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Deriving geodesic equation

  1. Feb 9, 2014 #1
    I am trying to derive the geodesic equation by extremising the integral
    $$ \ell = \int d\tau $$
    Now after applying Euler-Lagrange equation, I finally get the following:
    $$ \ddot{x}^\tau + \Gamma^\tau_{\mu \nu} \dot{x}^\mu \dot{x}^\nu = \frac{1}{2} \dot{x}^\tau \frac{d}{ds} \ln \left| \dot{x}_\nu \dot{x}^\nu \right| $$
    where ## \dot{x}^\tau \equiv \frac{d x^\tau}{ds} ## and ##s## is a parameter. Now I get the geodesic equation if the right-hand side vanishes, and the only way that happens is if ## \dot{x}_\nu \dot{x}^\nu ## is constant. Now the question is why is it constant?
     
  2. jcsd
  3. Feb 9, 2014 #2

    stevendaryl

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    Well, [itex]s[/itex] is an arbitrary parameter. We are free to choose any parametrization we like. One particular parametrization, which is possible for slower-than-light geodesics, is to let

    [itex]ds = \sqrt{|dx^\mu dx_\mu|}[/itex]

    For this choice, [itex]|\frac{dx^\mu}{ds} \frac{dx_\mu}{ds}| = 1[/itex]
     
  4. Feb 9, 2014 #3

    Bill_K

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    The geodesic equation takes its usual form only when s is chosen to be an affine parameter. For a timelike geodesic this means s must be a linear function of proper time, s = aτ + b where a and b are constants. If s is not an affine parameter, the geodesic equation has the extra term you mentioned.
     
  5. Feb 9, 2014 #4
    Thanks Bill, Wikipedia too says a similar thing, but what exactly is an affine parameter?
     
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