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Math Help - Geodesics

  1. #1
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    Geodesics

    .
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  2. #2
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    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.
    Thanks from Plato13
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