[SOUND] [MUSIC] So we have obtained the following solution of GR equations or motions for the matter content and the consideration. dr squared 1- 2 kappa M of r divided by r- r squared d omega squared. Where I remind you, M(r) = 4 pi integral from 0 to r, dx times x squared rho(x). And d new over dr new is a gravitational potential. dr is defined from the falling equation, 2 kappa M(r) + 8 pi r cube kappa p(r) divided by r [r- 2 kappa M(r)]. And there is also the condition of the stability of this solution is that, r is greater than or equal to 2 kappa M(r), so there is no horizon. This guy is always has appropriate sign and the [INAUDIBLE] of condition for the aerodynamic stability of the star follows from this equation. And the conserved energy momentum conservation and is given by this equation. (p + rho) kappa M(r) + 4 pi r cubed kappa p(r) divided by r [r- 2 kappa M(r)]. So now, well, as a side remark let me state that this equation on the relativistic limit. When p is much smaller than rho and kappa M(r) is much smaller than r this equation reduces to dp over dr. Approximately, equal to rho kappa M(r) divided by r squared. Okay, now, let me stress a few points about this solution. First of all, when r is greater than, bigger the radius of the star where p = 0 and the rho = 0. This just reuses to the Schwarzschild part of the metric. And in this region, when we can solve this equation, and the result of this solution is nothing. But that exponent of nu = 1- 2 kappa M(R) divided by r.. So this equation indeed, again, reduces to the Schwarzchild solution outside of the star. And this is another revelation of the Birkhoff's theorem. The theorem that declares that static, spherically, symmetric solution in result matter. With 0 energy momentum tensor is always Schwarzschild. And I remind you that outside of the star, energy momentum tensor is 0. So now, to go further. So the solution becomes different from Schwarzschild only inside the star, as you notice. Where the energy momentum tensor is not 0 anymore. So to go farther, we have to specify the equation of state. So we have to specify the dependence p(rho). The simplest equation of state is incompressible fluid. Incompressible fluid is when rho is constant = rho 0, constant, is independent of r. In this case, it is easy to integrate this equation. And to find that M(r) whose function is just 4 pi over 3r cubed times rho 0. And one can solve this equation for r less than R. The solution of this equation is given by the falling function p(r). p(r) is rho 0 times ratio square root of 1- 2, sorry, kappa.2 times kappa M divided by R- square root of 1- 2 kappa M times r squared divided by r cube. And divided by square root of 1- 2 kappa M r squared divided by r cubed- 3. 1- 2 kappa M R where M here, M, which is a total mass of the star, so it's just volume of the star times rho 0. Now, from this equation, one can see that the pressure in the origin. p at r equals to 0, blows up, goes to infinity, if R = 9 over 4 kappa M. If this is abate, so from this, one can see, well, this is a rather unphysical situation. That the pressure inside the body of the star blows up. So we don't want to have this sedation, this is unphysical. It doesn't realize in real situations. So we expect that the maximal mass of the star with a given radius is given by, 4 divided by 9 kappa R. So if mass exceeds this value, the star becomes unstable, because the pressure inside it blows up. So the star becomes just unstable and this is not realized in real situation. Now, everything is ready to draw the Penrose Carter diagram for spherical star solution. So we have to separate the relevant two dimensional part of the metric which as one can guess is just this one. dt squared- dr squared 1- 2 kappa M(r) divided by r. So and we represent this as follows. So this is just exponent up to conformal factor, we represent it like this. dt squared, well, I mean, it's exact transformation of the conformal factor. And I just separated the conformal factor. dr squared divided by exponent of nu(r) 1- 2 kappa M(r) divided by r. And now, recall that this is everywhere regular quantity. The one which is standing in denominator. So we represent this as follows. Exponent of nu(r) times dt squared- dr bar star squared. So this is nu tortoise coordinate so this we noted as dr start bar squared. And to find it, one has to take the integral with this functions. But it is important that by proper adjustment of the integration constant. We can make such that r bar star will be ranging from 0 to + infinity. Notice, that the regular tortoise coordinates, the ones we have already encountered are ranging from- infinity to + infinity. But in that case, the denominator was singular at the horizon. Here we don't have a horizon. This thing never gets singular, this thing never gets singular. So we can adjust it to be arranging like this, and this is important. Now, dealing with these quantities and making the change to the coordinates t +- r bar star. And making the same transformation as we did at the lecture on Penrose diagrams. One can easily find that the Penrose diagram for this spacetime is just as follows. Is just half of the square, half of the square. Where we have the following situation. We have this is skrye +, this is skrye -. This is asymptotic infinity, this is past infinity, I'm sorry, past infinity. This is future infinity and this is a body of the star, so this is r = R. So this is internal side of the star. Compare this to the Penrose diagram of the Minkowski space in the coordinates t and r. Where r is greater than 0 is like this, it's just half of the square. And you remember, that for the t and x, where x is ranging from- infinity to + infinity, we had this Penrose diagram. So for this case, we have only half of it. And for this case, similarly to this situation, we encounter this diagram. So as r goes to infinity, r star bar becomes approximately = r. So this diagram is just coincided asymptotic infinity. But inside, closer to the body of the star, they become a bit different. But geometry of the Penrose diagram is more or less the same. So cashing surface here, this is constant time. Constant surface looks like this and how you call it, the world line of an observer which is fixed above the star is something like that. So this is just r constant surface. So the Penrose diagram is just obvious in this situation. And not much different from the Minkowski space. [SOUND] [MUSIC]