# Proper time in circular orbit

1. Jun 17, 2013

### johne1618

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 Schwarzchild radius.

Is this right and how would one show it using the Schwarzchild metric and the geodesic equation for a circular orbit?

2. Jun 17, 2013

### Bill_K

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.

3. Jun 17, 2013

### johne1618

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