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Weak field general relativity

  1. Mar 7, 2006 #1
    I was asked to do a problem about the gauge freedom one has in the weak field limit in general relativity. I am given a coordinate transformation [tex]x^{\mu '} = x^{\mu} - \zeta^{\mu}[/tex]. Now I have to show how the perturbation of the metric transforms under this coordinate transformation.

    The result should be:
    [tex] h_{\mu\nu}^{new} = h_{\mu\nu}^{old} - \zeta_{\mu ,\nu} - \zeta_{\nu ,\mu}[/tex]

    The most simple way to do this, i thought, was to just use the transformational law for tensors.

    so: [tex]h_{\mu ' \nu '} = \partial_{\mu '}x^{\mu} \partial_{\nu '}x^{\nu} h_{\mu \nu}[/tex]

    Using the fact that we can neglect the product of the derivatives of the small perturbation, I find that:

    [tex]h_{\mu ' \nu '} = h_{\mu\nu} - (\partial_{\mu '}\zeta^{\mu} + \partial_{\nu '}\zeta^{\nu})h_{\mu\nu}[/tex]

    From this point I don't know how to get rid of the metric factor in the second term, because if I can lose it, i think i would be able to come to the correct form that is wanted.

    Can someone give me some hints? :)
     
    Last edited: Mar 7, 2006
  2. jcsd
  3. Mar 8, 2006 #2

    dextercioby

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    It's more rigurous if you start with the Pauli-Fierz action and do the canonical analysis and determine the irreducible abelian infinitesimal gauge transformations for the P-F fields [itex] h_{\mu\nu} [/itex].

    Just a thought.

    Daniel.
     
  4. Mar 8, 2006 #3

    Galileo

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    If you start from [itex]g_{\mu \nu}=\eta_{\mu \nu}+h_{\mu \nu}[/itex],
    then the tranformation law gives:

    [tex]g'_{\mu \nu}=\partial_{\mu} x^{\alpha}\partial_{\nu} x^{\beta}g_{\alpha \beta}[/tex]
    The left hand sides gives [itex]g'_{\mu \nu}(x')\approx \eta_{\mu \nu}+h_{\mu \nu}'(x)[/itex] and you can work out the right hand side explicitly.
     
    Last edited: Mar 8, 2006
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