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Solving the Geodesic equations for a space

  1. Dec 7, 2011 #1
    I'm in an intro course and my shaky ability to solve differential equations is apparent.

    How would you go about solving

    [itex]\ddot{r}[/itex]-r[itex]\ddot{\theta}[/itex]=0

    [itex]\ddot{\theta}[/itex]+[itex]\frac{1}{r}[/itex][itex]\dot{r}[/itex][itex]\dot{\theta}[/itex]=0

    It might be obvious. They're the geodesic equations for a 2d polar coordinate system (if i'm correct).
     
  2. jcsd
  3. Dec 8, 2011 #2

    pervect

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    The second equation is missing a factor of 2. I'm not sure exactly how you derived the geodesic equations, I'm hoping that if I remark that [tex]\Gamma^{\theta}{}_{{r}{\theta}} = \Gamma^{\theta}{}_{{\theta}{r}} = \frac{1}{r}[/tex] you'll see what you omitted. If not, you'll need to write down your derivation.

    After this correction, the second equation is equivalent to saying a geodesic conserves angular momentum, i.e

    [tex]\frac{d}{dt} \left( r^2 \, \dot{\theta} \right) = 0 [/tex]

    If you expand this via the chain rule you'll get

    [tex]r^2 \ddot{\theta} + 2\,r\,\dot{r}\,\dot{\theta} = 0[/tex]

    which is equivalent to the corrected version of eq 2 (just divide both sides by r^2). This should allow you to separate the variables and solve the equatinos without too much difficulty by solving for [itex]\dot{\theta}[/itex] as a function of r and the conserved, constant angular momentum - which you can also regard as a constant of integration.
     
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