 Okay, so let's continue so that our plan for today, we can then have a slightly shorter the utility class. The utility is that I'm going faster than light. Actually, let me just perform, let me start the class of five minutes before we need to do it. Because it's clearly very important today. Okay, so as you know, so this is just for fun. As you know, there's this upright step that is created in the bottom. They produce new signals here. They measure new signals there. Okay, and then there are also details about how they, how they compute when they're produced, how they compute when they measure it, how they compute the distance it's on. And there may well be problems in all of this. Let's take a look at how we're going to tell us a lot at 4 o'clock today. But the really surprising thing about it is that if you believe that results, what they did is they measure the distance, they measure the time, they compute at L by 18. And they get an unforeseen speed of light in a one plus, and I don't even remember. That's minus five. So you might wonder, so you might wonder, suppose this was true, so we're talking about these neutrinos, right? So you might wonder, if this was true, do we have to throw away all the physics as we want to? Okay, so most likely, as I think I will tell us today, there's an experiment. But suppose it's true. Suppose it's true. Would we have to give up physics as it is understood today? Now, you might think, well, let's think about it. Suppose special relativity is correct in platelets. Then it would be impossible for anything to go faster than that. And the world goes flat. Then it would be impossible for anything to go faster than that. Okay, so if everything was just flat, it's special relativistic, I think. If the experiment was correct, we'd have to rewrite a hundred years of physics. Okay. Now, you might wonder, well, what about, suppose this was coagulated. Suppose there's more coagulated space-time than we thought for some reason than the other. And it's the effect of this coagulation that accounts for this extra-fast propagation. It's not, not including geometry, that's why it's so fast. And the problem with that explanation is that if that were the case, then anything, including light, would take the same amount of time to go from one spot to another because, you know, because of the equivalent. So that looks like it can't be fixed. But there is another answer that, just to give a thorough opportunity for answers like that, that might totally decrease, you know, might be correct without requiring to explore all of physics as we understand. So taking this scenario from a paper somebody wrote a couple of days ago, I can see the compster and he's basically done the same on the previous papers, right? The idea is very simple. Suppose we live in a high, you know, world in more than four dimensions. And there's a brain. What is a brain? It's some special surface in this four-dimension space. On which most of the fields of the sample are localized. So space-time is, let's say, five dimensions. But there's a four-dimensional brain. You know, it's like a rubber sheet setting. And all of the particles in the standard model, almost all of the particles in the standard model are vibrations of the rubber sheet. So as far as these particles are concerned, you don't really see the fact that space is five dimensions. Okay? But suppose that the neutrino is a particle that's not organized in this brain. But probably it's in the mesh language. Then it could be that a neutrino right out of the pseudo-dimension and dark health is lost in something that is constrained to stay on this surface. How does this happen? Let me make it a little more precise. Suppose the metric of your space. Okay? Whereas the s-square is equal to e to the power of, let's say, 2 a over r. So r is the fifth core. So the coordinates are r and the four-space-time coordinate is equal to x. Suppose the s-square is e to the power of 2 a over r. And minus h of r dT-square. Okay? Plus vxx squared. Plus e to the power of 2 v of r. Now, suppose this was the case. Suppose this was the case. And now you ask. And suppose that at this place, we say r equals, I don't know, zero. It's some distinguished value for r. Let's say zero is the bridge. And the h of r, zero, is equal to one. So is there anything that's stuck to the bridge? Moves to the speed of light. But something that's stuck to the bridge. Okay? It could go down, propagate here. It could pass through this propagation. But it would have to be, it's supposed to be propagating just on the x-section. You're going around the, you're going around the local now, right here. Sorry. Yeah. The equation in your way is h of r dT is equal to dx. Or vx by dt is equal to h of r. But if this space was something like this, by h of r, somewhere where r was not zero, it becomes larger than one, then something down here can propagate faster than the effective speed of light for this bridge. And you can imagine a path that does this. Comes back up here. Faster than any excitation that's overrides on this surface. Could reach the same point. Now this proposal is way out for a number of different reasons. I mean, let me just say one of these reasons. I say out of this because, sir, ideally it's like one in 10,000 that the experiment is correct. But if it is one in 10,000, and we get that one chance is one out, then there's an opportunity for a new outside. So it's very important. Almost certainly it's wrong, but it's supposed to be. So, fine. So let's look at this. Now, what are the problems with this? Problems with this scenario are very different. As Amor will explain today, this is not the first time neutrino speeds have been measured. They've been measured in a supernova experiment, supernova that was seen on Earth in 1987. Neutrinos and light gave on the same time from the supernova as far as we can tell. And the distance from the supernova is so large that if this difference has been what these guys say, then the neutrino should have come before the light by years. So it's not an experiment. The response is when the neutrino came from the supernova and much lower energy than the neutrino that came from the formation of the ground. So perhaps how far the neutrinos go down in the space, this is just a couple of pictures of what that means. How far the neutrinos go down in the space depends on the energy. And that higher energy things go further down. It's a bit weird. The other equally big objection