 I understand. So we discussed in the last class, we discussed gravitational radiation in the last class. So we discussed the most general solution of the linearized Einstein equation to the maximum. Just to remind you, but you remember we had this, we worked with this combination, yeah? It was h mu alpha minus half the h delta mu alpha. We put the gauge condition, del mu sine mu alpha 0. And then we found that this gauge condition did not completely fix coordinate, variance. We fixed it only up to a coordinate transformations, Killing vector fields, coordinate transformations labeled by Killing vector fields such as box of sine mu 0. We computed the, we computed our mu, but that was equal to half of box of sine mu. And then what we did was to, was to parametrize the solutions of these equations. Solutions of the equations of the, oh. Solutions of the Einstein equations that vary in any period of time, I think, in a way is box h mu 0. But not all of them are physical. Firstly, we have this gauge condition that eliminated four h's. And we took, we looked at the special case, so a case of wave struggling in the x minus range. So we looked at a wave that was 4, for instance, even 5. And it's unordinary. It's a relation structure between the study. Okay. And we found that this condition here was del minus of sine minus alpha equals 0. If you're aware of this problem, del minus is because of factor of k minus. It's only got sine minus of alpha. Okay. Then we realized that, we realized that this kind of coordinate change allowed us to transform h's in the following way. So we had delta h alpha minus equals delta alpha h minus plus del minus. So this, notice that delta of h times del plus is equal to plus h minus plus del minus. I should, delta of h. So h is equal to delta h alpha beta times beta beta. And this involves twice h plus minus. Yes, I suppose this is h minus minus. Changes by del minus h minus plus del minus h. So apart from h minus minus, the only components, or all other components changed by del minus of h, whatever. The only component that has no zero contribution to the trace is del minus h plus. Okay. And number times. Number. We have to work on this number one. The trace will get a contribution to quantity of del minus h plus. We have to check the number. It will then matter square. Okay. But the important thing was that, and of course, so there was a change, there was a shift in this h. And there was a shift in h minus anything, which meant h plus anything. Now h plus minus, or psi plus minus, which is the same as h plus. Now we moved to the sides. Okay. So the trace of psi was related to the trace of h. Psi plus minus was already set to zero. So psi plus everything else could be set to zero. And in addition, because there was a shift in h, which is the same as the shift in the determinant. The trace of psi, because that is the trace of psi. Okay. So we worked finally in psi alpha, alpha equal to zero. We also set psi minus anything. This equals psi plus anything. This equals zero. Okay. So what we had was that all the components of our solution had all their indices transverse to the plane of propagation of the wave. Transverse to both the x plus and x minus dimension. So transverse to both time, and where is the direction in which the wave is. So it was, so there's, we actually described by some two cross two matrix. There's two cross two matrix of stresses. That was the solution. Okay. So actually we had two degrees of freedom representing the propagation of these waves. And as we discussed last time, these were split to each other. Any questions or comments about this? Well, at some point, I'm going to compute the energy of this wave. And you'll get the land origins. So let's do it like so. We're going to perform that a little bit when we actually compute the radiation. Of course, I won't. What we want to do is to, so we've understood now how I set equations here in linearized manner in the absence of water sources. Okay. Now what we want to do is to understand how the metric field reacts to the presence of sources. Okay. Now, in the first case, in an expansion around the non-relative sequence. Okay. So something we're going to see is that, as we will see in this lecture or the next, that gravitational radiation happens in order v to the 5. Okay. So up to what of v to the 4? We've got the response of Einstein's equations to moving particles. Okay. It is to just correct the conservative Newtonian quotient by additional terms. So our first goal, the first task we need to set ourselves is to write down the Lagrangian, that captures the first correction in Newton's equations for a bunch of particles moving around. Okay. And one higher order, we will then try to compute the gravitational radiation emitted by this moving object. But what we've got is a bunch of particles moving around. We want to write down, in today's class, we want to write down the Lagrangian of this to a much higher order than the Newton's equation. So first let's remind us something, and remind the sense of the Newton's equation. Okay. So suppose you've got a bunch of particles on M, especially the positions. R, M, I. Okay. So in Newton's equation, the Lagrangian of the system of particles is simply given by kinetic energy. So some M, N, or N, I'll try that one. And it's potential energy, so some of them, okay? So the potential energy is attractive, so it's negative. So that's the class, sum over M, N, K, M, N, M, the power M. It's Lagrangian. And it comes out of general relativity, of course, in a very straightforward way, and we've already understood. Let's see that once again. So what's the Lagrangian for a bunch of particles moving around in the space? This is the Lagrangian that we started out with pictures with. Okay. So this Lagrangian is M times N times the X, sum over N, which is the same way as sum over N. N times the square root of the X mu, the X mu, when this refers to the displacements of the N. Or DT. Or is equal to sum over N, M. The table, if we move to writing it, the table DT is 1 minus, oh, and I should have reminded myself. 