 And we wrote down a form of the metric that took the form and now I'll stick to Landau-Liefschitz's notation. So, in general we could write it as h of rt dr squared plus t of rt sin squared theta d phi squared d theta squared. Plus l of rt dt squared plus a of rt dr dt. Okay? We argued this last class to change the names of the functions that this matter. Okay. Now to proceed, you see theta and phi have physical meaning, meaning they are the coordinates on a round two-sphere. We've used the spherical symmetry to distinguish coordinates on the two-sphere. However, the coordinates r and t are so far completely arbitrary. This tells, yes? There are four functions because ten in the metric, four are two from the symmetric. Two from? Two from the symmetric. Okay. Let's see how we would come. Suppose we looked at rt theta and phi. The ten functions that you have are the various elements of this matrix. Since it's symmetric, we'll just look at this and, you know, not have things here. Okay? Now what elements do we have? We have dr squared, we have dt squared. Now, we have a d theta squared and a d phi squared. We have no d theta d phi. All of these three have been replaced by one. That is the coefficient behind this guy. We're just not going to allow anything that is crossed between r and theta, sorry, r and t and phi inside. So that kills four here. Those are the cross terms between r and theta, r and phi, t and theta, t and phi. And we've killed two additional ones here by saying that of the three possibilities in the theta phi, in the theta phi space, that's just one. Okay? So that's the ten minus four minus two. But you see, so far what we've done is not work with any particularly distinguished coordinate in the rt sense. Now, you can ask what coordinate redefinition freedom do we have in our hand? Because we've got full coordinate redefinition freedom. If we do any redefining of either involving theta and phi, that will change this distinguished part of the metric to something else. We'll be using some different coordinates on this sphere. We don't want to do that. We want to use particular coordinates on this sphere. And any coordinate freedom that if we redefine r and t in a theta and phi dependent way, it will give you spurious dependencies on theta and phi. That's not convenient for our problem. So we want to keep this form of the metric. So what we are going to try to do is to ask what coordinate freedom do we have left that preserves this form of the metric? And the answer to that question is obvious. Can somebody tell me what it is? Rotations just don't change the metric at all. We want to look at more generally coordinate transformations that will preserve this form of the metric but change the values of these functions a, k, l and r. Scaling of r. Scaling of r. That's one possibility, but that's a very minor possibility. Function of r. Right. Any field redefinition which is r is equal to some function of r prime and t prime. And t is equal to another function of r prime and t prime. Well, that will change what this particular function is just by putting what r is as function of it. And we'll change this part of the metric. But we'll present this form clearly. So by adopting this form we have essentially broken our difumofism invariance down to difumofism invariance in two dimensions. Okay. So now we have this coordinate definition freedom of r and t and we can use that to further simplify a metric. Okay. So now we're going to do two things. The first thing we're going to do is to... So since we've got two functions in our hand, two coordinate freedoms in our hand, we can use it to set two of these functions to something nice. Okay. The first thing we're going to do is to set this guy to zero. The second thing we're going to do is after having done that adopt the convenient definition of what r is. So we will just set k is equal to r square. So we use coordinate redefinition ambiguity to say that our new r coordinate is whatever came behind this thing. The new r coordinate square is whatever came behind this. So that has now given a physical interpretation to this r coordinate. The r coordinate for pi times r square is the physical size of the two-sphere at that location of r. Physical area of the two-sphere at that location of r. Okay. So we use a physical criterion to give meaning to r and then we use the further field redefinition ambiguity, the t field redefinition ambiguity after that to get rid of this term. Okay. Now you could ask when you always do that, postpone such questions. I mean certainly the local patching. There are topological obstructions with this concept. Okay. So what this leaves us with is a metric of the form ds squared is equal to and then we do some renaming again to stick to Landau-Liefschitz conventions. ds squared is equal to e to the power nu dt squared minus r squared dt theta squared plus sine squared pi d phi squared. Minus e to the power lambda dr squared. Well we've just taken sine squared and we've just renamed these functions. Okay. Whatever comes behind dt squared we call e to the power nu. Whatever comes behind dr squared we call e to the power lambda. And here we are committing to implicitly committing to the assumption that this coefficient will be positive whereas this coefficient will be negative because an exponential can't change sine. Please. The RT term is not there because it was translation. Where? The what? RT and r squared. R. R and dr dt. Dr dt. That's just a choice of coordinates. Coordinates. Coordinates. We've got two functions. We've got two coordinate redefinitions left. Here we had a dr dt. We're going from here and here we get rid of the dr dt by using our two coordinate freedoms to set two conditions. Okay. The two conditions are a, the radius is whatever appears behind dt theta squared plus d phi squared and b is the dr dt term is given. Clear? Okay. So now this is our form of the metric. Now can somebody tell me is there any further field redefinition ambiguity left here? A coordinate redefinition ambiguity left here. There is one. What? e to the power lambda is equal to what? Well first how do you say e to the power lambda equals one? Lambda in general is a function of r and t. So you have to tell me what coordinate transform I can do. Can I redefine