 was the metric of flat space, the unit 3 square, the unit utility A is A square, ok. Or d s square is equal to minus A square of A square into minus d plus square plus d s 3 square. Now, these two A's are the same thing because the coefficient of what multiplies is equal to minus d s 3 square, ok. So, it is the same A, it is just that you are using that there is a coordinate change going between them. And let us write out this coordinate change to the middle of this picture, I mean d e j by e d A, right now. Now, I am going to be a little more careful than last class. And sometimes we use this coordinate system and sometimes the other one, some things are easier and some things are easier, ok. So, I am going to do the following, lambda which is the only ever used, well, if the guy here should only ever use this one, ok. And so, I am going to follow the notation that prime is equal to d by d, ok. In a bigger lambda which is well dot will be d by d, ok. So, from this fact we find that A prime of A dot is equal to 1 by A and alpha prime, alpha prime, right. So, this dot is d by d t which is d eta by d t times d by d a, d eta, so by the j, ok. This will be true for every one. And it will be useful when we move between the three points, I cannot make it on the right side, on the top, ok. So, we started last class trying to find the equations of motion for A and epsilon, the energy density in the fluid stress tensor that we talked about a lot in this class, ok. So, let us continue with this, if we sum this up. If you remember, we had an expression for r 0 0 which was equal to 3 by 8 to 4 into A prime square minus A eta. And this expression is correct, whether the space is x 3, flat x or infinity, ok. But, and then we also had an expression for r i j which was equal to delta i j times something. Now, the something was, so ok, this is equal to delta i j times minus 1 by 8 to the 4 r i j is equal to minus 1 by 8 to the 4 delta i j and then times for the sphere we had 2 A square plus A prime square plus A eta prime. If you remember, these terms came from the intrusive curvature of the sphere, this would have been the three dimensional r i j of the spatial metric of the sphere, we derived that. But, because these terms came about again because of the 0, because gamma is equal to 0, ok. Now, clearly if we were dealing with a flat space guy, this would be 0, this would be replaced by 0 and if we were dealing with a negatively curved guy, this would be replaced by minus 1, like minus 2, ok. And it turns out that these terms are affected, by the way because it turns out, ok. So, if we are doing this in more generality, we keep it track of or we try to do all cases at once, you just put a 2 k A square again, where k is plus 1 to the sphere, 0 for the flat space and minus 1 for the Euclidean atmosphere, ok. Once we have done that, these equations are everything we do from now on, we will be correct for all the three cases, ok. This is the only, this place is the only place in which the original tensor knows about the difference between the sphere and the flat space. This is Euclidean atmosphere. Now that we have got this, next thing we are going to do is to compute the Ritchie scale. The Ritchie scale R, which is R 0 0 plus R i i. So, let us compute that. So, that is equal to 3 by 8 to the 4 into A prime square minus A A prime, that should be R 0 0. Now from here, we get a factor of 3, we are tracing all these patients. So, that is that everywhere. Everything is minus this A to the 4 and then there is minus 2 k A square minus A prime square minus A, this cancels this. So, we get 3 by 8 to the 4 into actually 6 by 8 to the 4 and then the minus sign A A double prime plus k A square for the Ritchie scale. Now that we have got these things in hand, what we need to do, whether we are forced to check, we need to derive the Einstein equations. There are effectively two equivalent Einstein equations. One of the Einstein equations is all the 0 0 Einstein equations and the other one is either the 1 1 over 2 2 over 3 3, they are all the same because everything is provoking at that time. So, we can conveniently look at these two different Einstein equations. I first looking at the equation that you get from 0 0 and second looking at just the trace, the trace of Einstein's equations. That will be some linear combination of 0 0 0. So, let me write down the trace equation first. So, Einstein's equation is R minus R by 2 R to the nu minus R by 2 g mu nu is equal to 8 by 15 nu. Take the trace of this, we find minus R is equal to 8 by k. You get minus R because you can argue and then you get minus 4 R by 2 because the trace of g is 4. So, let us also remember the stress tensor form t mu nu is equal to epsilon plus p mu mu mu mu mu minus p g mu nu. So, in particular we find that t is the trace of this guy. You get plus 1 from here, minus 4 from here. So, it is equal to epsilon minus 3. Let us immediately write down Einstein's equation that we have from here. This equation is minus R is equal to 8 by k times t. So, we get minus 6 by 8 to the 4 a a double prime plus a a square. But there is a minus, there is a minus here. So, that equals plus is equal to 8 by k times times t and t was epsilon minus 3. Now, we go to the second equation. The second equation is a plane from the other zero to the zero component. So, let us keep