1 minus N squared. And then expanding to the relative to VM squared. And then there was the fact that this thing was contracted, sorry. So this thing was contracted with cheese. And to the relative order, to the order of interest, the only thing that contributed was when G00 hits this 1. Okay. So this was 0, 0 minus VM squared. And we had learned that, we're studying high respect equations, that G00 was equal to 1 plus 2 phi, where phi is just a pseudonymic potential. Okay. So by doing the same, expanding the square root, keeping terms only the first order and potential and second order in the velocity. Remember, these have been treated as the same order, because of the Newtonian theory, velocity squared and potential are the same order. Scanning to get the J-potential energy is freely exchanged in proportion. We just recover this action. What we want to do in this lecture is to go beyond, beyond the first Newtonian equation. We want to write down the ground in the calculus, the first correction to this equation. Okay. So in order to do that, I think this Newtonian theory, we're going to try to understand how general reality reacts to point particles. Okay. So now these point, the, the, yeah, of course there's an idealization in every particle. Microscopic particles are the same. We, we weren't zero toward an expansion in size. So the first question we want to write, if we want your answer, is what is the stress energy tensor? Okay. What is the stress energy tensor of a point particle with some opportunity? Okay. So, well, this thing, it's going to be some invariant expression. This object that we write down has to be some invariant expression. Okay. And, oh, by the way, is your problem saturated? Solution. Solution. By today tomorrow. Yes. So, yeah. I can't remember. Did we have the problem about the charge current? Yeah. So, this thing is going to be very similar to that. You see, you want to write down the stress tensor of a bunch of particles moving around. So you can write down an expression that is similar to the charge current in the field. You know, it's dictated that the factor is going to be that we need the variance. Okay. That we need the stress tensor to be non-zero only when the particle is. Where it initially does that first in the factor that's non-zero only when the particle is. It actually does this in the vector. We're going to get these two indices, right? Okay. So we've got to get these two indices, right? So we've got to have dx mu and dx mu. So two velocities. Okay. Okay. And to be safe that we're doing something invariant, let's use the parameterization of this particle to be the proper diagonal of this particle. So, S. Yes. It's equal to the square root of g mu invariant expression. Okay. I wanted to emphasize one thing. This expression is less nice than the expression for the charge current. It's less nice because of the expression for the charge current. There was one derivative and one derivative. So though you could have chosen dS as your integration measure and S as your parameter of the particle, that expression was completely invariant parameterization invariant. We could have chosen arbitrary lambda. Where is it? Where? It's just a proper diagonal. The length of the, the invariant length of the particle in this matrix. Okay. So... There should be some... Thank you. Expression that should reduce when the particles at rest to just having energy and no momentum at that point and clearly that's a trick. Okay. Because when the particles at rest S is the same as t. So the only non-zero derivatives here are in the t-damage. So t00 is the only non-zero value. These two values are equal to one. And then we just get this delta function. It's a spatial delta function. The time part is killed by doing the integral over S which in that case is just t. So that would tell us that t00 is spatial delta function at the origin. tij t0i0 and this is now an indelible generalization. But I want to emphasize that here we needed to use the proper time parameter transition unlike for the charged particle where we could use any parameter transition because the final expression was parameterization invariant. So here we want to use... replace one of these dS's by d lambda and put a d lambda here and the other one is already a dS. Okay, so the matrix. Okay, now we should also ask about this delta 4x. Now, in order to get an invariant expression this delta 4x should be an invariant delta function. So it should be a delta function such that delta 4x square root g f of x is equal to x because this is the invariant value. How do I mean by this delta function? We call it delta 2. If we want to write it in terms of just an ordinary delta function a delta function such that ordinary integral d4x takes to give 1 then we could rewrite this as dmium is equal to m dx mu by dS dx mu by dS ordinary delta function divided by square root g. Okay, this is the version that will be most accurate. This is the stress tensor for a particle and stress tensor for the integration of particles just some zone. Is it clear? This is, you know, the reasonable expression