Stokes theorem under covariant derivaties?

[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:
[tex]\int_{\Sigma}\sqrt{-g}\nabla_{\mu} V^{\mu} d^nx= \int_{\partial\Sigma}\sqrt{-g} V^{\mu} d^{n-1}x[/tex]

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

[tex]\int_{\Sigma} \nabla_{\mu} V^{\mu} d\omega=\int_{\partial \Sigma} \sqrt{-g} V^{\mu} \omega[/tex]

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?

 

[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

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:
[tex]\int_{\Sigma}\sqrt{-g}\nabla_{\mu} V^{\mu} d^nx= \int_{\partial\Sigma}\sqrt{-g} V^{\mu} d^{n-1}x[/tex]

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

[tex]\int_{\Sigma} \nabla_{\mu} V^{\mu} d\omega=\int_{\partial \Sigma} \sqrt{-g} V^{\mu} \omega[/tex]

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?

 

[Entropia]

Stokes' theorem is a fundamental result in differential geometry that relates the integral of a differential form over a manifold to the integral of its exterior derivative over the boundary of the manifold. In the context of GR, this theorem can be extended to covariant derivatives, which are the generalization of partial derivatives to curved spacetime.

In your GR book, it claims that the integral of a covariant divergence reduces to a surface term. This is indeed true and can be seen as a consequence of Stokes' theorem. However, the notation used in your equation may be a bit confusing. Let's break it down step by step.

First, let's define some terms:
- \Sigma: a region in spacetime
- \partial\Sigma: the boundary of \Sigma
- \sqrt{-g}: the square root of the determinant of the metric tensor, which is a factor that appears in the integration measure in GR
- V^{\mu}: a vector field on \Sigma
- d^nx: the integration measure on \Sigma, which is a differential form of degree n

Now, let's look at the first equation:
\int_{\Sigma}\sqrt{-g}\nabla_{\mu} V^{\mu} d^nx= \int_{\partial\Sigma}\sqrt{-g} V^{\mu} d^{n-1}x
This equation is a manifestation of Stokes' theorem, where the integral over the boundary of \Sigma is equal to the integral of the exterior derivative of a differential form over \Sigma. In this case, the differential form is \sqrt{-g}\nabla_{\mu} V^{\mu} and its exterior derivative is \sqrt{-g} V^{\mu} d^{n-1}x. This can be seen by expanding the exterior derivative and using the definition of the covariant derivative.

The d^{n-1}x term in the second integral is the integration measure on the boundary, which is a differential form of degree n-1. This is because the boundary of \Sigma is a hypersurface, which has one less dimension than \Sigma. In general, the integration measure on a hypersurface will have one less degree compared to the integration measure on the bulk manifold.

As for your second equation:
\int_{\Sigma} \nabla_{\mu} V^{\mu} d\omega=\int_{\partial \Sigma} \