# Kerr Black Hole

## Kerr

The spacetime metric is, in Boyer-Lindquist coordinates,

$ds^2 = -\frac{\Lambda^2}{\rho^2} (dt - a\, \sin^2\theta \, d\phi)^2 + \frac{\sin^2\theta}{\rho^2} [ (r^2 + a^2) d\phi - a\, dt]^2 + \frac{\rho^2}{\Lambda^2} dr^2 + \rho^2 d\theta^2$

where

$\begin{array}{rcl} \Lambda^2 & = & r^2 - 2Mr + a^2 \\ \rho^2 & \equiv & r^2 + a^2\cos^2\theta \\ a & \equiv & J/M \end{array}$

When split into 3+1 (space + time), the nonzero components of the extrinsic curvature are given by

$K_{r\phi} = \frac{(r \partial_r \Omega - \Omega) M a \sin^2 \theta}{\Lambda \sqrt{\rho^6 \Omega}}$

and

$K_{\theta \phi} = \frac{(\partial_{\theta} \Omega) M a r \sin^2 \theta}{\Lambda \sqrt{\rho^6 \Omega}}$

where $\Omega \equiv (r^2 + a^2) \, \rho^2 + 2 M a^2 r \sin^2 {\theta}$.

In horizon-penetrating (Kerr-Schild) coordinates, the four-metric is:

$ds^2 = (\eta_{\mu \nu} + 2 H \ell_\mu \ell_\nu) dx^\mu dx^\nu,$

where $H$ is a scalar function, and $\ell_\mu$ a null (in Minkowski space) vector field. We can write these for an arbitrarily directed spin vector:

$\ell_\mu = \left( 1, \frac{r^2 \vec{x} - r \vec{a} \times \vec{x} + (\vec{a}\cdot\vec{x})\vec{a}}{r (r^2 + a^2)} \right)$

where

$H = \frac{M r^3}{r^4 + (\vec{a}\cdot\vec{x})^2}$

## Kerr-Newman

On top of mass and spin a black hole can be characterized by a third parameter for charge. The Kerr line element given by

$ds^2 = -\frac{\Lambda^2}{\rho^2} (dt - a\, \sin^2\theta \, d\phi)^2 + \frac{\sin^2\theta}{\rho^2} [ (r^2 + a^2) d\phi - a\, dt]^2 + \frac{\rho^2}{\Lambda^2} dr^2 + \rho^2 d\theta^2$

Becomes the Kerr-Newman spacetime for the inclusion of charge given the modification of:

$\Lambda^2 \equiv r^2 - 2Mr + e^2 + a^2$

There are three important mathematical surfaces for this line element, the static limit and the inner and outer event horizons. The static limit is the outermost place something can be outside the outer horizon with a zero angular velocity. It is

$r_s = M+\sqrt{M^2 - a^2 cos^2 \theta -e^2}$

The event horizons are coordinate singularities in the metric where $\Lambda = 0$. The outer event horizon is at

$r_{+} = M+\sqrt{M^2 - a^2 -e^2}$

and the inner horizon is at

$r_{-} = M-\sqrt{M^2 - a^2 -e^2}$

An external observer can never see an event at which something crosses into the outer horizon. A remote observer reckoning with these coordinates will reckon that it takes an infinite time for something infalling to reach the outer horizon even though it takes a finite proper time till the event according to what fell in.

## Kerr-Newman Equatorial Geodesic Motion

The exact equations of equatorial geodesic motion for a neutral test mass in a charged and rotating black hole's spacetime are

$\frac{dt}{d\tau}=\frac{\gamma \left(r^2 +a^2 +a^2 \frac{2M}{r}-a^2 \frac{e^2}{r^2}\right)-al_{z} \left(\frac{2M}{r}-\frac{e^2}{r^2}\right)}{r^2 -2Mr + a^2 +e^2}$
$\frac{d\phi}{d\tau}=\frac{l_z \left(1-\frac{2M}{r}+\frac{e^2}{r^2}\right)+\gamma a\left(\frac{2M}{r}-\frac{e^2}{r^2}\right)}{r^2 -2Mr + a^2 +e^2}$
$\frac{1}{2}\left(\frac{dr}{d\tau}\right)^2 +V_{eff} =0$
$V_{eff} = -\frac{M}{r}+\frac{e^2}{2r^2}+\frac{1}{2}\frac{l_z ^2}{r^2}+\frac{1}{2}\left(1-\gamma ^2\right)\left(1+\frac{a^2}{r^2}\right)-\left(\frac{M}{r^3}-\frac{e^2}{2r^4}\right)\left(l_z -a\gamma \right)^2$

where $\gamma$ is the conserved energy parameter, the energy per mass of the test mass and $l_z$ is the conserved angular momentum per mass for the test mass.

