[MUSIC] So, to move further, and to solve Einstein equations, let us make the following change of [INAUDIBLE]. Let us change dt to a of eta d eta. Then, our metric that we are discussing will acquire this form. [d eta squared- dr squared 1- kr squared- r squared d omega squared]. And the metric has this conformal factor, and is conformal to this metric. This eta is called conformal time. Unlike this t, this eta is conformal time. We actually have encountered this eta in the lecture on Oppenheimer-Snyder collapse. That is exactly that eta which we have seen there. So, then, from the equation 3, a dot squared + k = 8 pi kappa rho a squared, which is (0,0) component of Einstein equation. We obtain the following relation. We obtain that dt = plus, minus, because we have to extract square root, times the da square root of 8 pi kappa over 3, a squared rho- k. Well, we have to find rho through a, plug it here, and then integrate, so to say. And plus and minus corresponds to the solutions that are related to each other by time reversal. Now using this relation and this relation, we find that eta, eta is plus, minus da over a square root of 8 pi kappa over 3a squared rho- k. Now we're going to find from here a of eta and then from here t of eta. And so our goal is to find a of eta and then t of eta from these two equations. And to use here the relation between rho and a for six different situations, for k = -1, 0, 1 and omega = 0, one-third. So for two types of equations of state and three types of special sections. So let us start with the case of k = 1 and omega = 0. This is a case actually we have encountered in the lecture on Oppenheimer-Snyder collapse. Then from the fact that rho times a cube is constant, this is the case as we remember for this situation. We denote this constant as M over 2 pi squared. We use here a bit different notation from the lecture in Oppenheimer-Snyder collapse. So to say this M is the total mass of the universe, because this is related to the total mass. And it's so to say remains constant through the evolution. So, if we plug this into here and take the integral, we obtain, in this case, that a of eta = 2 kappa M divided by 3 pi (1- cosine of eta). And from this relation, we can find that t of eta, of course, we have fixed the integration constant here and here to fix this. That sets up the initial conditions. 2 kappa M over 3 pi times eta- sine eta. So one can see this is a solution we already encountered in the case of Oppenheimer-Schneider collapse. And then the physical meaning of eta is now obvious, etc., etc. So now for small eta, this solution describes the following situation. For small t and small eta, so when we start the evolution of the universe, a of t, as follows from here, is proportional to t to the power 2 over 3. Then, so evolution starts at eta = 0 when a = 0, a = 0. Then, as eta goes by, then as eta goes by, as eta = pi. And hence t = t 2 Kappa M over 3. The universe reaches its maximal size, = 4 kappa m divided over 3 pi. And then shrinks back. So we have the following situations, the scale of the universe expands from zero and then shrinks back. So this is eta = 0, eta = pi eta = 2 pi. So and this is a. That's the situation we encounter for this case. So the universe expands and then shrinks back. This is what we encounter in this case. So we continue our [INAUDIBLE] of the metric ds squared a of eta squared times d eta squared- dr squared 1- kr squared- r squared d omega squared. Where dt = a of eta, d eta. This is the relation between conformal time and regular time. And we'd restrict our consideration to the case of plus sign here. da square root of 8 pi kappa over 3a squared rho- k. And also hence to the plus sine for the eta da over a square root of 8 pi, k kappa over 3 a squared rho- k. So we have already considered the case when k = 1 and omega, the relation between rho and p is equal to zero. In that case, all sections, so we continue now our consideration with the case of k = 1 and omega equal to one-third. This case again, one more clarification to the previous situation and to this one. For this case, I remind you that every special section in this metric is a sphere, three sphere. And we have encountered the situation that the size of this three sphere, starts its expansion from zero radius, gets to some final value and then shrinks back. In this case, we again will encounter this situation with a little bit different behavior of the radius. So because in this case, unlike the previous situation, in this case we have rho multiplied by a4, which is equal to constant rather than a cube. Let us denote this constant as 3a1 squared over 8 pi kappa. And this is constant. This is for convenience reasons. If we plug this constant into here, then we immediately obtain that a of eta is a1 sine eta. And t of eta = a1 (1- cosine of eta). So as we start at eta = 0, t = 0, a is 0. So, again, the universe starts its expansion, starts its expansion with zero size. Expands to a finite size as eta goes by, at eta = to pi over 2. It gets finite size and then