# Oppenheimer-Snyder metric: how to integrate it to recover the total mass-energy?

Weinberg recaps dustball collapse in comoving coordinates, where the energy-momentum tensor is

$$T^{\mu\nu}(t) = \rho(t)\left(\begin{array}{l} 1&0&0&0\\ 0&0&0&0\\ 0&0&0&0\\ 0&0&0&0 \end{array}\right)$$

from which he derives the metric

$$d\tau^2 = dt^2-Q^2(t)\left(\frac{1}{1-kr^2}dr^2 + r^2d\Omega^2\right)$$

$$r$$ ranges from $$0$$ to $$a$$ throughout time, as we have comoving coordinates. The initial, uniform proper density is $$\rho_0$$, and increases with time: $$\rho(t)=\rho_0/Q^3(t)$$. $$Q(t)$$ shrinks from $$1$$ to $$0$$.

Now, with spherical symmetry, we can integrate the mass-energy of shells to get the total mass-energy.

It seems the integral should be

$$m(t)=\int_0^a\frac{\rho_0}{Q^3}4\pi (Qr)^2\sqrt{\frac{Q^2}{1-kr^2}}\,dr=\frac{\rho_0}{Q^3}V_\textrm{proper}$$

When I do this, then m(t) is constant in time, but the resulting $$m$$ is somewhat larger than the mass-energy $$M$$ that an external observer would measure, the $$M$$ in the Schwarzschild metric. When the ratio $$r_0/r_S$$ (initial radius over Schwarzschild radius) is large, the ratio $$m/M$$ approaches 1 from above.

How to make sense of this?

How to get total mass-energy $$M$$ from some integral of either the metric or $$T^{\mu\nu}$$?

• The notion of mass in GR is not well understood. There are quite a number of definitions as you can see here: en.wikipedia.org/wiki/Mass_in_general_relativity For k=0 FRW metric, you can use Penrose's Quasi Local Mass definition, it will give consistent result Commented Aug 27, 2021 at 14:00
• I edited the question to clarify that it is total mass-energy that I am interested. Also note that for a dustball, $k>0$ Commented Aug 27, 2021 at 14:15
• Hmm understood. So there is no simple explanation for why $m(t)\neq M$, deep down it is arises from the fact that gravitational energy is non-local in GR ... I believe someone else can clarify this point. I will check which definition for mass should be applicable for this system. Commented Aug 27, 2021 at 14:35

The result should be that the ADM/Bondi mass is the same as the parameter $$M$$ appearing in the O-S solution. The discrepancy you report is because you're not using a valid method for finding the mass. Conceptually, your method doesn't work because you're just trying to integrate the rest mass of all the matter fields, but there is also mass-energy stored in the gravitational field itself. Note that if you applied your method to the Schwarzschild metric, which is a vacuum solution, you'd get zero.
• Yeah. So are there any better definitions for mass which can be evaluated inside the dust ball (i.e. not going into asymptotic picture) which can account for the discrepancy b/w $m(t)$ and $M$? Commented Aug 27, 2021 at 16:25