 So let's continue. In last class, we had completed the next set that we would write to. We had this second-order action for gravity. This is gravity up. An action, an accurate up-to-order v to the 4. Or you, it's called a square. Gravity, so you try and find the answer. And then we look at it for 2. So the ground-trip is equal to sum over m a p squared by 2 a 1 plus 3 sum over b. k r a p is a k m p by 2 squared r a p. And this b is not equal to k. Sum over sum over a m a p a to the 4 by h plus a, not equal to b, k m a m p by 2 r a p minus sum over a, sum over b, k m p by 7th of b a dot b p plus b dot n a p. Sum over a b c, not another thing, not really a, not equal to b and a, not equal to c. k squared m a m p into c divided by 2 r a p r. Let's just look at the terms in the same set. Oh, one by one. So let's see if we can identify. Some of the terms are obvious that you're familiar with. What's this term? Classical identity. What's this term? The correction is when you do 1 minus b squared, you get 1 minus b squared by 2 plus b into the 4 by 8. Half, sorry, the expansion of the square root has half in a minus half over 2 factorial. So, ah, this is all. It's so nice. This term is for the plus, plus half, but the minus comes from here. Here, the minus gets squared, but the expansion is squared. So this term here should be the same as this term, except it's 4 by 8 instead of the square root. So these two terms just, it's what you're going to predict from special identity. All right. What's this term? Potential. It's in the action, so it's minus the Newtonian potential. That's my plus sign. And we're doing the sum overall in a not equal to b instead of summing over base. That's why the initial factor root. All right. So this term here is completely wrong. Okay. Meaning terms like this. These terms here represent velocity dependent. So these terms here represent velocity dependent. It tells you that the corrections to Newton's law, due to the first corrections, give you an order of order b to the 4, because they've got one topic of Newton's constant, which remembers always 2 to the order b squared. Okay. Times explicitly squared. Okay. And they're velocity dependent forces. They're the same 1 over r type forces, but they're velocity dependent. Okay. And finally, this term here represents a 1 over r square kind of force. It's not velocity dependent, it's just a 1 over r square kind of force, but it's also 3 bodies. It's a force that depends, it's not that you don't see with 2 bodies. It's even in 3 different bodies. Okay. Okay. And this term here, this term here is of the basic Newtonian form. And I wonder if you've heard of this. But this term is same. You know, this term is basically, it's basically saying that when things move, the effect of gravitational mass is not quite M. See, you can absorb this term into this term if you wrote down a shift in gravitational mass, which was MA plus, MA to 1 plus number times 1 squared. By choosing the number correctly, this term you're going to do, you're going to do like this now. It's interesting that number is in columns of 3. Let's do that. Let's do that. So what have we got here? A i equal to B, M a and B, B a squared by 2, or MA. Now, suppose in this expression, we replaced MA by curly MA, which is MA plus in a 1 plus theta i equal to theta, A squared. So what do we get? We get a factor of 2, so that we cancel the 2 now. We get theta times k times MA dA squared MB by 2R AB. The 2 was cancelled because I could have chosen the B squared either from the first term of sentence. Comparing with this, we see that theta is 3 by 2 and the sign is plus. Oh, well, where is the half here, which meant to me? No, do you think this by 2 is the same as this by 2? But that by 2 is more to me. All right. There is no by 2 here because there was a choice of which one to take the B squared. That was your answer. Okay? So this term here tells you that in a velocity-independent way, the first order, the effective mass as far as gravitational purpose is concerned, it gets shifted to MA in a 1 plus 3 by 2. Now, the council is, this is a pretty tame term. It's a change in the effective mass in the velocity-independence. Okay? These are interesting velocity-independent forces and this is interested in interesting three-body forces. Of course, if you were to be able to, if it was consistent to be able to continue this to higher and higher order, it would get more and more interesting in length. But as we're now going to see in this class, that it's not consistent to go to other three-body forces. In fact, any questions or comments about this? We're going to talk about this gravitational radiation, mass of source, mass of source gravitational radiation. And the basic point is this one, you see, we've been trying to, we've been trying to find the response in the last class, we tried to find the response of the gravitational beam to say, you know, there's a constant, to a set of particles. And we did this under the assumption that the variation in space and the variation in time of these things were very different. The things vary over very long periods of time, but the distance is over the one in space. So the basic assumption was that of this, of this anisotropy between space and time. So this is reasonable to the solar system, where the variation in time divided by the variation in space is of things. A measure of that is the velocity of how fast things are moving. So the velocity is small, means things vary over very long distance to a long interval in time for every interval in space. Now, however, you see, it may be true that your matter particles have this probability that they vary over time intervals that are long compared to space intervals. They're going as long lessons. But it's not necessarily true that the field set up, the gravitational field set up, by such a matter particle, will also have this probability done so far, it was true. Because if you look at the variation there, it will be, we worked out in the last two classes. And in every order, in the variation expansion, the kind of equation we were satisfied, if we were solving, it was 10 squared of some unknown to see the equation. 