is that there's fabulous experimental evidence that electrons don't exceed. You know, electrons, neutrinos are hard to measure. Neutrinos are brilliant butterscotch. You have them moving around in accelerators all the time. You measure other properties all the time. There's fabulous experimental evidence that electrons don't move faster than that. Now, for those of you who know what's a particle physics, you know that under the SC2 left gave symmetry. Electrons and neutrinos have all been out there. And so had that gave symmetry been an unbroken symmetry of the real world, then whatever the electron was doing, whatever the neutrino was doing, the electron would also have to be new. The difference is that the neutrino goes down here where the electron stays new. And the possible way out for this is, electro-ring symmetry is not an exact symmetry of the real world. It's broken by the Higgs mechanics. And perhaps the same thing that breaks this... that does the Higgs in, gives you differences of speeds between the electron and the neutrino. No, I have to say that just just this laying analysis we've done so far, well, then it makes it sound like a conspiracy theory. Right? I mean, so it's a little like the claim that, I don't know, the U.S. intelligence agency has been behind it. It's not impossible, but it sounds like one in a very good way. It sounds beginning to sound to me like that, because every other piece of data doesn't agree with this. But then you're trying to find some way of explaining it. Okay, but who knows? Okay, so the reason I said all this is to say that well, suppose this experiment is correct. You see, in physics as we understand that, there is something so sacrosanct about the local light, but there is nothing sacrosanct about global light. You have a solution of general relativity, such that at infinity it has flat four-dimensional states. Then you have such that it's basically an implicit of a flat four-dimensional state. Then you've got global and internet, things properly on the speed of light, kind of being the six-dimensional measures. But if you come up with a solution of general relativity, the principles of general relativity, certainly do not imply that the vacuum of your theories, universe 11, has global and internet. This would be a situation in which it doesn't, because if you're trying to make the T, E, and X Lorentz transformation, it would have worked. It would have been a symmetry of the metric, and H could have worked. But it isn't. It's perfectly consistent with the basic principles of general relativity. So I just thought I would say this, that there are two possibilities, either all of them, suppose this experiment is correct, either all of physics, it's a basic formality of physics, after we've grown up, or that the world is somehow more interesting, you know, we could possibly accommodate this within general relativity, rather than on the special one. Okay. This was all, of course, my entertainment, just because we're hearing so much about this experiment. I just wanted to say how astrophysicist might think about this issue, even though it's great, but most likely it's not. Okay. Any questions or comments? Any other weird answers? No. No. No. No. No. No. No. No. No. No. No. It doesn't work. Okay, so if there's a problem with the battery, what could it be different from? So now that we've had the fun, that's a difficult question. Okay. The main thing I wanted to discuss today is the notion of, it's the notion of energy and momentum in general. The basic idea would easily be extended to any language. There are many problems with the analysis. If you are going to get a module that will actually work, 15. Everything you will have another problem. But it makes the point that it doesn't necessarily require throwing out basic principles. It just reflects that there is a more complicated attitude than you would think. Now, let's remind ourselves of the Einstein equations. So the Einstein equations work with a question. So the Einstein is J8 by G. To remind you of another fact that we proved in the course of the right-hand side, it's true of identity. The thing that we proved was that the left-hand side is true of an identity in the dead eye of R, I and J from the top. G, I and J is R. This is not true from the dynamics. It's just true for any lecture configuration. It's an identity. This is something we checked in our lecture. And so this equation, or just by taking it, is the equation that's true that automatically implies the conservation of the stress sets. It implies that I, T, I, J is independently we checked that T, I, J satisfies this equation. So it was all a nice existence. Sorry. And so it might seem that conservation of stress energy is built into the basic structure of J. So it would be easy to find a definition for energy and momentum in J's relativity. It's getting stressed. That's it. It's greater. And it should not be conserved by Gauss's. The answer to that is no. This is no because this is not an ordinary expression. We found an expression, a simplified expression for the divergence of the symmetric bits. The really simple expressions were for the divergence of the symmetric bits. Okay? Well, it was just putting a square root G inside. And a 1 by square root of G outside. The expression for a symmetric bit, so it was more complicated. And I remind you, the expression for a symmetric bit so was that this is equal to 1 by square root minus L by L X K of T, K, I, square root of minus G. This is exactly like the word in the next lecture. But then there's an additional case. Minus half L G K L by L X I. Okay, let's not use L because of this thing. Okay, so the only free index here is I. This is an identity that is true for any subject. An identity that didn't have this term. Then if you take, if you take square root minus G of this divergence and integrate it, you know, if we take this equation, this term was not there. Then we take this equation, multiply by square root minus G and integrate over all space. That's this equation. This square root minus G gets the square root of minus G. Okay? And we will get the integral over total derivative. And so we would then be able to associate a conserved energy momentum tensor associated with the integral over space like slice of T K normal to the space like, normal to the slice that we integrate. Okay? So this term spoils this formal conservation of the stress tensor leading to a conserved quantity. It spoils us