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Timelike Geodesic and Christoffel Symbols

  1. May 18, 2010 #1
    1. The problem statement, all variables and given/known data

    How do I show the following metric have time-like geodesics, if [tex]\theta[/tex] and [tex]R[/tex] are constants

    [tex]ds^{2} = R^{2} (-dt^{2} + (cosh(t))^{2} d\theta^{2}) [/tex]

    2. Relevant equations

    [tex]v^{a}v_{a} = -1[/tex] for time-like geodesic, where [tex]v^{a}[/tex] is the tangent vector along the curve

    3. The attempt at a solution

    First, I write it as the Lagrangian

    [tex] L = -R^{2}\dot{t}^{2} + (cosh(t))^{2} \dot{\theta}^{2} = -R^{2}\dot{t}^{2} [/tex]

    as [tex]\theta[/tex] is a constant.

    How do I proceed to show that this indeed gives us a time-like geodesic.

    Could someone also tell me if I have computed the Christoffel symbol components correctly? My result is

    [tex] \Gamma^{t}_{\theta \theta} = 0-sinh(t) \times cosh(t) [/tex]

    [tex] \Gamma^{\theta}_{t \theta} = tanh(t) [/tex]

    and all other components vanish.


    P.S. How do I type minus sign? It doesn't seem to work if I have left the 0 out at above.
    Last edited: May 18, 2010
  2. jcsd
  3. May 18, 2010 #2
    Maybe I'm not understanding the problem correctly, but if you're holding [itex]\theta[/itex] constant, couldn't you just use [itex]d\theta = 0[/itex]?
  4. May 18, 2010 #3


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    [tex] \Gamma^{t}_{\theta \theta} = -sinh(t) \times cosh(t) [/tex]
    Works fine for me :confused:
    Anyway, I get the same thing as you for the Christoffel symbols except without the minus sign. I could have made a mistake but you might want to recheck your calculations.
  5. May 20, 2010 #4


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    I don't see the need to compute the Christoffel coefficients at all. The solution to the Euler-Lagrange equations will be a geodesic, so if [itex]R[/itex] and [itex]\theta[/itex] are constants, you want to solve

    [tex]\frac{\partial L}{\partial t}-\frac{d}{d\tau}\frac{\partial L}{\partial \dot{t}}=0[/tex]

    (Where I'm using [itex]\tau[/itex] to represent your affine parameter...i.e. [tex]\dot{t}=\frac{dt}{d\tau}[/tex] )
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