 the speed of light in a particular medium. Sharon got my attention. Stop. Yeah. You know, Sharon, you know, normally when something's moving, it cannot support being in the sleep at all, energy and momentum conservation for the people. Okay? However, there are special, if you've got an object that's moving faster than the speed of light, the kind of light is changing. And it's possibly a possibility to support this limit stuff in a process that is available at Sharon co-variation. So what these guys, for less demonstrates, at least within the standard model, these neutrinos that are moving, that if they were moving faster than the speed of light, would have been Sharon co-variation. Actions could be charged faster. Right? Okay. So the Sharon co-variation would be through the interactions in this case, and could actually be a Sharon co-variation of E plus and E minus bits. But the important thing is not, it's just an N and M, the word is used as an N and M. The important thing is that something moving faster than the speed of light, put it and move it slow down spontaneously by a mission of particles. And then they calculated how long the, what fraction of the flux would survive from CERN to Francis. So in fact, it was a negligible fraction. Okay? So that point is that it's just, unless you give up much more than the speed of light, it's just not a sense, it's just an explanation of what's made of it. That this thing moves faster than the speed of light. Well, I don't know. This sounds right to me. It sounds like the six-metre must be normal. Okay. So that's the video. The bossy is done. This paper was on Fridays, so if you looked up the head, DHR kind of flashed out the board. I don't remember the video. It's a very good paper. It's two or three pages. It actually turns this estimate from CERN configuration around to analyze other experiments that have not seen, that have not seen things going faster than the speed of light but have also not seen the attenuation. You know, you've centered neutrality out from here. It's reached there. Because of that, if it was moving faster than the speed of light, there would be CERN configuration. Only a fraction would reach there. So you use this other experimental data to put a bound. But how much the deviation of neutrality from the speed of light could possibly be and they get an answer like that by mistake. It's purely experimental. I mean, sir, if you accept a little theory in between, basically they say, but this experiment is not consistent. That can be checked right now. The number of neutrinos, the fraction of neutrinos are reached. Yeah. So you're using the CERN configuration. I'll give you a check here with what the data is. Exactly. So in this grand SASO experiment, the number that reaches is consistent without any generative. It's without generative. Yes. It's very... Exactly. So this is not a thing that's a good experimentalist at all, I think. Okay. So you know, from that point of view, the neutrinos were admitted here. They reached there. Okay. The part of this new paper is that had they been moving faster than the speed of light, they couldn't all have reached there. So it's not self-consistent. This explanation is not self-consistent. Unless we start giving up other pieces of physics. Yeah. Yeah. Rather than giving up high speed of light. No, but you see, we're not giving the option here of giving up the speed of light. Take the option of giving up the speed of light. Can you find an consistent explanation without giving up other parts of physics? No. I think we've been cornered into finding that the experiment is strong. All that, you know, we've it's the last two, three hundred years of physics displaced. Okay. Excellent. So let's continue. Just to remind you, in the last class, we were discussing the LADL engines. Stress energy tensor. Stress energy pseudo tensor. And we had a series of expressions which I want you all to write down all over again. For many of you. We first have this object, the four index object, lambda i k l m. So let's call it alpha because my i is in the air. Which is equal to 1 by 16 by a into minus g i k for m minus g alpha which we define as three index object which we call h i k alpha which was equal to del by del xm of lambda in terms of this three index object h we had the conserved quantity. The conserved quantity was del by del xl of x alpha of h i k. This actually we had called minus g of this g. Minus g. That's d i k total. And this conserved quantity of in the equation. So this quantity here of in the equation del by del x alpha h i k alpha was an a symmetric i and a and b conserved in the equation del by del xk of del by del x alpha h i k alpha two properties followed just directly to this end. These are just my entities involving algebra and half of this existence. But the the interesting thing was that this d total in a frame in which when g alpha beta beta and which first derivatives of an intervention. Okay. Now I would like to derive this view last time last class where we didn't get the time. And so I take like fifteen or twenty minutes by trying to do it here. Basically I just it's not perfect. So I'm going to put this in your next problem. So we have now the next problem set by Monday. People we want the first problems. Okay. Okay so let's let's make you know let's really try to have today was the deadline I last suggested to you. I don't see any problems that's here. Let's really try to have it by next Monday. Okay. By the middle of this week we'll send you new problems. One of the problems will be showing this this thing after me. Just add it to the ground. It should be done either in the classical problem or in the problem in there. Okay. Yes. Third question about this. Yes. So you write down del alpha not quantity to zero. Yes. Sorry del alpha. I mean del x. Yes. I mean del y is equal to zero. This calls the way to know words in a flat space. Now you think this is just Just answer. Okay. I'm sorry. I mean the interpretation you were discussing in your course was not words in a flat space. No. This is an this is an this is an this is an this is an this is an this is an this is an this is an this is an this is an this is an space-time, then when you integrate a child, you get a conserved child. So that conservation law holds in dozen probability flat space-time. Yes, yes, basically what you are doing is you are deciding the dynamics of some code space-time using some quantity in the flat space-time. Yeah, you see what we are doing is this, we are looking at dynamics that is going on in space-time, dynamics could be complicated, the space-time