collinear fields in SCETII. That's where we're going, but we'll leave that to next time.
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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. In this short section, you're going to look at the weak interaction a little bit more, and specifically discuss neutral currents. We looked in some detail at charged currents-- specifically, the interaction with quarks. So here, I'm going to look at the Z boson specifically, and the weak interaction via the neutral current. So studying those two processes here, where there is an electron and a positron through some process including a Z boson and a photon and resulting in a muon and an anti-muon-- those processes have been studied in great detail at SLAC and at CERN, at the SLC, and the Large Electron-Positron Collider. So if we want to calculate the cross-section and study the cross-section of the center of mass energy, we see a number of interesting effects. At low energies, and at very large energies, the cross-section runs with 1 over the energy squared. But at the mass of the Z boson, we see this enormous resonance here. The cross-section at the resonance from the Z boson is about 200 times that of just a photon exchange. So this allows you to study the Z boson with great precision at those colliders. You have sizable cross-section when you are in electron-positron colliders. And then you can, with precision, look at, what is the rate into a muon/anti-muon? What is the rate into a quark/anti-quark? And so on. And you can study the mass, the width of the Z boson with an enormous level of precision. Again, so I will not go into too much detail here. And please have a look at chapter 9.6 in Griffiths, for example. But there's many other resources where you can learn more about [? neutral ?] currents. Neutral currents, electroweak neutral currents are specifically important in the study of neutrinos, as we will discuss more in the lectures as well.
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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: Yeah, let's show the answer. All right. For a special treat, special benefit-- I'm not sure what to call it-- something special later in the class, does someone want to tell me why that's the right answer? AUDIENCE: Well as we did last lecture, we crossed off the k2 because that's the slow step. So it doesn't matter as much. CATHERINE DRENNAN: Yeah. And so when you get rid of that, when the k minus 1 is really fast compared to k2, k2 is very small compared to that. So you get rid of it. And that simplifies the expression. So temperature-- today's lecture is largely about temperature. But we're going to tie in all sorts of other things we've learned over the course of the semester. So I'm very excited. Some of my very favorite topics are coming back today. So effective temperature, we talked about this a little bit when we were talking about making bread, adding your baking soda and about rates and why you put things in the oven. And we talked about it has to do with also the spontaneity, whether you change the temperature and what your delta H and delta S are. But we also talked about rates and that increasing the temperature often increases the rate. So many of us have observed this. You increase the temperature, you increase the rate. But today we're going to talk about how you can quantitatively say how much the rate might be changed-- the rate constant might be changed if you increase the temperature. So in 1889, it was a wonderful day for Arrhenius. He had been trying to plot different values of rate constants versus temperature to see what would happen. And then he tried natural log of the rate constant k versus inverse temperature. And he got a straight line. And scientists always get very excited when your data falls on a straight line. It means you figured out some relationship between two values. So here is the Arrhenius plot. And we'll introduce some terms. So again, we're plotting the natural log of k versus 1 over temperature, 1 over inverse kelvin over here. And here is our plot of this straight line that Arrhenius found. Natural log of k on the y-axis versus 1/T gives you a slope then of minus the activation energy-- so E to the little a is activation energy-- over R, our good friend the gas constant. And then the y-intercept over here is the natural log of something called A. We have a lot of different A's in this particular unit of kinetics. This one is factor A, sometimes called the Arrhenius factor, A for Arrhenius. It has a couple of different names. But importantly, it has the same units as k, our rate constants. So what this plot told Arrhenius back in 1889 was that rate constants vary exponentially with inverse temperature. So this was the first kind of connection between rate constants and temperature that could be used to kind of come up with quantitative numbers. So factor A, this Arrhenius factor, and the activation energy, E to the sub a, depend on the reaction being studied. So they have to be measured for the particular reaction. So it's a clicker competition. So we'll have a bunch of clicker questions today. And why don't you tell me if you think factor A is temperature dependent? All right. 10 more seconds. Yep. The answer is no. And so what is this factor A? If we think about this in terms of the plot, what it is is the rate constant when 1/T is equal to 0. Because it is the y-intercept. And when 1/T is equal to 0, what has to be true about T? Yeah. Infinitely large. So factor A is the rate constant at an infinitely large temperature, at a huge temperature. So it's the fastest that particular reaction could ever go at this infinitely huge temperature. So that's what factor A is. And of course, we can't plug things in and say, how fast is this at an infinitely huge temperature? So conveniently, you can get that value out of plotting your data. You measure a bunch of rate constants versus temperature. And you plot it this way, and you can calculate this sort of maximum rate constant possible for this reaction if you had this infinitely huge temperature. What about activation energy? Do you think that's temperature dependent? All right. 10 more seconds. So no, it isn't. And so if we sort of just think about it back here, the answer is no. Again, you plot the rate constants over all these temperatures to get one value out of the slope. So it is largely independent of any kind of temperature. You get one activation energy for the reaction in question. But it does depend on the reaction, for sure. There isn't just one value for this for everything. It depends on the materials. All right. So let's look at some other ways that we can express the Arrhenius equation. So we have the Arrhenius equation written as a straight line. And we can also do something very, very simple to it, which is switch these two terms. And that gives us what is officially known as the Arrhenius equation. So natural log of k equals natural log of A, this Arrhenius or factor A, minus the activation energy over the gas constant times temperature. And of course, if you want to solve for k, you take the inverse log of both sides. And so then k is going to be equal to this factor A-- again, measured for every reaction in question-- e to the minus Ea, the activation energy, over RT. So here is our equations. These will be on equation sheets, so you don't have to memorize them. But you already, if you've been studying your equation sheet for exam four, realize that there's a lot of symbols that are very similar in these units. So we have a, in nuclear chemistry, activity. So just keep in mind what equation is what. When you're talking about nuclear decay, you don't have your activation energy term. So that should give you somewhat of a clue. So one of the challenges in the final is making sure you know which equation is which. So let's look at an example now where if we're given some of the information, we can solve for other things. So an example, this is classes at lunch time. So we can think about the hydrolysis of an average lunch of sucrose to form a molecule of glucose and a molecule of fructose as part of the digestive process. So some of you might have eaten already. You might be digesting sucrose-- it comes in many different forms-- right now. So using the information about activation energy, we can predict what the rate constant is going to be at a different temperature. So some kind person measured the activation energy for this digestive process at 108 kilojoules per mole and also figured out that the k observed, observed rate constant, for this reaction is 1.0 times 10 to the minus 3 per mole or per at normal body temperature. And so now we're asked to calculate what that rate constant should be at lower temperature, 35 degrees Celsius, somewhat below body temperature. So how are we going to do that? Well, let's remind ourselves of our equations, our Arrhenius equation. So we know what Ea is. We were not told what our Arrhenius factor A is. We don't know that. But we do know one of the rate constants at one of the temperatures. So if we combine these two equations, then we have an appropriate number of variables to solve for the rate constant at the new temperature. So let's combine those two equations. So we have natural log of k2, our rate constant, at temperature 2 minus the natural log of the rate constant at temperature 1, which can also be expressed as natural log rate constant 2 over rate constant 1. And our natural log of the Arrhenius factor drops out. And so we have minus Ea, our activation energy, over our gas constant times 1 over the temperature minus 1 over the first temperature. So we can put in our values that we're given. So we have our first rate constant down here, the rate constant at the first temperature. We put in our temperatures, making sure we convert them to kelvin because our gas constant is in kelvin. And also, we are going to convert our activation energy from kilojoules to joules because we want to cancel our units. So now we can cancel joules. We can cancel moles. And we can cancel kelvin. And so that gives us a new rate constant, 7.6 times 10 to the minus 4 per mole or per second. And