 0, because this is the delta, delta is 1 of 0. This is a very useful trick to remember. Whenever you differentiate the inverse of a metric, where one of its indices are contracted with a metric, you can transfer the derivative of the metric at the price of a minus sign. The next thing we remember is that this quantity can be put in terms of diamonds. It should be conveniently constant. So we go very constant to see if the metric is the assertion that del mu of GAA is equal to gamma mu A alpha G alpha A plus gamma mu A alpha G A alpha. Because the derivative is to the left hand side, then we give you the expression to go very del symmetric, which must match. In fact, these are the relations we used to solve for gamma del. So we can now make these expressions and replace them with the metric with these. So what do we get? So we get minus alpha of mu A alpha G alpha A. Plus gamma mu alpha A G A times G AM times whatever we had there. Now this was simply this. So this was now gamma with G AM was gamma EO. Take a look now. Thank you. So that was from the first step. That's also right. That's the first step. So from the second step, that's the easiest. Minus mu of mu. So we get plus gamma mu A alpha G alpha A plus gamma mu alpha A G A alpha times gamma A mu B. And plus what we got from here, which was plus G A A gamma A mu alpha gamma alpha mu B minus G A A gamma A mu alpha gamma alpha mu B. So this is a full set of first step. Now if I can write, these two should cancel against two of these. But look back and see what could be the answer. So let's see. What does this have the property? This has the property that you lower an upper index of a gamma and that index is A. So this is like that and this is like that. So let's compare this step. See here, we take an upper index of a gamma and lower it to make A. And you contract away what is lower in the same and that's mu and mu. Here I take the upper index of a gamma, lower it to make A, contract away mu and mu. This step comes with a plus and this step comes with a minus. Okay. This clearly cancels with this and then of course the mu versus mu also gets this. This cancels with this. So we left with, from the first step of the terms, we left with mu A gamma. Now I'll replace an alpha and I'll replace capital A by beta. Beta, I'm writing another step. Gamma, alpha, mu B. Gamma beta mu B minus gamma. Now this step, mu A and gamma beta mu B. The whole thing multiplied by G. Is this correct? Now this term has all the symmetries in it. Firstly, let's look at A goes to B. Clearly, if A goes to B, you pick up a minus. Because this term means it goes to B. Is this correct? Okay. So symmetry of, anti-symmetry under A goes to B is manifest. Symmetry under pairs is also manifest. And mu goes to B. This term under the pair flip, this term will become gamma B mu gamma mu A minus gamma mu B gamma mu A. Symmetry of the pairs is also manifest. Symmetry of symmetry is also manifest. Why is it manifest? You see, what is this term? This term says you keep A fixed. Okay? Which one goes here and the other goes there? You keep A fixed, so I keep A fixed. Which one goes here and the other goes there? Clear and cyclical permutations of one. Can't see that. We have gave two things. We've got another expression for the curvature tensor, which I've written only in parts, but I'm going to ask you to put together it is. This is the second derivative and this first derivative, please. Okay? The curvature tensor is this plus this. It's a nice expression. It makes many properties of the curvature tensor manifest. And we use that to verify symmetric properties, as well as cyclic assumptions. Any questions about this before we proceed to the Bianchi area? I'm sorry, people. This lecture in the previous one may seem a little boring because it's so algebraic. But I think it's important for us to go through this in class because you shouldn't be left with the feeling that there is anything mysterious about this stuff. This stuff is totally simple, as you see. It's just mindless algebra with nothing to it. However, if you just said that you can check that, I don't know about you, but I somehow feel sometimes, I don't know, can I check it? It looks pretty complicated. I don't know how it looks, OK? So that's the purpose of doing the algebra, not because I'm sure all of you would be able to do it yourselves. But I think we should do it in class anyway, basically, for the purpose of demystification, just to see that it's totally mindless, you know, an idiot, OK? So here, it's just totally straightforward. Everything that we want to do is totally straightforward. Also, let's get some sense of how to do these index manipulations so that when you have to do it yourself, you know. So we're going to, you know, bite, you know, we do a little more of this. We bite the bullet, and just, you know, we're going to exercise ourselves with strength in our muscles by doing the index manipulations. Even though it's a little boring, just to get that. OK? We come to the more exciting stuff. Just imagine doing all of this before you knew that it was correct. It's not totally, not really, you know, you're totally in the dark. You've got this crazy idea that space-time was dynamical. You know, every second day, you're probably thinking, this is totally crazy, I must take a crack for it. And then you have to do all the same index manipulations and stuff, it's totally remarkable that that happened. OK? OK, let's go on. OK, now there is another important identity which the analogy with field