[SOUND] [MUSIC] So, the obtained energy momentum tensor, t mu nu equal to rho u mu u nu plus p u mu u nu minus g mu nu. This energy momentum tensor, we will want to use to solve that problem that I described at the very beginning of this lecture. So, we want to consider spherically symmetric ideal bowl, where inside it we have energy momentum those are equal to this and outside t mu nu equals to 0. So the radius of the bowl is at R. So, we expect in such a situation which is time independent, we expect to get static metric. Whose components are independent of time, and it is diagonal. Static means invariant under translations and time reversal, and we expect that this in this situation, is spherically symmetric so rho is a function of r only. P is a function of R only, also. So, it depends only on the radius and also the metric ansi for the metric. We assume use the same one as it was used in the lectures before where we have been discussing Schwarz's solution. So, the difference is that energy momentum tensor is not zero in Einstein equations. So in the end it is like this, minus exponent of lambda of r dr squared Minus r squared omega squared. And this is metric on the unit two dimensional sphere. So, these are functional form. So this problem is absolutely static and very symmetric. With this metric, and that with this metric times that we obtain the equations we have already seen in the lecture four. So the equations as follows. We write them in a bit different manner, and in that lecture, nu and lambda were, in principle, functions of time. Now, we consider them as time independent. So, the equations we obtain as follows. But this is just R 0 0- 1/2 R, so this is just this component of the tensor and scalar. And for this metric, this is (e to the (-lambda))/r lambda prime + 1/(r squared) (1- e to the -lambda). So [COUGH] this is the first thing. -8 pi kappa. P of R is equal to R one one minus half R, which is, again, for this metric. Minus lambda divided by R, mu prime 1 over r square, 1 minus exponent of lama and finally, let me start writing here because its a long equation 8 pi cuppa p of r which is R2 2- 1/2R =- e nu + lambda/2/2 d/dr exponent of nu- lambda over 2. Lambda 2 over r nu prime plus exponent of- lambda/2r lambda prime. So these are the equations. They are non trivial part of Einstein equations. The rest of the equations either follow from this or trivially satisfied, it means that they just establish that 0 equal 0, trivial relation. And probably we have to stress that nu prime is just d over dr of nu. And lambda prime is d over dr of lambda. So, let me just start with the consideration of this equation. It depends only lambda as one can see and can be easily solved, easily solved. The solution of this equation, the solution for lambda. Well, first of all, it can be represented like this, one over r squared d over d r, easy exercise, r 1 minus lambda is equal to eight pi kappa rho, which is a function of r only. As a result, the solution equation is that this guy is just 1 minus 2 kappa, some function of (r), which I'm going to specify here in a moment. So it's like this, where M(r) Is the following thing. It's four pi integral from zero to r dx times x squared times rho of x plus constant of integration. So, but we should have that this guy is regular everywhere. We want to have metric to be regular everywhere. Hence this constant should be zero, constant of integration. That's a first thing. Second, we don't want to have a horizon because it's a star. So, I mean, it's not a black hole, and no way inside it. Density such that we have a horizon. So, we want to have this to be greater than 0, which imposes the relation on rho, as follows on here, because 2 copper m of r should be smaller than r. So, this is a restriction on row as follows from here. So these are the restrictions we have from the regularities, and there are more restrictions that we're going to describe now. We are looking for the solution as follows nu (r) dt squared- e lambda (r) dr squared plus r squared d omega squared. And we already have found that this guy- lambda (r) is just 1 minus 2 kappa M (r) / r. Where M of r is 4 Pi integral from zero to r dx times x squared rho of x. And to derive this, we have used the fact that rho is a function of r only. So, and obviously rho(r) is equal to 0, r to R. And, as a result we obtain that outside of the star, this part of the metric, is reduced to the Schwarz shift one. In fact, mass becomes M R, which is so called total mass of the star, dr times r squared rho of r. But one has to have in mind that actually the total mass. Real mass of the star, real mass of the star should be integral over the body of the star over the volume form times the three dimensional determinant of the three dimensional metric. Times row of r. So, this is a determinant of this part of the metric. And this is just volume form. So, this integral, this integral is reduced to 4 pi from the angle integration times integral of 4 from 0 to big r. Dr times r squared rho(r) divided by square root (1- 2 kappa M(r) / r). So, this contribution comes from here, and notice that this, the difference between this formula and this formula, is due to this fact, and this M is always greater. M real minus M is always greater than 0. And this is nothing, but the gravitational binding energy which sustains the star. So, this is so far, now, we have to fix this quantity. To fix it, let us go on with the second equation that we have derived previously, I mean with this equation. Negative 8 pi kappa p(r) equals negative e minus lambda/r nu' + 1\r squared. One minus exponent of minus lambda. So now we know this and this quantity, from here, and we can find nu. The equation for nu is as follows, that d nu over dr Is 2 kappa M of r plus 8 pi r cube kappa p divided by r and r minus 2 kappa M of r. So in non-relalivisticly, let us understand this equation. To understand this equation, let us point out that, in a non-relalivistic limit p is much less than rho. So here we can make like this neglect this and compare with this non-relalivistic limit. And also, 2 kappa m is much, even kappa m, is much less than r, so we all again can make like this German comparison with this, as a result, the equation that we obtain is d new over dr Is nothing but 2 kappa M (r) divided by r squared. And this is nothing, but the Poisson equation, this is nothing non-relativistic Poisson equation, spherically symmetric version of the Poisson equation, surely, for the gravitational potential. So nu/2, nu/2, is nothing but the gravitational potential, rotational potential. So the physical meaning of this contribution is clear. It's just relativistic generalization of the rotational potential. Remember that in the lecture describing Schwarzschild black hole, the g00 factor was playing the role of 1 + 2v. In the relativistic limit, so this is the similar situation here we encounter and that helps to go further. Now, instead of the last equation, we have three equations. The first equation gave us to define lander. This is the second equation, there was a third equation, that equation was looking quit complicated Instead of using it, we can use the continued equation. The conservation of the energy momentum tensor. Remember, that we have in this third lecture, we have explained that G R equations of motion, and this equation, are not independent equations. They fall from each other. So, some of, instead of using some of the equations, Einstein's equation, we can use this one. We're going to use this opportunity, this equation is simpler. It has this form. Well, let me write it here. It has this form, 2(dp/dr) =- (p + rho) d nu / dr. So this equation follows from this one. And if we use this in here, we obtain the following equation for the pressure. It has the following form, d p over d r is equal to minus p, minus p, plus rho times kappa m of r plus 4 pi r cubed times kappa rho of r divided by r times r minus 2 kappa m of r. This equation is so-called Tulmann-Oppenheimer-Volkov equation, which describes hydrodynamic equilibrium viewing for the star for the helium equilibrium of the star. We are now going to describe the solution for this equations and specify the equation of state for the liquid. To go farther, we need to do that. [SOUND] [SOUND]
