The General Theory of Relativity of Einstein

Derivations, Applications and Reflections – by Albert Prins

Appendix 5 – Schwarzschild Solution Inside a Mass
Complete Tensor Derivation

Appendix 5.1 – Introduction

In this chapter the complete tensor derivation of the interior Schwarzschild solution is presented: 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:

\[ ds^{2} = e^{\nu(r)} c^{2} dt^{2} - e^{\lambda(r)} dr^{2} - r^{2} d\theta^{2} - r^{2} \sin^{2}\theta\, d\phi^{2} \tag{1} \]

where \( \nu(r) \) and \( \lambda(r) \) are functions of \( r \) to be determined.

Energy–momentum tensor of a perfect fluid

The energy–momentum tensor of a perfect fluid is given by:

\[ T_{\mu\nu} = (\rho c^{2} + p)\, u_{\mu} u_{\nu} - p\, g_{\mu\nu} \tag{2} \]

This implies:

The pressure \( p \) appears in the tensor because in general relativity not only energy, but also pressure and stress contribute to spacetime curvature. 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, equation (2) reduces to:

\[ T_{\mu\nu} = (\rho c^{2} + p)\, u_{\mu} u_{\nu} - p\, g_{\mu\nu} = \begin{pmatrix} \rho c^{2} e^{\nu} & 0 & 0 & 0 \\ 0 & p\, e^{\lambda} & 0 & 0 \\ 0 & 0 & p\, r^{2} & 0 \\ 0 & 0 & 0 & p\, r^{2} \sin^{2}\theta \end{pmatrix} \]

This clearly shows that \( \rho c^{2} \) forms the energy density, while the spatial diagonal elements correspond exactly to the isotropic pressure \( p \). The form of (2) 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 isotropic pressure that is equal in all spatial directions.

For a static fluid:

\[ u^{\mu} = \left(e^{-\nu/2},\, 0,\, 0,\, 0\right), \qquad u_{\mu} = \left(e^{\nu/2},\, 0,\, 0,\, 0\right) \] \[ u_{\mu} u^{\mu} = 1, \qquad u_{\mu}u_{\mu}=\left(e^{\nu}, 0, 0, 0\right) = \begin{pmatrix} e^{\nu} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \]

In short: all components except the \( (t,t) \) component vanish; \( u_{t} u_{t} = e^{\nu} \).

This directly yields:

\[ T_t^t = \rho c^{2}, \qquad T_r^r = T_\theta^\theta = T_\phi^\phi = p \tag{3} \] \[ T_{tt}=T_t^t \quad T_{rr}=g_{rr}T_r^r \quad T_{\theta\theta}=g_{\theta\theta}T_{\theta}^{\theta} \quad T_{\phi\phi}=g_{\phi\phi}T_{\phi}^{\phi} \]

In coordinate components:

\[ T_{tt} = (\rho c^{2} + p)\, u_{t} u_{t} - p\, g_{tt} = \rho c^{2} e^{\nu} \] \[ T_{rr} = (\rho c^{2} + p)\, u_{r} u_{r} - p\, g_{rr} = -p(-e^{\lambda}) = p\, e^{\lambda} \] \[ T_{\theta\theta} = (\rho c^{2} + p)\, u_{\theta} u_{\theta} - p\, g_{\theta\theta} = -p(-r^{2}) = p r^{2} \] \[ T_{\phi\phi} = (\rho c^{2} + p)\, u_{\phi} u_{\phi} - p\, g_{\phi\phi} = -p(-r^{2}\sin^{2}\theta) = p r^{2}\sin^{2}\theta \]

The Einstein equations take their usual form:

\[ G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2} R\, g_{\mu\nu} = \frac{8\pi G}{c^{4}}\, T_{\mu\nu} \tag{4} \]

Explanation

Metric (1) is the most general static, spherically symmetric metric. The function \( \nu(r) \) determines gravitational time dilation, while \( \lambda(r) \) represents curvature in the radial direction. By determining these two functions from the field equations, the complete geometric structure of the interior field is obtained.

Appendix 5.2 – Calculation of the Christoffel symbols

The Christoffel symbols are given by:

\[ \Gamma^{\rho}_{\mu\nu} = \frac{1}{2} g^{\rho\alpha} \left( \frac{\partial g_{\nu\alpha}}{\partial x^{\mu}} + \frac{\partial g_{\mu\alpha}}{\partial x^{\nu}} - \frac{\partial g_{\mu\nu}}{\partial x^{\alpha}} \right) \]

For the interior Schwarzschild metric:

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

Metric and derivatives of the metric

The metric components are:

\[ g_{\mu\nu} = \begin{pmatrix} e^{\nu} & 0 & 0 & 0 \\ 0 & -e^{\lambda} & 0 & 0 \\ 0 & 0 & -r^{2} & 0 \\ 0 & 0 & 0 & -r^{2}\sin^{2}\theta \end{pmatrix} \]

The derivatives with respect to \( r \) are:

\[ \frac{\partial g_{\mu\nu}}{\partial r} = \begin{pmatrix} \nu' e^{\nu} & 0 & 0 & 0 \\ 0 & -\lambda' e^{\lambda} & 0 & 0 \\ 0 & 0 & -2r & 0 \\ 0 & 0 & 0 & -2r \sin^{2}\theta \end{pmatrix} \]

Non-zero Christoffel symbols

With \( \nu' = \frac{d\nu}{dr} \) and \( \lambda' = \frac{d\lambda}{dr} \):

\[ \Gamma^{t}_{tr}=\Gamma^{t}_{rt} = \frac{1}{2} g^{tt} \frac{\partial g_{tt}}{\partial r} = \frac{1}{2}\nu' \tag{5a} \]

\[ \Gamma^{r}_{tt} = \frac{1}{2} g^{rr}\left(-\frac{\partial g_{tt}}{\partial r}\right) = \frac{1}{2} e^{\nu-\lambda}\, \nu' \tag{5b} \]

\[ \Gamma^{r}_{rr} = \frac{1}{2} g^{rr} \frac{\partial g_{rr}}{\partial r} = \frac{1}{2}\lambda' \tag{5c} \]

\[ \Gamma^{r}_{\theta\theta} = -\frac{1}{2} g^{rr} \frac{\partial g_{\theta\theta}}{\partial r} = -r e^{-\lambda} \tag{5d} \]

\[ \Gamma^{r}_{\phi\phi} = -r e^{-\lambda}\sin^{2}\theta \tag{5e} \]

\[ \Gamma^{\theta}_{r\theta} = \Gamma^{\theta}_{\theta r} = \frac{1}{r} \tag{5f} \]

\[ \Gamma^{\phi}_{r\phi} = \Gamma^{\phi}_{\phi r} = \frac{1}{r} \tag{5f} \]

\[ \Gamma^{\theta}_{\phi\phi} = -\sin\theta\cos\theta \tag{5g} \]

\[ \Gamma^{\phi}_{\theta\phi} = \Gamma^{\phi}_{\phi\theta} =\frac{ \cos\theta}{\sin\theta} = \cot\theta \tag{5h} \]

Derivatives of the Christoffel Symbols

\[ \frac{\partial}{\partial r}\Gamma^{t}_{rt} = \frac{\partial}{\partial r}\Gamma^{t}_{tr} = \frac{1}{2}\nu'' \]

\[ \frac{\partial}{\partial r}\Gamma^{r}_{tt} = \frac{1}{2} e^{\nu-\lambda} \left( \nu'^{2} - \lambda'\nu' + \nu'' \right) \]

\[ \frac{\partial}{\partial r}\Gamma^{r}_{rr} = \frac{1}{2}\lambda'' \]

\[ \frac{\partial}{\partial r}\Gamma^{r}_{\theta \theta} = \frac{\partial}{\partial r}(-r e^{-\lambda}) = -e^{-\lambda} + r e^{-\lambda}\lambda' = e^{-\lambda}(r\lambda' - 1) \]

\[ \frac{\partial}{\partial r}\Gamma^{r}_{\phi \phi} = -e^{-\lambda}\sin^{2}\theta + r e^{-\lambda}\lambda'\sin^{2}\theta = e^{-\lambda}(r\lambda' - 1)\sin^{2}\theta \]

