Appendix 5 – Schwarzschild Solution Inside a Mass
Full tensor derivation
Appendix 5.1 – Introduction
In this chapter, the complete tensor derivation is given of the internal Schwarzschild solution: the solution of the Einstein field equations for a static, spherically symmetric mass with constant density \( \rho \).
We work in Schwarzschild coordinates with the metric:
\begin{equation}
ds^{2}
= e^{\nu(r)} c^{2} dt^{2}
- e^{\lambda(r)} dr^{2}
- r^{2} d\theta^{2}
- r^{2} \sin^{2}\theta\, d\phi^{2}
\label{eq:R01}
\end{equation}
where \( \nu(r) \) and \( \lambda(r) \) are functions of \( r \) that must be determined.
Energy-momentum tensor of a perfect fluid
The energy-momentum tensor of a perfect fluid is given by:
\begin{equation}
T_{\mu\nu}
= (\rho c^{2} + p)\, u_{\mu} u_{\nu}
- p\, g_{\mu\nu}
\label{eq:R02}
\end{equation}
Here:
- The first term \( (\rho c^{2} + p) u_{\mu} u_{\nu} \) represents the contribution of moving energy + pressure.
- The second term \( -p\, g_{\mu\nu} \) ensures consistency with Lorentz invariance and isotropy of a perfect fluid.
- \( \rho \) = mass density (in the rest frame of the fluid).
- \( \rho c^{2} \) = energy density of the matter.
- \( p \) = isotropic pressure.
- \( u^{\mu} = \dfrac{dx^{\mu}}{d\tau} \) = four-velocity of the matter, with normalization \( u_{\mu} u^{\mu} = 1 \).
- \( g_{\mu\nu} \) = metric tensor.
The pressure \( p \) appears in the tensor because in general relativity not only energy, but also pressure and stress contribute to the curvature of spacetime. Pressure is a form of energy per volume and must therefore be included in the total energy content of the system.
In an ideal (isotropic) rest frame of the fluid, (\ref{eq:R02}) reduces to:
\begin{equation}
T_{\mu\nu}
= (\rho c^{2} + p)\, u_{\mu} u_{\nu}
- p\, g_{\mu\nu}
=
\begin{pmatrix}
\rho c^{2} e^{\nu} & 0 & 0 & 0 \\
0 & p\, e^{\lambda} & 0 & 0 \\
0 & 0 & p\, r^{2} & 0 \\
0 & 0 & 0 & p\, r^{2} \sin^{2}\theta
\end{pmatrix}
\end{equation}
This clearly shows that \( \rho c^{2} \) represents the energy density and that the spatial diagonal elements correspond exactly to the isotropic pressure \( p \). The form of (\ref{eq:R02}) follows directly from the requirements of isotropy and Lorentz invariance: the term \( (\rho c^{2} + p)\, u_{\mu} u_{\nu} \) represents the energy and momentum density along the direction of motion, while the term \( -p\, g_{\mu\nu} \) represents the isotropic pressure that is equal in all spatial directions.
For a static fluid:
\begin{equation}
u^{\mu} = \left(e^{-\nu/2},\, 0,\, 0,\, 0\right),
\qquad
u_{\mu} = \left(e^{\nu/2},\, 0,\, 0,\, 0\right)
\end{equation}
\begin{equation}
u_{\mu} u^{\mu} = 1,
\qquad
u_{\mu}u_{\mu}=\left(e^{\nu}, 0, 0, 0\right) =
\begin{pmatrix}
e^{\nu} & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}
\end{equation}
In short: all elements except the \( (t,t) \)-component are zero; \( u_{t} u_{t} = e^{\nu} \).
It follows directly that:
\begin{equation}
T_t^t = \rho c^{2},
\qquad
T_r^r = T_\theta^\theta = T_\phi^\phi = p
\label{eq:R06}
\end{equation}
\begin{equation}
T_{tt}=T_t^t \quad T_{rr}=g_{rr}T_r^r \quad T_{\theta\theta}=g_{\theta\theta}T_{\theta}^{\theta}
\quad T_{\phi\phi}=g_{\phi\phi}T_{\phi}^{\phi}
\end{equation}
In coordinate components:
\begin{equation}
T_{tt}
= (\rho c^{2} + p)\, u_{t} u_{t} - p\, g_{tt}
= \rho c^{2} e^{\nu}
\end{equation}
\begin{equation}
T_{rr}
= (\rho c^{2} + p)\, u_{r} u_{r} - p\, g_{rr}
= -p(-e^{\lambda})
= p\, e^{\lambda}
\end{equation}
\begin{equation}
T_{\theta\theta}
= (\rho c^{2} + p)\, u_{\theta} u_{\theta} - p\, g_{\theta\theta}
= -p(-r^{2})
= p r^{2}
\end{equation}
\begin{equation}
T_{\phi\phi}
= (\rho c^{2} + p)\, u_{\phi} u_{\phi} - p\, g_{\phi\phi}
= -p(-r^{2}\sin^{2}\theta)
= p r^{2}\sin^{2}\theta
\end{equation}
The Einstein equations are as usual:
\begin{equation}
G_{\mu\nu}
= R_{\mu\nu}
- \frac{1}{2} R\, g_{\mu\nu}
= \frac{8\pi G}{c^{4}}\, T_{\mu\nu}
\label{eq:R12}
\end{equation}
Explanation
The metric (\ref{eq:R01}) is the most general static, spherically symmetric metric. The function \( \nu(r) \) determines the time dilation in the gravitational field, while \( \lambda(r) \) describes the curvature in the radial direction. By determining these two functions from the field equations, we obtain the full geometric structure of the interior field.
Appendix 5.2 – Computation of the Christoffel symbols
The Christoffel symbols are given by:
\begin{equation}
\Gamma^{\rho}_{\mu\nu}
=
\frac{1}{2} g^{\rho\alpha}
\left(
\frac{\partial g_{\nu\alpha}}{\partial x^{\mu}}
+ \frac{\partial g_{\mu\alpha}}{\partial x^{\nu}}
- \frac{\partial g_{\mu\nu}}{\partial x^{\alpha}}
\right)
\end{equation}
For the internal Schwarzschild metric:
\begin{equation}
ds^{2}
=
e^{\nu(r)} c^{2} dt^{2}
-
e^{\lambda(r)} dr^{2}
-
r^{2} d\theta^{2}
-
r^{2} \sin^{2}\theta\, d\phi^{2}
\end{equation}
Metric and derivatives of the metric
The metric components are:
\begin{equation}
g_{\mu\nu} =
\begin{pmatrix}
e^{\nu} & 0 & 0 & 0 \\
0 & -e^{\lambda} & 0 & 0 \\
0 & 0 & -r^{2} & 0 \\
0 & 0 & 0 & -r^{2}\sin^{2}\theta
\end{pmatrix}
\end{equation}
The derivatives with respect to \( r \) are:
\begin{equation}
\frac{\partial g_{\mu\nu}}{\partial r} =
\begin{pmatrix}
\nu' e^{\nu} & 0 & 0 & 0 \\
0 & -\lambda' e^{\lambda} & 0 & 0 \\
0 & 0 & -2r & 0 \\
0 & 0 & 0 & -2r \sin^{2}\theta
\end{pmatrix}
\end{equation}
Non-zero Christoffel symbols
With \( \nu' = \frac{d\nu}{dr} \) and \( \lambda' = \frac{d\lambda}{dr} \):
\begin{equation}
\Gamma^{t}_{tr}=\Gamma^{t}_{rt}
= \frac{1}{2} g^{tt} \frac{\partial g_{tt}}{\partial r}
= \frac{1}{2}\nu'
\label{eq:R17}
\end{equation}
\begin{equation}
\Gamma^{r}_{tt}
= \frac{1}{2} g^{rr}\left(-\frac{\partial g_{tt}}{\partial r}\right)
= \frac{1}{2} e^{\nu-\lambda}\, \nu'
\label{eq:R18}
\end{equation}
\begin{equation}
\Gamma^{r}_{rr}
= \frac{1}{2} g^{rr} \frac{\partial g_{rr}}{\partial r}
= \frac{1}{2}\lambda'
\label{eq:R19}
\end{equation}
\begin{equation}
\Gamma^{r}_{\theta\theta}
= -\frac{1}{2} g^{rr} \frac{\partial g_{\theta\theta}}{\partial r}
= -r e^{-\lambda}
\label{eq:R20}
\end{equation}
\begin{equation}
\Gamma^{r}_{\phi\phi}
= -r e^{-\lambda}\sin^{2}\theta
\label{eq:R21}
\end{equation}
\begin{equation}
\Gamma^{\theta}_{r\theta}
= \Gamma^{\theta}_{\theta r}
= \frac{1}{r}
\label{eq:R22}
\end{equation}
\begin{equation}
\Gamma^{\phi}_{r\phi}
= \Gamma^{\phi}_{\phi r}
= \frac{1}{r}
\label{eq:R23}
\end{equation}
\begin{equation}
\Gamma^{\theta}_{\phi\phi}
= -\sin\theta\cos\theta
\label{eq:R24}
\end{equation}
\begin{equation}
\Gamma^{\phi}_{\theta\phi}
=
\Gamma^{\phi}_{\phi\theta}
=\frac{ \cos\theta}{\sin\theta}
= \cot\theta
\label{eq:R25}
\end{equation}
Derivatives of the Christoffel symbols
\begin{equation}
\frac{\partial}{\partial r}\Gamma^{t}_{rt}
=
\frac{\partial}{\partial r}\Gamma^{t}_{tr}
=
\frac{1}{2}\nu''
\end{equation}
\begin{equation}
\frac{\partial}{\partial r}\Gamma^{r}_{tt}
=
\frac{1}{2} e^{\nu-\lambda}
\left(
\nu'^{2} - \lambda'\nu' + \nu''
\right)
\end{equation}
\begin{equation}
\frac{\partial}{\partial r}\Gamma^{r}_{rr}
=
\frac{1}{2}\lambda''
\end{equation}
\begin{equation}
\frac{\partial}{\partial r}\Gamma^{r}_{\theta \theta}
=
\frac{\partial}{\partial r}(-r e^{-\lambda})
=
-e^{-\lambda} + r e^{-\lambda}\lambda'
=
e^{-\lambda}(r\lambda' - 1)
\end{equation}
\begin{equation}
\frac{\partial}{\partial r}\Gamma^{r}_{\phi \phi}
=
-e^{-\lambda}\sin^{2}\theta
+ r e^{-\lambda}\lambda'\sin^{2}\theta
=
e^{-\lambda}(r\lambda' - 1)\sin^{2}\theta
\end{equation}
Derivatives of the Christoffel symbols
For the angular components:
\begin{equation}
\frac{\partial}{\partial r}\Gamma^{\theta}_{r\theta}
=
\frac{\partial}{\partial r}\Gamma^{\theta}_{\theta r}
=
\frac{\partial}{\partial r}\Gamma^{\phi}_{r\phi}
=
\frac{\partial}{\partial r}\Gamma^{\phi}_{\phi r}
=
\frac{\partial}{\partial r}\left(\frac{1}{r}\right)
=
-\frac{1}{r^{2}}
\end{equation}
\begin{equation}
\frac{\partial}{\partial \theta}\Gamma^{\theta}_{\phi\phi}
=
\frac{\partial}{\partial \theta}(-\sin\theta\cos\theta)
=
-(\cos^{2}\theta - \sin^{2}\theta)
\end{equation}
\begin{equation}
\frac{\partial}{\partial \theta}\Gamma^{\phi}_{\theta\phi}
=
\frac{\partial}{\partial \theta}\Gamma^{\phi}_{\phi\theta}
=
\frac{\partial}{\partial \theta}\left(\frac{\cos\theta}{\sin\theta}\right)
=
-\frac{\sin^{2}\theta + \cos^{2}\theta}{\sin^{2}\theta}
=
-\frac{1}{\sin^{2}\theta}
\end{equation}
Explanation
The derivation of these symbols follows directly from the definition:
\begin{equation}
\Gamma^{\rho}_{\mu\nu}
=
\frac{1}{2} g^{\rho\sigma}
\left(
\frac{\partial g_{\nu\sigma}}{\partial x^{\mu}}
+
\frac{\partial g_{\mu\sigma}}{\partial x^{\nu}}
-
\frac{\partial g_{\mu\nu}}{\partial x^{\sigma}}
\right)
\end{equation}
Since the metric depends only on \( r \), only derivatives with respect to \( r \) remain.