this stuff on the board. Look, I have to give you a minute to look at and copy down whatever you want. We are going to clear this and all of this because unfortunately we can't stuff on the board. Oh, I am sorry. We also need an other zero. I lost something to me. The second equation we want is this zero-zero component of Einstein's equation. So, let us write the static weight without the zero-zero minus half delta zero-zero so that is 1 times R is equal to the time the t zero-zero components 8 by 8 k times epsilon. You see that this just becomes epsilon because we are looking at fluid flows in which U mu is 1 zero zero. We are looking at a fluid that is stationary in this coordinate system to maintain the isometric, the symmetries. So, the zero-zero component of this resonance and whether it is or not it makes no difference. Oh, I am sorry. Now, I am working in, I should probably be kept, I was thinking of working in this coordinate system. Then, I am easy to know what it makes So, half zero-zero is equal to 8 by k times t zero-zero. Now, let us compute this. So, we have t zero-zero. First, let us remind ourselves what U zero is. You see, U zero squared A squared is equal to, from the fact that U is a normalized component. And U is equal to the fact that g zero-zero is x squared. Is this clear? So, our metadunium is equal to U zero-zero zero-zero where U zero-squared A squared is equal to 1. So, U zero is equal to, U zero with an upper is equal to or is equal to 1 by k. What about U zero with a lower? Somebody correct? U zero with a lower is equal to 8. In many ways. You see, this one, the simplest is, you take U zero with an upper which is 1 by 8 and lower it. In fact, you select it, which is the A squared. So, A squared is 1 by 8. Now, t zero-zero is equal to U zero U zero epsilon plus p minus p in the delta zero-zero. So, just waste what it is. But U zero is equal to what? p cancels the guarantee. So, yeah, just relax. So, it was correct what I did. It was correct because it is much lower. It could not be correct because it is a lower set. Excellent. So, now let somebody please read up our zero-zero. 4 a number, 3 by 8 to the form, right square. Minus 8 theta by 4. Okay, correct. Minus r, 4 what? I don't know, r is equal to, I'm sorry. Somebody please read up our zero-zero. Yes sir. It was minus of less? Yes sir. R is equal to minus since we are making it less. Then, we have, we want r by 2. So, we make the 6 by 8 to the form. Into? Into. So, take it. Into equal to? Into. Into 2 times plus this. This, just k. So, we get 3 this is equal to a to the power of a prime square plus a plus k a square is equal to a to the power of a prime square plus a plus k a square is equal to a to the power of a prime square plus k a square we proceed actually let me for the two three things I want to do first of these is to take the equations well actually this is not the basis in which the equations are normally obtained the basis in which they are normally obtained is this equation as it is but then subtracting twice of this equation from this equation ok. So, that is what I am going to do I am going to obtain a new equation by taking this minus 2 times this to make the combination here epsilon plus that of epsilon plus 2. So, let me do that here. So, I want this minus 2 times this. So, what will I get because the two everything becomes 6. So, we get 6 by e to the power 4 and now I have a e double prime plus k a square minus k a square minus a prime 2 prime square is equal to minus a to the power of k a to the power of regarding this and this is the fundamental equation in that regard this and this as you will see actually this equation this is simple equation ok. So, I am going to clear this and write this. Now, we are not doing all proceeding pre obtain the equations in terms of derivatives respective rather derivatives respective. Now, what I need to do is to employ the simple chain every time this you can say in alpha prime I will replace it by 8 times alpha prime ok. So, let me do that. So, how about this equation this equation is very simple it becomes 3 by e to the power 4 into and a square I can take an a square of this a square into a dot square plus k a square plus k is equal to a by k epsilon. Equation a little more involved because it was the same ok. So, what do we do? So, this a prime square is the same thing. So, we get 6 by e to the power 4 then we get write this as a and this was d by d tau d by d tau of a is equal to a e by d t of a a dot dot let us say minus a square a dot. Now, this dot this one term where you get dot acting here this one term where you get the dot acting here where you get the dot acts here that cancels this term. So, it is only the double dot term to survive. So, that is 6 by a to the power 4 into 6 by a square a double dot by double dot a is equal to a to the power k epsilon plus and yes this is a this is 8 and ok. So, now maybe a need all these equations write these equations once again. So, the fundamental equations are prime square plus a square which is equal to 3 by a square a dot square plus a a by a epsilon x by a a double dot which is equal to 6 by a to the power 4 a a double prime square is equal to minus a a by k epsilon. Now, we got the basic equation of cosmology and we are about to start studying that ok except for one more little thing. I have