Gauss's law for potentials

Remarks

Proof idea: first take $\mu = \delta_y$ with $y \notin \partial\Omega$. If $y \notin \overline{\Omega}$, the function $x \mapsto \abs{x - y}^{2-n}$ is harmonic on a neighbourhood of $\overline{\Omega}$, and the flux is $0$ by the divergence theorem (or Green's first identity with $u = 1$). If $y \in \Omega$, remove a small ball about $y$; the flux through the small sphere is $(2-n) r^{1-n} \cdot \sigma_n r^{n-1} = -(n-2)\sigma_n$. The general case follows by integrating in $y$ against $\mu$ and exchanging the integrals, which is allowed because $\partial\Omega$ stays at positive distance from the support of $\mu$. In $\R^3$ the constant is $-4\pi$: with the gravitational field $g = \nabla(G U^\mu)$ pointing towards the masses, the flux of $g$ out of a closed surface is $-4\pi G$ times the enclosed mass. The hypothesis on $\partial\Omega$ is needed: for a point mass at a point where $\partial\Omega$ is smooth, the flux counts only half of that mass.

Going deeper

For a student

This is the integral form of the distributional equation $\Delta U^\mu = -(n-2)\sigma_n \mu$, obtained by integrating over $\Omega$ and using the divergence theorem. It gives a quick proof of the shell theorem by symmetry: for a radial mass distribution, the flux through a sphere fixes the radial derivative of the potential. The same argument with the logarithmic kernel in the plane gives the constant $-2\pi$.