Conformal Transformation of the metric (General Relativity)

[chronnox]

Homework 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 $$\overline{\nabla}_{a}k_{b}=\nabla_{a}k_{b}-C^{r}_{ab}k_{r}$$

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 I am lost , can anyone help me on this?

 

[Jhero]

Hi there,

Your approach is on the right track, but there are a few mistakes in your calculations. Let me walk you through the correct solution:

First, let's start with the killing equation for the metric g_{ab}:

\nabla_{a}k_{b}+\nabla_{b}k_{a}=0

Next, we can use the relation for the conformal transformation of the connection coefficients:

C^{r}_{ab}=\omega^{-1}(\delta^{r}_{a}\nabla_{b}\omega+\delta^{r}_{b}\nabla_{a}\omega-g_{ab}g^{rc}\nabla_{c}\omega)

Substituting this into the equation for \overline{\nabla}_{a}k_{b}+\overline{\nabla}_{b}k_ {a}, we get:

\overline{\nabla}_{a}k_{b}+\overline{\nabla}_{b}k_{a}=\nabla_{a}k_{b}-C^{r}_{ab}k_{r}+\nabla_{b}k_{a}-C^{r}_{ba}k_{r}

Notice that the first two terms cancel out by the killing equation for g_{ab}. So we are left with:

\overline{\nabla}_{a}k_{b}+\overline{\nabla}_{b}k_{a}=-C^{r}_{ba}k_{r}

Next, we can use the relation for the conformal transformation of the metric:

\overline{g}_{ab}=\omega^{2}g_{ab}

This gives us the following relation for the connection coefficients:

C^{r}_{ab}=\frac{2}{\omega}(\delta^{r}_{a}\nabla_{b}\omega+\delta^{r}_{b}\nabla_{a}\omega-g_{ab}g^{rc}\nabla_{c}\omega)

Substituting this into the equation for \overline{\nabla}_{a}k_{b}+\overline{\nabla}_{b}k_ {a}, we get:

\overline{\nabla}_{a}k_{b}+\overline{\nabla}_{b}k_{a}=-\frac{2}{\omega}(\delta^{r}_{b