Appendix 5.3 – Ricci tensor components
As found in Chapter 2.14.2 (First attempt with the Ricci tensor), the definition is:
\begin{equation}
R_{\mu\nu}
=
R^{\sigma}_{\mu\sigma\nu}
=
\frac{\partial \Gamma^{\sigma}_{\mu\nu}}{\partial x^{\sigma}}
-
\frac{\partial \Gamma^{\sigma}_{\mu\sigma}}{\partial x^{\nu}}
+
\Gamma^{\sigma}_{\lambda\sigma}\Gamma^{\lambda}_{\mu\nu}
-
\Gamma^{\sigma}_{\lambda\nu}\Gamma^{\lambda}_{\mu\sigma}
\end{equation}
Full derivation of \( R_{tt} \)
We set \( \mu = \nu = t \). Compute the four terms separately.
1. First term: \( \frac{\partial \Gamma^{\sigma}_{tt}}{\partial x^{\sigma}} \)
Only \( \sigma = r \) contributes, since all other \( \Gamma^{\sigma}_{tt} = 0 \).
With (see above):
\begin{equation}
\Gamma^{r}_{tt}
=
\frac{1}{2} e^{\nu - \lambda}\, \nu'
\end{equation}
we obtain:
\begin{equation}
\frac{\partial}{\partial \sigma}\Gamma^{\sigma}_{tt}
=
\frac{\partial}{\partial r}\Gamma^{r}_{tt}
=
\frac{\partial}{\partial r}
\left(
\frac{1}{2} e^{\nu - \lambda}\, \nu'
\right)
=
\frac{1}{2} e^{\nu - \lambda}
\left(
\nu'' + \nu'^{2} - \lambda'\nu'
\right)
\end{equation}
2. Second term: \( -\frac{\partial \Gamma^{\sigma}_{t\sigma}}{\partial t} \)
Since the metric is static (no \( t \)-dependence), this term is zero:
\begin{equation}
-\frac{\partial}{\partial t}\Gamma^{\sigma}_{t\sigma} = 0
\end{equation}
3. Third term: \( \Gamma^{\sigma}_{\lambda\sigma}\Gamma^{\lambda}_{tt} \)
Only \( \lambda = r \) gives a non-zero \( \Gamma^{\lambda}_{tt} \). Thus the term becomes:
Computation of the term \( \Gamma^{\sigma}_{\lambda\sigma}\Gamma^{\lambda}_{tt} \)
We have:
\begin{equation}
\Gamma^{\sigma}_{\lambda\sigma}\Gamma^{\lambda}_{tt}
=
\Gamma^{\sigma}_{r\sigma}\,\Gamma^{r}_{tt}
\end{equation}
First we compute:
\begin{equation}
\begin{aligned}
\Gamma^{\sigma}_{r\sigma}
&=
\Gamma^{t}_{rt}
+
\Gamma^{r}_{rr}
+
\Gamma^{\theta}_{r\theta}
+
\Gamma^{\phi}_{r\phi}
=
\frac{1}{2}\nu'
+
\frac{1}{2}\lambda'
+
\frac{1}{r}
+
\frac{1}{r}
\\ &\quad
=
\frac{1}{2}(\nu' + \lambda') + \frac{2}{r}
\end{aligned}
\end{equation}
Thus:
\begin{equation}
\Gamma^{\sigma}_{\lambda\sigma}\Gamma^{\lambda}_{tt}
=
\left[
\frac{1}{2}(\nu' + \lambda') + \frac{2}{r}
\right]
\cdot
\frac{1}{2} e^{\nu - \lambda}\nu'
\end{equation}
\begin{equation}
=
\frac{1}{4} e^{\nu - \lambda}\nu'(\nu' + \lambda')
+
e^{\nu - \lambda}\frac{\nu'}{r}
\end{equation}
4. The fourth term: \( -\Gamma^{\sigma}_{t\lambda}\Gamma^{\lambda}_{t\sigma} \)
We must sum all non-zero products \( \Gamma^{\sigma}_{t\lambda}\Gamma^{\lambda}_{t\sigma} \). The only non-zero Christoffel symbols with a time index are:
\begin{equation}
\Gamma^{\sigma}_{t\lambda} = \frac{1}{2}\nu',
\qquad
\Gamma^{r}_{tt} = \frac{1}{2} e^{\nu - \lambda}\nu'
\end{equation}
By symmetry, there are two equal contributions:
\begin{equation}
-\Gamma^{\sigma}_{t\lambda}\Gamma^{\lambda}_{t\sigma}
=
-\Gamma^{t}_{r t}\Gamma^{r}_{t t}
-
\Gamma^{t}_{t r}\Gamma^{r}_{t t}
=
-2\left(\frac{1}{2}\nu'\right)
\left(\frac{1}{2} e^{\nu - \lambda}\nu'\right)
=
-\frac{1}{2} e^{\nu - \lambda}\nu'^{2}
\end{equation}
5. Summing all contributions
We now combine the four terms (the second term was zero):
\begin{equation}
\begin{aligned}
R_{tt}
&=
\frac{1}{2} e^{\nu - \lambda}
\left(
\nu'' + \nu'^{2} - \lambda'\nu'
\right)
+
\frac{1}{4} e^{\nu - \lambda}\nu'(\nu' + \lambda')
\\ &\quad
+
e^{\nu - \lambda}\frac{\nu'}{r}
-
\frac{1}{2} e^{\nu - \lambda}\nu'^{2}
\end{aligned}
\end{equation}
After simplification:
\begin{equation}
R_{tt}
=
\frac{1}{2} e^{\nu - \lambda}
\left(
\nu'' + \frac{1}{2}\nu'^{2} - \frac{1}{2}\lambda'\nu' + \frac{2\nu'}{r}
\right)
\end{equation}
Thus:
\begin{equation}
\boxed{
R_{tt}
=
\frac{1}{2} e^{\nu - \lambda}
\left(
\nu'' + \frac{1}{2}\nu'^{2} - \frac{1}{2}\lambda'\nu' + \frac{2\nu'}{r}
\right)
}
\end{equation}
Full derivation of \( R_{rr} \)
The Ricci component:
\begin{equation}
R_{rr}
=
\frac{\partial \Gamma^{\sigma}_{rr}}{\partial x^{\sigma}}
-
\frac{\partial \Gamma^{\sigma}_{r\sigma}}{\partial r}
+
\Gamma^{\sigma}_{\lambda\sigma}\Gamma^{\lambda}_{rr}
-
\Gamma^{\sigma}_{r\lambda}\Gamma^{\lambda}_{r\sigma}
\end{equation}
We set \( \mu = \nu = r \). Compute the four terms separately.