this is a lower rate at a lower temperature. And in fact, that is often true. You have a lower rate at a lower temperature. And this is one of the reasons why it's a really good idea to keep your body temperature at body temperature. And around this time of year, people come in to MIT from warmer climates. And they do not have appropriate clothing. And so your body does not work well in the cold. The rates slow down. You're not digesting things. Your body is really not doing anything at the rate it's supposed to be doing these reactions because they're nicely tuned to the appropriate body temperature. So go out and buy a winter coat. So this now, this equation that we just derived shows us the relationship between rate constants and temperature. So again, natural log k2 over k1 minus the activation energy over the gas constant and our temperature term. And if we look at this equation, we'll realize that if we have a very large value for this activation energy, this Ea term here, that's going to mean that the rate constants are very sensitive to temperature. If this is small, there won't be a big difference between k1 and k2. But if the activation energy is really big, there will be a big difference between k1 and k2. The rate constants will be very sensitive to temperature. All right. So now let's think about rates and temperature. And let's think of a cold temperature. Let's think, for example, of liquid nitrogen temperatures. What do you think happens to an enzyme? You have enzymes in your digestive process that are hydrolyzing sucrose that you might have had for lunch. What do you think happens to enzymes if you put them at liquid nitrogen temperature? They will slow way down. One would say they would stop working entirely. They would just sort of be, we call it frozen. And in fact, I use this all the time in my research. So one thing we do, we study the structures of proteins. And so we grow these lovely protein crystals. And what we can do, we take these-- a lot of times, the proteins or enzymes in the crystals are active. They're able to convert reactants to products. And so we can soak in reactants and then take our crystals and dunk them in liquid nitrogen and then determine the structure of that. And if we do it at different time points, you can actually kind of walk through the mechanism of the enzyme. Watch what happens at various different states. So this is a very common thing that's used. We like to do things at liquid nitrogen to see things sort of-- we sort of pause it when it's dumped in liquid nitrogen. It stops what it's doing. And we can capture its structure. What do you think about non-enzymatic reactions? Do you think they're also going to be slowed down? I don't know. Should we do an experiment and find out? AUDIENCE: Yeah. CATHERINE DRENNAN: Yeah, I think we should. So we now have-- let's bring out our demo TAs. We now have glow sticks. Glow sticks are in fact chemicals in there that when you break them, you have a chemical reaction. So we're going to do this. And we're going to break them and watch them glow. Let's break them and put them in here first. And then we're going to cool them down and see what happens to the chemical reaction. Here, let's have a little help over here. Please break some of these. AUDIENCE: How do you break them? CATHERINE DRENNAN: You break-- see, you just kind of snap them and then shake them up. And I think we'll bring down the lights so you can all see these pretty well. So now the chemical reaction is going. And so we can see the chemical reaction. That's what's wonderful about glow sticks. You can observe the chemical reaction because it causes it to glow. Move so everyone can see. Now we're going to see what happens to the chemical reaction when it slows again. When the chemical reaction is slowed down, you won't see it anymore. So it'll stop glowing. That should be good. We'll do all of them, and we'll see. It'll take a little bit of time. But it should not be too long. And hopefully, it'll slow down faster than the liquid nitrogen destroys the plastic cups. Yeah, I think you can see it with the first one now. Mary, why don't you come down here? We have something else for you to do. Let's put some liquid nitrogen in here. Let's bring up the lights because our glowing has stopped. So another thing that liquid nitrogen changes the properties. So Mary will demonstrate to you what happens to a flower. And just like right now, the flower does this. But let's put it in liquid nitrogen and see what happens. Give it a decent soak. And now smash it. [LAUGHTER] Thank you very much, our lovely assistant. I will brush off my computer later. We have more flowers, and apparently there's a broom somewhere. So if we finish class, for some lucky other individuals, you can come down and smash the rest of the flowers. So liquid nitrogen really changes the property of things. Sometimes it's used by dermatologists to remove warts or other things. They'll dab some liquid nitrogen on you. And I always feel like paying a lot of money to go have that done-- I have liquid nitrogen around my lab. But then I'm like, yeah, I probably have enough liquid nitrogen burns on my figures already that I don't want. Crystallographers who use this a lot, they'll often be in a situation where you have this liquid nitrogen. And it's sort of dripping on you. But you have your crystal, and if you let go, then you'll destroy your crystal experiment. So you'll have liquid nitrogen sitting in your palm. And you're like, OK, I'm going to get a burn. Or I'm going to lose my crystal. And so you decide how important that is. And so sometimes a crystallographer will walk up to you and go, see that scar? That's 1.2 angstrom data, baby, right there. So we suffer for our science sometimes. So liquid nitrogen is very-- changes the property. Cold things are different than warm things. Buy a winter coat. So everything slows down in the cold, in terms of elementary reactions. Our rate constants slow down. So now let's think about a reaction and what's called a reaction coordinate and consider what's happening in a reaction. And then we're going to come back to thinking about the effect of temperature, where temperature really makes a difference. We'll quiet down a little. So considering the reaction coordinate, reaction coordinate the reactions of-- you bring your reactants together and form your product. So two things coming together to form something else, that is a reaction. And as you go along in the reaction, that's your reaction coordinate. And we're going to also introduce this term of activation complex or transition state. So two molecules can collide. Two molecules colliding, bimolecular. But every time those two molecules come together, they're not necessarily going to form a product. Why? So only when that collision energy is greater than some critical energy-- which is sometimes called Emin, this sort of minimum energy to get this reaction to go, or as I like to call it, the activation energy, Ea, that's what we've been talking about, our activation energy-- will you get a reaction. So you need to have enough energy in your two things coming together. Those two things coming together need to have a critical energy for them to react. So why is this true? You need to have that critical energy. But why? Why is this necessary? So it's necessary because before this reaction takes place, even if it's a very happy reaction, things need to occur. So the two things coming together need to often distort. Bonds might have to be broken. And new bonds need to be formed. And while that is occurring, while there is distortion of the bond or bonds are breaking, you need to have some potential energy to do that. You need some energy to make that reaction go. So the potential energy of the system increases first while these distortions are happening. So the encounter then between them, you form some activated complex, or what's known as a transition state. And that activated complex, it can go on to form a molecule. Or from that activated complex, the two molecules may just go apart again. So what determines whether they're going to go on happily to form their complex or going to depart from each other, never to form a larger molecule? And the thing that determines the fate is that critical energy. So only those molecules with that sufficient energy to allow for those bond distortions and rearrangements will be able to go on and make this. And you can think about this, I think, in terms of a couple, a relationship, that there's always some work that has to go into it. It's never perfect. And if you put in this effort and this work, you can go on. And if you don't really have it in you and you're like, ah, you walk away. You don't have that critical energy. You don't have that special overcome, that activation energy, and you go back. So let's just take a look at some molecules checking each other out and figuring out if they have what it takes, if they have that critical energy. Let's watch and see what happens. Here they come. They're finding each other. They're circling each other. And oh my goodness. They had the critical energy necessary. And they formed a bigger yellow molecule. There they go. What a happy ending. So only those molecules that have that critical energy can go on and form their complex. And so here is where temperature comes into play. Here is where temperature becomes really important. Because they need to have a certain amount of energy. And there's a relationship between the kinetic energy of molecules and their temperature. So let's take a look at this plot over here. So this is, on one axis, fraction of molecules. And on the other axis, we have kinetic energy. So if you're at low temperature over here, most of your molecules are going to have a low kinetic energy. But it will tail off. And there will be some molecules that will have a higher energy. And if this line here represents that minimum energy, that critical energy or that activation energy, in blue here we have those low temperature molecules. Only a very small fraction of those will have the energy to react, will be able to overcome that minimum energy, that critical energy