10 squared was a spatial, was purely spatial, purely spatial depth. The reason for that was to form. The equations we were solving, the reason for that was that we only had to solve the constraint Einstein equation. If you remember at some point in this course, we discussed that Einstein's equations could be divided into the constraint equations. Those that have a zero and the dynamic equations those that don't. And we discussed how the constraint equations never had terms of two time derivatives. They could have terms with one time derivatives and actually saw terms with one time derivatives. But those terms with one time derivatives were set to zero by a choice of each. So, in the end, working in an appropriate gauge, since we only had to solve the Einstein constraint equation, we were always left to solve an equation of this form. Now, an equation of this form determines things time slice by time slice. Given what's on the right-hand side of one time slice, it determines something else at the same time space. And so, if things are varying slowly, if the right-hand side is varying slowly, the solution will be varying slowly. That's clear, right? Because it determines things time slice by time slice. So, if things are there, what do you get as a function of t that will vary slowly if your source is very small? All of this is a consequence of the fact that if so far, never you just solve more of the constraint equations. And solve the dynamical Einstein equation. Then what kind of equation are we going to have? What kind of equation is going to arise? The kind of equation that will arise as we've seen in our study of gravitational waves, and we will see more. Now, is del squared just dT squared minus dT squared plus del squared gradient squared, dT squared minus gradient squared of unknown is equal to what can you say about solutions to this equation? Well, what you can say about solutions to this equation is that you already know from your courses that a course on electric equation. You know what happens when you solve equations from this. Suppose you've got a source that is distributed however you want, you know, let's say we were back to electric. You had a charged part of me was moving with some free and they're not relativistic. Okay? But with some frequency the frequency of omega. What you would get is a long distance that's all in this equation. It's an electric wave. In electric wave, if we have like the pi of omega d minus x. And on this solution the variation the variation of the solution in space is of same order as the variation of the solution in time. Although you started with a source that kind there was very long length scales in time and only short scales in space. The solution that you get when you start solving the equations would include free configurations that have variation of the same spatial distance as this very long time distance. Once you get to the order that such fields are produced this whole idea of solving the equations in an expansion which is based on the separation of scales between space and time starts breaking up. So this whole idea which was a very nice idea is an inherently approximate because because essentially of electric wave what we said in the lecture because in this case it's actually gravitational breakage. Okay? Now once we have understood this you can ask me to see I've got an order for you to start seeing these equations. Well, already from what I've said it's clear that once we start solving the dynamical Einstein equations in some non-trivial context we're going to start seeing these equations. Now the dynamical Einstein equations the IG equations precisely if you remember the equations we didn't need to solve in the last case because we never wanted to determine HIG to order beautiful because its effect from the Lagrangian is always HIG times VIBJ So if HIG is to jump to order V to the 4, contributes to the Lagrangian to order V to the 6 Okay? But had we needed to go to that order and had we needed to go to order V to the 6 we would have to solve for HIG to order V to the okay? And that problem as we're going to see now as we've already motivated and we're going to see now in detail has this problem of breaking the spacetime scale separation okay? So this idea of trying to correct Newton's equations in an Lagrangian essentially starts breaking the order of HIG to the 6th and 9th because that's the first order in which we solved the equation which has to get, which has an an extension of the time limit and that's the first order in which we start producing gravitation equation okay? Now we'll produce a field at order V to the 4 which will affect the Lagrangian to order V to the 6th is that action that will involve one field and some part of the is that okay? action that will involve one field and some part of the okay? I'm not going to try to do more you're saying how do you correct that is that okay right? now I'm going to just give up this effective action because once you start including the values you're solving Einstein's equation okay? you've got Einstein's equations you know the thing that we were doing before was integrating all the field and degrees of freedom and keeping only the particle in degrees of freedom once we go to order V to the 6th that is an inconsistent the Lagrangian starts involving essentially the degrees of freedom which he starts seeing from the he usually like to be necrated which is the same thing including the speed of degrees of freedom which is the same thing as making the approximation in which we were working in the previous classes is this correct? okay so now we're not going to warm this fact very much you know okay so the difference idea was essentially warm because the Lagrangian exists but now let's try to characterize this so what we're going to do is the following very imagine that we're working in the same in the same approximate in the same approximations we were in previous class everything matter fields are moving slowly but now we're going to try to do more than compute the effect of matter matter particles in themselves we're going to try to characterize the radiation field set up okay so because before starting this solving process let me very briefly remind you of what we discussed few classes ago but free okay if you remember we we looked at an object which is h nu nu minus 24 mu dou a is delta nu nu h alpha we worked in a gauge in which we worked in a gauge in which we delta nu psi nu 0 we noted that this gauge did not completely fix what it left unfixed one of its transformations labeled by labeled by the left vectors such that box squared by 0 so unfixed imposing this gauge condition our only was simply half box so then what we did was just solve the equation box