1. First term: \( \frac{\partial \Gamma^{\sigma}_{rr}}{\partial x^{\sigma}} \)
The only non-zero \(\Gamma^{\sigma}_{rr} \) component is:
\begin{equation}
\Gamma^{r}_{rr} = \frac{1}{2}\lambda'
\end{equation}
Thus:
\begin{equation}
\frac{\partial}{\partial x^{\sigma}}\Gamma^{\sigma}_{rr}
=
\frac{\partial}{\partial r}\Gamma^{r}_{rr}
=
\frac{1}{2}\lambda''
\end{equation}
2. Second term: \( -\dfrac{\partial \Gamma^{\sigma}_{r\sigma}}{\partial r} \)
First we compute the sum:
\begin{equation}
\begin{aligned}
\Gamma^{\sigma}_{r\sigma}
&=
\Gamma^{t}_{rt}
+
\Gamma^{r}_{rr}
+
\Gamma^{\theta}_{r\theta}
+
\Gamma^{\phi}_{r\phi}
\\ &\quad
=
\frac{1}{2}\nu'
+
\frac{1}{2}\lambda'
+
\frac{1}{r}
+
\frac{1}{r}
=
\frac{1}{2}(\nu' + \lambda') + \frac{2}{r}
\end{aligned}
\end{equation}
Differentiate with respect to \( r \):
\begin{equation}
\frac{\partial}{\partial r}\Gamma^{\sigma}_{r\sigma}
=
\frac{1}{2}(\nu'' + \lambda'')
-
\frac{2}{r^{2}}
\end{equation}
Thus:
\begin{equation}
-\frac{\partial}{\partial r}\Gamma^{\sigma}_{r\sigma}
=
-\frac{1}{2}(\nu'' + \lambda'')
+
\frac{2}{r^{2}}
\end{equation}
3. Third term: \( \Gamma^{\sigma}_{\lambda\sigma}\Gamma^{\lambda}_{rr} \)
Only \( \lambda = r \) contributes, since \( \Gamma^{\lambda}_{rr} \) is non-zero only for \( \lambda = r \).
\begin{equation}
\Gamma^{\sigma}_{\lambda\sigma}\Gamma^{\lambda}_{rr}
=
\Gamma^{\sigma}_{r\sigma}\Gamma^{r}_{rr}
\end{equation}
We use again:
\begin{equation}
\Gamma^{\sigma}_{r\sigma}
=
\frac{1}{2}(\nu' + \lambda') + \frac{2}{r}
\end{equation}
\begin{equation}
\Gamma^{r}_{rr}
=
\frac{1}{2}\lambda'
\end{equation}
Thus:
\begin{equation}
\Gamma^{\sigma}_{\lambda\sigma}\Gamma^{\lambda}_{rr}
=
\left[
\frac{1}{2}(\nu' + \lambda') + \frac{2}{r}
\right]
\cdot
\frac{1}{2}\lambda'
=
\frac{1}{4}(\nu' + \lambda')\lambda'
+
\frac{\lambda'}{r}
\end{equation}
4. Fourth term: \( -\Gamma^{\sigma}_{r\lambda}\Gamma^{\lambda}_{r\sigma} \)
We sum all non-zero products:
\begin{equation}
\Gamma^{\sigma}_{r\lambda}\Gamma^{\lambda}_{r\sigma}
=
\Gamma^{t}_{rt}\Gamma^{t}_{rt}
+
\Gamma^{r}_{rr}\Gamma^{r}_{rr}
+
\Gamma^{\theta}_{r\theta}\Gamma^{\theta}_{r\theta}
+
\Gamma^{\phi}_{r\phi}\Gamma^{\phi}_{r\phi}
\end{equation}
Substituting:
\begin{equation}
=
\left(\frac{1}{2}\nu'\right)^{2}
+
\left(\frac{1}{2}\lambda'\right)^{2}
+
\left(\frac{1}{r}\right)^{2}
+
\left(\frac{1}{r}\right)^{2}
=
\frac{1}{4}\nu'^{2}
+
\frac{1}{4}\lambda'^{2}
+
\frac{2}{r^{2}}
\end{equation}
Thus:
\begin{equation}
-\Gamma^{\sigma}_{r\lambda}\Gamma^{\lambda}_{r\sigma}
=
-\frac{1}{4}\nu'^{2}
-
\frac{1}{4}\lambda'^{2}
-
\frac{2}{r^{2}}
\end{equation}
5. Summing all terms
We now combine the four contributions:
\begin{equation}
\begin{aligned}
R_{rr}
&=
\frac{1}{2}\lambda''
-
\frac{1}{2}(\nu'' + \lambda'')
+
\frac{2}{r^{2}}
+
\frac{1}{4}(\nu' + \lambda')\lambda'
\\ &\quad
+
\frac{\lambda'}{r}
-
\frac{1}{4}\nu'^{2}
-
\frac{1}{4}\lambda'^{2}
-
\frac{2}{r^{2}}
\end{aligned}
\end{equation}
The terms \( +\frac{2}{r^{2}} \) and \( -\frac{2}{r^{2}} \) cancel. Also \( \frac{1}{2}\lambda'' - \frac{1}{2}\lambda'' = 0 \).
We obtain:
\begin{equation}
R_{rr}
=
-\frac{1}{2}\nu''
+
\frac{1}{4}(\nu' + \lambda')\lambda'
+
\frac{\lambda'}{r}
-
\frac{1}{4}\nu'^{2}
-
\frac{1}{4}\lambda'^{2}
\end{equation}
Simplify:
\begin{equation}
\boxed{
R_{rr}
=
-\frac{1}{2}\nu''
+
\frac{1}{4}\nu'\lambda'
+
\frac{1}{4}\lambda'^{2}
+
\frac{\lambda'}{r}
-
\frac{1}{4}\nu'^{2}
-
\frac{1}{4}\lambda'^{2}
}
\end{equation}
The \( \lambda'^2 \) terms cancel:
\begin{equation}
\boxed{
R_{rr}
=
-\frac{1}{2}\nu''
+
\frac{1}{4}\nu'\lambda'
+
\frac{\lambda'}{r}
-
\frac{1}{4}\nu'^{2}
}
\end{equation}
or:
\begin{equation}
R_{rr}
=
-\frac{\nu''}{2}
-\frac{\nu'^2}{4}
+\frac{\nu'\lambda'}{4}
+\frac{\lambda'}{r}
\end{equation}
By factoring out \( -\tfrac{1}{2} \), we obtain the conventional form:
\begin{equation}
\boxed{
R_{rr}
=
-\frac{1}{2}
\left(
\nu'' + \frac{1}{2}\nu'^2 - \frac{1}{2}\nu'\lambda' - \frac{2\lambda'}{r}
\right)
}
\end{equation}
Full derivation of \( R_{\theta\theta} \)
The definition:
\begin{equation}
R_{\theta\theta}
=
\frac{\partial \Gamma^{\sigma}_{\theta\theta}}{\partial x^{\sigma}}
-
\frac{\partial \Gamma^{\sigma}_{\theta\sigma}}{\partial x^{\theta}}
+
\Gamma^{\sigma}_{\lambda\sigma}\Gamma^{\lambda}_{\theta\theta}
-
\Gamma^{\sigma}_{\theta\lambda}\Gamma^{\lambda}_{\theta\sigma}
\end{equation}
We take \( \mu = \nu = \theta \). We again split into four terms.
1. First term: \( \dfrac{\partial \Gamma^{\sigma}_{\theta\theta}}{\partial x^{\sigma}} \)
Only \( \sigma = r \) contributes, since:
\begin{equation}
\Gamma^{\sigma}_{\theta \theta} = -r e^{-\lambda}
\end{equation}
Thus:
\begin{equation}
\frac{\partial}{\partial x^\sigma}\Gamma^{\sigma}_{\theta \theta}
=
\frac{\partial}{\partial r}\Gamma^{r}_{\theta \theta}
=
\frac{\partial}{\partial r}(-r e^{-\lambda})
=
-e^{-\lambda} + r e^{-\lambda}\lambda'
\end{equation}
2. Second term: \( -\dfrac{\partial \Gamma^{\sigma}_{\theta\sigma}}{\partial x^{\theta}} \)
Only \( \sigma = \phi \) contributes, since (see equation (\ref{eq:R25})):
\begin{equation}
\Gamma^{\phi}_{\theta\phi} = \cot\theta
\end{equation}
Thus:
\begin{equation}
-\frac{\partial}{\partial \theta}\Gamma^{\phi}_{\theta\phi}
=
-\frac{\partial}{\partial \theta}(\cot\theta)
=
\frac{1}{\sin^{2}\theta}
\end{equation}
3. Third term: \( \Gamma^{\sigma}_{\lambda\sigma}\Gamma^{\lambda}_{\theta\theta} \)
For \( \lambda = r \):
\begin{equation}
\Gamma^{r}_{\theta \theta} = -r e^{-\lambda}
\end{equation}
We again use:
\begin{equation}
\Gamma^{\sigma}_{r\sigma}
=
\frac{1}{2}(\nu' + \lambda') + \frac{2}{r}
\end{equation}
Thus:
\begin{equation}
\Gamma^{\sigma}_{\lambda\sigma}\Gamma^{\lambda}_{\theta\theta}
=
\Gamma^{\sigma}_{\sigma r}\Gamma^{r}_{\theta\theta}
=
\left[
\frac{1}{2}(\nu' + \lambda') + \frac{2}{r}
\right]
(-r e^{-\lambda})
\end{equation}
Expanding:
\begin{equation}
=
-\frac{1}{2} r e^{-\lambda}(\nu' + \lambda')
-
2 e^{-\lambda}
\end{equation}
4. Fourth term: \( -\Gamma^{\sigma}_{\theta\lambda}\Gamma^{\lambda}_{\theta\sigma} \)
We first write the full sum of all possible products:
\begin{equation}
\Gamma^{\sigma}_{\theta\lambda}\Gamma^{\lambda}_{\theta\sigma}
=
\Gamma^{t}_{\theta t}\Gamma^{t}_{\theta t}
+
\Gamma^{t}_{\theta r}\Gamma^{r}_{\theta t}
+
\Gamma^{t}_{\theta \theta}\Gamma^{\theta}_{\theta\ t}
+
\Gamma^{t}_{\theta \phi}\Gamma^{\phi}_{\theta t}
\end{equation}
\begin{equation}
+
\Gamma^{r}_{\theta t}\Gamma^{t}_{\theta r}
+
\Gamma^{r}_{\theta r}\Gamma^{r}_{\theta r}
+
\Gamma^{r}_{\theta \theta}\Gamma^{\theta}_{\theta r}
+
\Gamma^{r}_{\theta \phi}\Gamma^{\phi}_{\theta r}
\end{equation}
\begin{equation}
+
\Gamma^{\theta}_{\theta t}\Gamma^{t}_{\theta \theta}
+
\Gamma^{\theta}_{\theta r}\Gamma^{r}_{\theta \theta}
+
\Gamma^{\theta}_{\theta \theta}\Gamma^{\theta}_{\theta\theta}
+
\Gamma^{\theta}_{\theta \phi}\Gamma^{\phi}_{\theta\theta}
\end{equation}
\begin{equation}
+
\Gamma^{\phi}_{\phi t}\Gamma^{t}_{\theta \phi}
+
\Gamma^{\phi}_{\theta r}\Gamma^{r}_{\theta \phi}
+
\Gamma^{\phi}_{\theta\theta}\Gamma^{\theta}_{\theta\phi}
+
\Gamma^{\phi}_{\theta\phi}\Gamma^{\phi}_{\theta\phi}
\end{equation}
For the static spherically symmetric metric, almost all terms vanish. The only non-zero contributions are:
\begin{equation}
\Gamma^{\theta}_{\theta r} = \frac{1}{r},
\qquad
\Gamma^{r}_{\theta\theta} = -r e^{-\lambda},
\qquad
\Gamma^{\phi}_{\theta\phi} = \cot\theta.