needed. Now if you're at high temperature, more molecules have a higher kinetic energy. And if you see over here shaded in orange, way more molecules have that critical energy, have the energy necessary to distort those bonds and go on to make a molecule. So here's where temperature is important. So kinetic energy, pass it on. So let's draw some reaction coordinates and think about what's going on. Yeah. AUDIENCE: For this idea of collision, could it also apply to the association reactions? Because that's [INAUDIBLE] molecule. Or are [? they still into ?] each other? [INAUDIBLE] CATHERINE DRENNAN: Yeah. So if you're talking about the forward reaction and the backward reaction of things, whether it's two molecules that are coming together to form things or molecules breaking apart, there's a critical energy both directions. And we'll see that, actually. It's a good question. That kind of leads us in to our reaction coordinate diagram. So most of this is in your notes. There's a few things that are not. But let's look at a reaction coordinate diagram. And by that, I mean we have Potential Energy, or PE, on one side. And on the other axis, we have what's called just the reaction coordinate. So if you're asked to draw a reaction coordinate diagram on a problem set or an exam, asking for potential energy versus reaction coordinate. Now our reactants are going to have a particular amount of potential energy. And in this case, it's up here. And our products are going to have some potential energy down here. And there will be a difference in energy between these, our delta E. But these reactants aren't just going to be able to go to products without overcoming some kind of critical energy. So before they can go down there, they need to go up here. So we have the activation energy, f for forward direction. Also, if you're down in products and you want to go back to reactants, there's also a barrier that you have to overcome. And it's not just this. It's all the way up here. And so this would be the activation energy of the reverse reaction. And this dashed line on the top is our transition state or our activated complex. Both are two expressions that are used kind of interchangeably. So this transition states, some kind of weird mixture where they've come together, and they're sort of breaking and distorting. It's not our final products. That's down here. But you need to go up in energy, overcome an activation energy barrier. So the transition here, you need to go up, and then you go down. So now let's think about all of these energy terms. And what's true is that the change in energy-- this change in energy between reactants and products is equal to our activation energy for our forward reaction minus the activation energy of the reverse reaction. And this delta E over here can be measured from a calorimetry experiment. And you might recall back when we were talking about thermodynamics that E is kind of closely related to delta H. So we had this equation at one point that delta H is equal to delta E plus the change in PV. And we said that for gases, there's about a 1% to 2% difference between delta H and delta E. But for solids and liquids, it's pretty much the same. So you can think about this delta E like you were thinking about delta H, for the most part. So now for this, if we have some of the numbers-- if you know activation energy in the forward and the reverse, you can calculate delta E. If you know delta E and one of the activation energies, you can calculate the other. So I'll tell you for this particular reaction what these values are. And I was going to bring colored chalk, and somehow that didn't happen. So the activation energy in this is not in your notes. So I'll write it down. Is 132 kilojoules per mole. And I'll also put it here, a little bigger, 132 kilojoules per mole. And for the reverse reaction, it's 358 kilojoules per mole and minus 358 kilojoules per mole. And so if we do the math here, we will calculate that delta E is equal to minus 226 kilojoules per mole. So would you expect that to be an endothermic or exothermic reaction? Exothermic. Right. And we can put that value also in here, minus 226 kilojoules per mole. So what's really important here, even though it's exothermic reaction, it's not going to just go without having those molecules have some critical energy necessary to overcome this activation energy barrier. You still need energy. You need to go up here before you can go down there. So just sort of thinking about this in layman's terms, this idea of these activation energy barriers, for me whenever I sit down to write a grant or write a paper, I think, wow. I sit down at my computer, have my cup of coffee. I know that we're about to get scooped on this data. So I really should write this paper, a little stressed about it. Then I think, wow. I could write a lot better if my office was clean. Now I hate cleaning my office. But compared to writing a paper, it's pretty good. I don't mind doing it. So then clean the office. Then walk the dog. Maybe do later some laundry. Lots of things happen. And then eventually, you hear from the collaborators. They're like, I need this paper now. And you're like, oh man. And so that stress gets you over that activation energy barrier, just catapults you over. Or you wait to a grant-- you start six months early. And then the week before it's due, suddenly you have that nervous energy that gets you over that. So many of you have probably experienced this. And you know, if you need some extra help with an activation energy barrier to do the rest of those extra problems for exam four, you can come talk to me. And I'll stress you out and get you right over that activation energy barrier. But the thing to keep in mind is that look at this slope in this case down here. So often it's just that little something to get you over. And then it's smooth sailing. So you can work these games with yourself. You've just got to get over it and know that once you're over that barrier, it's all going to be good. So you can get yourself over. You should never forget about activation energy barriers. There's always a barrier. Am I right? There's always some kind of barrier. Anything you're going to do that's worthwhile has some barrier associated with it. So now let's think about the results of these barriers, again, coming back to this idea of temperature. So for an elementary reaction-- again, that's a reaction that occurs exactly as written. It's a step in an overall reaction mechanism. There's always a barrier, always a barrier. Barrier is always positive. It's always there. And so if you increase the temperature, you're always going to increase the rate of that reaction. It's always going to help get over that barrier. But increase the temperature, increase the rate. But for an overall reaction, temperature is a little more complicated to predict. So if you increase the temperature, it's not always as clear what's going to happen to the rate. Because you can have a lot of different steps involved. It can be exothermic and endothermic. A lot can be happening. Changing the temperature, you're changing k. So to think about an overall reaction, we need to understand reaction mechanisms. And that's really convenient because we just talked about reaction mechanisms on Monday. So you all know how to write out reaction mechanisms. So let's just practice. So for this example, we have two molecules of N O plus O2 going to 2NO2. Step one, fast and reversible. N O plus N O, going to an intermediate. Then the intermediate is reacting with oxygen going to two molecules of our product. So we can write the rate of formation of product based on the slow step, or if we didn't know the slow set, on the second step or the last step. So two molecules, again, two molecules are being formed. k2 times the concentration of the intermediate times the concentration of O2. But we have an intermediate. So we need to solve for the intermediate in terms of rate constants, reactants, and products. But now we're told we have a fast step followed by a slow step. So we know how to do this. And remember, when you have a fast, reversible step followed by a slow step, this first step approximates an equilibrium reaction. There is not much of the intermediate that's being siphoned off in the slow step. So pretty much, when you form the intermediate, it's going back and forth just like you have in an equilibrium situation. And just to emphasize this, I'll share a little picture here of the beach in the summer when you don't need your winter coat. And here we have my daughter and her best friend. And they are trying to empty the ocean with this princess bucket. And one can ask the question, is this going to affect the tides? Is the ocean going to be different? But the rate at which my daughter and Aiden fill up their buckets and bring them over here, that's a really slow step compared to everything that's going on in the ocean. So when you have a very slow step like a kid and a long distance with a princess pail, you don't really need to worry about that. And you can think about the fast step as being an equilibrium. This is not being siphoned off enough to worry about it. And so that makes it easier to solve for your intermediate because then you can just do it by an equilibrium expression. So in the clicker question, tell me how I can do that. AUDIENCE: I think you're good. CATHERINE DRENNAN: OK. 10 more seconds. So let's just take a look at that. So again, we're writing our equilibrium expression for the first step, products over reactants. And then we can rearrange it to solve because the product here is our intermediate. So we can solve for the intermediate in terms of K1 and our reactant. And then we can substitute it in. And that gives us this. So we had the 2k2 here. Now we have our K1 and our N O squared. And here we have our oxygen there. So now we have a rate law that doesn't include any intermediates. I'm just going to put that back up here if you didn't get it written down. Now let's think of what happens with temperature. So this is an elementary rate constant, little k2. So you increase the rate when you increase temperature. And here is our expression