wondered here there is something odd. But first thing what are the variables? Variables can be thought of as epsilon you know epsilon you know the pressure and ease. But two variables we have got two equations so from this we might have thought that look it is a bit odd that we did not have to use the matter equations. We just use Einstein's equations and we are going to ask equations to solve everything including the matter why is that? Good so why is it true that we did not have to use why did not we have to use the flu equations that is the equation of state that is just some statement of what kind of matter there is also an equation of motion which is the conservation of the stress. Why do we have to use that? Because that is automatic. The stress tensor on the right hand side of Einstein's equations must be conserved it is consequence of Einstein's equations. So, I would like to see that I would like to see that more explicitly. So, I mean what if I had a motion of equation of motion. Why? In addition to the conservation of stress. And some other equation of motion. Then you will have to impose this. Then there will be new variables in the fluid. The whole fluid would not be captured by just 5 percent. That would let's say be a charge density. So, you have to be charged. And then you would have to impose the equation then you would be able to see that as a second equation. That would not be enforced by Einstein's equations. Now, Einstein's equations just impose the four energy conservation. In this case, the conservation equation is imposed by the form of Gminu. No, Gminu was the form of Gminu does not mean that it is conserved. It is imposed by the fact that you put on the right hand side of Einstein's equations. So, in some sense you are getting some part of the animation of sharing Einstein's equations. Yes, yes. By putting on the right hand side of Einstein's equations. Because Einstein's equations are inconsistent unless the stress tensor is in the equation. Okay? Now, I am going to see this a little more directly. So, I am going to just write out the equation of conservation of the stress tensor. Okay? And try to see if I can show that this, that that follow. I will show you. Okay? So, what is the equation of the conservation of the stress tensor? Where is the, in your first problem, hey, I have a really problem set. Give me your second problem set. This has been second problem in the last week. Maybe one more. Maybe we will have one last problem set as an example. We need to have some sort of evaluation that we vary. We want to have a real thing set. Maybe just a problem set, which you have two or three years. Okay? Fine. Yeah, the conservation of the stress tensor. So, now, let's remember what the equation of the conservation of the stress tensor was. We started our discussion of the energy moment of pseudo tensor. And the equation was, when you look it up, you know, what is gradient of a symmetric state? So, it's 1 by square root of minus g, l, u, square root of minus g, d, u. And then that's because of x addition. The term that may cause the energy moment to be difficult to define. G alpha beta, l by l, x, u, x, d, u. Work this out for the particular case in hand. I mean, I'm going to work it out. Let's say in this coordinate system. Okay? So, now, in this coordinate system, T00 is equal to x. So, now, as you see, Tij is equal to p. Let's work this out. We've got epsilon and p, both of which depend only on t. So, the whole stress tensor is a function only on t. So, the only derivative, also, the matrix is a function only on t. So, the only derivative that's not 0 is the unit. Okay? So, this term, mu has to be t. Therefore, mu has to be t. Therefore, mu has to be t. So, this term is 1 by a cube, that's square root g, and dp, d by dk, of a cube epsilon. The temperature must be constant for, because it's a matrix. Yes, it must be in space. Okay? Minus half, once again, the only not 0 derivative is the g derivative. But, g00 is independent of time. So, it's a certain gij that this matters. So, that's d by dp of gij times d by dp. Gij is a d times d by d. So, the ij of 1 of 1,000 is d. You raise it with one more, and you get the last sign. So, gij had a negative, and negative is square. Gij had a negative is square. The ij, also has a negative. The ij of 1 of 1,000 is dp. So, this is the sign. We have three such signs. Very good. We have three such signs, we get extra. Thank you. Now, this we can make d by d, a d by d. Canceling the two. Let's look at this. Suppose, we let the d by dp act on a cube. We get exactly the same term as this, except that we have p instead of epsilon. And epsilon is nothing. So, this is equal to d a by dp. Actually, the sign epsilon is very important. Let me see what we have got. Epsilon minus p. Okay, let's epsilon plus p. Okay. One by a, three by a plus a. d epsilon by d. Let's check what I have written out. I have written as epsilon plus p. The ij of both lower was minus p. It was plus p. Let's look at it. Remember, T mu mu is equal to epsilon plus p minus p j mu. What dictated this minus sign? It was dictated by the fact that it is a 0, 0 component p s. Now, there's no freedom in changing the sign. So, that's the case. So, this speaker is minus gij. And so, it's plus with both of them. Because gij is equal. Raising once makes it minus. Raising twice