\end{equation}
Substituting gives:
\begin{equation}
\Gamma^{\sigma}_{\theta\lambda}\Gamma^{\lambda}_{\theta\sigma}
=
\left(-r e^{-\lambda}\right)\left(\frac{1}{r}\right)
+
\left(\frac{1}{r}\right)\left(-r e^{-\lambda}\right)
+
\cot^{2}\theta
\end{equation}
\begin{equation}
=
-2 e^{-\lambda} + \cot^{2}\theta
\end{equation}
Therefore:
\begin{equation}
\boxed{
-\Gamma^{\sigma}_{\theta\lambda}\Gamma^{\lambda}_{\theta\sigma}
=
2 e^{-\lambda} - \cot^{2}\theta
}
\end{equation}
Relevant non-zero products (compact)
-
Radial contribution:
\begin{equation} -\Gamma^{r}_{\theta \theta}\Gamma^{\theta}_{\theta r} = -\left(-r e^{-\lambda}\right)\left(\frac{1}{r}\right) = e^{-\lambda}. \end{equation} -
Angular contribution:
\begin{equation} -\Gamma^{\phi}_{\theta\phi}\Gamma^{\phi}_{\theta\phi} = -\cot^{2}\theta. \end{equation}This term combines with the contribution from step 2:\begin{equation} -\frac{\partial}{\partial \theta}\Gamma^{\phi}_{\theta\phi} = \frac{1}{\sin^{2}\theta}, \end{equation}and with the identity:\begin{equation} \frac{1}{\sin^{2}\theta} - \cot^{2}\theta = 1, \end{equation}so that the total angular contribution yields exactly the well-known spherical constant \(1\).
In summary: the radial term contributes \(2e^{-\lambda}\), while the angular terms together yield \(1\), which fits exactly into the standard derivation of \(R_{\theta\theta}\).
5. Sum all contributions (radial + angular parts)
We now combine all terms:
\begin{equation}
\begin{aligned}
R_{\theta\theta}
&=
\left(-e^{-\lambda} + r e^{-\lambda}\lambda'\right)
+
\frac{1}{\sin^{2}\theta}
+
\left(
-\frac{1}{2} r e^{-\lambda}(\nu' + \lambda')
-
2 e^{-\lambda}
\right)
\\ &\quad
+
\left(
2 e^{-\lambda} - \cot^{2}\theta
\right)
\end{aligned}
\end{equation}
The angular parts combine to:
\begin{equation}
\frac{1}{\sin^{2}\theta} - \cot^{2}\theta = 1
\end{equation}
Thus:
\begin{equation}
R_{\theta\theta}
=
1
-
e^{-\lambda}
+
r e^{-\lambda}\lambda'
-
\frac{1}{2} r e^{-\lambda}(\nu' + \lambda')
\end{equation}
Simplifying:
\begin{equation}
\boxed{
R_{\theta\theta}
=
1
-
e^{-\lambda}
\left(
1
+
\frac{r}{2}(\nu' - \lambda')
\right)
}
\end{equation}
Full derivation of \( R_{\phi\phi} \)
The definition:
\begin{equation}
R_{\phi\phi}
=
\frac{\partial \Gamma^{\sigma}_{\phi\phi}}{\partial x^{\sigma}}
-
\frac{\partial \Gamma^{\sigma}_{\phi\sigma}}{\partial \phi}
+
\Gamma^{\sigma}_{\sigma\lambda}\Gamma^{\lambda}_{\phi\phi}
-
\Gamma^{\sigma}_{\phi\lambda}\Gamma^{\lambda}_{\phi\sigma}
\end{equation}
We take \( \mu = \nu = \phi \). Again, we split into four terms.
1. First term: \( \dfrac{\partial \Gamma^{\sigma}_{\phi\phi}}{\partial x^{\sigma}} \)
The non-zero Christoffel symbols are:
\begin{equation}
\Gamma^{r}_{\phi \phi} = -r e^{-\lambda}\sin^{2}\theta,
\qquad
\Gamma^{\theta}_{\phi\phi} = -\sin\theta\cos\theta
\end{equation}
Thus:
\begin{equation}
\frac{\partial \Gamma^{r}_{\phi \phi}}{\partial x^\sigma}
=
-e^{-\lambda}\sin^{2}\theta
\left(1 - r\lambda'\right)
\end{equation}
\begin{equation}
\frac{\partial \Gamma^{\theta}_{\phi\phi}}{\partial \theta}
=
-(\cos^{2}\theta - \sin^{2}\theta)
\end{equation}
Combined:
\begin{equation}
\frac{\partial \Gamma^{\sigma}_{\phi \phi}}{\partial x^\sigma}
=
-e^{-\lambda}\sin^{2}\theta\,(1 - r\lambda')
-
(\cos^{2}\theta - \sin^{2}\theta)
\end{equation}
2. Second term: \( -\dfrac{\partial \Gamma^{\sigma}_{\phi\sigma}}{\partial \phi} \)
All Christoffel symbols are independent of \( \phi \), so:
\begin{equation}
-\frac{\partial}{\partial \phi}\Gamma^{\sigma}_{\phi\sigma} = 0
\end{equation}
3. Third term: \( \Gamma^{\sigma}_{\sigma\lambda}\Gamma^{\lambda}_{\phi\phi} \)
Non-zero contributions come from \( \lambda = r \) and \( \lambda = \theta \):
\begin{equation}
\Gamma^{r}_{\phi \phi} = -r e^{-\lambda}\sin^{2}\theta,
\qquad
\Gamma^{\theta}_{\phi\phi} = -\sin\theta\cos\theta
\end{equation}
We again use:
\begin{equation}
\begin{aligned}
\Gamma^{\sigma}_{\sigma r}
=
\Gamma^{t}_{tr}
+
\Gamma^{r}_{rr}
+
\Gamma^{\theta}_{\theta r}
+
\Gamma^{\phi}_{\phi r}
&=
\frac{1}{2}\nu'
+
\frac{1}{2}\lambda'
+
\frac{1}{r}
+
\frac{1}{r}
\\ &\quad
=
\frac{1}{2}(\nu' + \lambda') + \frac{2}{r}
\end{aligned}
\end{equation}
\begin{equation}
\Gamma^{\sigma}_{\sigma\theta}
=
\Gamma^{t}_{t \theta}
+
\Gamma^{r}_{r \theta}
+
\Gamma^{\theta}_{\theta \theta}
+
\Gamma^{\phi}_{\phi \theta}
=0+0+0+\cot{\theta}
\end{equation}
\begin{equation}
\Gamma^{\sigma}_{\sigma\lambda}\Gamma^{\lambda}_{\phi\phi}
=
\Gamma^{\sigma}_{\sigma r}\Gamma^{r}_{\phi\phi}+
\Gamma^{\sigma}_{\sigma \theta}\Gamma^{\theta}_{\phi\phi}
\end{equation}
\begin{equation}
=\left(
\frac{1}{2}\nu'
+
\frac{1}{2}\lambda'
+
\frac{2}{r}
\right)\,
\Gamma^{r}_{\phi\phi}
+
\left(\cot{\theta}\right)\,
\Gamma^{\theta}_{\phi\phi}
\end{equation}
Thus:
\begin{equation}
\Gamma^{\sigma}_{\sigma\lambda}\Gamma^{\lambda}_{\phi\phi}
=
\left[
\frac{1}{2}(\nu' + \lambda') + \frac{2}{r}
\right]
\left(
-r e^{-\lambda}\sin^{2}\theta
\right)
+
\cot\theta\left(-\sin\theta\cos\theta\right)
\end{equation}
Expanding:
\begin{equation}
=
-e^{-\lambda}\sin^{2}\theta
\left(
\frac{r}{2}(\nu' + \lambda') + 2
\right)
-
\cos^{2}\theta
\end{equation}
4. Fourth term: \( -\Gamma^{\sigma}_{\phi\lambda}\Gamma^{\lambda}_{\phi\sigma} \)
The non-zero products are:
\begin{equation}
\Gamma^{r}_{\phi\phi}\Gamma^{\phi}_{\phi r}
=
\Gamma^{\phi}_{\phi r}\Gamma^{r}_{\phi\phi}
=
\left(\frac{1}{r}\right)\left(-r e^{-\lambda}\sin^{2}\theta\right)
=
-e^{-\lambda}\sin^{2}\theta
\end{equation}
\begin{equation}
\Gamma^{\theta}_{\phi \phi}\Gamma^{\phi}_{\phi\theta}
=
\Gamma^{\phi}_{\phi \theta}\Gamma^{\theta}_{\phi\phi}
=
(\cot\theta)(-\sin\theta\cos\theta)
=
-\cos^{2}\theta
\end{equation}
Now all relevant terms:
\begin{equation}
\Gamma^{\sigma}_{\phi\lambda}\Gamma^{\lambda}_{\phi\sigma}=
\Gamma^{r}_{\phi\phi}\Gamma^{\phi}_{\phi r}
+
\Gamma^{\theta}_{\phi \phi}\Gamma^{\phi}_{\phi\theta}
+
\Gamma^{\phi}_{\phi r}\Gamma^{r}_{\phi\phi}
+
\Gamma^{\phi}_{\phi \theta}\Gamma^{\theta}_{\phi\phi}
\end{equation}
The first and third terms are equal, as are the second and fourth terms. Thus:
\begin{equation}
\Gamma^{\sigma}_{\phi\lambda}\Gamma^{\lambda}_{\phi\sigma}
=
-2e^{-\lambda}\sin^{2}\theta
-
2\cos^{2}\theta
\end{equation}
Thus, the resulting fourth term is:
\begin{equation}
-\Gamma^{\phi}_{\lambda\sigma}\Gamma^{\sigma}_{\phi\lambda}
=
2 e^{-\lambda}\sin^{2}\theta
+
2\cos^{2}\theta
\end{equation}
5. Sum all contributions
We now combine everything:
\begin{equation}
R_{\phi\phi}
=
-e^{-\lambda}\sin^{2}\theta(1 - r\lambda')
-
(\cos^{2}\theta - \sin^{2}\theta)+
\end{equation}
\begin{equation}
-
e^{-\lambda}\sin^{2}\theta
\left(
\frac{r}{2}(\nu' + \lambda') + 2
\right)
-
\cos^{2}\theta
+
2 e^{-\lambda}\sin^{2}\theta
+
2\cos^{2}\theta
\end{equation}
The angular terms simplify to:
\begin{equation}
-(\cos^{2}\theta - \sin^{2}\theta)
-\cos^{2}\theta
+2\cos^{2}\theta
=
\sin^{2}\theta
\end{equation}
The radial terms combine to:
\begin{equation}
R_{\phi\phi}
=
\sin^{2}\theta
\left[
1
-
e^{-\lambda}
\left(
1
+
\frac{r}{2}(\nu' - \lambda')
\right)
\right]
\end{equation}
Thus, the final form is:
\begin{equation}
\boxed{
R_{\phi\phi}
=
\sin^{2}\theta\,
\left[
1
-
e^{-\lambda}
\left(
1
+
\frac{r}{2}(\nu' - \lambda')
\right)
\right]
}
\end{equation}
Alternatively written as:
\begin{equation}
\boxed{
R_{\phi\phi}
=
\sin^{2}\theta
\left[
1
+
e^{-\lambda}\frac{1}{2}
\left(
r\lambda' - r\nu' - 2
\right)
\right]
}
\end{equation}
Comparison with Rθθ
We previously found that:
\begin{equation}
R_{\theta\theta}
=
1
+
e^{-\lambda}\frac{1}{2}
\left(
r\lambda' - r\nu' - 2
\right)
\end{equation}
Therefore, it immediately follows that:
\begin{equation}
\boxed{
R_{\phi\phi}
=
\sin^{2}\theta\, R_{\theta\theta}
}
\end{equation}
This is exactly what one expects for a spherically symmetric metric: the \( \phi\phi \)-component is the \( \theta\theta \)-component multiplied by \( \sin^{2}\theta \).
