Vaidya metric

Example

Use units where the gravitational constant and the speed of light equal $1$. Let $m \colon \R \to \R$ be a smooth function. In coordinates $(v, r, \theta, \phi)$ with $r > 0$, the ingoing Vaidya metric is $$g = -\left(1 - \frac{2m(v)}{r}\right) dv^2 + 2\, dv\, dr + r^2\left(d\theta^2 + \sin^2\theta\, d\phi^2\right).$$ For constant $m$ this is the Schwarzschild metric in ingoing Eddington–Finkelstein coordinates.

Check: a computation of the Ricci curvature gives one nonzero component, $$R_{vv} = \frac{2\, m'(v)}{r^2},$$ and the scalar curvature is $0$. So the Einstein equations $G_{ab} = 8\pi T_{ab}$ hold with $$T_{ab} = \frac{m'(v)}{4\pi r^2}\, \ell_a \ell_b, \qquad \ell = dv.$$ The covector $\ell$ is null, so this is null dust: radiation falling inwards at the speed of light. The null energy condition holds exactly when $m'(v) \ge 0$, that is, when the mass grows as radiation falls in. The outgoing version, with $u$ in place of $v$ and $-2\, du\, dr$, describes a star that loses mass by emitting radiation.