Models of spherical stars

In Newtonian theory the stability of stars is described in terms of hydrostatic equilibrium. In the framework of General Relativity this is similar, leading to the Oppenheimer-Volkoff equation.

Components of the Einstein tensor

The stellar interior will have a metric of the form

\[\dd{s}^2 = -e^\nu \dd{t}^2 + e^{\lambda} \dd{r}^2 + r^2 \qty( \dd{\theta}^2 + \sin[2](\theta) \dd{\phi}^2 )\]

where \(\nu\) and \(\lambda\) are functions of \(r\). The Ricci tensor will have components

\[\label{eq:193} R_{tt} = \half e^{\nu-\lambda} \qty( \nu'' + \half \nu'^2 - \half \nu' \lambda' + \frac{2}{r} \nu')\]
\[\label{eq:201} R_{rr} = - \half \qty( \nu'' + \half \nu'^2 - \half \nu' \lambda' - \frac{2}{r} \lambda')\]
\[\label{eq:202} R_{\theta \theta} = 1 - e^{- \lambda} \qty( 1 + \frac{r}{2} ( \nu' - \lambda') )\]
\[\label{eq:203} R_{\phi \phi} = R_{\theta \theta} \sin[2](\theta)\]

with all other components zero. The metric is orthogonal, so

\[\begin{split}\begin{aligned} g^{tt} &= - e^{- \nu} \\ g^{rr} &= e^{- \lambda} \\ g^{\theta \theta} &= r^{-2} \\ g^{\phi \phi} &= (r^2 \sin[2](\theta) )^{-1} \end{aligned}\end{split}\]

again, with all other terms zero. The Ricci and metric tensors are orthogonal, so the Ricci scalar has the expression

\[ \begin{align}\begin{aligned} \label{eq:204} R = g^{\mu \nu} R_{\mu \nu} = g^{tt}R_{tt} + g^{rr} R_{rr} + g^{ \theta \theta} R_{\theta \theta} + g^{\phi \phi} R_{ \phi \phi}\\after making the appropriate substitutions, we find\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned}\begin{split} \label{eq:205} \begin{split} R = - e^{-\lambda} \qty[ \qty( \nu'' + \half \nu'^2 - \half \nu' \lambda' ) + \frac{\nu' - \lambda'}{r}]\\ {}+\frac{2}{r^2} \qty[1 - e^{- \lambda} \qty( 1+ \frac{(\nu' - \lambda')r}{2})] \end{split}\end{split}\\The fully covariant Einstein tensor is\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned}G_{\mu \nu} = R_{\mu \nu} - \half g_{\mu \nu} R\\so\end{aligned}\end{align} \]
\[\begin{split}\begin{aligned} G_{tt} &= \frac{e^{\nu}}{r^2} \qty[ 1+ e^{-\lambda} (r \lambda' -1) ] \\ G_{rr} &= \frac{\nu'}{r} - \frac{e^{\lambda}}{r^2} \qty( 1 - e^{- \lambda}) \\ G_{\theta \theta} &= r^2 e^{- \lambda} \qty[ \frac{\nu''}{2} + \frac{\nu'^2}{4} - \frac{\nu' \lambda'}{4} + \frac{\nu'-\lambda'}{2r}] \\ G_{\phi \phi} &= \sin[2](\theta) G_{\theta \theta} \end{aligned}\end{split}\]

Components of the energy-momentum tensor

For a perfect fluid the energy-momentum tensor, in fully covariant form has components

\[T^{\mu \nu} = (\rho + P) u^{\mu} u^{\nu} + P g^{\mu \nu}\]

for \(\rho\) the mass-energy density, and \(P\) the pressure, while \(u^{\mu}\) are the four velocity components of a fluid element. Seeking a static solution has \(u^{i} = 0, i = \set{1,2,3}\), so from the geodesic equation

\[ \begin{align}\begin{aligned} \label{eq:206} g_{tt}(u^t)^2 = -1 \implies u^t = e^{- \frac{\nu}{2}}\\thus\end{aligned}\end{align} \]
\[\begin{split}\begin{aligned} T_{tt} &= \rho e^{\nu} \\ T_{rr} &= P e^{\lambda} \\ T_{\theta \theta} &= P r^2 \\ T_{\phi \phi} &= P r^2 \sin[2](\theta) \end{aligned}\end{split}\]