strength suggests. And that analogy goes this way. As you know, field strength away a beyond the time limit. This is your familiar with the context of it. In a U-unfairly, you know that it's true that d mu, that if you take the cyclical sum of d alpha f mu nu and cyclically, it gets 0. Why is that? If we do f mu nu, f mu nu is d alpha del mu a nu minus d alpha del mu a nu. And clearly, the cyclical sum of this is the cyclical sum. The cyclical sum of this is this. This is just, these two are symmetric. The only question is which one goes here. Cyclically, you sum over all three possibilities. Yeah, but where do you sum over all three possibilities? So, yeah, the famous thing that you know in electric magnetism. And the similar identity is true now in the study of non-abelian gauge theory when you replace the derivative with what's the non-abelian covariant derivative. Now, this is not a constant non-abelian gauge theory, so we want to talk about that. That is, just to suggest that the analogy with the non-abelian gauge theory suggests that a similar identity should be true for the curvature. So when you replace derivatives, of course, by covariant derivatives. Makes no sense to have an identity with ordinary derivatives, but if it's true in one column, it just won't be true in another. That's also when we do the gauge there. In the non-abelian theory, there's covariant derivatives. So, this is the suggestion. This is the suggestion that comes up. So, the suggestion is let's check whether an identity of the form is true that r a b, and now, you see, we should keep the first two indices fixed because they're like the matrix indices. Okay? So, mu mu covalent derivative. Okay, this notation I haven't used so far in the class. Landau-Lichitz uses a many texts used very often. Covalent derivative, ordinary derivative is denoted by comma. Covalent derivative is denoted by semicolon. Okay, so is it true that this r a b phi mu semicolon mu plus r a b phi... Okay, but we might as well do it with lower indices. Why does it matter whether we do lower indices, somebody? Because if you're taking the derivative, why does the difference between, ordinarily, you see, the difference between these two expressions, upper and lower, the difference between these two expressions, upper and lower, is not true here because there's a derivative. So, g lower and g upper, you know, you have to differentiate often doing the lower and upper. But for the special case of lowering the upper and raising to the metric, it doesn't matter because covalent derivative in the metric is same. So, you can furiously raise and lower indices through derivatives, through covalent derivatives because of this wonderful problem. Yeah, so since we can do it as well for... might as well do it. So, is this true? It's a question, right? To actually verify this in full detail, it's, you know, quite an algebraic exercise. But we're going to prove it using a trick. Okay? And the trick, which is a very useful trick, and we've used repeatedly, we could also have used this trick to check that the commutative to covalent derivatives was a curvature. I think we checked last time. But I think we want to do it for the first exercise. Okay? And this... But I know you should start that exercise using that trick. You might want to go ahead and do that. Okay? Here, I'm going to use the trick. And the trick calls up all of us. It says, well, look, this is a tensor equation. Okay? So, if it's true in one covalent system, it's true in any covalent system. Now, is there any, especially in nice coordinates? Yeah, the locally Euclidean coordinates. As we've proved, we can always, at a point, move to a coordinate system where G is eta and the first derivatives of G value. Okay? Fine. We'll use that coordinates. It's probably more convenient for me to use the... Sorry. After having made these things a little lower, I think we'll use the upper one. Okay. Okay? So let's remember what I was doing, for RABs. RAB, mere mu. Now, once again, I've been writing down without looking at the answer, just so that, you know, because it's useful to be able to do and... So, remember what is this? It's del mu of gamma nu. And gamma nu was R gamma AB minus del H mu of gamma nu. Del mu of gamma AB mu. Minus tau. Minus gamma gamma nu. I'm not going to report this right at the indices, because I won't be there. Why would I need that? Because what I'm going to be doing is taking covariant derivatives in this special frame. Gamma's vanish. Now, gamma's vanish means derivative of gamma's vanish. But this is going to product with two gammas when you take the derivative. One of them will always vanish. We don't need to... Look at these things. Is this correct? Because one of them, one of them you can't say... That's not... The derivative is not zero, but the derivative. Gamma's vanish. Okay? Second thing, because we're in this coordinate frame, covariant derivative second ordinary derivative. Gamma's vanish. It's pretty convenient, right? Okay? So, this expression in the special coordinate system, the expression that we had, simply becomes... So, let's write it out. So, simply becomes cyclically superior, cyclic of del between phi mu and mu. Del phi, del mu. Okay? How do I write this? d2 by d xr other gammas. A v mu gamma mu phi minus gamma A v mu mu phi. Cyclically, mu, mu and phi. Obviously, yes. Because what do you do? You permute to take mu. Which of these three is the guy that's special? That comes with b? So, there's one double mu here, one double mu here, one