Remark
The expressions derived here are precisely the standard results from the literature. After substituting
\begin{equation}
e^{-\lambda} = 1 - \frac{2Gm(r)}{c^{2}r}
\end{equation}
one obtains the usual Einstein equations for a static star,
from which, among other things, the mass function \( m(r) \) and the TOV equation follow.
Ricci scalar
The Ricci scalar is:
\begin{equation}
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}
\end{equation}
We use:
\begin{equation}
g^{\mu\nu}
=
\begin{pmatrix}
e^{-\nu} & 0 & 0 & 0 \\
0 & -e^{-\lambda} & 0 & 0 \\
0 & 0 & -\frac{1}{r^{2}} & 0 \\
0 & 0 & 0 & -\frac{1}{r^{2}\sin^{2}\theta}
\end{pmatrix}
\end{equation}
Since
\begin{equation}
R_{\phi\phi} = \sin^{2}\theta\, R_{\theta\theta},
\qquad
g^{\phi\phi} = -\frac{1}{r^{2}\sin^{2}\theta},
\end{equation}
it follows that:
\begin{equation}
g^{\phi\phi} R_{\phi\phi}
=
-\frac{1}{r^{2}\sin^{2}\theta}
\cdot
\sin^{2}\theta\, R_{\theta\theta}
=
-\frac{1}{r^{2}} R_{\theta\theta}
=
g^{\theta\theta} R_{\theta\theta}
\end{equation}
Thus, the angular contributions combine elegantly:
\begin{equation}
g^{\theta\theta}R_{\theta\theta} + g^{\phi\phi}R_{\phi\phi}
=
2 g^{\theta\theta} R_{\theta\theta}
\end{equation}
Ricci scalar from the components
We use the derived components:
\begin{equation}
R_{tt}
=
\frac{1}{2} e^{\nu-\lambda}
\left(
\nu'' + \frac{1}{2}\nu'^2 - \frac{1}{2}\lambda'\nu' + \frac{2\nu'}{r}
\right)
\end{equation}
\begin{equation}
R_{rr}
=
-\frac{1}{2}
\left(
\nu''
+ \frac{1}{2}\nu'^2
- \frac{1}{2}\nu'\lambda'
- \frac{2\lambda'}{r}
\right)
\end{equation}
\begin{equation}
R_{\theta\theta}
=
1
-
e^{-\lambda}
\left(
1 + \frac{r}{2}(\nu' - \lambda')
\right)
\end{equation}
\begin{equation}
R_{\phi\phi}
=
\sin^{2}\theta\, R_{\theta\theta}
\end{equation}
Ricci scalar
The Ricci scalar is:
\begin{equation}
R = g^{\mu\nu} R_{\mu\nu}
= g^{tt}R_{tt} + g^{rr}R_{rr} + 2 g^{\theta\theta} R_{\theta\theta}
\end{equation}
With:
\begin{equation}
g^{tt} = e^{-\nu},\qquad
g^{rr} = -e^{-\lambda},\qquad
g^{\theta\theta} = -\frac{1}{r^{2}}
\end{equation}
Thus:
\begin{equation}
R
=
e^{-\nu} R_{tt}
-
e^{-\lambda} R_{rr}
-
\frac{2}{r^{2}} R_{\theta\theta}
\end{equation}
Substituting the components gives:
\begin{equation}
\boxed{
R
=
e^{-\lambda}
\left(
\nu'' + \frac{1}{2}\nu'^2 - \frac{1}{2}\lambda'\nu'
+ \frac{2}{r}(\nu' - \lambda')
\right)
+
\frac{2}{r^{2}}(e^{-\lambda} - 1)
}
\end{equation}
Ricci tensor components
The relevant components of the Ricci tensor for the internal Schwarzschild metric are:
\begin{equation}
\boxed{
R_{tt}
=
\frac{1}{2} e^{\nu-\lambda}
\left(
\nu'' + \frac{1}{2}\nu'^2 - \frac{1}{2}\lambda'\nu' + \frac{2\nu'}{r}
\right)
}
\end{equation}
\begin{equation}
\boxed{
R_{rr}
=
-\frac{1}{2}
\left(
\nu''
+ \frac{1}{2}\nu'^2
- \frac{1}{2}\nu'\lambda'
- \frac{2\lambda'}{r}
\right)
}
\label{eq:R133}
\end{equation}
\begin{equation}
\boxed{
R_{\theta\theta}
=
1
-
e^{-\lambda}
\left(
1 + \frac{r}{2}(\nu' - \lambda')
\right)
}
\end{equation}
\begin{equation}
\boxed{
R_{\phi\phi}
=
\sin^{2}\theta\, R_{\theta\theta}
}
\end{equation}
Ricci scalar
\begin{equation}
\boxed{
R
=
e^{-\lambda}
\left(
\nu'' + \frac{1}{2}\nu'^2 - \frac{1}{2}\lambda'\nu'
+ \frac{2}{r}(\nu' - \lambda')
\right)
+
\frac{2}{r^{2}}(e^{-\lambda} - 1)
}
\label{eq:R136}
\end{equation}
These components form the basis for the Einstein equations inside a star, from which the mass function \(m(r)\), the pressure equation, and ultimately the Tolman–Oppenheimer–Volkoff equation follow.
Appendix 5.4 — Einstein equations explicitly
Substituting (\ref{eq:R133}) and (\ref{eq:R136}) into the Einstein equation
\begin{equation}
G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2}R\, g_{\mu\nu}
= \frac{8\pi G}{c^{4}} T_{\mu\nu}
\label{eq:R137}
\end{equation}
yields three independent equations.
(i) tt-component
\begin{equation}
G_{tt}
=
R_{tt}
-
\frac{1}{2} R\, g_{tt}
=
\frac{8\pi G}{c^{4}}\, \rho c^{2} e^{\nu}
\end{equation}
Substituting:
\begin{equation}
G_{tt}
=
\frac{1}{2} e^{\nu-\lambda}
\left(
\nu'' + \frac{1}{2}\nu'^2 - \frac{1}{2}\lambda'\nu' + \frac{2\nu'}{r}
\right)+
\end{equation}
\begin{equation}
-
\frac{1}{2} e^{\nu}
\left[
e^{-\lambda}
\left(
\nu'' + \frac{1}{2}\nu'^2 - \frac{1}{2}\lambda'\nu'
+ \frac{2}{r}(\nu' - \lambda')
\right)
+
\frac{2}{r^{2}}(e^{-\lambda} - 1)
\right]
\end{equation}
\begin{equation}
G_{tt}
=
\frac{1}{2} e^{\nu-\lambda} \frac{2\lambda'}{r}
-
\frac{1}{2} e^{\nu} \frac{2(e^{-\lambda}-1)}{r^2}
\end{equation}
\begin{equation}
=
\frac{\lambda'}{r} e^{\nu-\lambda}
-
\frac{1}{r^2} e^{\nu-\lambda}
+
\frac{1}{r^2} e^{\nu}
\end{equation}
\begin{equation}
=
\frac{1}{r^2} e^{\nu-\lambda} (r\lambda' - 1)
+
\frac{e^{\nu}}{r^2}
\end{equation}
\begin{equation}
= \frac{e^{\nu}}{r^2} \left( e^{-\lambda}(r\lambda' - 1) + 1 \right)
= \frac{8\pi G}{c^4} \rho c^2 e^{\nu}
\end{equation}
Thus:
\begin{equation}
\frac{1}{r^2} \left( 1 + e^{-\lambda}(r\lambda' - 1) \right)
= \frac{8\pi G}{c^2} \rho
\end{equation}
Or in total derivative form:
\begin{equation}
\boxed{
\frac{1}{r^{2}}
\frac{d}{dr}
\left[
r\left(1 - e^{-\lambda}\right)
\right]
=
\frac{8\pi G}{c^{2}}\, \rho
}
\label{eq:R146}
\end{equation}
Expanding (\ref{eq:R146}) gives:
\begin{equation}
\frac{1}{r^2}\left(1 - e^{-\lambda} + r e^{-\lambda}\lambda'\right)
=
\frac{1}{r^2}\left(1 + e^{-\lambda}(r\lambda' - 1)\right)
=
\frac{8\pi G}{c^2}\rho
\end{equation}
(\ref{eq:R146}) is therefore identical to the previous expression.
