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Linearized gravity

  1. Nov 1, 2009 #1
    Why and how does from g[tex]_{ab}[/tex]=[tex]\eta_{ab}[/tex]+h[tex]_{ab}[/tex] follow
    g[tex]^{ab}[/tex]=[tex]\eta^{ab}[/tex]-h[tex]^{ab}[/tex]?

    Don't get the latex right, first equation all indices are supposed to below, second equation all are supposed to be up.

    thanks
     
  2. jcsd
  3. Nov 1, 2009 #2
    I think that you will arrive soon at this property.

    If you work with GR, such a metric splitting does not mean introducing a flat space-time since the R is not zero whatever variable change you do.
     
  4. Nov 1, 2009 #3

    haushofer

    User Avatar
    Science Advisor

    Because then

    [tex]
    g^{ab}g_{bc} = \delta^a_c
    [/tex]

    up to first order. Maybe it helps to write

    [tex]
    g_{ab} = \eta_{ab} + \epsilon h_{ab}
    [/tex]

    then we would like that [itex]g_{ab}g^{bc}=\delta_a^c[/itex] . So up to first order in epsilon you can easily see that [itex]g^{bc}=\eta^{bc} - \epsilon h^{bc}[/itex]
    does the job for you:

    [tex]
    g_{ab}g^{bc} = ( \eta_{ab} + \epsilon h_{ab}) (\eta^{bc} - \epsilon h^{bc})
    = \delta_a^c + \epsilon(h_{ab}\eta^{bc} - h^{bc}\eta_{ab}) + O(\epsilon^2)
    [/tex]

    Now, you can raise the indices of h with eta because we are working at linear order in epsilon; every correction to it would give higher order epsilon terms. So we get

    [tex]
    g_{ab}g^{bc} = \delta_a^c + \epsilon (h_a^c - h_a^c) + O(\epsilon^2) = \delta_a^c + O(\epsilon^2)
    [/tex]

    You could say that the minus-sign in [itex]g^{ab}[/itex] is for cancelling the two factors linear in epsilon in order to get [itex]g_{ab}g^{bc}=\delta_a^c[/itex]. Ofcourse, if you go up higher in order epsilon, say epsilon squared, then all the terms quadratic in epsilon have to cancel to maintain the identity [itex]g_{ab}g^{bc}=\delta_a^c[/itex].
     
  5. Nov 1, 2009 #4
    thanks so much, haushofer! crystal-clear now
     
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