from concluding that we have conserved energy momentum from cost. Okay? And it's just not true that the integral of the stress tensor, the matter stress tensor dotted with a normal is a conserved quantity. I mind what's going on here. Does that mean that the general derivative is incompatible with conservation of energy and conservation of momentum, which are very basic, you know, of principles of physics? And the answer is no. The thing is that you're just trying to incorrectly identify what your energy momentum tensor is. Okay? And intuitively the reason, the thing that you're really incorrectly is trying to associate all the energy momentum in your system with the matter. Now in general relativity, we have gravitational waves in the sense. The waves are no matter, but they can be re-energy. So the thing that's going to be conserved is the energy of the matter, plus the energy of the gravitational field. So it would have been strange had we been able to get a conservation law just from the matter stress tensor. The question that we're going to ask is how do we find what is the correct expression? How do we find the correct expression for the conserved energy momentum of the matter stress tensor? Okay? It is the question. We follow the calculations. First thing we're going to do is to study this equation in a frame at one point. In a frame with the first derivative to symmetry. In this case, this equation just boils down to Tk of del k i of z. See now? If we're in a quantum system at a point in which our first derivative to symmetry occurs, we can take the square root g out. So that would be different from this thing, my first derivative to symmetry. And this term is 0 because of the first derivative. We were commenting that in such a frame, the conservation of the stress tensor would re-express it, but we're commenting that the conservation of the stress tensor would re-express all the Einstein equations. It becomes an identity. It doesn't use equations at all. In such a frame, in a frame in which the derivatives of g vanish at a little point, this equation plus the use of the Einstein equations should just be an identity. Let me say this again. In an arbitrary frame, this equation is an identity. But in a frame in which the first derivatives of the metric vanish, this equation should be an identity. So we can prove that del i, with the upper lower end of the equation, d with the lower upper, so del i of r i d. My name is half g i d r. We must be through that in a frame in which first derivative to symmetry vanish at that point, this equation should be an identity. In the true, we should be able to see it directly. And we're going to try to determine that. Now, before we even do any calculations, because how could such an equation be true as an identity? And that is very simple to do. Suppose this quantity, r i k minus half g i k r, was equal to lambda i k m, del by del x, the lambda sub-expression, such that lambda is symmetric in i m k. And lambda is anti-symmetric in k m. Suppose we could find an expression of this form. Then let me take the k derivative of this. That would be zero because we have del k, del m, and lambda k m. If you want to see this at the anti-symmetry. Remember we dealing with ordinary derivatives, rather than covariant derivatives. Because we're working at a point where first derivatives are electric, because the way in which it could work and does not be the way in which it can work. So our goal, our algebraic goal for the next 10 minutes is to find an expression to show this and to find an explicit expression for this lambda i k. Is this clear? Any questions or comments? But let me tell you where we're going even before we do it. Why do you require these? This has to be symmetric between i and k, because all these expressions are symmetric. At least the derivative has to be symmetric. Actually that's the more I can say. This derivative is symmetric between i and k, not necessarily k. But it's important that lambda itself is anti-symmetric in k m. Because that conservation of stress tensor will come identity, because l by l is k, l by l is m, which is lambda i k m would be identity easy, because of the anti-symmetric. So we don't require that lambda itself is anti-symmetric between i and k. We require that the derivative of lambda is. But we do require that lambda itself is symmetric between k and k. And that's how it's going to work. Unfortunately it's going to be a little algebraic to say that. So we have to roll up our speaks into the algebraic. Before we start doing the algebraic, let me show you. Once this is true, once this is true, we're going to get an expression such that what we're going to do is the following. What we're going to do is to identify the stress energy tensor, the true stress energy tensor of a space time. The thing we have to integrate if we want to find the energy moment. With this expression, I'm working in a frame in which the first derivative of g dash. See how clever this is. This is really clever. Because this is you have to see the expressions for this to really make sense to you. We're going to write them. But this lambda will be built out of some derivatives of the expression involving the immeasurable tensor. Some saying that it's immeasurable tensor. Now, that expression, whatever it is, is magnetically. Maybe I should make this point after we get to the end. Let's first start with the derivative. I'm going to write down where we are heading towards, what we want to show, and then we will write a little faithful answer by the instructions. What I'm going to demonstrate to you is the one thing. Let me not call this lambda, let me call it. Alright, I'll call it lambda. If she calls it, it's down. It's a very awesome thing to use because it reminds you of the magnitude of that space. Okay, now what we're going to show is the following. We're going to show that this expression, we're going to find the volume. We're going to define h i k n is equal to l by l i k l is equal to and is equal to some factors of what we are understanding here. Normalization. g i k g l f g i l and in our special frame with this definition of lambda and this definition of h, we're going to show that l by l x a h i k a minus g h is almost the theta except that there's a matter of g between them. The g has to matter in the special frame. This is what we're going to try to show actually. Now before I go through the algebra which will take us some time to read for you. Let me