double phi here. Similarly, yes. There's one double mu here, one double mu here, one double phi here. You can't respect what the camera is. This class is not really important. One by fourth of that is... Thank you. Let me do the rest of it. It doesn't see until that part. Yeah, I mean, top part is very... What is this? Reader issues. And this part is completely... Yeah, and this part is completely... Okay, so then we can use this. Entry drop test, as well as... This is the end game. Little algebraic exercise I want to do is the part. If you remember, in the last class, we defined a Ritchie scale. Ritchie scale was A, B, mu, nu. Contract first to third, seventh to fourth. So, first to third, seventh to fourth. We also defined the Ritchie test completely identically that we derived. And from that compute an interesting property that relates derivatives of that Ritchie scale and the Ritchie scale. So, let's rewrite this identity. So, the identity was R, A, B, mu, nu, phi, plus R, A, B, uh, phi, nu, mu, plus R, A, B, uh, phi, mu, nu. This equals zero. In this identity, what I'm going to do is to contract indices like this. A with mu, A with mu and mu. Okay? Because it's a lovely property of covariance derivatives commuting with the metric, you can fearlessly contract through derivatives. So, this term gives us R, not my data file. R, covariance derivative. Okay? What do we get from the other one? Contract A with mu, so, and B with phi. This term gives minus R mu, nu, covariance derivative of mu. Because contract B with phi gives if you're supposed to contract either first or third, or second or fourth to get a plus. If you contract second or third, you get a minus. Either side. Either side. We should contract mu with mu. B with mu. Sorry. I should be contracting B with mu and, uh, this way. So, this is a plus. Plus R, uh, mu, phi, mu. Well, again. So, we should contract A with mu. Then the second. Start with, uh, mu, mu, phi, phi, R. Sorry. I'm cycling people with it. Cycling people. Phi, mu, nu. And one more cycle to go. Nu, phi, mu. Sorry. Let's look at it. So, B with, I'm supposed to contract B with mu. So, B with mu, that's this. And A with phi. So, A with mu. A with mu, huh. Exactly. So, this A is a minus. Because A with mu gives minus. R, B, phi, and then contract B with mu. And this one also, yeah, so we're supposed to contract B with mu. That means a minus. Minus R, mu, phi. Which tells you, mu, phi, mu, is equal to half of R. But you can rewrite this as, in the form that will be of importance to us. As minus delta mu, phi, R, to a very derivative respect to mu. Probably I think in the notation we've been using before. Delta mu, R mu, phi. Minus, oh, half minus delta mu, phi, R, R, R, 2. We finally finished all the basic algebra we want to go through. That's just all. We finally finished all the basic algebra that we want to go through. And now we're, you know, where we're going to start studying this. That's the grunge that will generate the dynamics of the relativization equation. So the grunge that we're going to suggest is the volume. We're going to suggest the volume. We're going to suggest the grunge. 1 over, minus 1 to 16 phi k. This course we're working in units in which the speed of light is set to 1. But we'll keep our balance. 1 over 16 phi k times integral square root minus g R plus 1 over 16 phi k square root minus g multiply this 1 over 16 phi k square root minus g R plus matter. Again, this is the grunge that we're going to suggest. Why are you trying to make up the second derivative of mu? Why not consider the derivative of mu in this case? Why are we taking our field as... Well, you've seen that if we do form f mu nu, we get from Christophelsen as we saw in this lecture. We get the coverage. Now, if you make an f mu nu f mu nu kind of term, that will be an action that is forth-ordered in that context. Why do you... I mean, why do you consider g mu as a field instead of... We can consider gamma as a field. Suppose you try to consider gamma as your field. How many gammas would you have? You'd have 40 of them. Okay? How many g's are there? They're 10. So, gammas are highly redundant. Since they're in A as well, so we can always base... You can. See, what... I mean, it's not that it's impossible. It's that... Okay, let me say this. See, what we're going to do is to write down... Let's try to write down the most general action that we can that is generally coordinated. Now, what you're suggesting is why not write down an action that's not performed. In the end, the question is none of our variables. The question is, why have you chosen this form of variable? You can work in any variable at this point. That's not the hard part. The question is that the analogy with gauge theory suggests that that would be a more accurate question. There are many reasons why... This would not be... You see, if you start looking at gravitational waves, I mean, it is the components of the metric that I've looked at. If you start looking at quantizations of small fluctuations, you're going to have great trouble in the sense of this theory because it's... It is in some sense more accurate. You know, however you think of it. But that's not the important point. Those are finite dimensions. That's much simpler. And the simpler part is that, look, you see, first, before we do anything else, let's do some dimensions. Action. Okay? And action must be dimensions. Okay? Now, what are the dimensions of... What are the dimensions of this square root g-out? So how do we do it? We might