What is the assumption of conservation of rest mass in gravitational collapse?

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?

• 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? Commented Feb 13, 2020 at 8:03
• I meant how can I arrive at the same result, calculation-wise. Commented Feb 14, 2020 at 23:59

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 $$T_{00}=\rho$$ $$$$\require{cancel} \rho dV= d \mu\\ \\\ \\\ \rho\sqrt{b^2R^{4}\sin^2\theta} \cancel{d\mu} d\theta d\phi = \cancel{d\mu} \implies \rho (4\pi R^2 b) = 1$$$$