Sunday, April 14, 2013

Horseshoes, hand grenades but not black hole event horizons: Being close to the horizon is not good enough

I do not currently really understand the math of general relativity, but based on some stated rules about the behavior of light around black holes I've created a model in my mind that I would like to write down and then explore and/or destroy.

Start with an event horizon:
In general relativity, an event horizon is a boundary in spacetime beyond which events cannot affect an outside observer. In layman's terms it is defined as "the point of no return" i.e. the point at which the gravitational pull becomes so great as to make escape impossible. 
Outside the event horizon of black hole is another interesting boundary - the photon sphere:
photon sphere is a spherical region of space where gravity is strong enough that photons are forced to travel in orbits. The radius of the photon sphere, which is also the lower bound for any stable orbit, is:
r = \frac{3GM}{c^{2}}
which is one and half times the Schwarzschild radius. 

Also important about the photon sphere:
As photons travel near the event horizon of a black hole they can escape being pulled in by the gravity of a black hole by traveling at a nearly vertical direction known as an exit cone. A photon on the boundary of this cone will not completely escape the gravity of the black hole. Instead it orbits the black hole.
And finally:
There are no stable free fall orbits that exist within or cross the photon sphere. Any free fall orbit that crosses it from the outside spirals into the black hole. Any orbit that crosses it from the inside escapes to infinity. No unaccelerated orbit with a semi-major axis less than this distance is possible, but within the photon sphere, a constant acceleration will allow a spacecraft or probe to hover above the event horizon.

Putting the above together, I have drawn this diagram:

In this figure each lettered arrow represents a possible initial photon position and initial trajectory, and a statement of how I understand it will behave.  Photon A starts arbitrarily close to, but outside of the photon sphere, with an initial trajectory heading in.  Photon B starts arbitrarily close to, but outside of the event horizon with an initial trajectory directly away from the center.  Photon C starts arbitrarily close to, but inside of the event horizon with an initial trajectory directly away from the center..

What is unclear to me is what is the intermediate behavior of C?  It is fairly well supported that photon C will not escape from the black hole, but my question is whether it will cross the event horizon and travel through the region between the event horizon and the photon sphere before re-crossing the event horizon on its way back to the singularity.  This statement was my original hypothesis, but working through the next example convinced me of the opposite.  Basically, I wanted to devise a reasonable "story" about the paths of photons B & C such that there would be a smooth transition in the behavior regardless of how close to the event horizon one was (neglecting quantum effects).

Consider the hypothesis in which C never crosses the event horizon - despite being arbitrarily close and with an initial trajectory directly at / normal to the even horizon.  In order to not cross the event horizon photon C's path must be sharply curved.  Now consider Photon B which is an arbitrarily small distance outside the event horizon.  In order for spacetime to be continuous, photon B must also have a sharply curved path.  Here is a rough schematic:

Since the initial position of photon B and photon C are arbitrarily close, their initial trajectories are also arbitrarily similar - but they are different.  After making the sharp turn, Photon B is spiraling outward, while photon C after making the sharp turn is spiraling inward.


  1. You have certainly posed a very deep and thought provoking question. Even people who do understand the math may be unable to answer it as it involves both general relativity and quantum mechanics.GR presumes effects as deterministic while QM presumes effects as probabilistic.
    I presume that B would have a chance to barely escape but with only an infinitesimal amount or it's energy once it reaches a great distance while C would come infinitesimally close to an escape. The path of neither photon could be presume as that would require both an exact knowledge of position and moment which is forbidden due to the uncertainty principal. The event horizon only has great significance to a distant observer but not to an observer local to the event horizon.

  2. That's a good point about taking quantum mechanics into account. This was definitely a thought exercise in which I was just thinking about pure general relativity - definitely not the correct model - but still a long stretch for me to try to wrap my brain around.

    I agree with your description of the energies of B & C. Looking back on this now, it seems that a useful shorthand is to think of the event horizon as roughly analogous to a turning point (local maximum) in a potential energy diagram. In classical mechanics, a particle that is infinitesimally on one side will travel in the opposite direction from a particle that is infinitesimally on the other side. I understand the event horizon is not a maximum in a potential energy diagram :) but it's a useful analogy.