Electrovacuum solution

Remarks

Example: the Reissner–Nordström metric with charge $Q \ne 0$ and the radial field $F = \frac{Q}{r^2}\, dr \wedge dt$ is an electrovacuum solution. Clause 1 holds because $F = d(-Q\, dt / r)$ is exact, and $\star F$ is a multiple of $Q \sin\theta \, d\theta \wedge d\phi$, which is closed. Clause 2 is a direct computation of the Ricci tensor.

Non-example: the interior Schwarzschild metric is not an electrovacuum solution. Its source is a perfect fluid with positive pressure, whose stress–energy tensor has nonzero trace, while the trace $T^a{}_a$ of the electromagnetic stress–energy tensor is always $0$.

Equivalent form: since $T_{ab}$ is trace-free in dimension four, the scalar curvature vanishes, and clause 2 becomes $$R_{ab} = 2\left(F_{ac} F_b{}^{c} - \frac{1}{4} g_{ab} F_{cd} F^{cd}\right),$$ where $R_{ab}$ is the Ricci curvature of $g$. If $F = 0$ were allowed, the definition would reduce to a vacuum solution.