# Orbiting a BH

1. Apr 10, 2010

### Medium9

I do have a question that bothers me for some days now. I'm not too sure how qualified it is, but my mind needs to be eased here.

I understand, that it may be taken as a fact, that BHs cause enough gravity to make photons "vanish" behind the event horizon. So like with any stellar object, there needs to be a breaking point, a balance between "this particle will escape" and "this particle will fall down"; an orbit. So if photons escape BHs at one distance, but fall into it at a closer one, there necessarily has to be a distance where photons will actually orbit a BH, or at least circle around it for considerable time. Of course, that area would be RIGHT above the event horizon.

If this assumption is accurate, then any matter that is pulled in needs to cross this orbital zone, and there should be a chance for this matter to be hit by such an orbiting photon - of which I could imagine quite many to be there. This then should either cause heat, or scattered radiation directly, which should then be measurable.

Has this been considered and attempted to either calculate or measured by someone? If so, are there any noteworthy results? (Preferably in terms that a non-physicist has a chance to understand

Medium9

2. Apr 10, 2010

### Nabeshin

Well you hit something on the head here! The event horizon is located at a distance r=2M (in units where c=G=1) from the singularity. Now, if you imagine putting a photon just outside the EH, it will have to be directed completely radially outward in order to balance the tendency to fall in -- it will hover. This is not an orbit, and not what we want. It turns out that the distance a photon is able to orbit at is r=3M, 3/2 the schwarzschild radius of the black hole. Indeed, at least for a non-charged, non-rotating black hole, photons travel in a circular orbit at this radius. The orbit, however, is unstable. Any deviation from the orbit will lead to the photon either escaping to infinity or falling into the hole.

Some other information you might be interested in:
The closest distance at which a massive particle can be in a stable circular orbit is r=6M. Any closer, and the orbit is unstable (small deviations lead to big deviations). You have to orbit progressively faster and faster below r=6M in order to stay in the unstable circular orbit, though. So as you approach r=3M, the requisite velocity for the orbit approaches the speed of light!

3. Apr 11, 2010

### Medium9

First of all: If photons closer than 3/2 of the schwarzschild radius orbit, wouldn't they always fall in if they got any closer? Why than is the schwarzschild radius smaller than that? What escapes between those two thresholds?

Also, my intention to where I wanted to go with this question stands open. If there are orbiting photons (unstable ones that soon deviate enough are still there for quite some time I guess), and matter gets hit by them, shouldn't we be able to detect BHs as dimly flashing areas? I ask, because so far I've always only learned that BH detection can only be done through interpretation of lensing effects, so that BHs without much bright background may stay undetected forever. It would have been quite a nice way to detect BHs more directly.

4. Apr 11, 2010

### Nabeshin

First, to answer the 2nd part of your question. Black holes are easily recognizable by their accretion disks (inspiraling matter which heats up and gives off characteristic x-ray radiation). The observation you propose is fundamentally no different, but extraordinarily more difficult to carry out (these events would be far dimmer). Lensing effects do give us a view of black holes without accretion disks, but without matter near the hole, it's doubtful your photon scheme would work either.

You misunderstood. Photons closer than 3/2 the schwarzschild radius do not orbit, they either fall in or go out. They do not need to fall in necessarily because they can be directed more outwards than inwards, in which case they escape to infinity. Like I said, you can imagine a photon right at the event horizon needs to be directed completely radially outwards to avoid falling in. Now imagine moving slightly and slightly further away. The critical angle the photon needs to be pointed at to avoid falling in becomes greater and greater (measured with respect to the radial line) until at r=3m it is 90 degrees, which is the condition for a circular orbit! Hopefully this gives you a decent picture in your head, I wish I could draw one for you. If you understand this, the first part of your question should make sense now.

5. Apr 11, 2010

### Medium9

Ah okay, I think I'm getting there. My mistake was to assume the circular orbit at the event horizon in the first place, for which there actually was no reason, I see that now.

Also, I'm feeling a bit dumb right now for the second part of my question. I was aware of accretion discs, but I simply didn't consider that with the absence of matter to make up one, there is just as few matter that could fall in, producing the flashes I assumed. Wow, that took embarrassingly long for me to realize...