again that tells us about that change. So now for the equilibrium constant, the effect of temperature depends on whether the reaction is exothermic or endothermic. And does anyone remember what the name of the equation is that tells us about temperature effect with equilibrium constants? Yes. The van 't Hoff equation. I told you that some time in the future I'd ask you for the name of this equation. And that time is now. You don't really need to know it. But it's just fun. There's not many names for equations in general chemistry. Look how similar these equations are. These are very similar equations. So we have the natural log of rate constants here. Here we have the natural log of equilibrium constants. Here we have our activation energy. Here we have delta H. So if the reaction is exothermic, if we increase the temperature, what happens to our equilibrium constant? Does it increase or decrease? You can just yell it out. It decreases. Now let's think about what happens here. So our kobs now has an elementary rate constant term, which is going to increase with temperature. And it has an equilibrium constant term, which is going to decrease with temperature for an exothermic reaction. So we have these going in opposite directions. So overall, we want to think about the magnitude of our activation energy term for our rate constants. And we want to think about the magnitude of our delta H term in terms of the equilibrium constants. So for the particular reaction in question, you can look up. The activation energy is a small number. And delta H, again, it's exothermic. And it's a big number. So then if you're thinking about that, if Ea is small and positive-- Ea is always positive. There is always a barrier, always positive. Then the rate constant is only going to increase a little bit. But delta H is a big negative number, so the equilibrium constant is going to decrease a lot. So for this particular reaction, increasing the temperature actually decreases the kobs. And so for any reaction, it depends on the magnitude of Ea and the magnitude and the sign of delta H. So again, large Ea means very sensitive to temperature. Large delta H means that the equilibrium constant is very sensitive to temperature. For an elementary rate constant, it's always going to increase with temperature. Because Ea is always positive. There is always a barrier to overcome. Temperature always increases a rate constant. But for an equilibrium constant, it can increase or decrease because delta H isn't always positive. Like the activation energy barrier, it can be positive or negative. So the magnitude of delta H, how big a number it is, tells you about the magnitude of the change, how much the equilibrium constant will change. Will k1 and k2 be almost like each other or really, really different from each other, those equilibrium constants? And the sign of delta H, whether it's positive or negative, tells you the direction of the change. Will it increase or decrease? So I just want to show where we're going with this because it's just super exciting. Yes. We're back to Le Chatelier. This is what I was so excited about. Remember, Le Chatelier told us, when we apply a stress to the system, the system responds in such a way to minimize that stress. So if we increase the temperature, according to Le Chatelier, what direction will the reaction shift, in the exothermic or the endothermic direction? What is it? The endothermic. So you increase the temperature. You add heat. It shifts in a direction to minimize that or to use up that heat. So we'll end with one last clicker question. And we'll finish this concept on Monday. But we want to think about how Le Chatelier and what we've known already applies here. So why don't you tell me which of these diagrams is exothermic and which is endothermic? Because we're going to tie this all back to temperature and Le Chatelier. AUDIENCE: We don't have that question. CATHERINE DRENNAN: We don't have that question. OK. Then we will not do that question. AUDIENCE: [INAUDIBLE] the question. CATHERINE DRENNAN: OK. So I'll just give you the answer to that. And then we'll see who the winners are. So that one is endothermic. And this one is exothermic. And we'll finish the rest of this on Monday. But for now, can you tell us who the winner is? No, it's not coming up. AUDIENCE: It's Dan. CATHERINE DRENNAN: Top two. AUDIENCE: Dan and Jay. CATHERINE DRENNAN: Dan and Jay have beat out the other folks. So we know who's in the playoffs. All right. Friday exam. See you Monday in class to finish the handout. All right. Let's take 10 more seconds on the clicker question. So let's take a look at this. 56% do you have the right answer. So rate constants always increase with increase of temperature. That's our little k's. Because there is always some activation energy barrier to overcome. There is always a positive barrier. Whereas, for equilibrium constants, K, you can have a reaction that's exothermic or endothermic. And so that will change. It either increases or decreases because delta H can be positive or negative. But activation energy barriers are always positive. There's always a barrier. Always a barrier to doing anything that's important. I've got to calm down. Le Chatelier tells me that when a stress is applied to the system, I should respond in such a way to minimize the stress. It's a very calming rule in chemistry. Reactions don't like stress. And they'll respond in a way to minimize it. So we've been talking about Le Chatelier for a lot of the semester. And we've talked about the effect of temperature on reactions previously. So if you add heat to a reaction, it will want to respond in such a way to minimize the heat or absorb the heat. So it shifts in the endothermic direction. So we've already talked about this. And nothing is new here today. But today, I'm going to give you a different way to rationalize why that happens. So we're not changing what happens. We're just going to come up with a new way to rationalize the why. So we ended last time with these two diagrams. And we identified one as endothermic and one as exothermic. So these, again, are our reaction coordinate diagrams. We have PE on one side. What does PE stand for again? Potential energy. It stands for other things too, but in a reaction coordinate diagram, that's what it is. And then the reaction coordinate going this way. So we go from reactants to products along the reaction coordinate. So here in this endothermic reaction, we have our reactants down here, our products up here, and a very large activation energy barrier for the forward direction, a much smaller activation energy barrier for the reverse direction. With the exothermic reaction here, we have a smaller activation energy barrier for the forward direction and a bigger one for the reverse direction. So let's look at the one equation that will be on the exam that will not be on your equation sheet. I think this is-- for this unit, this is the one that will be not on your equation sheet. This change in energy is equal to the activation energy for the forward direction the of reaction minus the activation energy for the reverse reaction. And again, this is our delta E here. And this is our delta E here. Delta E is like delta H. If you're talking about a gas, it's 1% to 2% different. If you're talking about a solid or a liquid, it's negligibly different. So if we look at this endothermic reaction here, we have a big forward activation energy barrier minus a small reverse activation energy barrier. So that's going to give us a positive value for delta E. And it'll be an endothermic reaction. And if we look at this equation again and fill it in for the exothermic reaction, we have a small barrier in the forward direction, a big barrier in the reverse direction, and a negative value for delta E and delta H. It's an exothermic reaction. Now let's think about what happens when we increase the temperature. So if we increase temperature, it's a lot easier to overcome this forward activation energy barrier, the big one. And that's going to shift it to products. And this is because the small activation energy barrier, it's not that hard to get over something that's a small activation energy barrier. It's not hard. But if it's a really big barrier, it's hard to get over that barrier. You need a lot of kinetic energy. Increasing the temperature will give that kinetic energy to overcome that barrier. Remember that when molecules come together, bonds are distorted and formed. The potential energy goes up first. And only those molecules with that critical energy that can overcome that activation energy barrier can go on to products. Increase the temperature, that allows you to overcome the big barrier. Now if we look in this side, this is a small barrier. They weren't really having-- molecules were probably not having a huge amount of trouble with that barrier. The barrier that was hard was for the reverse direction. So now you increase the temperature of an exothermic reaction. It's easier to overcome this big barrier, the reverse barrier. And you have a shift toward reactants. So this is what we had seen before. You increase the temperature of an endothermic reaction. You go to products. You increase the temperature of an exothermic reaction. It shifts to reactants. So these are the same things we saw before. But now there's a new rationalization behind it. Now we can think about this in terms of activation energy barriers. So the important points to remember, big activation energy barrier, rate constant very sensitive to temperature. If you have a big barrier, increasing the temperature makes a big difference. If you have a small barrier, doesn't really matter that much. Most of your molecules can get over that small barrier. So big barrier, increasing the temperature, more molecules can go. And so you're going to shift in the direction of the big barrier.