if we do those solutions we know specializing in some harmonic solutions that move in a particular direction of functions of x minus d which we call x minus we noted that our gauge condition told us that psi minus mu was equal to 0 and then we used our unfixed coordinate freedom to also set psi plus mu to 0 and psi 1 plus psi 3 3 okay so what remained was just so anything about either a plus or a minus was able to be inside all that remained was psi in the 2 and 3 directions so psi from being a 4 plus 4 matrix we get 2 cross 2 matrix and this 2 cross 2 matrix the trace was 2 so the trace was 2 cross 2 matrix has a symmetric of course like h so there were two components that's the off diagonal guide and the trace was back this which can trace the two harmonization of our harmonization so this thing was about the discussion of magnetic radiation of gravitational radiation in the absence of sources any quick any questions of harmonization we also promised a problem on your problem set you have to say energy of this question or comments so so a gravitational wave you should think of as a wave of the matrix so so let's suppose we have a gravitational wave passing this way y and z 2 and 3 coordinates this way so h is being turned on in these two directions so let's look at the two organizations one organization is h y y minus h z the other organization is h y in the let's look at the first organization so suppose you get a sine wave for h y y minus z z z when h y y plus z z z so so what you say is that h y y is equal to h z z minus h z z y y plus z z z okay and you've got a sine wave so what that's doing is giving you an increase in distance in y but a decrease in distance in z and then at the next trough of the wave it'll be a decrease in distance in y and an increase in distance in z so that's what it's doing to geometry so suppose you know a bunch of particles they're just sitting there you don't have to work it out but roughly speaking what they would do is that when the particles that are here in the y direction go like this the particles in the z direction go like this and then they would be at the other polarization that you could this would turn on some sort of shift so is there something for one particle what? just one particle what? where you know one particle it's like a frame this is the question that you would ask if you want to ask a physical question so the kind of question that you would ask is why is the difference in distances between particles for one particle it's a frame between particles particles are always following this geodes you know I'm saying like the geodes they're changing yes they're changing but I mean in this coordinate system the system can't disappear the way the way it's propagating the direction and I don't know that's why it should be along the propagation that's right that's right this is very similar to the equilibrium where E and B fields are common they're actually common ok the other one turns out each xy they'll take each y's now that's a slightly odd thing see the thing that we turned on was was just taking distances and changing them but each y's is a component of the metric that is not there in that space ok so now suppose we have a wave at each y's what would that mean ok so ok so suppose we've got h, y, z so suppose our metric is ds squared is equal to dy squared plus ds squared plus h, y, z so what does this space look like you see what this is doing is that it's changing angles now the way to say it is that it's basically let me say this before let's look at this metric this guy has just got some matrix is this and let's call this guy epsilon epsilon x ok now let's take this metric and dive into this we in this coordinate system this metric looks complicated but if I look at the coordinate system in which it looks ok um yeah so this is just identity this is the first boundary fix ok so to diagonalize it to go to the eigenvalues of the first boundary fix ok and these eigenvalues are like 1 1 and 1 minus so basically and the these eigenvectors are those so basically in these directions we'll do exactly what we saw in the in a shrink here or flat here and in shrink here it's flat here in shrink here it's flat it's just like 45 degree rotation of the previous thing ok excellent um ok now let's move on now we want to understand what happens not just in empty flat space but in other sources ok so the equation that we're going to try to solve is not the eigenstand equations in um in the absence of that the eigenstand equations with that so the equation we're going to try to solve is our union is equal to d mu nu minus d mu nu and it's where it's 4 pi k it's 8 pi k 8 pi k we'll do all the same things that we did ok we'll at least in the far field region we'll work to linearize it out at h when we're far away from the sources ok we'll adopt the same gauge that we did 4 ok and so this guy once again will become half so we get half box into union and then on the right hand side we get now so so what are we doing what we're doing is working to a given order in this case order b to the 4 so what we're trying to do is to calculate the dynamic h i t is to the 4 that's the key and on the right hand side of this equation you might have thought well we've set up things so that we don't want to do this problem because on the right hand side of the problem there's already a source that's order b to the 4 and then there's nu to the 4 ok so what we get here is something that's that's automatically order b to the 4 when we try to solve this equation this is great it's a bit inaccurate the most textbooks ignore this the possibility ok but not like that and the reason that it's a bit inaccurate is the problem what we should really be doing is solving an extensive equation and taking every source that's order b to the 4 to the right hand side now it's true that k times h i j is the order b to the 4 you could ask when you take r i j and keep terms that are quadratic in h could you also get terms that are order b to the 4 because you could for instance get a term that is like h 0 0 which is order b square as we know times h i h i j which is leading order is also order b as the left i j so in this equation it's dangerous to directly substitute t i j on the right though that's correct in many situations and many textbooks suggest that let's do that you'd land on your feet if you did in most situations but if you were being really careful as we are going to do because we are