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2. ## Re: Geodesics

let $\displaystyle \gamma(u,t)$ be a smooth map from a rectangle range $\displaystyle [0,1]\times[a,b]$ of $\displaystyle R^2$ to M. And denote
$\displaystyle U=\gamma_u$ and $\displaystyle T=\gamma_t$ as the coordinate vectors
For each u define
$\displaystyle L(u)=\frac{1}{p}\int_a^b \langle T, T \rangle^{p/2} dt$
Then $\displaystyle \frac{dL}{du}=\frac{1}{p}\int_a^b \frac{\partial}{\partial u} \langle T, T \rangle^{p/2} dt$
=$\displaystyle \frac{1}{p}\int_a^b \frac{p}{2}\langle T, T \rangle^{p/2-1}\frac{\partial}{\partial u} \langle T, T \rangle dt$
=$\displaystyle \frac{1}{p}\int_a^b \frac{p}{2}\langle T, T \rangle^{p/2-1} 2 \langle D_U T, T \rangle dt$
=$\displaystyle \int_a^b \langle T, T \rangle^{p/2-1} \langle D_U T, T \rangle dt$
And since $\displaystyle [T,U]=0$ we have $\displaystyle D_U T = D_T U$ hence
$\displaystyle \langle D_U T, T \rangle = \langle D_T U, T \rangle = T\langle U, T \rangle - \langle U, D_T T \rangle$, hence
$\displaystyle \frac{dL}{du}=\int_a^b \langle T, T \rangle^{p/2-1} (T\langle U, T \rangle - \langle U, D_T T \rangle) dt$
When u=0 the curve $\displaystyle \gamma_0(t)$ has unit speed so $\displaystyle \langle T, T \rangle=1$ we have
$\displaystyle \frac{dL}{du} |_{u=0} = \langle U, T \rangle |_a^b - \int_a^b \langle U, D_T T \rangle dt$
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.