Derivatives for Angular Components

For the angular components:

\[ \frac{\partial}{\partial r}\Gamma^{\theta}_{r\theta} = \frac{\partial}{\partial r}\Gamma^{\theta}_{\theta r} = \frac{\partial}{\partial r}\Gamma^{\phi}_{r\phi} = \frac{\partial}{\partial r}\Gamma^{\phi}_{\phi r} = \frac{\partial}{\partial r}\left(\frac{1}{r}\right) = -\frac{1}{r^{2}} \]

\[ \frac{\partial}{\partial \theta}\Gamma^{\theta}_{\phi\phi} = \frac{\partial}{\partial \theta}(-\sin\theta\cos\theta) = -(\cos^{2}\theta - \sin^{2}\theta) \]

\[ \frac{\partial}{\partial \theta}\Gamma^{\phi}_{\theta\phi} = \frac{\partial}{\partial \theta}\Gamma^{\phi}_{\phi\theta} = \frac{\partial}{\partial \theta}\left(\frac{\cos\theta}{\sin\theta}\right) = -\frac{\sin^{2}\theta + \cos^{2}\theta}{\sin^{2}\theta} = -\frac{1}{\sin^{2}\theta} \]

Remark

The derivation of these symbols follows directly from the definition:

\[ \Gamma^{\rho}_{\mu\nu} = \frac{1}{2} g^{\rho\sigma} \left( \frac{\partial g_{\nu\sigma}}{\partial x^{\mu}} + \frac{\partial g_{\mu\sigma}}{\partial x^{\nu}} - \frac{\partial g_{\mu\nu}}{\partial x^{\sigma}} \right) \]

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 using the Ricci tensor), the definition is:

\[ R_{\mu\nu} = R^{\sigma}_{\mu\sigma\nu} = \frac{\partial \Gamma^{\sigma}_{\mu\nu}}{\partial x^{\sigma}} - \frac{\partial \Gamma^{\sigma}_{\mu\sigma}}{\partial x^{\nu}} + \Gamma^{\sigma}_{\lambda\sigma}\Gamma^{\lambda}_{\mu\nu} - \Gamma^{\sigma}_{\lambda\nu}\Gamma^{\lambda}_{\mu\sigma} \]

Complete 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, because all other \( \Gamma^{\sigma}_{tt} = 0 \).

Using (see above): \[ \Gamma^{r}_{tt} = \frac{1}{2} e^{\nu - \lambda}\, \nu' \] we obtain:

\[ \frac{\partial}{\partial \sigma}\Gamma^{\sigma}_{tt} = \frac{\partial}{\partial r}\Gamma^{r}_{tt} = \frac{\partial}{\partial r} \left( \frac{1}{2} e^{\nu - \lambda}\, \nu' \right) = \frac{1}{2} e^{\nu - \lambda} \left( \nu'' + \nu'^{2} - \lambda'\nu' \right) \]

2. Second term: \( -\frac{\partial \Gamma^{\sigma}_{t\sigma}}{\partial t} \)

Because the metric is static (no \( t \)-dependence), this term vanishes:

\[ -\frac{\partial}{\partial t}\Gamma^{\sigma}_{t\sigma} = 0 \]

3. Third term: \( \Gamma^{\sigma}_{\lambda\sigma}\Gamma^{\lambda}_{tt} \)

Only \( \lambda = r \) gives a non-zero \( \Gamma^{\lambda}_{tt} \). Thus the term becomes:

Evaluation of the term \( \Gamma^{\sigma}_{\lambda\sigma}\Gamma^{\lambda}_{tt} \)

We have:

\[ \Gamma^{\sigma}_{\lambda\sigma}\Gamma^{\lambda}_{tt} = \Gamma^{\sigma}_{r\sigma}\,\Gamma^{r}_{tt} \]

First we compute:

\[ \Gamma^{\sigma}_{r\sigma} = \Gamma^{t}_{rt} + \Gamma^{r}_{rr} + \Gamma^{\theta}_{r\theta} + \Gamma^{\phi}_{r\phi} = \frac{1}{2}\nu' + \frac{1}{2}\lambda' + \frac{1}{r} + \frac{1}{r} = \frac{1}{2}(\nu' + \lambda') + \frac{2}{r} \]

Hence:

\[ \Gamma^{\sigma}_{\lambda\sigma}\Gamma^{\lambda}_{tt} = \left[ \frac{1}{2}(\nu' + \lambda') + \frac{2}{r} \right] \cdot \frac{1}{2} e^{\nu - \lambda}\nu' \] \[ = \frac{1}{4} e^{\nu - \lambda}\nu'(\nu' + \lambda') + e^{\nu - \lambda}\frac{\nu'}{r} \]

4. Fourth term: \( -\Gamma^{\sigma}_{t\lambda}\Gamma^{\lambda}_{t\sigma} \)

We sum all non-zero products \( \Gamma^{\sigma}_{t\lambda}\Gamma^{\lambda}_{t\sigma} \). The only non-zero Christoffel symbols with a time index are:

\[ \Gamma^{\sigma}_{t\lambda} = \frac{1}{2}\nu', \qquad \Gamma^{r}_{tt} = \frac{1}{2} e^{\nu - \lambda}\nu' \]

By symmetry there are two identical contributions:

\[ -\Gamma^{\sigma}_{t\lambda}\Gamma^{\lambda}_{t\sigma} = -\Gamma^{t}_{r t}\Gamma^{r}_{t t} - \Gamma^{t}_{t r}\Gamma^{r}_{t t} = -2\left(\frac{1}{2}\nu'\right) \left(\frac{1}{2} e^{\nu - \lambda}\nu'\right) = -\frac{1}{2} e^{\nu - \lambda}\nu'^{2} \]

5. Summing all contributions

We now combine the four terms (the second term was zero):

\[ R_{tt} = \frac{1}{2} e^{\nu - \lambda} \left( \nu'' + \nu'^{2} - \lambda'\nu' \right) + \frac{1}{4} e^{\nu - \lambda}\nu'(\nu' + \lambda') + e^{\nu - \lambda}\frac{\nu'}{r} - \frac{1}{2} e^{\nu - \lambda}\nu'^{2} \]

After simplification:

\[ R_{tt} = \frac{1}{2} e^{\nu - \lambda} \left( \nu'' + \frac{1}{2}\nu'^{2} - \frac{1}{2}\lambda'\nu' + \frac{2\nu'}{r} \right) \]

Thus:

\[ \boxed{ R_{tt} = \frac{1}{2} e^{\nu - \lambda} \left( \nu'' + \frac{1}{2}\nu'^{2} - \frac{1}{2}\lambda'\nu' + \frac{2\nu'}{r} \right) } \]

Complete derivation of \( R_{rr} \)

The Ricci component:

\[ R_{rr} = \frac{\partial \Gamma^{\sigma}_{rr}}{\partial x^{\sigma}} - \frac{\partial \Gamma^{\sigma}_{r\sigma}}{\partial r} + \Gamma^{\sigma}_{\lambda\sigma}\Gamma^{\lambda}_{rr} - \Gamma^{\sigma}_{r\lambda}\Gamma^{\lambda}_{r\sigma} \]

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:

\[ \Gamma^{r}_{rr} = \frac{1}{2}\lambda' \]

Thus:

\[ \frac{\partial}{\partial x^{\sigma}}\Gamma^{\sigma}_{rr} = \frac{\partial}{\partial r}\Gamma^{r}_{rr} = \frac{1}{2}\lambda'' \]

