Dr Phill’s Science Made Simple

Relativity

Schwarzschild Outer Metric

In 1915, the same year as the publication of the Field Equations, the German physicist Karl Schwarzschild produced the first exact solution. He did this while in hospital having developed a fatal skin condition while serving in the German army on the Russian front in World War I.

The Schwarzschild Outer Metric is an exact solution to the Field Equations for the vacuum of space around a static and symmetric body. The metric must be independent of time and the spherical \(\theta\) and \(\phi\) coordinates.

Define the metric

The metric can be written as:

\[ds^2 = A(r)dt^2 - B(r)dr^2 - r^2d\theta^2 - r^2 \sin^2\theta d\phi^2\]

Where \(A\) and \(B\) are unknown functions of \(r\). The solution must become the Minkowski metric as \(r \rightarrow \infty\) and as \(M \rightarrow 0\).

The metric and its inverse are now defined to be:

\[ g_{\alpha\beta} = \begin{bmatrix} A & 0 & 0 & 0\\ 0 & -B & 0 & 0\\ 0 & 0 & -r^2 & 0\\ 0 & 0 & 0 & -r^2 \sin^2 \theta \end{bmatrix}\]

\[ g^{\alpha\beta} = \begin{bmatrix} \frac{1}{A} & 0 & 0 & 0\\ 0 & -\frac{1}{B} & 0 & 0\\ 0 & 0 & -\frac{1}{r^2} & 0\\ 0 & 0 & 0 & -\frac{1}{r^2 \sin^2 \theta} \end{bmatrix}\]

Calculate the Christoffel Symbols

The Christoffel symbols are:

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

Calculate Derivatives

Calculate the non-zero partial derivatives of the metric.

\[\frac{\partial g_{tt}}{\partial r} = \frac{dA}{dr} = A'\]

\[\frac{\partial g_{rr}}{\partial r} = -\frac{dB}{dr} = -B'\]

\[\frac{\partial g_{\theta\theta}}{\partial r} = -2r\]

\[\frac{\partial g_{\phi\phi}}{\partial r} = -2r\sin\theta\]

\[\frac{\partial g_{\phi\phi}}{\partial\theta} = -2r^2\sin\theta\cos\theta\]

There are four sets of Christoffel symbols, one set for each of \(t, r, \theta, \phi\).

Calculate t Sysmbols

For the \(t\) component, \(\lambda = t\).

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

The only non-zero \(g^{t\rho}\) term is \(g^{tt}\), so \(\rho = t\).

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

The third term is zero as there are no non-zero time derivatives. This leaves two non-zero symbols \(\mu = t, \nu = r\) and \(\mu = r, \nu = t\).

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

Calculate r Symbols

For the \(r\) symbols, \(\lambda = r\), and again for all non-zero symbols \(\rho = r\).

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

Let \(\mu = t\).

\[\Gamma^{r}_{t\nu} = \frac{1}{2}g^{rr}\left(\frac{\partial g_{tr}}{\partial\nu} + \frac{\partial g_{\nu r}}{\partial t} - \frac{\partial g_{t\nu}}{\partial r}\right)\]

The only non-zero term is the third term with \(\nu = t\).

\[\Gamma^{r}_{tt} = -\frac{1}{2}g^{rr}\frac{\partial g_{tt}}{\partial r} = \frac{A'}{2B}\]

Let \(\mu = r\).

\[\Gamma^{r}_{r\nu} = \frac{1}{2}g^{rr}\left(\frac{\partial g_{rr}}{\partial\nu} + \frac{\partial g_{\nu r}}{\partial r} - \frac{\partial g_{r\nu}}{\partial r}\right)\]

There only non-zero symbolsis \(\nu = r\).

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

It should be apparent that for \(\mu = \theta\) then the only symbol is for \(\nu = \theta\).

\[\Gamma^r_{\theta\theta} &= -\frac{1}{2}g^{rr}g_{\theta\theta,r} = -\frac{r}{B}\]

Likewise for \(\mu = \nu = \phi\).

\[\Gamma^r_{\phi\phi} &= -\frac{1}{2}g^{rr}g_{\phi\phi,r} = -\frac{r\sin^2\theta}{B}\]

Calculate \(\theta\) Symbols

For the \(\theta\) component, \(\lambda = \theta\), and again for all non-zero symbols \(\rho = \theta\).

\[\Gamma^{\theta}_{\mu\nu} = \frac{1}{2}g^{\theta\theta}(\frac{\partial g_{\mu\theta}}{\partial\nu} + frac{\partial g_{\nu\theta}{\partial\mu} - frac{\partial g_{\mu\nu}}{\partial\theta})\]