shrinks back to zero size. All this consideration correspond to plus sign. Minus sign would correspond to the time reversal of this solution. The same is true for the case of k = 1 and omega = 0. And let me say one more thing that for small t, when t is of order zero for the very beginning of the expansion, it's not hard to see that a of t is proportional to square root of t. And rho is proportional to 1 over t squared as follow from the formulas. Let us continue with the case of hyperbolic special sections. In this case, k = -1 and we start our consideration with the omega = 0. So in this case, every special section is a hyperboloid. And for this case, we encounter that r times a cube is constant. Let us denote this as 3a 0 over 4 pi kappa. Well, this is just a notation, this is constant. The reason for the notation, well for example, the reason for this notation is just because a1 appears over a amplitude here and the same we'll encounter here. If we use here this and plug it here, we obtain that a (eta) = hyperbolic sine, a0, this is a0, times hyperbolic sine, cosine, of eta- 1. And t (eta) is a0 times hyperbolic sine of eta- eta. Well, once we have got this from this equation, we can find t of eta. Now, we encounter the following situation that the universe starts its expansion from zero size and continues at eternally. It doesn't shrink back, unlike this iteration of the closed universe. And again, it starts at homogeneous expansion from zero size at t = 0. And at the beginning of the expansion, when t is very small, a is proportional to the t two-third. In fact, it's actually similar to the case of k = 1 and omega = 0. The expansion also starts with such a behavior, and we going to continue with the other cases, no. So, we continue our discussion of this space time for several different situations. We already discussed the case of k=1 and both omega=0 and omega = one-third. And we already discussed the case of k = -1 and omega = 0. We continue our discussion with the case of k = -1 and omega = one-third. In this case, we encounter that rho times a4 is constant. We defined this constant to be 3a1 squared over 8 pi kappa = constant. And then we obtained plugging this into here and then into here. We obtained that a(eta) = a1 hyperbolic sign of eta, and t(eta) = a1 times hyperbolic sine of eta- 1. Again, we encounter the eternal expansion of open universe. Open because for this case, for this case special sections, hyperboloids, not spheres, they are open. So eternal expansion, [COUGH] and it starts also with zero a at eta = 0. We have a = 0, t = 0. And actually, at the beginning of the expansion, when t is very small, a(t) is proportional to the square root of t. You see, as a beginning of the expansion in every case for the radiation and for the dust, we encounter the same kind of behavior. I mean, for the dust, it's the same independently of the type. And for the radiation, it's the same independently of the type. But they're different. For the dust, differs for the radiation. Now, for the case when k = 0, and omega = 0, one can directly integrate this. Directly integrate this, as one can see. And in this case, well, first of all, we encounter rho a cubed = constant. And then, as a result from the integration from here, we encounter then a(t) = constant times t two-third. So for this case, this is always like that, this relation. For the other cases, such a behavior for the dust is only at the beginning of the expansion. And finally, when k = 0 and omega = one-third, we, as usual, have rho a to the fourth = constant. Using this directly here in this equation, we obtain that a(t) = some constant times square root of t. Again, always we have this relation for this case. While for the radiation, for the case of k = to 1 and k = to -1, such a relation is valid only at the beginning of the expansion. Now let me say a few words that at present our universe, with a good precision corresponds to the case K=0 and omega = 0 up to one important issue that recently, well, several comments are in order. First of all, at the beginning of the expansion of the universe, we expect that the predictions say there was an error when this was valid. And moreover, there was an exponential inflationary expansion. There are signs saying that. And also, now, we see that the situation is not quite [INAUDIBLE]. Because there is a non-zero cosmological constant, so-called doc energy, which describes exponential expansion. Again, in all this lecture we have assumed that lambda cosmological constant is 0, while inflationary expansion corresponds to the case when lambda is not equal to zero. And this situation will be described in the next lecture, not in this one. So, this inflationary expansion is important to describe our present universe, apparently observational data show that lambda is not zero but very small. And also there are some signs that at the beginning of the universe at early stages lambda was big and not zero. And that is original inflationary expansion and that will be discussed in the next lecture. [MUSIC]