for the stress tensor for a function. Any questions about this one? It's clearly nice and invariant. It's a tensor. It's a tensor. There's an interesting question about the conservation of the stress tensor which I'll postpone to the discussion of the video. You guys checked that the charge pattern of this tensor does appear without the homework. Should you expect this thing to be good? Exactly. Okay, we'll discuss this later. We'll come back to the conservation of the stress tensor a little later. You must remember that the gravitational field carries energy by which it can be charged. Yeah? That's so. The source is just a gene. It's one index. Yes. It's a two-index. Can you explain what the effect is whether or not you think twice? Why is it a two-index or what? How it's being two-indexed or the effect of the line? Yeah, the effect of the line is very important. The fact that it's two-indexed is a very good question. The fact that it's two-indexed tells you that the thing that it's a source four is a spin two-index. It's basically just like the fact that the current was one index of two-index. The thing that it was a source four was a spin one-index. Good. This is a spin. This is a source for a spin two-index. Now, spin two objects, or it's a very, very small negation. You see, quantum mechanics, it's very difficult to make a unitary Lorentz invariant theory of a massless, of a particle with a massless degree of freedom that has spin greater than 0. And the reason for that is the following. You see, when you try to do canonical quantization of a spin negation, let's say spin one. You impose computation relations A mu A nu. Let's say A mu dot A nu is equal to a what? If you want to stick with Lorentz in there, the thing that it would naturally impose is equal to E w. The computation relations that you would naturally impose are position with momentum. Let's say you just call it pi. I think you would naturally impose is equal to E w. You see, what you're doing here is that suppose you get the sign so that it's the standard sign for the spatial bits. Because E w has opposite sign for time compared to space, you get the wrong sign for the time you get. And if you try to track through what the last quantum theory is, what the last quantum theory is that either it makes the energy unbounded from below, it makes your theory not unbounded. If you impose the appropriate conditions to make your energy unbounded from below, it gives rise to states of things you don't know when these times are going up or both of your stages are going down. So if you take a theory with massless particles spin one higher, and you just try to do the ninth quantization, you try to build the quantization of the Lagrangian that you don't get. At least the ninth thing you do doesn't give you a theory that says either it has negative norm states or it has energy unbounded from below. Now how do we actually make theory based on particles with massless particles with spin? We basically, what we do is endow these things to the gauge invariance. The gauge invariance allows you to kill the time degree of freedom. It's sufficient to kill the negative norm states and give you a well-defined theory. You would have the same problem if you try to make a theory of a spin two part. In fact, you have four legumes. It's anything in the z. No, actually it's z. The problem I think is true. So once again, if you try to just make a theory of a particle with spin two, you would naturally be led to introduce a gauge invariance in this system. But you see, gauge invariance, you see the gauge invariance of a system is tightly tied to be connected, coupling something to a consultant. You see, suppose in electromagnetism you couple A mu to A mu tends to A mu. If this action is to be gauge invariant, that is, you're going to get the same answer for this plus or a gauge transformer, it must mean that this d mu lambda times d mu is nothing. Others have worked with this integrated by parts, but it must then be that this is just zero. So gauge invariance is tightly coupled to the conservation of the current that you couple the gauge beam to. Now, so suppose you try to build a theory of a spin two particle. You try to build a theory of a spin two particle. And then you try to couple it to, you must try to couple, and you're wanting to coordinate these elements. You must try to, you must try to couple it to a conserved current. But precisely the point that I will find later in the beginning of this class, maybe that this current is not always being conserved, it is the rise to trouble. And the way you try to fix this trouble is by including nonlinear terms to fix the conservation of the current. And that's one way of thinking of how Einstein's theory exists. You keep trying to, you know, complete the theory in a nonlinear way so as to make everything consistent. This is one way of trying to understand from particle physicists view, from particle physicists view that this guy's theory arises. This has never been worked out, what I'm talking about has never been worked out completely satisfactorily. But it has been most completely tried by Feynman in this book that is the way for Feynman's lectures on gravitation. So he tries to develop, in those lectures what he tries to do is to develop general relative sequence photography. Let's try to make a theory of spin two particle, that is unitary, you know, based on the consistency conditions that you see you can't stop there. Okay, the lectures start out very nicely and then get very confusing by going to lecture. I don't think he was satisfied with