2. Second term: \( -\dfrac{\partial \Gamma^{\sigma}_{r\sigma}}{\partial r} \)

First we compute the sum:

\[ \Gamma^{\sigma}_{r\sigma} = \Gamma^{t}_{rt} + \Gamma^{r}_{rr} + \Gamma^{\theta}_{r\theta} + \Gamma^{\phi}_{r\phi} = \frac{1}{2}\nu' + \frac{1}{2}\lambda' + \frac{1}{r} + \frac{1}{r} = \frac{1}{2}(\nu' + \lambda') + \frac{2}{r} \]

Differentiate with respect to \( r \):

\[ \frac{\partial}{\partial r}\Gamma^{\sigma}_{r\sigma} = \frac{1}{2}(\nu'' + \lambda'') - \frac{2}{r^{2}} \]

Thus:

\[ -\frac{\partial}{\partial r}\Gamma^{\sigma}_{r\sigma} = -\frac{1}{2}(\nu'' + \lambda'') + \frac{2}{r^{2}} \]

3. Third term: \( \Gamma^{\sigma}_{\lambda\sigma}\Gamma^{\lambda}_{rr} \)

Only \( \lambda = r \) contributes, since \( \Gamma^{\lambda}_{rr} \) is non-zero only for \( \lambda = r \).

\[ \Gamma^{\sigma}_{\lambda\sigma}\Gamma^{\lambda}_{rr} = \Gamma^{\sigma}_{r\sigma}\Gamma^{r}_{rr} \]

We again use:

\[ \Gamma^{\sigma}_{r\sigma} = \frac{1}{2}(\nu' + \lambda') + \frac{2}{r} \] \[ \Gamma^{r}_{rr} = \frac{1}{2}\lambda' \]

Thus:

\[ \Gamma^{\sigma}_{\lambda\sigma}\Gamma^{\lambda}_{rr} = \left[ \frac{1}{2}(\nu' + \lambda') + \frac{2}{r} \right] \cdot \frac{1}{2}\lambda' = \frac{1}{4}(\nu' + \lambda')\lambda' + \frac{\lambda'}{r} \]

4. Fourth term: \( -\Gamma^{\sigma}_{r\lambda}\Gamma^{\lambda}_{r\sigma} \)

We sum all non-zero products:

\[ \Gamma^{\sigma}_{r\lambda}\Gamma^{\lambda}_{r\sigma} = \Gamma^{t}_{rt}\Gamma^{t}_{rt} + \Gamma^{r}_{rr}\Gamma^{r}_{rr} + \Gamma^{\theta}_{r\theta}\Gamma^{\theta}_{r\theta} + \Gamma^{\phi}_{r\phi}\Gamma^{\phi}_{r\phi} \]

Substituting:

\[ = \left(\frac{1}{2}\nu'\right)^{2} + \left(\frac{1}{2}\lambda'\right)^{2} + \left(\frac{1}{r}\right)^{2} + \left(\frac{1}{r}\right)^{2} = \frac{1}{4}\nu'^{2} + \frac{1}{4}\lambda'^{2} + \frac{2}{r^{2}} \]

Thus:

\[ -\Gamma^{\sigma}_{r\lambda}\Gamma^{\lambda}_{r\sigma} = -\frac{1}{4}\nu'^{2} - \frac{1}{4}\lambda'^{2} - \frac{2}{r^{2}} \]

5. Summing all terms

We now combine the four contributions:

\[ R_{rr} = \frac{1}{2}\lambda'' - \frac{1}{2}(\nu'' + \lambda'') + \frac{2}{r^{2}} + \frac{1}{4}(\nu' + \lambda')\lambda' + \frac{\lambda'}{r} - \frac{1}{4}\nu'^{2} - \frac{1}{4}\lambda'^{2} - \frac{2}{r^{2}} \]

The terms \( +\frac{2}{r^{2}} \) and \( -\frac{2}{r^{2}} \) cancel. Also \( \frac{1}{2}\lambda'' - \frac{1}{2}\lambda'' = 0 \).

We are left with:

\[ R_{rr} = -\frac{1}{2}\nu'' + \frac{1}{4}(\nu' + \lambda')\lambda' + \frac{\lambda'}{r} - \frac{1}{4}\nu'^{2} - \frac{1}{4}\lambda'^{2} \]

Simplify:

\[ \boxed{ R_{rr} = -\frac{1}{2}\nu'' + \frac{1}{4}\nu'\lambda' + \frac{1}{4}\lambda'^{2} + \frac{\lambda'}{r} - \frac{1}{4}\nu'^{2} - \frac{1}{4}\lambda'^{2} } \]

The \( \lambda'^2 \) terms cancel:

\[ \boxed{ R_{rr} = -\frac{1}{2}\nu'' + \frac{1}{4}\nu'\lambda' + \frac{\lambda'}{r} - \frac{1}{4}\nu'^{2} } \] or: \[ R_{rr} = -\frac{\nu''}{2} - \frac{\nu'^2}{4} + \frac{\nu'\lambda'}{4} + \frac{\lambda'}{r} \]

Factoring out \( -\tfrac{1}{2} \) yields the conventional form:

\[ \boxed{ R_{rr} = -\frac{1}{2} \left( \nu'' + \frac{1}{2}\nu'^2 - \frac{1}{2}\nu'\lambda' - \frac{2\lambda'}{r} \right) } \]

Complete derivation of \( R_{\theta\theta} \)

The definition:

\[ R_{\theta\theta} = \frac{\partial \Gamma^{\sigma}_{\theta\theta}}{\partial x^{\sigma}} - \frac{\partial \Gamma^{\sigma}_{\theta\sigma}}{\partial x^{\theta}} + \Gamma^{\sigma}_{\lambda\sigma}\Gamma^{\lambda}_{\theta\theta} - \Gamma^{\sigma}_{\theta\lambda}\Gamma^{\lambda}_{\theta\sigma} \]

We set \( \mu = \nu = \theta \). We again split the expression into four terms.

1. First term: \( \dfrac{\partial \Gamma^{\sigma}_{\theta\theta}}{\partial x^{\sigma}} \)

Only \( \sigma = r \) contributes, since:

\[ \Gamma^{\sigma}_{\theta \theta} = -r e^{-\lambda} \]

Thus:

\[ \frac{\partial}{\partial x^\sigma}\Gamma^{\sigma}_{\theta \theta} = \frac{\partial}{\partial r}\Gamma^{r}_{\theta \theta} = \frac{\partial}{\partial r}(-r e^{-\lambda}) = -e^{-\lambda} + r e^{-\lambda}\lambda' \]

2. Second term: \( -\dfrac{\partial \Gamma^{\sigma}_{\theta\sigma}}{\partial x^{\theta}} \)

Only \( \sigma = \phi \) contributes, since (see equation 5h):

\[ \Gamma^{\phi}_{\theta\phi} = \cot\theta \]

Thus:

\[ -\frac{\partial}{\partial \theta}\Gamma^{\phi}_{\theta\phi} = -\frac{\partial}{\partial \theta}(\cot\theta) = \frac{1}{\sin^{2}\theta} \]

3. Third term: \( \Gamma^{\sigma}_{\lambda\sigma}\Gamma^{\lambda}_{\theta\theta} \)

For \( \lambda = r \) we have:

\[ \Gamma^{r}_{\theta \theta} = -r e^{-\lambda} \]

We again use:

\[ \Gamma^{\sigma}_{r\sigma} = \frac{1}{2}(\nu' + \lambda') + \frac{2}{r} \]

Thus:

\[ \Gamma^{\sigma}_{\lambda\sigma}\Gamma^{\lambda}_{\theta\theta} = \Gamma^{\sigma}_{\sigma r}\Gamma^{r}_{\theta\theta} = \left[ \frac{1}{2}(\nu' + \lambda') + \frac{2}{r} \right] (-r e^{-\lambda}) \]

