Circular Orbit Four-Velocity and Schwarzschild Metric

johne1618
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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

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

where r_0 is the Schwarzschild radius.

Is this right and how would one show it using the Schwarzschild metric and the geodesic equation for a circular orbit?
 
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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.
 
Bill_K said:
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.

Thanks very much - I've managed to derive it with your help (V'(r)=0 condition) and some notes from the internet.
 
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