this. I mean, I don't think he worked it out. Sir, can you toss it in when you have just a electromagnetic energy to start one px mu ds? What? Px mu ds. In the card, yes. Sir, can I ask you how much the friction how is the wave propagation in this case because of quadratic two px mu ds? Well, wave propagation is not affected very much. As we've seen, the equation for the gravitational wave at an appropriate px is the same thing, it's del squared or h is equal to z. The thing that is affected is the interaction of matter with the sources. You see, one big difference, for instance, is that the interaction is always attractive between spin and repulsion. That you don't have repulsion is always attractive. And there are many other differences. The differences that you're asking are the difference between Einstein's theory and the Maxwell's theory. All the differences are the differences in the end between spin to repulsion. When you're doing what important difference then it's inconsistent between spin to theory between. The sensible thing to make the Indian do concern is that it moves around the G.R.S. What moves around the G.R.S. Particle moves around the G.R.S. The theory will not be conserving the ordinary end. But the gauge invariance, if you wanted this gauge invariance in an arbitrary background thing, you would want it to be conserved with an ordinary end. The theory, give it up. You see, because what you want, this is a story that requires some development. You know what? Maybe if you're interested in a second to put this in a problem. Step by step, try to do that. Let's not get into it here. But what I mean is that, look, suppose you try to write on linearized coordinate transformations of each. This is the thing that we want. So suppose we try to write down the theory that is in very tender H mu nu goes to H mu nu plus 10 mu zeta nu plus 10 mu zeta nu I'm going to try to write that. So the Lagrangian I'm going to write it down. I'm going to try to write that. I'm going to write the Lagrangian of the form H mu nu T mu. For this to be consistent, just like we talked about in electromagnetism, it would have to be that the unilaterally-awaited equation. Then in unilaterally-awaited equation, the expression for the charge current, the unilaterally-awaited equation like this has an identity. This is something that you will see in your problems. Again, this expression does not include conservation as an identity. It's related to the fact that you will see it. Because it does not obey this conservation as an identity, it does obey in all straight lines. Let's put it in the source. The GD6 are not straight lines. It curves very tight. Once you've got a source, when you get ordered by order of H, once you put in here some sort of source, once you put in a source, what you're doing is taking this Lagrangian. Let's say that you take the Lagrangian for a free particle. It's just a Lagrangian. Let's say that we're working with an expansion of the line spacing. So you take this Lagrangian in paper of the S plus the coupling of the source. Right? So this team of new images is written in the terms of the particle, but it couples to the source. It couples straight. Now if you write down the equation of the portion of this Lagrangian, the solution will not be joining straight lines. Because there's a part from there in this straight line stuff, but there will be something else. And what it would of course be is just a linearization of the GDC equation. So now, as an exact statement however, once you take into account the deviations of these parts, this stress tensor is not conserved. It's covariously conserved. It's not ordinary. So that makes this thing inconsistent. It makes this action not Asian bearable. No, no, no. We're trying to do something else. We just written that up. You see, what we're trying to do is to write down the theory of spin 2. We're going to far-field. But then what we're trying to do is to write down the theory of the spin 2 particle. We're trying to write down a theory that has GDC equations. And we're trying to see if a linear if we can do this in a linearized way. So this is a natural expression for what the Asian bearage means. We're trying to see whether the linearized theory with these Asian bearages is consistent. And the answer is not exactly. Because it would have been consistent had this been conserved as an identity. But that is not conserved as an identity. That's the point. So unlike the Maxwell theory. So from this point of view, this explains the difference between having a consistent theory, a linear theory of spin 1 particle, but not a consistent linear theory of spin 2 particle. Okay? So in order to make it consistent, we're going to have to somehow modify the gauge of consummation codes. Order by order to try and make the book the gauge in there. And of course Einstein's theory gives you that. We've seen that this is one solution. That you take the stress tensor and put it on the right-hand side of Einstein's equations. But we are completely coordinated in that equation of motion. So we have a solution. Only it's not linear. But unless there are impressive questions, let's move on. Excellent. So let's move on. So we've got our expression for our stress tensor. Now, what do we want to do? We want to solve Einstein's equations. As we saw at the beginning of this class. You see if we're interested in the Lagrange to order V to the 4 then to what order do we need to solve all the Einstein's equations? Well, can you say that we need G 0 0 to order because there will be the term that is 1 times G 0 0 in the