This gives three symbols:

\[\begin{align*} \Gamma^\theta_{\theta r} &= \Gamma^\theta_{r\theta} = \frac{1}{2}g^{\theta\theta}g_{\theta\theta, r} = \frac{1}{2r^2}2r = \frac{1}{r}\\ \Gamma^\theta_{\phi\phi} &= -\frac{1}{2}g^{\theta\theta}g_{\phi\phi,\theta} = \frac{1}{2r^2}2r^2\sin\theta\cos\theta = -\sin\theta\cos\theta \end{align*}\]

Calculate \(\phi\) Symbols

For the \(\phi\) component, \(\lambda = \phi\), and for all non-zero symbols \(\rho = \phi\).

\[\Gamma^{\phi}_{\mu\nu} = \frac{1}{2}g^{\phi\phi}(g_{\mu\phi,\nu} + g_{\nu\phi,\mu} - g_{\mu\nu,\phi})\]

This gives four symbols:

\[\begin{align*} \Gamma^\phi_{\phi r} &= \Gamma^\phi_{r\phi} = \frac{1}{2}g^{\phi\phi}g_{\phi\phi,r} = \frac{1}{2r^2\sin^2\theta}2r\sin^2\theta = \frac{1}{r}\\ \Gamma^\phi_{\theta\phi} &= \Gamma^\phi_{\phi\theta} = \frac{1}{2}g^{\phi\phi}g_{\phi\phi,\theta} = \frac{1}{2r^2\sin^2\theta}2r^2\sin\theta\cos\theta = \cot\theta \end{align*}\]

So, the Christoffel symbols are:

\[\begin{align*} \Gamma^{t}_{rt} &= \Gamma^{t}_{tr} = \frac{A'}{2A}\\ \Gamma^r_{tt} &= \frac{A'}{2B}\\ \Gamma^r_{rr} &= \frac{B'}{2B}\\ \Gamma^r_{\theta\theta} &= -\frac{r}{B}\\ \Gamma^r_{\phi\phi} &= -\frac{r\sin^2\theta}{B}\\ \Gamma^\theta_{\theta r} &= \Gamma^\theta_{r\theta} = \frac{1}{r}\\ \Gamma^\theta_{\phi\phi} &= -\sin\theta\cos\theta\\ \Gamma^\phi_{\phi r} &= \Gamma^\phi_{r\phi} = \frac{1}{r}\\ \Gamma^\phi_{\theta\phi} &= \Gamma^\phi_{\phi\theta} = \cot\theta \end{align*}\]

Construct the Ricci Tensor

The Ricci tensor is defined to be:

\[R_{\mu\nu} = \Gamma^\lambda_{\mu\nu,\lambda} - \Gamma^\lambda_{\lambda\mu,\nu} + \Gamma^\lambda_{\lambda\eta}\Gamma^\eta_{\mu\nu} - \Gamma^\lambda_{\mu\eta}\Gamma^\eta_{\lambda\nu}\]

Due to symmetry the Ricci tensor has just the four trace elements:

\[R_{\mu\mu} = \Gamma^\lambda_{\mu\mu,\lambda} - \Gamma^\lambda_{\lambda\mu,\mu} + \Gamma^\lambda_{\lambda\eta}\Gamma^\eta_{\mu\mu} - \Gamma^\lambda_{\mu\eta}\Gamma^\eta_{\lambda\mu}\]

To calculate \(R_{tt}\), \(\mu = t\) expand each term and eliminate zero symbols. Only derivatives of \(r\) exist.

\[\begin{align*} R_{tt} &= \Gamma^\lambda_{tt,\lambda} - \Gamma^\lambda_{t\lambda,t} + \Gamma^\lambda_{\lambda\eta}\Gamma^\eta_{tt} - \Gamma^\lambda_{t\eta}\Gamma^\eta_{\lambda t}\\ &= \Gamma^r_{tt,r} + \bigg(\Gamma^t_{tr} + \Gamma^r_{rr} + \Gamma^\theta_{\theta r} + \Gamma^\phi_{\phi r}\bigg)\Gamma^r_{tt} - 2\Gamma^t_{tr}\Gamma^r_{tt}\\ &= \bigg(\frac{A'}{2B}\bigg)' + \bigg(\frac{A'}{2A} + \frac{B'}{2B} + \frac{1}{r} + \frac{1}{r}\bigg)\frac{A'}{2B} - 2\frac{A'}{2A}\frac{A'}{2B}\\ &= \frac{A''}{2B} - \frac{A'B'}{2B^2} + \frac{A'^2}{4AB} + \frac{A'B'}{4B^2} + \frac{A'}{rB} -\frac{A'^2}{2AB}\\ &= \frac{A''}{2B} - \frac{A'}{4B}\bigg(\frac{A'}{A} + \frac{B'}{B}\bigg) + \frac{A'}{rB} \end{align*}\]

