I have been reading up a paper on gravitational collapse, where a particular equation $$ 4\pi \rho R^2 b = 1 $$ means the assumption of conservation of rest mass. Here the spacetime interval is described by this line integral $$ ds^2 = a^2(\mu,t)c^2dt^2 - b^2(\mu,t)c^2d\mu^2 - R^2(\mu,t)c^2d\Omega^2$$ and $T^{1}_{1} = T^{2}_{2} = T^{3}_{3} = P$, $T^{0}_{0} = -\rho(c^2 + \epsilon)$ I'm not sure how the equation $ 4\pi \rho R^2 b = 1 $ came to be, can anyone explain?

  • $\begingroup$ On the first page it says: "We neglect pair production and annihilation,and the interaction of the fluid with external fields so that rest mass is conserved." Can you explain why this does not answer your question? $\endgroup$
    – TimRias
    Commented Feb 13, 2020 at 8:03
  • $\begingroup$ I meant how can I arrive at the same result, calculation-wise. $\endgroup$ Commented Feb 14, 2020 at 23:59

1 Answer 1


Since the paper already mentions that they are using "$\mu$" as the rest mass between the point labelled and the center, which also appears in the context of metric tensor, the whole expression becomes quite trivial to evaulate $$\frac{d m}{dt} = \frac{d \mu}{dt} = 0 \implies dm = d\mu$$

Rewriting the expression in terms of $T_{00}=\rho$ \begin{equation}\require{cancel} \rho dV= d \mu\\ \\\ \\\ \rho\sqrt{b^2R^{4}\sin^2\theta} \cancel{d\mu} d\theta d\phi = \cancel{d\mu} \implies \rho (4\pi R^2 b) = 1 \end{equation}


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