following good people you should be the other one if we take this equation this equation was true though for all components all the new components max equations so in particular it was also true for it was also true for the 0 0 component because you know we are adopting a kind of age that mixes what happens in i j and 0 0 now in the 0 0 we're pretty safe by just replacing the right hand side why are we safe? because t 0 0 is order 1 Newton's constant gives you order b so the right hand side for 0 0 is order b anything quadratic in h's will be at least order b to the 4 this as far as the equation is concerned let me represent by now you know what tau mu is tau mu is whatever you get by taking i j equation and keeping all terms of order in the form as when mu nu or i j one such term is k times t i j but that could be other terms all terms that will lead to keeping leading order for the mu nu of interest for i j that would be order b to the 4 that's what we eventually will do but here leading order at 0 0 would be just keeping terms of order b to the 4 and that's a unique such term a quadratic or higher order in h's it's automatically order b to the 4 so this equation what we can in these gauges accurately assert we want to calculate is this a gravitational variations this h i j and now we can do a few manipulation that allow us to move from solving this equation to solving h i j in the gauges that we are interested okay and the manipulations go they're simple and they go smooth look because our psi obeys this gauge condition and that's crucial you link what's happening in h 0 0 to what's happening in h i j by the gauge condition because our psi is obeying this gauge condition we can now take this equation and derive the equation and derive the equation to this tau mu so let's do that so let's write down what let's write down the equation psi so we get del squared by 2 of course h mu nu minus half h g mu is equal to 8 into tau mu nu minus half tau eta mu nu this is the equation multiply by eta mu nu and subtract half of that equation but this is now a sine mu nu so we can take del of 2 of this is equal to here by a gauge condition so that implies that d mu of tau mu minus tau by 2 or it implies that the effective source obeys that equation very similar to the gauge equation del mu nu mu minus del mu tau by 2 when tau is the trace of time over the stress energy tensor but tau is tau so tau is more or less the stress energy tensor and you know in terms of tau with stress energy tensor and if tau is the stress energy tensor then this equation would be obvious it would be the conservation of the stress energy tensor because tau can be more or less discrete so now this combination gives you back the stress energy tensor let's take that suppose we took this object and subtracted trace by 2 times eta mu nu trace of this object by 2 times eta mu nu we would get back the stress energy let's take what's the trace of that object the trace of this object is trace b minus 2 trace d so it's minus trace d so if you subtract half of the trace of that object then you get half trace d as you are getting back t mu okay so word tau with stress tensor this equation would simply have asserted the conservation of stress tensor okay actually the stress tensor is covalent and that is the subtlety the earlier kind of argument but tau by definition the thing that's on the right hand side is actually conserved okay so whatever tau is we don't do things about it A that satisfies this equation this is the conservation type equation and B that tau 0 0 is the equation now we are going to use these two things to process what happens to H I J following okay so let's do it so in order to do that we use okay firstly firstly what's the solution so before we do all of this suppose we got this form of the solution then you know what the solution is for you know that H mu mu is equal to 1 over R mu mu when this R is at is the spatial distance for the cause of the spatial distance that is you take the source let it a little light ray something that moves at the speed of light from the return of the spatial distance you know what I mean right so R is the spatial distance for the point that is connected to that direction point x is proportional to 1 over the distance of the source but 1 over the distance when 1 over the distance of the source at the time at which it was when a light ray could have gone from the source to the point where you are that represents that's represented by this the delta function that tells you the source over all spatial and time points with the delta function that tells you to pick out the time so that so that a light ray could have gone from there to the spatial point at the time that you are reaching okay so that's the solution plus with what with some constants that's the solution wait this is time because I equate this time in the end x minus y is a quite complicated however restrict to sources that are moving slowly compared to the speed of light if we restrict the sources that are moving slowly then you can essentially you know the variation of let me say let me say two things see there are two things that are interesting firstly this distance r here depends on where your source is okay you can extend the source and the distance from source to where you are depends on where your source is but it's clear that if this distance here the distance between where you are and where the source is is very large then that difference is only about the 1 over r squared because you tailor expand you take something that's about the 1 over r minus r I mean it would be about this far r squared plus 2 r dot e you tailor expand that the leading order guy at the 1 over r piece she said agency extended source when you're looking at a field very far away that's ignorable as well as the 1 over r part of the field it's concerned because it's important for the 1 over r squared okay however the fact of when you receive what you receive this delay because of this delta function in time that's not necessarily ignorable however easy ignorable everything is moving slow compared to this you see because there's time delay this guy is moving very slowly compared to the speed of light the time delay between here and here is basically negligible so what I'm saying is something that you might find I've said that in a situation in which we're looking at a field very far compared to the size of the field I think that's what I'm radiating and