assign length dimensions to length and time because c is equal to 1. Okay? Now, ds squared is equal to metric times dx dx. The components will always have dimension in them. That tells you that the metric is dimensionless. So the only dimensions associated with the problem are in derivatives. derivative carries one inverse length dimension, which is something that I will follow in the quantum field theory language referred to as a mass dimension. So you often say h bar equal to 1 in such an inverse length dimension. That's just a term. Inverse length dimension. Okay? Now, if somebody tells me what is the dimension of square root g-out? What's the dimension of... One by L squared. One by L squared, exactly. One by L squared. Because it's two derivatives of dimension. It's either two derivatives of g or gamma times gamma. Each gamma is one derivative. Okay? So one by L squared is this that. Now, if this may be dimensionless, what was the dimension of this? One by L squared. Because this whole thing is like one by L squared. So k has to also be one by L squared. So that cancels. So L squared goes here. Okay? So the dimension of this k, whatever it is, is equal to is the dimension of k No, no, I got it wrong. Sorry. I missed something else. There's d for x. Yeah. Okay, sorry. So what's the dimension of k? Let's repeat that. L squared. Because this is one by L squared. Cancels two of the four length squares. There's length squared remaining. k has to be length squared. Okay. So in such an action here, there is a length scale associated with... There's a length scale associated with this new fundamental constant of k. Which, of course, will be related to new instance. Okay. Now, you see, now let's suppose we make the following constant. Then there is a single length scale. There's a single interesting length scale associated with gravitating dynamics. Actually this constant will be false because the constant will have to be constant. But let's first make this constant and see how far we can go. Suppose that every term in the action is weighted by the same length scale. Suppose this was the case. Then, how do we go behind this? Well, so let's suppose I'm replacing an upright k as equal to L p squared. L p is called a blanket. In the real world, it's said to be about minus than it can say. Okay. Let me write this as L p squared. If I had known like this, what power of length scale would it be? Zero. Okay. This will come with zero and it might have even higher number of terms. Let's say an odd to the four. An odd to the four term will come with what power of L p. Right. So, if we had plus, for instance, odd to the power of four, squared minus g, before x, it will come with like L p squared in the numerator. Now, suppose it is true as it turns out to be the case in the real world. The L p is a very small number. Okay. Suppose L p is a very small number. Then, for physics at some length scale capital L, let's estimate the relative importance of all these terms of nature. Okay. For physics of length scale L, this term will work out whatever its mass dimension is, its length dimension is, which is L p squared. So, this term will contribute like L squared by L p squared. This term will be staying there. And this term will be L p squared by L squared. If we're interested in looking at the motion of the sun around the earth, the earth around the sun, I'm regressing five hundred years. If we're interested in looking at the motion of the earth around the sun, and we'll do the earth sun distance. L p is a fixed number. So, the question is, the layer we are suppose, that there is some unknown dynamics of fundamental scale, that is generating all terms that are allowed by symmetry. Okay. Let's assume that a grid term that is allowed by symmetry is then an action. Four distances. Okay. Four distance at distances long compared to L p. Which is the most important one. Some suspicious reason for that. But you know, when you start to contemplate theory, such arguments can be being quite precise. This is a bit suspicious, because quantum theory doesn't quite work the correct way. But these arguments are made precise and formalized in what's called the Wilsonian renormalization, true. And the interesting thing is that these dimensional analysis kind of arguments often are very good guide to physics. Even when we take into account complicated coupling, basically certainly. I mean, there's nothing to check but if that's just true. The important mechanism. Okay. So, what? What? What the? What the? We've concluded that suppose your excellent question, the real answer to your excellent question about why is such a term is probably a fact. There's some as-here unknown theory, it's a fundamental landscape. Just quantum mechanical. Okay. Which is, when you take that theory, whatever it is, and go down to low energy compared to this blank scale. Okay. You get some effective action that has all the possible terms that are out by symmetry. I mean, actually the terms will be dictated by the dynamics of that theory. But in the absence of knowing what that theory is, we will expect all possible things. Okay. Now, observationally we see LP is extremely small. For this reason, at lens scales is large compared to LP. It's the first term that is of maximum dynamics. Okay. For those of you who are familiar with the study of quantum field theory, this is the general law that terms of lowest mass dimension govern long distances. Okay. This sounds like a very beautiful story. It sounds like I have very plausibly