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Math Help - Boundary terms and divergence theorom

  1. #1
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    Boundary terms and divergence theorom

    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.
    Last edited by ppyvabw; August 24th 2007 at 06:30 PM.
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  2. #2
    Super Member Rebesques's Avatar
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    Everything is surely ok if it has a compact support. But in this case it works fine, as it leaves "zero trace" near infinity.
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  3. #3
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    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.
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  4. #4
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    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.
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  5. #5
    Super Member Rebesques's Avatar
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    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?
    Last edited by Rebesques; August 25th 2007 at 02:11 AM. Reason: old age
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  6. #6
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    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
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  7. #7
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    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.
    Last edited by ppyvabw; August 25th 2007 at 12:15 PM.
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  8. #8
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    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
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  9. #9
    Super Member Rebesques's Avatar
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    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})??
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  10. #10
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    I did do

    Quote Originally Posted by ppyvabw View Post
    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})
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  11. #11
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    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.
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  12. #12
    Super Member Rebesques's Avatar
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    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...
    Last edited by Rebesques; August 25th 2007 at 04:10 PM. Reason: F* apple and their crappy products. Arghhhhhh
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  13. #13
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    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?
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  14. #14
    Super Member Rebesques's Avatar
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    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?
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  15. #15
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    lol, im not, I told you that here.

    Quote Originally Posted by ppyvabw View Post
    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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