Einstein’s equations

In fully covariant form

\[ \begin{align}\begin{aligned}G_{\mu \nu} = 8 \pi T_{\mu \nu}\\so\end{aligned}\end{align} \]
\[\begin{split}\begin{aligned} \label{eq:207} \frac{e^{\nu}}{r^2} \qty[ 1+ e^{-\lambda} (r \lambda' -1) ] &= 8 \pi \rho e^{\nu} \\ \label{eq:208} \frac{\nu'}{r} - \frac{e^{\lambda}}{r^2} (1-e^-\lambda) &= 8 \pi P e^{\lambda} \\ \label{eq:209} r^2 e^{-\lambda} \qty[ \frac{\nu''}{2} + \frac{\nu'^2}{4} - \frac{\nu' \lambda'}{4} + \frac{\nu'-\lambda'}{2r}] &= 8 \pi P r^2 \end{aligned}\end{split}\]

Solution of the first Einstein equation

By cancelling the \(e^{\nu}\) term in equation ([eq:207]), and rearranging,

\[ \begin{align}\begin{aligned} \label{eq:210} \dv{r} \qty[ r(1-e^{-\lambda})] = 8 \pi \rho r^2\\Then introducing the mass function, :math:`m(r)`, defined as\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned} \label{eq:211} \dv{m}{r} = 4 \pi \rho r^2 = \half \dv{r} \qty[r (1-e^{-\lambda})]\\and integrating,\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned} \label{eq:212} r(1-e^{-\lambda}) = 2m +C\\where :math:`C` is a constant of integration, equal to zero unless the\end{aligned}\end{align} \]

star is singular at \(r=0\). Thus

\[\label{eq:213} e^{-\lambda} = 1 - \frac{2m}{r}\]

The Oppenheimer-Volkoff equation

Rearranging equation ([eq:208]), with some straight-forward algebra,

\[ \begin{align}\begin{aligned} \label{eq:214} \dv{\nu}{r} = e^{\lambda} \qty[8 \pi P r + \frac{1}{r} (1-e^{-\lambda})] = 2 \qty[\frac{4 \pi P r^3 + m}{r(r-2m)}]\\Then, considering conservation of mass-energy,\end{aligned}\end{align} \]
:math:`tensor{T}{^{alpha

beta}_{;beta}} = qty[(rho +P) u^{alpha}u^{beta} + P

g^{alpha beta}]_{;beta}= 0`, then

\[ \begin{align}\begin{aligned}\begin{split} \label{eq:215} \begin{split} (\rho + P)_{;\beta} u^{\alpha} u^{\beta} + (\rho+P)(u^{\alpha})_{; \beta} u^{\beta} \\ + (\rho+P) u^{\alpha}(u^{\beta})_{; \beta} + P_{,\beta} g^{\alpha \beta} + Pg^{\alpha \beta}_{; \beta} = 0 \end{split}\end{split}\\This is infact a set of four equations, as :math:`\alpha` is a free\end{aligned}\end{align} \]

index, so with just \(\alpha \equiv r\),

\[ \begin{align}\begin{aligned}\begin{split} \label{eq:215} \begin{split} (\rho + P)_{;\beta} u^{r} u^{\beta} + (\rho+P)(u^{r})_{; \beta} u^{\beta} \\ + (\rho+P) u^{r}(u^{\beta})_{; \beta} + P_{,\beta} g^{r \beta} + Pg^{r \beta}_{; \beta} = 0 \end{split}\end{split}\\But since :math:`u^r=0` two of these terms vanish, and the derivatives\end{aligned}\end{align} \]

of the metric tensor are all zero, so the last term drops out too. We then have the simplification

\[ \begin{align}\begin{aligned} \label{eq:216} ( \rho +P)(u^r)_{;t} u^t + \dv{P}{r} g^{rr} = 0\\then\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned} \label{eq:217} u^r_{;t} = \Gamma^r_{tt} u^t = \half \nu' e^{\nu-\lambda} e^{-\nu/2} = \half e^{-\lambda} \nu' e^{\nu/2}\\Thus\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned} \label{eq:218} \half(\rho + P) e^{-\lambda} \dv{\nu}{r} + e^{- \lambda} \dv{P}{r} = 0 \implies \dv{\nu}{r} = - \frac{2}{(\rho+P)} \dv{P}{r}\\and using this to eliminate :math:`\nu'` from equation ([eq:214]),\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned} \label{eq:219} \dv{P}{r} = - \frac{(\rho + P)(4 \pi P r^3 + m)}{r(r-2m)}\\which is the *Oppenheimer-Volkoff* equation. In the weak-field limit,\end{aligned}\end{align} \]

\(P \ll \rho\) implies \(4 \pi P r^3 \ll m\), and the metric will be almost flat, so \(m \ll r\), so this simplifies to

\[ \begin{align}\begin{aligned} \label{eq:220} \dv{P}{r} = - \frac{\rho m}{r^2}\\which is the Newtonian hydrostatic equilibrium equation.\end{aligned}\end{align} \]

Solving the O-V equation

There are three unknown functions in the O-V equation, \(P(r)\), \(\rho(r)\), and \(m(r)\), but the latter two are related, so we need an additional relation, an equation of state, to link all three, in the form

\[ \begin{align}\begin{aligned}P(r) = P(\rho(r))\\For a fluid which is in local thermodynamic equilibrium there is always\end{aligned}\end{align} \]

a relation between pressure, density, and entropy, of the form

\[ \begin{align}\begin{aligned}P=P(\rho,S)\\In most astrophysical situations we can regard :math:`S` as constant.\end{aligned}\end{align} \]

In practice, to solve this system we need boundary conditions.

