ADM Mass is a useful measure of a system. It is often defined (Wald 293)

$$M_{ADM}=\frac{1}{16\pi} \lim_{r \to \infty} \oint_{s_r} (h_{\mu\nu,\mu}-h_{\mu\mu,\nu})N^{\nu} dA$$

Where $s_r$ is two sphere with a radius going to infinity.

However, I have also seen the following definition (EDIT: for example, Page 147 of "A Relativists Toolkit"):

$$M_{ADM} = - \frac{1}{8\pi} \lim_{r \to \infty} \oint_{s_r}(H-H_0)\sqrt{\sigma}d^2\theta$$

Where $H_0$ is the extrinsic curvature of $s_r$ embedded in flat space and $H$ is the extrinsic curvature of $s_t$ embedded on a hypersurface of the spacetime.

It is not overly clear to me how these two are related. Obvious we can get the same differential with the follow definition.


But more so than that it doesn't seem clear why those extrinsic curvatures should be equal to those metric derivatives (times the constant term of $-1/2$). Presumably the fact that we are taking the limit out to $r \to \infty$ is important as we are assuming that the metric should be asymptotically flat (or at least asymptotically constant metric) and these quantities might be reduced to some similar form in the limit.

Any insight?

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    $\begingroup$ Could you tell us more about the second expression; I have not seen it before. Maybe source an example where they use it? $\endgroup$
    – levitopher
    Commented Mar 15, 2012 at 15:55
  • $\begingroup$ Page 147 of "A Relativists Toolkit" $\endgroup$ Commented Mar 16, 2012 at 3:34
  • $\begingroup$ Maybe discussion on p469 of Wald can help? especially E.2.46 and 2.47. $\endgroup$
    – Demian Cho
    Commented Mar 19, 2012 at 17:46

3 Answers 3


First, let's try and clean up the MTW expression a little bit by getting rid of the manifestly coordinate variant stuff, by introducing a proper contraction, a proper integral over the surface at infinity, and the unit normal $r^{a}$ to the surface at infinity:

$$M_{ADM}={16\pi}\oint\sqrt{\gamma}\,\,d^{2}x \gamma^{ab}r^{c}\left(\gamma_{ac,b}-\gamma_{ab,c}\right)$$

This is only valid, however, not just only for in asymptotically flat spaces, but actually only for asymptotically Cartesian coordinates. You can prove this to yourself by calculating the ``ADM Mass'' of the flat 3-metric in polar coordinates:

$\begin{align*} 16\pi M_{ADM}&=\lim_{r\rightarrow\infty}\oint r^{2}\sin\theta \gamma^{ab}\left(\gamma_{ra,b}-\gamma_{ab,r}\right)d\theta d\phi\\ &=\lim_{r\rightarrow\infty}4\pi r^{2}\left(\gamma^{rr}\gamma_{rr,r}-\gamma^{ab}\gamma_{ab,r}\right)\\ &=\lim_{r\rightarrow\infty}4\pi r^{2}\left(0-\frac{4}{r}\right)\\ &=-\infty \end{align*}$

Obviously, this is wrong, and just an artifact of the way that spherical coordinates behave at infinity. To fix this, we need to always subtract the divergence from the ADM mass of flat spacetime from the ADM mass of the coordinate system in question. So, that's the origin of the $H$ and $H_{0}$ terms. The $H$ term is the extrinsic curvature of the space in question, while $H_{0}$ is the extrinsic curvature of flat spacetime. Now, it's just a matter of showing that the expression inside the integral is equal to $H$.

Typically, I define the extrinsic curvature as $\gamma^{ab}\nabla_{a}r_{b}$. So, let's go on an adventure:

$\begin{align*} H&=\gamma^{ab}\nabla_{a}r_{b}\\ &=\gamma^{ab}\left(\partial_{a}r_{b}-\Gamma_{ab}{}^{c}r_{c}\right)\\ &=\gamma^{ab}\partial_{a}r_{b}-\frac{1}{2}\gamma^{ab}\gamma^{cd}\left(2\gamma_{ad,b}-\gamma_{ab,d}\right)r_{c}\\ &=\gamma^{ab}\partial_{a}\left(\gamma_{bc}r^{c}\right)-\gamma^{ab}r^{c}\left(\gamma_{ac,b}-\frac{1}{2}\gamma_{ab,c}\right)\\ &=\delta^{a}{}_{c}\partial_{a}r^{c} + \gamma^{ab}r^{c}\gamma_{bc,a}-\gamma^{ab}r^{c}\left(\gamma_{ac,b}-\frac{1}{2}\gamma_{ab,c}\right)\\ &=\partial_{a}r^{a}-\frac{1}{2}\gamma^{ab}r^{c}\left(\gamma_{ac,b}-\gamma_{ab,c}\right) \end{align*}$

Now, we note that in any coodinate system adapted so that the coordinate $r=constant$ that determines the surface is chosen for one of the coordinates, we have, necessarily $r^{a}=(1,0,0)$, we note that $\partial_{a}r^{a}=0$, and we thus conclude that the two expressions for ADM mass are equivalent.

