Jonathan Scott said:
The result which I'd previously understood was that the time dilation factor effectively resulted in exactly double the binding energy decrease (for reasons previously mentioned) but adding in the pressure compensated for one of those, giving the local mass minus the binding energy, matching the Newtonian view.
Let's look at this for the Komar mass integral I posted earlier. In order to obtain expressions that we can actually solve analytically, I'm going to assume (highly unrealistically) that the density ##\rho## is constant, independent of ##r##. For this case, there are known closed-form solutions for ##g_{tt}##, ##g_{rr}##, and ##p## as functions of ##r##, which are (note that ##R## is the radial coordinate of the surface of the object):
$$
g_{tt} = \frac{1}{4} \left( 3 \sqrt{1 - \frac{8}{3} \pi \rho R^2} - \sqrt{1 - \frac{8}{3} \pi \rho r^2} \right)^2
$$
$$
g_{rr} = \frac{1}{1 - \frac{8}{3} \pi \rho r^2}
$$
$$
p = \rho \frac{\sqrt{1 - \frac{8}{3} \pi \rho r^2} - \sqrt{1 - \frac{8}{3} \pi \rho R^2}}{3 \sqrt{1 - \frac{8}{3} \pi \rho R^2} - \sqrt{1 - \frac{8}{3} \pi \rho r^2}}
$$
Notice that the denominator of ##p## is equal to ##2 \sqrt{g_{tt}}## and the first term in the numerator (and the second term in the denominator) is equal to ##1 / \sqrt{g_{rr}}##; so we can write the Komar mass integral as (note that we have expanded out the numerator and canceled terms):
$$
M = \int_0^R 4 \pi r^2 dr \sqrt{g_{rr} g_{tt}} \rho \frac{2}{2 \sqrt{g_{rr} g_{tt}}} = \int_0^R 4 \pi r^2 \rho dr = \frac{4}{3} \pi \rho R^3
$$
Now, in order to assess the gravitational binding energy, we need to know what to compare ##M## to. What we want is what the total mass of the matter forming the object would be if it were all moved out to rest at infinity and widely separated. If you think about it, it should be evident that this total mass is given by the following integral:
$$
M_0 = \int_0^R 4 \pi r^2 dr \sqrt{g_{rr}} \rho = \int_0^R 4 \pi \rho \frac{r^2}{\sqrt{1 - \frac{8}{3} \pi \rho r^2}}
$$
This doesn't have a handy closed-form solution as it stands, but we can see that it is larger than ##M##, and if we go to the weak field limit where ##\frac{8}{3} \pi \rho r^2 << 1##, so that we can expand the square root in the denominator, we obtain
$$
M_0 \approx \int_0^R 4 \pi r^2 \rho dr \left( 1 + \frac{4}{3} \pi \rho r^2 \right) = \frac{4}{3} \pi \rho R^3 + \frac{16}{15} \pi^2 \rho^2 R^5 = M \left( 1 + \frac{3}{5} \frac{M}{R} \right)
$$
So the binding energy in this limit is approximately ##3/5## of the "naive" Newtonian value of ##M / R## times the rest mass.
Now we need to assess the impact of the pressure term. The obvious way to do that is to evaluate the integral
$$
M_1 = \int_0^R 4 \pi r^2 dr \sqrt{g_{rr} g_{tt}} \rho = \int_0^R 4 \pi r^2 dr \rho \frac{1}{2} \frac{3 \sqrt{1 - \frac{8}{3} \pi \rho R^2} - \sqrt{1 - \frac{8}{3} \pi \rho r^2}}{\sqrt{1 - \frac{8}{3} \pi \rho r^2}} = \frac{1}{2} \left( 3 M_0 \sqrt{1 - \frac{2M}{R}} - M \right)
$$
In the weak field limit, we can expand the square root and discard terms beyond first order in ##M / R##, and we obtain
$$
M_1 = \frac{1}{2} M \left[ 3 \left( 1 - \frac{2}{5} \frac{M}{R} \right) - 1 \right] = M \left( 1 - \frac{3}{5} \frac{M}{R} \right)
$$
So in this limit, yes, we can view the Komar mass integral as the outcome of the following process: take the mass ##M_0## at infinity, bring it all into a spherical shape with radius ##R##, which in the absence of pressure would have mass ##M_1##, and then support it against its own gravity with pressure, which adds back half of the difference ##M_0 - M_1## to obtain a final mass ##M##. The only difference is the factor ##3/5##.
There is probably a slick way to generalize this result by using the equation for hydrostatic equilibrium, but I'm not up to trying it right now.
