I'm taking a course in GR and I want to determine for which radius a photon can move in a circular orbit in the gravitational field produced by a massive spherical body (Schwarzild's metric).

What I have done is to calculate the tangent vector for an arbitrary geodesic of this metric and make the norm equal to $0$ because a massless particle must move in a null-like geodesic. This gives me $r_0=2GM$, the Schwarzild radius.

Then the problem ask me to calculate the period of the photon in this orbit for the coordinate time and for an observer in a point of that orbit.

I was wondering how to do this. Physically, as the radius that I've got is the Schwarzild radius, I expect the solution for the coordinate time to be $\infty$ and some value for an observer in the orbit.

To calculate the period should I apply the classical formula for the angular velocity of a particle $\omega=v/r$ and then say that $T=2\pi/\omega$ or isn't this a valid approximation?

Is there any way to get this information from the null-like geodesic besides calculating the Cristoffel symbols and then the velocity in coordinates $(t,r,\theta, \phi)$?

  • $\begingroup$ Did you also set $\dot r=0$? It would be helpful for us to see the whole calculation that you've done. $\endgroup$
    – Ryan Unger
    Jan 17, 2016 at 23:25
  • $\begingroup$ Yes, I set $\dot{r}=0$ and $\theta=\pi/2$ (which can be done choosing a suitable reference frame). I will edit with details of the calculations later today. $\endgroup$
    – S -
    Jan 18, 2016 at 8:04
  • $\begingroup$ For future readers: (part of) the calculation can be found here. $\endgroup$
    – Ryan Unger
    Jan 19, 2016 at 15:44

1 Answer 1


My previous calculations had a mistake (I set a factor to 0 that I shouldn't have set). Now I got the right answer that it is that the photon might orbit for a value $r=3M$ (where I have considered $G=1$). Also I've found a nice way to calculate the period.

To calculate the period (for the coordinate time) I have done the following:

$$t = \int_0^{2\pi} d \phi = \int_0^{2\pi} \frac{\dot{t}}{\dot{\phi}} d \phi$$

And then it is possible to express this integral using the following integrals of movement and the value of the radius obtained previously

$$\left( 1 - 2M/r \right) \dot{t} = E$$

$$r^2 \dot{\phi} = L$$

Where $L$ and $E$ are constants.


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