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Metric Connection from Geodesic Equation

  1. Mar 31, 2012 #1
    For the following two-dimensional metric

    [tex]ds^2 = a^2(d\theta^2 + \sin^2{\theta}d\phi^2)[/tex]

    using the Euler-Lagrange equations reveal the following equations of motion

    [tex]\ddot{\phi} + 2\frac{\cos{\theta}}{\sin{\theta}}\dot{\theta}\dot{\phi} = 0[/tex]
    [tex]\ddot{\theta} - \sin{\theta}\cos{\theta}\dot{\phi}^2 = 0[/tex]

    Using the general geodesic equation form [itex]\ddot{x}^{\alpha} + \Gamma^{\alpha}_{\beta\gamma}\dot{x}^\beta\dot{x}^{\gamma}=0[/itex], we infer that the equations derived describe geodesics. This shows that the only non-zero terms of the metric connection [itex]\Gamma^{\alpha}_{\beta\gamma}[/itex] are

    [tex]\Gamma^{\theta}_{\phi\phi} = -\sin{\theta}\cos{\theta},\quad \Gamma^{\phi}_{\theta\phi} = \Gamma^{\phi}_{\phi\theta} = \frac{\cos{\theta}}{\sin{\theta}}[/tex]

    My problem is comprehending where the factor of two has gone for the [itex]\Gamma^{\phi}_{\phi\theta}[/itex] term. Is it due to fact that it's a coefficient of a mixed derivative, why is that?
     
  2. jcsd
  3. Mar 31, 2012 #2

    tom.stoer

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    Science Advisor

    You have to sum over theta and phi, so the geodesic equation contains the two terms

    [tex]\Gamma^{\phi}_{\theta\phi}\dot{\theta}\dot{\phi} + \Gamma^{\phi}_{\phi\theta}\dot{\phi}\dot{\theta}[/tex]
     
  4. Mar 31, 2012 #3
    Of course! Thank you, it's very clear now.
     
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