makes it minus. Okay. So, this is correct. But this here has a plus. Because the minus here cancels the minus. So, this is correct. So, this is the equation that follows from the conservation of the, this is the equation that follows directly from the conservation of the stress. So, now we've got this equation. So, now we've got three different equations. This equation. This equation. And that equation. The three equations are two variables. So, either we've got a consistent set of equations. Or they're dependent. Okay. They're not dependent equations. Of course, the actual situation is that they are independent equations. And how do we see that? Well, in order to see that, what I'm going to do is to show that I can deal with this equation. And this equation. I can derive this equation. So, I'm going to do two equations. This guy. And 3 by a squared. a dot squared plus k. Is equal to 8 by a epsilon. And use this to derive the 32. Just to check that. Okay. Okay. So, how do I do that? Well, straight forward. I take this equation. So, I take this equation to the a squared of the right hand side. And that's it. So, I get 3 into 8. So, 6 a dot a double dot. Is equal to 8 by k. d by d of epsilon squared. So, that does. That gives me an equation for a double dot. So, I have this equation with, what I have to do is to use this equation now. Yeah. So, now, let me substitute this again. Okay. So, this is equal to 8 by a d epsilon by d into a squared. So, this is equal to 8 by k epsilon into d a squared by d j. Okay. So, that becomes 16 by k into a d into a a dot. And here, I use this to relate this to. So, this is equal to, well, let me keep the 8 by k. So, this is equal to 8 by a into, this was 2 a 2 epsilon a a dot. And here, I have this guy is equal to minus 3 by a. So, that's, this takes this a a dot in epsilon plus 2. Yeah. So, the a a dot comes out. 8 by a a dot in a 2 epsilon minus 3 epsilon. So, it's minus epsilon plus minus epsilon minus 3. Put a minus of epsilon. So, exactly. That is you. But, that, we can forget about this place given the second Einstein equation. Just use the equation that came from the conservation of energy momentum. Plus this equation, which was the 0 0 Einstein equation. Okay. And that gives you all the information that you want. Because the space part of the Einstein equations obtained can be derived from these two. Okay. Now, we can adopt these two as our fundamental equations. Okay. Now, with these equations in hand, how do you go around, go about solving the problem? Okay. So, it's very straightforward. This equation looks like it has time derivatives. But, that's actually a phase. You can't study these. It's a differential equation that allows you to determine a in terms of epsilon. So, we write this equation as 3 by a integral dA by A. So, that's d log A. So, log A. So, that's 3 by, that's 3 log A is equal to minus integral dF. So, that's an equation of state. You need to know the equation of state. Once you have that equation of state, you can do the integral on the right-hand side. And that gives you A as a function of epsilon and epsilon as function of A. Once we have this in hand, then we take the second equation. That's this. And solve it for, solve it for, solve it as follows. We find what dA by, this is now like an equation in classically mechanics. And I think that, in terms of the potentials, you use the equal tricks. So, when you use the tricks, you write this as, you first solve it. So, you got it is equal to 8 by kA square epsilon by 3. Now, we're using k for 2. This is, this is 0, 1, minus 1. Let's make this a g. Let's make this one. 8 by g by 3 epsilon square. And then, minus k square. In the absence of minor, the equation of state and the iris simply goes up there. Very good. This is for the common logic, right? Yeah. Yes. So, what does that tell you? Then we should just look at the equation here. Take the dA down here. This tells us that because epsilon just plays 0, dA by dA, d epsilon by dA is 0. Which means epsilon doesn't change. Which we know. Epsilon is the basic property of the vacuum. It doesn't change because of the vacuum. Right? So, in that particular case, you don't have to go through all of this. You just replace epsilon by the constant value in every lambda. And p by n is the same constant. Minus 1. Okay? Excellent. And so, this can now be integrated by dA by square root of 8 by g e squared epsilon by 3 minus k integral as equal to d. That's good. The answer is like in the system. Okay. So, let me first do the realistic results. Let me write the equation of conservation of stress tensor. Yeah? So, that's 3 dA. Well, let me write this as 3 by a epsilon plus p is equal to minus p epsilon. This is not the energy moment. Energy momentum conservation equation. Okay? So, the solution was log A is equal to minus d epsilon by epsilon plus p A by root by g by 3 e squared epsilon minus k is equal to d plus n. This is what the solution was in terms of the dynamic equation that referred to the dynamic data. So, it must be the same. It's the same. It's the same. And it's not the same. And so, this is close to, this is an absolute relation. There's your A is in terms of this thing. Resonance to what coordinates we're using. Everybody's using the table. But this of course changes depending on whether we're using the A or E term because this