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Homework Help: Conformal Transformation of the metric (General Relativity)

  1. Nov 8, 2007 #1
    1. The problem statement, all variables and given/known data
    I need to prove that if two metrics are related by an overall conformal transformation of the form [tex]\overline{g}_{ab}=e^{a(x)}g_{ab}[/tex] and if [tex]k^{a}[/tex] is a killing vector for the metric [tex]g_{ab}[/tex] then [tex]k^{a}[/tex] is a conformal killing vector for the metric [tex]\overline{g}_{ab}[/tex]

    2. Relevant equations

    killing equation
    killing conformal equation

    3. The attempt at a solution

    i think i need to show that
    [tex]\overline{\nabla}_{a}k_{b}+\overline{\nabla}_{b}k_ {a}=(k^{r}\nabla_{r}a(x))\overline{g}_{ab}[/tex]

    which as far as i understand is the killing conformal equation for the metric [tex]\overline{g}_{ab}[/tex]

    so using the relation [tex]\overline{\nabla}_{a}k_{b}=\nabla_{a}k_{b}-C^{r}_{ab}k_{r}[/tex]

    where [tex]C^{r}_{ab}[/tex] are the connection coefficients for the conformal transformation, i.e., if [tex]\overline{g}_{ab}=\omega^{2}g_{ab}[/tex] then:

    [tex]C^{r}_{ab}=\omega^{-1}(\delta^{r}_{a}\nabla_{b}\omega+\delta^{r}_{b}\nabla_{a}\omega-g_{ab}g^{rc}\nabla_{c}\omega)[/tex] if i substitute this in [tex]\overline{\nabla}_{a}k_{b}+\overline{\nabla}_{b}k_ {a}[/tex]

    and use killing equation for the metric [tex]g_{ab}[/tex] i obtain:


    which is not the conformal killing equation for [tex]\overline{g}_{ab}[/tex] so im lost , can anyone help me on this?
  2. jcsd
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