Two metrics $ds^2 = g_{\mu\nu} dx^\mu dx^\nu$ and $d\tilde s^2 = \tilde g_{\mu\nu} dx^\mu dx^\nu$ in the same coordinate system are said to be conformally related if ̃
$\tilde g_{\mu\nu} (x) = \psi(x)g_{\mu\nu}(x)$
for some positive function $\psi(x)$ of the coordinates.
I am trying to prove that if the curve $x^\mu(\lambda)$ is a null geodesic of the metric $ g_{\mu\nu}$ , it is also a null geodesic of the metric ̃$\tilde g_{\mu\nu}$ . (By this it's meant the same physical curve in the spacetime manifold M , with an affine curve parameter which is a suitable function of $\psi$, in general not coinciding with $\lambda$.) That is, conformal spaces share their null geodesics.
The following fact can be used:
whenever a curve $x^\mu(\omega)$ satisfies:
$\frac{d^2x^\mu}{d\omega^2}+\Gamma^\mu_{\nu\lambda}\frac{dx^\nu}{d\omega}\frac{dx^\lambda}{d\omega}=\phi \frac{dx^\mu}{d\omega}$
for some function $\phi$ on $M$ , it also describes a geodesic (obviously, in a non-affine parametrization) in the sense that there is a curve parameter $s = s(\omega)$, depending on the function $φ$, such that $x^\mu(\omega(s))$ viewed as function of $s$ satisfies the usual geodesic equation.
I have not been able to get to anything meaningful, how should I do this?
Let us write $\psi = e^{2f}$, so that $\tilde{g} = e^{2f}g$. Let $\gamma$ be a null-geodesic for $g$. This means that $g(\gamma',\gamma') = 0$ and $\nabla_{\gamma'}\gamma' = 0$.
By the formula derived in this answer, we have \begin{align} \tilde{\nabla}_{\gamma'}\gamma' &= \nabla_{\gamma'}\gamma' + df(\gamma') \gamma' + df(\gamma') \gamma' + g(\gamma',\gamma') \mathrm{grad}(f) \\ &= 0 + 2 df(\gamma') \gamma' +0 \\ &= 2df(\gamma') \gamma'. \end{align} Therefore, $\tilde{\nabla}_{\gamma'}{\gamma'} = 0$ if and only if $\gamma'$ is everywhere in the kernel of $df$. Hence, for a generic conformal metric $\tilde{g} = e^{2f}g$, $\gamma$ is not a geodesic for $\tilde{g}$, even though it is a null curve.
However, we can reparametrize $\gamma$ in order to produce a null-geodesic for $\tilde{g}$. Here is the trick: let $\varphi$ be any reparametrization of the interval of definition of $\gamma$ and consider $\tilde{\gamma} =\gamma\circ \varphi$. Then $\tilde{\gamma}' = \varphi'\cdot(\gamma'\circ \varphi)$. It is already a null-curve. Let us derive an equation on $\varphi$ in order for $\tilde{\gamma}$ to be a geodesic.
By the linearity of the Levi-Civita connection in the lower entry, it is sufficient that $\varphi$ is such that $$ \tilde{\nabla}_{\gamma'\circ \varphi} (\varphi'\cdot (\gamma'\circ \varphi)) = 0. $$ Note that \begin{align} \tilde{\nabla}_{\gamma'\circ \varphi} (\varphi'\cdot (\gamma'\circ \varphi)) &= \varphi''\cdot (\gamma'\circ \varphi) + \varphi' \tilde{\nabla}_{\gamma\circ\varphi}(\gamma \circ \varphi)\\ &= \varphi'' \cdot (\gamma'\circ \varphi) + \varphi' \cdot 2df(\gamma'\circ\varphi) (\gamma'\circ \varphi) \\ &= (\varphi'' + 2df(\gamma'\circ\varphi)\varphi')(\gamma'\circ \varphi), \end{align} so that it is sufficient to solve the differential equation $\varphi'' + 2df(\gamma'\circ \varphi) \varphi'= 0$, which can be done easily in coordinates.