[SOUND] So again, we are gluing this Schwarzschild metric outside of the body of the star, To their metric inside the body of the star which is equal to detail squared- a squared d chi squared, sine squared d omega squared. And we have described this metric in the following sense. That this a and tau as functions of parameter eta behave like this. Cosine of eta and tau (eta/g) is ao/2 (eta + cosine of eta ) and eta is ranging from 0 to pi. And that eta equals to 0, the collapse begins. As eta reaches pi the body of the star shrinks to 0 size. So now we want to glue these metrics to find the other parameters of the metric. So within this metric, the surface sigma, the surface of the star, sigma, is just one of the surfaces inside a free sphere. So the one whic corresponds to chi equals chi 0, so the metric on this surface is just ds minus square, so it's a world surface in this case, the metric on this world surface is just dj squared minus a squared of t sign squared chi 0 d omega squared. So this is, Metric on the surface sigma. Restricted to the surface sigma. Over which chi is fixed. At the same time, the metric on this surface sigma From the point or this metric is just ds+ squared sigma equals to 1 minus rg divided by r dt over d tau squared. -1 divided by 1- rg over r, (dR over d tau) squared times d tau squared. -R squared (tau) d omega squared. To glue this metric to this one, to equate them, they should be equal because they are the same metric on the same surface. One is induced from this one, the other one induced from this one. To equate them we have to have the following relation, that R(tau) = a(tau) times sine chi 0. And 1, so this is just equates this to this. So basically, this sets the radius. And 1- rg over R dT over d tau squared minus 1 divided by 1- rg over R. dR over d tau squared equals to 1 So this other relations, basically knowing this, knowing this we know R of t, and knowing R of t from this equation we can find t of tow. This way, we will find the volume of the surface sigma. Well, this equation can be solved for T of tau, T as a function of tau, so the solution is obvious, so T.=, well for T. Can be solved of course, is just square root of R.squared +1-rgor divided by 1-rg or r. And well again, T.. And R.. This is justT. And this is just R.. I hope it's obvious. So this equation, if we know r of tow from here, we can find t of tow by other integration. Well, this can be done, but let us just stress here that R, as R goes through rg. So as the surface of the star approaches it's rotational radius, we can [INAUDIBLE] find the solution of this Equation. So first of all this term becomes approximately zero and can be neglected with respect to this term. Because R is a function of proper time, as a function of [INAUDIBLE] is not zero As a function of proper time is not 0 in the vicinity of the horizon, so the surface goes with nonzero velocity. So we can neglect this in this limit. And this equation approximately acquires the following form, dT is equal approximately minus dR divided by 1- rg over r. Minus sign here, of course, appears because due to the fact that the dR is less than 0 because the surface of the star is shrinking as dt as the time goes by as dt goes To reach infinity. Due to this we have a minus sign here. This can be solved approximately that R as a function of T is approximately equal to R G times 1, plus exponent of minus T divided by G. So this is as T goes to infinity. As T goes to infinity because as R Goes to rg, T as R big goes to rg T goes to infinity. So we encounter the following situation that despite the fact that it takes finite proper time for the star surface to cross the radius, to cross the horizon, so it takes finite amount of this for this to cross rg. But it takes infinite amount of time for outside, from the point if you're an outside observer, for this star surface to cross the horizon. So star surface [INAUDIBLE] is slowing down and approaching the surface, their short shield radius. That is the situation we have encountered already for the particles in the lecture 6. So again, what we are doing, we are gluing this metric to this metric. Second derivative of this metric and this metric are related to the richer tensor, and are fixed, by Einstein equations of motion. It means that this second derivative of this metric is related to the energy,momentum tensor in the corresponding part of this space time. By gluing the metric itself, this and this, we were able to find A of tau through the parametric relation. From gluing, well, by solving, actually, an equations here, we found this A of tau. By gluing this to this from this, we can find R of tau, and T of tau. And we explicitly presented the solution for T of tau in there, infinity. As this time or this time goes to infinity. So this is so far. What remains to be done is to glue first the derivative of this metric to the first derivative of this metric but it demands some differential geometry for the surfaces of the space times, which goes beyond the scope of our lectures. But the result of this gluing is pretty much all this and can be understood on general physical grounds. The result of that gluing just leads to the fact that rg, this rg standing here is related to that what is standing there as follows, two kappa. 4 pie over 3 R cubed of tau times R of tau constant. Well, what happens is that why this is constant is because we have obtained that a cubed of tau times row of tau is constant. Well, what do we see here? That this quantity standing here remains constant, as it should be as we explained on general grounds at the beginning of the lecture, and it's appropriately related to the mass of the ball, because this is the volume of the ball times its radius. And this is 2 kappa, as it should be. So perfect. We have obtained the perfect relation between two metrics. Now, we are ready to draw the Pennrose-Carter diagram for this metric. So basically what do we have here? We have the following situation so outside of the sigma this is what we call sigma tau so the surface of the sigma of tau. So we have the following situation we have the outside of we have basically have the same situation as this is shear shield space time so for this shield metric. To remind you, we have there in cross coordinates we have the V and U, so we have the singularity r equals to zero, this occurs one into the singularity. We have the surface sigma. So the surface sigma has the following. This is sigma, and it has the following behavior. Before t = 0, so this part is the same as we have encountered in the previous lecture for the star, and after t = 0, we have this collapse described by this situation. So this is sigma, this is sigma. So this metric is applicable here in Cruzco coordinates, we have this and in here, we have to glue what we haven't counted in the previous lecture. And here, in this region, we have to glue diagram following from this metric in here. So in here from ds minor squared, so to say. So let us draw what we obtained. Before drawing that, let us draw diagram for this guy. To do that, one can establish the following fact that this metric can be represented like this. Exactly with the same parameter eta, as we have encountered before in this lecture. So, this is compact. This is compact, because eta and kappa are ranging in compact regions. We can drop off this metric and consider this spacetime itself. But we have to bear in mind that f is ranging from 0 to pi, which chi is ranging from 0 to chi0. chi0 is Remember the k0 defines us the surface of the start in this metric. As a result this is just a part of the square,where this corresponds to k0. In all Gluing this to what we, so before here, I mean here in this region we have what we have encountered in the previous lecture for the star. Here we have what we have for the same as for the Schwarzchild for the black hole. And here we have this guy. In all, we obtain the following picture. The following diagram. Diagram for this guy has the following form. So [SOUND] So this is the region inside the star. This is sigma, the surface of the star. This is the moment t = 0. So below this region we have the same diagram which coincides with the same diagram as for the star. And this is r equals to rg. And this is r equals to 0. So this is singularity. This is square plus, this is square minus. So this is a Penrose diagram for the collapsing star. For the collapsing star. Perfect, now we see that for this case We do not have white hole we have a black hole. We have a black hole we have event horizon we have the region here from which no one can escape the script was. So everyone appear finding himself here will inevitably end up this similarity. But we are considering time dependent solution There is, of course, time reversal of this solution. Time reversal of this solution is given by a of eta equals to a zero over 2(1 minus the sine of cos eta. And = 2 a0 over 2 so we have this illusion where there was a plus sign here and now we have this solution with ranging from 0 to pi. So that solution and plus sign as describing the collapse. This solution is describing the reverse process, f = zero, we have zero size of the ball and it inflates, becomes bigger and if at the moment equals P, pie, something stops this inflation. This inflation and the radius remains fixed after that moment. For this solution, we obtained just the time reversal of this diagram which is just the flip of this diagram over the horizontal line. So this is the situation we have predicted in the previous lecture. Saying that for every solution of Einstein's equations we have its time reversal in case the solution is time dependent. [MUSIC]