Boundary terms and divergence theorom

**ppyvabw** · August 24th 2007, 02:30 PM

Say I have a vector field $k^{\mu}$ defined on a space such as $R^{2,p}$ Actually, it is AdS space im working on but I don't think that matters for what I want to ask. I assume that I can write the coordinates as (time,radial, p-angles)

Now if I take the integral of the divergence of the field

$\int \partial_{\mu}k^{\mu}dx^{p+2}$ then by the divergence theorem this gives zero if K dies of sufficiently quickly at spatial (and temporal)infinity i.e. (r --->infinity)

Everything works as I want it to if I take it to die of as 1/r or quicker. Unless I have made a mistake, then any other possibility doesn't work like I think it should.

Can any one reassure me about this.

**Rebesques** · August 24th 2007, 02:50 PM

Everything is surely ok if it has a compact support. But in this case it works fine, as it leaves "zero trace" near infinity.

**ppyvabw** · August 24th 2007, 05:29 PM

lol, sorry but I don't have a clue what you just said.

I made my 'assumption' to make the answer do what I hoped it would and am relying on the words 'sufficiently quickly' But I still have to justify why I have assumed that a 'fall off' of 1/r is enough to ensure the divergence integrates to zero over spacetime.

**ppyvabw** · August 24th 2007, 05:35 PM

I mean, $1/sqrt(r)$ gives zero at infinity, but for my purposes it doesn't work and I am hoping it's because it doesn't satisfy the 'sufficiently quickly' condition.

**Rebesques** · August 25th 2007, 01:06 AM

Ofcourse it doesn't. Just saw you have time+radial+space=2+p dimensions.

I mean the divergence theorem applies when your vector field is zero near infinity (="compactly supported") over space+time. Are you interested in radial dimensions only? If yes, remember that you must have a finite integral over time also.

Now the divergence cannot go to zero slower than $r^{p+2}$ . Can you see why?

**ppyvabw** · August 25th 2007, 04:49 AM

Nope, I cant see why. Are you saying 1/r doesn't work?

I can safely assume the integral vanishes at infinite time in my problem. It's the r bit that is the problem

**ppyvabw** · August 25th 2007, 05:03 AM

My expression for K at large r is
$<br /> <br /> K^{\mu}=X^{\mu}r^{p-2{\lambda}-2}(-\partial_{t}\widetilde{\varphi}\partial_{t}\varphi + \widetilde{\varphi}\varphi r^{2}+\partial_{\Omega}\widetilde{\varphi}\partial _{\Omega}\varphi+r^{2}m^{2}\widetilde{\varphi} \varphi)<br />$

Where $\varphi$ depends only on t and Omega , and X is a vector, possibly with some t and omega dependance I think but no r dependance. $\varphi$ can be safely assumed to die at temporal infinity sufficiently quickly. Now I want to arrange p and $\lambda$ so that $\int \partial_{\mu}k^{\mu}dx^{p+2}=0$ . The second and third fourth terms have the highest power of r. These are what I am thinking I should arrange to be 1/r, and then it does what I want it to.

I should also say, that the factor of sqrt(g) where g is the metric determinant is included in K, so there need not be any sqrt(g) in the integral, and it is not the covariant divergence that I'm interested in as K is a tensor density, not a tensor.

**ppyvabw** · August 25th 2007, 08:35 AM

Oh, and by the way. X is a killing vector, so its covariant derivative is zero by killing's equation, and I am hoping that $p-2 \lambda =-1$ where $\lambda$ depends p and m and this condition will impose the correct restrictions on m.

I am sort of managing to convince myself that I am right

**Rebesques** · August 25th 2007, 02:27 PM

My expression for K at large r is
...can be safely assumed to die at temporal infinity sufficiently quickly.

Why didn't you say so in the first place? This makes things easier. One last thing: Sure $\partial_{\mu}=(\partial_t\,\partial_r,\partial_{\ Omega})$ ??

**ppyvabw** · August 25th 2007, 02:40 PM

I did do

Originally Posted by ppyvabw

if K dies of sufficiently quickly at spatial (and temporal)infinity i.e. (r --->infinity)

Yes Im sure $\partial_{\mu}=(\partial_{t}, \partial_{r}, \partial_{\Omega})$

**ppyvabw** · August 25th 2007, 02:51 PM

I am kind of thinking the integral is 'seperable' and can just do the integral over r seperately, the point being that any terms involving 1/r or greater will integrate to something like $r^{+something}$ which is infinity at r = infinity.

**Rebesques** · August 25th 2007, 03:03 PM

Yeap, actually

$\int \psi(t,r,\Omega) {\rm dx}^{p+2}=\int_{0}^{\infty}\left(\int_0^{\infty}\l eft(\int_{S_r}\psi(t,r,\Omega){\rm d\Omega}\right){\rm dr}\right){\rm dt}$

where $S_r$ is the p-sphere of radius r. And the order of integration cannot be changed.

Now, there's something else I am worried about...

**ppyvabw** · August 25th 2007, 03:12 PM

No, $\varphi$ independant of r, Im thinking

$\int_{t} dt \int_{\Omega_p} d{\Omega} \int_r \partial_{\mu}K^{\mu} dr$ .

The expression for K that I gave is only the boundary behaviour. That expression is singular at r= 0 which is clearly wrong
Why can't you change the order of integration in your expression?

**Rebesques** · August 25th 2007, 04:02 PM

Man, you are letting me know one thing at a time

'Course you cannot change the order of integration, what is S_r if you haven't fixed r?

**ppyvabw** · August 25th 2007, 04:49 PM

lol, im not, I told you that here.

Originally Posted by ppyvabw

My expression for K at large r is

Where $\varphi$ depends only on t and Omega .

$\Omega$ are just angles like the sperical polars $(r, \theta, \phi)$ except there's many more angles, which is why I am thinking they can be seperated out. I don't think r and omega are related in anyway.

Not to worry, there's a lot of details that would take forever to write down. Thanks anyway.

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