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Proper time in circular orbit

  1. Jun 17, 2013 #1
    If an object is orbiting on a circular time-like geodesic path around a mass then the Wikipedia claims that the first component of its four-velocity is given by

    [tex]\frac{dt}{d\tau} = \frac{1}{\sqrt{1-\frac{3}{2}\cdot \frac{r_0}{r}}}[/tex]

    where [itex]r_0[/itex] is the Schwarzchild radius.

    Is this right and how would one show it using the Schwarzchild metric and the geodesic equation for a circular orbit?
     
  2. jcsd
  3. Jun 17, 2013 #2

    Bill_K

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    Yes it's right, but a bit messy to derive. It comes partially from the Schwarzschild gravitational potential and partially from the orbital motion. You write the orbital equation in the form (dr/dφ)2 = V(r). Then the conditions for a circular orbit are V(r) = V'(r) = 0, and this gives, among other things, your result. More details upon request.
     
  4. Jun 17, 2013 #3
    Thanks very much - I've managed to derive it with your help (V'(r)=0 condition) and some notes from the internet.
     
    Last edited: Jun 17, 2013
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