assume that we've completed this algebraic existence. And let me draw the consequence of it. Because that's the important thing. The algebra we have to do is we teach it in a class but that's the volume. Let's first extract what okay now look. Firstly, let's do this. Firstly we want to show that we want to check that all these equations are consistent. The first thing I want to check is that we find something that is symmetric in i k a. It just wouldn't be true that this equation was correct unless the left hand side was symmetric. So let's check that. So what we have is del by del x a of h i k a del by del x a of h i k a and h i k a. So that's del by del x m of you know like this factor. Symmetric minus g i k g l m minus g i l g k a I mean h i k a Now I want to check whether this thing is symmetric between when this quantity is symmetric in whether this quantity is symmetric in i m k okay so let's see this is obvious because a and m are dummy variables so this term is obviously symmetric and this term is not obvious but after the symmetry wasn't true until we've taken the event. So it was not true that h was symmetric between i and g. This is a general thing. Those of you who have studied with permutation group know it's difficult to both make things symmetric in some indices and antisymmetric in other indices and this is a discharge. If you start symmetrizing one set of indices and then after that antisymmetrizing another set of indices ruins the symmetrization in an h that itself is symmetric. But the derivative isn't that much. The second point is that we want to show that this quantity is manifestiveness. What we want to do is choose one of these variables that say del by l x i where we want to take del by l x i of this and see that it's 0. But all that that requires is checking that we have antisymmetry between i and let's say h that this h that the h itself was antisymmetric between i and h. So from here we have del by del xm of numbers minus g by dA. I want to show that it's antisymmetric this index and one of these indices. Let's see k suppose I change k and l if I change k and l that's k right? You see suppose I change k and l let's just do with the change h i of l k h i of l k is equal to del by del xm of minus g g i where I have k I change to l where I have l I change to k g k a m minus g i k g l m and this term this term, this term, this term the rest of minus I change to s. I've managed to show you the algebra that leads to the stick and that's a big supposition that we have to get the algebra right? Suppose I manage to show you this algebra then we have established that quite in general there is what would be such? We have established that there is a quantity h such that d by dx d by dx a of h i k gives you a stretch tensor that's automatically conserved moreover, this quantity reduces to the matter of stretch tensor in a frame in which first derivatives of metric branch in a frame in which first derivatives of metric branch where in the locally falling element we can't distinguish that frame from just that space locally so the matter the full stretch energy tensor of gravity should have the probability that it gets no gravitational contribution in a frame in which first derivative of metric branch so what we've done now is to find an expression that has two good properties one that it reduces to the matter energy stretch energy tensor in a frame in which in a locally intermediate frame and two that it is manifestly always conserved it's manifestly always conserved because we take l by l x i on it you can see it so now you can use Gauss's law you can now use Gauss's law to define a charge associated with this with this name sir and that charge will automatically be conserved but Gauss's law need to apply on the region yes but we have to say only a point where we have made the derivative okay good this is a very important and let me say we have established what have we done we established that this quantity is equal to the matter stress tensor only in a frame in which first derivative of metric branch however the good properties of this quantity the fact that the fact that the stress tensor is symmetric and the fact that the stress tensor is not covalently but ordinarily conserved applies to everything in the check let me say this again what we are going to show using algebra is that in a frame in which the first derivatives of the metric vanish the matter stress tensor reduces to this expression now that we have got the expression we are working the fact that I am going to define a stress tensor what language we can define a stress tensor by this formula by this formula we can show that there is derivative between 0 and that yes so let me say it is true with the matter stress tensor if the derivative is 0 and that okay now I am going to define that by we then h r k a is equal to minus g d v full okay inquiry frame this is my definition of the full local stress energy pseudo stress I say pseudo stress is scary and other things because we are not demanding the fact over this is the definition this definition satisfies the good property that in a frame in which the first derivatives of the metric vanish the definition reduces to by Einstein's equations it gives you the matter stress tensor but the definition is a definition it is a definition for all frames for this to work it is necessary that all the consistency properties the properties that we want of this thing then it gives you a symmetric stress tensor and that it is conserved otherwise it is not interesting work independent of using the first derivative of the metric vanish but we just check it in our check of these two properties we know where I use the first derivative for the matter stress energy tensor that has been covariantly conserved yes is it a consequence of that that this the ordinary derivative of this is 0 and that is well you see the ordinary derivative of this is 0 it is just an identity we define something and for an arbitrary metric the ordinary derivative of that is 0 but you know you can find many identities that out of the form some derivative of something is 0 most of that may not be interesting this one is interesting however because this quantity in a frame in which the first derivative of the metric vanishes reduces to what it should reduce to namely the matter stress energy so this quantity deserves at least provisionally deserves more chance of deserving the name stress energy tensor of general relativity it is