(ii) rr-component
\begin{equation}
G_{rr}
=
R_{rr}
-
\frac{1}{2}R\, g_{rr}
=
\frac{8\pi G}{c^{4}}\, p\, e^{\lambda}
\end{equation}
Substituting:
\begin{equation}
G_{rr}
=
-\frac{1}{2}
\left(
\nu''
+ \frac{1}{2}\nu'^2
- \frac{1}{2}\lambda'\nu'
- \frac{2\lambda'}{r}
\right)
+
\end{equation}
\begin{equation}
+
\frac{1}{2} e^{\lambda}
\left[
e^{-\lambda}
\left(
\nu'' + \frac{1}{2}\nu'^2 - \frac{1}{2}\lambda'\nu'
+ \frac{2}{r}(\nu' - \lambda')
\right)
+
\frac{2}{r^{2}}(e^{-\lambda} - 1)
\right]
\end{equation}
\begin{equation}
\begin{aligned}
&= -\frac{1}{2}\left(\nu'' + \frac{1}{2}(\nu')^2 - \frac{1}{2}\lambda'\nu' - \frac{2\lambda'}{r}\right)
\\ &\quad
+ \frac{1}{2}\left[\left(\nu'' + \frac{1}{2}(\nu')^2 - \frac{1}{2}\lambda'\nu' + \frac{2(\nu' - \lambda')}{r}\right)
+ \frac{2(e^{-\lambda} - 1)}{r^2} e^{\lambda}\right]
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
&= -\frac{1}{2}\left(\nu'' + \frac{1}{2}(\nu')^2 - \frac{1}{2}\lambda'\nu' - \frac{2\lambda'}{r}\right)
\\ &\quad
+ \frac{1}{2}\left(\nu'' + \frac{1}{2}(\nu')^2 - \frac{1}{2}\lambda'\nu' + \frac{2(\nu' - \lambda')}{r}\right)
+ \frac{e^{-\lambda} - 1}{r^2} e^{\lambda}
\end{aligned}
\end{equation}
\begin{equation}
= \frac{1}{2} \left( \frac{2(\nu' - \lambda')}{r} + \frac{2\lambda'}{r} \right)
+ \frac{1 - e^{\lambda}}{r^2}
= \frac{8\pi G}{c^4} \rho \, e^{\lambda}
\end{equation}
\begin{equation}
= \frac{\nu'}{r} + \frac{1}{r^2}(1 - e^{\lambda})
= \frac{8\pi G}{c^4} \rho \, e^{\lambda}
\end{equation}
\begin{equation}
= \frac{\nu'}{r} e^{-\lambda} + \frac{1}{r^2}(e^{-\lambda} - 1)
= \frac{8\pi G}{c^4} \rho
\end{equation}
Or in standard form:
\begin{equation}
\boxed{
\frac{\nu'}{r}
+
\frac{1}{r^{2}}
\left(
1 - e^{\lambda}
\right)
=
\frac{8\pi G}{c^{4}}\, p\, e^{\lambda}
}
\label{eq:R156}
\end{equation}
This is the second independent Einstein equation.
(iii) θθ-component
\begin{equation}
G_{\theta\theta}
=
R_{\theta\theta}
-
\frac{1}{2}R\, g_{\theta\theta}
=
\frac{8\pi G}{c^{4}}\, p\, r^{2}
\end{equation}
Substituting the components:
\begin{equation}
\begin{aligned}
G_{\theta\theta}
&=
1 - e^{-\lambda}
\left(1+ \frac{r^2}{2}(\nu' - \lambda')\right)
\\ &\quad
- \frac{1}{2}(-r^2)\left[e^{-\lambda}
\left(
\nu'' + \frac{1}{2}(\nu')^2 - \frac{1}{2}\lambda'\nu' + \frac{2(\nu' - \lambda')}{r}\right)
+ \frac{2(e^{-\lambda} - 1)}{r^2}
\right]
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
&=
1 - e^{-\lambda}
\left(1+ \frac{r^2}{2}(\nu' - \lambda')\right)
\\ &\quad
+ \frac{r^2}{2}\left[ e^{-\lambda}
\left(
\nu'' + \frac{1}{2}(\nu')^2 - \frac{1}{2}\lambda'\nu' + \frac{2(\nu' - \lambda')}{r}\right)
+ \frac{2(e^{-\lambda} - 1)}{r^2}
\right]
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
&=
1 - e^{-\lambda}
\left(1+ \frac{r^2}{2}(\nu' - \lambda')\right)
+ \frac{r^2}{2} e^{-\lambda}
\left(
\nu'' + \frac{1}{2}(\nu')^2 - \frac{1}{2}\lambda'\nu'+\frac{2(\nu' - \lambda')}{r}
\right)
\\ &\quad
+ r e^{-\lambda}(\nu' - \lambda')
+ (e^{-\lambda} - 1)
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
&=
1 - e^{-\lambda}
\left(1+ \frac{r^2}{2}(\nu' - \lambda')\right)
\\ &\quad
+ \left[\frac{r^2}{2} e^{-\lambda}
\left(
\nu'' + \frac{1}{2}(\nu')^2 - \frac{1}{2}\lambda'\nu'
\right)-1
+ e^{-\lambda}\left(1+r(\nu' - \lambda')\right)
\right]
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
&=
\frac{1}{2} r^2 e^{-\lambda}
\left(
\nu'' + \frac{1}{2}(\nu')^2 - \frac{1}{2}\lambda'\nu'
\right)
+1 - e^{-\lambda}
\\ &\quad
-\frac{r}{2}e^{-\lambda}(\nu'-\lambda')-1
+e^{-\lambda}+re^{-\lambda}(\nu'-\lambda')
\end{aligned}
\end{equation}
\begin{equation}
G_{\theta\theta}
=
\frac{1}{2} r^{2} e^{-\lambda}
\left(
\nu'' + \frac{1}{2}\nu'^2 - \frac{1}{2}\lambda'\nu'
+ \frac{1}{r}(\nu' - \lambda')
\right)
=
\frac{8\pi G}{c^{4}}\, p\, r^{2}
\end{equation}
Thus:
\begin{equation}
\boxed{
\frac{1}{2} e^{-\lambda}
\left(
\nu'' + \frac{1}{2}\nu'^2 - \frac{1}{2}\lambda'\nu'
+ \frac{1}{r}(\nu' - \lambda')
\right)
=
\frac{8\pi G}{c^{4}}\, p
}
\label{eq:R164}
\end{equation}
Explanation
The three equations (\ref{eq:R146}), (\ref{eq:R156}) and (\ref{eq:R164}) are not independent. Equation (\ref{eq:R164}) follows mathematically from (\ref{eq:R146}), (\ref{eq:R156}) and the energy conservation equation:
\begin{equation}
\nabla_{\mu} T^{\mu}_{r} = 0
\end{equation}
Therefore, it suffices to use (\ref{eq:R146}) and (\ref{eq:R156}) to derive \( \lambda(r) \), \( \nu(r) \), and \( p(r) \).
Appendix 5.5 — Integration of the first equation
We start with equation (\ref{eq:R146}):
\begin{equation}
\frac{1}{r^{2}}
\frac{d}{dr}
\left[
r\left(1 - e^{-\lambda}\right)
\right]
=
\frac{8\pi G}{c^{2}}\, \rho
\end{equation}
Multiply by \( r^{2} \):
\begin{equation}
\frac{d}{dr}
\left[
r\left(1 - e^{-\lambda}\right)
\right]
=
\frac{8\pi G}{c^{2}}\, \rho\, r^{2}
\end{equation}
Integrate over \( r \):
\begin{equation}
\int
\frac{d}{dr}
\left[
r\left(1 - e^{-\lambda}\right)
\right] dr
=
\int
\frac{8\pi G}{c^{2}}\, \rho\, r^{2}\, dr
\end{equation}
Thus:
\begin{equation}
r\left(1 - e^{-\lambda}\right)
=
\frac{8\pi G}{c^{2}}\, \rho\, \frac{r^{3}}{3}
\end{equation}
Or:
\begin{equation}
\boxed{
1 - e^{-\lambda}
=
\frac{8\pi G}{3c^{2}}\, \rho\, r^{2}
}
\end{equation}
Result of the integration
From equation (\ref{eq:R171}):
\begin{equation}
e^{-\lambda(r)} = 1 - \frac{8\pi G}{3c^{2}}\,\rho\, r^{2}
\label{eq:R171}
\end{equation}
With the definition of the enclosed mass:
\begin{equation}
m(r)
=
4\pi \int_{t}^{r} \rho\, r'^{2}\, dr'
=
\frac{4\pi}{3}\rho\, r^{3}
\end{equation}
we obtain the standard form:
\begin{equation}
\boxed{
e^{-\lambda(r)} = 1 - \frac{2Gm(r)}{c^{2}r}
}
\label{eq:R173}
\end{equation}
The function \( m(r) \) is the enclosed mass within radius \( r \). For \( r = R \), one has \( m(R) = M \), so that the internal solution smoothly matches the external Schwarzschild metric.