r? No. Because r has physical meaning. Can I redefine t? Yes, nobody can stop me from redefining t. If I redefine it in a way that depends on r, that will generate a dt dr term. Okay. However, if I redefine it in a way that depends only on t, all it will do is change this new here. Okay. So we've got some coordinate freedom left. And the coordinate freedom effectively allows us to take nu to nu plus some function of t. Where that function is the logarithm of the derivative of the times field redefinition. Is this clear? Okay. So two metrics that differ only by nu, differing by function of t, but otherwise equal are the same metrics. Just differ by a coordinate redefinition. We're not going to actually fix a coordinate choice for t just yet. We will do that in the solution as we get to it. Okay. But it's important to remember this at this point. Is this clear? Okay. With this kept in mind, we've now completely fixed coordinate freedom. This is a lesson of how we work with general relativity. The formalism of general relativity is completely coordinate invariable. But to solve a given problem, you want to set up equations. Equations have to be set up in some coordinates. Sometimes mathematicians manage to solve problems without writing down equations in any coordinates. I don't understand how they do it. Okay. But physicists like equations that you can see. And you won't choose a particular choice of coordinates. So what you usually tend to do is tend to try to search for coordinates that are as well adapted to the problem as possible. Okay. Suppose you worked in a completely coordinate invariant way. You said I'll work with these arbitrary coordinates. Then when you try to solve equations, you will find more solutions than physical solutions. Because in addition to having the solutions that you want, you will also have coordinate transforms of those solutions. Okay. And then you'll have to disentangle which parts are physical and which parts have trivial. Okay. So it's often convenient to fix coordinate freedom in as good a way as possible at the beginning of a problem. Okay. Now we've done that. Now we have to solve Einstein's equations. Okay. So now this is the painful thing. Just because it's the first time, we're going to do this quite explicitly in class. We're going to try to solve these equations quite explicitly in class. The painful part of this is computing all the crystals. But let's do it anyway. Okay. So we'll roll up our sleeves and start computing crystals. Okay. So let's first start with... Okay. Let's use Landau-Liefschitz for guidance. Let's see which are the easiest ones. Let's first do the Christoffel symbols that are purely in the sphere directions. Okay. So let's do gamma, theta, theta, theta. Okay. The metric is completely diagonal. So this is equal to g, theta, theta times theta derivatives of g, theta, theta without zero. Because g, theta, theta is not theta. Okay. So this is equal to zero. What about gamma, theta, theta, phi? Okay. This involves either phi derivatives of g, theta, theta or theta derivatives of g, theta, theta. There's no g, theta, phi. There's no phi derivative of g, theta, theta. So that's zero. So let's... All the zeros will put... So we've got clear that theta, theta, phi is equal to zero. Okay. Now let's look at gamma, theta, phi, phi. Is this going to be zero? No. Because? Exactly. There is a term which is... Now let's be careful. Minus half... Oh, sorry. I'll put the minus later. This is theta, theta times the piece that has the theta derivative of phi, phi. That comes with a minus sign. Del theta of g, phi, phi. Okay. g, phi, phi was sine squared theta times r squared. So that's equal to two... So the two cancels this. So sine, theta, cos, theta, r squared with a minus sign. Okay. Let's record it. Let's see if we can get this whole thing right ourselves. Okay. So we'll record this. Do you agree with me? If you spot an error, please tell me. We'll also compare with lambda. Okay. And so this was gamma, theta, phi, phi was equal to this. Okay. So now we've dealt. Okay. What about... Let's do everything with the theta. Okay. So what about gamma, theta, t, t? This is either involves... There's no t, theta. This either involves g, t, t with respect to theta, which is zero. Or... Or nothing else. So that's zero. Okay. What about gamma, theta, t, r? Amma, theta... Okay. The only non... Okay. Let's speed it up. What could be non-zero? With the theta, what could be non-zero? We've dealt with one thing here. Okay. There's one more thing that could be non-zero. What could it be? Somebody said, theta, r. You see, suppose we take gamma, theta, theta, r. Then that will involve the term that's an r derivative of g, theta, theta. This will be equal to... Oh, and I forgot here the g, theta, theta up. Oh, I put that. No, no, no. g, theta, theta up. I messed up. Sorry. Let me just do this again. Sorry. Let me do this again. This one was g, theta, theta times minus half times del theta of g55. Ah. So the r square cancelled. There's an r square in here, an r square in here, and that cancels. So sorry. Okay. Now what about this guy? This guy has a g, theta, theta up. So that's a 1 by r square. So we get 1 by 2 r square times the r derivative of g, theta, theta, which is minus 2 r. Another thing we messed up under the sign. That was a minus, because g, theta, phi was... So minus 2 r. So that is equal to minus 1 by r. Okay. So g, theta, theta, r is minus 1 by r. Any other things with theta on top that could be non-zero? I don't think so. Because the only metric components involving theta that are non-zero are g, theta, theta. And then you need an r derivative. Or you could have a theta derivative and only phi, phi. Okay. Great. So let's rub these out. We've dealt with everything with theta on the top. Now let's deal with everything with phi on the top. Which gamma is with phi at the top could be non-zero. So firstly, anything involving a phi derivative is zero, because phi doesn't appear in the metric. Okay. So that phi has to be one of the metric components. The only metric component involving a phi is g, phi, phi. What does g, phi, phi depend on? It depends on theta and an r. So there are two things that could be non-zero. Okay. So there's gamma, phi. There's phi, r. And gamma, phi, phi, theta. Okay. Now