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Let $M$ be a 3-manifold with positive definite metric $g$, and let $S\subset M$ be an oriented 2-surface. For $x\in S$ let $K(x)$ be the Gaussian curvature and $H(x)$ be the mean curvature of $S$ at the point $x$. Gauss-Bonnet tells us exactly what $\int_S K\,\mathrm{d}S$ is $-$ it is the Euler characteristic of $S$ (up to proportionality). Can a similarly simply interpretation be put to $\int_S KH\,\mathrm{d}S$?

I'm particularly interested in the case where $S$ has spherical topology.

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    $\begingroup$ Why do you ask? Where does this quantity come from? $\endgroup$
    – Igor Rivin
    Commented Oct 3, 2017 at 14:12
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    $\begingroup$ I'm studying apparent horizons ($S$) in general relativity. These are codimension-2 surfaces in a pseudo-Riemannian 4-manifold whose second fundamental forms are traceless. In a canonical Hamiltonian formalism, observable physical quantities are defined on a 3-dimensional Cauchy surface (which is $M$ here). Taking the Poisson bracket of two easy to physically interpret quantities gives me this integral, the physical interpretation of which I would like to ascertain. $\endgroup$
    – Josh Kirklin
    Commented Oct 3, 2017 at 14:20
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    $\begingroup$ There certainly cannot be a topological interpretation, as your expression depends on the extrinsic curvature, and is not scale invariant for round spheres in $\mathbb{R}^3$. Mean curvature is also only defined up to sign (choice of normal). $\endgroup$
    – Willie Wong
    Commented Oct 3, 2017 at 19:41
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    $\begingroup$ If there is a good interpretation it would be specifically in the context of GR. Is the factor multiplied by any normalization constants (like total area?) Knowing exactly how the quantity scales can bring in some dimensional analysis arguments for the physical interpretation. Of course, it may possibly more obvious if you can just show us where this term comes up exactly. $\endgroup$
    – Willie Wong
    Commented Oct 3, 2017 at 19:44
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    $\begingroup$ If you have a variation $S_t$ of $S$ with normal speed $V$, then $\partial_t \operatorname{Area}(S) = -\int H V dS$. So $\int_S KH dS$ is (minus) the variation of area under the Gauss curvature flow. $\endgroup$
    – Paul Bryan
    Commented Oct 5, 2017 at 5:19

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