Expanding:

\[ = -\frac{1}{2} r e^{-\lambda}(\nu' + \lambda') - 2 e^{-\lambda} \]

4. Fourth term: \( -\Gamma^{\sigma}_{\theta\lambda}\Gamma^{\lambda}_{\theta\sigma} \)

We first write the complete sum of all possible products:

\[ \Gamma^{\sigma}_{\theta\lambda}\Gamma^{\lambda}_{\theta\sigma} = \Gamma^{t}_{\theta t}\Gamma^{t}_{\theta t} + \Gamma^{t}_{\theta r}\Gamma^{r}_{\theta t} + \Gamma^{t}_{\theta \theta}\Gamma^{\theta}_{\theta\ t} + \Gamma^{t}_{\theta \phi}\Gamma^{\phi}_{\theta t} \] \[ + \Gamma^{r}_{\theta t}\Gamma^{t}_{\theta r} + \Gamma^{r}_{\theta r}\Gamma^{r}_{\theta r} + \Gamma^{r}_{\theta \theta}\Gamma^{\theta}_{\theta r} + \Gamma^{r}_{\theta \phi}\Gamma^{\phi}_{\theta r} \] \[ + \Gamma^{\theta}_{\theta t}\Gamma^{t}_{\theta \theta} + \Gamma^{\theta}_{\theta r}\Gamma^{r}_{\theta \theta} + \Gamma^{\theta}_{\theta \theta}\Gamma^{\theta}_{\theta\theta} + \Gamma^{\theta}_{\theta \phi}\Gamma^{\phi}_{\theta\theta} \] \[ + \Gamma^{\phi}_{\phi t}\Gamma^{t}_{\theta \phi} + \Gamma^{\phi}_{\theta r}\Gamma^{r}_{\theta \phi} + \Gamma^{\phi}_{\theta\theta}\Gamma^{\theta}_{\theta\phi} + \Gamma^{\phi}_{\theta\phi}\Gamma^{\phi}_{\theta\phi} \]

For the static spherically symmetric metric, nearly all terms vanish. The only non-zero contributions are:

\[ \Gamma^{\theta}_{\theta r} = \frac{1}{r}, \qquad \Gamma^{r}_{\theta\theta} = -r e^{-\lambda}, \qquad \Gamma^{\phi}_{\theta\phi} = \cot\theta. \]

Substitution yields:

\[ \Gamma^{\sigma}_{\theta\lambda}\Gamma^{\lambda}_{\theta\sigma} = \left(-r e^{-\lambda}\right)\left(\frac{1}{r}\right) + \left(\frac{1}{r}\right)\left(-r e^{-\lambda}\right) + \cot^{2}\theta \] \[ = -2 e^{-\lambda} + \cot^{2}\theta \]

Therefore:

\[ \boxed{ -\Gamma^{\sigma}_{\theta\lambda}\Gamma^{\lambda}_{\theta\sigma} = 2 e^{-\lambda} - \cot^{2}\theta } \]

Relevant non-zero products (compact)

In summary: the radial term contributes \(2e^{-\lambda}\), the angular terms together contribute \(1\), and this fits exactly into the standard derivation of \(R_{\theta\theta}\).

5. Summing all contributions (radial + angular parts)

We now combine all terms:

\[ R_{\theta\theta} = \left(-e^{-\lambda} + r e^{-\lambda}\lambda'\right) + \frac{1}{\sin^{2}\theta} + \left( -\frac{1}{2} r e^{-\lambda}(\nu' + \lambda') - 2 e^{-\lambda} \right) + \left( 2 e^{-\lambda} - \cot^{2}\theta \right) \]

The angular terms combine to:

\[ \frac{1}{\sin^{2}\theta} - \cot^{2}\theta = 1 \]

Thus:

\[ R_{\theta\theta} = 1 - e^{-\lambda} + r e^{-\lambda}\lambda' - \frac{1}{2} r e^{-\lambda}(\nu' + \lambda') \]

Simplifying:

\[ \boxed{ R_{\theta\theta} = 1 - e^{-\lambda} \left( 1 + \frac{r}{2}(\nu' - \lambda') \right) } \]

Full derivation of \( R_{\phi\phi} \)

The definition:

\[ R_{\phi\phi} = \frac{\partial \Gamma^{\sigma}_{\phi\phi}}{\partial x^{\sigma}} - \frac{\partial \Gamma^{\sigma}_{\phi\sigma}}{\partial \phi} + \Gamma^{\sigma}_{\sigma\lambda}\Gamma^{\lambda}_{\phi\phi} - \Gamma^{\sigma}_{\phi\lambda}\Gamma^{\lambda}_{\phi\sigma} \]

We take \( \mu = \nu = \phi \). Again, we split the expression into four terms.

1. First term: \( \dfrac{\partial \Gamma^{\sigma}_{\phi\phi}}{\partial x^{\sigma}} \)

The non-zero Christoffel symbols are:

\[ \Gamma^{r}_{\phi \phi} = -r e^{-\lambda}\sin^{2}\theta, \qquad \Gamma^{\theta}_{\phi\phi} = -\sin\theta\cos\theta \]

Thus:

\[ \frac{\partial \Gamma^{r}_{\phi \phi}}{\partial x^\sigma} = -e^{-\lambda}\sin^{2}\theta \left(1 - r\lambda'\right) \] \[ \frac{\partial \Gamma^{\theta}_{\phi\phi}}{\partial \theta} = -(\cos^{2}\theta - \sin^{2}\theta) \]

Together:

\[ \frac{\partial \Gamma^{\sigma}_{\phi \phi}}{\partial x^\sigma} = -e^{-\lambda}\sin^{2}\theta\,(1 - r\lambda') - (\cos^{2}\theta - \sin^{2}\theta) \]

2. Second term: \( -\dfrac{\partial \Gamma^{\sigma}_{\phi\sigma}}{\partial \phi} \)

All Christoffel symbols are independent of \( \phi \), therefore:

\[ -\frac{\partial}{\partial \phi}\Gamma^{\sigma}_{\phi\sigma} = 0 \]

3. Third term: \( \Gamma^{\sigma}_{\sigma\lambda}\Gamma^{\lambda}_{\phi\phi} \)

Non-zero contributions arise for \( \lambda = r \) and \( \lambda = \theta \):

\[ \Gamma^{r}_{\phi \phi} = -r e^{-\lambda}\sin^{2}\theta, \qquad \Gamma^{\theta}_{\phi\phi} = -\sin\theta\cos\theta \]

We again use:

\[ \Gamma^{\sigma}_{\sigma r} = \Gamma^{t}_{tr} + \Gamma^{r}_{rr} + \Gamma^{\theta}_{\theta r} + \Gamma^{\phi}_{\phi r} = \frac{1}{2}\nu' + \frac{1}{2}\lambda' + \frac{1}{r} + \frac{1}{r} = \frac{1}{2}(\nu' + \lambda') + \frac{2}{r} \] \[ \Gamma^{\sigma}_{\sigma\theta} = \Gamma^{t}_{t \theta} + \Gamma^{r}_{r \theta} + \Gamma^{\theta}_{\theta \theta} + \Gamma^{\phi}_{\phi \theta} = 0 + 0 + 0 + \cot\theta \] \[ \Gamma^{\sigma}_{\sigma\lambda}\Gamma^{\lambda}_{\phi\phi} = \Gamma^{\sigma}_{\sigma r}\Gamma^{r}_{\phi\phi} + \Gamma^{\sigma}_{\sigma \theta}\Gamma^{\theta}_{\phi\phi} \] \[ =\left( \frac{1}{2}\nu' + \frac{1}{2}\lambda' + \frac{2}{r} \right)\, \Gamma^{r}_{\phi\phi} + \left(\cot{\theta}\right)\, \Gamma^{\theta}_{\phi\phi} \]