let's compute this guy. So this guy, we need... Firstly, we have a g, phi, phi. Then we have a plus half. Aren't we messing up a lot? I'm going to come back and look at the signs of this. Remember that the g inverses also have signs. We were messing up on that. I'm going to come back and look at the signs of the top two ones again, but let's do this correctly. So there's g, phi, phi. There's a half. And then there is del r of g, phi, phi. Signs cancel between these two. Okay. This thing is like r square. So the two factors says that this is one by r. Okay. Now I'm going to go back to the earlier ones and this is one by r. Here it was minus and here it was plus. Everything else cancels. Right, because we had g, phi, phi at the top and then g, phi, phi at the bottom differentiated with respect to r. So everything here doesn't depend on r cancels. So it's basically a logarithmic derivative of g, phi, phi with respect to r. A derivative of log g, phi, phi with respect to r. So that's a one by r. Okay. Now what about this guy? Here it's very similar except that what we're differentiating is the sine square theta. Right? So we get sine theta cos theta divided by sine square theta with cot theta. With no sine, I believe. Tell me if you use cot theta. Let me now just check against Landau-Liffith. 3, 2, 3. 3 is presumably phi. 3, 2, 3 is cot theta. Good. But he doesn't have a 3. Sorry, just wondering. 3, 3, 3. I suppose it's 1. 3, 3, 1. He doesn't report 3, 3, 1. I think he's just messed up. Okay, he doesn't report it. He says all the rest are 0. This is not 0. Okay. It's probably messed up. We'll see. Okay. And let's check these two. So this is what he calls 2. So 2, 3, 3. Does he have 2, 3, 3 at least? 2, 3, 3. Minus sine theta cos theta. Excellent. And 2, 2, 1. 2, 2, 1 is 1 by r. Excellent. Okay. So apart from hopefully his rather than r messed up. With this guy, we agree with that. Now let's continue. Okay. So we've done everything with an angular coordinate on top. Now let's take the things with r on top. So gamma r. Now it's more complicated because everything depends on r. Okay. So r could be either the derivative or it could be the metric. So suppose we want the terms that are non-zero because r is the derivative. There are many such things. There's gamma r, r, r. Okay. There's gamma r, t, t. There's gamma r, pi, pi. There's gamma r, theta, theta. Let's get all of these. So let's look at this first. So this thing is minus half, sorry, it's plus half. g r, r. Del r, g r, r. Right? So that is equal to now g r, r was minus e to the power, lambda, the minus cancels between the two. So this is equal to lambda prime by two. By prime, I will mean r direct. Lambda prime by two. Okay. When we do our next problem in class, I will just write down first-off examples. Okay. Okay, but once we'll calculate them. Okay. What about gamma? Then there's t, t, for instance. Okay. In this case, it's minus, this is equal to minus half, g r, r times del r of g, t, t. Okay. Now let's get the signs carefully. This guy comes with a minus, but this guy has a plus. So the minus goes away. Okay. So this is equal to e to the power, nu minus lambda into a nu dash. Into the power nu minus lambda into nu dash by two. Dash is derivative with respect to r. Okay. What about gamma r, t, t, t? Okay. So that is g r, r times del r of g, t, t, t with a minus half. Okay. So that is equal to e to the power minus lambda, the minus signs between these cancels. So there's a minus here, r by just r. r times e to the power minus e to the power minus lambda. The minus sign goes away? No, I don't think so. There were three minus signs. There was a minus sign for the inverse metric, the minus sign for the metric, and the minus sign because we're using the minus part of the gamma. Okay. And there was a gamma r phi phi. This is extremely similar. It's just that you multiply the space signs with it. Okay. But we're not even through with the, are we through? Let's see. I think we're through, maybe. So all that remains, we've taken into account all terms that come about because we're taking r derivatives with an r on top. But there are also the terms that could come about because we're differentiating a part of the metric which has an r index. The only part of the metric that has an r index is g r, r. Ah, so there's one more thing. So we could have, so there's one more thing. What could it be? No, but yeah. So it could be r, r, t because g r, r depends on t. Right? What could that be? So that could be plus half g r, r into del t of g r, r. Times cancel. And so this is equal to half lambda dot. But dot means derivative to this vector time. Okay. Great. So we've dealt with everything involving an upper r. All that remains are things that involve an upper t. That's actually easier than upper r because the angular coordinates, the coefficient of that depends only on r square and not t square. Okay? So let's get the remaining guys. Okay? So there's gamma t, t, t of course, which was equal to plus half g t, t, del t, g t, t, which is equal to, signs cancel, so it's equal to nu dot by 2. Full symmetry. Lambda prime by 2, nu dot by 2. Okay. Then there was gamma t, r, r, which was equal to minus half g t, t into del t, g, r, r, okay, which is equal to what? Exactly. So first the signs, this minus cancels a minus sign here. So it's plus. And the derivative is t derivative of r, r. So it's lambda dot. So we get half e to the power lambda minus nu into lambda dot. Okay? Again, look at the symmetry between this and this. Flip under r goes to t and nu goes to lambda. Okay? What about gamma t theta theta? Zero. Because g theta theta doesn't depend on t. Okay? Only remaining guy, there's one remaining guy. What could it be? Now, okay, we've dealt with all that come about because we're taking t derivatives. But what about the term with the t component? Exactly. This is the last nonzero guy. So what is it? It's plus half g t, t into del r, g t, t. Okay? So that's equal to nu prime by 2. Okay? Yeah, I suppose it's worth checking all these. Yeah, it's probably worth checking all these. Okay. Fine. So let's see. So firstly, the guy is with one on top. So gamma 11 is lambda prime by 2. Correct? Gamma 110. 