  • $\begingroup$ The last line of the derivation of H seems to be incorrect. Namely, $\gamma^{ab}r^c\gamma_{bc,a}-\gamma^{ab}r^c\gamma_{ac,b}=0$, so it is not clear where the factor of 1/2 comes from in the last line multiplying the first summand in the parenthetical. Further, even at fixed radius, $\partial_a r^q=1$, but this is an overall constant that can be subtracted by $H_0$. Still, one should find $\endgroup$
    – alphanzo
    Commented Apr 27, 2016 at 18:10
  • $\begingroup$ Edit: ...One should find that $\nabla_a r^a=\partial_a r^a+\frac12 r^c\gamma^{ab}\partial_c\gamma_{ab}$, so there is a missing term in this derivation. Also, I meant $\partial_a r^a=1$, but cannot edit due to 5 minute rule. $\endgroup$
    – alphanzo
    Commented Apr 27, 2016 at 18:17
  • $\begingroup$ @alphanzo: You're right, and looking at it, I'm almost certain the missing term has to do with the difference between the three-metric and the four-metric. The christoffel symbols should be expended as $g_{ab}$ terms, not $\gamma_{ab}$ terms, but I dont' have th etime to fix it righ tnow. $\endgroup$ Commented Apr 27, 2016 at 18:22
  • $\begingroup$ I think even with $g_{ab}$ in place of $\gamma_{ab}$ the terms still cancel. My guess is that it must have to do with the asymptotic equivalence, since that extra term goes like $\gamma^{rr}\partial_r g_{rr}$ which goes to zero at infinity like $1/r^2$. $\endgroup$
    – alphanzo
    Commented Apr 27, 2016 at 18:30
  • $\begingroup$ @alphanzo: I know for a fact that it is unrelated to anything asymptotic. The formula is true for finite-area surfaces. $\endgroup$ Commented Apr 27, 2016 at 19:08

Your second expression, when the integral is not taken at infinity, is usually known as either the Brown-York or Hawking-Horowitz (quasilocal) mass. It is closely related to the Liu-Yau mass. For a finite radius sphere, it is not equal to the ADM integral in general.

In the original paper of Brown and York, they asserted but did not provide (as far as I can tell) a detailed proof of the asymptotic behaviour of the quasilocal mass. But if you consider an asymptotically flat space-time and take asymptotic Euclidean coordinates, a direct computation should show that the $K_0\to 0$ as a sphere goes to "infinity" (since coordinate spheres in Euclidean space has mean curvature decaying to 0 as the radius increases), and the decay conditions on the metric and the second fundamental form of the spatial slice should guarantee that the difference between $K$ and the coordinate derivative expression tends to zero. (Remember that the space-time mean curvature can be computed from the mean curvature of the spatial slice embedded in the space-time together with the mean curvature of the two sphere embedded in the spatial slice.)

Similar computations were also performed by Brewin.

The convergence of the Brown-York mass to the ADM mass is also recovered in this paper of Miao, Shi, and Tam.

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    $\begingroup$ The integrals are usually evaluated with respect to the coordinate system. In this business you first fix an asymptotically flat coordinate system of an end (so think of parametrizing the region near infinity by $\mathbb{R}^3$ (with a compact center region removed). The $s_r$ are the spheres in the coordinate system, and $dA$ are the area elements of the sphere in the coordinate system. $\endgroup$ Commented Apr 5, 2022 at 14:05
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    $\begingroup$ The asymptotic flat condition requires that the actual metric falls off like $1/r$, and so the coordinate system area element will converge (when you take large enough balls) to the one given by the metric, so it doesn't really matter that much which one you use. @Steve $\endgroup$ Commented Apr 5, 2022 at 14:06
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    $\begingroup$ Does your chart go from (domain in $\mathbb{R}^n$) to (domain in manifold) or the other way? The integration is being carried out over the sphere as a subset of domain in $\mathbb{R}^n$. (Everything in sight is done using coordinates.) $\endgroup$ Commented Apr 5, 2022 at 19:20
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    $\begingroup$ $\Phi_k$ is a diffeo, so you can use it (or its inverse) to pull back volume forms. And the integrals respect the diffeo. So I don't really know what you mean by "I only know how to integrate a Riemannian volume form over a Riemannian manifold". I mean, the whole point of coordinate charts is so that you can have explicit formulas for computing "abstract" things... $\endgroup$ Commented Apr 5, 2022 at 23:01
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    $\begingroup$ Those are the functions corresponding to the components of the metric tensor expressed in the chart/coordinates, as well as the coordinate functions. $\endgroup$ Commented Apr 5, 2022 at 23:57

The question is given as a practice problem in Poisson's book "Relativists Toolkit" (p. 159, problem #7) along with some helpful hints.

The tensor $\gamma_{ab}$ (which is not a tensor at all) must be defined in the following way: Let $h_{ab}$ be the induced 3-metric on the spacelike hypersurface $\Sigma$, and let on its boundary, $\partial \Sigma$, be difined the two-metric $\sigma_{ab}$. Then define $$ \gamma_{ab}=h_{ab}-h^{(0)}_{ab} $$ where $h^{(0)}_{ab}$ is the induced metric in the Minkowski limit (the asymptotic condition gives $\sigma_{ab}=\sigma^{(0)}_{ab}$).

Further, define the derivation $D_a$ to be the covariant derivative associated to the flat metric $h^{(0)}_{ab}$. Then, $$ M_{ADM}=\frac1{16\pi}\oint_{\partial \Sigma}\sqrt{\sigma}d^2\theta\ r^a(D^b\gamma_{ab}-D_a\gamma) $$ which is shown by direct calculation to equal $$ M_{ADM}=-\frac1{16\pi}\oint \sqrt{\sigma}d^2\theta\ r^ch^{(0)ab}\ \partial_c\gamma_{ab} $$ and this is the same as $M_{ADM}=-\frac1{8\pi}\oint \sqrt{\sigma}d^2\theta\ (H-H_0)$ through a straighforward comparison. (Specifically, by evaluating $H=\partial_a r^a+\frac12 r^ch^{(0)ab}\partial_c h_{ab}$)


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