Appendix 5.6 — Energy Conservation and the TOV Equation
1. Energy Conservation
The conservation equation
\begin{equation}
\nabla_{\mu} T^{\mu}_{r} = 0
\end{equation}
gives:
\begin{equation}
\boxed{
\frac{dp}{dr}
=
-\frac{1}{2}(\rho c^{2} + p)\, \nu'
}
\label{eq:R175}
\end{equation}
2. Eliminating \( \nu' \) using equation (\ref{eq:R156})
From (\ref{eq:R156}):
\begin{equation}
\frac{\nu'}{r} e^{-\lambda}
+
\frac{1}{r^{2}}(e^{-\lambda} - 1)
=
\frac{8\pi G}{c^{4}}\, p
\end{equation}
Solving for \( \nu' \):
\begin{equation}
\nu'
=
\frac{
\frac{8\pi G}{c^{4}}\, p\, r
+
\frac{1}{r}\left(1 - e^{-\lambda}\right)}
{e^{-\lambda}}
\end{equation}
Now use (\ref{eq:R173}):
\begin{equation}
e^{-\lambda} = 1 - \frac{2Gm}{c^{2}r},
\qquad
1 - e^{-\lambda} = \frac{2Gm}{c^{2}r}
\end{equation}
Substitution gives:
\begin{equation}
\nu'
=
\frac{
\frac{8\pi G}{c^{4}}\, p\, r
+
\frac{1}{r}
\frac{2Gm}{c^{2}r}}{1 - \frac{2Gm}{c^{2}r}}
\end{equation}
\begin{equation}
=\frac{
\frac{8\pi G}{c^{2}}\, p r^{3}
+
2Gm}
{c^2r^2\left(1 - \frac{2Gm}{c^{2}r}\right)}
=
=\frac{2G\left(m+
\frac{4\pi}{c^{2}}\, p r^{3}\right)
}
{c^2r^2\left(1 - \frac{2Gm}{c^{2}r}\right)}
\end{equation}
Rewritten:
\begin{equation}
\boxed{
\nu'
=
\frac{2G}{c^{2}}
\frac{m + 4\pi r^{3}p/c^{2}}{r^{2}\left(1 - \frac{2Gm}{c^{2}r}\right)}
}
\label{eq:R181}
\end{equation}
3. The Tolman–Oppenheimer–Volkoff Equation
Combining (\ref{eq:R175}) and (\ref{eq:R181}) gives:
\begin{equation}
\frac{dp}{dr}
=
-\frac{1}{2}(\rho c^{2} + p)\, \nu'
\end{equation}
\begin{equation}
\frac{dp}{dr}
=
-(\rho c^{2} + p)
\frac{G\left(m + 4\pi r^{3}p/c^{2}\right)}{c^{2} r^{2}\left(1 - \frac{2Gm}{c^{2}r}\right)}
\end{equation}
Thus the TOV equation reads:
\begin{equation}
\boxed{
\frac{dp}{dr}
=
-\frac{(\rho c^{2} + p)G\left(m + 4\pi r^{3}p/c^{2}\right)}
{c^{2} r^{2}\left(1 - \frac{2Gm}{c^{2}r}\right)}
}
\label{eq:R184}
\end{equation}
This is the relativistic equilibrium equation for a static, spherically symmetric star. Together with the mass function
\begin{equation}
m'(r) = 4\pi r^{2}\rho
\end{equation}
this forms the complete system for the internal structure of a star.
Explanation
Equation (\ref{eq:R184}) describes the mechanical equilibrium between gravity (inward) and the pressure gradient (outward). For constant density \( \rho = \text{constant} \), this equation is exactly solvable.
Appendix 5.7 — Solution for Constant Density
With
\begin{equation}
m(r) = \frac{4\pi}{3}\rho\, r^{3}
\end{equation}
the TOV equation (\ref{eq:R184}) becomes:
\begin{equation}
\frac{dp}{dr}
=
-
\frac{\rho c^{2} + p}{3c^{2}}
\,
\frac{4\pi G r \left(\rho + 3p/c^{2}\right)}
{1 - \frac{8\pi G}{3c^{2}}\rho r^{2}}
\label{eq:R187}
\end{equation}
Substitution: pressure as a dimensionless variable
Define:
\begin{equation}
x(r) \equiv \frac{p(r)}{\rho c^{2}}
\qquad\Longrightarrow\qquad
p = \rho c^{2} x,
\quad
dp = \rho c^{2}\, dx
\label{eq:R188}
\end{equation}
Substitute this into ((\ref{eq:R187})).
Left-hand side
\begin{equation}
\frac{dp}{dr}
=
\rho c^{2}\, \frac{dx}{dr}
\end{equation}
Right-hand side
\begin{equation}
\rho + \frac{3p}{c^{2}}
=
\rho(1 + 3x),
\qquad
\rho c^{2} + p
=
\rho c^{2}(1 + x)
\end{equation}
Thus equation (\ref{eq:R187}) becomes:
\begin{equation}
\frac{dp}{dr}= -\rho c^2 (1 + x)
\cdot
\frac{4\pi G r \rho(1 + 3x)}{3c^2 (1 - \beta r^2)}
\end{equation}
\begin{equation}
= -\frac{4\pi G \rho}{3} \,
\frac{r(1 + x)(1+3x)\rho c^2}{c^2 (1 - \beta r^2)}
\end{equation}
\begin{equation}
= -\frac{4\pi G \rho}{3} \,
\frac{r(1 + x)(1+3x)\rho }{1 - \beta r^2}
\end{equation}
\begin{equation}
\rho c^{2}\frac{dx}{dr}
= -\frac{4\pi G \rho}{3} \,
\frac{r(1 + x)(1+3x)\rho }{1 - \beta r^2}
\end{equation}
where
\begin{equation}
\beta \equiv \frac{8\pi G}{3c^{2}}\rho.
\end{equation}
Simplify:
\begin{equation}
\frac{dx}{dr}
=
-
\frac{4\pi G \rho}{3c^{2}}
\,
\frac{r(1+x)(1+3x)}{1 - \beta r^{2}}
\end{equation}
Now introduce:
\begin{equation}
\frac{\beta}{2} \equiv \frac{4\pi G}{3c^{2}}\rho
\end{equation}
so that:
\begin{equation}
\frac{dx}{dr}
=
-
\frac{\beta}{2}\,
\frac{r(1+x)(1+3x)}{1 - \beta r^{2}}
\end{equation}
Separation of variables
We rewrite:
\begin{equation}
\frac{dx}{(1+x)(1+3x)}
=
-
\frac{\beta}{2}\,
\frac{r\, dr}{1 - \beta r^{2}}
\end{equation}
This is the form that can be directly integrated:
\begin{equation}
\boxed{
\int \frac{dx}{(1+x)(1+3x)}
=
-
\frac{\beta}{2}
\int
\frac{r\, dr}{1 - \beta r^{2}}
}
\end{equation}
The left-hand side requires partial fractions; the right-hand side yields a logarithm.
Integration of both sides
We start with:
\begin{equation}
\frac{dx}{(1+x)(1+3x)}
=
-\frac{\beta}{2}\,
\frac{r}{1 - \beta r^{2}}\, dr
\end{equation}
Left integral — partial fractions
We write:
\begin{equation}
\frac{1}{(1+x)(1+3x)}
=
-\frac{1}{2}\frac{1}{1+x}
+
\frac{3}{2}\frac{1}{1+3x}
\end{equation}
Integrate term by term:
\begin{equation}
-\frac{1}{2}\int \frac{dx}{1+x}
=
-\frac{1}{2}\ln(1+x) + C_{1}
\end{equation}
\begin{equation}
\frac{3}{2}\int \frac{dx}{1+3x}
=
\frac{3}{2}\cdot \frac{1}{3}\ln(1+3x)
=
\frac{1}{2}\ln(1+3x) + C_{2}
\end{equation}
Thus together:
\begin{equation}
\int \frac{dx}{(1+x)(1+3x)}
=
-\frac{1}{2}\ln(1+x)
+
\frac{1}{2}\ln(1+3x)
+
C_{3}
\end{equation}
Or more compactly:
\begin{equation}
\int \frac{dx}{(1+x)(1+3x)}
=
\frac{1}{2}
\ln\!\left(
\frac{1+3x}{1+x}
\right)
+
C_{3}
\end{equation}
Right integral
\begin{equation}
-\frac{\beta}{2}\int \frac{r}{1 - \beta r^{2}}\, dr
\end{equation}
Substitution:
\begin{equation}
u = 1 - \beta r^{2},
\qquad
du = -2\beta r\, dr
\end{equation}
\begin{equation}
r\, dr = -\frac{du}{2\beta}
\end{equation}
Thus:
\begin{equation}
-\frac{\beta}{2}
\int
\frac{r}{1 - \beta r^{2}}\, dr
=
-\frac{\beta}{2}
\left(
-\frac{1}{2\beta}
\int \frac{du}{u}
\right)
=
\frac{1}{4}\ln u
\end{equation}
Thus:
\begin{equation}
-\frac{\beta}{2}\int \frac{r}{1 - \beta r^{2}}\, dr
=
\frac{1}{4}\ln(1 - \beta r^{2})
\end{equation}
Equating both integrals
\begin{equation}
\frac{1}{2}
\ln\!\left(
\frac{1+3x}{1+x}
\right)
=
\frac{1}{4}\ln(1 - \beta r^{2})
+
C_{4}
\end{equation}
Multiply by 2:
\begin{equation}
\ln\!\left(
\frac{1+3x}{1+x}
\right)
=
\frac{1}{2}\ln(1 - \beta r^{2})
+
C_{5}
\end{equation}
Exponentiate:
\begin{equation}
\left(
\frac{1+3x}{1+x}
\right)
=
C_{6}\,
\sqrt{1 - \beta r^{2}}
\end{equation}
The physically relevant branch is the positive root:
\begin{equation}
\boxed{
\frac{1+3x}{1+x}
=
C_{7}\,\sqrt{1 - \beta r^{2}}
}
\end{equation}
We define:
\begin{equation}
\alpha(r) \equiv \sqrt{1 - \beta r^{2}}
\end{equation}
Determination of the integration constant \(C_{7}\)
We had the integrated relation:
\begin{equation}
\frac{1+3x}{1+x} = C_{7}\,\alpha(r)
\end{equation}
Now use the boundary condition at the surface:
\begin{equation}
p(R) = 0 \quad\Rightarrow\quad x(R) = 0
\end{equation}
Substitution gives:
\begin{equation}
\frac{1+0}{1+0} = C_{7}\,\alpha(R)
\end{equation}
\begin{equation}
1 = C_{7}\, a
\end{equation}
where
\begin{equation}
a \equiv \alpha(R) = \sqrt{1 - \beta R^{2}}
= \sqrt{1 - \frac{2GM}{c^{2}R}}
\end{equation}
Thus:
\begin{equation}
\boxed{
C_{7} = \frac{1}{a}
}
\end{equation}
Solving for \(x(r)\)
We had:
\begin{equation}
\frac{1+3x}{1+x} = \frac{\alpha(r)}{a}
\end{equation}
Solve this for \(x(r)\):
\begin{equation}
1 + 3x = \frac{\alpha(r)}{a}(1+x)
\end{equation}
\begin{equation}
1 + 3x = \frac{\alpha}{a} + \frac{\alpha}{a}x
\end{equation}
\begin{equation}
3x - \frac{\alpha}{a}x = \frac{\alpha}{a} - 1
\end{equation}
\begin{equation}
x\left(3 - \frac{\alpha}{a}\right)
=
\frac{\alpha - a}{a}
\end{equation}
\begin{equation}
\boxed{
x(r) = \frac{\alpha(r) - a}{3a - \alpha(r)}
}
\label{eq:R228}
\end{equation}
Pressure profile \(p(r)\)
Since \(p = \rho c^{2} x\), it follows:
\begin{equation}
\boxed{
p(r)
=
\rho c^{2}\,
\frac{\alpha(r) - a}{3a - \alpha(r)}
}
\label{eq:R229}
\end{equation}
where:
\begin{equation}
\alpha(r) =\sqrt{ 1 - \beta r^{2}},
\qquad
\beta = \frac{8\pi G}{3c^{2}}\rho,
\qquad
a = \alpha(R) = \sqrt{1 - \frac{2GM}{c^{2}R}}
\end{equation}
Checks
- At the surface: \( r = R \Rightarrow \alpha(R) = a \Rightarrow p(R) = 0 \)
- At the center:
\( r = 0 \Rightarrow \alpha(0) = 1 \Rightarrow \)
\begin{equation} p(0) = \rho c^{2}\,\frac{1 - a}{3a - 1} \end{equation}finite as long as \(3a > 1\).