could be complicated. But we are assuming that all the complication is localized. So that far away is space-time becomes flat. Okay? And defining an energy momentum, energy and momentum for the system with reference to that flat space-time, with respect to a reference to a flat space-time and energy. So can we do this like we have dynamics of momentum and then you see it as a fixed momentum against some dynamics. Fixed momentum and with, yeah, added to the dynamics on that fixed momentum. The dynamics inside about that fixed momentum. It's kind of a thing we do quite routinely in everything we do. You see, for instance, the first time we were doing electromagnetism. And you had electrons roaming around, but never going off to infinity. Okay? They can go very far, but never, let's take it beyond this. Okay? Then the electromagnetic field and infinity would die off at most one by hour of this radiation field. So at infinity, the electromagnetic field goes to zero. Okay? So it's very similar to that. But you have the dynamics of the metric itself, the dynamics of the metric. And you are saying dynamics of the metric can be seen as in a way that we have a fixed metric at some point. A fixed metric and infinity are infinity and some field are. Mathematically speaking, it's just fixed metric. And infinity. Not in the middle. We're not referring, we're not requiring any particular fixed space-time. What we did is put in boundary conditions. You know, we're solving Einstein's equations subject to some boundary conditions. The boundary conditions are that that metric becomes flat. Okay? It's a reasonable boundary condition in a situation in which you don't have any sources for the garbage added. Just like if these are Maxwell's equations. With a boundary condition, the gauge field goes to zero at infinity. That's a reasonable boundary condition providing you don't have sources of charge. But it's important that we're not asking for the space to be flat at infinity. Just at infinity. Okay? This is a reasonable thing to do. It's what we always do when we do in our field. You see, because we're going to solve some equations of motion. So let's talk. We're going to solve some equations of motion. We need to be able to evaluate the space in there. So suppose you were in our computer. You wanted to solve some equations of motion. You discretized space, so you put two points into this. You do this, you know, because you're a computer, you have to aim the space. So that's some endpoint to the space. Okay? Now, how do you evaluate the derivatives here? You can't understand the boundary condition. Okay? So to make the problem of the initial value problem well defined, you need a boundary condition at the edge of your space. Now, we sometimes hide this boundary condition by taking space to the infinity peak. But of course, boundary condition is still there. There's some implicit boundary condition that things are going on. It's something we routinely do when we solve it, when we look at a field theorem in any system, double by differential equations. At the edge of our space, we need boundary equations. Okay? And when we take the space to be arbitrary, but sometimes very physically, very natural. And these are boundary equations. That's all we're doing. We're not doing both. We're not doing both. These boundary conditions are what's called the requirement that space can be asymptotically flat. That allows ways to property. That allows ways to property because ways down like one by r. Any way that is sourced by something low is true. The only way we could not dial this is if it was sourced by something infinitely low. That would violate your boundary conditions. Yes. So if you have boundary conditions, go with the one itself. Yes. Then how do you go over it? Yes. So the question of finally defining energy on a closed manifold, so that's it. I don't know how to do it. I don't know how to do it. I don't think anybody has a sense of what you're doing. So this definition is really limited to asymptotically flat space. That's the extent that an infinity could be flat first. It can be extended to many other situations. That's the extent that the infinity becomes anti-dissiduous space. If you have another situation of interest, you can try to extend it. But note that from the fact that finite energy configurations, you might think I've been trying to disturb what goes on in infinity by bumping a lot of energy into the system. But if we're seeing how things behave, the energy of the system is measured by the fall-off of this age field. As we go to infinity, and the fall-off is slow. I mean, it's one by one time for infinity. So it's you would need, if you start with the space-time, that's the best way to do it. You would get out of that situation. You would need an infinite amount of energy. This is a self-consistent thing to do. It's not the most general thing to do. But it's sufficient for a certain class of course. Okay? Now, if you don't have a boundary, it's very possible that there is no interesting notion of energy in some accident sense. And the reason for that is that, you see, what is energy? Energy is the quantity conserved quantity due to time translation. Now, wherever for an energy, you need some time. In general, there's no distinguishable element. There's no particularly interesting value. But maybe a boundary condition is good. Now, so if you put the boundary conditions that the space-time has to totally flat, you've got something distinguished happening at the boundary. There's a nice time coordinate at the boundary. And that can be used as a crotch on which to define energy. Okay? But if you don't have anything like this in space-time, it's closed. No. There's no particularly natural notion of energy. Certainly it's not. But you could have this free-time passage. But, you know, if you've got a closed space-time, it could be that there are other conserved quantities. But it's far from clear that you want to call that energy. But let's go on. So, please note this down. Okay. The last thing that I wanted to say that I didn't get the time to say, or maybe I said it with the other speaker, is the following. We went through these arguments that the stress energy and momentum measured by this stress tensor is the same for every choice of internal coordinates. We also went through the arguments that every choice of internal coordinates that leads to flat space-time fixed. And that these things transform like Lorentz vectors under a change of coordinates that transform the flat space coordinates in a Lorentz, in a boost map. The last thing I wanted to say that