To calculate \(R_{rr}\), \(\mu = r\) expand each term and eliminate zero symbols. Only derivatives of \(r\) exist.

\[\begin{align*} R_{rr} &= \Gamma^\lambda_{rr,\lambda} - \Gamma^\lambda_{r\lambda,r} + \Gamma^\lambda_{\lambda\eta}\Gamma^\eta_{rr} - \Gamma^\lambda_{r\eta}\Gamma^\eta_{\lambda r}\\ &= \Gamma^r_{rr,r} - (\Gamma^t_{rt,r} + \Gamma^r_{rr,r} + \Gamma^\theta_{r\theta,r} + \Gamma^\phi_{r\phi,r}) + (\Gamma^t_{tr}\Gamma^r_{rr} + \Gamma^r_{rr}\Gamma^r_{rr} + \Gamma^\theta_{\theta r}\Gamma^r_{rr} + \Gamma^\phi_{\phi r}\Gamma^r_{rr}) - (\Gamma^t_{rt}\Gamma^t_{rt} + \Gamma^r_{rr}\Gamma^r_{rr} + \Gamma^\theta_{r\theta}\Gamma^\theta_{r\theta} + \Gamma^\phi_{r\phi}\Gamma^\phi_{\phi r})\\ &= - (\Gamma^t_{rt,r} + \Gamma^\theta_{r\theta,r} + \Gamma^\phi_{r\phi,r}) + (\Gamma^t_{tr}\Gamma^r_{rr} + \Gamma^\theta_{\theta r}\Gamma^r_{rr} + \Gamma^\phi_{\phi r}\Gamma^r_{rr}) - (\Gamma^t_{rt}\Gamma^t_{rt} + \Gamma^\theta_{r\theta}\Gamma^\theta_{r\theta} + \Gamma^\phi_{r\phi}\Gamma^\phi_{\phi r})\\ &= -\bigg(\frac{A'}{2A}\bigg)' + \frac{2}{r^2} + \frac{A'B'}{4AB} + \frac{B'}{rB} - \bigg(\frac{A'}{2A}\bigg)^2 - \frac{2}{r^2}\\ &= -\frac{A''}{2A} + \frac{A'^2}{2A^2} - \frac{A'^2}{4A^2} + \frac{A'B'}{4AB} + \frac{B'}{rB}\\ &= -\frac{A''}{2A} + \frac{A'}{4A}\bigg(\frac{A'}{A} + \frac{B'}{B}\bigg) + \frac{B'}{rB} \end{align*}\]

To calculate \(R_{\theta\theta}\), \(\mu = \theta\) expand each term and eliminate zero symbols. Only derivatives of \(r\) and \(\theta\) exist.

\[\begin{align*} R_{\theta\theta} &= \Gamma^\lambda_{\theta\theta,\lambda} - \Gamma^\lambda_{\theta\lambda,\theta} + \Gamma^\lambda_{\lambda\eta}\Gamma^\eta_{\theta\theta} - \Gamma^\lambda_{\theta\eta}\Gamma^\eta_{\lambda\theta}\\ &= \Gamma^r_{\theta\theta,r} - \Gamma^\phi_{\theta\phi,\theta} + (\Gamma^t_{tr} + \Gamma^r_{rr} + \Gamma^\theta_{\theta r} + \Gamma^\phi_{\phi r})\Gamma^r_{\theta\theta} - 2\Gamma^\theta_{\theta r}\Gamma^r_{\theta\theta} - \Gamma^\phi_{\theta\phi}\Gamma^\phi_{\theta\phi}\\ &= - \bigg(\frac{r}{B}\bigg)' + \csc^2\theta - \frac{r}{B}\bigg(\frac{A'}{2A} + \frac{B'}{2B} + \frac{1}{r} + \frac{1}{r}\bigg) + 2\frac{1}{r}\frac{r}{B} - \cot^2\theta\\ &= -\frac{1}{B} + \frac{rB'}{B^2} + \frac{2}{B} - \frac{r}{2B}\bigg(\frac{A'}{A} + \frac{B'}{B}\bigg) - \frac{2}{B} + 1\\ &= -\frac{1}{B} - \frac{r}{2B}\bigg(\frac{A'}{A} - \frac{B'}{B}\bigg) + 1 \end{align*}\]

