[MUSIC] In this lecture, we're going to start with the discussion of a stable star solution. A solution of Einstein equation which describes a spherically symmetric, stable star. In literature, this solution is referred as interior solution. We assume that the star is a ball of matter, ball of matter. And inside this ball, we have non-zero energy-momentum tensor, T mnu is not 0. Outside it, T mnu is 0. And everywhere we assume that lambda, cosmological constant, is 0, everywhere. So to have a stable star there, so far we have encountered several different choices for the energy-momentum tensor. And one of the peculiar energy-momentum tensor was a energy-momentum tensor of dust. But that one is just a collection of particle which do not interact with each other and do not create any pressure. So that one is not suitable for the problem that we are considering, because we need to have such a matter which creates some pressure. On general physical grounds, we need to have something which creates a pressure such that it doesn't allow this body to collapse. Because gravity is attractive force, so everything is tending to shrink while pressure pushes it away to make it stable. So we start with the definition of the energy-momentum tensor for the perfect fluid, the one which describes a matter with a pressure. So for that fluid, for the perfect fluid, energy-momentum tensor is as follows. Where x bar is a full vector, which is t and x. And so it is equal to rho, which is a function of x. In general, u mu, which is a four-velocity field of the fluid. U mu and U nu is a four-velocity field plus pressure times U mu of x U nu of x- a G mu nu, which is a metric tensor of X. So why this is pressure, and why this is energy density is going to be obvious soon. Now we start to explain why this is an energy-momentum tensor for the perfect fluid. But recall that we have already encountered when p is equal to 0, this is the energy-momentum tensor of a dust that we have encountered already. But now let us consider this energy-momentum tensor in the reference frame where the species, where the fluid is static. In such a reference frame, U mu of x is nothing but (g 00 -1/2, 0, 0, 0). The reason for having this is because we have to have u mu u mu = 1, which follows from the dissonation of the fall velocity. So this is a form of the vector for velocity in the reference frame where the fluid is static and in such a reference frame, the T mu nu, the metrics of the tensor, let me write this as g nu alpha t mu alpha. So we want to consider the metrics of this stanza. It's not hard to see the metrics of this stanza has the following form, it's diagonal as it follows from here. It's diagonal with content like this. Rho- p- p- p- p. Already from this form of the tensor, of the energy momentum tensor. It should be apparent for those who have studied classical electrodynamics and special theory of relativity already. This form tells us that this is energy density and this is pressure. But, let us do a bit more. I mean, we're going to give more explanations whether this is a energy momentum tensor for the fluid, for relativistic fluid. So, let us consider the conservation of this energy momentum tensor. The conservation law for this energy momentum tensor is as follows. And, for this particular form, for this particular form, if we subsidized it there, it requires the following form, this consideration. So U nu D nu, U nu + D mu, rho + p U mu, U nu- D mu P. So this is conservation equation, and now if we multiply both sides of this equation on u nu, and use the fact that U nu times U mu = 1. And as follows from this, we have that U nu D mu U nu = 0. So using this and this, after multiplication of this equality on U nu, we obtain the following equation, that d mu acting on rho, plus p, U mu- U mu, D mu, p, and opening this up, we obtain that this is D mu of rho, U mu + p D mu U mu. So, we obtain this equation, this is a relativistic, covariant. Covariant, extension of the standard matter conservation law, which has the falling form. Rho mu U is equals to 0. So this just a continuity equation. This is a continuity equation in flat space and this is a relativistic covariant extension of this equation. Indeed, first of all, if we go to covariantisation, this derivative becomes covariant, so we obtain this part. And also, in addition we have to have this button, the presence of pressure, in the relativistic limit for the following reason. In the non relativistic limit, pressure is always much less than the energy density for the same reason why non-relativistic kinetic energy is much less than m c squared. So from the relativistic energy due to mass. So this is for the same reason it's smaller than rho, so in order to deduce that limit we would neglect this and we will obtain this equation which in three dimensional notations is nothing but dt rho + divergence of rho v = 0. So the obtained equation has obvious interpretation of the relativistic covariant extension of the continuity equation. So from the condition of the conservation of this energy momentum tensor + p (U mu U nu- g mu nu). From the condition of the conservation of this energy momentum. Then there, we have obtained these two equations. Which one falls from the other. This equation U mu D mu U nu + D mu ((rho + p) U mu) U nu- DuP. So this is equal to 0, and the equation that D mu (rho u mu) + p D mu U mu = 0. So the last equation is covariant relative extension of the continuous equation. So, if we use this equation here, we obtain the following relation, that (rho+p) U mu D mu U nu = (delta nu alpha- U alpha U nu ) D alpha P. So this, if one looks to the metrics of this tensor, it is diagonal 0,1,1,1 hence this is a projector on the spatial. Special part of the space time and we declare that this is a relativistic co-variant extension of the Euler equation. Euler equation has the following form, Euler equation is nothing but rho (dt V + V ) differential of V = -p. So this is just relativistic dictation of this equation, the Euler equation is nothing but Newton's Second Law, ma = F, attributed to the unit volume. Indeed, if we divide by volume both sides of this equation, and recall that pressure is nothing but the force attributed per unit area, and this is mass attributed per unit volume. And this is nothing but acceleration which is dv / dt, but v is a velocity field, so it depends of t, x of t. So if we apply this, we obtain that this is (dt V + di V) dt xi. Well, this is nothing but dt (V + V) the gradient of V. So, this equation is Euler equation, which has obvious meaning, this equation is its relativistic extension. And, finally, the Euler equation, this equation, can be written in this form. It is dt rho vi plus dj Tij = 0, for Tij = rho vi vj + p delta ij. So this equation is equivalent to this one. And perhaps I have to stress, just finally, that this is equivalent to this. I mean, this is just tensor notation for the gradient vector. Anyway, so all in all, what I have said should convince one that this tensor is a tensor describing perfect fluid. And what remains to be done to specify the fluid is to give one the equation of state. Equation of state is the equation relating rho and p. So, equation relating the pressure and density. If we specify this, we specify the fluid completely with this tensor. [SOUND]