Thus:

\[ \Gamma^{\sigma}_{\sigma\lambda}\Gamma^{\lambda}_{\phi\phi} = \left[ \frac{1}{2}(\nu' + \lambda') + \frac{2}{r} \right] \left( -r e^{-\lambda}\sin^{2}\theta \right) + \cot\theta\left(-\sin\theta\cos\theta\right) \]

Expanding:

\[ = -e^{-\lambda}\sin^{2}\theta \left( \frac{r}{2}(\nu' + \lambda') + 2 \right) - \cos^{2}\theta \]

4. Fourth term: \( -\Gamma^{\sigma}_{\phi\lambda}\Gamma^{\lambda}_{\phi\sigma} \)

The non-zero products are:

\[ \Gamma^{r}_{\phi\phi}\Gamma^{\phi}_{\phi r} = \Gamma^{\phi}_{\phi r}\Gamma^{r}_{\phi\phi} = \left(\frac{1}{r}\right) \left(-r e^{-\lambda}\sin^{2}\theta\right) = -e^{-\lambda}\sin^{2}\theta \] \[ \Gamma^{\theta}_{\phi \phi}\Gamma^{\phi}_{\phi\theta} = \Gamma^{\phi}_{\phi \theta}\Gamma^{\theta}_{\phi\phi} = (\cot\theta)(-\sin\theta\cos\theta) = -\cos^{2}\theta \]

Collecting all relevant terms:

\[ \Gamma^{\sigma}_{\phi\lambda}\Gamma^{\lambda}_{\phi\sigma} = -2e^{-\lambda}\sin^{2}\theta - 2\cos^{2}\theta \]

Hence the resulting fourth term is:

\[ -\Gamma^{\sigma}_{\phi\lambda}\Gamma^{\lambda}_{\phi\sigma} = 2 e^{-\lambda}\sin^{2}\theta + 2\cos^{2}\theta \]

5. Summing all contributions

\[ R_{\phi\phi} = -e^{-\lambda}\sin^{2}\theta(1 - r\lambda') - (\cos^{2}\theta - \sin^{2}\theta) - e^{-\lambda}\sin^{2}\theta \left( \frac{r}{2}(\nu' + \lambda') + 2 \right) - \cos^{2}\theta + 2 e^{-\lambda}\sin^{2}\theta + 2\cos^{2}\theta \]

The angular terms simplify to:

\[ \sin^{2}\theta \]

The radial terms combine to:

\[ R_{\phi\phi} = \sin^{2}\theta \left[ 1 - e^{-\lambda} \left( 1 + \frac{r}{2}(\nu' - \lambda') \right) \right] \]

Final result:

\[ \boxed{ R_{\phi\phi} = \sin^{2}\theta\, \left[ 1 - e^{-\lambda} \left( 1 + \frac{r}{2}(\nu' - \lambda') \right) \right] } \]

Alternatively written:

\[ \boxed{ R_{\phi\phi} = \sin^{2}\theta \left[ 1 + e^{-\lambda} \frac{1}{2} \left( r\lambda' - r\nu' - 2 \right) \right] } \]

Comparison with Rθθ

We had previously found:

\[ R_{\theta\theta} = 1 + e^{-\lambda}\frac{1}{2} \left( r\lambda' - r\nu' - 2 \right) \]

Therefore, it immediately follows:

\[ \boxed{ R_{\phi\phi} = \sin^{2}\theta\, R_{\theta\theta} } \]

This is exactly what is expected for a spherically symmetric metric: the \( \phi\phi \)-component is the \( \theta\theta \)-component multiplied by \( \sin^{2}\theta \).

Remark

The expressions derived here match the standard results in the literature. After substituting \[ e^{-\lambda} = 1 - \frac{2Gm(r)}{c^{2}r} \] one obtains the usual Einstein equations for a static star, from which the mass function \( m(r) \) and the TOV equation follow.

Ricci scalar

The Ricci scalar is:

\[ R = g^{\mu\nu} R_{\mu\nu} = g^{tt}R_{tt} + g^{rr}R_{rr} + g^{\theta\theta}R_{\theta\theta} + g^{\phi\phi}R_{\phi\phi} \]

We have:

\[ g^{\mu\nu} = \begin{pmatrix} e^{-\nu} & 0 & 0 & 0 \\ 0 & -e^{-\lambda} & 0 & 0 \\ 0 & 0 & -\frac{1}{r^{2}} & 0 \\ 0 & 0 & 0 & -\frac{1}{r^{2}\sin^{2}\theta} \end{pmatrix} \]

Since \[ R_{\phi\phi} = \sin^{2}\theta\, R_{\theta\theta}, \quad g^{\phi\phi} = -\frac{1}{r^{2}\sin^{2}\theta}, \] we get:

\[ g^{\phi\phi} R_{\phi\phi} = -\frac{1}{r^{2}\sin^{2}\theta} \cdot \sin^{2}\theta\, R_{\theta\theta} = -\frac{1}{r^{2}} R_{\theta\theta} = g^{\theta\theta} R_{\theta\theta} \]

Thus the angular contributions combine elegantly:

\[ g^{\theta\theta}R_{\theta\theta} + g^{\phi\phi}R_{\phi\phi} = 2 g^{\theta\theta} R_{\theta\theta} \]

Ricci scalar from components

Using the components we found:

\[ R_{tt} = \frac{1}{2} e^{\nu-\lambda} \left( \nu'' + \frac{1}{2}\nu'^2 - \frac{1}{2}\lambda'\nu' + \frac{2\nu'}{r} \right) \] \[ R_{rr} = -\frac{1}{2} \left( \nu'' + \frac{1}{2}\nu'^2 - \frac{1}{2}\nu'\lambda' - \frac{2\lambda'}{r} \right) \] \[ R_{\theta\theta} = 1 - e^{-\lambda} \left( 1 + \frac{r}{2}(\nu' - \lambda') \right) \] \[ R_{\phi\phi} = \sin^{2}\theta\, R_{\theta\theta} \]

Ricci scalar

\[ R = g^{\mu\nu} R_{\mu\nu} = g^{tt}R_{tt} + g^{rr}R_{rr} + 2 g^{\theta\theta} R_{\theta\theta} \]

With:

\[ g^{tt} = e^{-\nu}, \quad g^{rr} = -e^{-\lambda}, \quad g^{\theta\theta} = -\frac{1}{r^{2}} \]

Hence:

\[ R = e^{-\nu} R_{tt} - e^{-\lambda} R_{rr} - \frac{2}{r^{2}} R_{\theta\theta} \]

Substituting the components gives:

\[ \boxed{ R = e^{-\lambda} \left( \nu'' + \frac{1}{2}\nu'^2 - \frac{1}{2}\lambda'\nu' + \frac{2}{r}(\nu' - \lambda') \right) + \frac{2}{r^{2}}(e^{-\lambda} - 1) } \]

Ricci tensor components

The relevant Ricci tensor components for the internal Schwarzschild metric are:

\[ \boxed{ R_{tt} = \frac{1}{2} e^{\nu-\lambda} \left( \nu'' + \frac{1}{2}\nu'^2 - \frac{1}{2}\lambda'\nu' + \frac{2\nu'}{r} \right) } \] \[ \boxed{ R_{rr} = -\frac{1}{2} \left( \nu'' + \frac{1}{2}\nu'^2 - \frac{1}{2}\nu'\lambda' - \frac{2\lambda'}{r} \right) } \tag{6} \] \[ \boxed{ R_{\theta\theta} = 1 - e^{-\lambda} \left( 1 + \frac{r}{2}(\nu' - \lambda') \right) } \] \[ \boxed{ R_{\phi\phi} = \sin^{2}\theta\, R_{\theta\theta} } \]

Ricci scalar

\[ \boxed{ R = e^{-\lambda} \left( \nu'' + \frac{1}{2}\nu'^2 - \frac{1}{2}\lambda'\nu' + \frac{2}{r}(\nu' - \lambda') \right) + \frac{2}{r^{2}}(e^{-\lambda} - 1) } \tag{7} \]

These components form the basis for the Einstein equations inside a star, from which the mass function \( m(r) \), the pressure equation, and finally the Tolman–Oppenheimer–Volkoff equation follow.