motivated that given that we have general coordinate and variance and so on, Einstein's action should be the right action. But there's a big j-pick hole in this, in this, in summary point. If you take my logic to completion, there is one more term you can add to this action that is even more important at long distances. Which one? Root G. Root G. What's wrong with this G? Why power of LP will come with 1 by LP to the power of 4? This is what you will expect. This will affect most of all. Okay. Okay. So, of course, the term by itself is totally true. Not yet. This will give you two different equations of motion. I mean inconsistent equations. Now, if you add this to this, what you can find is that the only solutions are universes which are curved and let's say 90. So, if you took the logic that I was applying for you and took it through a completion you would relate that we have a universe whose curvature scale is 10 to the power minus 33 centimetres, which unfortunately is totally wrong. Is 1, 100? Okay. Sure of its suffice, sophisticated, this is what the question of why this prediction of gravity is totally wrong. Why this prediction? It's not prediction of gravity. It's prediction of being 90. What you're saying is knowing nothing else, you predict that all terms allowed by symmetry are generated. And that's how it works pretty well. We get the second term. That's somehow not the first term. Probably what must be the case is that once we understand the correct theory of gravity, presumably the correct quantum theory of gravity, it will help us understand why this cosmological constant term is not generated, was actually generated any smaller. Now, until the end of the 1990s everyone had every, most theoretical set as you, that since the cosmological constant is so small and blank units, clearly the fact that the universe is big that the cosmological constant can't be, is approximately is less than or equal to the inverse size of the universe. So what, that if such a term appears it's not having surprised by 1 over L, 3 to the 4, but 1 over Hubble scale. Size of the universe. There's a huge distance between delta minus 3 and size of the universe. So since this, since I was a huge hierarchy, I think most theoretical physicists, certainly when I started my PhD, it was commonly known that the cosmological constant 0. However, it's got observational evidence that this is probably not the case. Okay. So now there are two mysteries. First question is, why isn't it 0? And the second question is, why isn't it 0? In the sense that it's really 0. Nobody had a good idea for why that was the case. But at least 0 is a good number. You can imagine explaining 0. But now, it's got even worse. How are you going to explain that it's not 0? But enormously small. Okay. Right. So this argument that I gave you, there's this nice beautiful dimension analysis argument that sounds so nice, has this big gaping hole in it. And I would say this, this big gaping hole is one of the important elephant in the room in theory, in fundamental theory. Because every, every way you have of trying to understand the gravitational theory from a fundamental point of view, would predict that the universe was curved at a much smaller lens scale than we see it. And that's obviously not. Okay. So it's a startling gaping fact that fundamental physicists have failed to account for, for 50 years, 60 years. And you know, it's one of these, it's one of these things. Now, you know, Planck was guided by, for instance, the failure of black body radiation. That's because little benefits, you know, specific curves. More detail, you have the better. But this is one of these things. It's an obvious contradiction between what we think we know how to predict and what we see. And it's resolution is awaited. Okay. This is a very important question we get. Now, this is the cause and I don't know what the answer to this. It's a cosmological question. It's a cosmological question. It's a very, a lot of very disquieting talk. You know, something to try to explain it by saying, look, but what's called the anthropic question. Which says that, look, if the universe was very, was very, very highly curved, then, you know, you wouldn't have galaxies, you wouldn't have matter, you wouldn't have people living inside planets to observe the universe. So, maybe somehow, for some reason or the other, the eternal inflation, there's a whole ensemble of universes. Most of which are very highly curved. Once in a while, you come about to the universe, it's a big curvature. And the question is, why are we so incredibly lucky as to live in that one? It's like, because we couldn't have lived in any other. Okay. Now, this may be a certain, sophisticated, just because, it's a cute argument, but I don't think it's business. You know, once you start getting into these kind of discussions, you're going very near to kind of things, you know, philosophies. And that has led us nowhere. You know, what we want is something with equations, which we can predict with, which we can make progress with, and if this is the best we can do, I would say, we're giving up on business. Okay. People won't even let to such disparaging arguments, because they haven't been able to think of anything better. Okay. And I think that's true. And it's your job. You know, you are the fresh new generation. You guys have to come up with something. The right crazy idea that will help us understand why this is happening. Okay. What would this be? I don't know. How do I explain this? Okay. But