5
$\begingroup$

I am reading a paper, where an integral of a divergence over a closed surface is used without proof.

$\oint_S [\nabla \cdot \vec{v}(\vec{r})] d\vec{s} = 0$,

where $\vec{v}$ is tangential to the surface ($\vec{v}(r)\cdot \vec{n}(\vec{r}) = 0$)

I have looked at vector calculus identities and Green theorems and can't seem to find the expression I need. Any suggestions?

$\endgroup$
0
$\begingroup$

This is a direct consequence of the divergence theorem. For a vector field $X$ on an oriented $n$-dimensional Riemannian manifold $(M, g)$, the divergence theorem states that $$ \int_M (\operatorname{div} X) dV_g = \int_{\partial M} g(X, N) dV_{\tilde g}$$ where $N$ is the outward-pointing normal vector at the boundary, $dV_g$ is the Riemannian volume form, and $dV_{\tilde g}$ is the induced volume form on the boundary. For a surface embedded in Euclidean space, we use the metric induced by the pullback of the inclusion $i:S \to \mathbb{R}^k$, i.e., $i^*g$ where $g = \delta_{ij}dx^idx^j$ is the usual Euclidean metric. Then $dV_g = ds$ where $ds$ is the area element, and $dV_{\tilde g} = dt$ where $dt$ is a length element. Since the surface in your question is closed, the boundary $\partial S$ is empty and the right-hand side integral is $0$.

If $M$ is not orientable, the divergence theorem still holds if you replace $dV_g$ and $dV_{\tilde g}$ with the respective densities $d\mu_g$ and $d\mu_{\tilde g}$. These densities are nothing but local volume forms on different patches of the manifold glued together with a partition of unity.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I'm sorry, I do not see how it follows. Please provide a detailed proof $\endgroup$ – Aleksejs Fomins Oct 6 at 12:47
  • $\begingroup$ Also, there seems to be a mistake. The LHS of your divergence theorem equation should refer to an integral of a volume, but you use the symbol $ds$ in there, which you call "the area form of the surface". This is inconsistent with the standard definition of the theorem $\endgroup$ – Aleksejs Fomins Oct 6 at 12:53
  • $\begingroup$ It is not a mistake. The divergence theorem applies to manifolds of all dimensions, not just 3-dimensional manifolds. I have rewritten it in the general form to avoid confusion. Also note that "closed surface" means a compact surface without boundary, so $\partial S = \emptyset$ is part of the definition. $\endgroup$ – abhi01nat Oct 6 at 16:54
  • $\begingroup$ Also, the wiki page you linked to also states what I have said, under the "Generalisations" section. $\endgroup$ – abhi01nat Oct 6 at 16:58
  • $\begingroup$ I am sorry for assuming it was a mistake, but thanks anyway for cleaning up the notation. My current knowledge of differential geometry is not sufficient to fully comprehend this proof. I will attempt to come back to it when I have some free time to fully understand it. $\endgroup$ – Aleksejs Fomins Oct 6 at 21:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.