action. However, we will only need G 0 I to order V to the what? 7. So we will try to take this action G mu nu dx mu dx mu Remember, every dx I is a V. G 0 I we want this whole action to order V to the 4 whatever we need G 0 I to No. G 0 I to V Q. But the other one will have to be the Einstein's equations So that's what we will be to order V to the Q What about G I J? That is all. So if we need to take the Lagrange and write to order V to the 4 we need to be able to correctly solve for G 0 0 to order V to the 4 G 0 I to order V to the Q G I J to order V to the 4 So that's the task we are going to set ourselves We are going to take this stress tensor and plot this stress tensor into Einstein's equations and try to use it to solve for to solve the equations Try to use it to solve the equations accurately to to order V to the 4 for G 0 0 V Q for the G 0 I and V square for G I J So let's first start with G 0 0 Let's first try to evaluate G 0 0 in a more human way than the equation But firstly let me process this equation this equation in more detail You see what we've got here integral over 4 space we've got a delta function over 4 space and then integral over 1 point so now integral is cancelled by delta functions So it would be nice to express this as a 3-dimensional equation because we are looking at slowly moving particles time especially what we are going to try to do is to kill the integral to kill the delta function over time So I will give you back what we do the n is equal to the x mu by ds the x mu by ds integral and then I write ds by dt that is integral dt d4 of x minus x mu of s Now this d4 this d4 delta function can be written as d delta function 3 of x minus x i minus x i of s times delta 0 and delta of x 0 minus x 0 Now that we are going to explicit dt we just kill the integral So what we left with is m dx mu by ds dx mu by ds times ds by dt times delta 3 of x i minus x i of s Take one of these ds by dt into this ds and then this for instance and we still have nothing to divide in this equation So this doesn't look as covariant as the previous expression but it is more convenient to calculate it right This is another expression for the stress tensor this is the formula expression in language Let's use this expression to evaluate t0 Now we are going to evaluate all the objects that we need We need t0 to order d4 So and actually even before the solution What was the solution to Einstein's equations to order v squared Now we have already seen what g0 0 was talking about in v squared but we actually know more we also know actually what g0 i and gj are to order v squared How do we know that? We know this because suppose we take the Schwarzschild metric and we take that solution and expand it to keep only those terms So let's write down the Schwarzschild metric because ds squared is equal to minus the plus dd squared is equal to 1 minus 2 phi 2 phi 6 Thank you 1 minus 2 phi plus dr squared 1 minus 2 phi plus r squared is between phi and being order v squared So if we take this metric and expand it to order v squared ds squared is equal to dp squared Well as you can see So vx being vx being eta nu and plus plus correction minus 2 phi d squared plus 2 phi Remember we are treating phi as being order v squared so we don't need to do the expansion to any higher order So now the Schwarzschild metric of course was derived for a particle at least and it's very easy to get the exact metric for our particle and arbitrary motion but that's not even necessary for what we want because you say suppose something is moving in the velocity field then it can take this metric but it doesn't take this metric and boost it Now because the metric is boosted in magnet so the velocity of the particle of course is magnet it can change ways and that's why we call it v cube because phi is already order v squared and if you boost with velocity v you become another factor from the fact that we know the Schwarzschild metric we know in great generality that at least in some appropriate coordinates the metric of the metric in response to a stress tensor up to order v squared is given by the metric which we may want to rewrite now let's make it path easy so suppose we write beta as path easy and vx squared is equal to dx nu dx nu then u is phi t squared and this dr squared we just write it as dx i squared is equal to dx squared is equal to minus 2 phi 2 squared plus 2 phi dx i and move to the coordinates multiple system there are many particles but there is no natural coordinate system not natural coordinate at the origin this phi obeys the equation del squared phi is equal to del squared phi is equal to summation over all other particles so this we already this is the that that this is the metric this is the metric this is the metric caused by a a collection of particles up to order v squared what we are going to try to do is to correct this to get the metric to order v to the fourth okay so let's move so now let's say that we evaluate this t0 here so what is t0 0 so t0 0 0 we just have to evaluate here so this is equal to m times now this as x0 by ds by dx so that's 1 so we write this as 1 square root of 1 minus v squared square root of g alpha beta dx alpha v x beta divided by dt times 1 times 1 then we lower the metric twice so we have g0 0 because the square root of g and this delta 3 what about the terms that are g0 i what about the abstract difference okay and all of those vanish the order that we are interested in the question is what we have is an expression of t and when I write it down here now it's an expression of t0 0 so I take the terms which are g0 t0 