where things also move very slow compared to the speed of light we can just pull this radial position pull this radial position in that non-valid realistic far field level minus 4 pi by r integral tau mu mu of y at the retarded time and you know that time effect in this r if you relate to this hd mu field this is a spatial integral of some stress to stress energy that's a source now from here on it's simple algebraic manipulations that give us an answer let's do these manipulations what we want to do is to relate tau ij this integral of tau ij what we want to do is to the integral of just something involving the integral of tz that tau is zero which we know is the stress okay so we do some interesting manipulations we first do these two equations we separate the two equations that we have at the top the thing that was essentially now I'm really sorry I've just switched back to Landau I'm really sorry about this the equation I drew was del box h mu nu is equal to what was it two is tau mu nu but Landau Lushitz works for the equation works for the equation in psi mu okay so let me call the tau that I'm dealing with so far tau hat so every where that we use tau in your notes in the last ten minutes that's tau hat tau mu nu is defined as tau hat mu nu minus half tau eta mu nu mu nu is naively just the stress that's the thing that follows from our gauge equation which was a complicated equation for tau hat the one that we wrote down is simply del mu tau mu okay so the Landau equation is better we should use it's just a basis change we take the equations I had in mind but this is the equation we used because we want to follow that equation sorry about that from now when I say tau this is working it's not the tau we have written the tau we have written before was dotted this is okay sorry about that okay so now we are going to do some algebraic processing and the algebraic processing goes as follows first it is tau mu nu equation and write that down in the cases that so we have one of the three in this is the space index one of the three index is the time that's what the time index is so that's del i tau 0 i minus del 0 tau 0 0 okay this would have been plus had the contracted index of tau being upper but it's minus because there's a minus sign in this one okay that's one of the equations the second equation is again where the three index is the space index that's del i tau ji minus del del 0 tau ji 0 0 separating the other classes that one equation and two indices where the three index by the time is space now what do we get in this well look at the following notation suppose we are interested in del by del x 0 del by del t of integral tau alpha 0 x beta integrated over all the space for this quantity I can use this equation alphas let's call it del let's call it ji let's call it beta so this equation on this quantity I can use this equation so del by del x 0 has an actual excite so that's okay for this guy I use this equation so that becomes integral del i let's call it del m tau g m x i x 3 y now this I can rewrite as del m of tau g m x i x 3 y minus del if you move together tau g m del tau i m and therefore tau g i because m is there to go by is del time m now this tau whatever it is will go to 0 and infinity matter sources assume to be concentrated and the remaining parts of tau are like non-linearities in H which are very small than infinity so this tau goes to 0 and infinity so as far as the integral over infinity is at infinity this becomes surface integral and infinity again okay so let's select this so we've concluded that because of this concentration equation del by del x 0 tau g 0 x i d 3 y is equal to minus tau g i d 3 y now on the right hand side we've got something symmetric between g and i left hand side it's maximum I mean clearly we could have let's say g and i so most of the symmetric equation this formula is half of n by del x 0 tau j 0 x i plus tau i 0 x j d 3 y is equal to minus tau g i these are all spaces doesn't matter but let's do it the way you do So, that is the progress. Firstly, we have related this integral tau ij over space. You think that we want, you think that is the source for the h ij. Two is something involving one 0 and x and one j. However, we can do even better. We can relate it to something entirely involving two 0 and x. So, now what we will do is take this object and relate it to an integral of tau 0 0. Now to do that, we just play the same game. So, the game we should play is let us look at del by del 0 of integral of tau 0 0, if we want. And then we use this equation. That is the same thing as integral. Wait, when did I want the x i? Yeah, sorry tau 0 0 x i x j. Once again, del by del x 0 does not act on x i x j. We use this equation. So, that gives us del i, let us not talk about del m tau 0 m x i x j d 3 y. Let us say j. We ignore the surface terms by integration by quads. So, the other terms that we get are, so ignoring surface terms, we get tau 0 m and delta n i. So, that is tau 0 i x j plus tau 0 j x i to the minus sign. Substitute this equation. And this equation. What we find? We find that integral tau i j d 3 y is equal to d 2 by d x 0 square integral tau 0 0 x i x j. The minus is cancelled, maybe we need a half factor. So, as long as this source term obeys the, the, the consolidation equation, which was a necessary consequence of our choice of each. That source term, whatever it is, it relates the i j source to the 0 0 source in the scalar way. But the 0 0 source we know to be approximation that we are interested in working is simply t 0 0. The lowest order stress tensor for a matter stress tensor, which is simply, what we conclude is that the lowest order tau i j d 3 y is equal to half sum over a m e x i a x j a. And then, second time then, naturalizations, eighth particle application x i. In this, so now it is quite nice manner. We have got a nice formula for the effective source on the right hand side of the equation for h i j. In terms of what the matter distribution is doing, some moment of the matter distribution, the moment of the mass distribution of your object. And now they, these guys like to put some fancy formula. So, this is jazz. You see, what we have got here, let us, let me define the formula. Let me define t i j as what they call the quadrupole tensor of the, of the matter distribution. So, let me define that as sum over masses of integral over density of integral over x i x j, the factor of 3 minus, minus, minus, delta i j x 