I didn't at least emphasize enough is that the expression for the expression for energy momentum is very simple in the sense of as follows. Suppose I wanted to compute P i. So what am I going to do? I'm going to take this expression and this conserved expression and dot it with respect to the normal to my space-time vector, a space-time surface and I'm going to integrate it. So what I'm supposed to do is to take h i alpha. Oh no, we won't be able to use the answer. Then there's alpha, so that's del by del x alpha and do this integral. This object here is normal to the volume of my space-time surface just to be completely concrete. What is this object? This object is epsilon beta phi 1, phi 2, phi 3 d x phi 1 d x phi 2 d x phi 3 1 by 3 factorial. These are the three infinitesimal displacements that we have on the three surfaces. This epsilon is just the object whose components are one as you. There's no matrix in this thing. This is the quantity that you're guaranteed to be conserved to. This quantity itself is a derivative. You can write a more convenient expression to the quantity. Instead of writing it as an integral over all the space, we can once again use Gauss's theorem. To rewrite this, as we do h i beta alpha and then epsilon beta alpha d x phi 2 by 3 factorial psi phi d x psi d x phi where this is now an area element on the boundary of the space-like manifold that we chose. We've got the space-like manifold. We've got our space-like manifold that runs through it. This integral is taken on the space-like manifold. But the space-like manifold has boundary in infinity. It's a two-dimensional equation. This integral is taken on the two-dimensional equation. What this makes clear is that the notion of energy and momentum in general relativity is very similar to the notion of charge in electron. See, in electrodynamics, if you wanted to measure the charge in your space-time, you could do two things. Either you could do an integral over space and totaling up every bit of charge that there is in space. Or you could do something more similar. You could just take a huge surface surrounding your space and measure the flux of the heat through that surface. From Gauss, there's a lot of these two things we can set. The second thing is the more simple thing to do. You don't have to volume it down. You just have to sit at infinity. Sit at infinity in all the charges. Because charge leaves its mark in an electron. So really that's the way you should think of the notion of energy and momentum. Whatever is happening in the center, some complicated dynamics, leaves its mark at infinity. And this is the mark at infinity that measures the energy and momentum of the space-time. So energy and momentum in general relativity can be measured sitting entirely at infinity. And it's measured by the deviation of the momentum from flux. The first deviation of the momentum from flux. That's what this H will measure. Because you see, had G just been eta, all the derivatives that go into defining H would be z. So it's the first deviation from eta that matters. And roughly speaking, how we're going to get something finite? How we're going to get something finite? Well, we're doing an integral over a big two-sphere. The area of the two-square, two-square scales like ours. Now, H, the components of the first deviation away from flux space, will die out at infinity like 1 over r, as you see from Schwarzschild metric. But this quantity is not just the deviation of the metric. It's a derivative of the deviation. But at 1 over r squared. So the 1 over r squared in this would cancel the r squared in the volume element, giving something finite. So this formula tells you that space times in which you've got 1 over r formulas are away from flat-space metric, are of finite energy. These are the space times that we will call asymptotically finite. Their finite energy excitations are of flat-space. Space times with lower than that follow, 1 over r squared follow, are sort of zero energy. And as suspicious, if you get such a space time, you should start searching to see if you can find the coordinate transfer transformation at the beginning, exactly. Because at least with reasonable matter content, there's always nothing you can do with no way in which you can matter in and not have energy. Of course, if you have matter content, it's funny if you have negative energy density. But at least with reasonable matter content. And space times which fall off slower than 1 over r, suppose you have 1 over square root power 4, you're not dealing with asymptotically finite space. The space time infinity is not flat. It's not a finite energy excitation with flat-space. It may be an interesting solution, but it's not asymptotically finite. Fine. I think this is all I wanted to say about the energy movement in pseudotensor. So there are two things that we have not derived from clubs. The first thing is the algebra of delixes that gives us this. We talked about how to do it, and we'll give you a problem. The second thing that we have not derived from clubs is the actual expression for the pseudotix. Meaning, when we wrote this as T total, and then T total minus T matter gives you the gravitational part. What is the gravitational part? I'm not going to ask you to do that. It's very interesting. What is the gravitational part? There's nothing in principle. It's clear what to do. So many space times can have the same h, right? The same h at infinity? The same integral. Can we write the dynamics of the whole space? No. This is the conserved quantity. But there's no way that even the other quantities describe the dynamics of the field of nerves and profiteers. When you say... I'll be good. Good question. First, if you ask, can you write just in terms of energy movement? Certainly. Energy movement is conserved. What's happening in space? It's totally unrefected. It just gives you a number that stays steady as the dynamics progresses. Okay. Now, the next question you're asking is, can you construct what's going on in space time from knowing various fall-offs of the field? This is a very analogous to the question of an electromagnetic. Suppose I gave you a knot. Suppose I gave you an electromagnetic. The expansion of the field at infinity in a power series of 1 by r. From each term there encodes additional data about what things are doing in the middle. The first term encodes the charge, the next term dipole, the next term. What concerns something similar there? What do you need? It's a degree of formality. Can you write down the dynamics of G and L? On asymptotically flat space time, when you're dealing