# The Unapologetic Mathematician

## The Divergence Theorem

I’ve been really busy with other projects — and work — of late, but I think we can get started again. We left off right after defining hypersurface integrals, which puts us in position to prove the divergence theorem.

So, let $S$ be a hypersurface with dimension $n-1$ in an $n$-manifold $M$, and let $F$ be a vector field on some region containing $S$, so we can define the hypersurface integral $\displaystyle\int\limits_SF\cdot dS$

And if $F$ corresponds to a $1$-form $\alpha$, we can write this as $\displaystyle\int\limits_S\langle\alpha,*\omega_S\rangle\omega_S$

where $\omega_S$ is the oriented volume form of $S$ and $*\omega_S$ is a $1$-form that “points perpendicular to” $S$ in $M$. We take the given inner product and integrate it as a function against the volume form of $S$ itself.

A little juggling lets us rewrite: $\displaystyle\int\limits_S*\alpha$

where we take our $1$-form and flip it around to the “perpendicular” $n-1$ form $*\alpha$. Integrating this over $S$ involves projecting against $\omega_S$, which is basically what the above formula connotes.

Now, let’s say that the surface $S$ is the boundary of some $n$-dimensional submanifold $E$ of $M$, and that it’s outward-oriented. That is, we can write $S=\partial E$. Then our hypersurface integral looks like $\displaystyle\int\limits_{\partial E}*\alpha$

Next we’ll jump over to the other end and take the divergence $\nabla\cdot F$ and integrate it over the region $E$. In terms of the $1$-form $\alpha$, this looks like $\displaystyle\int\limits_E(*d*\alpha)\omega=\int\limits_E(d*\alpha)$

But Stokes’ theorem tells us that $\displaystyle\int\limits_{\partial E}*\alpha=\int\limits_E(d*\alpha)$

which tells us in our vector field notation that $\displaystyle\int\limits_SF\cdot dS=\int\limits_E\nabla\cdot F\,dV$

This is the divergence — or Gauss’, or Gauss–Ostrogradsky — theorem, and it’s yet another special case of Stokes’ theorem.

November 22, 2011 - Posted by | Differential Geometry, Geometry

## 8 Comments »

1. Yay! You’re back! Comment by Joe English | November 22, 2011 | Reply

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