To calculate \(R_{\phi\phi}\), \(\mu = \phi\) expand each term and eliminate zero symbols. Only derivatives of \(r\) and \(\theta\) exist.

\[\begin{align*} R_{\phi\phi} &= \Gamma^\lambda_{\phi\phi,\lambda} - \Gamma^\lambda_{\phi\lambda,\phi} + \Gamma^\lambda_{\lambda\eta}\Gamma^\eta_{\phi\phi} - \Gamma^\lambda_{\phi\eta}\Gamma^\eta_{\lambda\phi}\\ &= \Gamma^r_{\phi\phi,r} + \Gamma^\theta_{\phi\phi,\theta} + \bigg(\Gamma^t_{tr} + \Gamma^r_{rr} + \Gamma^\theta_{\theta r} + \Gamma^\phi_{\phi r}\bigg)\Gamma^r_{\phi\phi} - 2\Gamma^\phi_{\phi r}\Gamma^r_{\phi\phi} - \Gamma^\phi_{\phi\theta}\Gamma^\theta_{\phi\phi}\\ &= - \frac{\sin^2\theta}{B} + \frac{r\sin^2\theta B'}{B^2} - \cos^2\theta + \sin^2\theta -\frac{r\sin^2\theta}{B}\bigg(\frac{A'}{2A} + \frac{B'}{2B} + \frac{1}{r} + \frac{1}{r}\bigg) + 2\frac{1}{r}\frac{r\sin^2\theta}{B} + \cos^2\theta\\ &= \sin^2\theta\bigg(-\frac{1}{B} + \frac{rB'}{B^2} - \frac{r}{2B}\bigg(\frac{A'}{A} - \frac{B'}{B}\bigg) + 1\bigg)&&\\ &= \sin^2\theta R_{\theta\theta} \end{align*}\]

The Ricci tensor is:

\[\begin{align*} R_{tt} &= \frac{A''}{2B} - \frac{A'}{4B}\bigg(\frac{A'}{A} + \frac{B'}{B}\bigg) + \frac{A'}{rB}\\ R_{rr} &= -\frac{A''}{2A} + \frac{A'}{4A}\bigg(\frac{A'}{A} + \frac{B'}{B}\bigg) + \frac{B'}{rB}\\ R_{\theta\theta} &= -\frac{1}{B} - \frac{r}{2B}\bigg(\frac{A'}{A} - \frac{B'}{B}\bigg) + 1\\ R_{\phi\phi} &= \sin^2\theta R_{\theta\theta} \end{align*}\]

Construct the Ricci Scalar

The Ricci scalar is:

\[R = g^{\mu\nu}R_{\mu\nu}\]

\[\begin{align*} R &= \frac{R_{tt}}{A} - \frac{R_{rr}}{B} - \frac{R_{\theta\theta}}{r^2} - \frac{R_{\phi\phi}}{r^2\sin^2\theta}\\ &= \frac{A''}{AB} - \frac{A'}{2AB}\bigg(\frac{A'}{A} + \frac{B'}{B}\bigg) + \frac{2}{r^2B}(1 + \frac{rA'}{A} - \frac{rB'}{B} - B) \end{align*}\]

Construct the Einstein Tensor

The Einstein Tensor can now be constructed.

\[G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2}Rg_{\mu\nu}\]

\[\begin{align*} G_{tt} &= -\frac{A}{r^2B}(1 - r\frac{B'}{B} - B)\\ G_{rr} &= \frac{1}{r^2}(1 + r\frac{A'}{A} - B)\\ G_{\theta\theta} &= \frac{r^2}{2B}\bigg(\frac{A''}{A} + \frac{A'}{rA} - \frac{B'}{rB} - \frac{A'(AB)'}{2A^2B}\bigg)\\ G_{\phi\phi} &= G_{\theta\theta}\sin^2\theta \end{align*}\]

Construct the Field Equations

The Einstein tensor is zero because the stress energy tensor \(T_{\mu\nu} = 0\) in a vacuum.

