[MUSIC] So in this lecture, we continue our study of the black hole and star like solutions. In particular in this lecture, we're going to study so called Oppenheimer-Snyder collapse of a star into black hole. So we assume the following situation. That before, that for t less than 0, we have an ideal ball. Which is standing at the radius R0. And there is inside it a nonzero density and nonzero pressure. Say pressure is due to thermonuclear reactions that are going inside the black hole. And for t greater than 0, the thermonuclear fuel is spent entirely, and the collapse starts. So the ball starts to shrink. And well, before t equals to 0, the solution is exactly the same one as we have been considering in the previous lecture. After the moment t equals to 0, we consider the approximation that was used by Oppenheimer and Snyder that p = 0. So we have a pressureless dust, which means that energy momentum tensor is rho times u mu u nu. And we assume that the dust here is spatially homogeneous. But due to the collapse, its density is time dependent. So this is a function of proper time only and nothing else. And outside of the star, always we can see that T mu = 0 and lambda everywhere is 0. Cosmological constant is 0. Well, we consider ideal spherical symmetry. And even during the collapse process, we assume that there is no any violation of homogeneity of the dust distribution inside a star. And ideal spherical symmetry. Of course, this is a very crude approximation. Because any violation of the spherical symmetry or the homogeneity of the dust distribution. Dust density inside the star will grow with time due to tidal forces. If there is some violation of the spherical symmetry due to tidal forces. Because this part of the body is attracted to the center stronger than this part while this perturbation is going to grow. This is due to tidal forces, of course. So we assume some very crude approximation. But then this crude approximation, we have the following situation. Because of this ideal spherical symmetry outside of the body of the star. We always assume to have Schwarzschild metric due to Birkhoff's theorem. So outside of the of the star, we always assume the metric. Plus corresponds to that, this is a metric outside of the star. We always assume to have the following metric. dt squared- dr squared / 1- rg / r, this one,- r squared d omega squared. And rg = 2 kappa M, where the M is the mass of the star. The total mass of the star. So due to the ideal spherical symmetry, there is no gravitational radiation, no gravitational radiation. Why? The argument behind this fact is the same as we have been discussing at the very end of the previous lecture. So let me repeat this argument again. Assume that you have an ideal spherical body. With the homogeneous electric charge distribution inside it. And assume that this ball of the charge starts to shrink, acceleratively shrink. In such a way that the homogeneity of the charge distribution and the spherical symmetry is not violated. Because this ball, independently of its radius, creates the same Coulomb field outside of its body. We have nonzero electric field but zero magnetic field. As a result, there is no radiation. Despite the fact that during this shrinking, there is an accelerative motion of electric charge. Despite the fact that we have an accelerative motion of the electric charge. There is no radiation in such an ideal situation. The reason for that is the following. That to have electromagnetic radiation, one has to have at least dipole moment changing in time. While for the ideal spherical symmetry, not only dipole, but all the momenta. All the multiple momenta with respect to the center of the ball as zero. With respect to the center of the ball as zero. So this is a situation with electromagnetic radiation. With the gravitational radiation, the situation is even more refined in the sense, well first of all, there is no radiation also. Even can be guessed within Newtonian approximation. Because independently of the radius of such a ball. Outside of it, it would create a Newtonian limit, just Newtonian potential. Independently of the same Newtonian potential corresponding to the same mass. As if it is situated in the very center of the ball. So this is the essence, again, of some kind of revelation of the Birkhoff's theorem. That independently of the size of this guy, if the spherical symmetry is never violated. We have outside of the ball the same Schwarzschild metric with fixed mass. Intuitively should be understood that if there is no energy fluxing out. The total energy of this ball remains constant. And the total energy of this ball is just its mass. Okay, that's the situation we have. But at the same time, the surface, the radius of this ball is the function of time. The radius of the surface of this ball is a function of time. So we assume that the world-surface of the surface of the ball is as follows. It's the following vector. So it's [T(tau), sigma(tau)]. And we should stress that throughout this lecture. We assume that we denote by the same sigma. By the same sigma, we denote two different things. Both the world-surface of the surface and the surface also. So world-surface And surface of the bowl at the same moment. I hope it will not cause confusion. So, in spherical code and this guy occupies all the angles theta and phi, which defined this metric, and R as a function of proper time at the initial moment of time t0 is R0. Which is greater than rg, such that we have an actual start rather than a black hole. So the radius of the star initially is greater than its gravitational radius. This is the conditions. Inside the star, inside here, during the collapse process, in this moment. Before t=0, the solution is the same as It was in the previous lecture. But after t=0 here we have the following metric. ds- squared, so inside the star, minus for the inside. dtao squared- a squared of tao( d chi squared + sine square of chi d omega squared). So this is, what is this? This is a metric. This metric describes at every given moment, at every dt=0, Tau=const. So at every given moment of time it describes three sphere. So this is a two sphere, S squared, this is two sphere. Here also, S squared two sphere. And this is total metric is metric on the unit 3 sphere. So this factor, this metric, together with this factor describes 3 sphere with the radios a square of tau, I'm sorry. And this radius is changing in time such that a gets smaller as time goes by. And R of tau