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The ADM mass is expressed in terms of the initial data as a surface integral over a surface $S$ at spatial infinity: $$M:=-\frac{1}{8\pi}\lim_{r\to \infty}\int_S(k-k_0)\sqrt{\sigma}dS$$ where $\sigma_{ij}$ is the induced metric on $S$, $k=\sigma^{ij}k_{ij}$ is the trace of the extrinsic curvature of $S$ embedded in $\Sigma$ ($\Sigma$ is a hypersurface in spacetime containing $S$). and $k_0$ is the trace of extrinsic curvature of $S$ embedded in flat space.

Can someone explain to me why ADM mass is defined so. Why is integral of difference of traces of extrinsic curvatures important?

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    $\begingroup$ Related question $\endgroup$
    – Trimok
    Nov 21, 2013 at 10:52
  • $\begingroup$ The short answer is that this term is a boundary term of the ADM hamiltonian, and the bulk term vanishes. $\endgroup$ Nov 21, 2013 at 17:47

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