PeterDonis said:
given the formula I posted in #262, deriving the result should be pretty straightforward. I'll put that in a separate post.
Just to close this out, here's the derivation of the ADM mass for a Schwarzschild black hole. We want to show that the ADM mass of a Schwarzschild BH equals the M parameter that appears in the metric. To keep the two distinguished, I'll use M_{ADM} for the ADM mass.
The exact version of the formula I posted in #262 is
M_{ADM} = \frac{1}{16 \pi} \lim_{S \rightarrow i^{0}} \int_{S} g^{ij} \left( g_{ik, j} - g_{ij, k} \right) n^{k} d S
where I have put back in the constant factor 1/16pi at the front, and I have also changed index notation to make it clear that the metric g is the 3-metric of a slice of constant time, in this case Schwarzschild coordinate time (which I didn't make clear before). If the metric is diagonal, which it will be in the coordinates we will use, we can simplify this a bit to
M_{ADM} = \frac{1}{16 \pi} \lim_{S \rightarrow i^{0}} \int_{S} g^{ii} \left( g_{ik, i} - g_{ii, k} \right) n^{k} d S
which reduces the number of terms we have to sum since i and k only range over 3 indexes (and as we'll see, k ends up ranging over only one).
One thing to note about the above formula is that it is not in covariant form; indeed, as the paper I got it from points out, there is no covariant formula involving first derivatives of the metric except the trivial g_{ab;c} = 0. So the above formula is only valid for a restricted class of coordinate systems, which the paper calls "asymptotically Euclidean" coordinates. The paper doesn't define exactly what those are, but I suspect that what is meant is really "asymptotically *Cartesian*" coordinates. To see why, let's first try computing the ADM mass of flat Minkowski spacetime in spherical coordinates; the outward normal to a 2-sphere is just n^{r} = 1, with other components zero, and we find that only two nonzero terms survive in the integral:
M_{ADM} = \frac{1}{16 \pi} \lim_{S \rightarrow i^{0}} \int_{S} \left[ g^{\theta \theta} \left( - g_{\theta \theta, r} \right) + g^{\phi \phi} \left( - g_{\phi \phi, r} \right) \right] n^{r} d S = \frac{1}{16 \pi} \lim_{S \rightarrow i^{0}} \int_{S} \frac{2}{r^{2}} \left( - r^{2} \right)_{,r} dS = \frac{1}{16 \pi} \lim_{S \rightarrow i^{0}} \int_{S} - \frac{4}{r} dS
Taking the integral is now easy since the integrand is a function of r only; the integral just contributes a factor of 4 \pi r^{2}, the area of the 2-sphere, which gives
M_{ADM} = \lim_{S \rightarrow i^{0}} - r
The limit is now just a limit as r -> infinity, so we get minus infinity as the ADM mass of flat spacetime in spherical coordinates. Obviously this is not the right answer.
We could fix this by using Cartesian coordinates, but the Schwarzschild metric looks really ugly in those coordinates. An easier way is to just evaluate the ADM mass of Schwarzschild spacetime in appropriate spherical coordinates, and then correct the result by "subtracting off" the above value for flat spacetime. This basically corresponds to adjusting the "zero point" of energy to compensate for the behavior of spherical coordinates at infinity.
I said "appropriate spherical coordinates" just now. I'm not sure if standard Schwarzschild coordinates are actually "appropriate" in this sense, since they are not isotropic. In any case, I've done the computation instead in isotropic coordinates, where it is straightforward. The line element is
ds^{2} = K(r)^{4} \left( dr^{2} + r^{2} d\theta^{2} + r^{2} sin^{2} \theta d\phi^{2} \right)
where K(r) = \left( 1 + \frac{M}{2r} \right). The unit outward normal to a 2-sphere at r is then n^{r} = 1 / K(r)^{2}, with all other components zero. Again, only two nonzero terms survive in the sum (the terms with i = r cancel each other out):
M_{ADM} = \frac{1}{16 \pi} \lim_{S \rightarrow i^{0}} \int_{S} \left[ g^{\theta \theta} \left( - g_{\theta \theta, r} \right) + g^{\phi \phi} \left( - g_{\phi \phi, r} \right) \right] n^{r} d S = \frac{1}{16 \pi} \lim_{S \rightarrow i^{0}} \int_{S} \frac{2}{K^{4} r^{2}} \left( - K^{4} r^{2} \right)_{,r} \frac{1}{K^{2}} d S = \frac{1}{16 \pi} \lim_{S \rightarrow i^{0}} \int_{S} \left( \frac{4M}{K^{3} r^{2}} - \frac{4}{K^{2} r} \right) d S
As above, the integral just contributes a factor of 4 \pi r^{2}, and we have:
M_{ADM} = \lim_{S \rightarrow i^{0}} \left( \frac{M}{K^{3}} - \frac{r}{K^{2}} \right)
Again, the limit is now just a limit as r -> infinity, and in this limit K -> 1. As noted above, we have to correct the "zero point" of energy by subtracting off the result we got above for flat spacetime, so we have
M_{ADM} = \lim_{r \rightarrow \infty} \left( \frac{M}{K^{3}} - \frac{r}{K^{2}} + r \right)
The last two terms cancel in the limit, and we have M_{ADM} = M, as desired.
