After the Field Equations of General Relativity were published in 1915, several exact solutions were discovered
within a few years. Physicists strove to find an exact solution for a rotating star or black hole. This proved
to be very difficult.
In the case of the solutions for stationary charged, or uncharged, stars, the metric
is trace \(g_{ij} = 0\;if\;i \ne j\). This makes integration of the field equations relatively simple.
If the star is rotating then the metric has off-diagonal terms. This makes the integration of the field equations
almost impossible. Like many great developments in recent years, an ansatz has unlocked great discoveries.
It took almost fifty years before Roy Kerr and others found a solution.
The solution is for a body of mass \(M\) with angular momentum \(J\). In 1965 the solution was
extended to a body with charge \(Q\). This is now known as the Kerr-Newman metric.
I am going to provide the Kerr metric in different forms, prove that they satisfy the field equations and
then work back through the derivation.
Notation
The solutions make use of several variables that are used to define the metric. Some of these make useof the
so-called natural constants where \(c = 0\) and \(G = 0\). I will state both forms.
The angular momentum per unit mass \(a\).
If natural units are used the \(c\) is omitted from the denominator.
There is a \(\Delta\) term that is based on the distance \(r\) from the centre of the body.
\[\Delta = r^2 - 2Mr + a^2\]
The terms don’t have the same units, so natural units are being used. Add the units to give agreement.
\[\Delta = r^2 - \frac{2GM}{c^2}r + a^2\]
This contains the Schwarzschild radius \(r_s\).
\[r_s = \frac{2GM}{c^2}\]
\[\Delta = r^2 - r_sr + a^2\]
If the body has charge \(Q\) then.
\[\Delta = r^2 - r_sr + a^2 + Q^2\]
There is also a \(\Sigma\) term.
\[\Sigma = r^2 + a^2\cos^2\theta\]
The Kerr metric has been published in several equivalent forms.
\[ds^2 = -\left(1 - \frac{2Mr}{\Sigma}\right)^2dt^2 + \frac{\Sigma}{\Delta}dr^2
+ \Sigma d\theta^2 + \frac{(r^2 + a^2)^2 - a^2\Delta\sin^2\theta}{\Sigma}\sin^2\theta d\phi^2
- \frac{4aMr}{\Sigma}\sin^2\theta dtd\phi\]
The numerator of the fourth right hand side term can be expanded to eliminate \(\Delta\).
\[(r^2 + a^2)^2 - a^2\Delta\sin^2\theta = r^4 + 2a^2r^2 + a^4 - a^2(r^2 - r_sr + a^2)\sin^2\theta
= (r^2 + a^2)\Sigma + 2r_sra^2\sin^2\theta\]
This is used in some examples.
The metric in full using the +--- notation.
\[ds^2 = \left(1 - \frac{r_sr}{r^2 + a^2\cos^2\theta}\right)^2dt^2 - \frac{r^2 + a^2\cos^2\theta}{r^2 - r_sr + a^2}dr^2
- (r^2 + a^2\cos^2\theta)d\theta^2 - \frac{(r^2 + a^2)^2 - a^2(r^2 - r_sr + a^2)\sin^2\theta}{r^2 + a^2\cos^2\theta}\sin^2\theta d\phi^2
+ \frac{2ar_sr}{r^2 + a^2\cos^2\theta}\sin^2\theta dtd\phi\]
Metric Tensor
\[g_{tt} = \left(1 - \frac{r_sr}{r^2 + a^2\cos^2\theta}\right)^2 =
\left(\frac{r^2 - r_sr - a^2\cos^2\theta}{r^2 + a^2\cos^2\theta}\right)^2\]
\[g_{rr} = -\frac{r^2 + a^2\cos^2\theta}{r^2 - r_sr + a^2}\]
\[g_{\theta\theta} = -(r^2 + a^2\cos^2\theta)\]
\[g_{\phi\phi} = -\frac{(r^2 + a^2)^2 - a^2(r^2 - r_sr + a^2)\sin^2\theta}{r^2 + a^2\cos^2\theta}\sin^2\theta\]
\[g_{t\phi} = g_{\phi t} = \frac{2ar_sr}{r^2 + a^2\cos^2\theta}\sin^2\theta\]
The inverse metric tensor.