110 is lambda dot by 2. No, right. Gamma 110 is lambda dot by 2. Correct? Gamma 122 is minus r e to the power minus lambda. Gamma 133 is minus r sine squared theta e to the power minus lambda. Correct? And gamma 100 is nu prime by 2 e to the power nu minus lambda. Correct? Okay, so all that remains is this. So gamma 011 lambda dot e to the power lambda minus nu. Correct? Gamma 000, nu dot by 2. Correct? And gamma t r t, so 010 we want. 010 is nu dot by 2. Perfect. So apart from the fact that lambda issues is one less gamma than we do, we've agreed with him completely. Okay, excellent. So now we know how we've computed all our... all our... Christoffels. Okay, but now in order to get Einstein's equations, we're not yet done. We need to compute curvature. Okay. Now, if you were really working out a physics problem, you know what you would do is to go to Mathematica. You'd find a package that takes a metric and computes curvatures for you. And that's what you would do. In your life, you should see how Mathematica does it. Okay. Yeah. Okay. Yeah. In this one, deciding the sine of what is the sine of dt square and dr square, how do we do this fix? Okay. So we're thinking of t as a time-like coordinate. Okay? So in flat space, we would have had minus dt... sorry, plus dt set. So we're looking at that by continuity. However, it's a very good question because as you will see in the black hole metric, this goes wrong at some point. So if you're near enough to being flat-spaced, it can't go wrong. You know, because this was plus one. These were minus ones. Okay? So you have to be very far from the flat-space metric in order for it to possibly go wrong. In dozen some cases, but we'll ignore that possibility for now and correct for it when we do. It's a good question. So I think we fixed another choice of coordinates. By another choice of coordinates? But you see, it could be that this... the coefficient of dr square, for instance, will become positive. Science cannot be changed. R is the thing, you know, R is physical. It's the thing behind the two-sphere. And how do you know that that's the coefficient of that? In this case, but in general, if your dt term has a negative sign, then it can be changed by a charge of 30 seconds. The signature of spacetime cannot be changed. The fact is that one minus and three pluses cannot be changed. But of course, which one is minus and which one is plus? You can do just by renaming, right? If you call R, T and T are... In the blackboard case, inside, doesn't the signature change? No, it doesn't change. But the choice of coordinates can add to the other side. Well, you can choose to call what is R. You can choose to call it T. Okay? Except that then you'll have T squared behind the two-sphere metric. So once we've fixed coordinates like this, we've removed much of our coordinate redefinition of the gate. Then the question of what the sign behind this term is and what the sign behind this term is is a physical question. These things. These things can flip physically. Good. Coming from there to here. Yes. We have fixed K and A. Yes. And then we are also fixing the sign of K and R. So is it not like that gamma goes to gamma plus and T goes to gamma? Gamma goes to gamma as well? Mu. The sign fixing... No, no, no. This doesn't change the sign. But in an exponential, if you change a function, it doesn't change the sign. But then putting it to the exponential somehow, how many of these are negative? Now, we've made an assumption by putting it to the exponential with this sign. You see, the most general thing was some function that could have been positive or negative. And they've given up some freedom by saying that this is an exponential because that fixes it to be positive. That freedom... So that freedom, like I said, is true in flat space. I mean, function-wise, is it not true something? Like, we had only four functions. No, no. That's a discrete thing. It doesn't count as a function. It's a... But this is, you know, more... It's an assumption. It's not a coordinate definition kind of freedom. We're assuming the metric is of the form that this thing is positive and these three are negative. And this is an assumption that is either true or false. You see, when we solve equations, at some point, the assumption will turn out to be false. In which case, we'll have to do better. Okay? But we're not using some coordinate definition. Physical assumption. Okay. Okay, great. Other questions, comments? Now... Now we want to set and compute Einstein's equations. Okay. So... Basically, in order to do that, we need the components of the curvature that are non-zero. Okay? Now I'm not going to compute the zero ones for you. Okay? I'll leave that as an exercise. It's too exhausting to do that. Okay. But what I'm going to do is compute all the non-zero ones, all the ones that will give us our equations out. Okay. So, first I'll just list them and I'll ask you to verify that you get nothing else. The ones that give you something, you know, are r11. So, what we call rr. Now, of course, once you know rrt, you can compute rtr because once you lower the curvature, the Ricci tensor is symmetric. So, there's also rtr, but I'm not going to write that because it's not new. Okay? And then you also have r theta theta and r phi phi. Okay? These are the only non-zero curvature Ricci components. This is the claim. Okay? Maybe we'll check one that's not of this form at the end. But at the moment, I'm first going to sit and calculate these five components. Do you have an interpretation for these components? I was like, are we a new alpha, beta? No. I know how to interpret this, but... There's probably something to say about this, but I don't know. I don't have... There's no way I think of it like that. There probably is someone who has a good way of thinking of it. I don't know. So, now, let me start by computing, for instance, rr, rr. Okay? Now, let's remember what this thing is. Okay? So, it was, in terms of gammas, it was gamma firstly r, let's... So, firstly, let's write it in terms of rs. Riemann. So, it was gamma. We contract first and third and get second and fourth. Or maybe we contract second and fourth. Okay. So, suppose we contract second and fourth. So, we have a and then you've got... So, this is... We're going to contract second and fourth. Or... That's what rr is. You can choose to contract either