I haven't said all of this. Maybe I should sweep it out of the college. I do. A view that for some reason, this argument which does not work in predicting that a big first term works in giving you a second term and all other terms will be highest and best. That has an extremely good confirmation from the expert. At least from the expert, we know that it's true that at reasonable landscapes, the term that governs interactions is square root j. And had it not been for this elephant in the room, they would have been very reasonable argument. Okay. We should take an old argument with many pinches of salt because of the elephant. But let's move on. Okay. So this is Einstein's action. This is the action you don't have without this one. And we're going to invest in this course we're going to study. We're going to study this action. And then what we want to do is understand this action a little bit. So now we have two parts to this action. What we're going to do is, of course, we're going to take the action and get it moving to motion by varying this action with respect to the action. Please. Yes? Yes. Other terms also? Then all terms are equally important. So if they're interested in physics, those others, then using just plain Einstein's action is completely understandable. F-R theory is a guarantee. But those are very specific theories of that. Because they keep some terms from knowledge. Now, you know, if you just try to keep all possible terms, you'll more or less lose predictability. So basically what this is telling you is that once you get a length scale small compared to the order of the prime scale, then you know you need the correct quantum theory of gravity that tells you what the right action is. Crude symmetry analysis is not correct. You need to know what the right theory is. Once we start trying to understand the right sense of feedback or something like that, we need that. That's a great way to understand quantum. Unfortunately we're not going to be able to do that because I don't know how to understand this. Okay? But these are great questions. Excellent. So, let's move on. We're going to make this action and vary it with respect to the metric to get the Einstein equation to motion. Now, we vary it in two parts. There's a variation of minus square root of gr and then there's a variation of square root of g in the magnetic field. Both terms are important. Okay? When you start by starting the variation of square root of g in the magnetic field, let me take an example. Let me take the max value of the the max value of the gun in the magnetic field. Okay? There are minus 1 by 4, for instance, f mu mu f mu which I'll write explicitly as minus 1 by 4 square root minus g f alpha beta f mu mu alpha beta g mu. Let me take this function and vary it with respect to the metric field and let me choose the metric with the upper components of my basic equation. Then g mu alpha and g mu alpha. Yes, that's true. With respect to the metric field. Okay. So, there are two sources of variation. Because we're contracting like this and varying with respect to g mu. So, delta s is equal to minus 1 by 4. One term is minus square root of g f alpha beta f mu. Okay? And then we get up. We'll make it half because there are two such terms. Okay? Or minus 1 by 4 out. And write 2. And this term has this mu and mu contracted. It doesn't. And then delta g alpha. That's one such term. Okay? There are two possibilities. They're both the same. That's clear. We should also vary g inside the determinant. So, we have to learn how to take variation of the determinant. So, let's learn. Remember, is that you can take a determinant and write it as g. What you remember is this. That there is a name for the coefficient in a determinant. Suppose you've got a determinant. You take the determinant and look at the coefficient of a particular element in that term. There's a name for that coefficient. It's called the mind element. Okay? You see, a determinant cannot have any element in the matrix at higher than linear order. So, all terms are either if we fix on a given element in the matrix. All terms in that determinant, either of all is zero. I independent of that element or of order one. The coefficient of the order one piece, the path that is linear at that element, is the minor of that. Okay? So, let me, suppose I've got a matrix g alpha beta. Let me denote by the minor of g alpha beta. Let me denote it as m r. This is the minor. This tells us that the variation of a determinant. This, what is that? That's equal to the variation of the determinant with respect to each element times the variation of that element. I'll call it m. Is this clear? Because the derivative of the determinant with respect to that element is the minor. And that begins the check. This is del determined by del g alpha. That is equal to del g by del. And this is just a derivative check. Is this clear? Now, this is a nice formula but it's in terms of this irritated minor. Okay? This minor is related to some other object of the matrix. It's inverse. So, g inverse is equal to m in transpose. In transpose. Divided by g. And how do you know that? You know that because if you now multiply by g on this, when you take the sum, you get a minor to give you the determinant instead of giving you delta. So, this is a familiar formula from the study of matrices. So, we can rewrite this at least for some time. I would have to think through the transpose part but we don't care because we're dealing with symmetric matrices. We can rewrite this in