Conformally related metrics have the same null geodesics

[etotheipi]

Homework Statement:: i) If $\bar{g} = \Omega^2 g$ for some positive function $\Omega$, show that $\bar{g}$ and $g$ have the same null geodesics.

ii) Let $\psi$ solve $g^{ab} \nabla_a \nabla_b \psi + \xi R \psi = 0$. Determine $\xi$ such that $\bar{\psi} = \Omega^p \psi$ for some $p$ solves the equation in a spacetime with metric $\bar{g} = \Omega^2 g$ if $\psi$ solves the equation in a spacetime with metric $g$.
Relevant Equations:: N/A

A conformal transformation doesn't change the null cones of the metric, so if $n^a$ is a null vector of $g$ then it is also a null vector of $\bar{g}$. So it's necessary to show that $n^a \bar{\nabla}_a n^b = 0 \implies n^a \nabla_a n^b = \alpha n^b$ where $\nabla_a$ and $\bar{\nabla}_a$ are the covariant derivative operators adapted to $g$ and $\bar{g}$ respectively. We have\begin{align*}

\bar{\Gamma}^i_{kl} &= \frac{1}{2} \bar{g}^{im} (\partial_l \bar{g}_{mk} + \partial_k \bar{g}_{ml} - \partial_m \bar{g}_{kl}) \\

&= \frac{1}{2} \Omega^{-2} g^{im}(\Omega^2 \left[ \partial_l g_{mk} + \partial_k g_{ml} - \partial_m g_{kl}\right] + 2\Omega [g_{mk} \partial_l \Omega + g_{ml} \partial_k \Omega - g_{kl} \partial_m \Omega]) \\

&= \Gamma^{i}_{kl} + \Omega^{-1} (\delta^i_k \partial_l \Omega + \delta^i_l \partial_k \Omega - g^{im} g_{kl} \partial_m \Omega)

\end{align*}It follows that\begin{align*}

n^a \bar{\nabla}_a n^b &= n^a (\partial_a n^b + \bar{\Gamma}^b_{ac} n^c) \\

&= n^a\left( \partial_a n^b + \Gamma^b_{ac} n^c \right) + n^a n^c \Omega^{-1} \left( \delta^b_a \partial_c \Omega + \delta^b_c \partial_a \Omega - g^{bm}g_{ac} \partial_m \Omega \right) \\

&= n^a \nabla_a n^b + n^a n^c \Omega^{-1} \left( 2 \delta^b_a \partial_c \Omega - g^{bm}g_{ac} \partial_m \Omega \right)

\end{align*}I can't see how to tidy up the right hand side; a hint would be appreciated. Thanks!

 

[Orodruin]

The last term in the parenthesis of the last expression becomes zero when contracted with $n^a n^c$ due to $n$ being a null vector. The other term in the parenthesis is proportional to $n^b$ (also when contracted with $n^a n^c$).

Edit: You'll obtain
$$
n^a \bar\nabla_a n^b = n^a \nabla_a n^b + n^b n^a \partial_a \ln(\Omega^2).
$$

 

[etotheipi]

Thanks! I see, $n^b n^a \partial_a \mathrm{ln}(\Omega^2) = 2 \Omega^{-1} n^b n^a \partial_a \Omega =2\Omega^{-1} n^c n^a \delta^b_c \partial_a \Omega$ which under the replacement $a \leftrightarrow c$ gives the first term in my OP. And as you said the second contains $n^a n^c g_{ac} = n^a n_a = 0$ since $n$ is null, so vanishes. And that's it, because we can identify the scalar $\alpha = n^a \partial_a \mathrm{ln}(\Omega^2)$.

I'll have a go at part ii) probably tomorrow.