6.3: Oppenheimer-Volkoff Equation of Hydrostatic Equilibrium

a. Schwarzschild Metric

For reasons that are obvious by now, much of the initial progress in general relativity was made by considering highly symmetric metrics which simplify the Einstein tensor. So let us consider the most general metric which exhibits spherical symmetry. This is certainly consistent with our original assumption of spherical stars. If we take the usual spherical coordinates r, θ, φ, and let t represent the time coordinate, then the distance between two points in this spherical metric can be written as \[d s^2=-e^{\lambda(r)} d r^2-r^2 d \theta^2-r^2\left(\operatorname{Sin}^2 \theta\right) d \phi^2+\frac{e^{\alpha(r)} d t^2}{c^2}\label{6.2.1}\]

where \(\lambda(\mathrm{r})\) and \(\alpha(\mathrm{r})\) are arbitrary functions of the radial coordinate r. We must also make some assumptions about the physics of the star in question. This amounts to specifying the stress energy tensor.

Consistent with our assumption of spherical symmetry, let us assume that the material of the star has an equation of state which exhibits no transverse strains, so that all the off-diagonal elements of the stress energy tensor are zero and the first three spatial elements are equal to the matter equivalent of the energy density. The fourth diagonal component is just the matter density so \[\mathbf{T}^{11}=\mathbf{T}^{22}=\mathbf{T}^{33}=-\frac{P}{3 c^2} \quad \mathbf{T}^{44}=\rho\label{6.2.2}\]

This is equivalent to saying that the equation of state has the familiar form \[\mathrm{P}=\mathrm{P}(\rho)\label{6.2.3}\]

Now if we take the metric tensor specified by equation \ref{6.2.1}, and go through the operations specified by equations \ref{6.1.2} through \ref{6.1.8}, and sum over the three spatial indices because of the spherical symmetry, then the Einstein field equations become \[\begin{aligned}
& e^{-\lambda}\left(\frac{\alpha^{\prime}}{r}+\frac{1}{r^2}\right)-\frac{1}{r^2}=\frac{8 \pi G}{c^2} \frac{P}{c^2} \\
& e^{-\lambda}\left(\frac{\lambda^{\prime}}{r}-\frac{1}{r^2}\right)+\frac{1}{r^2}=\frac{8 \pi G}{c^2} \rho
\end{aligned}\label{6.2.4}\]

Here the prime denotes differentiation with respect to the radial coordinate r. This solution must hold through all space, including that outside the star where \(\mathrm{P}=\rho=0\). If we take the boundary of the star to be where r = R, then for r > R we get the Schwarzschild metric equations \[\begin{aligned}
& e^{-\lambda(r)}\left(\frac{1}{r} \frac{d \alpha(r)}{d r}+\frac{1}{r^2}\right)-\frac{1}{r^2}=0 \\
& e^{-\lambda(r)}\left(\frac{1}{r} \frac{d \lambda(r)}{d r}-\frac{1}{r^2}\right)+\frac{1}{r^2}=0
\end{aligned}\label{6.2.5}\]

which have solutions \[e^{-\lambda(r)}=1+\frac{A}{r} \quad e^{-\alpha(r)}=B\left(1+\frac{A}{r}\right)\label{6.2.6}\]

where A and B are arbitrary constants of integration for the differential equations and are to be determined from the boundary conditions. At large values of r, we require that the metric go over to the spherical metric of Euclidean flat space, so that \[\operatorname{Lim}_{r \rightarrow \infty} e^{\lambda(r)}=\operatorname{Lim}_{r \rightarrow \infty} e^{\alpha(r)}=1\label{6.2.7}\]

and B = 1. A line integral around the object must yield a temporal period and distance consistent with Kepler's third law, meaning that A is related to the Newtonian mass of the object. Specifically, \[A=-\frac{2 G M}{c^2}\label{6.2.8}\]

which has the units of a length and is known as the Schwarzschild radius.

b. Gravitational Potential and Hydrostatic Equilibrium

Since \[e^{\alpha(r)} \simeq 1+\alpha(r) \simeq 1+\frac{2 G M}{c^2 r}\label{6.2.9}\]

we know that \[\alpha(r)=\frac{2 \Omega}{c^2}\label{6.2.10}\]

where Ω is the Newtonian potential at large distances. The parameter \(\alpha(\mathrm{r})\) then plays the role of a potential throughout the entire Schwarzschild metric. So we can solve the first of equations \ref{6.2.4} for its spatial derivative and get \[\frac{d \Omega}{d r}=\frac{G\left[M(r)+4 \pi r^3 P / c^2\right]}{r\left[r-2 G M(r) / c^2\right]}\label{6.2.11}\]

This is quite reminiscent of the Newtonian potential gradient, except (1) that the mass has been augmented by a term representing the local "mass" density attributable to the kinetic energy of the matter producing the pressure and (2) that the radial coordinate has been modified to account for the space curvature. Now even in a non-Euclidean metric we have the reasonable result \[\nabla P=-\tilde{\rho} \nabla \Omega\label{6.2.12}\]

where \(\tilde{\rho}\) is the total local mass density so that the matter density, \(\rho\), must be increased by \(P/c^2\) to include the mass of the kinetic energy of the gas. [For a rigorous proof of this see Misner, Thorne, and Wheeler⁴ (p601)]. Combining equations \ref{6.2.11} and \ref{6.2.12}, we get \[\frac{d P}{d r}=-\frac{G\left(\rho+P / c^2\right)\left[M(r)+4 \pi r^3 P / c^2\right]}{r\left[r-2 G M(r) / c^2\right]}\label{6.2.13}\]

This is known as the Oppenheimer-Volkoff equation of hydrostatic equilibrium, and along with the equation of state it determines the structure of a relativistic star.