this case as g alpha beta. That's the inverse of g. That's the g alpha. A formula for variation of the determinant. Now, we're interested in varying this determinant. We're interested in varying this determinant. We're interested in varying this determinant with respect to the upward. Can you tell me what the right formula would be? You see, just from the fact that if you contract these, you get a number. You just transfer this delta. Yeah? We expect a minus. Minus delta g alpha beta. G alpha beta. Just parenthetically, you might be wondering what's making the symmetry between the metric and the inverse. You see, from the point of view of the determinant, the determinant of g is one of the determinant g's. Okay? So, let me call g prime as debt g inverse. Okay? So, of course, it must be true that delta g prime is given by this formula, where everything is replaced by g prime. Equal to g alpha beta delta g alpha beta g prime. Can you see that these two are the same formulas? They're the same, because g prime is one like g. Delta of one by g is equal to g alpha beta delta g alpha beta by g. Now, defensive. So, that implies that delta g is equal to minus delta one by g. It's minus one by g. G g alpha beta delta g. So, we have another way of getting this information. What is it? When is your next class? 11. 11. 11. Okay. So, let's add that to the other class. So, what we got is minus one by four. And now, we get, now, so far, what have we got? Delta g is equal to g alpha delta minus g alpha beta g alpha beta g. What we want is delta square root of minus. Okay? So, delta of square root of minus g is equal to minus half by square root of minus g times delta g is equal to minus half times minus delta g alpha beta g alpha beta into g by square root of minus g. Now, I'll take this minus in here and write this as minus half into delta g alpha beta and g alpha beta square root of minus g. Let's put everything together. Putting everything together what we get. We get that delta that changes the action is minus one by four integral square root minus g and then minus half f mu nu minus half g alpha beta f mu nu f mu nu plus alpha theta f theta beta theta into f by g. I want you to know this. The first thing is that so, let me write down Einstein's equations. This will be one of the contributions. It's the coefficient of delta g variation if you have an electromagnetic field. But in more generality I'm going to make a couple of definitions. Fine, right? The by definition is going to be associated with this is by definition going to be called half times square root of minus g times t alpha beta delta g alpha. So, whatever you get by varying the matter part of the action with respect to electromagnetic field it always takes this form. This is necessarily true from general point of view. Square root g times the tensor and delta g alpha beta. That tensor whatever it is you're going to be defined as a stress energy tensor. So, for the electromagnetic field we've concluded that t alpha beta is equal to minus 1 by 4 f alpha theta f beta theta minus half g alpha beta. So, this is just an example. In general, whatever you get by varying the matter field we're going to call the stress energy. Okay, now there is a very important property of the stress energy tensor. Okay, we're going to now try to prove stress energy tensor is basically up to one subtlety, the same thing as the nother column for space time transitions. And because of that the basic property of a nother column is that it gives us a conserved I mean that there's conserved column that's what's here. Okay, and so while we're going to prove to you is that this stress energy tensor that comes from an arbitrary matter field is conserved obeys a conservation technique. Okay, and prove the very simple we had in the class for this. It's very simple to require to almost know how to do that. I think, how does it go? Let's see. You see, consider an action that is minus G, minus G and then I think lambda is some function of some matter field. The matter field could be a mu like in electricalness. I make a coordinate transformation on this action. The action remains in there. How does coordinate transformation happen? A coordinate transformation changes the matrix. But a coordinate transformation also changes the matter field. Okay? But on shell but we independently know that from the equations of motion for the matter field the variation of the action with respect to arbitrary changes in the matter field to first order that change punishes by the equations of motion. Two zeroes. The first zero is kinematic. Any change that changes both the metric as well as the matter field in a way that is a coordinate transformation does not change the action. The second zero is dynamic. On the solution to the equations of motion and arbitrary variation of the action with respect to just the matter field is balanced. What do we conclude? We conclude that on a solution to the equations of motion the variation of the action with respect to a coordinate change just on the metric must vanish. How does a coordinate change act on the metric? Okay, when we have a small change in coordinates how does this act on the metric? Now, a minute what kind of thing we talked about in the first couple of cases? You see, a change in coordinates acts in two ways. It acts inside the attribute of the metric and it acts by transforming the imprecise of the metric. Let me put this in a problem set which we will definitely get next week. Okay? Let me put in a problem set the following