\[g^{tt} = \left(1 - \frac{r_sr}{r^2 + a^2\cos^2\theta}\right)^{-2}\]
\[g^{rr} = -\frac{r^2 - r_sr + a^2}{r^2 + a^2\cos^2\theta}\]
\[g^{\theta\theta} = -(r^2 + a^2\cos^2\theta)^{-1}\]
\[g^{\phi\phi} = -\frac{r^2 + a^2\cos^2\theta}{((r^2 + a^2)^2 - a^2(r^2 - r_sr + a^2)\sin^2\theta)\sin^2\theta}\]
\[g^{t\phi} = g^{\phi t} = \frac{r^2 + a^2\cos^2\theta}{2ar_sr\sin^2\theta}\]
Calculate the Christoffel Symbols
\[\Gamma^{\lambda}_{\mu\nu} = \frac{1}{2}g^{\lambda\rho}\left(\frac{\partial g_{\mu\rho}}{\partial\nu}
+ \frac{\partial g_{\nu\rho}}{\partial\mu}
- \frac{\partial g_{\mu\nu}}{\partial\rho}\right)\]
Calculate Partial Derivatives
We need all of the non-zero partial derivatives of the metric tensor. As none of the terms are time
dependent, all \(g_{\mu\nu,t} = 0\).
\[\frac{\partial g_{tt}}{\partial r} = -2\left(\frac{(2r - r_s)(r^2 + a^2\cos^2\theta) - 2r(r^2 - r_sr + a^2\cos^2\theta)}{(r^2 + a^2\cos^2\theta)^2}\right)
\left(\frac{r^2 - r_sr + a^2\cos^2\theta}{r^2 + a^2\cos^2\theta}\right)
= -2r_s\frac{(r^2 - a^2\cos^2\theta)(r^2 - r_sr + a^2\cos^2\theta)}{(r^2 + a^2\cos^2\theta)^3}\]
\[\frac{\partial g_{tt}}{\partial\theta} =
4a^2r_sr\sin\theta\cos\theta\frac{r^2 - r_sr + a^2\cos^2\theta}{(r^2 + a^2\cos^2\theta)^3}\]
\[\frac{\partial g_{rr}}{\partial r} =
\frac{2r(r^2 - r_sr + a^2) - (2r - r_s)(r^2 - a^2\cos^2\theta)}{(r^2 - r_sr + a^2)^2} =
\frac{a^2(2r - r_s)(1 + \cos^2\theta)}{(r^2 - r_sr + a^2)^2}\]
\[\frac{\partial g_{rr}}{\partial\theta} = -\frac{2a^2\sin\theta\cos\theta}{r^2 - r_sr + a^2}\]
\[\frac{\partial g_{\theta\theta}}{\partial r} = 2r\]
\[\frac{\partial g_{\theta\theta}}{\partial\theta} = 2a^2\sin\theta\cos\theta\]
RETURN TO THE OTHERS LATER
t symbols have the cross term and phi
Calculate the \(r\) Symbols
In this cse \(\lambda = r\) and \(\rho = r\).
\[\Gamma^{r}_{\mu\nu} = \frac{1}{2}g^{rr}\left(\frac{\partial g_{\mu r}}{\partial\nu}
+ \frac{\partial g_{\nu r}}{\partial\mu}
- \frac{\partial g_{\mu\nu}}{\partial r}\right)\]
There are two symbols \(\mu = \nu = r\) and \(\mu = \nu = \theta\).
\[\Gamma^{r}_{rr} = \frac{1}{2}g^{rr}\frac{\partial g_{rr}}{\partial r} =
-\frac{1}{2}\frac{r^2 - r_sr + a^2}{r^2 + a^2\cos^2\theta}\frac{a^2(2r - r_s)(1 + \cos^2\theta)}{(r^2 - r_sr + a^2)^2} =
-\frac{1}{2}\frac{a^2(2r - r_s)(1 + \cos^2\theta)}{(r^2 + a^2\cos^2\theta)(r^2 - r_sr + a^2)}\]
\[\Gamma^{r}_{\theta\theta} = -\frac{1}{2}g^{rr}\frac{\partial g_{\theta\theta}}{\partial r} =
r\frac{r^2 - r_sr + a^2}{r^2 + a^2\cos^2\theta}\]
Calculate the Ricci Tensor
Start with \(R_{rr}\) find which gamma and gamma' are required.