but there are also the terms which are of the form g 0 exactly for instance what about this d0 by g d0 0 is it true but you see from this metric here the 0 i components of t of the 0 i components of g started all of it the components started all of it t i 0 also starts all of it let me it looks to me now that I should keep this better we should certainly don't keep these terms now what we are going to do is to compute g0 i to the cubic order at some point in this calculation so if we check whether it's actually vanish really that would be good I'm going to just proceed with the calculation ignore this thing following like that which gets and if we come back by the end of the calculation to check whether this was set consistent I mean not at all I'm going to discuss this one it's the calculator of g0 i so g0 i let me just calculate what g0 is the boost in this metric just to see if something unexpected happens let me take this metric here and boost it in the dx that we should fix so we get let's say that t is equal to t prime plus dx and x equal to x prime that would be t okay so we so this metric here was by the end of the scientific this had a minus because of this this had a minus and minus so that became a plus so this was also totally clear we have t square plus dx square it's the Lorentz boost but I'm going to take it to first order the right we have the Lorentz boost but I'll set it to order dx yeah we'll compare only one of the terms of order v cubed see this thing was all this g0 i it's going to be an order v cubed i terms of order v to the fourth I'll compare d0 i with terms of order v to the fifth we're only interested in terms of order v cubed generated in g0 i means boost so we can ignore this okay so what are we going to get we're going to get we're going to get some 2 phi vt dx term the number is going to be 4 phi maybe 8 phi but it will be clearly getting this term so it's clearly a term in the metric which is a order v cubed that's v it's clearly a term in the metric that's order v cubed and the question you could ask is does it summarize the context sorry one minute this is embarrassing let me proceed to the calculation ignoring this term why no this is 1 take the term g0 0 that's just 1 and this is order v it looks to me like it's order v to the fourth it looks to me to be consistent to no wait I'm probably thinking wrong just give me one I'm probably thinking the wrong thing you know what we're probably interested in is it actually g0 0 we need order v squared you see I'm sorry t0 0 the 0 t0 0 was order 1 it was m times delta function and that gave rise to a g0 0 that's m divided by r the Newtonian potential which in dynamics is order v squared I'm sorry guys okay so in order to get g0 0 accurate to order v to the fourth we would only need to keep t0 0 accurate to order v squared because the solving of the equation the solving that will you know that will pull out the that will give the greens function effectively gives you the additional power of v squared as we've already seen so let me say again in order to get this I'm sorry about that in order to get the Newtonian solution all we need to do is to keep t0 0 to order we want to go to 1 higher order v squared about the solution so we should keep t0 0 to 1 higher we should keep all components of the stress tensor to 1 higher order than we than we did before okay so this thing is because it's order x so this would have been inconsistent if I was interested in calculating t0 0 to order v to the fourth I'm only interested in calculating t0 0 to order v squared sorry about that anyway now we can now we can go ahead so we just need this expression yeah okay so now because we just need this expression we know what to do we already know what this is and because we only know what to do we already know what this thing is to order v squared we're going to write it by square root of 1 plus 2 pi minus v squared that's the sixth question um then we get sorry this should have been g0 0 to the square we lower this to the least so we get g0 0 to the square we plug in from here so there's 1 minus 2 5 the whole thing squared then we go to square root of the metric the square root of the metric um the square root of the metric this is the product this this and this and uh to this what sorry square root of the metric is not we will do from this part, we get 1 minus 2 5 from the time part. But then 1 plus 2 5 from the space part. And this thing is a cube. It's in the direction. And then it gets going. OK. So that is equal to 1 plus 2 5. OK. So that's where we took a metric here. Let me go instead of function. So let's put it all together. So putting it all together, we get m into square root of 1 plus 2 5 minus b squared. We probably want an overall minus sign, just to match the variable. 