# 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

2. [...] people learn with that name in calculus courses. Ironically, unlike the fundamental theorem and divergence theorem special cases, Stokes’ theorem only works in dimension , where the differential can take us [...]

Pingback by The Classical Stokes Theorem « The Unapologetic Mathematician | November 23, 2011 | Reply

3. [...] what if it was well-defined, enough to take the integral inside the unit sphere at least? Then the divergence theorem tells us that the integral of the divergence through the ball is the same as the integral of the [...]

Pingback by Gauss’ Law « The Unapologetic Mathematician | January 11, 2012 | Reply

4. [...] this with the divergence theorem like last time, we conclude that there is no magnetic equivalent of “charge”, or else [...]

Pingback by Gauss’ Law for Magnetism « The Unapologetic Mathematician | January 12, 2012 | Reply

5. [...] the divergence theorem tells us that the first term [...]

Pingback by Ampère’s Law « The Unapologetic Mathematician | January 30, 2012 | Reply

6. [...] the left-hand sides we can use the divergence theorem, while the right sides can simply be [...]

Pingback by Maxwell’s Equations (Integral Form) « The Unapologetic Mathematician | February 2, 2012 | Reply

7. [...] by evaluating it over the solid ball of radius and taking the limit as goes off to infinity. The divergence theorem says we can [...]

Pingback by Energy and the Electric Field « The Unapologetic Mathematician | February 14, 2012 | Reply

8. [...] can pull the same sort of trick last time to make the second integral go away; use the divergence theorem to convert [...]

Pingback by Energy and the Magnetic Field « The Unapologetic Mathematician | February 14, 2012 | Reply