The Problem and its Solution

We will present, in the case of a spherically symmetric shell of liquid fluid with constant energy density, the exact solution of the Einstein field equations of General Relativity [#!DiracGravity!#],

  $$R_{\mu}^{\;\nu}-\frac{1}{2}\,R\,g_{\mu}^{\;\nu} = -\kappa\,T_{\mu}^{\;\nu},$$ (1)

where $\kappa=8\pi G/c^{4}$, $G$ is the universal gravitational constant and $c$ is the speed of light. Under the conditions of time independence and of spherical symmetry around the origin of a spherical system of coordinates $(t,r,\theta,\phi)$, the Schwarzschild system of coordinates, the most general possible metric is given by the invariant interval, written in terms of this spherical system of coordinates,

  $$ds^{2} = \,{\rm e}^{2\nu(r)}c^{2}dt^{2} - \,{\rm e}^{2\lambda(r)}dr^{2} - r^{2}\left[d\theta^{2}+\sin^{2}(\theta)d\phi^{2}\right],$$ (2)

where $\exp[\nu(r)]$ and $\exp[\lambda(r)]$ are two positive functions of only $r$. As one can see, in this work we will use the time-like signature $(+,-,-,-)$, following [#!DiracGravity!#]. Under these conditions the matter stress-energy tensor density $T_{\mu}^{\;\nu}$ on the right-hand side of the equation is diagonal, and given by the four diagonal components $T_{0}^{\;0}(r)=\rho(r)$, where $\rho(r)$ is the energy density of the matter, and $T_{1}^{\;1}(r)=T_{2}^{\;2}(r)=T_{3}^{\;3}(r)=-P(r)$, where $P(r)$ is the pressure, which is isotropic, thus characterizing a fluid.

Since under these conditions $R_{\mu}^{\;\nu}$ and $T_{\mu}^{\;\nu}$ are both diagonal, there are just four non-trivial field equations contained in Equation (1). In addition to these four field equations we have the consistency condition

  $$D_{\nu}T_{\mu}^{\;\nu} = 0,$$ (3)

which is due to the fact that the combination of tensors that constitutes the left-hand side of the Einstein field equation satisfies the Bianchi identity of the Ricci curvature tensor. Writing these equations explicitly in the chosen coordinate system, one finds that the component equations involving $T_{2}^{\;2}(r)$ and $T_{3}^{\;3}(r)$ turn out to be identical, so that we are left with the set of four equations, including the consistency condition,

$$\left\{\rule{0em}{3ex}1-2\left[r\lambda'(r)\right]\right\} \,{\rm e}^{-2\lambda(r)} = 1-\kappa r^{2}\rho(r),$$ (4)

$$\left\{\rule{0em}{3ex}1+2\left[r\nu'(r)\right]\right\} \,{\rm e}^{-2\lambda(r)} = 1+\kappa r^{2}P(r),$$ (5)

$$\left\{\rule{0em}{3ex} r^{2}\nu''(r) - \left[r\lambda'(r)\right]\left[r\nu'(r)\right] \right. \hspace{7em}$$

$$\left.\rule{0em}{3ex} + \left[r\nu'(r)\right]^{2} + \left[r\nu'(r)\right] - \left[r\lambda'(r)\right] \right\} \,{\rm e}^{-2\lambda(r)} = \kappa r^{2}P(r),$$ (6)

$$\left[\rho(r)+P(r)\right] \nu'(r) = -P'(r),$$ (7)

where the primes indicate differentiation with respect to $r$. Next, it can be shown that Equation (6) can be obtained from the others, being in fact a linear combination of the derivative of Equation (5) and of Equations (4), (5) and (7). If we denote Equations (4) through (7) respectively by $E_{t}$, $E_{r}$, $E_{\theta}$ and $E_{c}$, we have that

  $$E_{\theta} = \frac{1}{2} \left[ -r\nu'(r)\left(E_{t}-E_{r}\right) + rE'_{r} + \kappa r^{2}E_{c} \right].$$ (8)

This leaves us with a set of just three differential equations to solve. In addition to this, we will assume that we have an energy density $\rho(r)=\rho_{0}$ which is constant as a function of $r$ within the shell of fluid matter, thus characterizing a liquid fluid. The equations that we propose to solve are therefore those given in Equations (4), (5) and (7). It is important to note that, in this way, we are left with a system of just three first-order differential equations. Therefore, the discussion of boundary conditions can be limited to the discussion of the behavior of the functions involved, thus eliminating the need for any discussion of the behavior of their derivatives.

We will assume that the matter consists of a spherical shell of liquid, located between the radial positions $r_{1}$ and $r_{2}$, meaning that we will have an inner vacuum region within $(0,r_{1})$, a matter region within $(r_{1},r_{2})$, and an outer vacuum region within $(r_{2},\infty)$. This means that we will have for $\rho(r)$ and $P(r)$

$$\rho(r) = \left\{ \begin{array}{lcl} 0 & \mbox{for} & 0\;\leq r<r_{1}, \\ ... ... & r_{1}<r<r_{2}, \\ [3ex] 0 & \mbox{for} & r_{2}<r<\infty, \end{array}\right.$$ (9)

$$P(r) = \left\{ \begin{array}{lcl} 0 & \mbox{for} & 0\;\leq r\leq r_{1}, \\ [3ex] 0 & \mbox{for} & r_{2}\leq r<\infty. \end{array}\right.$$ (10)

The function $P(r)$ within the matter region is, of course, one of the unknowns of our problem. In addition to this, we have the boundary conditions for $P(r)$ at the two interfaces, in the limits coming from within the liquid,

$$P(r_{1}) = 0,$$ (11)

$$P(r_{2}) = 0,$$ (12)

since these constitute a requirement in any interface between fluid matter and a vacuum. The remaining boundary conditions are those requiring the continuity of $\lambda(r)$ and $\nu(r)$ across the interfaces, and the asymptotic conditions leading to the Newtonian limit at radial infinity.