first and second or fourth. Okay. So, now, this is equal to r a r b g b a. Now, we want to know what this guy is. So, what was it? This was b, r. We got the sign right this time. Minus gamma r a r, b. Gamma, r a alpha gamma alpha b r or r b. Just a minute. Yeah, r b doesn't matter. Just one minute. Let me... Possibly, let me just remind myself first. a, b, c, d is what? The other terms are it's gamma times gamma. So, this gamma so gamma c a, b gamma d sorry. Hello? Hi. Yes? Yes, yes, yes. Sorry, this one. Hi, definitely, I'm teaching a class now. I'll call you in the afternoon. Okay. Yeah. So, this is how it goes, right? Sorry. So, it's gamma c a gamma c a theta gamma d theta b minus gamma d a theta gamma c a theta b. So, yeah. So, that's the rule. Let me write this down again now. So, it's... This was... Sorry about this, people. This is r a. What was this? Okay. So, we get gamma b times gamma a gamma a it was this thing. So, gamma r gamma b minus gamma b gamma r and then we have to put in the indices. So, that's r c c c a Okay. I think I finally got it right. You agree? So, it's gamma r gamma b minus gamma b gamma r in the sense of matrix multiplication and these are the matrix indices. So, r and a r and a and c c c c. Okay. Fine. So, this is the formula for gamma r sorry for for this thing and then, of course, we've got to take this whole thing and contract it with GBA. Hello. Okay. It's the insurance for my car. Okay. Okay. So, okay. Excellent. So, so, so now we let's compute this guy. Okay. So, firstly, let's look at these two terms. All terms are a gamma r and gamma r and gamma r and gamma r and gamma r. Okay. This one is gamma r and we have an r derivative of it. Okay. And we've got a b so these these two have to be the same. So, what gamma r depend on r? Okay. Well, all of these depend on r. But, these two have to be the same. So, all four of these contribute. Okay. Painful, painful. Okay. Right. So, okay. So, let's write it down. So, from here, what do we get? We get g r r gamma r r r r plus g t t gamma r t t gamma r plus g t t t t t gamma r t t t t r plus g phi phi gamma uh phi phi gamma r. That's this term. That's it. We've completely dealt with this term. We'll put in what these are. Okay. What about this term? Now, this term has a gamma r r and the derivative has to be with respect to my neighbors here. So, that's much fewer terms. Uh, what it could be is gamma r r and an r derivative. So, one term there is minus and then this would be g r r g r r gamma r r r r comma r which is going to cancel this. Okay. And the other term that it could be is uh, this guy. Okay. Uh, so if we had that guy, this would be t, so this would have to be t. Okay. So, we would have plus g t t uh, gamma r um, t r comma t. Okay. So, we've dealt with this guy and this guy. Okay. What remains are these two guys. Okay. So, now let's look at this guy. This guy is gamma r and r something. Okay. Um, so the only option is either c is r or c is t. Let's look at the case c is r. If c is r, then this guy is gamma r and we have a equals b and that's a lot of possibilities. Okay. Uh, so uh, so we get plus gamma r r r um, gamma r r r with a g r r plus gamma r. Okay. Let's keep gamma r r out common. Okay. Let's, I'll rewrite this. So, that's plus gamma r r r times uh, g r r gamma r r r plus g t t comma r t t plus g theta theta comma r theta theta plus g g phi phi uh, comma r phi phi. That was this term where, that was this term when c was r. The other possibility is that c is t. Okay. If c is t, then uh, uh, and we want both equal, then the only thing that it could be is t t or r r. Right? So we get, no, it's just a little bit of a plus. So plus um, g r r gamma r r t gamma t r r plus g t t comma r r t gamma t t t. That's dealt with this term. Now we want this term. Let's look at that term. Um, in this term what do we have? We've got one r at the top and one r, uh, down there. And then, these two are the same as these two. Okay. So now we're going to ask, what are the possibilities for these pairs? The, yeah, because b and a are contracted. Right? Yeah, so b is the same as a. Okay. Sorry about this. We're almost getting there. Maybe we won't. Wow. Okay. I think we will possibly have this trend to compute one of these. Let's, let's see. Okay. So um, um, now b c and b c have to be the same. Um, now that restricts possibilities a lot. Um, let's see. One possibility, of course, is b c, b c and uh, uh, c a, they're both r's. Okay. So we get minus gamma r, r, r, gamma r, r, r, r. And then the up, the g r, r is upstairs. What are the possibilities do we have? Let's see. One possibility, let's exhaust the r t's first. What about r t? That's a possibility, right? So, suppose b is t and c is r. Okay. Um, so we get minus, uh, it be, so g t, t, gamma, r, t, r, gamma, r, b is, we did b is t and c is r, but whatever b is r and t is, c is t. That's also a possibility. Um, and that will give us minus, so, gamma r, r, gamma r, r, gamma t, r, r. What about b c above t's? B and c above t, that's also possible. So minus, minus, gamma r, b and c above t is a t t, gamma r, t, r, t, g t, t. Uh, say more clearly, I don't understand this. C term. This term? Yeah, I understand, but we're looking at cases one by one. We've looked at the possibilities for b and c being between r and t. Okay. Uh, now, uh, now, now what remains? Um, that involves at least one angular coordinate. But, there's never, there's never anything that has one angular coordinate. You have at least two angles. If you have one angular coordinate, you have two. Okay. So, b and c could take various combinations of the angles. Okay. So, uh, wait, wait, what are you saying? That, uh, r, what are you saying? In terms of, no, are you saying we missed some terms? Which are these? Just, just say in terms of c. This is r, what is b going to be? R r. I see that. Which is theta? Now c is, these two are equal. C is theta. Then b is also theta? I see. b and a are r, and c is theta. This is r, r, theta. How is that non-zero? I don't think there's anything left. Yeah. Okay. So, all that remains is that, I'm sorry about this, people. This is a very boring class, I know. Okay. But, let's finish this. Okay. Right. So, all that's left is b, c, and c, are both angular coordinates. Okay. So, what do we get? Let's, gamma, r, let's