10.5 The interior structure of the star

Combined with Eq. (10.30) for dm/dr and an equation of state of the form of Eq. (10.25), this gives three equations for m, ρ, and p. We have reduced to a subsidiary position; it can be found from Eq. (10.27) once the others have been solved.

General rules for integrating t he equations

Since there are two first-order differential equations, Eqs. (10.30) and (10.39), they require two constants of integration, one being m(r = 0) and the other p(r = 0). We now argue that m(r = 0) = 0. A tiny sphere of radius r = ε has circumference 2π ε, and proper radius |grr|1/2ε (from the line element). Thus a small circle about r = 0 has ratio of circumference to radius of 2π |grr|−1/2. But if spacetime is locally flat at r = 0, as it must be at any point of the manifold, then a small circle about r = 0 must have ratio of circumference to radius of 2π . Therefore grr(r = 0) = 1, and so as r goes to zero, m(r) must also go to zero, in fact faster than r. The other constant of integration, p(r = 0) := pc or, equivalently, ρc, from the equation of state, simply defines the stellar model. For a given equation of state p = p(ρ), the set of all spherically symmetric static stellar models forms a one-parameter sequence, the parameter being the central density. This result follows only from the standard uniqueness theorems for first-order ordinary differential equations.

Once m(r), p(r), and ρ(r) are known, the surface of the star is defined as the place where p = 0. (Notice that, by Eq. (10.39), the pressure decreases monotonically outwards from the center.) The reason p = 0 marks the surface is that p must be continuous everywhere, for otherwise there would be an infinite pressure gradient and infinite forces on fluid elements. Since p = 0 in the vacuum outside the star, the surface must also have p = 0. Therefore we stop integrating the interior solution there and require that the exterior metric should be the Schwarzschild metric. Let the radius of the surface be R. Then in order to have a smooth geometry, the metric functions must be continuous at r = R. Inside the star we have

grr = 1 − 2m(r) −1 r

and outside we have

grr = 1 − 2M −1 . r

Continuity clearly defines the constant M to be

This is exactly the same as the equation of hydrostatic equilibrium for Newtonian stars (see Chandrasekhar 1939), a fact which should not surprise us in view of our earlier interpretation of m and of the trivial fact that the Newtonian limit of ρ is just the mass density. Comparing Eq. (10.44) with its progenitor, Eq. (10.39), shows that all the relativistic corrections tend to steepen the pressure gradient relative to the Newtonian one. In other words, for a fluid to remain static it must have stronger internal forces in GR than in Newtonian gravity. This can be interpreted loosely as indicating a stronger field. An extreme instance of this is gravitational collapse: a field so strong that the fluid’s pressure cannot resist it. We shall discuss this more fully in § 10.7 below.

10.6 E x a c t i n t e r i o r s o l u t i o n s

In Newtonian theory, Eqs. (10.30) and (10.44) are very hard to solve analytically for a given equation of state. Their relativistic counterparts are worse.1 We shall discuss two interesting exact solutions to the relativistic equations, one due to Schwarzschild and one by Buchdahl (1981).

The Schwarzschild constantdensity interior solution

To simplify the task of solving Eqs. (10.30) and (10.39), we make the assumption

This replaces the question of state. There is no physical justification for it, of course. In fact, the speed of sound, which is proportional to (dp/dρ)1/2, is infinite! Nevertheless, the interiors of dense neutron stars are of nearly uniform density, so this solution has some interest for us in addition to its pedagogic value as an example of the method we use to solve the system.

We can integrate Eq. (10.30) immediately:

where R is the star’s as yet undetermined radius. Outside R the density vanishes, so m(r) is constant. By demanding continuity of grr, we find that m(r) must be continuous at R. This implies

1If we do not restrict the equation of state, then Eqs. (10.44) and (10.39) are easier to solve. For example,

we can arbitrarily assume a function m(r), deduce ρ(r) from it via Eq. (10.30), and hope to be able to solve Eq. (10.44) or Eq. (10.39) for p. The result, two functions p(r) and ρ(r), implies an ‘equation of state’ p = p(ρ)

by eliminating r. This is unlikely to be physically realistic, so most exact solutions obtained in this way do not interest the astrophysicist.