Reissner–Nordström metric

Example

Use geometrized Gaussian units, where the gravitational constant, the speed of light and the Coulomb constant equal $1$. Fix a mass $M > 0$ and a charge $Q \in \R$, and set $$f(r) = 1 - \frac{2M}{r} + \frac{Q^2}{r^2}.$$ On the region where $f(r) > 0$, with coordinates $(t, r, \theta, \phi)$, the Reissner–Nordström metric and its field are $$g = -f(r)\, dt^2 + f(r)^{-1} dr^2 + r^2\left(d\theta^2 + \sin^2\theta\, d\phi^2\right), \qquad F = \frac{Q}{r^2}\, dr \wedge dt.$$ The pair $(g, F)$ is an electrovacuum solution.

Check of Maxwell's equations: $F = dA$ with $A = -\frac{Q}{r}\, dt$, so $dF = 0$. The only nonzero components are $F_{rt} = -F_{tr} = Q/r^2$. Since $g^{tt} g^{rr} = -1$, one gets $F^{tr} = -F_{tr} = Q/r^2$, and with $\sqrt{-\det g} = r^2 \sin\theta$, $$\partial_r\left(\sqrt{-\det g}\, F^{tr}\right) = \partial_r(Q \sin\theta) = 0.$$ So the source-free equation $\nabla_a F^{ab} = 0$ holds, and the field is a Coulomb field of charge $Q$. The Einstein equations follow from a computation of the Ricci curvature, which gives $R_{ab} = 2(F_{ac}F_b{}^c - \frac14 g_{ab} F_{cd}F^{cd})$.

For $Q = 0$ the metric is the Schwarzschild metric. If $Q^2 < M^2$, then $f$ has two zeros $r_\pm = M \pm \sqrt{M^2 - Q^2}$, an outer event horizon $r_+$ and an inner Cauchy horizon $r_-$. If $Q^2 = M^2$ the two coincide (the extremal case), and if $Q^2 > M^2$ there is no horizon and the singularity at $r = 0$ is naked.