The Unapologetic Mathematician

Mathematics for the interested outsider

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.

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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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