# Can a person at a photon sphere of a black hole decide where the black hole is?

A person at the photon sphere of a black hole will observe the following:

• The black hole surface will cover exactly half of the visible sky (say, the left half), and cosmic horizon will cover the other half.

• Looking in any direction between these hemispheres the person will see oneself.

• The person will observe no centrifugal force however fast he moves. Any object the person thowing to the left will experience the centrifugal force pulling it further left, while any object thrown to the right also will experience the centrifugal force pulling it to the right. Approaching the black hole in such circumstances may look quite like actually escaping it to the crew.

• The space to the right as well the space to the left seemingly includes objects of any length and dimensions, the further objects are, the more they are time-dilated and length-contracted.

Is there any way for a spacecraft commander to decide (by an experiment conducted in a small$^1$ amount of time) which hemisphere is the sky and which hemisphere is black hole in such circumstances?

--

$^1$ Here small means small compared to characteristic time scales of the problem.

-
What about tidal forces? Of course you can make those negligible by choosing a suitably small observer or a suitably large black hole. – rob May 5 '14 at 14:10
I don't see why the "black hole surface will cover exactly half of the visible sky". – DavePhD May 5 '14 at 14:12
@rob tidal forces also will be symmetric. – Anixx May 5 '14 at 14:13
@DavePhD because all light rays coming from the left are originated on the BH surface (if not on the closer objects) while all light rays from the right hemisphere come from the cosmic horizon. If the observer is closer to the BH than the photon sphere, the BH will cover greater part of the sky. At BH surface (event horiizon) the BH covers all sky. – Anixx May 5 '14 at 14:16

The orbit at $1.5r_s$ is only possible for massless objects: the closest orbit possible for massive objects is $3r_s$.

So if you are in free fall simple wait a (very) short time and you'll hit the singularity and the question will be decided for you.

If you are not in free fall, e.g. you are firing some form of rocket motor to hover at $1.5r_s$, then the situation is asymmetric because you will need to fire your motor towards the black hole.

Either way, it will be obvious which side is which.

-
hmm can you please provide any link about such minimum orbit for massive objects and a proof that it is so different? – Anixx May 5 '14 at 15:27
Also note the following: any object (including massive) that crosses the photon sphere from inside to outside will escape to infinity without firing any engines. To me that indicates that in a limit of very slow crossing speed, the escape to infinity time will take a lot of time. – Anixx May 5 '14 at 15:31
@Anixx PDF: eagle.phys.utk.edu/guidry/astro616/lectures/lecture_ch18.pdf discusses both particle orbits and photon orbits in Schwarzschild metric, and I think also some discussion of Kerr metric further down. Re second comment: in this case it will take a long time to get to infinity, yes, but the initial speed is quite high, so it won't take very long to get "far away". – Kyle Oman May 5 '14 at 16:00
@Kyle an object that crosses the photon sphere from inside at ANY speed will escape to infinity. Even if the speed is arbitrary small. – Anixx May 5 '14 at 16:05
@Anixx the question is how do you get to move that way. If you are falling in from the outside, you are doomed to the singularity; if you have rockets, you have to fire them towards the singularity (and burn a lot of fuel) to get to cross the boundary. – Davidmh May 5 '14 at 16:24