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?
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:
(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:
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.