gets smaller as times go by. So we have two different metrics inside the star body and outside and we have to glue them across this surface. So first of all, second derivity for this metrics are fixed by the Einstein equations of motion and are related to the energy momentum inside and outside of the body of the star. We have to glue the values of the metric, across the shell and their first derivatives. That's what we're going to do now. And we will find the relation between the parameters of this metric, including this radius. And this parameters of this metric. So outside the star body, we have the flowing metric. And the Schwarzschild metric. And inside the star body, we have this metric. d tau squared -r squared times the metric on the unit 3 sphere. Where chi is running from 0 to pi. Then this metric covers the C sphere. And A is a function of tau, only due special homogeneity. So we know more or less everything about this metric, I hope. And now we're going to study this metric a bit more closely. Well, I should stress that we're going to encounter this metric in the lecture number 11. And we will study more intensively then. Now, we just give a few ideas of how to work with this metric. So, it's metric components are g tao tao, is one. Obviously, g chi chi= -a squared, g theta theta = -a squared sine squared ki. And finally g phi phi=-a squared sine squared ki sine squared theta. So one can easily calculate Christoffel symbols for this metric. The relevant for our discussion here are the following ones. Gamma 0 ij is a.ag twiddle ij. And gamma i 0 j equals to a dot a delta ij. Where i and j run from 1,2,3. And a.= da/ d tau. And here J ij twiddle is just the metric on the unit 3 sphere. So it was just this metric, it's equal to D chi squared+ sine squared chi d omega squared. And, of course, there are non-zero components Hence. Due to this metric, we have non-zero components of the Christoffel symbols like this. Which are just the same as for unit 3 sphere. And we don't' need the explicit value in this lecture. So we discard them throughout the discussion here. We give this metric in great detail, again, in the 11th lecture. So all we need is to know this component, so the Christoffel symbol. All the other components either trivial it follows due to symmetry. So the Christoffel symbol from this ones, or just 0. So as a result, due to this values of three special symbols the zero zero component of the Einstein equations of motion. So it was like this. 3/a squared 2. A dot squared + 1. So this is just equal to this for this metric and that. And also due to the fact that t min nu is equal to rho of tau, u mu, u nu, in this metric, the dust is fulfilling a free fall. So it's basically in the coordinates, when d tau is fixed, the dust is stationary. Well, it means that u mu for velocity in this metric is just 1, 0, 0, 0 for the dust. And as result, the only known T l, non 0 component of the energy momentum tensor in this frame is T0 = rho of tau. As a result, we have the following equation. So, for this metric inside the star, we have the following. This Einstein equation, this component of the Einstein equation Has the following form as a result. Rho a squared, these components of the Einstein equations, because we have this as 0, are trivial. So they just establish trivial relations 0 = 0. And as we explained in the lecture three, instead of some of the components of the Einstein equation, we can use energy momentum conservation. So we're going to use it here, in the present situation, for the metric tensor that we have. It has the following form. (Rho u mu, u nu) + gamma mu beta mu rho u beta u nu- gamma beta nu mu rho u mu u beta. And we have four equations for each value of mu. We have four equations. And hence, if nu is equal to 1, 2, 3, 1, 2, 3, we have trivial, again we get trivial relations as 0 = 0. While for nu = 0, we obtain the following equation that rho dot + 3 a dot over a rho = 0, and where rho dot is the time differential of rho. And as a result, this equation is equivalent to d tau of rho a cubed = 0. So, which is obvious equation, which states that, as a is shrinking, as the radius of the sphere is shrinking, its density of the dust inside this sphere behaves as a constant times a cubed. So reducing, density is increasing, radius is decreasing. And as the volume decreases, the density is increasing. If we fix this constant, as a result, we obtain the following. Rho a cubed = 3/8 pi kappa a0. And as a result, the solution of this equation, the solution of this equation can be represented as follows, parametrically represented as a(eta) = a0/2 (1 + cosine eta) while tau(eta) = a0/2 (eta + sine eta). Again, we're going to study this metric in greater detail in 11's lecture. And we'll understand the physical meaning of this parameter eta, which runs from 0 to pi. So eta is running from 0 to pi. And the physical meaning of this parameter will be studied in greater detail in that lecture. But now we can just show that this indeed solves this equation. In fact, it's not hard to see from here that d tau/d eta is nothing but a. We just stuff the differentiation. As a result, from here one can see that da/d tau, which is da/d eta d eta/d tau = da/d eta, 1/a. Using this, this equation can be represented. So this equation can be represented in the following form. Da over d eta squared + a squared = a 0a. This is nothing but oscillator type equation. And one can easily show that this guy solves this oscillator type equation. Now what does this solution describe? It describes the following situation that 1 eta = 0. Eta = 0. Tau corresponds to 0 also, which corresponds in this coordinate, which corresponds to the time = 0. This is a moment of time when the collapse begins. This is a moment of time when all the thermonuclear fuel is spent. Pressure is switched off, and the collapse starts. At this moment, a(eta) = a0, as can be seen from here. But as eta reaches pi, as eta reaches pi, a goes to 0. So the sphere shrinks to 0, shrinks to the singularity. And this is reached within finite proper time. This tau goes to pi a0/2. So within finite proper time, the surface of this sphere crosses the horizon, crosses r = rg. We didn't yet see that. We're going to see that in a moment when we'll glue these two metrics. And the body of the star shrinks to zero size within finite proper time and reaches the singularity. So this is the solution describing the collapse. Now, we are ready to glue this metric to this metric. [MUSIC]
