# Affine parameterization of null geodesics

How does one find an affine parameter for a null geodesic? I found this advice on planetmath.org:

Take s as an arbitrary parameter;

Set $$u^\mu=\frac{dx^\mu}{ds}$$ Then $$u^\mu \nabla_\mu u^\nu = f(s)u^\nu$$

If the RHS is zero, s is affine; if not, s is not affine. Does this make sense? It strikes me that to find an affine parameter all you need is the equations of the curve $x^\mu (s)$ and their derivatives $u^\mu$. Is this true? I was wondering if we need to know the metric tensor $g$ also? I thought one needs to write the geodesic equation $$\frac{d^2 x^\mu}{ds^2}+\Gamma^\mu_{\nu \sigma}\frac{dx^\nu}{ds}\frac{dx^\sigma}{ds}=0$$ and check if its RHS is zero or not. But then we also need to know the $\Gamma$s or we need to know the metric tensor $g$ if we rewrite the above geodesic equation in terms of $g$ instead of $\Gamma$s.

## 1 Answer

What you are missing is that $\nabla_\mu$ is the covariant derivative (aka Levi-Civita connection) associated to the metric $g$. It already encodes in it the connection coefficients. So hidden in the notation $\nabla_\mu$ is the knowledge on the metric you need.

That said, if you know that $x^\mu(s)$ is a geodesic curve, and you know the tangent vectors $u^\mu(s)$, you can actually find an affine re-parametrisation without too much of the underlying geometry. What you are looking for is a function $s = s(t)$ such that $x^\mu(s(t))$ is affinely parametrized by $t$. If you consider its inverse function $t = t(s)$, and let $v^\mu$ be the tangent vector relative to the parameter $t$, you can by chain rule reason that

$$v^\mu(t(s)) t'(s) = u^\mu(s)$$

Inserting this into the equation you have

$$t'(s)^2 (v^\mu\nabla_\mu v^\nu)(t(s)) + t''(s) v^\mu(t(s)) = f(s) t'(s) v^\mu(t(s))$$

By assumption $t$ is an affine parametrisation, so the transport of $v$ by itself vanishes. So you end up having to solve the second order ODE

$$t'' = t' f \implies \log (t') ' = f \implies t' = \exp \int f \implies t = \int \exp \int f$$

So... while the geodesic equation itself must encode geometric data (the connection coefficients), the algorithm for finding an affine parameter when given an arbitrarily parametrized geodesic curve only involves solving an ODE and the geometric data can be conveniently forgotten.

• I'm a bit confused, I thought that if a curve is parametrized as $u\mapsto \gamma(u)$, then $u$ is said to be an affine parameter if $g_{\gamma(u)}(\dot{\gamma}(u),\dot{\gamma}(u))=\mathrm{constant}$. At least this is the case in the Riemannian setting. In the Lorentzian setting, if we have a null curve $\gamma$ then $g_{\gamma(u)}(\dot{\gamma}(u),\dot{\gamma}(u))=0$. So by this definition any parametrization is affine. – Aerinmund Fagelson Jan 31 '18 at 12:03
• @AerinmundFagelson: this is one place where your Riemannian "definition" fails. In Riemannian geometry, the condition $g(\dot{\gamma}, \dot{\gamma})$ being constant along a geodesic is equivalent to $\nabla_{\dot{\gamma}} \dot{\gamma} = 0$. This equivalence is false for null curves in Lorentzian geometry. Hence affine parametrization in general pseudo-Riemannian geometry uses the more precise definition $\nabla_{\dot{\gamma}} \dot{\gamma} = 0$. (For non-affine parametrizations, you allow $\nabla_{\dot{\gamma}} \dot{\gamma} \propto \dot{\gamma}$ to define geodesics.) – Willie Wong Jan 31 '18 at 17:06