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Conformal invariance of null geodesics

  1. Sep 24, 2011 #1
    Hi, folks. I hope this is the right forum for this question. I'm not actually taking any classes, but I am doing self-study using D'Inverno's Introducing Einstein's Relativity. I have a solution, and I want someone to check it for me.

    1. The problem statement, all variables and given/known data
    Prove that the null geodesics of two conformally related metrics coincide.


    2. Relevant equations
    Conformally related metrics: [itex]\overline{g}[/itex]ab = [itex]\Omega[/itex]2gab
    Null geodesics: 0 = gab(dxa/du)(dxb/du)


    3. The attempt at a solution
    I define the parameter u = [itex]\frac{1}{2}[/itex][itex]\Omega[/itex]2. Thus [itex]\frac{du}{d\Omega}[/itex] = [itex]\Omega[/itex].

    Now, I use the chain rule on the null geodesics equation:
    0 = [itex]\Omega[/itex]2gab(dxa/d[itex]\Omega[/itex])[itex]\frac{d\Omega}{du}[/itex](dxb/d[itex]\Omega[/itex])[itex]\frac{d\Omega}{du}[/itex]
    0 = [itex]\Omega[/itex]2gab(dxa/d[itex]\Omega[/itex])(dxb/d[itex]\Omega[/itex])([itex]\frac{du}{d\Omega}[/itex])-2
    0 = [itex]\Omega[/itex]2gab(dxa/d[itex]\Omega[/itex])(dxb/d[itex]\Omega[/itex])[itex]\Omega[/itex]-2
    0 = gab(dxa/d[itex]\Omega[/itex])(dxb/d[itex]\Omega[/itex]), which is the null geodesics equation with the new parameter [itex]\Omega[/itex].

    So is this a legitimate proof of the coinciding of null geodesics of conformally related metrics?
     
  2. jcsd
  3. Sep 29, 2011 #2
    Bump; can anyone help me?
     
  4. Sep 29, 2011 #3

    diazona

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    Hmm... is u a parameter for the geodesic? If so, I don't think you can relate it to Ω like that, because the proof should work for arbitrary conformal transformations including (for example) those where Ω is a constant.
     
  5. Sep 29, 2011 #4

    Dick

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    No, it's not right. 0 = gab(dxa/du)(dxb/du) isn't the geodesic equation. It's just says that it's a null curve. And those are obviously conformally invariant. Not all null curves are null geodesics. If you want to see the right way to do it look at Appendix D in Robert Wald's book General Relativity.
     
  6. Sep 30, 2011 #5
    OK, I see now that I should be using the equation:
    d2xa/ds2 + [itex]\Gamma[/itex]abc(dxb/ds)(dxc/ds) = 0

    Unfortunately, I am coming up with
    d2xa/ds2 + {[itex]\Gamma[/itex]abc + ([itex]\delta[/itex]ac[itex]\partial[/itex]b[itex]\Omega[/itex] + [itex]\delta[/itex]ab[itex]\partial[/itex]c[itex]\Omega[/itex] - gadgbc[itex]\partial[/itex]d[itex]\Omega[/itex]) / [itex]\Omega[/itex]}(dxb/ds)(dxc/ds) = 0
    as the transformation. I don't see how that is invariant unless d[itex]\Omega[/itex] = 0, and I don't see any reason to make that assumption.

    I think I will try looking for that Wald book you mentioned.

    Update: Wow, thanks for that tip, Dick. The explanation in the Wald book really cleared things up. It starts by using the affine geodesic equation,
    Xa[itex]\nabla[/itex]aXb = 0,
    and uses a relation between covariant derivatives that I did not find in the Inverno book:
    [itex]\overline{\nabla}[/itex]aXb = [itex]\nabla[/itex]aXb + TbacXc

    I find how Tbac transforms conformally and get this equation:
    Xa[itex]\nabla[/itex]aXb = 2XaXb[itex]\nabla[/itex]a(ln [itex]\Omega[/itex]) - gbdgacXaXc[itex]\nabla[/itex]d(ln [itex]\Omega[/itex])

    Since it is a null geodesic, gacXaXc = 0, so the equation reduces to:
    Xa[itex]\nabla[/itex]aXb = 2XaXb[itex]\nabla[/itex]a(ln [itex]\Omega[/itex])

    The non-affine geodesic equation is Xa[itex]\nabla[/itex]aXb = [itex]\lambda[/itex]Xb, so I just define [itex]\lambda[/itex] = 2Xa[itex]\nabla[/itex]a(ln [itex]\Omega[/itex]), and the geodesic equation is satisfied. This is why the Inverno book gave the hint that both equations need not be affinely parametized.
     
    Last edited: Sep 30, 2011
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