## Kerr-Newman Polar Geodesic Motion

The exact equations of polar geodesic motion for a neutral test mass in a charged and rotating black hole's spacetime are

$\frac{dt}{d\tau }=\gamma \left(\frac{a^2 +r^2}{a^2 +e^2 +r^2 - 2Mr}\right)$
$\frac{1}{2}\left(\frac{dr}{d\tau}\right)^2 + \frac{-\frac{M}{r}+\frac{e^2}{2r^2}}{1+\frac{a^2}{r^2}}= \frac{\gamma ^2 -1}{2}$

where $\gamma$ is the conserved energy parameter, the energy per mass of the test mass.

## Reissner-Nordstrom

In the case that the rotation parameter is $a=0$, the Kerr-Newman spacetime reduces to the exact solution to Einstein's field equations for a charged black hole:

$ds^2 = -\left(1-\frac{2M}{r}+\frac{e^2}{r^2}\right)dt^2 + \frac{dr^2}{\left(1-\frac{2M}{r}+\frac{e^2}{r^2}\right)} + r^2 d\theta ^2 + r^2 sin^2 \theta d\phi ^2$

## Reissner-Nordstrom Geodesic Motion

The equations of geodesic motion for a neutral test mass in in a Reissner-Nordstrom spacetime with respect to time according to the test mass $\tau$ yield

$\frac{dt}{d\tau} =\frac{\gamma}{\left(1-\frac{2M}{r}+\frac{e^2}{r^2}\right)}$

where $\gamma$ is a constant of the motion called the energy parameter, the energy per mass for the test mass.

$\frac{d\phi}{d\tau} =\frac{l_z}{r^2 sin^2 \theta}$

where $l_z$ is the conserved angular momentum per mass for the test mass.

$\frac{d}{d\tau}\left(r^2 \frac{d\theta}{d\tau}\right) = r^2 sin\theta cos\theta \left(\frac{d\phi}{d\tau}\right)^2$

and finally

$\frac{\gamma^2 - 1}{2} =\frac{1}{2}\left(\frac{dr}{d\tau}\right)^2 -\frac{M}{r}+\frac{e^2}{2 r^2}+\frac{1}{2}\left(\frac{l_z ^2}{r^2 sin^2 \theta}+r^2 \left(\frac{d\theta}{d\tau}\right)^2 \right)\left(1-\frac{2M}{r}+\frac{e^2}{r^2}\right)$

For radial motion this reduces to

$\frac{dt}{d\tau} =\frac{\gamma}{\left(1-\frac{2M}{r}+\frac{e^2}{r^2}\right)}$
$\frac{\gamma^2 - 1}{2} =\frac{1}{2}\left(\frac{dr}{d\tau}\right)^2 -\frac{M}{r}+\frac{e^2}{2 r^2}$

## Reissner-Nordstrom Isotropic Coordinates

The Reissner-Norstrom solution can be expressed in isotropic coordinates under the transformation [1]

$r=\rho \left(\left(1+\frac{M}{2\rho }\right)^2 - \left(\frac{e}{2\rho }\right)^2 \right)$
$ds^2 = -\left(\frac{1-\left(\frac{M}{2\rho }\right)^2 +\left(\frac{e}{2\rho }\right)^2 }{1+\frac{M}{\rho}+\left(\frac{M}{2\rho}\right)^2 - \left(\frac{e}{2\rho}\right)^2 }\right)^2 dt^2 +\left(1+\frac{M}{\rho}+\left(\frac{M}{2\rho}\right)^2 - \left(\frac{e}{2\rho}\right)^2 \right)^2 \left(d\rho ^2 + \rho ^2 d\theta ^2 + \rho ^2 sin^2 \theta d\phi ^2 \right)$

## References

[1]*Relativity - Chapter 7 equation 7.1.1