# Metric for accreting spherically symmetric spacetime?

I know this may be an over-simplification of the system but can we treat the metric below as something that represents accretion?

$$\mathrm ds^2 = - \left(1-\frac{2m(t)}{r}\right) \mathrm dt^2 + \left(1-\frac{2m(t)}{r}\right)^{-1}\mathrm dr^2 +r^2 \mathrm d\Omega^2$$

where $$m(t)$$ is taken to increase with $$t$$. Correct me please if I'm wrong, but I interpret this as a Schwarzschild spacetime where the mass of the central object is increasing (for example through accretion).

• That's the Vaidya metric. – Slereah Apr 9 at 18:29
• For collapse of a cloud of material particles, see the Oppenheimer-Snyder model. However, the interpretation of this is complicated and controversial. – Ben Crowell Apr 9 at 23:04

To see this, take the function $$m(t)$$ to have the form $$m(t)=M+\delta M \, H (t)$$, where $$H(t)$$ is a step-like function with the properties $$H(t)=\begin{cases} 0,& t < -\epsilon \\ 1, & t> + \epsilon \end{cases},$$ for some small $$\epsilon$$. The metric then is Schwarzschild metric with the mass $$M$$ for $$t<-\epsilon$$ and Schwarzschild metric with the mass $$M+\delta M$$ for $$t>\epsilon$$, and so it satisfies vacuum Einstein field equations for $$|t|>\epsilon$$ while around $$t=0$$ Einstein tensor is non-zero. But we could not interpret this region of spacetime as having any sort of realistic matter, since the slice $$t=0$$ is spacelike (for $$r > 2 M$$). In other words, the mass in the OP's metric propagates with superluminal velocities.
A more realistic yet quite simple model of an accreting spacetime is the (ingoing) Vaidya metric that corresponds to a null dust matter: $$ds^{2}=-{\Big (}1-{\frac {2M(v)}{r}}{\Big )}dv^{2}+2dvdr+r^{2}(d\theta ^{2}+\sin ^{2}\theta \,d\phi ^{2}),$$ which is obtained from the Schwarzshild solution written in Eddington–Finkelstein coordinates, by replacing constant $$M$$ with a function $$M(v)$$. The difference from OP's metric is that here the matter propagates at precisely the speed of light along null ingoing hypersurface $$v=\mathrm{const}$$.