Conformal Transformation of the metric (General Relativity)

1. Nov 8, 2007

chronnox

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 $$\overline{g}_{ab}=e^{a(x)}g_{ab}$$ and if $$k^{a}$$ is a killing vector for the metric $$g_{ab}$$ then $$k^{a}$$ is a conformal killing vector for the metric $$\overline{g}_{ab}$$

2. Relevant equations

killing equation
killing conformal equation

3. The attempt at a solution

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

which as far as i understand is the killing conformal equation for the metric $$\overline{g}_{ab}$$

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

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

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

and use killing equation for the metric $$g_{ab}$$ i obtain:

$$\overline{\nabla}_{a}k_{b}+\overline{\nabla}_{b}k_{a}=-k_{a}\nabla_{b}a(x)-k_{b}\nabla_{a}a(x)+(k^{r}\nabla_{r}a(x))g_{ab}$$

which is not the conformal killing equation for $$\overline{g}_{ab}$$ so im lost , can anyone help me on this?

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