Can a black hole free from an accretion disk collect photons in the photon sphere in a prolonged stable orbit?

Could enough photons be in orbit to shade the actual black hole with light?

Would a rogue black hole appear to have an atmosphere made of photons from the light emitted from other distant stars entrapped and released?

  • $\begingroup$ See en.wikipedia.org/wiki/Photon_sphere $\endgroup$
    – PM 2Ring
    Commented Feb 16, 2019 at 23:05
  • $\begingroup$ @PM2Ring The question is more tricky as it seems. The orbit of such photons are unstable. To calculate their mean or typical stability requires the combined usage of quantum mechanics and general relativity. $\endgroup$
    – peterh
    Commented Feb 26, 2019 at 7:23
  • $\begingroup$ @peterh Yes, it's tricky. FWIW, I linked that article before the question was edited to mention the photon sphere. $\endgroup$
    – PM 2Ring
    Commented Feb 26, 2019 at 7:51
  • $\begingroup$ I very badly want to ignore GR and QM and just guess the answers to OP's query are: no, no, and no. $\endgroup$ Commented Feb 26, 2019 at 8:23
  • 2
    $\begingroup$ @userLTK Yes. I think a more exact calculation would use quantum field theory-based heuristics, calculating the uncertainity of the position of the photon. Here doesn't the atom size matter, but the wavelength of the photon, so probably also the result would depend on its wavelength. I suspect, photons with bigger energy, thus with smaller wavelength, would remain longer on the equilibrical orbit. $\endgroup$
    – peterh
    Commented Feb 26, 2019 at 17:56

1 Answer 1


For any massive object the gravitational potential energy is given by Newton's law:

$$ V(r) = -\frac{GMm}{r} $$

The gravitational potential energy is due to the attractive gravitational force, but for an orbiting object there is also a (fictitious) centrifugal force pushing it outwards. If we calculate the potential energy due to the centrifugal force and add it to the gravitational potential energy we get an effective potential energy:

$$ V_\mathrm{eff}(r) = -\frac{GMm}{r} + \frac{L^2}{2mr^2} \tag{1} $$

where $L$ is the angular momentum, which is a constant for an orbiting object (because angular momentum is conserved in a central field). If we plot this graph for the Earth-Moon system we get a graph like this:

Effective potential

(this comes from my answer to the question Could we send a man safely to the Moon in a rocket without knowledge of general relativity? on the Physics SE)

Note that there is a minimum in the potential energy curve, and this minimum gives the radius of the stable orbit. Note also that because it's a minimum if we displace the object away from the minimum it will fall back towards the minimum again i.e. this is a stable orbit.

Now, for light we cannot simply use Newtonian mechanics because light is massless, but we can do the calculation using general relativity. The details are a bit intimidating, but we end up with an effective potential just as described above. For light the effective potential turns out to be:

$$ V_\mathrm{eff}(r) = \sqrt{1 - \frac{2GM}{c^2r}}\frac{L}{r} \tag{2} $$

And if we graph this it looks like this:

Light effective potential

This looks very different from our first graph, and it's different because light is massless and only ever travels at the same speed of $c$. The graph of $V_\mathrm{eff}$ for light has a maximum not a minimum. The maximum corresponds to the position of a circular orbit, just like the minimum in the first graph, and we find the radius of this circular orbit is given by:

$$ \frac{r}{r_\mathrm s} = 1.5 $$

where $r_\mathrm s$ is the Schwarzschild radius. This radius gives the position of the notorious photon sphere.

But for light $V_\mathrm{eff}$ has a maximum not a minimum. That means if we displace the light by even the tiniest distance from this maximum it will lower its potential energy by moving either inwards or outwards. The orbit at $1.5r_\mathrm s$ is unstable and the tiniest perturbation will cause the light to spiral into the black hole or away from it. This means we cannot accumulate light in the photon sphere as the question asks. Any attempt to put light into this orbit is doomed to failure as even the tiniest perturbation (e.g. from other objects orbiting the black hole, or even from passing gravitational waves) will destabilise the orbit and the light will be lost.

  • $\begingroup$ @PM 2Ring left a comment on the stable photon orbit in the Kerr black hole. What do you think? $\endgroup$
    – Muze
    Commented Feb 27, 2019 at 15:17
  • $\begingroup$ @Muze which comment? Do you mean In case it isn't obvious, we can't see the photons that are in the photon sphere, only those that happen to spiral out of it & head our way. $\endgroup$ Commented Feb 27, 2019 at 15:28
  • 3
    $\begingroup$ @Muze that describes a rotating black hole, i.e. the Kerr geometry. That gets mind bendingly complicated because the rotation of the hole sweeps the light sideways as it orbits. However the photon sphere is still unstable so you'd never be able to build up any accumulation of light there. $\endgroup$ Commented Feb 27, 2019 at 15:35
  • 1
    $\begingroup$ @Muze the photon sphere in a Kerr black hole is actually an oblate ellipsoid not a sphere, but it is still a 2D surface and orbits in it are still unstable. In fact this was discussed on the physcs SE though the answer is rather mathematical. $\endgroup$ Commented Feb 27, 2019 at 15:41
  • 1
    $\begingroup$ @JohnRennie great answer! $\endgroup$
    – Max0815
    Commented Feb 27, 2019 at 23:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .