Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Christoffel symbols of Schwarzschild metric with Lagrangian

  1. Sep 4, 2016 #1
    So the Schwarzschild metric is given by

    ds2= -(1-2M/r)dt2 + (1-2M/r)-1dr2+r22+r2sin2θ dφ2

    and the Lagragian is ##{\frac{d}{dσ}}[{\frac{1}{L}}{\frac{dx^α}{dσ}}] + {\frac{∂L}{∂x^α}}=0##

    with L = dτ/dσ. So for each α=0,1,2,3 we have

    ##{\frac{d^2 x^1}{dτ^2}}=0## for Minkowski spacetime

    also the geadesic equation is ##{\frac{d^2 x^a}{dτ^2}}+Γ^a_{bc}{\frac{dx^b}{dτ}}{\frac{dx^c}{dτ}}=0##

    So what happens in the Schwarschild metric and how can I find the Christoffel symbols from the Schwarschild metric and the geodesic equations?
    Last edited: Sep 4, 2016
  2. jcsd
  3. Sep 4, 2016 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    For slower-than-light geodesics, you can use an effective Lagrangian (which is basically the square of the one you use):

    [itex]\mathcal{L} = \frac{1}{2} g_{\mu \nu} U^\mu U^\nu[/itex]

    where [itex]g_{\mu \nu}[/itex] is a component of the metric tensor, and [itex]U^\mu = \frac{dx^\mu}{d\sigma}[/itex]. This is slightly easier to work with than the one you are using (although it's a little work to see that they lead to the same equations of motion). The Lagrangian equations of motion for this lagrangian are:

    [itex]\frac{d}{ds} \frac{\partial \mathcal{L}}{\partial U^\mu} - \frac{\partial \mathcal{L}}{\partial x^\mu} = 0[/itex]

    This gives the following form of the geodesic equation:

    [itex]g_{\mu \nu} \frac{dU^\nu}{d\sigma} + [\frac{\partial g_{\mu \nu}}{\partial x^\lambda} - \frac{1}{2} \frac{\partial g_{\lambda \nu}}{\partial x^\mu}] U^\lambda U^\nu = 0[/itex]

    Unfortunately, the quantity in square brackets (call it [itex]Q_{\mu \nu \lambda}[/itex]) is not exactly equal to the Christoffel symbol [itex]\Gamma_{\mu \nu \lambda} \equiv g_{\mu \alpha} \Gamma^\alpha_{\nu \lambda} = \frac{1}{2}(\frac{\partial g_{\mu\nu}}{\partial x^\lambda} + \frac{\partial g_{\mu\lambda}}{\partial x^\nu}- \frac{\partial g_{\lambda \nu}}{\partial x^\mu})[/itex]. However, if you compute [itex]Q_{\mu \nu \lambda}[/itex], then you can find [itex]\Gamma_{\mu \nu \lambda}[/itex] by symmetrizing:

    [itex]\Gamma_{\mu \nu \lambda} = \frac{1}{2} [Q_{\mu \nu \lambda} + Q_{\mu \lambda \nu}][/itex]
    Last edited: Sep 4, 2016
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted