For a simulation, I want to compute the path that light follows near a black hole.

Non-relativistically, a massive point particle in a central newtonian gravitational field follows either an ellipse, a parabola, or a hyperbola. Is the same true relativistically for light around a black hole? A problem I see with this, is that while particles gain velocity when approaching a black hole, a photon gains energy instead. So do photons behave differently?

  • $\begingroup$ Nothing to add, I want to do the same. Relativistic raytracer computing the distortion of the stellar background when viewed in the vicinity of a black hole. Might be a fun exercise in GPU or Cell programming. But computing the photon trajectories got me stumped. $\endgroup$ Mar 29, 2012 at 15:42

2 Answers 2


Yes, photons behave differently, which I'm sure is no surprise! Any graduate book on GR will derive the orbits for a Schwarzchild metric. In my copy of "A first course in general relativity" by Bernard F. Schutz the orbits are calculated in chapter 11. Cribbing from this, for photons the orbit is:

$$ \left(\frac {dr}{d\lambda}\right)^2 = E^2 - \left(1 - \frac {2M}{r}\right) \frac {L^2}{r^2} $$

so you get an effective central potential:

$$ V^2(r) = \left(1 - \frac{2M}{r}\right) \frac {L^2}{r^2} $$

$L$ is the $p_\phi$ component of the four momentum and is constant. For particles the first equation would give you $dr/d\tau$ but proper time is always zero for a photon hence the use of the affine parameter $\lambda$ instead, where $\lambda$ is defined by $p^r$ = $dr/d\lambda$. The $r$ and $\phi$ are the Scharwzchild co-ordinates i.e. as seen by an observer at infinity.

  • $\begingroup$ I've implemented this into a simulation, and the orbits look pretty good. However I've noticed that if I set M to 0, I get an effective central potential of $V(r) = \frac{|L|}{r}$, while I would've expected the centrifugal potential, $V(r) = \frac{L^2}{r^2}$, so that the light travels in a straight line. Am I missing something? $\endgroup$
    – Hannesh
    Apr 2, 2012 at 16:19
  • $\begingroup$ Ignore that last comment, I'm using the formula for the orbit now, which works great $\endgroup$
    – Hannesh
    Apr 2, 2012 at 22:19

For massive point particles, the general relativistic trajectory can be much different from the classical one. The classical Newtonian gravity requires that the particle trajectory be given by a conic section: in particular, for escaping orbits (hyperbola or parabola) the trajectories in space cannot self-intersect. If you model a black hole (or a really heavy star) using the Schwarzschild metric, the general relativistic picture allows for a particle to come in from infinity, make any number of loops around the massive body, and then escape. The solutions to the geodesic/free-fall motion are much more complicated then the Newtonian picture. (You can see some pictures on p 215 of these lecture notes. Another picture disallowed in Newtonian gravity is that of the asymptotic orbit: a particle can approach from infinity and gradually settle into an unstable circular orbit around the gravitating body. In Newtonian gravity if the particle comes from far away it must follow a hyperbolic or parabolic orbit and fly away.

For photons, the situation is also different from the massive particle case (assuming a Schwarzschild like background). In the case of the massive particles there are both stable and unstable circular orbits around the gravitating body. For the photons, the only bound orbit is a single unstable orbit at the so-called "photonsphere". Aside from those unstable circular orbits, there are no other bounded trajectories for photons. However, one can still have the "scattering" trajectories where a photon comes in from far away, loops around the star arbitrarily many times, and then fly away, as well as the asymptotic trajectories where the photons come in from far away and gradually approach the circular orbit at the photonsphere.


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