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https://ocw.mit.edu/courses/5-08j-biological-chemistry-ii-spring-2016/5.08j-spring-2016.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. JOANNE STUBBE: So that's where we're going in terms of Module Six. We introduced this the last time. And this is the required reading that's also been posted-- and we started this last time. This introductory lecture is talking about metals, in general-- the chemical properties of metals, and why we have these kinds of metals in our bodies. Why we're using these kinds of metals. And focusing then on metal homeostasis, in general. And then we will move over into complete focus on iron for the next three-- so, subsequent three lectures. OK. So here we are-- here we are with the periodic table. And we're going to be focusing on transition-- the transition metals, which are most of the places where you see the chemistry that you're familiar with. And this just sort of tells you the relative abundance of the metals in our bodies. OK? And we talked about last time-- sort of an introduction to the different kinds of chemistries that we can have. And we talked about iron transport, reversible oxygen binding, and then we were at the place for electron transfer with nitrogen fixation. Again, this is a cursory overview with signaling, where I introduced the fact that you have calcium. And also you can have zinc and copper signaling, which people didn't realize until recently. Zinc is worked on extensively by the Lippard Lab and copper is worked on by other people-- the Chang Lab at Berkeley We will see we're going to have regulation at the transcriptional and the translational level-- that's true for all metals-- and that many kinds of reactions can happen. I'm only going to focus on the reactions that we're going-- that are related to iron in the course of this module. We had gone through reversible oxygen binding, and at the end of the last lecture we were focused on the amazing diversity of metallocofactors. These are some of my favorite metallocofactors. Most of you, I think, are not exposed to this. You sort of know there's an interesting cofactor, but I haven't really thought about how these cofactors work. And if you look at this one-- at the end, we were talking about-- you have-- this is the active cofactor formed-- found in the enzyme nitrogenase, which does an eight-electron reduction of nitrogen to ammonia plus hydrogen. And there are many enzymatic systems that use multi electrons, OK? And that's an active area from the chemical point of view, as well. How do you control multi electron oxidation and reduction, and what is the multi electrons versus single electrons get you? That's a hot area of chemistry now in the bio-inorganic world. And I think most intriguing is this little carbon in the middle. That's a carbon minus 4. How does it get there? Where does it come from? That should intrigue you and-- actually, where is even the reactive species? Where does nitrogen bind to do the reduction? Hydrogenase is another active area of research, now-- in energy. People thinking about how do you do-- how do you use catalysts to do oxygen evolution, here? Or hydrogen reduction or oxidation? And people who have taken inspiration from enzymes called hydrogenases, and they come in many flavors. You have nickel iron, iron iron, iron only. And we know quite a bit about the actual chemistry. The rate constants for turnover are amazingly fast, and so people are trying to do that in little devices nowadays, using this as an inspiration to generate these kinds of catalysts. And what do you see unusual about this iron cluster? I'm digressing, but I think this is a good thing for you to know. I wouldn't expect you to remember the details, but what's unusual about this cluster? Anybody see anything from a chemical perspective that's unusual? AUDIENCE: You mean, like all of the ligands on-- JOANNE STUBBE: Yeah. Look at the ligands. What's unusual ? AUDIENCE: Is there a carbon with five bonds? JOANNE STUBBE: There a what? AUDIENCE: Carbon with five bonds? JOANNE STUBBE: I don't see any carbons with five bonds. OK, yeah. So, OK. That's not what I want you-- so again, it depends on what the bonding is. But what is unusual about the carbon with five bonds? AUDIENCE: It's attached to both. JOANNE STUBBE: Yeah. So that's not what I mean, though. What is unusual about the ligand? AUDIENCE: It's carbon monoxide. JOANNE STUBBE: Yeah, it's carbon monoxide. What do you know about carbon monoxide? It kills you, right? OK. So how do how the heck do we have organisms that have carbon monoxide ligands, right? And we all have carbon monoxide detectors in our house because it binds to our heme proteins and kills us. What's the other thing that's unusual about the ligand environment, here? What's the other ligand that's unusual? AUDIENCE: Cyanide. JOANNE STUBBE: Cyanide. That also kills you. So automatically as chemists, you ought to be intrigued by where the heck did these things come from, and how do you prevent it from killing the organism? At the same time, is this able to use these ligands to actually do chemistry? And if you think about transition metal chemistry-- we won't talk about this, but we will see-- what are the oxidation states of iron that you're most familiar with? AUDIENCE: Two, three-- JOANNE STUBBE: Two and three. OK? And then what we'll see is four happens transiently, but we also go to iron zero's. So we have a wide range of redox spanning chemistry by altering the ligands. And that's going to be one of the take home messages from the four lectures. OK? And this is, I think, totally amazing. We now have an atomic resolution structure without the metals being destroyed, which normally happens when you put a metal into an X-ray beam. The electrons reduce the metal and you don't end up with the cluster you think you're going to be getting. And what you hear see here is you have four manganeses and a calcium. And again, this is a multi electron process where, in order to go from water to oxygen uphill, you need to have light. And a lot of people are focused on that now, in terms of energy production and chemical catalysts that can mimic these kinds of reactions. But what we're going to be focusing on now in the case of the iron is the iron cofactors. And most of you have seen these iron cofactors before. This is just a few of the iron sulfur cofactors. Where have you seen them before? What part of biochemistry in your introductory courses have you seen these before? And then I'll tell you where we're going to see it again. Nobody has ever seen them before? No? AUDIENCE: A lot of single-- a lot of single electron transfer-- JOANNE STUBBE: Yeah. So it's single electron transfer. Where? In respiration. Hopefully you all did that as a basic introductory part. You have iron-sulfur clusters all over the place, and these are the clusters that were found in the prebiotic world. So they are incredibly interesting. What we're going to be focused on are four iron, four sulfur clusters. That's a key-- that's a key component that allows us to sense iron in humans. And so we'll come back to the four iron, four sulfur cluster later on. And all of these-- I just leave you to think about, where do these things come from? You just think you throw in iron and molybdenum and you have a cofactor that looks like that? The answer is no. The other thing that I just wanted to introduce you to because this is a very active area of research in our department-- the Drennan lab-- and a lot of my former students have been working on enzymes called radical SAM enzymes. What does SAM normally do? S-Adenosyl mithionine? What does that normally do in biology? AUDIENCE: Does it methylate something? JOANNE STUBBE: Yeah, it methylates. So, you know-- here is that you have something that's activated for nucleophilic attack-- that's what it normally does. We now know there are 130,000 reactions that involve SAM that doesn't do a methylation. You do reductive cleavage of the carbon methyl bond to form this carbon methyl-- carbon sulfur bond to generate a radical. And you do really complex free radical chemistry. For example, 50% of all the methane gas in the environment comes from a radical SAM enzyme that cleaves the phosphorus carbon bond. If you look at the antibiotic resistance problem we have now, there's methylation in the active site-- the A site of the ribosome that you guys talk about-- that prevents five different antibiotics from binding. And it doesn't involve methylation in the standard form, it involves complex free radical chemistry. So I'm not going to say any more than that, but radical chemistry is taking off. I mean, there's all kinds of unusual chemistry that peak chemists didn't think was possible before, and we're-- every time we study another system, we learn something new and exciting from a chemical perspective. OK. So now what I really want to do is sort of get more into the nitty gritty. And so one of those-- I showed you the periodic table. We have manganese, we have iron, we have copper. Why were those chosen? And in part, those are chosen because it reflects-- the metals reflect earth's history. And one of the-- so, one of the things in the geochemical-- what we know about the geochemical production of the earth over the eons since its first-- since a big bang, anyhow. Here's the earth's core. And this is taken from the article by Fry and Reed. And I think it sort of-- this and then the next slide I'm going to be showing you-- sort of, I think, places in perspective why we're using iron and copper and zinc in almost all the enzymes we see inside of ourselves. So the earth's core is here. And then we have-- so we have the inner core, and we have the outer core. These two cores are 80% iron. OK? We then have the mantle. And then we have the crust. And the crust has-- the fourth most abundant metal is iron. But you also have other things in the crust-- aluminum, calcium, silicon. Why are we using carbon and not silicon, if this is the most-- most abundant-- one of the most abundant elements in the earth's core? And this article sort of goes in and discusses those kinds of issues. Making you think about what you learned in freshman chemistry about the periodic table. Iron. Iron is everywhere. The most abundant element in terms of mass is iron. OK? And so, iron, you might expect from this description, to be front and center. And in fact, it is front and center. And so the other thing I think you