1. So, it is this thing with the additional factor of 3 and the trace. So, this is the source in general. We are interested particularly in the source for the traceless spatial components of h, because those are the guys that in the gate of interest, we give us electromagnetic radiation. So, for those guys, this source is the same as this source, because I mean up to a number, because the trace part of this energy matches. So, in terms of this, we have a formula which is that h mu t equal to, let us check out the factors 2, minus 2 pi into 3 or 0. Firstly, we have a half from here, we had a, in our formula for, from the Green's function, we had a 4 pi, we had a 4 pi k. Now, 4 pi k became, oh, where the k is just, the formula in inverting this Green's function was psi is equal to minus 4 integral source divided by r. You can derive that if you want, but anyway this is the formula. So, there is a minus 4 between minus 2, because of this half, and then we are going to develop by a factor of 3, because this d here is 3 times this m a x. And the removal of the trace is basically irrelevant for the traceless degree. So, the net upshot is the formula, that this is the formula for the gravitational ray, radiated by a direction of particles, radiated by a collection of particles that are moving slowly far away from the particles, and part of this that are moving slowly. And competitive filter, okay? This object here is clearly a quarter v to the fourth, because it has one factor of mu equals to, which we decided out is this square and And two time derivatives that will make these things to velocities. So what we've concluded now is that the v to the 4 component of the spatial part of the metric is a radiation field, one that we've computed, and a formula for the amount of radiation released. So what you guys are going to do when your problem is set is to do various things. First, you will compute the energy of the radiation, energy loss of the radiation in terms of derivatives of these four different places. Second, you will take an example in which you've got a planet moving around the sun and compute how much energy it loses. How much energy it loses due to gravitational radiation, you compute the distribution of that energy very far away and various other things. Okay, you know what this Nobel Prize that was given in 1997 or so? For the indirect detection of gravitational radiation. And you had, what was that, a pulsar, I think, orbiting a very massive star on a black hole or something like that. It's orbit was measured very precisely. And they could measure precisely enough to see that it was slowing down, just losing energy. And that rate of loss of energy was precisely working for it. But that is like a jazz dive version of this formula, that we will choose. So this formula tells you how much things radiate, by radiating they use energy. As far as the effective motion of particles is concerned it's dissipated. It's spiraling towards it. Planets and this is very important. Okay, fine. So we may have a number of little problems of gravitational radiation. This is all I wanted to say about the subject of the question. So, okay, good. So these marks are, marks the end of the path of our course, which is dealing with, well, round one on the structure of general relativity. And on the small detail of Newtonian observation theory. So far what we've done is described the theory, the motivation for the theory, we've described the theory. And then basically, if you think about it, what we've done is look at how we get the Newtonian limit and how we get small corrections away from it. And how we get small effects like gravitational radiation. And this is the qualitative thing about it, okay. But we've always worked in the limit in which H is smaller. It's quite close to flat space, quite close to Newtonian ideas. Okay, and also the rest of the course, what we're going to do is is depart from this typical motion. You see, general relativity is fun. And very important in physics, because we've done it for two reasons. Firstly, because it gives you precision, ways of calculating, things that almost got right by Newton's theory. Secondly, because it predicts entirely new phenomena like gravitational radiation. But thirdly, because there are regimes in which relativistic effects are even ignoring issues of radiation. And not small perturbations on the Newtonian theorem. General relativistic effects give you entirely new phenomena, okay. That really do not have a counterpart. The two broad areas of this kind of physics is the study of cosmology. In a general relativistic scheme, you can't start that study without using general relativity, okay. And the study of black vertices. Now the first of these, the study of cosmology is a very active observational science, okay. And it's also very active today. I mean, the theory of inflation is being developed even as we speak. You know, it's an old idea, it's been years old, but there are many things to do about it. It's not going to clear whether it's conveying or form it's conveying. Okay, it's a very active observational, it's a very active thing. And in some sense, you know, in the broad structure of the scheme of physics today, it's more important. However, the study of black holes has an enormous theoretical view for various reasons that we can see. Because black holes are the most perfect objects. They constitute an entirely out of the elements of space standards. Something enormously theoretical about the study of black holes. And generations of physicists have come up with a conviction that may be right or wrong that the study of black holes will teach us something fundamental. Okay, so black holes, by the way, are also observational objects. You know, you observe them in the center of the galaxy, you observe. But they, in the observational universe, they're a bit of a size. It's not, they're small. However, theoretically, they're very interesting objects. So these are the two things that we're going to try to understand through the rest of this course. We're going to try to do use general relativity to study cosmology, and use general relativity to study black holes. Okay, so let's start that energy. So we're first going to start with cosmology. Okay, black holes will be more useful for us. Okay, so it's been odd to study cosmology for 10 minutes. Any questions about it? Any discussion about it? What is the study of cosmologies? Cosmology is the study of the