Take \(P=P_0\) and \(m=0\) at \(r=0\), and integrate outwards to \(P=0\) at the surface of the star, where \(r=R\) and \(m=M\), where \(M\) is the mass constant in the exterior Schwarzschild metric. This then allows us to find \(\nu\) and \(\lambda\) to form a complete expression for the metric inside the star. The effect of GR compared to Newtonian mechanics will be to steepen the pressure gradient within the star.

An exact solution for a star with constant density

Suppose that the density is constant, and \(\rho=\rho_0\) (implying, rather concerningly, an infinite sound speed!), and integrating equation ([eq:211]), and retrieve

\[ \begin{align}\begin{aligned} \label{eq:221} m(r) = \frac{4}{3} \pi \rho_0 r^3\\This can be substituted into the O-V equation, giving\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned} \label{eq:222} \dv{P}{r} = - \frac{4}{3} \pi r \frac{(\rho_0+P)(\rho_0 + 3 P)}{(1 - \frac{8 \pi \rho_0 r^2}{3})}\\thus\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned}\begin{split} \begin{aligned} \label{eq:223} \frac{\dd{P}}{(\rho_0 +P)(\rho_0 + 3P)} &= \frac{1}{2 \rho_0} \qty[\frac{3 \dd{P}}{(\rho_0 + 3P)} - \frac{\dd{P}}{(\rho_0 +P)}] \\ &= - \frac{4 \pi}{3} \frac{r \dd{r}}{\qty(1 - \frac{8 \pi \rho_0 r^2}{3})}\end{aligned}\end{split}\\and integrating both sides,\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned} \label{eq:224} \log( \rho_0 + 3P) - \log(\rho_0 + P) = \half \log(1 - \frac{8 \pi \rho_0 r^2}{3}) + C\\for a constant :math:`C`, which can be written\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned} \label{eq:225} \frac{\rho_0 + 3P}{\rho_0+P} = A \qty( 1 - \frac{8 \pi \rho_0 r^2}{3})^{1/2}\\when :math:`r=0` we have :math:`P=P_0`, so we can express :math:`A` in\end{aligned}\end{align} \]

terms of density and central pressure,

\[ \begin{align}\begin{aligned}A = \frac{\rho_0 + 3 P_0}{\rho_0 +P_0}\\Then\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned}\begin{split} \begin{aligned} \label{eq:226} \frac{\rho_0 + 3P}{\rho_0 + P} &= \frac{\rho_0 + 3P_0}{\rho_0 + P_0} \qty( 1 - \frac{8 \pi \rho_0 r^2}{3})^{1/2} \nonumber\\ &= \frac{\rho_0 + 3P_0}{\rho_0 + P_0} \qty( 1 - \frac{2m}{r})^{1/2}\end{aligned}\end{split}\\at the surface of the star :math:`P=0` so the left hand reduces to\end{aligned}\end{align} \]

\(1\), so

\[ \begin{align}\begin{aligned} \label{eq:227} \frac{\rho_0 + 3P_0}{\rho_0 + P_0} \qty( 1 - \frac{2m}{r})^{1/2} = 1\\for :math:`M` the Schwarzschild mass, and :math:`R` is the coordinate\end{aligned}\end{align} \]

radius of the star. We can obtain an expression for \(P\) as a function of \(r\) by rearranging,

\[\label{eq:228} P_0 = \frac{\rho_0 \qty[ 1 - \qty( 1 - \frac{2M}{R} )^{1/2}]}{3 \qty( 1 - \frac{2M}{R})^{1/2} - 1}\]

Buchdahl’s Theorem

From equation ([eq:228]), it can be seen that as \(P_0 \to \infty\) when \(3 \qty(1- \frac{2M}{R})^{1/2} \to 1\), that is, when \(M/R \to 4/9\). Clearly there can be no static star with uniform density with a radius smaller than \(9M/4\), as they would require an infinite internal pressure. If we require the exterior metric to be well-behaved then stars with a radius less than \(2M\) should be excluded, as the Schwarzschild metric misbehaves at this point, with timelike intervals becoming spacelike, and vice versa.

As a result we can exclude the possibility of a uniform static star with these properties, and Buchdahl’s theorem is a rigorous proof of this.