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 $T^{\mu\nu}$?

  • $\begingroup$ 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 $\endgroup$
    – paul230_x
    Commented Aug 27, 2021 at 14:00
  • $\begingroup$ I edited the question to clarify that it is total mass-energy that I am interested. Also note that for a dustball, $k>0$ $\endgroup$
    – Travis Lee
    Commented Aug 27, 2021 at 14:15
  • $\begingroup$ 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. $\endgroup$
    – paul230_x
    Commented Aug 27, 2021 at 14:35

1 Answer 1


You're dealing with a spacetime that is asymptotically flat but not stationary. Under these conditions, there are two common measures of mass, the Bondi mass and the ADM mass. These two can differ in general if there is gravitational radiation, but the symmetry of the O-S spacetime guarantees that there can't be any radiation, so they are the same in this case.

The Bondi/ADM mass gives a measure of how much mass the object has according to a distant observer. The explicit expression is given by Wald, p. 293, as the limit of a certain integral over a sphere, as the radius approaches infinity. But the exterior metric of the O-S spacetime is the Schwarzschild metric. Therefore the integral you would be evaluating would be the one for the Schwarzschild metric. The only issue would be verifying that the external and internal metrics match properly, which is exercise 32.4 in Misner, Thorne, and Wheeler.

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.

Normally we expect these measures of mass in an asymptotically flat spacetime to be conserved. However, that is not necessarily the case when there is a timelike singularity, as there is in the O-S metric. Once the singularity forms, the spacetime is no longer globally hyperbolic, so we can't even really define Cauchy surfaces in order to talk about "when" we're measuring the mass.

  • $\begingroup$ FRW metric is not asymptotically flat. See for instance here : en.wikipedia.org/wiki/…. Also this metric is defined within the dust ball which has a finite radius, outside it is Schwarzschild. $\endgroup$
    – paul230_x
    Commented Aug 27, 2021 at 15:33
  • $\begingroup$ @KP99: The O-S spacetime is asymptotically flat. Asymptotic flatness only depends on the behavior of the spacetime at large distances, which is where O-S is Schwarzschild, not FRW. $\endgroup$
    – user312362
    Commented Aug 27, 2021 at 15:35
  • $\begingroup$ 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 $M$? $\endgroup$
    – paul230_x
    Commented Aug 27, 2021 at 16:25

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