dynamics of the universe as a whole. Okay, and at least in the first part of our study, we will be concerned, we will be concerned with the study of cosmology at lens scales that are very large compared to the lens scales of structure. Large compared to the lens scales of galaxy. No, if you look at the universe at short lens scales, of course it's a very complicated place. And this chair is here, it's not there. So the universe is not homogeneous. Things are different in specifications. Okay, our galaxy is here, it's not there. This is not homogeneous. But in a proper statistical sense. If you go to lens scales, very large compared to the lens scales of galaxy, clusters of galaxies, observationally, you see it's very good evidence that the universe is a pretty, on a broad average statistical sense, is a pretty, a pretty simple guess. Okay, it appears to be both homogeneous and isometric. Homo-genius means that it's the same at every point. But intuitively speaking, if you're here or you're there, you don't see the difference. There is a distinguished location. Isotropic means it's the same at every location. There is a distinguished axis. You could have something that's homogeneous, but not isotropic. For instance, in an experiment in which a pipe of water is flowing through a pipe, except that the boundary layer is basically homogeneous. It's the same at every point. Let's say it's homogeneous in that direction at least. But it's not isotropic. Okay, and you could have... These are different motions. Nurture is vice-appropriate. Both appear to be true. They appear to be true in the universe as we see it today. And anyway, we're going to start our discussion of cosmology with this assumption of homogeneity and isocardium. Okay, now... So to start our study of cosmology, what we're going to do is to try to say what could be metric of the universe? We're going to try to write down the metric of the universe that is homogeneous and isotropic to the extent that we understand those ideas in space. But we're not going to demand homogeneity and isotropy in space time. Because it's not true that the universe looks the same now as it did 5 billion years ago. As we will see in this observational evidence and as we will see through our study, it was hotter 5 billion years ago. So there was a distinguished... I mean, there's no translation in there. It's a time. There appears to be in space, but not in time. Okay? So, the kind of metric that we're going to write down for our universe is going to be the following. We're going to try that metric. There is ds squared is equal to e squared minus a squared of t. What is that actually as you've tried transition and innovation and so on? Cosmological models of the universe. Like the steady state theory. Yes. It's true. There were models historically that were at risk. But those models are very much aware of that by... The steady state theory is an odd theory. You know, the universe is expanding. If on the other... That is clear because we observe the engines. Galaxies are retreating away from you. And the further the galaxy is, the faster it's retreating away. That's an observational fact. Are you? So you can say it's expanding universe. If an expanding universe is firstly related, nothing is going to change in time. You have to have constant production. Because otherwise nothing is going to change. So theoretically, the steady state theory never really borrowed from that. Basically it was a theory that was postulated on the violation of conservation energy. And it's possible that the conservation energy is not there. And that's the process. There's no evidence. But secondly, make sure you look at the doubly map. It seems to show very clear evidence. You know, it's a really good agreement with the idea that there was, if not a big map, at least, something quite dramatic happening at the beginning and then rapid expansion. A dilution of the stress tensor associated, not a dilution of the stress tensor associated. So I think it's very far from the observational reality at the universe. You know, there was a sudden appeal to the idea that the universe is as it always was. In the informal lecture I gave you when we didn't have a camera. I don't know if you remember, but I told you about how Einstein introduced the cosmological constant. In order to try to make a static universe, but our universe just does not look like that. It may have a cosmological constant. It looks like it does, but it doesn't look steady. We'll discuss much of this issue. So the observational evidence, at least in the case of the universe, is not a translation. Okay. Sir, very much of the public expansion, you say that due to gravitational redshift, we see a shift in the wavelength, but how do you compare the velocity between the galaxies where we're following? You have to take it upon the curvature of the space, right? Yes. So what? How do you get the relative velocity between the galaxies that pass? You see, so what you observe is the wavelength of the radiation you see. And the intensity of the radiation you see. Let's suppose I have a simplified experiment. You knew the absolute brightness of the galaxy, the brightness of the supernova already. You measure the observed brightness. So you measure some shift in luminosity. You measure, but actually, that word, the universe, to be flat, that would give you the distance to go by the wall over the screen. You measure also the Doppler shift. The word, the universe, to be flat, that would give you the distance. You measure two observations. Now, how do you compare with actual, how do you compare with that? What's actually happening in a body? This procedure could give you velocity greater than that. This procedure could give you velocity greater than that. And it's just, these are words to put to things that you measure. The thing that you have to compare it with is what, this is what we're going to be doing over the next three months. What predictions you get out of the real model of the universe? So you make a model of the universe. We'll write down a metric for the universe. It'll be parametrized by a few parameters. In such a metric, I'll be able to predict for you Doppler shift versus luminosity of saucer. But the question that you're asking is pretty much yours, we have to set up the model first. In that model, we will not, you know, the world distance will be a bit imprecise. The world velocity will be a bit imprecise, but anything measure