I'm trying to prove that if the first variation of length vanishes then the curve $\gamma$ must be an affinely parameterised geodesic. In the following $T=\dot{\gamma}$.
So I've attacked the contrapositive, assuming that $\gamma$ is not an affinely parameterised geodesic. I've got the stage where
$$L'(s)\leq \int_{t_1}^{t_2} -g(T,T)^{-1/2}||\nabla_TT-(\nabla_T.T)/||T||^2||^2 dt$$
with $\nabla_TT\neq 0$ in the region of integration. I'd like to conclude that this is strictly negative, which is immediate if $\nabla_TT\not\propto T$. I can't seem to show this though. Does anyone have any ideas how to proceed?
If you want to see the early part of the argument look at p32 here. Cheers!
I've realised that what I was trying to prove is in fact false! The stipulation that $L'(s)=0$ is not strong enough to guarantee an affinely parameterised geodesic. The correct argument is as follows.
In the contrapositive one must assume that $\gamma$ is not a general geodesic, that is $\nabla_TT\neq T$ in some region $(t_1,t_2)$. Then when we get to the crux point in the question above we can immediately conclude that $L'(s)<0$ as required.