# Stokes theorem under covariant derivaties?

1. Oct 20, 2008

### tim_lou

in my GR book, it claims that integral of a covariant divergence reduces to a surface term. I'm not sure if I see this...

So, is it true that:
$$\int_{\Sigma}\sqrt{-g}\nabla_{\mu} V^{\mu} d^nx= \int_{\partial\Sigma}\sqrt{-g} V^{\mu} d^{n-1}x$$

if so, how do I make sense of the $d^{n-1}x$ term? would it be just a differential form?? so what about the following?

$$\int_{\Sigma} \nabla_{\mu} V^{\mu} d\omega=\int_{\partial \Sigma} \sqrt{-g} V^{\mu} \omega$$

is it true? It certainly doesn't seem so to me... since the proof of stokes' theorem heavily rely on the similarities between taking the boundary of a simplex and taking the derivative of a form. This certainly does not seem to be the case with covariant derivatives. Perhaps it is only true with taking divergences. How do I go by proving/reasoning it? (simply saying things in flat space generalize in things in curved space with ; replaced by , doesn't do it for me)

edit: i just can't figure out what the heck is wrong with the latex... someone might have to fix it for me. (something is clearly messed up about the current latex system... seriously how can my d^{n-1} x term have any syntax error?

Last edited: Oct 20, 2008
2. Oct 21, 2008

### schieghoven

One way to demonstrate the specific example you raised is via

V^{\mu}_{;\mu} = \frac{1}{\sqrt{-g}} \left( V^\mu \sqrt{-g} \right)_{,\mu}

(tex is playing up for me too.) You can prove this result from the definitions of the covariant derivative and the Christoffel symbols. The factors of \sqrt{-g} appear in just the right places that the integral will reduce to the standard Gauss' law in terms of partial derivatives.

The Stokes' theorem generalises to something stated in terms of differential forms and exterior derivatives. I don't know the subject well enough to recommend any particular books, but perhaps you can find something in a book on differential geometry.

Regards,
Dave

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