with one piece, are there vectors in the asymptotically flat space time? No. You have asymptotically flat space time? Yes. You want to write the dynamics of the curved G and L? Now you're writing down the different dynamics in the asymptotically flat space time. What is the second thing that you're trying to do? I'm trying to figure out quantities like H in terms of H. I'll write my dynamics in asymptotically flat space time. Which will reflect exactly my focus. But you want to write it everywhere or just at infinity? I mean H is defined by the fact that H is defined everywhere. H is defined everywhere. H is defined everywhere. So if you go H everywhere, it's like you're in a metric everywhere. I thought you were asking another question. Maybe this is what you were asking. I thought you were asking how can you look at what's happening at infinity and bring together what's happening at infinity? That's a very interesting question, okay? It has recently become the ADSEM decoders one. We intensively studied in asymptotically ADSEs, okay? And I would be very happy to tell you more about that. What is the problem, okay? So let me, what is the system? We have that ADS space that was the typical form of the R squared by R squared. R squared to the mu squared. The mu is equal to one to b. So this is a b plus one, that's the equation, say b as b plus one. Then the metric field at infinity has to follow off, like in these coordinates, the metric field has to go like, I don't know if that's it. Suppose that a g is equal to g ads plus h. And h has to follow off, like one over r power b minus, these four should be one over r. That's one over r, that's just one over r. Fluctuations, the normalizable ones for a master's degree. Okay, it's like the numbers have been different, similar, similar, okay. And just, you know, just like in flat space and ADS space, you can write down the energy and momentum of what's going on at the boundary. In ADS space, you can do better, you can write down locally conserved boundary stress states. Flat space also does this thing about the Brown-Yorke stress tensor, which can be made up. Okay, you can do better than just its conserved charge. It's basically, okay. This formula, by the way, sometimes for the ADM, it's a causal sense. Landau, which is available for ADM. But ADM stands for Ardenal, Teaser, Swissner, something, okay. And they have a more geometric instead of fun. More geometric way of thinking of this formula. But, you see, we didn't need to get involved with that. We just followed our nose and had a nice way of asking. Okay, other questions or comments? Otherwise, we need to move on. I want to talk about gravitational radiation. So, Omish, one more exercise of the problem said, I want people to evaluate the energy of the Schwarzschild's place then, using this formula, and check that it's energy in the mass. So, let's just, before we start in our discussion of black holes and cosmology, there are just two things, that's two additional things we're going to discuss. Both are important. And the first one we need to start discussing now, and that's the problem of gravitational radiation. So, one of the big differences between Einstein's theory of relativity and the Newtonian theory, is that there are degrees of freedom of gravitational field. And these degrees of freedom themselves will be excited in the absolute matter. Yeah, the energy momentum, you know, move around just like anything there. You know? And to rate, I mean, there's rate, of gravitational radiation. So, just like in the Maxwell theory, a charge moving around things, it's accelerating in any way, in its electromagnetic radiation. In the Einsteinian theory, a modest moving around accelerating in any way, triple-accelerated, as we will see, it's the third derivative of this, of this object we're going to look at. Okay? In its gravitational radiation. Now, this was always a cute fact about general relativity, but it has, over the last 15 years or so, it's become very, it's very, very clean, 20 years. So, it's become very, very clean work. There was even a Nobel Prize given for, for the measurement of the hallway. You have, let's say, some very heavy star, and let's say a pulsar that's moving around it. And then slowly spiraling towards the star, using energy because of gravitational radiation, you can calculate what the expected rate of it, and it matches that in God. So, this is one of the indirect checks of gravitational radiation in the Einsteinian theory. So, this is the kind of thing that we want to understand. Okay? So, the first thing we're going to start slow. The first thing we're going to do is to try to understand, and extend the equations in the absence of matter at the linearized angle. You know, we've got a new theory that's clearly one thing we're going to do. Take it at the linearized angle. This is the linearized solution. But, so, now, beginning our exploration of this linearized theory. So, in this linearized theory, we should seek out different ways. Okay? But beginning our exploration of this linearized theory, we start by setting a gauge. That is, we start by making the coordinate choice. Okay? So, even before that, so what are we doing? We're setting the electric beam. This g-minium is equal to w-minium, not a flat space. So, that's h-minium. Okay? And we're going to work in this lecture to first order h-minium. This statement does not fix a choice of coordinates. Because, well, as you have shown in your last question, I will have my next one today, the change in the metric due to a coordinate change to first order in that coordinate change is given by this... So, let's do the following. Let's say that we've got some coordinate system in which this is true. And then we make a small change in that coordinate system. That's the smallest parameter is the same one used for h. Okay? So, we make a coordinate change that is on this form and x mu is equal to x, right? Plus, yeah. But the same smallest parameter is h. This is an order h. Okay? And now we ask, how does it change h? So, clearly, anything in this covariate derivative that uses the fact that it's covariate in terms of the ordinary derivative knows about h, but it acts as eta, which is already a order h. So, that's the order of small and square. We know. So, the change is that the delta h, h mu mu, is just d mu times mu minus delta plus d mu times mu plus delta over x. So, there's h, d, l, yeah, look. Transforms and transformations according to this form. When you're dealing with h theory, if you're dealing with some redundancy, it's often a derivative thing. It's often greatly