will change. And let's see how that goes. Here we would have used this equation. A dot squared plus k a squared by e to the 4. Okay? So, we would have had an extra a to the 4 here and an extra a squared. So, effectively the a squared, this answer of course, but effectively there is an extra a squared in the denominator of the expression for a prime. Right? So, a prime would have had an extra a squared. And so, an extra a to the a prime squared would have had an extra a. So, we would have a pi a squared root of 8 pi g a squared. So, 9 5 3 k is equal to x. And this completes our formal listing of the equations of course. In the next few minutes we're going to try to understand physics. What this means? Any questions or comments? Then learn an awful lot about the behavior of these equations without actually solving them just by looking at them. Let's first look at this equation. Let me work in the dot. Okay? So, let's look at the equation. 6 8 I mean dot pi a pi k epsilon plus. We'll get what's the reason? For all different, in the case we can. Yeah. For three k's are different discrete choice. For all discrete choice, we can get all these. All district is not dependent on k. What's the reason? When you have k? It's your independent choice from what's the equation of state. For given k, we will be given k. For given k? Yes. It's your independent choice from what's the equation of state you're on with. It's a discrete choice chosen once and for all. And then, all right, but we'll be seeing that these three different values of k have different kinds of physics. So, in order to try to get a feel for this, we're going to start let's look at this equation. The first thing that we see from here is that as long as epsilon plus 3P, in epsilon there's an energy density. So, it's reasonable that it's possible. If the matter in question has positive pressure, it's true many kinds of equations of state are not. The cosmological constant with one sign does not have positive pressure. It does not have positive pressure. But if the matter in a question has positive pressure, then the right hand side is possible. Okay? Does there appear another matter? What? Does there appear another matter? No. No. No. Okay. How it passively collapses and collapses? It collapses. You're part of the evolution of the universe. Okay? You know, the universe is not staying still. Partly because things are crashing. But once there you seem to be my impression of that question. But the universe is never ending. Okay? Right. Yes, you're right. Yeah. Negative pressures are to do with the analysis. Negative pressures are to do with the fact that it wants to. Pressures like positive pressures. But just like positive pressures, you can keep something with negative pressure in negative pressure by holding its ends. It's like a positive pressure is a balance between war and strength. It wants to expand, but it can't because it wants to. Similarly, negative pressure can be held by me. But what about it? What? Neuron is actually a pillar of the universe. Negative pressure, right? Can we sort of map that? Yeah. And you know, the matter that we're going to consider, we'll have, we'll have more. You know, we've got one sign which has negative pressure. Yeah. Supporters for a moment, when dealing with the right-hand side, where epsilon plus degree is positive. So then this acceleration is negative. It's the in-cuter figure expected from the table. What? It gave it the impossible. K, we can, this K is... Which is really positive. Do you really mean what you expect from that? You know, it's like the universe is panning, but it's a decelerating point. It's panning, but it's decelerating. Now, suppose this is the case. What was the curve for A versus D? A versus K, because it's decelerating, must look something like this. Because it's finite today in this equation. Because the curve looks like this, then at some finite time in the past, A must be equal to Z. We'll take that time for everything to be the origin. We'll set the origin to be at that time, where A is. So we get rid of these constants and the equations by making this choice, where A is equal to zero, when G is equal to zero. Okay, let's look at the next thing. Let's look at this energy conservation equation. It was more interesting. It was D... What was it? D, A, Q, epsilon, 90, C. The original form of the equation somebody had given. The first equation we wrote down, the energy conservation equation that we first wrote down, this was the energy part, and what was it? The other part. Minus P8R, something like this. This is the number, right? Did I get the same weight? It's minus 3G. My very Q. Something like this. Adjust this. I want to make this in the sign. I want to make this in the sign. Suppose the left and right hand side will see it. And suppose the right hand side will see it. Okay? Then this will tell us that epsilon might take 1 over H. So epsilon is decaying as A increases, like 1 over H. Now, if the pressure is positive, we know it not as positive. If the pressure is positive, that's our second assumption. Just to get a feeling, we're doing various assumptions. Just to get a feeling for all these equations. And if the pressure is positive, then the right hand side is negative. Which means that epsilon has to decrease faster than one day. So epsilon, as A increases, epsilon