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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!

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up vote 2 down vote accepted

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.

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The standard way around this issue (e.g., Wikipedia) is to minimize the energy $\int \langle \dot \gamma, \dot\gamma \rangle$ instead of the length $\int \sqrt{\langle \dot \gamma, \dot\gamma \rangle}$. The energy functional is strictly convex, hence minimizer is unique. It turns out to be an affinely parametrized geodesic. The absence of square root also helps with differentiability. – user53153 Jan 15 '13 at 18:15

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