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Deriving equation of motion in General Relativity from Principle of Least Action.

  1. Dec 4, 2011 #1
    I'm going through Landau/Lifgarbagez's book II of theoretical physics. In it they have a derivation of the equation of motion from the principle of least action, however I don't understand one step.

    1. The problem statement, all variables and given/known data
    Derive the equation of motion:
    [itex]\frac{d^2x^i}{ds^2}+\Gamma^i_{kj} \frac{dx^k}{ds} \frac{dx^j}{ds}=0[/itex]
    Using the principle of least action:
    [itex]\delta S=-mc\delta\int ds=0[/itex]


    2. Relevant equations

    [itex]\Gamma_{i,kj}=\frac{1}{2}\left(\frac{\partial g_{ik}}{\partial x^j}+\frac{\partial g_{ij}}{\partial x^k}-\frac{\partial g_{kj}}{\partial x^i}\right)[/itex]

    3. The attempt at a solution

    [itex]\delta ds^2=2ds\delta ds = \delta(g_{ik}dx^i dx^k)=dx^i dx^k \frac{\partial g_{ik}}{\partial x^j}\delta x^j + 2g_{ik}dx^i d\delta x^k[/itex]

    Therefore

    [itex]\delta S = -mc\int\left\{\frac{1}{2}\frac{dx^i}{ds}\frac{dx^k}{ds}\frac{\partial g_{ik}}{\partial x^j}\delta x^j + g_{ik}\frac{dx^i}{ds}\frac{d\delta x^k}{ds}\right\}[/itex]

    which equals

    [itex]\delta S = -mc \int \left\{\frac{1}{2} \frac{dx^i}{ds} \frac{dx^k}{ds} \frac{\partial g_{ik}}{\partial x^j}\delta x^j - \frac{d}{ds}\left\{ g_{ik} \frac{x^i}{ds}\right\} \delta x^k \right\} ds[/itex]

    The step I don't understand is going from the second last line to the last line.
    Thanks
     
  2. jcsd
  3. Dec 4, 2011 #2

    dextercioby

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    Science Advisor
    Homework Helper

    Are you sure this is ok ? You missed the d there. Ok, he takes out d/ds for the whole term (this gives 0 upon integration, because of the limit/boundary conditions), then by partial integration he gets that term with -. This is a standard trick when deriving Euler-Lagrange eqns from the action functional.
     
  4. Dec 4, 2011 #3
    Thanks, that has made it clear for me.
     
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