say, theta, theta, gamma, theta, r, theta, g, theta, theta. Okay. Now, what are the other possibilities? The other possibilities are, gamma, suppose b is theta, c is phi. Is that non-zero? That's not non-zero. So, we can forget about it. Suppose, b is phi, c is theta. That's not non-zero. So, the other possibility is, b and c are both phi's. So, we get phi, phi, gamma, phi, r, phi, g, phi, phi. It will be absolutely a miracle if we get this right. I'm going to try for five minutes. Let's, let's, let's, let's, let's, let's, let's, if we may have to, this one? Okay, good. Firstly, what's this g, t, t? And this had a minus sign, let me see where it comes from. It comes from the, the second term from here. Yeah. Yeah. And therefore, term from the second term had a minus sign. Now the first thing we are going to try to do is to cancel terms if we can. So these two we cancel and out of the terms with derivatives involving gamma R. What about this one? T R, I suppose this was R, R T R, no okay. So now in this line this, this, this and this nothing cancels right? Okay. What remains is all these product terms. So now let's look at the product terms carefully. Firstly, gamma R R, gamma R R, R, G R R. This term cancels with this term. What about G R R, gamma R, T R, gamma T R R? G R R, gamma R, T R, gamma T R R. Beautiful. This cancels this. I suppose this was G T T. So what about G T T, gamma R, T R, gamma T T T? G T T, gamma R, T R, gamma T T T. G T T, gamma R, T R. It doesn't look like there's anything to cancel that. Okay. Anything else that could potentially cancel here? Okay, let's organize terms. Let's write down the terms by organizing them. And then we'll see if they, if they cancel and on. Okay. So first the second derivative terms. Let's put up this board. Okay, so the second derivative terms are G T T, gamma R, T T, gamma R, plus G theta theta, gamma R, theta theta, gamma R, plus G phi phi, gamma R, phi phi, gamma R, minus G T T, gamma R, T R, gamma T. Okay. So those are the second derivative terms. Now we get the product first derivative terms. Anything with a gamma R R R? Anything else with a gamma R R R anywhere? Right. So all terms with a gamma R R R will write. So gamma R R R times G T T, gamma R T T, plus G phi phi, theta theta, gamma R, theta theta theta, plus G phi phi, gamma R, phi phi. Great. So that's dealt with this term. Now let's see. Put this term there. Yes. Okay. Okay. So now what remains is this term and the remaining terms here. Okay. So we'll do the ones without angular coordinates first. So all of these have G T T's, right? G T T, G T T, G T T. So plus G T T into gamma R R T, gamma T T T, minus G gamma R, T R, gamma R, R T, minus gamma R T T, gamma T R T. Okay. And then what remains? That's this and this. So minus gamma R theta theta, gamma T R theta, G theta theta, minus gamma R phi phi, gamma phi R phi, G phi phi. So we've got at least a one blackboard expression for this one thing. I should have computed it easier. This is the toughest one because R appears everywhere. Okay. So I should have started with an angular one. It's a bit easier. Okay. So now we have to collect these things. Okay. Is it worth it? On the other hand, what was the point of doing this if we don't do it? Okay. Let's try. Okay. I'm sorry. Ten minutes more. Okay. So now I'm just going to try to write down what these things are. So that people help me with this. Gamma R T T, gamma R. Okay. So that's e to the power nu minus lambda. Okay. Maybe it's not worth it. Okay. It looks like I'll take till the end of the class and how about I ask you to compute this guy. Oh, the problem sets by the way already. We'll make sure to email it to you. The first problem set is 13 problems. That sounds like a lot but they're not difficult. They're not difficult. So they'll be emailed to you today. Okay. How about I ask you to complete this computation to correct and complete this computation in a problem set. Okay. From now on either we've done it wrong. In which case we'll never be able to fix it here. Or we've done it right and okay. Then you'll have an easy time in your problem set. Okay. Okay. I'm going to retreat to that. I'm sorry. Maybe there was some value to this because you see what there is to do. You also see it's tough to do. Okay. Yeah. So sorry. It's just too time consuming. I hadn't realized it would take this long. We just retreat. We'll retreat past this to say somebody has done this calculation. Mathematica can easily do it. I'm going to give you the answers. Okay. It's a mindless algebraic procedure. I'll give you the answers. Okay. Okay. So the answers go this way. Okay. So Einstein's equations when you work them out. Okay. Actually maybe there's an easier one. Just one note. You know what's really irritating is that Landau-Lieschitz has a simple calculation gives. Okay. Maybe they have clever ways of doing it. Okay. But fine. So the Einstein equations work out to the following. Eighth, yk divided by, okay, c we're not keeping. Trr is equal to minus e to the power minus lambda nu prime by r plus 1 by r squared plus 1 by r squared. This is one equation. Okay. The second equation is pi k t 22 is equal to 8 pi k t 22. That's theta, theta. The 55 is equal to minus half e to the power minus lambda into nu double prime plus nu prime squared by 2 plus nu prime minus lambda prime by r minus nu prime lambda prime by 2 plus half e to the power minus nu lambda double dot minus lambda dot squared by 2 minus lambda dot nu dot. Okay. This is two of the equations. Then we also get an equation from t 8 pi k t 00 is equal to minus e to the power minus lambda 1 by r squared minus lambda prime by r plus 1 by r squared. And the fourth, the last equation is that 8 pi k t 10 is equal to minus e to the power minus lambda dot by r. What I could do if you want, what I could do is email all of you a package of Mathematica. Okay. I'll do this. This is actually more useful than actually we get it. And email your package of Mathematica and ask you to use that to compute these equations and you submit the Mathematica, print out along with your. Okay. So this will be part of your, this will be your, this will be in your homework set. Okay. The package, the package I'll email you is written by a guy called Matthew Hedrick who was my first PhD student. That's