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.
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
Using the comma notation, where a comma subscript means a partial derivative of that variable, the Christoffel symbols are:
\[\Gamma^{\lambda}_{\mu\nu} = \frac{1}{2}g^{\lambda\rho}(g_{\mu\rho,\nu} + g_{\nu\rho,\mu} - g_{\mu\nu,\rho})\]
There are four sets of Christoffel symbols, one set for each of \(t, r, \theta, \phi\).
For the \(t\) component, \(\lambda = t\), and for all non-zero symbols \(\rho = t\).
\[\Gamma^{t}_{\mu\nu} = \frac{1}{2}g^{tt}(g_{\mu t,\nu} + g_{\nu t,\mu} - g_{\mu\nu,t})\]
The term \(g_{\mu\nu,t} = 0\) as nothing is time dependent. Substitute the value of \(g^{tt}\), gives:
\[\Gamma^{t}_{\mu\nu} = \frac{1}{2A}(g_{\mu t,\nu} + g_{\nu t,\mu})\]
\[\Gamma^{t}_{rt} = \Gamma^{t}_{tr} = \frac{A'}{2A}\]
For the \(r\) component, \(\lambda = r\), and for all non-zero symbols \(\rho = r\).
\[\Gamma^{r}_{\mu\nu} = \frac{1}{2}g^{rr}(g_{\mu r,\nu} + g_{\nu r,\mu} - g_{\mu\nu,r})\]
\[\begin{align*}
\Gamma^r_{tt} &= -\frac{1}{2}g^{rr}g_{tt,r} = \frac{A'}{2B}\\
\Gamma^r_{rr} &= \frac{1}{2}g^{rr}g_{rr,r} = \frac{B'}{2B}\\
\Gamma^r_{\theta\theta} &= -\frac{1}{2}g^{rr}g_{\theta\theta,r} = -\frac{r}{B}\\
\Gamma^r_{\phi\phi} &= -\frac{1}{2}g^{rr}g_{\phi\phi,r} = -\frac{r\sin^2\theta}{B}
\end{align*}\]
For the \(\theta\) component, \(\lambda = \theta\), and for all non-zero symbols \(\rho = \theta\).
\[\Gamma^{\theta}_{\mu\nu} = \frac{1}{2}g^{\theta\theta}(g_{\mu\theta,\nu} + g_{\nu\theta,\mu} - g_{\mu\nu,\theta})\]
This gives three symbols:
\[\begin{align*}
\Gamma^\theta_{\theta r} &= \Gamma^\theta_{r\theta} = \frac{1}{2}g^{\theta\theta}g_{\theta\theta, r} = \frac{1}{2r^2}2r = \frac{1}{r}\\
\Gamma^\theta_{\phi\phi} &= -\frac{1}{2}g^{\theta\theta}g_{\phi\phi,\theta} = \frac{1}{2r^2}2r^2\sin\theta\cos\theta = -\sin\theta\cos\theta
\end{align*}\]
For the \(\phi\) component, \(\lambda = \phi\), and for all non-zero symbols \(\rho = \phi\).
\[\Gamma^{\phi}_{\mu\nu} = \frac{1}{2}g^{\phi\phi}(g_{\mu\phi,\nu} + g_{\nu\phi,\mu} - g_{\mu\nu,\phi})\]
\[\begin{align*}
\Gamma^\phi_{\phi r} &= \Gamma^\phi_{r\phi} = \frac{1}{2}g^{\phi\phi}g_{\phi\phi,r} = \frac{1}{2r^2\sin^2\theta}2r\sin^2\theta = \frac{1}{r}\\
\Gamma^\phi_{\theta\phi} &= \Gamma^\phi_{\phi\theta} = \frac{1}{2}g^{\phi\phi}g_{\phi\phi,\theta} = \frac{1}{2r^2\sin^2\theta}2r^2\sin\theta\cos\theta = \cot\theta
\end{align*}\]
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 emements:
\[R_{\mu\mu} = \Gamma^\lambda_{\mu\mu,\lambda} - \Gamma^\lambda_{\lambda\mu,\mu} + \Gamma^\lambda_{\lambda\eta}\Gamma^\eta_{\mu\mu} - \Gamma^\lambda_{\mu\eta}\Gamma^\eta_{\lambda\mu}\]
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.
