# Black hole laser or condensate?

1. Jun 12, 2009

### tiny-tim

If you shine a beam of light at a black hole, most of it will either fall into the black hole, or will just graze past it, hardly being bent at all.

But any photon that just happens to have L/E (angular momentum per energy) = M√27 will orbit forever (or until it hits infalling matter), with period ever closer to 2πM√27, getting closer and closer to the photon sphere (r = 3M) but never quite reaching it.

Also, photons with L/E very slightly greater than M√27 will orbit a large number of times with almost the same period, before escaping to infinity.

(The above figures are for a non-rotating black hole, but I think the situation for a rotating one is substantially the same, except that there is a different radius of photon sphere for each angle of approach, and the orbits precess round the photon sphere.)

Now, I've never really understood lasers or condensates

but since starlight must be doing this all the time, producing photons orbiting with almost the same period, will photons (of the same wavelength) tend to become coherent, and either form a laser, or (is this another way of saying the same thing? ) form a state of matter just outside the photon sphere something like a Bose condensate?

2. Jun 13, 2009

### xepma

I like your way of thinking ;)

But I doubt that such a state will hold, due to tunneling effects and the uncertainty relation. Any disturbace will destroy such a state.

3. Jun 14, 2009

### Count Iblis

Well, considering that the non-linear Schroedinger equation describing a Bose condensate (a.k.a. the Gross-Pitaevski equation) is completely analogous to what the Maxwell equations are for photons, you could just solve the Maxwell equations in a Schwarzschild metric to see what happens.