over there, if you want to go off, observe luminosity, which is experimentally the measure key, versus Doppler shift, we should be able to clear the curve. And then you go check, but so you don't have to talk about the distance, you don't have to talk about the distance. Once you have the model, you can predict anything that you would measure. And we do it. This is the kind of metric we're going to assume described by the universe. But so far I haven't said much, but I can't make some assumptions. The assumptions I've made, other than this gs3, square, is a spatial metric. So the gs3 squared is a metric in spatial, in spatial coordinates. Doesn't it work like this? All time dependence in this metric, some overall pre-factor, multiplying the spatial metric. Moreover, I'm going to assume, I'm going to assume that this spatial metric is both homogeneous and extra. Great thing about this assumption is that there is a large body of mathematics that allows you to classify all completely symmetric spaces, which is the mathematical way of saying that space is homogeneous and extra. And that classification, given the signature of the space, in this case our space, completely Euclidean signature, pure distribution. Given the signature of that space, that classification is essentially a one-parameter classification. Depending on the curvature of the space, the Richie scale of the space, you give one number of the space. Then, for instance, the Richie scale of the space. And mathematics, plus the assumption of maximal symmetry, determines the spatial limit of the space. That also justifies putting the overall time dependence outside. Because, you know, the maximum limit of maximal symmetry, the only thing that can change is the one, as far as space is concerned, is the one number, which can be a option of time, of course. But, you know, the whole metric of the space is determined up to an overall present. In the next class, I may give you a few more details about this mathematical classification. But just a few books. It's not our intention to go into the class for the shape of the... The full metric. But the shape is you. We're assuming that the signature is you. No, as in the coordinates, you choose that minus infinity to plus infinity. No, no, everything. Just assume that locally, the eigenvalues of the metric has only three positive... the metric is three positive. And then you give a point. So, that's the signature of that. The shape of this one, finally a direct coordinates, dx squared, dx squared, dx squared. No, not okay. But one of the metaphors we're getting is an S3. So, what its topology is also, we'll come out of this. Isotropic doesn't imply an S3. That's right, that's right. But, for instance, it would be an R3. R3 is a beautiful, homogenous, isotropic manifold. It's an R3. S3 is one of these. But there's one. Let's get that hyperbolic. And these metaphors and the metaphors of the S3 takes the value of zero for flat space. A positive number, or a negative number. So, we're going to... Okay, so in the next class, I may give you some more details of this classification. But that's not anymore, okay? I'm going to ask you to take it from me. That a value is somehow mathematical, more details just so that we know under what mathematical conditions this theorem is. Then I'll call it theorem. It's very nicely discussed in Weinberg's paper. Okay, so if you're interested, you're curious about it. But the important thing for us is that these three metaphors, once you know that it's unique, it's very easy to construct these metaphors. Okay? One for you. That there's nothing in a good structure. So one of them is R for you. S is the square root of the X squared. Y squared is the same square root. And R is equal to zero. What are the other metaphors? The other obvious metaphor is the S3. The S3 is, what do you use every point? Is it the same? Isotropic. In terms of mathematics, we know what it is. Let's discuss it. Okay. So we've got both the S3. We have... So let's think of S3. So S3 can be thought of as X squared plus Y squared plus... Let's think of it as X1 squared plus X2 squared plus X3 squared plus X4 squared is equal to A squared. It can be thought of as a three-dimensional sub-minoffold. Okay? And there are two quadrants systems we will find useful in this system. The system is one in which we single out these three directions and write these three directions as a two-sphere times radius. And then we use as quadrants in this manifold the radius of this two-sphere and the angle of the two-sphere. So what am I saying? Let's say we write X2 as equal to sin theta let's say cos psi X3 is equal to... It will be more traditional if we single out these three. X1 is equal to r sin theta phi X2 is equal to r sin theta sin phi and X3 is equal to r. Of course, now you are going to say X4 is the derivative equation. X4 squared doesn't make a trick of X3 in these quadrants. Well, that's very simple. It's the metric of r4. So the metric of r4 is dx1 squared dx2 squared dx3 squared dx1 squared substitute these two here. So what do we get? Suppose we do the substitution of X1, X2, X3 right here. This is a decaract on the theta squared decaract on the r or on theta and phi. The part where it acts on theta and phi is clearly something. It's just a metric on an S2. This is an S2. That's why the argument terms in which one reacts on an r and the other reacts on theta and phi. Those vanish. What? Such a term under rotations, SO3 rotations of the S2 would be a vector with one index on the S2 and one index on the other side. But the metric that we have here is a rotational invariant. Does there exist a rotational invariant vector field in S2? Vector spherical harmonics always start at A. There's no n equals 0 in S2. There's no rotational invariant vector field in S2. So on general grounds because we wanted a ground but on general grounds you know there can't be a dr times d theta phi. So what's left? What's left is we could get the r, the acting on r. So here's what we'll do. We'll get dr squared and from the x then we will get this stuff squared plus this stuff squared plus this stuff squared. What is this squared plus this squared plus this squared? Last one. That was the reason for the characterization. That's why we put it this way. So we can dr squared from here and then in x4 we also have a d acting on x4. x4 is this stuff. So we'll get what? Plus half minus r r all things squared is squared minus r squared.