the notion of it is basically a notion basically a notion of localized stress tensor in general now we should take this with this notion too seriously we should take this notion too seriously because it is not a thing so for instance in one frame it can be that this thing reduces just to the matter stress 0 because there is no matter stress another frame it could be that it is not a thing just because we change coordinates we should take this the local definition of it is too seriously however what we are going to use is a crotch to integrate to get a global notion of energy which as I will argue we will be able to take 6 just one last crotch the fact that the ordinary derivative of this is 0 is independent of your conservation of stress energy it is an identity for any metric and it marches that conservation now it reduces to the statement by this statement that it is conserved now this thing is ordinarily conserved works for any metric field no mention of Einstein equation this is just a statement about George however by use of Einstein equations it reduces to the conservation of the matter stress energy tensor in a frame in which perspective this second statement is the thing that gives you some confidence that this quantity will be associated with the actual stress tensor actual stress energy tensor will be the classian acting on T mu is equal to 0 actually stress energy tensor will be derivative of T mu is equal to 0 that is the actual matter stress energy tensor the full stress energy tensor general relativity includes matter stress energy tensor plus other stuff the thing that physically deserves to be called the stress energy tensor the thing by which if you integrate you get the conserved energy and the conserved momentum of your system is not the matter stress energy because gravity also carries energy the gravitational wave in the claim whether you make the claim that this object T full defined in this way is what we are going to define as the stress energy tensor as we said stress energy tensor is very little however it is not important in gravity in fact it is 0 however what we are actually going to do is not what is so much about the local formula but try to get a formula for the globally conserved energy moment so when you are talking about conservation you need a time direction for it you need to say that the energy is 0 and that choice of time you are saying since your theory is co-efficient that kind of becomes a good question so suppose because where is the lambda that the question here is there are more variables of energy so how it becomes in that this quantity reduces to the magnet stress tensor we use that the first level use of the metric energy but in showing that age is conserved and that it that it is symmetric that del by del xa is symmetric we have not used it see just as a matter of algebra we make this definition of lambda we make this definition of h and we make this definition of t lambda in that way we are not using the derivatives of g and that no this is how we came upon in this definition but now forget when we came or came to it for a minute somehow at the other I define lambda this way then I define h this way and I define t full this way you may or may not agree that it is an interesting thing to do but you cannot stop me from doing these definitions so it does not matter now I will just put these definitions that this quantity is that this h full is symmetric between h and k without anybody using first level t vanishes but also demonstrating that del by del xk of this quantity vanishes and that came about because this quantity was anti-symmetric between a and k no way have we used in that demonstration I will show you the first example I mean that there is no other there is nothing to do with the first first example yes in the definition it is just a definition of anti-symmetric and the two properties that we prove that a defines a symmetric stress means it defines a conserved stress you may not like the word stress but depends on a symmetric quantity that we prove just directly without using any assumptions ok now this was mathematics hey mathematics cannot tell you what energy is in the system hey you have to put in some physics and where you put in the physics is this that we have shown that this definition reduces to the matter stress tensor in a frame where the first derivative is of g dash this is an exact according to what you would expect for the actual stress tensor from the equivalence principle because in a frame where the first derivatives of g vanish there are just in that local frame there are quote unquote no gravitational waves these equivalence principles like we've discussed this first derivatives ok so in that frame this should reduce to the matter stress tensor now the fact that you can't over all of space then choose such a frame that is your question means that you cannot this will not, it will not be possible to everybody got the simple matter stress tensor it's a definition it's a definition of the full stress energy tensor in general relativity that agrees with the matter stress tensor in special frames but in general it's more general minus the matter stress energy tensor there is something about the energy energy tensor ok now this this algebra I'm going to work out but the algebra that Rathias was asking about has been worked out in random equations now I'm not going to work out but I'm just going to now that you have this code you have a definition you can just compute t full ij minus t matter ij for t matter I can use the Einstein equations so after constant you replace it by R ij minus half R g ij and the right coefficients this is some expression that is made out entirely of the metronome because t full is made out of the metronome and this quantity is made out of the metronome moreover it's some quantity that vanishes this quantity some quantity that we know that vanishes when all first derivatives of the metric vanish ok now lastly lastly lastly it is a quantity that is that is obviously a two derivative expression because t full has two derivatives because in h there is a derivative and if t is not also R has two derivatives so this quantity is going to be some quantity that consists of two derivative expressions of the metric ok and that vanishes when the first derivative of the metric vanishes that tells you that you cannot have a term that is second derivative of the metric because if you have a term that is second derivative of