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Linearized Einstein Field Equations

  1. Sep 27, 2012 #1
    Hi everyone,

    Say that one can separate the metric of a space time in a background metric and a small perturbation such that [itex]g_{\alpha \beta}=g'_{\alpha \beta}+h_{\alpha \beta}[/itex], where [itex]g'_{\alpha \beta}[/itex] is the background metric and [itex]h_{\alpha \beta}[/itex] the perturbation.

    Computing the christoffel symbols one would get, to first order in the perturbation: [tex]\Gamma^\alpha_{\beta \gamma}=\Gamma'^\alpha_{\beta \gamma}+\frac{1}{2}(h^{\alpha}_{\beta,\gamma}+h^{ \alpha }_{\gamma,\beta}-h_{\beta \gamma}\hspace{.2mm}^{,\alpha}),[/tex]

    Then why, in this reference, in the text right after Eq.19.23, [itex]C^\alpha_{\beta \gamma}=\frac{1}{2}(h^{\alpha}_{\beta;\gamma}+h^{ \alpha }_{\gamma;\beta}-h_{\beta \gamma}\hspace{.2mm}^{;\alpha})[/itex], is written with covariant derivatives?

    Thank you
    Last edited: Sep 27, 2012
  2. jcsd
  3. Sep 27, 2012 #2


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    A difference between two connections is a tensor, which can be checked by explicitly writing down the transformation of this difference. Hence you'll need covariant derivatives, not partial derivatives. Of course, these covariant derivatives should follow from your definition of the connection and your C.

    So that is something which you should do first. Second, you should be very careful with lowering and raising indices underneath partial derivatives.
  4. Sep 27, 2012 #3
    Yes, you're absolutely right, at both things. I didn't care about the partial derivative when I raised my indexes and indeed I was being naive in the definition of C.

    Thank you very much you were a life (brain) saver!
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