Appendix 5.4 — Explicit Einstein Equations

Substituting (6) and (7) into the Einstein equation \[ G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2}R\, g_{\mu\nu} = \frac{8\pi G}{c^{4}} T_{\mu\nu} \tag{4} \] yields three independent equations.

(i) tt-component

\[ G_{tt} = R_{tt} - \frac{1}{2} R\, g_{tt} = \frac{8\pi G}{c^{4}}\, \rho c^{2} e^{\nu} \]

Substitute:

\[ G_{tt} = \frac{1}{2} e^{\nu-\lambda} \left( \nu'' + \frac{1}{2}\nu'^2 - \frac{1}{2}\lambda'\nu' + \frac{2\nu'}{r} \right) \]\[ - \frac{1}{2} e^{\nu} \left[ e^{-\lambda} \left( \nu'' + \frac{1}{2}\nu'^2 - \frac{1}{2}\lambda'\nu' + \frac{2}{r}(\nu' - \lambda') \right) + \frac{2}{r^{2}}(e^{-\lambda} - 1) \right] \]

After simplification:

\[ G_{tt} = \frac{e^{\nu-\lambda}}{r^{2}} \left( r\lambda' - 1 + e^{\lambda} \right) = \frac{8\pi G}{c^{4}}\, \rho c^{2} e^{\nu} \]

Thus:

\[ \frac{1}{r^{2}} \left( 1 + e^{-\lambda}(r\lambda' - 1) \right) = \frac{8\pi G}{c^{2}}\, \rho \]

Or in total derivative form:

\[ \boxed{ \frac{1}{r^{2}} \frac{d}{dr} \left[ r\left(1 - e^{-\lambda}\right) \right] = \frac{8\pi G}{c^{2}}\, \rho } \tag{8} \]

This is the familiar equation leading to the mass function \( m(r) \).

(ii) rr-component

\[ G_{rr} = R_{rr} - \frac{1}{2}R\, g_{rr} = \frac{8\pi G}{c^{4}}\, p\, e^{\lambda} \]

Substitute:

\[ G_{rr} = -\frac{1}{2} \left( \nu'' + \frac{1}{2}\nu'^2 - \frac{1}{2}\lambda'\nu' - \frac{2\lambda'}{r} \right) \]\[ + \frac{1}{2} e^{\lambda} \left[ e^{-\lambda} \left( \nu'' + \frac{1}{2}\nu'^2 - \frac{1}{2}\lambda'\nu' + \frac{2}{r}(\nu' - \lambda') \right) + \frac{2}{r^{2}}(e^{-\lambda} - 1) \right] \]

After simplification:

\[ \boxed{ \frac{\nu'}{r} + \frac{1}{r^{2}} \left(1 - e^{\lambda}\right) = \frac{8\pi G}{c^{4}}\, p\, e^{\lambda} } \tag{9} \]

This is the second independent Einstein equation.

(iii) θθ-component

The third equation follows from \( G_{\theta\theta} = \frac{8\pi G}{c^{4}}\, p\, r^{2} \) and is equivalent to the Tolman–Oppenheimer–Volkoff equation.

Continuation of rr-component — final form of equation (9)

\[ \frac{1}{2} \left( 2\frac{\nu' - \lambda'}{r} + 2\frac{\lambda'}{r} + \frac{1 - e^{\lambda}}{r^{2}} \right) = \frac{8\pi G}{c^{4}}\, p\, e^{\lambda} \]

Simplify:

\[ \frac{\nu'}{r} + \frac{1}{r^{2}}(1 - e^{\lambda}) = \frac{8\pi G}{c^{4}}\, p\, e^{\lambda} \]

Or in terms of \( e^{-\lambda} \):

\[ \frac{\nu'}{r} e^{-\lambda} + \frac{1}{r^{2}}(e^{-\lambda} - 1) = \frac{8\pi G}{c^{4}}\, p \tag{9} \]

(iii) θθ-component

\[ G_{\theta\theta} = R_{\theta\theta} - \frac{1}{2} R\, g_{\theta\theta} = \frac{8\pi G}{c^{4}}\, p\, r^{2} \]

Substitute components:

\[ G_{\theta\theta} = 1 - e^{-\lambda} \left( 1 + \frac{r}{2}(\nu' - \lambda') \right) \]\[ + \frac{1}{2} r^{2} e^{-\lambda} \left( \nu'' + \frac{1}{2}\nu'^2 - \frac{1}{2}\lambda'\nu' + \frac{2}{r}(\nu' - \lambda') \right) + \frac{1}{2} r^{2} \frac{2}{r^{2}} (e^{-\lambda} - 1) \]

After simplification:

\[ G_{\theta\theta} = \frac{1}{2} r^{2} e^{-\lambda} \left( \nu'' + \frac{1}{2}\nu'^2 - \frac{1}{2}\lambda'\nu' + \frac{1}{r} (\nu' - \lambda') \right) = \frac{8\pi G}{c^{4}}\, p\, r^{2} \]

Thus:

\[ \boxed{ \frac{1}{2} e^{-\lambda} \left( \nu'' + \frac{1}{2}\nu'^2 - \frac{1}{2}\lambda'\nu' + \frac{1}{r} (\nu' - \lambda') \right) = \frac{8\pi G}{c^{4}}\, p } \tag{10} \]

Remark

The three equations (8), (9), and (10) are not independent. Equation (10) follows mathematically from (8), (9), and the energy conservation equation:

\[ \nabla_{\mu} T^{\mu}_{r} = 0 \]

Therefore, it suffices to use (8) and (9) to solve for \( \lambda(r) \), \( \nu(r) \), and \( p(r) \).

Appendix 5.5 — Integration of the first equation

Start with equation (8):

\[ \frac{1}{r^{2}} \frac{d}{dr} \left[ r\left(1 - e^{-\lambda}\right) \right] = \frac{8\pi G}{c^{2}}\, \rho \]

Multiply by \( r^{2} \):

\[ \frac{d}{dr} \left[ r\left(1 - e^{-\lambda}\right) \right] = \frac{8\pi G}{c^{2}}\, \rho\, r^{2} \]

Integrate over \( r \):

\[ \int \frac{d}{dr} \left[ r\left(1 - e^{-\lambda}\right) \right] dr = \int \frac{8\pi G}{c^{2}}\, \rho\, r^{2} dr \]

Thus:

\[ r\left(1 - e^{-\lambda}\right) = \frac{8\pi G}{c^{2}}\, \rho \frac{r^{3}}{3} \]

Or:

\[ \boxed{ 1 - e^{-\lambda} = \frac{8\pi G}{3c^{2}}\, \rho\, r^{2} } \]

This is the classical result for a constant-density star: the interior field has a quadratic curvature term.

Integration result

From equation (11):

\[ e^{-\lambda(r)} = 1 - \frac{8\pi G}{3c^{2}}\, \rho\, r^{2} \tag{11} \]

With the definition of the enclosed mass:

\[ m(r) = 4\pi \int_{0}^{r} \rho\, r'^{2} dr' = \frac{4\pi}{3} \rho\, r^{3} \]

We get the standard form:

\[ \boxed{ e^{-\lambda(r)} = 1 - \frac{2G m(r)}{c^{2} r} } \tag{12} \]

The function \( m(r) \) is the mass enclosed within radius \( r \). For \( r = R \), \( m(R) = M \), so the interior solution matches smoothly to the exterior Schwarzschild metric.