can think about is solubilities and evolution of-- from the beginning, where we were in a completely anaerobic world. So here's the gaseous environment with oxygen. In the very beginning up to 2.4 billion years ago, it was completely anaerobic. OK? And so is that important? So if we go to 2.4 billion, it's anaerobic. So now, if you look at-- and this is-- where these data come from and where this model comes from, it-- obviously everything is a model. You can go back and read this in detail if you become interested-- maybe some of you might have had a geology course where you've discussed this before. But if you look over here, where do you see iron? OK, so iron is going to be the focus. Where is it compared to cobalt, nickel, manganese, all these other-- zinc, copper-- all these other transition metals? It's way up here. So it's most abundant under anaerobic conditions. What do you think the oxidation state is? So you just told me you had iron that you've commonly encounter is two and three. And that's correct. Everything-- you're going to encounter this over and over again. Hopefully you have encountered this before. What happens in an anaerobic world? What do you think the oxidation state for iron is? AUDIENCE: Maybe two? JOANNE STUBBE: Yeah. It's two. And I think this is incredibly important from a chemical perspective, because many enzymes we're going to see-- metals-- can catalyze reactions by polarizing carbonyls, for example. In an anaerobic world, you likely used iron two all the time. But what's going to happen when we get over here in an aerobic world? And so that's the key question. And do we see iron two used in that capacity? The answer is no, because in the presence of oxygen some other reaction out competes it. So that's why I'm introducing you to this. It's sort of-- I don't expect you to remember the details, but I think it's an interesting exercise to think about what happened when we transitioned from an anaerobic world-- and this is all in the ocean, and versus the atmosphere-- into an oxygen atmosphere. And this is 0.8 billion years later. And if you look at this, what happens-- and this is a period where they believe that you had a lot of H2S around. And remember, we just saw iron clusters with all these sulfides on them. Iron sulfur was in the prebiotic world. They can self assemble. They do all this kind of chemistry that-- until they knew about this radical SAM super family-- they thought was one electron, oxidation, and reduction. And nothing could be farther from the truth. Iron sulfur clusters play a key role, for example, in DNA replication. OK, so I think-- thinking about this and where these iron sulfur clusters came from, you provided some insight perhaps from looking at the geological record of what people think was occurring. So we went through a period where you had a lot of H2S. Concentrations of species have changed. And then we move into the aerobic world, and what happens here? So what happens to the iron? It's dramatically decreased. So when we go from the anaerobic to the aerobic, why does the iron-- what happens to the oxidation state of the iron? It gets oxidized to iron three. So we're changing in the oxidation state, and so we're going to have to deal with it. So I'm going to show you, this presents a major issue we face now, both as humans and as bacteria. If you look at this, what happens to copper and zinc-- if you believe this model? That the copper and zinc concentrations increase. And in fact, that becomes really important. Because if you look at the biological record, and you look at archae and bacteria that are much much, much older, what you see is-- you don't see that many copper catalyzed reactions in zinc, which has a really important role in humans, with zinc fingers. Doesn't play a role like that in bacterial systems. So I think this represents an interesting way to think about metal speciation, oxidation states, what ligands are going to be involved in what's happening-- and also what's happening in bacterial systems. The key thing that I want you to remember about this is that in the aerobic world-- so we now go from iron two to iron three. And what we'll see is the solubility properties of iron three are dramatically different, and that's something we're going to have to deal with. How do we get-- we talked about this last time-- how do we get iron out of a rock? OK, so that's an issue if you're a bacteria-- you have to figure that out. And bacteria have done some pretty cool things to figure that out. And so-- OK, so this, I think, also has-- we're going to be focusing on iron here-- important implications in terms of the chemistry. So in terms of being in an anaerobic world, we can use iron as a Lewis acid. OK? And so it can polarize a carbonyl. We'll come back to this in a minute. Nowadays, we almost never use iron two as a Lewis acid in biological systems. And why is that true? Because when we transitioned to the aerobic world, now we have this problem of-- that the iron three is what? It's insoluble. So that's one problem. And the second problem is that since it's insoluble, we can't use it. How do you get-- how do you get it to actually use it for chemistry? We're going to-- we're going to talk about that. How do you get it to look at chemistry. And then we have this issue of oxidation with oxygen, and this is going to lead us into module seven. So while in the very beginning, we used iron to do a lot of chemistry without oxygen around. We then moved into an oxygen-- oxyphilic world, and we have this issue of during this oxidation using oxygen as the oxidant-- what happens? You produce reactive oxygen species. OK. So, and then you have also the problem of insolubility. So you generated-- by making this transition into an oxyphilic world you're encountering two major problems that we're focused on. How do you deal with the insolubility problem and how do you deal with reactive oxygen species? And that's going to be-- following this module, we're going to talk about what happens with reactive oxygen species as a consequence of moving from an anaerobic to an aerobic world. I don't want to spend a lot of time on this, but I want to make sure that you understand there are some kinds of reactions that are really distinct from the reactions you meet in the organic world. And a lot of you-- we looked at the vitamin bottle, we learn a lot about flavins, we learn about pyridoxine, we learn about vitamin C-- all the vitamins we learn about. But we sort of ignore the metals on our bottle that's required for life. And so I don't want to spend a lot of time, but there are-- what are the general reactions? So I just want to say a little bit about general reactions. OK. And one of them is this idea of Lewis acid-- or Bronsted acid. And so what you can have is a carbonyl, and you can have a metal that can activate the carbonyl you for nucleophilic attack. OK? Where have we seen this before? We've seen this before in-- if you go back and you look in the glycolysis pathway, lots of times you use zinc to activate the carbonyl. Sometimes you use shift spaces, maybe. You remember that? In aldehyde dehydrogenase, aldehyde oxidate that converts aldehyde to an acid or reduces aldehyde to an alcohol-- they use zinc. OK? In the completely anaerobic world, people thought-- most of the time they probably used iron. That was one of the most-- that was much, much more prevalent than zinc. But then things-- so if you go way back and you find bacteria that lived in that period, they still might be using iron in catalysis. But now we almost never use iron two in catalysis, because of the issue of the redox chemistry. So now they're saying the-- so it's now, you know, your polarize this for a nucleophilic attack. You've seen this over and over again with the Claisen reaction, the Aldol reactions, et cetera. I'm not going to go through the details. Another place you see it-- and where have we seen this one? Again we have a metal-- and I'll just leave it in the plus two oxidation state. But what happens to the pKa of the water bound to a metal? And what happens is the pKa is dramatically reduced. You have two positive charges here, depending on the interact-- and that interaction's unfavorable. So the pKa becomes reduced on bonding to a metal. Where we've seen that before? We saw that in the cholesterol module. We didn't talk about the chemistry-- again, I come from the chemistry side of it so I find the chemistry the most interesting-- but it fits into the biology. Where have we seen this before? Anybody remember-- in cholesterol? Homeostasis? What happens in the Golgi when you want to go from the Golgi to the nucleus? AUDIENCE: A zinc-- JOANNE STUBBE: Yeah, we had a zinc protease. So that would be an example. An example of this would be in the cholesterol section. And I'm not going to talk about this in detail. I used to talk about this in a lot more detail, but you can see with different metals-- this is just an example of the first case I'm giving you-- the pKa's of the metal bound are reduced. Again, it's a play off. Those all with waters, those ligands. Every time you start changing the ligands or you change the oxidation state, these numbers change. OK? So you need to know a lot about the metal you're dealing with. So that's one place-- you've already seen all of this before. Whoops. So the second thing I want to very briefly talk about is the second kind of reaction-- which maybe many of you haven't seen before-- is electron transfer. OK. So this is basically oxidation reduction. And so clearly, you've seen oxidation reduction. So if we have some metal m in the n plus state, and we add an electron, it gets reduced. So to get to the reduced state, remember we need two half reactions-- something gets reduced, something else has to get oxidized. And what's different-- we've looked at redox cofactors-- and most of you have looked at a lot of redox cofactors in primary metabolism, like glycolysis of the pentose phosphate pathway or whatever-- what are the normal redox cofactors you encounter? The organic redox cofactors you encounter in biology? AUDIENCE: NAD. JOANNE STUBBE: Yeah, NAD-- NAD flavins. OK, so this chemistry always