Period of photons on circular orbits in Schwarzschild spacetime Consider a photon that is moving on a circular orbit at the radius $r = 3 \ GM/c^2$ in the Schwarzschild spacetime. How can we find the period of the orbit as measured by a stationary observer at infinity?
How can we find the period of the orbit as measured by a stationary observer located at the same
radius (i.e. at $r = 3\ GM/c^2$
) is $T = 6 \pi \ GM/c^3$
.
I am using Kepler's 3rd law but I'm not getting the exact result.
 A: You should start from the Schwarzschild metric
$$ds^2 = -\left(1 -\frac{r_s}{r}\right)c^2dt^2 +\left(1 -\frac{r_s}{r}\right)^{-1}dr^2 + r^2d\phi^2\ . $$
For light on a null geodesic, $ds^2=0$; and at a fixed radial coordinate (NB, not a "radius"), $dr=0$. This gives you an expression that can be integrated to give the time interval $\Delta t$ measured in Schwarzschild coordinate time - the time measured by an observer at rest at infinity.
But the proper time of an observer at a fixed radius of $3GM/c^2$ is not given by $\Delta t$. Writing the metric in terms of a proper time interval $d\tau$, with $dr=d\phi=0$
$$c^2d\tau^2 = \left(1 -\frac{r_s}{r}\right) c^2dt^2\ .$$
This gives you an expression for $\Delta \tau$ in terms of $\Delta t$ and the result you are looking for
A: 
Consider a photon that is moving on a circular orbit at the radius r=3 GM/c2 in the Schwarzschild spacetime. How can we find the period of the orbit as measured by a stationary observer at infinity?

For general static spherically symmetric metric
\begin{equation}
ds^2=g_{tt} dt^2-g_{rr} dr^2-g_{\phi\phi} d\phi^2.
\end{equation}
The so-called Keplerian frequency is defined as
\begin{equation}
\Omega\equiv\sqrt{\frac{g_{tt}’}{g_{\phi\phi}’}},
\end{equation}
where prime denotes a derivative on $r$.

How can we find the period of the orbit as measured by a stationary observer located at the same radius (i.e. at r=3 GM/c2 ) is T=6π GM/c3 . I am using Kepler's 3rd law but I'm not getting the exact result.

The locally measured orbital velocity is defined as
\begin{equation}
v\equiv\sqrt{\frac{g_{\phi\phi}}{g_{tt}}\frac{g_{tt}’}{g_{\phi\phi}’}}
\end{equation}
Because the orbit's circumference for local and distant observer is the same ($2\pi r$), the corresponding local angular velocity reads
\begin{equation}
\omega= \frac{v}{r}
\end{equation}
You can find it in https://arxiv.org/abs/2110.05764, see equations (18) and (19). Another useful help for your considerations could provide https://arxiv.org/abs/0812.1806.