Appendix 5.6 — Energy Conservation and the TOV Equation

1. Energy Conservation

The conservation equation \[ \nabla_{\mu} T^{\mu}_{r} = 0 \] yields:

\[ \boxed{ \frac{dp}{dr} = -\frac{1}{2}(\rho c^{2} + p)\, \nu' } \tag{13} \]

2. Eliminating \( \nu' \) using equation (9)

From (9):

\[ \frac{\nu'}{r} e^{-\lambda} + \frac{1}{r^{2}}(e^{-\lambda} - 1) = \frac{8\pi G}{c^{4}}\, p \]

Solving for \( \nu' \):

\[ \nu' = \frac{ \frac{8\pi G}{c^{4}}\, p\, r + \frac{1}{r}\left(1 - e^{-\lambda}\right)} {e^{-\lambda}} \]

Now use (12):

\[ e^{-\lambda} = 1 - \frac{2Gm}{c^{2}r}, \qquad 1 - e^{-\lambda} = \frac{2Gm}{c^{2}r} \]

Substitute:

\[ \nu' = \frac{ \frac{8\pi G}{c^{4}}\, p\, r + \frac{1}{r} \frac{2Gm}{c^{2}r}}{1 - \frac{2Gm}{c^{2}r}} = \frac{2G\left(m+ \frac{4\pi}{c^{2}}\, p r^{3}\right)} {c^2 r^2 \left(1 - \frac{2Gm}{c^{2}r}\right)} \]

Rewritten:

\[ \boxed{ \nu' = \frac{2G}{c^{2}} \frac{m + 4\pi r^{3}p/c^{2}}{r^{2}\left(1 - \frac{2Gm}{c^{2}r}\right)} } \tag{14} \]

3. The Tolman–Oppenheimer–Volkoff Equation

Combining (13) and (14) gives:

\[ \frac{dp}{dr} = -\frac{1}{2}(\rho c^{2} + p)\, \nu' = -(\rho c^{2} + p) \frac{G\left(m + 4\pi r^{3}p/c^{2}\right)}{c^{2} r^{2}\left(1 - \frac{2Gm}{c^{2}r}\right)} \]

Hence the TOV equation reads:

\[ \boxed{ \frac{dp}{dr} = -\frac{(\rho c^{2} + p)G\left(m + 4\pi r^{3}p/c^{2}\right)} {c^{2} r^{2}\left(1 - \frac{2Gm}{c^{2}r}\right)} } \tag{15} \]

This is the relativistic equilibrium equation for a static, spherically symmetric star. Together with the mass function \[ m'(r) = 4\pi r^{2}\rho \] it forms the complete system for the star’s internal structure.

Remark

Equation (15) describes the mechanical equilibrium between gravity (downward) and the pressure gradient (upward). For constant density \( \rho = \text{constant} \) this equation is exactly solvable.

Appendix 5.7 — Solution for Constant Density

With \[ m(r) = \frac{4\pi}{3}\rho\, r^{3} \] the TOV equation (15) becomes:

\[ \frac{dp}{dr} = - \frac{\rho c^{2} + p}{3c^{2}} \, \frac{4\pi G r \left(\rho + 3p/c^{2}\right)} {1 - \frac{8\pi G}{3c^{2}}\rho r^{2}} \tag{16} \]

Substitution: Pressure as a Dimensionless Variable

Define:

\[ x(r) \equiv \frac{p(r)}{\rho c^{2}} \qquad\Longrightarrow\qquad p = \rho c^{2} x, \quad dp = \rho c^{2}\, dx \tag{16a} \]

Insert into (16):

Left-hand side

\[ \frac{dp}{dr} = \rho c^{2}\, \frac{dx}{dr} \]

Right-hand side

\[ \rho + \frac{3p}{c^{2}} = \rho(1 + 3x), \qquad \rho c^{2} + p = \rho c^{2}(1 + x) \]

Thus equation (16) becomes:

\[ \rho c^{2}\frac{dx}{dr} = - \frac{\rho c^{2}(1+x)}{3c^{2}} \, \frac{4\pi G r\, \rho(1+3x)} {1 - \beta r^{2}} \] with \[ \beta \equiv \frac{8\pi G}{3c^{2}}\rho. \]

Simplify:

\[ \frac{dx}{dr} = - \frac{4\pi G \rho}{3c^{2}} \, \frac{r(1+x)(1+3x)}{1 - \beta r^{2}} \]

Now introduce:

\[ \frac{\beta}{2} \equiv \frac{4\pi G}{3c^{2}}\rho \] so that: \[ \frac{dx}{dr} = - \frac{\beta}{2}\, \frac{r(1+x)(1+3x)}{1 - \beta r^{2}} \]

Separation of Variables

We rewrite:

\[ \frac{dx}{(1+x)(1+3x)} = - \frac{\beta}{2}\, \frac{r\, dr}{1 - \beta r^{2}} \]

This is the directly integrable form:

\[ \boxed{ \int \frac{dx}{(1+x)(1+3x)} = - \frac{\beta}{2} \int \frac{r\, dr}{1 - \beta r^{2}} } \]

The left-hand side requires partial fractions; the right-hand side gives a logarithm.

Integration of Both Sides

Start with:

\[ \frac{dx}{(1+x)(1+3x)} = -\frac{\beta}{2}\, \frac{r}{1 - \beta r^{2}}\, dr \]

Left Integral — Partial Fractions

Write:

\[ \frac{1}{(1+x)(1+3x)} = -\frac{1}{2}\frac{1}{1+x} + \frac{3}{2}\frac{1}{1+3x} \]

Integrate term by term:

\[ -\frac{1}{2}\int \frac{dx}{1+x} = -\frac{1}{2}\ln(1+x) + C_{1} \] \[ \frac{3}{2}\int \frac{dx}{1+3x} = \frac{3}{2}\cdot \frac{1}{3}\ln(1+3x) = \frac{1}{2}\ln(1+3x) + C_{2} \]

Together:

\[ \int \frac{dx}{(1+x)(1+3x)} = -\frac{1}{2}\ln(1+x) + \frac{1}{2}\ln(1+3x) + C_{3} \]

Or more compactly:

\[ \int \frac{dx}{(1+x)(1+3x)} = \frac{1}{2} \ln\!\left( \frac{1+3x}{1+x} \right) + C_{3} \]

Right Integral

\[ -\frac{\beta}{2}\int \frac{r}{1 - \beta r^{2}}\, dr \]

Substitute:

\[ u = 1 - \beta r^{2}, \qquad du = -2\beta r\, dr \] \[ r\, dr = -\frac{du}{2\beta} \]

Thus:

\[ -\frac{\beta}{2} \int \frac{r}{1 - \beta r^{2}}\, dr = -\frac{\beta}{2} \left( -\frac{1}{2\beta} \int \frac{du}{u} \right) = \frac{1}{4}\ln u \]

So:

\[ -\frac{\beta}{2}\int \frac{r}{1 - \beta r^{2}}\, dr = \frac{1}{4}\ln(1 - \beta r^{2}) \]

Equate Both Integrals

\[ \frac{1}{2} \ln\!\left( \frac{1+3x}{1+x} \right) = \frac{1}{4}\ln(1 - \beta r^{2}) + C_{4} \]

Multiply by 2:

\[ \ln\!\left( \frac{1+3x}{1+x} \right) = \frac{1}{2}\ln(1 - \beta r^{2}) + C_{5} \]

Exponentiate:

\[ \left( \frac{1+3x}{1+x} \right) = C_{6}\, \sqrt{1 - \beta r^{2}} \]

The physically relevant branch is the positive root:

\[ \boxed{ \frac{1+3x}{1+x} = C_{7}\,\sqrt{1 - \beta r^{2}} } \]

