# Conformally related metrics have the same null geodesics

• I
• etotheipi

#### 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!

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
$$n^a \bar\nabla_a n^b = n^a \nabla_a n^b + n^b n^a \partial_a \ln(\Omega^2).$$

• strangerep and 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. 