simplified, like if you're going to interact with a teacher. Okay? And for the problem, I had the name in study gravitation deviation. But that's a very definite choice of tree. Why does it begin terribly complicated? shortly the first time, signing. The choice of h goes forward. You follow that no? After this rotation, two times y, years of pencils, mean square, mu mu, h mu mu, h by two eta we can... double integral. This is an h. I raise and lower. I use it in the lecture. H is a small fluctuation of a flat space and its indices are raised and lowered using metric investment. You should think of H as like an answer for you in flat space. What? We're going to find. We're going to find. We're going to study the equations. This is of course constraint by constraint. So one of the things we're going to do is to count how many filters we need to do this object and we're going to improve the gauge condition is del mu, psi mu. It's similar to a Lorentz gauge condition for individual conditions. Except we're proposing Lorentz gauge condition not on H mu itself, which you might have thought of as the most natural gauge cosider. Well, it's more than a gauge cosider on this one. Well, H, by the way, what is H? H is equal to H alpha. It's okay to choose coordinates. We're going to choose our freedom. There were four functions in our hand to set four conditions. Four conditions are labeled by, maybe one question immediately. One question immediately, does this condition completely fix coordinates? Okay, answer no. That's fine. Suppose we've got a coordinate. We make a coordinate choice. We make a coordinate change. Suppose we've got an H that gives rise to a psi in its aspects. We make a coordinate change and ask, can we find some sort of non-trivial coordinate changes that gives rise to aspects? Okay. So that's easy. All we have to do is determine what the change in psi is because of this coordinate change and then impose this condition on it. So firstly, what is the change in H? The change in H was given. So the change in delta psi is equal to minus of, let's put the indices as well, del mu plus del mu plus del mu plus, how has little H changed? Little H has just changed by del dot, by minus of del dot. Okay, so plus del dot, half the coordinate and little H changed by 2 del dot. Okay, now let's take this change guy and see how much this condition is changed. So I take this and dot it with del d. So the first term gives me minus del squared delta mu. What about the second term? Second term gives me minus del mu del dot eta. Third term gives me plus del mu of del dot eta. So these two terms, okay. So what we have to do is minus del squared of eta. So now we have to come to the following conclusion that the unfixed coordinate invariance in a new gauge is given by harmonic changes of variables. This is measure zero in the space of all coordinate variables because an arbitrary coordinate transformation is labeled by a function of four variables. A function that obeys an equation of motion, like the samonic condition, labeled by a function of three variables. So it's a very small amount of a coordinate variable that we have not fixed yet. But as always happens in both Einstein's theory and Maxwell's theory, this very small amount of unfixed coordinate variables will be very important. So we keep this in mind. So we fix coordinates up to two harmonic changes of variables. Okay, so now forgetting about the fact that unfixed coordinate invariance in a moment, you just go ahead and write it down. So what we want to do is to write down Einstein's equations. And that means Einstein's equations are just argument new to zero. You just want to write down an argument new to zero. But we're interested only in the argument new to a linear order in H. Now you remember R has two terms. The terms that are second derivative of the metric and the terms that are products of a function are croissants. Okay, the second terms are quadratic in H. So we need not. So we only want the first derivative of H. Okay, and the terms that are our first derivative, we've derived them at various points of the course. I'm just going to write it down. Okay, in those terms where you have second derivative of G, you replace G by H, and everywhere else G is in there. Because you've got to keep things to the start. So just write it down. In square H, del x A del x minus eta will be, okay, and then plus del square H theta by del by del x nu. Plus del square H nu theta del by del x of eta del by del x of mu. And then minus del square of H, H which is H that we talked about here, del x mu del x nu. I'll show you in this package. The reason that they made this particular gate choice is that they made an expression for our derivative of a case of difference with this derivative. In order to see that, let's look at these last three terms. You see, this term plus half of the, let's say the first term plus half of the third. The first term plus half of the third can be written as del x nu of del H theta H mu theta by del x theta minus del by del x mu of H by del x mu of del by del x mu of del by del x mu of del by del x mu of del by del x mu of psi theta mu of psi. And therefore, is equal to zero given our gauge equation. Exactly similar manipulation, this term plus the remaining half of the third, those are equations. It's just mu goes to mu. So this expression for our video, we should inject in those four terms. But this choice of gauge symbolizes the ability to do one thing. Is this clear? And therefore in the choice of gauge that we put in one, we do that our video is equal to del square H mu nu. But del square is just the ordinary Laplace equation in Laplace. So that's pretty simple, right? You know what I'm saying? It's all by instant equations of linearized level in this gauge. All we have to do is to solve Laplace's equation. You know, this is sort of the way it works. Each component of h media obeys the way it works. Let's continue to analyze this a little bit. One thing to notice is that the solutions to the equation of course and remaining gauge redundance are of the same form basically. Maybe I'm using it often. What I mean is the equations are a way box equals. When I say harmonic here, I mean box equals. There is nothing left. No, no, we're not uniquely the answer. Sorry, is that what you're asking? I did not. Okay, you know, the equation of motion is del inversion on H mu nu equal to zero. Unfixing coordinate transformations are also those with del inversion on eta mu is equal to zero. The counting that, is this clear? Yes. I got a change in the inverse of this. Exactly. And therefore