\[\begin{align*}
R_{\phi\phi} &= \Gamma^\lambda_{\phi\phi,\lambda} - \Gamma^\lambda_{\phi\lambda,\phi} + \Gamma^\lambda_{\lambda\eta}\Gamma^\eta_{\phi\phi} - \Gamma^\lambda_{\phi\eta}\Gamma^\eta_{\lambda\phi}\\
&= \Gamma^r_{\phi\phi,r} + \Gamma^\theta_{\phi\phi,\theta} +
\bigg(\Gamma^t_{tr} + \Gamma^r_{rr} + \Gamma^\theta_{\theta r} + \Gamma^\phi_{\phi r}\bigg)\Gamma^r_{\phi\phi} - 2\Gamma^\phi_{\phi r}\Gamma^r_{\phi\phi} - \Gamma^\phi_{\phi\theta}\Gamma^\theta_{\phi\phi}\\
&= - \frac{\sin^2\theta}{B} + \frac{r\sin^2\theta B'}{B^2} - \cos^2\theta + \sin^2\theta -\frac{r\sin^2\theta}{B}\bigg(\frac{A'}{2A} + \frac{B'}{2B} + \frac{1}{r} + \frac{1}{r}\bigg) + 2\frac{1}{r}\frac{r\sin^2\theta}{B} + \cos^2\theta\\
&= \sin^2\theta\bigg(-\frac{1}{B} + \frac{rB'}{B^2} - \frac{r}{2B}\bigg(\frac{A'}{A} - \frac{B'}{B}\bigg) + 1\bigg)&&\\
&= \sin^2\theta R_{\theta\theta}
\end{align*}\]
\[\begin{align*}
R_{tt} &= \frac{A''}{2B} - \frac{A'}{4B}\bigg(\frac{A'}{A} + \frac{B'}{B}\bigg) + \frac{A'}{rB}\\
R_{rr} &= -\frac{A''}{2A} + \frac{A'}{4A}\bigg(\frac{A'}{A} + \frac{B'}{B}\bigg) + \frac{B'}{rB}\\
R_{\theta\theta} &= -\frac{1}{B} - \frac{r}{2B}\bigg(\frac{A'}{A} - \frac{B'}{B}\bigg) + 1\\
R_{\phi\phi} &= \sin^2\theta R_{\theta\theta}
\end{align*}\]
Solve the Equations
The first two field equations reduce to:
\[\begin{align*}
1 - \frac{rB'}{B} - B &= 0\\
1 + \frac{rA'}{A} - B &= 0
\end{align*}\]
\[\frac{A'}{A} +\frac{B'}{B} = \frac{BA' + AB'}{AB} = \frac{(AB)'}{AB} = 0\]
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:
\[\frac{1}{B} - \frac{rB'}{B^2} = \frac{d}{dr}\bigg(\frac{r}{B}\bigg) = 1\]
\[\frac{r}{B} = r + k_2\]
Where \(k_2\) is a constant of integration.
\[B = \bigg(1 + \frac{k_2}{r}\bigg)^{-1}\]
\[A = c^2\bigg(1 + \frac{k_2}{r}\bigg)\]
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.
\[\frac{1}{2}mv^2 = \frac{GMm}{r}\]
Now substitute the value of \(v^2\).
\[\frac{kM}{r} = -\frac{2GM}{c^2r}\]
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 Schwarzschild Solution
The Schwarzschild outer solution becomes:
\[\begin{equation}
ds^2 = c^2d \tau^2 = \bigg(1 - \frac{2GM}{c^2r}\bigg)c^2dt^2 - \bigg(1 - \frac{2GM}{c^2r}\bigg)^{-1}dr^2 - r^2d\theta^2 -
r^2 \sin^2\theta d\phi^2
\end{equation}
\begin{equation}
ds^2 = c^2d \tau^2 = \bigg(1 - \frac{r_s}{r}\bigg)c^2dt^2 - \bigg(1 - \frac{r_s}{r}\bigg)^{-1}dr^2 - r^2d\theta^2 -
r^2 \sin^2\theta d\phi^2
\end{equation}\]
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!
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