involves one electron. So that's distinct. NAD, we've already talked about this, always involves hydride transfer-- two electrons and a proton. So this is one electron. OK. And so one electron. And if you have other things-- we could have proton coupled electron transfer. So PC is proton coupled electron transfer. And remember, we just saw the example of nitrogen getting reduced to ammonia. OK? You're doing an eight-electron reduction, but you've got to have protons. That involves proton coupled electron transfer. If you're converting water into oxygen, again, you've got to take care of the electrons and the protons. And if I get that far in the last module, my lab works on ribonucelotide reductases-- that makes a precursor to DNA. You would never think about radicals, at all, but that chemistry involves proton coupled electron transfer. So here is some of the most important reactions in biology, and you really haven't been exposed to what's unique about the chemistry. So what do we know that's unique about the chemistry? What do what do we know about rate constants for electron transfer? Anybody know anything? Fast, slow. What's different about electron versus hydride transfer? AUDIENCE: With the hydride transfer, you have to transfer an entire proton, versus-- JOANNE STUBBE: So you're transferring the proton, which-- what's the difference in mass between an electron and a proton? AUDIENCE: A lot. JOANNE STUBBE: Huge. It's 2,000-- 2,000 fold. So you remember, probably from introductory chemistry, when you think about electrons, you think about-- you think about quantum mechanics and quantum tunneling, as well as-- it can be-- electrons can function as both particles and waves. So they can function as waves and particles. And while I'm not going to spend a lot of time talking about this, this is a central reaction in the inorganic part of biochemistry that occurs in humans that you need to take into account. When things behave as waves, they can function quantum mechanically. And we have an expression called the Marcus equation, which allows us to calculate the rate constants. So we have a rate constant for electron transfer. And if we have some acceptor her and some donor-- so, all we're doing is redox chemistry. The question is, what governs the rate constants for electron transfer? Well, it could be the electronic overlap, so that's part of it. This is part of the Marcus equation. What else governs the-- what else governs the redox chemistry if you have a donor and acceptor? The reduction potential of the donor and acceptor. So you need to think about the reduction potentials. And what other factor governs the chemistry? Does anybody know? Around the metals. So you have to think about, how much energy does it take to go from iron two to iron three, or copper two to copper one? What other factor? What else happens to the metal during a reduction or an oxidation? AUDIENCE: A reorganization. JOANNE STUBBE: Yes, a reorganization. So it can change its geometry. And so the other factor is called lambda, and this is reorganization chemistry. And furthermore-- and we'll see this is important a little bit with the iron systems-- it doesn't just have to be the immediate coordination sphere of the metals. It can be the second coordination sphere, as well. So the whole protein is important, I think as hopefully most of you know by now. And we're not going to spend a lot of time on this, but I think this is something you need to think about-- the rate constants for electron transfer. They could be 10 to the eighth, 10 to the 10th per second. How does that compare with the rate constant for chymotrypsin? What's the turnover number for a protease? Anybody remember? OK. So a turnover number for a typical protease hydrolyzes-- like the cholesterol one hydrolyzes on an amine bond-- might be anywhere from 10 to 50 per second. OK. So how does that compare to this? Slow. Very slow. So again, the chemistry of electron transfer is quite distinct from most of the chemistry you've encountered, and so you need to know it exists because it's everywhere in biology. We don't spend that much time on it in this class, but it's a unique part of the chemistry associated with metals. OK, so the third thing I wanted-- the third kind of chemistry I want to very briefly look at is substitution reactions. OK. Now, in organic chemistry, what kind of substitutions reactions do you have? This is something hopefully you all remember from your organic, but what do you-- what do you have? What are the two basic reactions you learn about in the first semester of organic chemistry? AUDIENCE: SN1 and SN2. JOANNE STUBBE: Right. SN1, SN2. Associate or dissociate. Same thing in metals, OK? So you need to think about associative-- what does that mean? Dissociative. If you have something with four-- a metal with four ligands around it, you're going to add a ligand to get the reaction to go. That's associative. If you have something with four ligands around it, one of the ligands could dissociate, and you only have three ligands-- and that's the basis for getting that chemistry to go. And the reason-- the thing that I want to focus on and-- the thing I want to focus on is ligand exchange. So ligand exchange could occur by associative or dissociative mechanisms. Where have you seen ligand exchange in recitation? I think it was recitation four? You probably didn't think about it. I mean, we were doing something else. But the key to it working is ligand exchange rates. What about histamine tags? OK, so here you have a metal. What kind of a metal do you have on your column? A nickel. And the nickel is bound. But in order-- so you can hang-- how does your thing hang up? By ligand exchange. How does it come off? By ligand exchange. So an example of this is histamine tag chemistry. And another example that you've seen is magnesium. OK. What are the rate constants for ligand exchange with magnesium? Where do we see magnesium in biology? I'm spending too much time on this. But I actually think this is incredibly important. If you take home a few of these basic reactions, this is all you really sort of need to know to deal with metals and biological systems. Where's magnesium? Where do you find it? You find it on-- AUDIENCE: Phosphates. JOANNE STUBBE: On phosphates, yeah. So you have nucleotides-- like ATP would be an example. Whenever you have ATP, if you look at the charges of ATP-- we went through this in one of the recitations that I taught-- you never have these negative charges. It's always complex. Just something to neutralize it. And the major thing-- since magnesium is 10, 15 millimolar inside the cell-- it's always bound. But if you try to isolate magnesium through some kind of a column, what happens? The magnesium-- because of the rate constants for exchange-- falls off. So if you have something else in there that can out compete it-- like protons or something-- it's gone. You never look at-- it depends on the rate constants for exchange-- but you never see the metal bound to those small molecules. So this is rapid exchange. And we'll see in the case of iron rapid exchange-- and I'm going to show you a table with this-- but rapid exchange is also important. And why is that important? It's important because say you isolate a protein and you're putting it through a column. What-- if the ligands are coming off and on, what happens to the metal by the time you get it out the bottom of the column? There's no metal. So the issues with iron, which is everywhere, that catalyzes many, many, many kinds of reactions, is it's really hard to tell that there was a metal there inside the cell, because the iron dissociates during-- in the plus two state-- during protein purification. So what about-- what if I changed the oxidation from iron two to iron three? What do you think would happen to the exchange rate? AUDIENCE: Slow down a lot. JOANNE STUBBE: Yeah. So it would slow down a lot. Every metal-- every metal is different. Every set of ligands is different. But you need to think about exchange reactions, because they're all over the place in biology. Here's is an example that I took out of Lippard's book. I used to give a lot more data than this, but these give you the rate constants for exchange for iron. Here you can see iron two, iron three-- and these are all waters. OK? That you're never going to sight see inside the cell. You might have a few waters, but you have other ligands around. All of the exchange rates change with different ligands, so you need to think about that. And also magnesium-- 6 times 10 to the fifth per second. So it's exchanging really rapidly. And that really does govern-- you know, here we're doing protein purification here. We're trying to identify what the metal is. This is it made it really challenging to tell whether you ever had iron two bound to your protein. Sometimes you isolate zinc bound to your protein. And I'm going to show you-- because of the periodic table, zinc always out competes iron. So when you're purifying something and you have zinc contaminant in your buffers and stuff like that, you'll get the iron replaced with zinc and think you have a zinc protein. And you don't. You really had an iron protein, but because of ligand exchange, you don't know what the real active form of the protein is. This is something that's plagued this area for a long time, and it certainly plagues the area of the iron that we're going to be focused on. So let me see. I think I want to go up one more. All right. What do I want to say now? So the other thing I want to talk about is-- that's unique and distinct from what you see in solution-- all of this stuff happens in solution. That's where we learn. Just like with organic cofactors. We sort of study them, we learn how they work, then we take them into biological systems. We use that as a starting point for think about-- thinking about how the enzymes use these cofactors. And in fact, what you learned over here is exactly what you learn over here, except nature has figured out how to catalyze the reactions by a factor of 10 to the 12th faster. OK? So nature adds her-- adds her two cents worth on top of all the organic and inorganic chemistry we learn. And