- #1
chronnox
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Homework Statement
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]
Homework Equations
killing equation
killing conformal equation
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:
[tex]\overline{\nabla}_{a}k_{b}+\overline{\nabla}_{b}k_{a}=-k_{a}\nabla_{b}a(x)-k_{b}\nabla_{a}a(x)+(k^{r}\nabla_{r}a(x))g_{ab}[/tex]
which is not the conformal killing equation for [tex]\overline{g}_{ab}[/tex] so I am lost , can anyone help me on this?