# Unparametrized lightlike geodesics are invariant under conformal change. How does the parametrization change?

Let $(M,g)$ be a pseudo-Riemannian manifold and $\gamma: I \rightarrow M$ be a lightlike geodesic with $0 \in I$. Let $\hat{g}=e^{2 \sigma} g$ be a conformally equivalent metric. Maybe having to make $I$ a bit smaller, say $\tilde{I} \subset I$, there exists some change of coordinates $\tau: J \rightarrow \tilde{I}$, such that $\gamma \circ \tau$ is a geodesic with respect to $\hat{g}$ and has the same 1-jet, i.e. $\gamma(0)=\gamma \circ \tau (0)$ and $\gamma'(0)=(\gamma \circ \tau)' (0)$.

Question: What reparametrizations can possibly occur here?

Suggested answer: All intervals $J \subset \mathbb{R}$ with $0 \in J$ and all smooth, strictly monotonously increasing maps $\tau: J \rightarrow I$ can appear as a change of coordinates.

Let $\nabla$ and $\hat{\nabla}$ be the Levi-Civita connections for $g$ and $\hat{g}$ respectively. We have the formula

$\hat{\nabla}_X Y = \nabla_X Y + d\sigma (X) Y+ d\sigma (Y) X- g(X,Y) grad \sigma$.

This implies $$\begin{split} &\hat{\nabla}_{(\gamma \circ \tau)'(s)} (\gamma \circ \tau)' \\&=\nabla_{(\gamma \circ \tau)'(s)} (\gamma \circ \tau)' + 2 d \sigma((\gamma \circ \tau)'(s)) \cdot (\gamma \circ \tau)'(s) - \| (\gamma \circ \tau)'(s) \| grad \sigma \\&=\nabla_{(\gamma \circ \tau)'(s)} (\gamma \circ \tau)' + 2 d \sigma((\gamma \circ \tau)'(s)) \cdot (\gamma \circ \tau)'(s) \\&=\frac{\tau ''(s)}{\tau'(s)} \cdot (\gamma \circ \tau)'(s) + 2 d \sigma((\gamma \circ \tau)'(s)) \cdot (\gamma \circ \tau)'(s). \end{split}$$

$\gamma \circ \tau$ is a geodesic if and only if above term vanishes, i.e. if and only if $\frac{\tau''(s)}{\tau'(s)}+2 d \sigma ( (\gamma \circ \tau)'(s))=0$ for all $s \in J$. This is a second-order ODE. Because the the reparametrization should leave the 1-jet invariant, we have the initial condition $\tau(0)=0$ and $\tau'(0)=1$. By the Picard-Lindelöf theorem a locally unique solution $\tau : J \rightarrow \tilde{I}$ exists, maybe making $\tilde{I}$ again a bit smaller. The solution satisfies

$\tau(s)=\int_0^s e^{-2 \sigma(\gamma \circ \tau (s))} ds$.

We made $\tilde{I}$ sufficiently small, so that $\gamma$ is injective. Hence we may (smoothly) choose arbitrary positive values for $\sigma$ along $\gamma$.

Can we receive any possible monotonously increasing (smooth) function $\tau$ through that? I looks like it, but I cannot put my finger on it.

## 1 Answer

Yes, any smooth reparametrization can be obtained by a smooth conformal change of the metric, provided that the geodesic does not intersect itself. Assuming no self-intersections, the question is whether one can choose $\alpha:=\sigma\circ\gamma$ (the conformal factor along the geodesic) so that $$\frac{\tau''(s)}{\tau'(s)}+2 d \sigma ( (\gamma \circ \tau)'(s))=0.$$ This can be rewritten as $$\frac{d}{ds}[\log(\tau'(s))+2(\alpha \circ \tau)(s)]=0,$$ so that $\log(\tau'(s))+2(\alpha \circ \tau)(s)$ is a constant $c$. Evaluating at $s=0$ gives $$c = \log(\tau'(0))+2(\alpha \circ \tau)(0) = 2\alpha(0).$$ Now that we have $$\log(\tau'(s))+2(\alpha \circ \tau)(s)=2\alpha(0)$$ for all $s$, we can plug in $s=\tau^{-1}(t)$ to get $$\alpha(t)=\alpha(0)-\frac12\log(\tau'(\tau^{-1}(t))).$$ Given any change of parameter $\tau$ and any initial $\alpha(0)$, this gives a method to find the corresponding $\alpha$.