a package I'm particularly familiar with. Okay. Excellent. So now we've got these four equations and they look like big complicated equations and the first question is what are we going to do with them? Okay. So I'm sorry, we're hitting lunchtime. I know it's 10, 15 minutes. We can finish this. Okay. No, no, we want the equations on the board. So we want a nest. Okay. I'm clearing this part. So Mesh, we promised, I don't think you were there. We promised to email them Mathtense. One of the problems in the assignment would be to get these equations using Mathtense. Okay. Okay. Great. Now, now what do we do? You see, what we want to do is take these equations and solve it in the simplest situation. The simplest situation is when there is no stress tensor, no matter stress tensor around. It's the vacuum. So let's first solve it in that situation. Okay. So then in that situation, the left-hand side of all equations is zero. Now let's look to see if we can find a particularly simple equation. Okay. This equation and this equation are particularly simple because they involve only lambda. What we want to solve for is lambda as a function of r and t and nu as a function of r and t. Right? So we want a computer metric. Okay. Now, it's a good question. The initial data will in general be lambda and nu as a function at some initial point. Now r and t, we'll examine that in more detail. But so we're going to encounter something very interesting. That is that the solution is going to be unique in the absence of matter. So then it will not, you know, it's going to be like the fact that in a spherically symmetric potential. Yeah, yeah. There are no spherically symmetric waves basically. There are no spherically symmetric spin-two waves. Okay. It's something that you're familiar with in even electromagnetism. If you've got a completely spherically symmetric distribution of charge, no matter how it's falling in, no matter what the charge is doing, the outside is basically given by the good answer. If you preserve, if you preserve spherically symmetric, we're going to encounter something like this in this situation. So we will not have to answer your question until we start looking at infalling collapsing matter to form black holes, in which case we will return to your question. So what is wrong with having no matter and choosing to give it function use? Well, the only functions that will obey the spherically symmetric answers and solve the equations will be uniquely determined. Let's solve the equations that we see. Okay. The first equation we look at is this one. This tells us that lambda dot is here. Therefore, we conclude that lambda is equal to lambda of r. Okay. The next equation we look at is this one. This equation now is just an ordinary differential equation in lambda. First order ordinary differential equation and it's quite easily integrated. Okay. So let's write down what the equation is. So the equation is e to the power minus lambda 1 by r squared minus lambda prime by r plus 1 by r squared is equal to 0. What? Thank you. Okay. We will return to this equation in a moment. But even before that, let's also look at this equation. This is also a pretty simple equation and I'll write it below it. It's then minus e to the power minus lambda of prime by r plus 1 by r squared plus 1 by r squared is equal to 0. Okay. So even before integrating any particular equation, let me take these two equations and subtract them. You see, in subtracting them this term and this term cancels and this term and this term. So what we get is e to the power minus lambda and that we can throw away since it's overall common outside is basically we get lambda prime minus nu prime is equal to 0. Is this clear? Okay. Now, since we get lambda prime minus nu prime equal to 0, we can integrate this equation to get nu is equal to lambda plus a function of t. But you remember we had a residual coordinate invariance that allowed us to shift nu by a function of t, t coordinate redefinition. We'll use that freedom to set this function to 0. So we'll choose our coordinates so that nu is equal to lambda. Our remaining field redefinition ambiguity, coordinate redefinition ambiguity is used up in this manner. Okay. So we conclude very simply that nu is equal to lambda and both of them are functions only of r. Notice we have not assumed that the problem is static. We've not assumed that things aren't moving. So what are we doing? We've got, we're solving this problem in a region where there's no stress tensor. But there could be stress tensor somewhere else. You could have a cloud of dust say that's collapsing. But as long as it's spherically symmetric, where there is no dust? We have concluded that nu is equal to lambda and both of them are functions only of r. So we've got an essentially static solution. This is the statement that there are no spherically symmetric gravitational waves basically. Okay. And of course we presume spherically symmetric. Okay. Now what remains is to solve the equation. So let's take this equation. Okay. So we've got an equation that we want to solve and that's easy to solve because we can, you know, linear first order differential equations can always be solved by integration. Now you might think this is not a linear first order differential equation because it's complicated dependence on lambda, but that's a fake. Let's move to the variable e to the power minus lambda. Okay. So this thing is minus e to the power minus lambda by r squared. And then we had minus and minus, but I'll make it, so minus e to the power minus lambda prime by r. Okay. Plus 1 by r squared is equal to 0. So that's pretty simple. So now such an equation can always be solved by integrating factors and it's actually a very simple solution here. So let's just multiply the whole equation through by r squared. So we get an r here. And now we recognize that this is simply r times e to the power minus lambda prime with a minus is equal to minus 1. So we conclude that r e to the power minus lambda prime is equal to 1. Okay. And therefore r times e to the power minus lambda is equal to r plus c by integration. And therefore e to the power minus lambda is equal 1 plus c by integration. I'll write minus c for reasons that we'll see. So the physical case will be c is