decreases into your positive equation. At least as fast as one day. Now, why do I want to know this? I want to know this because of this equation. Because of the equation here. 3 by H squared into A dot squared plus K is equal to 8 by K epsilon. As A increases, let's put D squared. Epsilon A squared decreases. Because epsilon decreases at least as fast as one day. Okay? So the right hand side here decreases. Now, what does that mean for A dot? Let's look at the three possible cases. The first case is that K is equal to 0. Okay? At some point, the right hand side hits 3K, at which point A dot is 0. So the equation of the universe stops and then the universe turns back. A dot. Okay? And then the universe turns back and comes back and turns. Take into a lot of details. If K is positive, if K is positive that the pressure is also positive. So then epsilon plus 3K is also positive. It is a generically true thing that the universe will expand, go to maximum size, stop and come back to absolute. So K positive, along with the assumption that the pressure is positive, always gives rise to any universe that is. Okay? Along with the same assumption for pressure is positive. This, as this decreases, as it decreases A dot, so this thing goes to 0, A dot square goes to a fixed constant value. The value of the cancels is K. A dot square is minus 1 and A dot square is 1. So then it cancels minus 1. In this case, the expansion of the history of the universe is this. Expansion, but the pressure is nothing. When K is equal to 0, in this case, as this goes to 0, A dot goes to 0, but never reaches. So it's shhhhhh to the three kinds of orbits in 1 by 1 potential. K positive is like the elliptic orbit. You know, you throw chalk up and if this was the center of the earth, it would just lost it. K equal to 0 is then not like a hyperbolic orbit. Just with the escape velocity. It just makes it in. K negative is like a hyperbolic orbit. It's not a hyperbolic orbit. Now K negative is like a hyperbolic orbit. It makes infinity in some velocity. This is a little like that. I mean, it's just an energy. It's not particle or whatever. It's a little like that. All of these analysis however, has been made under the assumption that the pressure is positive. How is the pressure when negative is positive? You could well have acceleration rather than deceleration here, because the epsilon rest will be equal. And all of these analysis could completely change. This right hand side here, looking at positive, A neither. All so fast, and analysis I made here, that this goes to 0. As a unit of large, it will not be true. And this can completely change. Pressure is negative and in addition, there is a gravity. So, what would explain that acceleration? What explains the acceleration? That's a good question. You could have made more sense in the sign that appeared in the first one. This has bothered me a little bit. I didn't like reading. Let me just check where it points to the other side of the line. There's probably something. It's something about the gravity that responds to negative pressures and strains. Somehow it anti-gravitates in the presence of negative pressure. Something like that. I don't have a good answer here for this question. Let me think about it. It's a very important question. One should have a way of thinking of this. I have to say that when I was working on these equations last time, I expected all these results to come up with the opposite sign. Sign of pressure, precisely. This is the sign I got. This is the sign I got, yes. From the line, from the way of pressure. Now let me get back to you. Let me get back to you. By Thursday. I would like that. There's some way of saying it, I think. I think it's right. I think there's some way of saying it that rationalizes it, but it's failed. It's just, you know what I mean? Yeah, okay. I'll get back to you. Okay. Now, in order to get even better sense of what's going on, let's take a few particular equations of states. Equations of state and solve. Exalicy. So we're looking to at least some of the equations of states. Okay. So, let's take the A sub p is equal to 0. Equation of state of dust. Galaxies. Okay. In this case, I really think it's very simple. This equation here tells us that 3 epsilon by a is equal to minus p epsilon by a. Okay. So that tells you that 3 log a plus log epsilon is 0, which means that epsilon is equal to some constant alpha by a. This saturates the dick. You remember we said that epsilon must decrease at least as fast as 1 by a cube if the pressure is not negative. In this case, pressure is 0. So it's such and such. So energy density goes as 1 by a cube. Does that mean that energy continuity is constant? Yes. Yes. Well, I mean, you know, energy as we know is a slightly tricky concept. But you see, what's going on here is that what you've got is dust. Okay. So you just, let's say we've got n units of dust. And each of which has one unit rest mass. What's going on is that the number of galaxies has been concerted. And so the total rest mass has been concerted. What the energy of the space has got is not one of these asymptotically flat spaces for which we had this nice discussion over that direction. I don't know of a good way of defining the energy of the space or even if there is this closed universe. A large collection of galaxies will have a