本文演示从静态球对称时空推导TOV方程
预备知识：Schutz或MTW track1级别的相对论知识储备（甚至不需要）
0.1.1 Metric of Static Spherically Symmetric Stars
The metric of spherically symmetric spacetime is:
$$
\mathrm{d} s^2=g_{00} \mathrm{~d} t^2+2 g_{0 r} \mathrm{~d} r \mathrm{~d} t+g_{r r} \mathrm{~d} r^2+r^2 \mathrm{~d} \Omega^2
\tag{1.1}$$
We define a static spacetime to be one in which we can find a time coordinate $t$ with two properties: (i) all metric components are independent of $t$, and (ii) the geometry is unchanged by time reversal, $t \rightarrow-t$. (A spacetime with property (i) but not necessarily (ii) is said to be stationary.)
Condition (ii) has the following implication. The coordinate transformation $(t, r, \theta, \phi) \rightarrow$ $(-t, r, \theta, \phi)$ has $\Lambda^{\overline{0}}{ }_0=-1, \Lambda_j^i=\delta_j^i$, and we find:
$$
\left.g_{\bar{0} \bar{0}}=\left(\Lambda^0_{\bar{0}}\right)^2g_{00}=g_{00} \right.\
$$
$$
g_{\bar{0} \bar{r}}=\Lambda_{\bar{0}}^0 \Lambda^r_{\bar{r}} g_{0 r}=-g_{0 r}\
$$
$$
g_{\bar{r} \bar{r}}=\left(\Lambda^r {\bar{r}}\right)^2 g{r r}=g_{r r}
\tag{1.2}$$
by assuming $g_{00}<0$ and $g_{rr}>0$ $($ we shall see these conditions do hold inside stars $)$, denote: $g_{00}=-e^{2\phi}$ and $g_{rr}=e^{2\lambda}$
then solve Eq.$(1.2)$ for the metric of static spherically symmetric spacetime:
$$
\mathrm{d} s^2=-\mathrm{e}^{2 \Phi} \mathrm{d} t^2+\mathrm{e}^{2 \Lambda} \mathrm{d} r^2+r^2 d \Omega^2
\tag{1.3}$$
0.1.2 Einstein Tensors
We can show that for the metric given by Eq. $(1.3)$, the Einstein tensor has components:
$$
\begin{aligned}
&G_{00}=\frac{1}{r^2} \mathrm{e}^{2 \Phi} \frac{\mathrm{d}}{\mathrm{d} r}\left[r\left(1-\mathrm{e}^{-2 \Lambda}\right)\right] \\
&G_{r r}=-\frac{1}{r^2} \mathrm{e}^{2 \Lambda}\left(1-\mathrm{e}^{-2 \Lambda}\right)+\frac{2}{r} \Phi^{\prime} \\
&G_{\theta \theta}=r^2 \mathrm{e}^{-2 \Lambda}\left[\Phi^{\prime \prime}+\left(\Phi^{\prime}\right)^2+\Phi^{\prime} / r-\Phi^{\prime} \Lambda^{\prime}-\Lambda^{\prime} / r\right] \\
&G_{\phi \phi}=\sin ^2 \theta G_{\theta \theta}
\end{aligned}
\tag{1.4}$$
where $\Phi^{\prime}:=\mathrm{d} \Phi / \mathrm{d} r$, etc. All other components vanish.
0.1.3 Structure Equations
In this subsection we derive Tolman-Oppenheimer-Volkof equation from the $r$ component of conservation law along with the $(r,r)$ component of \textit{Einstein field equation}, and then an auxiliary equation from the $(t,t)$ component of Einstein field equation, and end with an introduction on the equation of state EoS
The stress-energy tensor of perfect fluid is given by:
$$
T^{\alpha \beta}=(\rho+p) U^\alpha U^\beta+p \eta^{\alpha \beta}
\tag{1.5}$$
alongs with the conservation law: $T^{\alpha \beta}_{; \beta}=0$, of which the $r$ component gives:
$$
(\rho+p) \frac{\mathrm{d} \Phi}{\mathrm{d} r}=-\frac{d p}{\mathrm{~d} r}
\tag{1.6}$$
Einstein field equation states: $G^{\alpha \beta}=8 \pi T^{\alpha \beta}$, by denoting: $m(r):=\frac{1}{2} r\left(1-\mathrm{e}^{-2 \Lambda}\right)$ and substitute Einstein tensors by eq$(1.4)$, the $(r,r)$ component of field equation reads:
$$
\frac{\mathrm{d} \Phi}{\mathrm{d} r}=\frac{m(r)+4 \pi r^3 p}{r[r-2 m(r)]}
\tag{1.7}$$
Now by combing Eq$(1.6)$ and Eq$(1.7)$, we can eliminate $\Phi$ and obtain:
$$
\frac{\mathrm{d} p}{\mathrm{~d} r}=-\frac{(\rho+p)\left(m+4 \pi r^3 p\right)}{r(r-2 m)}
\tag{1.8}$$
known as the Tolman-Oppenheimer-Volkov equation
In addition to Eq$(1.7)$, the $(t,t)$ component of Einstein field equation gives another structure equation:
$$
\frac{\mathrm{d} m(r)}{\mathrm{d} r}=4 \pi r^2 \rho
\tag{1.9}$$ where again we have denoted $g_{rr}$ by $\left(1-\frac{2 m(r)}{r}\right)^{-1}$.
To complete the struction equations requires a relation between energy density $\rho$ and pressure $p$, called the Equation of State (EoS):
$$
\rho = \rho(p)
\tag{1.10}$$
0.1.4 Boundary Conditions and General Solving Procedure
To solve for $m(r)$ and $p(r)$ from TOV equation Eq$(1.8)$ and auxiliary equation Eq$(1.9)$ along with EoS Eq$(1.10)$, we need two boundary conditions:
The first boundary on $m$ given by properties of spacetime, we state without a proof$($check Schutz if you want$)$:
$$
m(r=0)=0
\tag{1.11}$$
while the second arbitrarily chosen:
$$
p(r=0)=p_c
\tag{1.12}$$
STEP1: with these two boundary conditions providing two integral constants, we can obtain expressions for $m=m(r)$ and $p=p(r)$ simply by integration
STEP2: with condition $p(R)=0$ and expression obtained in STEP1, obtaining $R$; then obtain $M$ by $M=m(R)$
STEP3: to solve for $\Phi(r)$ from Eq$(1.6)$, which need another boundary condition on $\phi$, this can be obtained from continuity condition and Schwarzschild exterior metric which yields $g_{00}(r=R)=-\left(1-\frac{2 M}{R}\right)$, substitute $M$ and $R$ gives the boundary condition on $g_{00}(r=0)$, or equivalently, $\Phi(r=0)$