positive. So what is our solution? Our solution, even though we had to cheat them, sorry to get it, is ds squared is equal to, now dr squared was multiplied by e to the power lambda. So we get 1 minus c by r minus r squared d theta squared plus d phi squared. Okay. Sorry, again minus. Sine squared it. Plus dT squared in 1 plus, 1 minus squared because nu is equal to minus. Solution. I mean there's one parameter ambiguity from this, from this, from this constant c. It is otherwise a unique solution of gender relativity in the absence of matter assuming spherical symmetry. Okay. This plays the role of the unique Newtonian potential of a spherically symmetric mass distribution. Outside the mass the potential falls off like 1 by r. In fact it reduces to that in the Newtonian limit because we wanted to identify this thing as 2 phi. So c by r should be identified as, twice c by r should be identified as m by r. Okay. Or maybe k m by r because it's Newton's question. Okay. So this tells us that it's natural to replace this c by k m by r. I did this wrong. 2 phi which is equal to 2 k m by r should be identified with c by r. Minus 2 phi which is 2 k m by r should be identified with c by r. So c is equal to 2 k. Okay. At least in the Newtonian limit this constant can be thought of as, so the metric you might want to write as d s squared is equal to 1 minus 2 k m by r d t squared minus d r squared over 1 minus 2 k m by r plus r squared d theta squared plus n squared theta d phi squared. In the sequence what went into the thing was that you see suppose you've got a situation like the sun where there's matter inside but nothing outside. What we've done is accurately solve for the metric outside where there's matter. Basically you aren't of very symmetric solutions in our total absence of mass. Yeah. That every spherical symmetric solution other than the vacuum has a mass sitting at the center. Somehow or the other. We turn to this question more as we go on. But one thing about this is that this we've identified as m by, as the mass only with analogy with Newtonian limit and that could be wrong. It could be a non-linear function of the same. Okay. However this analogy actually gave us a right answer. As I had planned to discuss today but I'm sorry hijacked by the stupid algebra I should never want it. One of the knowledge we can make is that we are flipping the p r sin is for the real radius. What? I don't know. Sine is your d d squared d r squared sin. Huh? Flip when you. Go past a particular area. Go past a particular area. That's a good point. However in many real objects like the sun or the earth that radius happens well within where there's matter. So this doesn't have to happen. But just let me say that we'll start next time's class by do it over. We'll sometime in next time's class I will explain to you that general relativity does come equipped with a conserved energy and a conserved momentum. There's something we have not really seen yet. Okay. And that the energy and momentum of a space tank can be measured by doing some integral at infinity. It's called the ADM energy and the ADM momentum. Okay. And doing that integral at infinity for the space tank will identify the energy as m. So writing in this way correctly identifies m as the energy of the space tank. The conserved energy of the space tank. We will see that we haven't yet seen this. No gravitational waves are present. What? Prescription. Gravitational waves are present. Some energy? Yes. Yes. Yes. Yes. In general relativity there is no gauge invariant way no coordinate invariant way of localizing energy. However the question of what the energy of a space tank is is completely unambiguous. Okay. All time coordinates that reduce to the same thing so as we will see when we do this we will make the following assumptions. We are dealing with a space time that is completely arbitrary but an asymptote at infinity becomes flat. Now at infinity then we will adopt a canonical T coordinate. Any choice of coordinates which reduces to the same canonical choice at infinity will give you the same answer. So we will use a metric that at infinity reduces to the Minkowski space metric in standard Minkowski coordinates. Okay. That's the only thing we will assume about a coordinate system. And then with that assumption we will get a well-defined energy that will not depend on your choice of coordinates. Yes. The solutions we have found out for unit and lambda. Yes. Those are found out from first and final two equations. Yes. But those also should be consistent with that T theta delta d phi prime. Excellent question. Excellent question. So you say that this does not work out? Well, let's see. Firstly, everything here involves dots. So this is just zero. Now, new prime minus lambda prime is zero. Here, new prime lambda prime remains. And that cancels out with new prime. And that cancels out with new prime square. No, it adds because there's a minus. And new is minus lambda. No, new is lambda. No, new is lambda. New is lambda. Wait, wait, wait. I did something wrong somewhere. Our metric had one on top and one at the bottom. Let's subtract. I just wrote this wrong. New is minus lambda. Our metric had this thing and this thing inverse. Sorry, I just wrote that. Now let's check. What you want is that, so we had lambda. So where did we have our lambda? We had e to the power minus lambda was equal to 1 minus c by r. So lambda was equal to minus log of 1 minus c by r. And new was equal to plus log of 1 minus c by r. So these two terms here are just, for instance, lambda prime or new prime. Let's say new prime squared, those added up. And so we got 1 over 1 minus c by r into the derivative of minus c by r, which is equal to c by r squared. And this thing, the whole thing squared. What about this guy? Well, let's simplify this so that's equal to c squared by r to the 4 in 1 minus c by r, the whole thing squared. Now let's compute new double prime. So the first prime we had, so that was c by r squared over 1 minus c by r. And let's write this as c divided by r minus c, the whole by r. Now I'm going to differentiate. So what do we get? So differentiating gives 1 by r squared r minus c, the whole thing squared with a minus into the derivative of this guy. Which is 2r minus c. Doesn't look like the same.