Kerr–Newman metric

Example

Use geometrized Gaussian units. Fix a mass $M > 0$, a rotation parameter $a$ and a charge $Q$, and set $\Sigma = r^2 + a^2\cos^2\theta$ and $\Delta = r^2 - 2Mr + a^2 + Q^2$. The Kerr–Newman metric is the metric $g$ written in Boyer–Lindquist coordinates with these $\Sigma$ and $\Delta$, together with the electromagnetic potential $$A = -\frac{Q r}{\Sigma}\left(dt - a\sin^2\theta\, d\phi\right), \qquad F = dA.$$ The pair $(g, F)$ is an electrovacuum solution: $F$ solves the source-free Maxwell equations and $g$ solves the Einstein equations with the electromagnetic stress–energy of $F$.

Checks of the special cases follow. For $Q = 0$ the field vanishes and $g$ is the Kerr metric. For $a = 0$, $\Sigma = r^2$, the cross term vanishes, $A = -\frac{Q}{r} dt$, and $(g, F)$ is the Reissner–Nordström metric. For $a = Q = 0$ it is the Schwarzschild metric. Since $dF = d(dA) = 0$, half of Maxwell's equations hold automatically.

The parameters have a physical meaning: $M$ is the mass, $J = aM$ the angular momentum, $Q$ the charge, and the magnetic dipole moment is $Qa$. If $a^2 + Q^2 \le M^2$, the zeros of $\Delta$ are $r_\pm = M \pm \sqrt{M^2 - a^2 - Q^2}$, and $r_+$ is the event horizon. If $a^2 + Q^2 > M^2$ there is no horizon.