1 plus 2 5 means b squared in a 1 minus 2 5 whole thing squared over 1 plus 2 5. So let's see what we get from that. We get into 1 plus b squared by 2, of course. And then minus 2 5 plus minus 5 times delta. What is it? Remember this technique. So there's 4 from here and additional 2 from there. So that's minus 6. And then there's some sign. Let me get all this. No, no. Let me get all this. OK. So g 0 0 had all my signs up. OK. So let's go back to you. So g 0 0 was 1 minus 2 m by r. Now 5 is minus n. OK. So this is dt squared in 1 plus 2 5. OK. And dr squared plus dr squared over 1 plus 2 5. OK. So this guy should have been a plus 2 5 here. And this guy should have been a plus 2 5. And the bare dog is p squared minus dA sign. The metric without any connection. OK. So now let's go back here. This was a plus. This was indeed a plus. This is the same sign as before. But this was a minus. Now we get a 5. But if we look at the opposite sign. Where did I find it? Plus 6 minus 5. OK. And new 5 is what we want to look at. OK. Excellent. So this is the expression for t 0. We need t 0 0 to order v squared. Higher connections. You know that the connections in t 0 i and t i j. T 0 i and t i j are already order v and d squared respectively. Higher connections. So those would be order v cubed and v to the fourth. In inverting with a plus, it takes it to v to the fifth. v to the sixth respectively. We don't need those. OK. So let's see. Now summarize. So the summary is that g mu nu 1. The first connection. Until the first connection. It's minus v d squared plus v d squared 1 plus 25. Minus v xi squared in 1 minus 25. t 0 0 is equal to m into 1 plus v squared by 2 plus 5 1. OK. This, of course, has an obvious interpretation. It's just a kind of image. This is the mu nu 2. That's the next function. xi minus v d squared plus. OK. And t i 0 is simply m d i 1. I think we're going to need to have this. And t i j is equal to. Just by plugging into the general formula. We have a t mu nu. And start getting to the rest of the value of interest. OK. Excellent. Now, it's a data function. Nothing like just provided with my stupidity, but it's v to the fourth v squared. And then that was cleared up. And this is our stress tensor. But the patient's getting. So, you see, we know what the stress tensor is. Do the order of interest in terms of an unknown. We only interested in order v squared. And therefore, it's technically known. OK. So, i is equal to. To the order that we would be interested in plugging into the height instead. We know. 5 is equal to the sum of i minus k m d. This known stress tensor. Plug it into Einstein's equations. And compute the correct metric to the order of interest. Is this clear? Fine. Now, the next step is to, is to evaluate, is to evaluate the, the richer the richer the sum. Expand it to the order of interest. So, OK. Now, I want your advice almost. This kind of stuff, will you let me put, just go to the answer. OK. After quite, it gets a bit quiet. It's not going to be. Is this OK with everyone? Yes. OK. So, now we just get into. So, I tell you exactly what we need to do. And that, that, that, that, not the answer. OK. So, first I'm going to write down the formula. And then you're going to work with this. That's true. Write down the formula for r 0 0. And also the r 0 0. So, r 0 0. This is that our, like, N, N are run only over space. All of this is contracted with the metric of flat space. So, there's no difference between upper and lower indices, as far as the N is. So, if you do a calculation for r 0 0, and keep only those terms, keep only those terms that are of order V to the 4. Famous, this is what it is. Now, in doing this, we have to remember that h 0 i is order V Q. h i j is of order V squared. Actually, just the, the, the, the, the, the. Yes. Size favorite. h i j is delta h j times order V squared plus order V squared. And we remember that differentiating with respect to t increases the order of smallest number. Because, if you ask, well, what is it actually? It's, it's, it's, it's by del x. 2 del h N N by del x. 8. Multiply by this. And of course, h 0 0 is equal to V. Is of order V squared plus, we're going to use this, this, this is unknown V. Plus, order V to the 4. This is unknown. We use this equation to determine the unknown part of h 0 0 and order V to the 4. So, let's look at this equation in some detail. Look at this term, for instance. This term here is del h 0 i by del t. So, this is an order V to the 4 term. Because h 0 i by itself is order V. What about this term? This term here is also order V to the 4 because it's nothing more than V squared times 2 derivatives. And 2 back derivatives and therefore, they're order V. This term here is the order of the h 0 0 derivatives. So, we'll have a term which has the unknown part that ought to be to the 4. This term here is a product of 2 h's. This term is already order V squared. So, the only term that we'll have to do is h 0 0. Once again, there's a product of 2 terms. So, it's only h 0 0 order order V squared. So, where everything is quadratic term. So, basically, the unknown term is going to appear here. Everything with h 23 is known in this equation. So, the strategy is clear. We take this equation, we expand it including all terms of order V to the 4 or lower. Then we plug in all terms of t 0 0 on the right-hand side or the part of t 0 0. I'm going to do this. And solve for the sum of h 0 0 at order V to the 4. Strategy clear. Now, we're going to do this. In order to implement this, we can do something else. I'm going to set, I'm going to make a choice. So far, what we've done is linearized line equations without having chosen a gauge. The only implicit choice in the gauge is that the first order solution takes this point. Any change of coordinates that respects that statement is a coordinate invariance that's advanced. Now, in order to make life a little simpler, we're going to just set a choice of gauge. So, the choice of gauge that I make is the choice that del h 0 m by del x m minus half del h m m by del x plus. This is the combination that appears in the gauge. This sum, with this choice of gauge, this guy goes away. This guy survives. Okay. Now, in what remains, as we said, what we're supposed to do is substitute in what we get from the 0th order solution. I mean, what we have from Geffen. So, what is the substitution we're supposed to make? We're supposed to