very large collection of galaxies. Well, I think the only equation p is equal to sq. Then the slightly thin units of what I know, you know, you could think of it like maybe it's best to think of it as the limit of p is equal to epsilon to the alpha 5 and the limit alpha to sq. In that case, let's let's write it. So p is equal to sq. In that case, how would I know if you just associate a particular temperature with a particular energy density? Just like the usual formula. And you could ask what temperature is associated with this? It's something you're... I won't try to work it out here. We could easily... Here it's a well defined question. And then we could take the alpha goes to zero limit. What is the plausible answer? Either the temperature is not defined or the actual dust system we're looking at. In its local restaurant, there has no random motion. So I would suspect that if there's a well defined temperature, it's effectively in this situation. Either the equation of state is so similar or the temperature is in this particular case. Yes? Sir, in the limit of alpha going to zero and p going to zero... What? We wrote epsilon raised to alpha, right? Yeah. Oh. Thank you. I'm going to venture as p is equal to j. Yeah. So... In this situation, I think it's quite easy to actually work this out. But I won't try it. Okay. Great. So now let's... So we call epsilon equal to alpha by a cube. Okay, this is now the alpha. By the way, we can give alpha an interpretation along the lines that the hungation was suggesting. Because the rest mass of the system, total number of galaxies, okay, is as we suggested, is a cube in a 2 pi square root of x. So this is the total number of galaxies times the n-axis mass. So let's call that quantity wherever it is in. Total n-axis mass of our universe. So then, in this situation, what do we have? We have epsilon is equal to... So this constant, alpha, is m by 2 pi square root... So this is just the... So this is a three-dimensional sphere. We can work this out, but I think it's too fast. I think it's too fast. Okay. So epsilon is equal to... Yeah. So giving an interpretation to what that constant is, is this... Okay. Now let's put this relationship into this one. Okay. So we have integral dA over a square root of 8 pi g by 3 epsilon, which is m by 2 pi square times a cube. That's a square, so that's 1 by 8. Okay. Minus k. What is the mass corresponding to the total energy of the galaxy? Total rest energy. Total energy of the galaxies. They have the rotation. Yeah, it's the total rest energy associated with each galaxy. It's true. All the energy associated with the galaxy, that gives rise to some rest energy. But the energy of that collection, that I don't know how to define. Yes. Now let's look at this in cases. Suppose k is equal to 1. Okay. So suppose k is equal to 1. What do we get? We get... And let's call this whole thing something positive. Let's call the whole thing in on-screen. Okay. So let's get 8 pi, so 4 by 3 pi g. So k is equal to... So, okay, in general, this is dA by A, A naught squared, not that A naught squared c, we've got to call it A naught squared. A naught squared by A minus k, is equal to d. Is equal to d. This is... You know, I think that means you see a square root of 1 minus positive minus something. You make a trick to make substitution. Okay. So let's make the substitution Let's make the substitution. Okay. Now let's suppose k is equal to 1. Then we make the substitution. A is equal to A naught squared in a sine squared. And it turns out we can make theta by 2 by... Okay. So now let's see how this integral goes. So we have dA is equal to A naught squared. The 2 goes because sine theta by 2 cos theta by 2. A itself is A naught sine theta by 2 squared with this. Sorry. A was... I'm sorry, I'm sorry. A was this thing squared. So we got that right. dA was equal to A naught squared sine theta by 2 cos theta by 2. Yeah. Now this is A and then this becomes square root A. Square root A is A naught sine theta by 2 and then this quantity here is A naught cos theta by 2. So this here is parametric guys vanish. This is why we made the substitution. A naught is vanish. Okay. So we have d eta is simply equal to eta is theta up to some kind of constant. Theta in terms of A. Theta was A was equal to A naught squared into 2 minus cos theta. So it tells us that A is equal to A naught squared. This is what we know. Whatever it was. So 4 by 3 by gm by an addition of 2, 2 by 3 by m into 1 minus cos theta. This is the function of A. At eta equal to zero, we've chosen zero correctness. At eta equal to zero and eta equals 2 pi A is zero again. So as eta runs from zero to pi, the universe expands to its maximum. And then from pi to 2 pi, it shrinks back again. So this is our straights, this general analysis that we have. That I previously discussed. One more quick thing. Let's also determine the relationship between eta and t in this analysis. So we have v eta by v t which is equal to 1 by A. 1 by A is equal to 2 gm by 3 which implies that t is equal to integral of this to eta and then the integral of cos eta is the minus sign eta equal to 1 minus cos t. t pi by u or what? This type of problem. We know in our pattern, we know how A depends on eta. So let's start with A and t. So as eta goes from 0 to pi, that is just constantly increasing. But about the same