the metric in order for it to vanish when the first derivative has vanished you would have to multiply it by another derivative but that is three derivative not two derivative so the only possibility is a quadratic form in first derivatives of the metric but we know that first derivatives of the metric can be written in terms of the crystals so we regard it as a quadratic form of a quadratic expression to stop it ok now one can actually just work this out and I am not even good at actually I didn't even try to work this out because Landau we should say after a rather lengthy calculation we know when they say after the trivial calculation I swear to you again you get some expression that is quadratic in the stopper symbols it's a equation 96.8 of Landau equations we will never need that expression ok Landau equations but the important thing is some quadratic expression as I said as you will see we don't need that expression we can work it out this is the thing that we might want to associate this is the thing that we might want to call the stress energy tensor the gravitation because the full stress tensor is the stress tensor of matter however I emphasize again and again and again that this is a big thing because in the stopper symbols you can be in a frame where that gravitational stress that's a zero just when I make a change of coordinates you can make a normal thing basic beautiful property there is zero in one frame that's the transformation clause I leave you this is a pseudo tensor it has inovogeal terms it's transformation because inovogeal terms in their transformations the more that you might think when we've done a lot of work that we haven't done anything too interesting if you find some gravitational stress tensor or totally stress tensor it seems a bit ambiguous and why should anyone care about it quite a change in the stress tensor that's stifled minus stif whatever that matter is the thing that makes the class of this definition is the way we approached it was to ask when we started this class the serious part of this class the way we started this class was by was by asking look Einstein's equations if you really use Einstein's equations the covariant conservation of the stress tensor is an identity because it's an identity for the metric for the Richie tensor now in a frame in which the first derivatives of the metric vanish the covariant conservation of the Richie minus Gij by 2R becomes an identity must be an identity which reflects the ordinary conservation that quantity in that special frame must be identically orderly conserved then we ask how do we have a quantity that is just identically conserved there's two derivatives there are two indices and it's identically conserved so again that is that if this quantity is del by del xA of HRK A where you have an anti-semitic material then the conservation of this thing will be automatic so that's one way in which such an identity could look so we look for such a quantity and found it that's what the difference is so that's how we came upon the definition of this I haven't done the algebra that shows you that this is what I will do at some point unless we run to the faster better formula for lambda looks somewhat like epsilon delta xA in fact if you take two G's outside and then it becomes delta into delta minus delta delta into delta so that can they put as product here to coincide so is that the indication of where this formula comes from a good point there may be ways in which I don't know I don't know if you're saying that it sounds like it's coming out of forms coming out of anti-semitic objects very likely there's something in what you said but I don't know yes in any way energy of that it's like if you just take the variation what we're going to mean is when we define a new theorem even if we define some concept in the new theory it should satisfy two criteria the first criterion is that it should satisfy all the properties you want in this case it should satisfy conservation the second criterion is that it should reduce to your older definition of energy in moment when you reduce to the special limit that's when you talk in there so these are the two properties we need to demand with this that in the Newtonian element it reduces to our usual action of kinetic plus potential that's gravitational energy that's minus Gm by R it should reduce to that in the appropriate limit but more generally than Newtonian element it should be conserved the quantity that satisfies these criteria if you have a quantity that satisfies this it's a good candidate to call it is this clear there's no independent way of defining the special energy we're going to have a negative effect through noether charges you see noether charges are a bit trickier because you want to use time translation in variance or something but an arbitrary coordinate system does not have time to translate actually the variable is a variable in flow form that's true but there are an infinite number of deformations but the parametric are functions even so you see how many symmetries do we need for energy moment we need a full translation of this how many symmetries do you have and infinite number because arbitrary functional redefinitions of x arbitrary functional redefinitions of t so what you're saying now however you're on the right track there is an involved theory for people to do this next formula of how to define energy momentum that serves in general in various interesting generalistic contexts and the theory basically that's the formula we're just asking we won't go through that what you see even the Hamiltonian formulas the Hamiltonian formulas would work if you didn't correct but it's difficult because you want to somehow respect general coordinates but just let me get back to the previous question let me say the following you see in any theory in any gauge theory there are two kinds of gauge transformations those that vanish at infinity and those that do not it's a general principle of gauge theories that a few of you says that do not vanish at infinity define the gauge transformations that do not vanish at infinity if you want to say that are associated with global benzene charges whereas gauge transformations that do vanish fast enough at infinity are not now there's a way the way that this is processed is following to define your dynamical system you put some boundary conditions these boundary conditions are typically that the metric now are fast enough at infinity then