S.Carrol Exercise G.10: Proving Conformal Killing Vector

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To prove that a Killing vector k^a for the metric g_{ab} becomes a conformal Killing vector for the metric \overline{g}_{ab} under a conformal transformation, one must show that \overline{\nabla}_{a}k_{b} + \overline{\nabla}_{b}k_{a} equals (k^{r}\nabla_{r}a(x))\overline{g}_{ab}. The attempt involves substituting the relation for the covariant derivative of k_b in terms of the connection coefficients between g_{ab} and \overline{g}_{ab}. However, the resulting equation does not match the conformal Killing equation, leading to confusion. Clarification on the correct application of the Killing and conformal equations is needed to resolve the issue. The discussion emphasizes the importance of proper mathematical manipulation in proving the relationship between the metrics.
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1. The problem statement

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}

Homework Equations



killing equation
killing conformal equation

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_{c}

where C^{r}_{ab} are the connection coefficients relating the derivative operatrors for g_{ab} and \overline{g}_{ab}

i sustitute this in \overline{\nabla}_{a}k_{b}+\overline{\nabla}_{b}k_{a}

and using 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)+g_{ab}k^{r}\nabla_{r}a(x)

which is not the conformal killing equation so I am lost , can anyone help me on this?
 
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