things. How the t clock runs? How the t clock runs? Changes. Let's plot these things. What was A? A looked like this. On the other hand, t looks like it's constantly increasing. It starts off with 0 slope. Probably accelerates a lot of things. These two relations allow you, paradoxically, to determine I mean, is a parametric determination or A is a function of t? Or it would be messy if you tried to actually determine things out of it. This is one advantage of eta. It's often easier to solve the equation in eta. And then raise rest terms of that. Okay. Now, I wanted to have a quick comment. Maybe I do a K-negative case first. And then we'll have one of the quick comments. The K-negative case can be not exactly the same, okay? I'll just write it. Something I should do is check I got 2 gm by 3 pi tonight. So 2 gm by 3 pi. This is quite amazing. Let's just comment on the K-negative case. In that case it turns out that this relation goes to 2 gm by 3 pi into cos h to the power minus 2. Okay. And this goes to sin h to the power minus 2. Sine h to the power minus 2. Okay, so there's nothing about this in science. Okay. Of course. But qualitatively this is, of course, very different. As eta goes from eta equals 0, once again it starts at 0. But as eta goes to increases with our power. And you can also like I understand about the big idea of time this. You see, if this A increases exponentially fast at large eta in eta. But t also increases exponentially fast. And in fact with the same coefficient 2 gm by 3. Right? So that dA by dt asymptotically is 1. Exactly as we predicted. You remember, general grounds we predicted that dA by dt asymptotically is 1. Okay. So this is an explicit solution showing what's happening. If you have dust in an open universe. By the way, the universe k equals once for the closed universe. For obvious reasons. A spatial manifold is a spherical object. The universe with k equals minus once for the open universe. Also got all these reasons. You can knock out that spatial section. Okay. So the total amount of rest mass in the open universe is equal to the amount of mass in the galaxy. In the open universe is equal to just because the volume is reduced. Okay. And we can also work this out for the special case of k equals zero. Okay. That gives rise to power of type behavior. I'll leave that for you to work out as a basis. I'll leave you to work out as a basis. It's simpler than the integrals we've got here. Because in that case, this term is not there. So the integrals aren't even symmetric, they're just integrals of power. And so you see immediately that we get dA by E squared. So you get 1 by A as the integral. So it can go like 1 by A and A goes like 1 by A. Yeah. That's the case of dust. Now, very briefly do you have anything going on now? I just realized the next, we're missing two classes next week. Because Monday and Thursday are always. We have to do something about that. I may request you to come on. But anyway, I'll take a little extra. Thank you. Okay. So we discuss the case of dust. Very briefly let's talk about what you see. In the actual universe as far as we understand it today it looks like stress energy is a combination of three kinds of matter. Okay. Of dust which is a cold matter including dark matter which isn't a matter. Radiation. Which is an negligible fraction of what's going on today but it was a very important part of the universe in the early part of the universe like photon radiation. And it looks like cosmological constant. Okay. So that's why I wanted to take you through the analysis separately of dust radiation and cosmological constant before trying to put all these three together. Okay. So now we want to discuss the case of radiation. So when we put these things together the exchange between them also has to do with them. The exchange between them has to be taken into account. Yes. However cosmological constant doesn't exchange with any it's just constant. So it's the exchange between matter and radiation. Now matter and radiation couple once matter at the early part of the early age early epoch in the age of the universe up to a few hundred thousand years the universe was ionized. So the matter was ions of, you know, ionized hydrogen electrons and protons from here on and then the matter and the radiation were interacting very very immediately the important way. But after matter we ionized, sorry, we unionized, you know, we combined. Then the matter was essentially neutral and essentially the matter and the radiation have formed two non-interactive fluids then after. So it's true that during the era where things were ionized you have to work harder. But after recombination these were essentially two non-interactive fluids. Of course the cosmological constant always is non-interactive. So does this Well, if it's really a constant if it's just a lambda term it does not interact with it. If it's a what? But can't be sure that it really is a constant. Yes, but can't be sure. Well, you know, the question is whether you know what lambda, you need to give it dynamics. I'm asking, does that does any sort of dynamics come from here on? No, no, you see in the in the Lagrangian I have a general relativity. If lambda were anything but a constant it wouldn't have a break if you're more physical. The only way for lambda not to be a constant would be for