you look at all of your morphisms they are consistent with those boundary conditions but don't die off so fast and you have to find out what that means that they don't give you a conserved charge and thinking this problem through you would get energy momentum the way we talk about but it's a very involved technical analysis this problem by the way is very this way of thinking of it can be very useful because what I'm going to talk about is gravitational stress energy you know if you were asymptotically anti-decider space decider space you might want to define the same thing that would work you might want a general principle that will generalize many different backgrounds and there are many this is a matter of much research in the gravitational community there are 20 different ways of thinking of how we define stress tensors and the point is that in reasonable contents it's always the same so there is a way of thinking of it but it's very technical so does this video like science science fixing in the background very good let me just make a point so the question leads me to the make up so far all that I have told you about is leading you to dissatisfaction because I've defined a quantity but I've said that it doesn't really seem to be very feasible but now let me take this quantity and integrate it over a space like this because I take this quantity and integrate it over a space like this and I assume that I have infinity over the space like this my space is flat this is the assumption what that means I've made this today that the metric is flat space plus some small mutations I'll explain that but you can already guess right because you know from the Schwarzschild metric deviates from the ordinary metric like one by one that's the fact that all deviations from flat space and infinity are afforded over by one suppose I make this assumption then the quantity that I have defined which was this pseudo tensor locally now in this case there is a distinguished set of coordinates at infinity not in the workspace but at infinity at infinity the space time at infinity is flat so I've got a space time that at infinity is approximately flat like I've made that precise it's basically almost flat so there's a distinguished set of coordinates at infinity now demanding that you're working in these coordinates doesn't completely fix your coordinates because what transformations of flat coordinates do you have flat coordinates? Lorentz transformations so it fixes these coordinates up to Lorentz transformations but Lorentz transformations are linear difference and these Christophels symbols which do not transform as tensors under arbitrary difference do not transform as tensors under linear difference because the inaugural genius part is associated with the second derivative of the coordinate change so the quantity that we're going to get by so it's clear then from what I've been saying that in order to start getting something nice somehow it will refer to infinity and infinity is something nice there's a particular nice coordinate system not quite a particular nice coordinate system but a class of coordinate systems related by Lorentz constants just one infinity means that everywhere it should be in every direction it should go now the next thing about this is that now suppose I take this quantity this conserved stress answer and I just do the ordinary integral of it I do d i k d s k well this d s k is just the geometric volume of it it has no reference to the metric because d is an ordinary legal derivative I take this quantity and I integrate it over our space this quantity is guaranteed to be conserved by the usual class logins the volume element that is the determinant of the metric the volume element that is the determinant of the metric but you see what we have here is the quantity that sorry I didn't show you that more I cannot give this x a 85 what I am interested in doing is defining something that is conserved since this derivative makes no mention of the metric if I want to get something conserved I have to use just the ordinary Gauss law of ordinary Gauss law integration Gauss's law Gauss's theorem is a theorem about integrals you understand what I mean Gauss's theorem is a theorem about integrals it tells you that if you have got a total derivative then you can relate it you can relate the integral of that total derivative over a slice to the integral of some surface something okay just as an identity on integrals okay so I want to use that identity on integrals to get something conserved so I have one second so yeah we have in 4 minutes we are going to have to go for the first line but let me so now this quantity by the usual Gauss law argument gives you some firstly how many indices does it have A is not trapped A is not trapped with an i that's a 4 in the next one for each value of i by the usual Gauss law argument it is a conserved quantity and just remind yourself what is the what is the usual Gauss law argument the usual Gauss law argument tells you that nothing is happening at infinity which is what I am assuming then the integral, the surface integral of this what is equal to the volume of the thing but we assume that the surface is an infinity or a contract so this surface integral minus this surface integral minus this is equal to the volume integral of the divergence for the divergent's finishes so this surface integral is equal to this surface This is true of arbitrary space-like stuff, spaces-like stuff. Yes, by the fact that we do this volume integral. Yes. Instead of the, I mean, including the gene. Yes. This is the code space volume. Yes. It means that you've sort of changed your background. It's the same changed your what? Changed your background on which you're integrating this stress energy tensor. Maybe you can interpret it this way. We'll just take an algebraic thing. I want to define a quantity that will be conserved. This quantity will be conserved. Yes. Okay? Now, you might think it's really unnatural to do us, you know, some sort of surface integral without the volume factor. But remember that it's only an infinity that this thing will have me. Okay? And infinity, your space-time reduces to flat. So, there's no difference between the ordinary integral there and the flat integral. Okay? But here, you're just doing the ordinary integral because you, because you are a conserved quantity. We're not going to allow any part. You see, it's dangerous to use your biases because we're not doing