what is it that at this information? It's the protein environment. So the last thing that one really needs to think about is how do proteins tune metal properties? OK. So that's the big question. And we're going to spend a little bit of time talking about that. And to do that, I want to go back to the periodic table. OK, so again we're going to be focused on these metals. And what we see is there is a set of rules that inorganic chemists Irving and Williams-- many of you may have heard of the Irving-Williams series-- it sort of makes a prediction based on what you learned about transition metals in terms of ability to bind. If you compare all of these metals in the same oxidation state, in the same geometric environment. So one of the questions that we face is binding. And why is that important? Because inside the cell, we will see that copper binds much more tightly than manganese-- no matter what you do, that's true. And what's the basis of that? It's the atomic number, which changes the atomic radius-- it makes it smaller. It makes the ligands bind more tightly. So the problem is, when you're inside the cell-- if all these things were floating around inside the cell-- how do you control the metallation state inside the cell? So that's the key issue, and I'm going to give you-- I'm going to show you a little bit about how nature has figured out how to control all of this. It goes awry quite frequently, and that's-- how does it manifest itself? It manifests itself in disease. Just like we saw with cholesterol. So what we're going to see-- we are going to look at first row transition metals. In general, we'll see that manganese two binds less tightly-- if you look over there, you can see where we are in the periodic table. The atomic numbers increase less than nickel, less than copper. So here are-- here are our transition metals. And so what we see is the atomic numbers decrease-- increase. And the atomic radius decreases. And therefore what you see is over at this end, you have weak binding-- over at the manganese and iron end, we have weak binding. And over at this end, we have strong binding. So if you had a protein-- and I'm going to give you an example of this. There is a protein I'm going to show you that combined both copper and manganese. And you had equal amounts? Copper would always win-- by a lot. OK? So you need to study this, but you know-- you'd have to use 10,000 times more manganese to out compete the copper. OK? So this just shows you. So this is called the Irving-Williams series after the people who described this. And what they compared to get these numbers-- they are looking at all of these things in the plus two oxidation state. OK? And they're looking at it all in an octahedral environment, with six ligands around it. This is all plus two oxidation state, and all octahedral. Everybody remember octahedral? We have four equatorial ligands, and two axial ligands-- I'm not going to draw that out on the board. OK. So that's an issue. And the question then is, how do we deal with this issue? So here's our Irving-Williams series that I've given you here. But what do we do-- how do we deal with this inside the cell? The issue is that-- in vitro, you have an issue. And there's not much you can do about it, except control the relative concentration of the metals. Inside the cell, do you think it's easy to control the relative concentrations of metals? What do you think? Concentration is everything in biology, we just don't talk about it that much. Do you think it's easy to-- say I threw in the outside of a cell 15 millimolar copper. Do you think the cell could control that? Do you think it would all get taken in and then all of your enzymes would be loaded with copper? No. So you have to have a way to actually control all of that. There was a spectacular paper, I think, published a couple of years ago that sort of demonstrates this point. And so I'm going to give you this example, because I think it really-- it was published in 2009. So in vivo. So this would be, over here, in vitro-- sorry. In vitro. And we can't get rid of the in vitro part. That's the chemical properties of the molecule, we're stuck with them. So it depends-- in vivo, metallation depends on abundance. Can we control abundance? Absolutely, we can control abundance. You've already seen with cholesterol, you control abundance with transcription factors. That's one way. There are many ways. We're going to see-- that's one of two general ways that iron is controlled. What about speciation? I've already told you-- and we're going to come back to this with iron later on-- you know, are the metals all bound to waters? But you have ATP inside the cell. Could iron two be bound to ATP? Absolutely. So it's a question of competition and what the binding constants are, which is what are talking about in recitations this week. You know, if it's really weakly bound, then something else will out compete it. But you can purify-- you put ATP in a solution, you'll pick up iron all the time if you do atomic absorption on it. Because iron can easily bind to all the negative charges on ATP. And the other thing that you need to think about is location. So location is what we're going to be focused on in the example. And what do I mean by location? Even in a bacteria you have location, right? What are the two different compartments? You have a periplasm and you have the cytosol. We're going to be talking about periplasm, and you have cytosol. In us, we have much more complicated locations in metal homeostasis. We'll see when we get to the second lecture. Has a lot of issues it has to deal with, OK? Let's look at this example. And I'm not going to spend a lot of time on it, but let me just show you what you need to think about. And these workers we're interested in a cyanobacteria. And they wanted to find what was the protein that bound the most copper, and what was the protein that bound the most manganese. So we're looking at two extremes of the Irving-Williams series. So this group identified-- C-- I can never remember the acronym. CucA is the most abundant copper two binder. And they identified MncA. And the way they did it was pretty creative. If you're interested, you can go read the paper. It's the most abundant manganese binder. Both of these things are made in the cytosol-- both of the proteins are made in the cytosol of the cell. And what they found when they studied this system in more detail is the structures of the proteins and the ligands bound to the metals are exactly the same. So you have a beta barrel in both cases, and you have the same first coordination sphere. The first coordination sphere of the ligands directly bound to the metal. If you took these two proteins, and you wanted to load MncA-- this is manganese binder-- in the test tube, you would have to add 10,000 times more manganese than copper to get the manganese in there. So that, again, goes back to this-- I mean, it's going to be different for every system-- but it goes back to this question of controlling metallation inside the cell, which is extremely challenging to do. And we don't-- there's a major, in my opinion, on cell problem in biology. They're going to do this by localization. So let me just walk you through the kinds of experiments they did. So, both these two proteins-- the one that binds copper and the one that binds manganese-- are produced in the cytosol. It turns out that the one that binds manganese folds uniquely in the cytosol. And the cytosol, if you look at metal speciation, how much free-- you're going to learn about free today, or tomorrow in recitation-- how much free copper or zinc do you have-- do you think you have in the cells? Copper two and zinc two in the cell? Do you think you have a lot? A little. AUDIENCE: For copper, I know it's less than one percent. JOANNE STUBBE: Yeah, it's less-- yeah. It's tiny. Both copper and zinc bind extremely tightly to-- again, it's all about speciation, so it depends on what the ligands are inside the cell. And in fact, in the cytosol cyanobacteria, they have measured a micromolar of free manganese. And so again, this speaks to this question of is manganese readily oxidized? No. So you don't have to worry about reactive oxygen species with manganese. What happens is this protein folds in the cytosol of the cell. Comes off the ribosome. It picks up the manganese and folds. But its location is in the periplasm. How does it get to the periplasm? It gets-- there are two ways you can get proteins from the cytosol to the periplasm. One is through the Tat transporter. And Tat transfers-- it recognizes a couple of arginines. A little zip code-- we've seen zip codes over and over again-- which then takes it in the folded state into the periplasm. And the manganese, once it's in there, doesn't come out. Doesn't exchange. So there's something about the environment that does not allow exchange. So the manganese is placed into the protein in the cytosol. Now, what happens to the copper binding protein? In this case, as soon as it comes off the ribosome, it gets grabbed by a second kind of transporter. And this second kind of transporter transfers the unfolded protein through the plasma membrane. And it folds in the periplasm. And in the periplasm, I don't know what the ratio of copper to manganese is, but remember-- copper, by this model, out competes manganese by a lot. So even if you have manganese and copper in equal amounts, the copper will always win out. And so what happens here is the copper then binds. And so the copper is loaded in a different location than the manganese. So the way this organism-- this is just one solution-- I think a pretty creative solution-- to how you deal with the Irving-Williams series, which we're faced with all the time with the many, many metallocofactors we actually have inside the cell. I'm going to come back-- next time I'll talk about two more issues. I want you to be in tune with me when we move on in the iron world. And talk about sort of a big cartoon for metal homeostasis. Doesn't matter what the metal is-- any of these metals. And then we're going to move on and focus on iron.
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