there was no change if box of eta was zero. So coordinate changes that respect our gauge condition are those that obey box of eta. So once you've got into this gauge function, you want to know can you make another coordinate change that does not violate, continues to respect the gauge. Does not violate the gauge condition. And says yes, provided if and only if that coordinate change obeys box and coordinate function. Okay, now counting that function is already useful. Yes, we're looking at Einstein's equations in the absence of stress density. So Einstein's equation was R mu nu minus half or G mu nu is equal to the number next to mu. We're saying 3 mu is equal to zero. Okay, so R mu nu minus half R. But if this is zero, then we take the trace of this equation. We also compute R as z. Einstein's equation is as well written as R mu nu. This is not saying that R A B C D is equal to z. So we're not in that space. Do you remember we went through this counting? How many components in R mu nu is something? 4 divided by 2. How many components in R A B C D? 20. So you're setting 10 components of R mu nu equal to zero does not set the 20 components of R A B C D. Excellent. So let's, let's, let's, let's get it. So, so the counting now is already, the counting, the process is already. Why is it, okay? You see, nightly, if you just do nothing, you just stare at this, you say, wow, we've got 10 to mu polarizations. So x mu runs over 10 different indices and 10 different combinations. Each of which of these are equation. That's great. But 10 different waves. It's not right. Why is it not right? Of course, we've got nothing. We got this condition only after we imposed a gauge condition. This gauge condition kills all of those polarizations. So 4 out of 10 is good. However, note that 4 more are killed by this coordinate. By 4 more of pure gauge. Because you see, while the amount of pure gauge, this was a small fraction of the set of total number of functions. It is a finite fraction of the total number of solutions to the equation. Fixed, we imposed 4 conditions to fix the, fix the gauge transformation. We found we didn't completely manage to fix gauge transformation. But you know, the counting can never be totally wrong. To fix 4 conditions, to fix 4 functions, you must have done more or less something right, unless you weren't really setting 4 conditions. The fact that we did completely fix gauge transformations was reflected in the fact that some coordinate transformations left that gauge condition unchanged. But that coordinate transformation had to satisfy a differential equation. So it was not a function, not a free function of 4 variables. But good. Instead, we thought of, let's say, the free function of 3 variables. Because solutions is a differential equation labeled by initial data, which are free functions of 3 variables. Okay? So you might have thought that you might say that this is a negligibly small fraction of the net coordinate invariance we had. Because the net coordinate invariance was 4 functions of 4 variables. What we have to do is 4 functions of 3 variables. That's the infinitesimally small component of 4 functions of 4 variables. So you might think this will not play any important role in that matrix. However, you must be careful. Because equations of motion do exactly what we would all want. You see, in the space of all matrix, which is matrix of functions of 4 variables, not all matrix are allowed for equations of motion. Distinguished matrix are allowed. Those that are labeled by initial data, those that satisfy the equation of motion. And the equation of motion had precisely the same differential operator as that that given unfixed gauge invariance. So while the coordinate invariance that was unfixed was a negligibly small fraction of all functions of 4 space times. It's a very important fraction of 4's particular functions. There was a difference in the equation of motion. In fact, there were 4 functions. It's clear, let me say this in more detail, that we can use these four details to get to the 4 more equations. So the correct kind of thing is that there are 2 gravitational waves. There are 2 functions in space times, 2 remaining functions in space times. They are the way the equation works where it is. Is this clear? You have a problem, your first problem is asking you to work out things like this in a similar gauge. We basically work them out here. We want to make this a little more precise. So just understand things a little bit. Suppose we take a particular example of a solution or something. That is, suppose my 3 space coordinates were x, y instead of 36 space coordinates, then we have time. And I define a new set of 4, that is, I define x plus, equal to x plus 2 times 2, and x minus is equal to x minus 2. As you know very well, any function of x plus satisfies this wave equation. So I have a problem with the 2 gravitational waves. Formally this is the case because L squared in these coordinates becomes L2 squared plus L3 squared. Those are attached. But plus L plus L minus. So ds squared equal to ds2 squared equal to ds2 squared plus minus the wave equation. That is plus 2. There are 3 parameters of the solution. No, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no. They are not parameter 3 numbers. They are functions of 3 variables. No, 4 sets of functions. Each function is a function of 3 variables. This guitar is a function of 4 variables. But it always a differential equation. And that was a function of 3 variables. But there are 4 such functions. Yes. Okay, good. Yeah, so, because this is 2. So, eta plus 1 1. So, this is this one. So, this del squared operator looks like del 2 squared plus del 3 squared plus 2 del plus del minus. So, clearly anything that can do this function only of del plus is added. Because each of these 3 operators are there. Now, we can look at the waves that are the special case of waves that are functions of 3 variables. Okay, let's look at waves solutions like h mu nu is equal to mu of x minus. The wave is traveling to the left. It's a function of x minus t. Traveling in the speed of light to the left. And to the right. To the positive acceleration. Okay, such waves. Let us work out exactly what are the ancient issues. So, if this is true, it's also true that sine where psi is the thing that defines h minus tracer fraction is equal to sine mu nu of x minus. And what can we do to do this as we render to sine of h? Okay, so, let's first work out what our gauge condition means for this sine. Okay, so our gauge condition was that del mu sine mu alpha is equal to 0. But the only derivative