thing to show that under infinity as in the coordinate transformation given action has to be a problem. Suppose we have X mu is equal to X mu plus zeta mu. Part of the reason we have under the problem sets together is that your teaching assistant seems to have a right. Something. What? Somebody see that? Yeah. Okay, suppose we got what the change of this form? X mu is equal. Okay? Then up to a sign that I can't remember that we will get straight when we give you a problem set delta G mu, which is G prime minus G is equal to del mu of zeta mu minus del mu of zeta mu. But then it's covariant. This is not obvious from what I've told you. But it's true and you'll prove this. What do we find? You see, what we had was that by definition the variation of this action with respect to G, G alpha for arbitrary variation with respect to G alpha, beta first half square root minus G T alpha, beta, delta G alpha, beta. This is always true. But we've concluded from our logic that the variation with respect to those changes of the method that come from a coordinate transformation must vanish on solution sequence. Okay? So we've concluded that it must be that half square root minus G of T alpha, beta del alpha strongest zeta, beta plus del A in W Symmetrical. Now del beta, zeta, alpha is equal to 0. Now, this is symmetric. So these two terms are the same thing. So that implies that minus square root minus G T alpha, beta del alpha zeta, beta is equal to 0. How about 3 zeta, beta? How about you just use this to do something for this T alpha? So first thing we do is use the gene. We write this as square root minus G del alpha of T alpha, beta zeta, beta minus the alpha now we get minus square root minus G of del alpha, T alpha, beta beta, zeta, beta is equal to 0 because we go right there to the basic chain rule. Okay? And now this thing has a vector in it. And do you remember we have this nice formula for the divergence of a vector? We have covariant divergence del alpha of zeta any any vector is equal to 1 by square root minus G del alpha square root minus G theta we derive this in one of our analysis. Let's use that formula. So this term that makes that term the 1 by square root G cancels the square root G. So it becomes del alpha of square root minus G T alpha, beta zeta, beta and therefore this is a total net. So if we focus on zeta you know we take some action that's a boundary term we look at zeta as the banished of the boundary the quantum transfer is to complete the variance. We look at such quantum transformations the banished and the boundary are non-zero away from the set region only in a set region this is our banished in order to use Gauss's law and make it the surface integral this is an ordinary derivative Gauss's law works for ordinary derivatives it's just a studio statement of integration doesn't work for ordinary derivatives and in this case this very integral is converted into an ordinary derivative nicely canceling the square root G cancels otherwise we couldn't make this up Once we have that now everything seems to be banished so what we conclude is that this term is zero but this is zero for arbitrary Zeta, beta therefore it must be a del alpha of T alpha, beta the stress tensor defined through this definition as the variation of the matter part of the action with respect to the upper part of the metric as a property that it is necessarily automatically conserved is this clear where do we find Einstein's action we will find some equation of the source sometimes alpha beta that comes from there in the Einstein part of the action it's completely alpha that's clear but this T alpha beta is not something arbitrary it's automatically conserved now can you use that to guess what this is going to be why? look at this part you see if you have the equation that's true it must be true of the variation derot is on both sides remains true okay this thing as a property that it's covalent derot is zero see the way we get on the left hand side no we don't it's some constant right side now Einstein when he tried to run down the general result he made many false attempts and one of the problem that falls at them was to make run down the equation r alpha beta is equal to T alpha Einstein was not thinking he was just writing down equations of motion not thinking so much through the action so it just wrote down this equation and it took some time to realize that it was obviously wrong because it was inconsistent derotive on both sides now at the same time that Einstein was working you know these guys Hilbert and a few other people were you know were deeply investigating the geometry of arbitrary curve spaces and Hilbert and friends knew the stuff very well they knew of the right combination of curvature of that that vanished under governing derot they written it down before so the action the Einstein action is often called the Einstein Hilbert for various reasons Hilbert himself though was very modest about his contributions to general derot even when asked about whether he thought he'd done the important stuff he said something like well any child in Grottingham knew the combination of curvature that was vanished under governing derot that wasn't the important thing the real story was about that man in the element so in the next class people actually derive the Einstein part of the or the Hilbert part of the Einstein action by varying square root Gr then we will have Einstein's equations in our hand then we'll be ready sorry for being too busy see you guys tomorrow at the evening