.

2. ## Re: Geodesics

let $\gamma(u,t)$ be a smooth map from a rectangle range $[0,1]\times[a,b]$ of $R^2$ to M. And denote
$U=\gamma_u$ and $T=\gamma_t$ as the coordinate vectors
For each u define
$L(u)=\frac{1}{p}\int_a^b \langle T, T \rangle^{p/2} dt$
Then $\frac{dL}{du}=\frac{1}{p}\int_a^b \frac{\partial}{\partial u} \langle T, T \rangle^{p/2} dt$
= $\frac{1}{p}\int_a^b \frac{p}{2}\langle T, T \rangle^{p/2-1}\frac{\partial}{\partial u} \langle T, T \rangle dt$
= $\frac{1}{p}\int_a^b \frac{p}{2}\langle T, T \rangle^{p/2-1} 2 \langle D_U T, T \rangle dt$
= $\int_a^b \langle T, T \rangle^{p/2-1} \langle D_U T, T \rangle dt$
And since $[T,U]=0$ we have $D_U T = D_T U$ hence
$\langle D_U T, T \rangle = \langle D_T U, T \rangle = T\langle U, T \rangle - \langle U, D_T T \rangle$, hence
$\frac{dL}{du}=\int_a^b \langle T, T \rangle^{p/2-1} (T\langle U, T \rangle - \langle U, D_T T \rangle) dt$
When u=0 the curve $\gamma_0(t)$ has unit speed so $\langle T, T \rangle=1$ we have
$\frac{dL}{du} |_{u=0} = \langle U, T \rangle |_a^b - \int_a^b \langle U, D_T T \rangle dt$
Note the final result has nothing to do with p, so for any p, the energy function achieves extreme when the curve is a geodesic.