# Trouble with Stokes' theorem 4-dim

1. Aug 12, 2012

### crimsonidol

Well I am studying the book Relativisits' Toolkit by Eric Poisson and I am stuck in the part Stokes Theorem.

Working under 4 dim. spacetime. greek indices runs from 0 to 3 latin indices runs from 1 to 3
It defines surface element as

dƩ$_{\mu}$=ε$_{\mu\alpha\beta\gamma}$e$^{\alpha}_{1}$e$^{\beta}_{2}$e$^{\gamma}_{3}$d$^{3}$y

$\int_{V}$A$^{\alpha}_{;\alpha}$$\sqrt{-g}$d$^{4}$x=$\oint_{\partial V}$A$^{\alpha}$dƩ$_{\alpha}$

It says "think a nest of closed hypersurfaces foliating V with boundary ∂V forming outer layer of the nest.let x$^{0}$ constant on each one of these hypersurfaces with x$^{0}$=1 designating ∂V and x$^{0}$=0 , the zero volume hypersurface at the center of V."

$\int_{V}$A$^{\alpha}_{;\alpha}$$\sqrt{-g}$d$^{4}$x= $\int_{V}$($\sqrt{-g}$A$^{\alpha}$),$_{ \alpha}$d$^{4}$x
$^{*}$=$\int$dx$^{0}$$\oint$($\sqrt{-g}$A$^{0}$),$_{0}$+$\int$dx$^{0}$$\oint$($\sqrt{-g}$A$^{a}$),$_{a}$d$^{3}$x

$^{*}$=$\int$dx$^{0}$$\frac{d}{dx^{0}}$$\oint$($\sqrt{-g}$A$^{0}$)d$^{3}$x

$^{*}$=$\oint$($\sqrt{-g}$A$^{0}$)$_{0}$d$^{3}$x l$^{x^{0}=1 }_{x^{0}=0 }$
$^{*}$=$\oint$($\sqrt{-g}$A$^{0}$)$_{0}$d$^{3}$y
"$^{*}$=" means in the desired coordinate system.

$\oint_{∂V}$A$^{\alpha}$dƩ$_{\alpha}$ $^{*}$= $\oint_{∂V}$A^{0}$\sqrt{-g}$d$^{3}$y

What I did not get is probably quite easy but I'm stuck, I cannot figure it out.
$\int$dx$^{0}$$\oint$($\sqrt{-g}$A$^{a}$),$_{a}$d$^{3}$x part is equal to zero in the frame x^{0}=constant ,so called radial part ,and this part is angular part. I cannot get why it is equal to zero.Any help greatly appreciated.

Last edited: Aug 12, 2012