that's round zero of this particular wave is the minus del. Okay, so what we're getting is del. Everything is function of just one variable. So we know derivative by dot. So we're getting sine minus alpha dot. This doesn't mean things like e to pi p dot. Something. So, the derivative that thing is equal to zero, the thing itself is equal to zero. Because you pull out it. The derivative just gives you k plus. We're not interested at the moment in the special case, k plus. Okay, so as far as wave motion is concerned, in the setting sine minus alpha plus, let's see what this unfixed, what the unfixed gauge transformation does. The unfixed gauge transformation, how does it change? Well, we don't want to spoil the fact. Just as a choice, you know, our solution was a function only of x minus. We want to continue to have it as a function of x minus. Okay, so again if you look at the unfixed gauge transformation, zeta mu equals zeta mu. Do I got a zeta? No. Eta. Eta mu equals the eta mu of x minus. Okay, I'm going to look at those special categories for the transformation. Clearly they are zeta squared. Eta is equal to zero. Okay, so h. So what I want to see under such coordinate transformations is how h changes. Okay, so delta h mu. Delta h minus mu is equal to delta minus eta mu plus delta mu eta plus and we have two special cases. This is the only component that changes because the only non-zero derivative is the delta minus. So basically what this does is move h minus is to shift h minus over the lower, anything by an arbitrary function. There are four such categories. H minus plus, h minus minus and h minus two, h minus three. And they would be shifted by arbitrary amounts. Okay, now we use, now it's important that it shifts h minus to the lower. Because the metric in plus and minus is off the end. Can you see that h minus lower is like h plus with an upper. So what this coordinate change does is allow us to send h plus mu to anything arbitrary for all four components of mu. In terms of psi. Psi is an admixture of h and the trace of it. Once in psi that are moved by this coordinate change are those with a plus on top or the trace of it. Because the trace are mixed together. So using all of these comments. I now want to click the follow and you tell me if you believe this. You can use this unfixed page invariance to send h plus two, psi plus two, psi plus three, psi plus minus, psi two to minus psi three to three. You might ask me why you also try to send psi plus plus to zero. And the reason for that is that that was already simple. It's indelible. Because if you have minus on top that's equal to have like a plus at the bottom. Anything for the minus on top is already zero. That means anything for the plus at the bottom is already zero. So I don't want to put a plus at the bottom. That's an automatic. The one that's automatic is this trace at the bottom. And this you would allow it to send to zero even though it has no plus in it. Because psi is not h. It has the administer of the trace. If you find this a little confusing just go and check it at least. And just check how all these could pull and transform under them. We've explicitly looked at how h transforms. In explicitly right now how all these can pull and transform, we might have nothing not zero. We can use that to send these gays to zero. So we add up the schedules. Now what remains? Of the ten components of psi, four are consent to zero because of a gage condition. Four are consent to zero because of a new gage condition. Using up the unfixed gage condition. Okay? What's not zero? What's not zero is psi one. I just went to the bottom. This is the trace point. Psi minus psi three three. Traceless part of the map. Psi two three. Psi symmetric of course. I mean when both of you see that over the trace. So psi two three and psi three two is the same. These are the components that are not present. The components that in general psi and h would be different. But in the particular case that h is traceless or that psi is traceless psi h points there. But our conditions ensure that psi is traceless. You see the trace part in two and three is zero. And the other part, the other trace part which is psi plus minus psi minus plus matches. Okay? So we can now replace this by, if we want to, by h. So it's h two two minus h three three and h h two three. These are the not real ones. But we conclude it. We conclude that we're moving the x direction as two polarizations. The first polarization we've also given a convenient layman. The first polarization has h in the two three direction. Two and three is the plane perpendicular to motion. With h two three step. The second polarization, h two two minus section, two three step. Now these two components of polarization are not as different as you might think. Because they're related to each other under rotations. What's going on here is what we've got is if we effectively get this four dimensional problem to produce two dimensions. As a little group theory would tell us, must be the case. The little group of four dimensional masses particle is s o two. Okay? Now what we've got is something that's transforming some representation of s o two. This is the rotation that's transverse to the direction of them. Why? What representation? Well it's transforming like a traceless symmetric tens. You see what they are allowing? The most general components of the tensor that are traceless and symmetric. There are two components of that. We parameterize that by the of diagonal guy and the traceless part of the diagonal guy. But more invariant way, in a more invariant way, these modes are parameterized by the components of a traceless symmetric two dimensional rotation. Okay? And of course a traceless, since a trace is invariant under rotations. And since symmetry properties are invariant under rotations, well first a two dimensional plane transforms under rotations. Now the trace is invariant under rotations, so that's a single part to throw it away. Symmetric properties are invariant under rotations of symmetric traceless guy transforms in itself as a representation. Some sun, it gets mixed up by rotations. And it's an easy exercise to convince yourself that these two components mix into each other under two dimensional rotations. What is the general element of this? Let's look it. So let's see how there's a general element of the traceless guy, it's going to group at the end of the line. Actually it's even easier. Let's write it down as a matrix. Then as a matrix what we do is just cross theta minus theta plus theta and probably the transpose. See each of the indices is