1
$\begingroup$

In the book "A Relativistic toolkit - Eric Poisson", proof of Gauss Stokes theorem is given.This is the Gauss-Stokes theorem

enter image description here

In the proof they have tried to show that LHS=RHS in a particular cooridnate system. I haven't understood that how do we eliminate the second term in the second line. The explanation which is given in the book is that the integration is over a closed three dimensional surface and x^a are angular coordinates. How does this work?

$\endgroup$

2 Answers 2

1
$\begingroup$

I don't have enough reputation points to comment, but, to be honest, this is merely a partial answer and comment.

tsufli's answer makes Eric Poisson's suggestion more explicit, but I think both are a bit too fast, and require a fair bit of work to make fully explicit. (I do think the underlying intuition is correct, but I also think the argument is not particularly compelling as-is.)

Let's look at the suggested 2+1 dimensional analog. In that analog, the integral which vanishes is over a 2 dimensional surface of a sphere in three dimensions (not over the three-dimensional sphere itself). The divergence is a sort of "surface-divergence" in the two dimensional surface and not the three-dimensional divergence in space.

Stokes' theorem does not apply directly, because [1] there is a boundary in three dimensions (the surface itself) and [2] the divergence in Stokes' theorem is a 3-dimensional spatial divergence, not a "surface divergence."

Green's theorem is also not directly applicable. The dimensions are correct -- both for the surface and for the divergence, but it's strictly for a planar surface. To get a 2-dimensional surface with no boundary requires leaving the plane. There is something frequently called the "2D divergence theorem" or "curl theorem" but it's essentially Green's theorem and as stated and proved it's strictly planar.

I have been unable to find a treatment or proof of the needed theorem. I believe there is such a theorem and it's both correct and applicable. But a direct reference to Stokes or Green is not enough.

Note that the above is not entirely accurate. There is a post (and a very nice answer) that provides the required proof of the "Divergence Theorem for a surface in ℝ3." It's at https://math.stackexchange.com/questions/130081/does-the-divergence-theorem-work-on-a-surface . The only issue with that work is that its definition of "surface divergence" is not obviously (to me) the same as the (unsupplied) definition used implicitly here. Relating the definition there to the implicit definition of surface divergence here would "fill in the gap." I think that's do-able (even for me) and I'm working on it.

I may well be wrong about all this -- if so, I'd love to be corrected.

$\endgroup$
0
$\begingroup$

As suggested in the referenced book, Lets think in 1 dimension less - imagine foliating a ball with spheres, and $x_{0}$ parametrizes the radius.

The term which is claimed to vanish is $\int _{x_{0}=c}d^3x(\sqrt {-g}A^{a})_{,a}$. In our 2+1 dimensional picture, this is an integral over a sphere of a divergence. The sphere has no boundary, so by Stoke's theorem it is 0. The same thing happens in any dimension, if the $x_{0}=c$ solutions are 3 dimensional manifolds without boundary.

I recommend the appendix of Carroll's GR book for an explanation on Stokes and divergence theorem in GR.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.