The field equations are:

\[\begin{align*} -\frac{A}{r^2B}(1 - \frac{rB'}{B} - B) &= 0\\ \frac{1}{r^2}(1 + r\frac{A'}{A} - B) &= 0\\ \frac{r^2}{2B}\bigg(\frac{A''}{A} + \frac{A'}{rA} - \frac{B'}{rB} - \frac{A'(AB)'}{2A^2B}\bigg) &= 0\\ G_{\theta\theta}\sin^2\theta &= 0 \end{align*}\]

Solve the Equations

The first two field equations reduce to:

\[\begin{align*} 1 - \frac{rB'}{B} - B &= 0\\ 1 + \frac{rA'}{A} - B &= 0 \end{align*}\]

Subtracting gives:

\[\frac{A'}{A} +\frac{B'}{B} = \frac{BA' + AB'}{AB} = \frac{(AB)'}{AB} = 0\]

Integrating gives:

\[AB = k_1\]

Where \(k_1\) is a constant of integration. As the metric must conform to the Minkowski metric \(k_1 = c^2\), \(AB = c^2\).

The first equation can be written as:

\[\frac{1}{B} - \frac{rB'}{B^2} = \frac{d}{dr}\bigg(\frac{r}{B}\bigg) = 1\]

Integrating gives:

\[\frac{r}{B} = r + k_2\]

Where \(k_2\) is a constant of integration.

This gives.

\[B = \bigg(1 + \frac{k_2}{r}\bigg)^{-1}\]

Therefore.

\[A = c^2\bigg(1 + \frac{k_2}{r}\bigg)\]

The solution must become the Minkowski metric as \(r \rightarrow \infty\) and as \(M \rightarrow 0\). Hence the constant must be a multiple of the mass \(k_2 = kM\).

The \(A\) and \(B\) terms are in fact the Lorentz transformations.

\[1 + \frac{kM}{r} = 1 - \frac{v^2}{c^2}\]

\[\frac{kM}{r} = -\frac{v^2}{c^2}\]

This must agree with Newton’s law of gravity for a small object of mass \(m\) falling from an infinite distance. In this case the kinetic energy of the object is equal to the gravitational potential energy.

\[\frac{1}{2}mv^2 = \frac{GMm}{r}\]

\[v^2 = \frac{2GM}{r}\]

Now substitute the value of \(v^2\).

\[\frac{kM}{r} = -\frac{2GM}{c^2r}\]

\[k = -\frac{2G}{c^2}\]

The Schwarzschild radius is:

\[r_s = \frac{2GM}{c^2} \\ c^2r_s = 2GM\]

This is covered in more detail in Geodesics where the agreement with Newtons Law is proved.

The Christoffel Symbols

Now, the Christoffel symbols are:

\[\begin{align*} \Gamma^{t}_{rt} &= \Gamma^{t}_{tr} = \frac{r_s}{2r(r - r_s)}\\ \Gamma^r_{tt} &= \frac{GM(r - r_s)}{r^3}\\ \Gamma^r_{rr} &= \frac{-r_s}{2r(r - r_s)}\\ \Gamma^r_{\theta\theta} &= -(r - r_s)\\ \Gamma^r_{\phi\phi} &= -\sin^2\theta(r - r_s)\\ \Gamma^\theta_{\theta r} &= \Gamma^\theta_{r\theta} = \frac{1}{r}\\ \Gamma^\theta_{\phi\phi} &= -\sin\theta\cos\theta\\ \Gamma^\phi_{\phi r} &= \Gamma^\phi_{r\phi} = \frac{1}{r}\\ \Gamma^\phi_{\theta\phi} &= \Gamma^\phi_{\phi\theta} = \cot\theta \end{align*}\]

The Schwarzschild Solution

The Schwarzschild outer solution becomes:

\[ds^2 = c^2d \tau^2 = \bigg(1 - \frac{2GM}{c^2r}\bigg)c^2dt^2 - \bigg(1 - \frac{2GM}{c^2r}\bigg)^{-1}dr^2 - r^2d\theta^2 - r^2 \sin^2\theta d\phi^2\]

In terms of the Schwarzschild radius.

\[ ds^2 = c^2d \tau^2 = \bigg(1 - \frac{r_s}{r}\bigg)c^2dt^2 - \bigg(1 - \frac{r_s}{r}\bigg)^{-1}dr^2 - r^2d\theta^2 - r^2 \sin^2\theta d\phi^2\]

The solution has two singularities. The singularity ar \(r = 0\) is not an issue as the solution is only valid in the vacuum of space around a massive object.

The singularity at \(r = r_s\) has interesting consequences. The radius \(r_s\) is normally well inside the object and hence out of scope of the solution. If however, all of the mass is compressed into a volume less than the Schwarzschild radius, the Schwarzschild radius defines the event horizon of a black hole!