the things. Yeah, but I was saying this quantity, del X, H, I, K, can't be interpreted as a density-encourved space-query on which it is defined. I mean, a modulo all the problems of it not being a density. Yeah, that's true. You know, so we're not going to give too serious interpretation to the local contributions to the integral. Okay? What we're going to seriously look at is the final answer I think. Is this clear? Okay. From the fact that gammas transform as pseudo-tensors. That is, tensors under linear derivatives. It follows that this quantity transforms like a 4-vector under Lorentz transformations. Okay? So, 4-vector. That's because gammas transform under, as 4-vectors under Lorentz transformations. In the case of a contractor, okay? And it would be totally clear if you looked at Landau-Flüftschild's difficulty-directed expression. Okay? So, that's another positive hit. You know, when we're looking at something that reduces the flat space far away, for somebody who's interested in just an overall conserved quantity, I think you should have to care whether things go deep inside or outside. You've got an anchor flat space. So things should transform like under, you know, good representations of Lorentz transformations in that anchor flat space. This quantity is real. Now, the next question is, are we going to get the same answer? Are we going to get the same answer, sir, if we use different coordinate systems? Because in order to do this, in order to do this integral, I've said we have to use some coordinate system that goes to the flat space coordinate at infinity. But I'm telling you what is inside. Okay? And there are many different coordinate systems that go to the flat space. But they differ by different morphisms of the first kind. The kinds that don't correspond to each other. That's the dialogue. And the answer is of many Svf. Okay? And let me give you the answer. This is the argument. If you have coordinate systems, while I compute my stress energy, that's my energy momentum, on a particular space like slice of that, I've got some other coordinate systems. And I want to compute my energy momentum on some other space like slice of that. On a particular space like slice you choose and a particular coordinate system can't matter because it's uncertain. That's for sure. But I can do, I want to check that in the same answer here again. So what I can do is take this space like slice in the same coordinate system to the far past. Let's take the answer. Then I invent a new coordinate system. One which smoothly interpolates between the second coordinate system in the far past and the first coordinate system in the far future. Every coordinate system has stress energy density. So this new interpolating coordinate system is uncertain. But that tells you that the energy momentum defined in the second coordinate system is equal to the energy momentum in fact. It's a brilliant argument. You might think it's too slick for your taste. It's correct. Okay? So this quantity gives us an energy momentum that does not depend on which coordinate system we use. It only depends on what your coordinate system tends to at infinity. When we demand that there are 10 to a flat coordinate system. Of course, under coordinate systems the deferred infinity energy momentum does transfer. Like, no dance transfer. Exactly as it should. Please. That's a bigger class of coordinate system. Coordinate system going from one to the other. Yes. Which has the same power constant. The same far past as coordinate system too. Suppose I want some slice. I want to check that I have some size in coordinate system one. I get the same energy momentum as in coordinate system. Okay? What I'll do is consider some slice in the far past of coordinate system two. Then I'll embed the third coordinate system. Then interpolate between coordinate system two in the far past and coordinate system one in the far future. In the future. I'll just say it again. Since in any given coordinate system, say in coordinate system two it didn't matter where you took the slice. If doing it in the far past doesn't matter, you get the same answer. But now this new coordinate system, the factor energy momentum is conserved. Tells you that the conserved value in the first coordinate system is equal to the conserved value in the same. It's a tricky sounding argument. We show you. You buy it. It's totally good. It's the kind of argument you land on. It's the kind of argument that someone will land on. Brilliant. Is this totally clear? Sir, when you have the infinite i in the limit that space goes to infinity, it's flat. Yes. But these two slices which represent the same space line at infinity can be very, very important. No. What I'm going to assume is very, very important. What I want to prove is that two coordinate systems that are identical at infinity but differ inside give you the same energy. If you do two coordinate systems that are not identical at infinity, then what you get is the energy momentum will be the same. We already see that. Two things I want to check. Firstly, two coordinate systems that are identical at infinity but differ inside and give you the same value. Secondly, now if you take some coordinate transformation that does whatever you want inside but at infinity reduces to a linear coordinate. Then transforms like a form. Both these we prove. We prove because every expression of t-full is made up of pseudo-genesis. What's that again? Of course, of course, they can. They can. So you could listen. Most of them. Yes. The universe is probably such a thing. Such a place. Yes. This is the space. Then we are going to have to think hard about how to define energy in that space. But of course, you have a practical purpose as an argument. Suppose we want to define the energy of the Earth-Sun system. We don't need to. We can take infinity to be far away from the Sun. So for practical purposes, it's useful anyway. To the extent that you can approximate space-time as well. This is useful. But for some absolute statements. In principle, the statement, different space-time asymptotes get totally different. For instance, in the sister space, it's very problematic. The stress of energy. As we may discuss with you. But in the 19th century, it's very difficult. Because global sister space is a compact. Space-like slices of global sister space are not. Yeah. Why is that a problem? That's actually its own problem. Hey, I organized that. Neutrino talks. It's a big problem.