This is exactly the classical internal Schwarzschild solution for constant density.
Continuation
From equation (\ref{eq:R175}):
\begin{equation}
\frac{dp}{dr}
=
-\frac{1}{2}(\rho c^{2} + p)\,\nu'
\end{equation}
we can now integrate \( \nu'(r) \) to find the time component of the metric:
\begin{equation}
e^{\nu(r)}
\end{equation}
Derivation of the time component of the metric
We begin with equation (\ref{eq:R234}):
\begin{equation}
\frac{d\nu}{dr}
=
-\,\frac{2}{\rho c^{2} + p}\frac{dp}{dr}
\label{eq:R234}
\end{equation}
Use the chain rule:
\begin{equation}
\frac{d\nu}{dr}
=
\frac{d\nu}{dp}\,\frac{dp}{dr}
\quad\Rightarrow\quad
\frac{d\nu}{dp}
=
-\frac{2}{\rho c^{2} + p}
\end{equation}
Integration with respect to \(p\)
\begin{equation}
\nu
=
-2\ln(\rho c^{2} + p) + \ln C'
=
\ln C'(\rho c^2+p)^{-2}
=
\ln\!\frac{C'}{
(\rho c^{2} + p)^{2}
}
\end{equation}
Exponentiate:
\begin{equation}
e^{\nu/2}
=
\frac{C''}{\rho c^{2} + p}
\qquad (C''>0)
\end{equation}
Use the pressure solution (\ref{eq:R229})
According to (\ref{eq:R229}):
\begin{equation}
p(r)
=
\rho c^{2}\,
\frac{\alpha(r) - a}{3a - \alpha(r)}
\end{equation}
Thus:
\begin{equation}
\rho c^{2} + p
=
\rho c^{2}
\left(
1 + \frac{\alpha - a}{3a - \alpha}
\right)
=
\rho c^{2}\,
\frac{2a}{3a - \alpha(r)}
\end{equation}
Substitute into \(e^{\nu/2}\):
\begin{equation}
e^{\nu/2}
=
\frac{C''}{\rho c^{2} + p}
=
\frac{C''}{\rho c^{2}}
\frac{3a - \alpha(r)}{2a}
\label{eq:R240}
\end{equation}
Determination of the integration constant \(C''\)
At the surface \(r = R\):
\begin{equation}
p(R) = 0,
\qquad
\alpha(R) = a,
\qquad
e^{\nu(R)/2} = a =\sqrt{ 1 - \frac{2GM}{c^{2}R}}
\end{equation}
Substituting into (19b):
\begin{equation}
a
=
\frac{C''}{\rho c^{2}}
\frac{3a-a}{2a}
=
\frac{C''}{\rho c^{2}}
\end{equation}
Thus:
\begin{equation}
\boxed{
C'' = a\,\rho c^{2}
}
\end{equation}
Final result
\begin{equation}
e^{\nu/2}
=
\frac{a\rho c^{2}}{\rho c^{2}}
\,
\frac{3a-\alpha(r)}{2a}
\end{equation}
Since:
\begin{equation}
\alpha(r) =\sqrt{ 1 - \beta r^{2}},
\qquad
\beta = \frac{8\pi G}{3c^{2}}\rho,
\qquad
a = \sqrt{1 - \frac{2GM}{c^{2}R}}
\end{equation}
we obtain:
\begin{equation}
\boxed{
e^{\nu(r)/2}
=
\frac{3}{2}a - \frac{1}{2}\alpha(r)
=
\frac{3}{2}a - \frac{1}{2} \sqrt{1 - \beta r^{2}}
}
\label{eq:R246}
\end{equation}
Or explicitly:
\begin{equation}
\boxed{
e^{\nu(r)/2}
=
\frac{3}{2}a - \frac{1}{2}\sqrt{1 - \frac{8\pi G}{3c^{2}}\rho\, r^{2}}
}
\label{eq:R247}
\end{equation}
This is the time component of the internal Schwarzschild metric for a star with constant density.
Complete internal metric
The interior solution for a homogeneous sphere with radius \(0 \le r \le R\) is:
\begin{equation}
ds^{2}
=
\left(
\frac{3}{a} - \frac{1}{2}\alpha(r)
\right)^{2}
c^{2} dt^{2}
-
\frac{dr^{2}}{\alpha^2(r)}
-
r^{2}\left(d\theta^{2} + \sin^{2}\theta\, d\phi^{2}\right)
\label{eq:R248}
\end{equation}
where
\begin{equation}
\alpha(r) = \sqrt{1 - \frac{8\pi G}{3c^{2}}\rho\, r^{2}},
\qquad
a = \alpha(R) = \sqrt{1 - \frac{2GM}{c^{2}R}}.
\end{equation}
Use of the mass relations
Since:
\begin{equation}
M = \frac{4\pi}{3}\rho R^{3}
\quad\Rightarrow\quad
\rho = \frac{3M}{4\pi R^{3}},
\end{equation}
it follows:
\begin{equation}
\frac{8\pi G}{3c^{2}}\rho r^{2}
=
\frac{2GM}{c^{2}R}\frac{r^{2}}{R^{2}}.
\end{equation}
Substitution into (\ref{eq:R248}) yields:
\begin{equation}
\boxed{
ds^{2}
=\left(\frac{3}{2}\sqrt{1-\frac{2GM}{c^2R}}
-\frac{1}{2}\sqrt{1-\frac{2GM}{c^2R}\frac{r^2}{R^2}}\right)^2 c^2dt^2
-
\frac{dr^{2}}
{1 - \frac{2GM}{c^{2}R}\frac{r^{2}}{R^{2}}}
-
r^{2}\left(d\theta^{2} + \sin^{2}\theta\, d\phi^{2}\right)
}
\label{eq:R252}
\end{equation}
Remarks
This metric describes the spacetime inside a homogeneous sphere. For \(r = R\), continuity with the exterior Schwarzschild solution follows:
\begin{equation}
e^{-\lambda(R)} = 1 - \frac{2GM}{c^{2}R},
\qquad
e^{\nu(R)} = 1 - \frac{2GM}{c^{2}R}.
\end{equation}
Appendix 5.8 — Central pressure and Buchdahl limit
From \( \alpha(0) = 1 \), the central pressure is:
\begin{equation}
p(0)
=
\rho c^{2}
\frac{1 - a}{3a - 1}
\label{eq:R254}
\end{equation}
with
\begin{equation}
a = \sqrt{1 - \frac{2GM}{c^{2}R}}.
\end{equation}
When the denominator vanishes, the central pressure diverges:
\begin{equation}
3a - 1 = 0
\quad\Rightarrow\quad
a = \frac{1}{3}.
\end{equation}
This gives the Buchdahl limit:
\begin{equation}
\boxed{
\frac{2GM}{c^{2}R} = \frac{8}{9}
}
\label{eq:R257}
\end{equation}
This limit marks the maximum compactness for a stable, static configuration with constant density. If it is exceeded, hydrostatic equilibrium can no longer exist and the star collapses into a black hole.
Appendix 5.9 — Summary
- The full tensor derivation confirms the consistency of the internal solution with the Einstein equations.
- The mass function \(m(r)\), the pressure equation, and the time component of the metric follow logically from the TOV equation.
- The Buchdahl limit determines the maximum compactness of a constant-density star.
- The functions \( \nu(r) \) and \( \lambda(r) \) can be determined exactly for \( \rho = \text{constant} \).
- The pressure \( p(r) \) follows directly from the TOV equation and decreases continuously to \( p(R) = 0 \) at the surface.
- The central pressure diverges when the Buchdahl limit is reached:
\begin{equation} \frac{2GM}{c^{2}R} = \frac{8}{9}. \end{equation}
- The internal metric matches smoothly and continuously to the external Schwarzschild solution at \( r = R \).