Kerr metric

Example

Use units where the gravitational constant and the speed of light equal $1$. Fix a mass $M > 0$ and a rotation parameter $a \in \R$, and set $$\Sigma = r^2 + a^2 \cos^2\theta, \qquad \Delta = r^2 - 2Mr + a^2.$$ In Boyer–Lindquist coordinates $(t, r, \theta, \phi)$, on the region where $\Delta > 0$ and $\Sigma > 0$, the Kerr metric is $$g = -\left(1 - \frac{2Mr}{\Sigma}\right) dt^2 - \frac{4 M a r \sin^2\theta}{\Sigma}\, dt\, d\phi + \frac{\Sigma}{\Delta}\, dr^2 + \Sigma\, d\theta^2 + \left(r^2 + a^2 + \frac{2 M a^2 r \sin^2\theta}{\Sigma}\right)\sin^2\theta\, d\phi^2.$$ Its Ricci curvature vanishes, so it is a vacuum solution of the Einstein equations.

Checks: the coefficients do not depend on $t$ or $\phi$, so $\partial_t$ and $\partial_\phi$ are Killing vector fields: the metric is stationary and axisymmetric. For $a = 0$, $\Sigma = r^2$ and $\Delta = r^2 - 2Mr$, the cross term vanishes and $g$ becomes the Schwarzschild metric. For $M = 0$ it is flat Minkowski space in oblate spheroidal coordinates. The mass is $M$ and the angular momentum is $J = aM$, read off from the behaviour of $g$ for large $r$.

If $a^2 \le M^2$, the zeros of $\Delta$ are $r_\pm = M \pm \sqrt{M^2 - a^2}$. The surface $r = r_+$ is the event horizon and $r = r_-$ is an inner Cauchy horizon. Between $r_+$ and the surface $r = M + \sqrt{M^2 - a^2\cos^2\theta}$, where $g_{tt}$ changes sign, lies the ergosphere. The curvature blows up where $\Sigma = 0$, that is, on the ring $r = 0$, $\theta = \pi/2$. If $a^2 > M^2$ there is no horizon.