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Quick expression on geodesic equation

  1. Apr 7, 2015 #1
    Taken from Hobson's book:


    How did they get this form?

    [tex] \dot u^{\mu} = - \Gamma_{v\sigma}^\mu u^v u^\sigma [/tex]
    [tex] \dot u^{\mu} g_{\mu \beta} \delta_\mu ^\beta = - g_{\mu \beta} \delta_\mu ^\beta \Gamma_{v\sigma}^\mu u^v u^\sigma [/tex]
    [tex] \dot u_{\mu} = - \frac{1}{2} g_{\mu \beta} \delta_\mu ^\beta g^{\mu \gamma} \left(\partial_v g_{\sigma \gamma} + \partial_\sigma g_{v\gamma} - \partial_\gamma g_{v\sigma} \right) [/tex]
    [tex] = - \frac{1}{2} \delta_\mu ^\beta \delta_\beta ^\gamma \left(\partial_v g_{\sigma \gamma} + \partial_\sigma g_{v\gamma} - \partial_\gamma g_{v\sigma} \right) [/tex]
    [tex] = -\frac{1}{2} \left( \partial_v g_{\sigma \mu} + \partial_\sigma g_{v\mu} - \partial_\mu g_{v\sigma} \right) [/tex]

    Why are the first two terms zero?
  2. jcsd
  3. Apr 7, 2015 #2


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    Staff Emeritus
    Science Advisor

    Since [itex]u_\mu = g_{\mu \nu} u^\nu[/itex], then
    [itex]\dot{u}_\mu = \dot{g}_{\mu \nu} u^\nu + g_{\mu \nu} \dot{u}^\nu[/itex]

    Using the chain rule, or whatever the rule is, [itex]\dot{g}_{\mu \nu} = u^\sigma \partial_\sigma g_{\mu \nu}[/itex]. So we have:

    [itex]\dot{u}_\mu = u^\sigma u^\nu \partial_\sigma g_{\mu \nu} + g_{\mu \nu} \dot{u}^\nu[/itex]

    Using [itex]\dot{u}^\nu = - \Gamma^\nu_{\sigma \lambda} u^\sigma u^\lambda[/itex], we have:

    [itex]\dot{u}_\mu = u^\sigma u^\nu \partial_\sigma g_{\mu \nu} - g_{\mu \nu} \Gamma^\nu_{\sigma \lambda} u^\sigma u^\lambda[/itex]

    Finally, using [itex] g_{\mu \nu} \Gamma^\nu_{\sigma \lambda} = \Gamma_{\mu \sigma \lambda} = \frac{1}{2}(\partial_\sigma g_{\mu \lambda} + \partial_\lambda g_{\mu \sigma} - \partial_\mu g_{\sigma \lambda})[/itex], we can write this as:

    [itex]\dot{u}_\mu = u^\sigma u^\nu \partial_\sigma g_{\mu \nu} - \frac{1}{2} u^\sigma u^\lambda (\partial_\sigma g_{\mu \lambda} + \partial_\lambda g_{\mu \sigma} - \partial_\mu g_{\sigma \lambda}) [/itex]

    After some cancellations, you get their result.
  4. Apr 7, 2015 #3
    Brilliant, thank you!
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