make a solid substitution. We're supposed to make the substitution. h alpha beta is equal to minus 2 phi delta alpha beta. And h 0 0 is equal to 2 phi. Okay. So, these things are going to plug into here. So, let me, this is most systematic. Let's say that r 0 0 is equal to del i squared of h 0 0 4 by 2. This is the term that I want. Plus everything else. Firstly, there is a term here that is order of v squared. That's del i squared of h 0 0 by 2. So, let's take that out. Okay. So, that's plus half into del squared of minus 2 phi. What is the order of this term going to be? This is going to cancel the 0th order term in the stress tensor. This is just the statement that is. That this thing obeys the equations at order of v squared. Okay. We'll read the statement. But that, of course, is going to be sort of trivial. Everything that's interesting comes from these things. So, here what we do, we substitute h m n. And this becomes a del squared. So, this is now going to be at the science, right? That's half h. We have to be careful of the science because, in this case, it's two specialties. We don't have that. So, this is half. Then h m n, so space has minus 2 phi. And then this is del squared on 2 phi. That's this term. Then we have this term. So, that's minus 1 by 4. Then it's del i of 2 phi. The whole thing's squared. Then we get minus 1 by 4 of del h 0 by del n. So, it's del of 2 phi by del x m times the subject. So, that's minus. H contracted. So, that's 6 phi. Del by del x m of 6 phi. And now I've got the sign, right? Let's be a little bit here. That's about h alpha. That's a minus 2 phi. So, if I take this thing here, I get minus 6 phi. There's a minus here. Sorry. So, that's a plus. And here, I get h m n del x n. So, this is delta function. So, that's once again, and there's a minus here. So, that's minus del by del x m. I'm going to have quite a bit of that. From the unlikely event that the boundary algebra is correct, what's the answer? It's del i squared h 004 by 2 plus, okay, that's got that l squared. Del squared by 2 of minus 2 phi. Now, we've grouped together the two kinds of terms. There are the terms that are phi del squared phi, and then there are the terms that are del phi del squared. So, that's grouped together. So, this term is phi del squared phi, and that's the only such term. And everything else is the other variety. So, we've done this right. So, that's half minus 2 plus 2. So, minus 2 phi del squared phi. And now, these terms are the other sort. So, this gives us a minus 1. It's 2 phi here. So, this gives us a 4 into 2 divided by 4. So, 2 with a minus. So, it looks like minus 3 l phi of x squared. Now, now you should have plus 2 phi del squared phi minus 2 del squared. I'm enjoying what? Every touch. This should actually be plus 2. This should be minus 2. Let's see if we can figure that out. Okay. So, let's first look at the phi del squared phi term. It's actually being seen. So, it's only one term. That was this term, right? And we got, the 2 is clear. We got minus 2 because one of these k have a plus 2 phi and the other have a minus 2 phi. You see. Let's look here. h 0 0 is 2 phi. But h i t is also 2 phi. It's a relative sign that are off by minus sign. So, this is 2 phi. Okay. So, then this term here becomes this plus. This way, right? Because it just came from here, right here. Now, we need to, we're back to the drawing board for this case. This guy here has an easy term. It, the sign just totally cancels it. So, this guy clearly contributes minus 1. Minus del phi, though. That's clear. You see. This term here contributed a factor of 3 with a minus sign. This term here contributed a factor of 2 with a plus sign. So, the next contribution is 1 with a minus sign. Okay. 1 with a minus sign and there's a minus here. Each of these is 2 phi. So, it's 1 with a plus sign. Right? So, I got here 1 with a plus sign and 1 here with a minus sign. Almost. I just wanted to pick up that straight. Did I copy that to the popular, right? Let's, let's, let's take the two minutes to get this straight. Because even if I do it at home and then come back here it wouldn't make any sense for me to see this work here. That's fine. Can anyone see, can anyone see where I'm right now? This is twice. There's a nice image, not just an image. No, it's twice. Twice. Twice. It's so rich. So, there's real four or five. There's a full five there. Yeah. That, that, just about there. The sum? Yes. So, this term will be minus. This term will have a minus four. Yes. Minus four. That, that term, the final term that you wrote just about that. Ah, here? Yes. So, six times minus four times four. That's true. That's true. That's true. Yes. So, you take that out of this cancel. First of all, we get a minus four. That is two. The two phi outside is a cancel. So, that's two phi. Another two phi? Yes. Ah, but my problem is that we wrote this through the wrong signs. We wrote this as, ah, now we've agreed that h k k would space is plus. So, I agreed that we're getting a one here, but I think we're getting a plus one. Here, we're getting a minus. So, it's the relative sign between this term and this term that I wrote, that I missed. Now, let me just check this. It's probably the fact that our spatial metric has a minus one. Let's get that to the truth. You see? Yeah. You see? You see? You see? This is minus consistent with h alpha beta lower being a plus. Okay? So, we did this correctly. So, we correctly got this. Okay? But here, what we need was h to one upper one in every case. Okay? So, that sign is off. So, then this is correct. So, minus, now we get minus.