acting on like a vector of this. And because we're acting to the right, to transpose. So this transpose is theta minus theta, theta plus theta plus theta. The trace is clear. The trace is clear. And symmetry is also clear. Because you see vector of each index is the same. That's what it's like. But the fact that this matrix was symmetric. See what we're doing is we've got matrix and we're acting on it by our R, M, R transpose. Let's take the transpose of M. So M was equal to M transpose. We want to see the root of M is also equal to R transpose. Let's take the transpose of this side. We get R transpose, transpose of R, M transpose, R transpose is equal to itself. Because M is equal to M. So this action preserves both symmetries under stress. So it just mixes these two elements together. According to law. So if we call this guy equal to alpha, this thing here would be alpha minus alpha. And because this guy is equal to beta, it's on the multiply itself, that gives you the actual number. Basically the statement that the gravitational waves are spin two waves. Because they transform the tens. Each component of the tens has a vector. Can I spin one? This thing transforms to the two vectors together. I'm fine. Lastly, I want to say about wavelengths is about how to compute their energy. Now the energy of this wave can be computed in many ways. One convenient way is just to use this lambda equation sooner or later. The path that we have in the right, the path that is purely gravitational and you value it in this way. The path that you get by taking T total minus, minus t matter, because in this case t matter is the same. So you take the lambda equation sooner or later, the path that is right. Just erase the statement once, there we go. Just use our definitions for the lambda equation sooner or later, truncate to quadratic order, then the minimum order energy. You find energy and evaluate the energy. A straightforward exercise, we know what the sooner or later answer is. We've got a clear definition in terms of derivatives of the metric. We know what the metric is doing with the solution. It's a straightforward exercise to evaluate the sooner or later answer of the solution. Okay, I won't try to do it here. Umesh added to the problem set and I'll just write that here. Which things? Yes. Yes. Yes. Yes. What does it mean? The initial data actually is a linear combination of the quantity because the angles don't matter. Make the statement. The initial data, that's what I do. It's not as much as I'm experiencing. That's no problem. That's not a problem. What do you initial data in the order of the order? In the order of the order. The order of the order is not just in the means. We can go back to that. In the end our polarization is in tally plus minus. Two or three are just the same order. So these statements are true in the order of the checklist. Two plus minus is just a fancy way of saying primary belief. Yeah. I'm just to ease some algebra. I went to the same world system. That actually doesn't means presentation always takes a bit. That's confusing. You can just work it out. Okay. So, yes, I would try to write down the energy. So We do one thing by looking at 2 by a times h alpha beta of u, h beta alpha, when you take the randomization pseudodense up, specialise the terms quadratic, and set the meta-logistic sequence. Actually, nothing, nothing else. Just specialise it in something like that. Okay? And then evaluating this on this way, gives you the unique, the tensor of this way. It's very similar to the energy moment tensor, by the way. Okay? There will be one contribution from each polarisation, and it flies in some ways. Okay? Are you necessary to signal coordinates to a t equals 0 and that's it. No. How do you, by class? If you're working in some manner. Any coordinate, given a differential equation, you can set initial data. Now the tree slices. Say we've got a differential equation with derivatives and two formal variables, alpha and beta. There's nothing even posed about setting an initial condition on some alpha equals beta squared. Yeah, the theory of differential equations doesn't need you to set the boundary conditions at one of the variables of your equation. Yeah, yeah. Okay, I think we're basically, I'm a little slower than I hope to response. Okay? We're out in the course, we're at the beginning of October, so when is the turning? December. December. And you guys have some lab or something like that in December? Yes. Orange, you know my friend. So we've got two months to go. It's too bad. Okay, we have one more lecture radiation. What we've not talked about, so we've just talked about free gravitational radiation. What we've not talked about is how radiation is created by substance. You know we want the equivalent of the, where is the lamor from the electron. Okay? So we want to discuss that. And I would discuss one more thing in Landau-Lewisians that, before we talked about cosmology of course, and that was what the first connection to the Newtonian Lagrangian is. You know, if you've got a bunch of gravitating bodies and they're interacting with each other, Newton gives us a great regard. I think we wrote that down, something like in the many kinds of classical mechanics course. Skylight energy is a potential energy between each body and each other. Now, you could ask, generally, it completely changes, completely changes, the formalism includes electromagnetic radiations. But you could ask, is there some approximation scheme in which it creates the skew-drug in the couch? Now, once you start radiating of energy, because of radiation, then what a formalism creates? You just have to take a look at the universe. The degrees of freedom in the field. It's not good enough to draw attention. But as we will see, the radiation, in general, is so slow that it happens at order v to the power 5. That is, when I say v, it's like v by c. And the Newtonian Lagrangian involves v squared, so that's v to the 2. So there's a lot of room between 2 and 5. So until you reach 5 where there is no sense in which you can correct the Newtonian Lagrangian, you can ask, what is the effective correction due to general relativity of Newtonian Lagrangian? And this is a very interesting question, because it shows you how within the Newtonian way of thinking. So that's the second thing I want to do, because that's probably one or two more engines. After that, we're going to devote the rest of the time to more or less equally spaced to black hole physics and cosmology. And that's why we start having real fun, because these are the two real counter-day applications.