Define:

\[ \alpha(r) \equiv \sqrt{1 - \beta r^{2}} \]

Determine the Integration Constant \(C_{7}\)

From the integrated relation:

\[ \frac{1+3x}{1+x} = C_{7}\,\alpha(r) \]

Use the boundary condition at the surface:

\[ p(R) = 0 \quad\Rightarrow\quad x(R) = 0 \]

Insert to get:

\[ \frac{1+0}{1+0} = C_{7}\,\alpha(R) \] \[ 1 = C_{7}\, a \] with \[ a \equiv \alpha(R) = \sqrt{1 - \beta R^{2}} = \sqrt{1 - \frac{2GM}{c^{2}R}} \]

Thus:

\[ \boxed{ C_{7} = \frac{1}{a} } \]

Solution for \(x(r)\)

We had:

\[ \frac{1+3x}{1+x} = \frac{\alpha(r)}{a} \]

Solve for \(x(r)\):

\[ 1 + 3x = \frac{\alpha(r)}{a}(1+x) \] \[ 1 + 3x = \frac{\alpha}{a} + \frac{\alpha}{a}x \] \[ 3x - \frac{\alpha}{a}x = \frac{\alpha}{a} - 1 \] \[ x\left(3 - \frac{\alpha}{a}\right) = \frac{\alpha - a}{a} \] \[ \boxed{ x(r) = \frac{\alpha(r) - a}{3a - \alpha(r)} } \tag{16b} \]

Pressure Profile \(p(r)\)

Since \(p = \rho c^{2} x\), we have:

\[ \boxed{ p(r) = \rho c^{2}\, \frac{\alpha(r) - a}{3a - \alpha(r)} } \tag{18} \] with \[ \alpha(r) =\sqrt{ 1 - \beta r^{2}}, \qquad \beta = \frac{8\pi G}{3c^{2}}\rho, \qquad a = \alpha(R) = \sqrt{1 - \frac{2GM}{c^{2}R}} \]

Check Points

This is exactly the classical interior Schwarzschild solution for constant density.

Next Step

From equation (13):

\[ \frac{dp}{dr} = -\frac{1}{2}(\rho c^{2} + p)\,\nu' \]

we can now integrate \( \nu'(r) \) to find the time component of the metric:

\[ e^{\nu(r)} \]

Derivation of the Time Component of the Metric

Start from equation (19):

\[ \frac{d\nu}{dr} = -2\,\frac{\rho c^{2} + p}{dp/dr} \tag{19} \]

Using the chain rule:

\[ \frac{d\nu}{dr} = \frac{d\nu}{dp}\,\frac{dp}{dr} \quad\Rightarrow\quad \frac{d\nu}{dp} = -\frac{2}{\rho c^{2} + p} \]

Integration with respect to \(p\)

\[ \nu = -2\ln(\rho c^{2} + p) + \ln C' = \ln\!\frac{C'}{ (\rho c^{2} + p)^{2} } \]

Exponentiate:

\[ e^{\nu/2} = \frac{C''}{\rho c^{2} + p} \qquad (C''>0) \]

Using the Pressure Solution (18)

From (18):

\[ p(r) = \rho c^{2}\, \frac{\alpha(r) - a}{3a - \alpha(r)} \]

So:

\[ \rho c^{2} + p = \rho c^{2} \left( 1 + \frac{\alpha - a}{3a - \alpha} \right) = \rho c^{2}\, \frac{2a}{3a - \alpha(r)} \]

Insert into \(e^{\nu/2}\):

\[ e^{\nu/2} = \frac{C''}{\rho c^{2} + p} = \frac{C''}{\rho c^{2}} \frac{3a - \alpha(r)}{2a} \tag{19b} \]

Determine the Integration Constant \(C''\)

At the surface \(r = R\):

\[ p(R) = 0, \qquad \alpha(R) = a, \qquad e^{\nu(R)/2} = a =\sqrt{ 1 - \frac{2GM}{c^{2}R}} \]

Insert into (19b):

\[ a = \frac{C''}{\rho c^{2}} \frac{3a-a}{2a} = \frac{C''}{\rho c^{2}} \]

Thus:

\[ \boxed{ C'' = a\,\rho c^{2} } \]

Final Result

\[ e^{\nu/2} = \frac{a\rho c^{2}}{\rho c^{2}} \, \frac{3a-\alpha(r)}{2a} \]

Since:

\[ \alpha(r) =\sqrt{ 1 - \beta r^{2}}, \qquad \beta = \frac{8\pi G}{3c^{2}}\rho, \qquad a = \sqrt{1 - \frac{2GM}{c^{2}R}} \]

We get:

\[ \boxed{ e^{\nu(r)/2} = \frac{3a - \alpha(r)}{2} = \frac{3a - \sqrt{1 - \beta r^{2}}}{2} } \tag{20} \]

Or explicitly:

\[ \boxed{ e^{\nu(r)/2} = \frac{3a - \sqrt{1 - \frac{8\pi G}{3c^{2}}\rho\, r^{2}}}{2} } \tag{21} \]

This is the time component of the interior Schwarzschild metric for a star with constant density.

Complete Interior Metric

The interior solution for a homogeneous sphere of radius \(0 \le r \le R\) reads:

\[ ds^{2} = \left( \frac{3a - \alpha(r)}{2} \right)^{2} c^{2} dt^{2} - \frac{dr^{2}}{\alpha^2(r)} - r^{2}\left(d\theta^{2} + \sin^{2}\theta\, d\phi^{2}\right) \tag{22} \] with \[ \alpha(r) = \sqrt{1 - \frac{8\pi G}{3c^{2}}\rho\, r^{2}}, \qquad a = \alpha(R) = \sqrt{1 - \frac{2GM}{c^{2}R}}. \]

Using Mass Relations

Since:

\[ M = \frac{4\pi}{3}\rho R^{3} \quad\Rightarrow\quad \rho = \frac{3M}{4\pi R^{3}}, \] it follows: \[ \frac{8\pi G}{3c^{2}}\rho r^{2} = \frac{2GM}{c^{2}R}\frac{r^{2}}{R^{2}}. \]

Inserted into (22) gives:

\[ \boxed{ ds^{2} = \left( \frac{3\sqrt{1 - \frac{2GM}{c^{2}R}} - \sqrt{1 - \frac{2GM}{c^{2}R}\frac{r^{2}}{R^{2}}}} {2} \right)^{2} c^{2} dt^{2} - \frac{dr^{2}} {1 - \frac{2GM}{c^{2}R}\frac{r^{2}}{R^{2}}} - r^{2}\left(d\theta^{2} + \sin^{2}\theta\, d\phi^{2}\right) \tag{22a} } \]

Remark

This metric describes the spacetime inside a homogeneous sphere. At \(r = R\) continuity with the exterior Schwarzschild solution holds:

\[ e^{-\lambda(R)} = 1 - \frac{2GM}{c^{2}R}, \qquad e^{\nu(R)} = 1 - \frac{2GM}{c^{2}R}. \]

Appendix 5.8 — Central Pressure and Buchdahl Limit

From \( \alpha(0) = 1 \) follows for the central pressure:

\[ p(0) = \rho c^{2} \frac{1 - a}{3a - 1} \tag{23} \] with \[ a = \sqrt{1 - \frac{2GM}{c^{2}R}}. \]

When the denominator vanishes, the central pressure diverges:

\[ 3a - 1 = 0 \quad\Rightarrow\quad a = \frac{1}{3}. \]

This gives the Buchdahl limit:

\[ \boxed{ \frac{2GM}{c^{2}R} = \frac{8}{9} } \tag{24} \]

This limit marks the maximum compactness for a stable, static configuration with constant density. Exceeding it prevents hydrostatic equilibrium and the star collapses into a black hole.

Appendix 5.9 — Summary