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This is a bit of a complex question - example scenario:

X and Y (who happen to be immune from radiation and gravitation spaghettification, don't draw this into the equation) are some distance from a black hole. X approaches and orbits the black hole, outside the event horizon for some time, and then returns to Y, who has not moved. GR suggests that Y is now older than X.

X approaches again, at a deeper angle and therefore, closer to the even horizon. X completes several more orbits this time before returning to Y. Y is now significantly older than X.

This time, X gets so close to the event horizon, X is within a few nm of it, and travelling extremely close to the speed of light. X orbits for several weeks, before finally returning. This time, Several trillions of years have passed for Y.

As X approaches the event horizon, X must also be approaching the speed of light (speed of causality), therefore it is not possible for X to pass through, as X must be travelling at the speed of causality inside the event horizon, which breaks GR.

My hypothesis is, unless you are traveling directly toward the singularity as you pass the event horizon (no angle exists to allow this, until infinitely close to the event horizon) you must remain outside. Therefore, there is nothing in the black hole - the event horizon contains all information and energy that approaches it, and field strengths render any particle to it's component fields, which radiate out as hawking radiation, due to uncertainty principle ( the only way out is at an angle related to the uncertainty probability distance, which causes a specific frequency related to the curve of the event horizon).

This theory seems to meet established criteria (holographic theory, hawking radiation, conservation of information, GR cannot break down without singularity, the list goes on). It means that, at the birth of the black hole, only energy exists near the location of conception, bound by uncertainty to leave only at very narrow angles away from the center. uncertainty principle does not allow a singularity to exist, rather, energy wraps around an uncertain point, where uneven hawking radiation, spin and other factors prohibits an exact location of a black hole.

Are there any counter-arguments for this hypothesis?

Russell King
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  • Possible duplicates: https://physics.stackexchange.com/q/21319/2451 , https://physics.stackexchange.com/q/202935/2451 and links therein. – Qmechanic Sep 27 '18 at 21:37
  • Thanks - the question is aimed at establishing what counter-arguments may exist against my hypothesis, such that I can research further, not to disprove the hypothesis. I will re-word the question. – Russell King Sep 27 '18 at 22:09
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    It's perfectly fine for X to travel faster than light relative to Y due to spacetime curvature. For the same reason, galaxies outside of our cosmological horizon are traveling faster than light relative to us. – Dmitry Brant Sep 27 '18 at 22:37
  • @DmitryBrant I do not argue with that, I understand this notion. My hypthesis is X cannot travel faster than causality (faster than any interaction it has with another particle can propagate) – Russell King Sep 27 '18 at 22:44
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    ...causality relative to what? – Dmitry Brant Sep 27 '18 at 22:45
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    If we're fixing our reference frame on Y, then sure, X will never cross the event horizon. But if we travel alongside X, then we'll pass the event horizon (relative to Y) without anything particularly interesting happening. – Dmitry Brant Sep 27 '18 at 22:49
  • My hypothesis is that it is impossible to travel slower than the speed of light relative to X, even if you are within 0 plank lengths of X, at the EH. Any particle travelling with X instantly surpasses the speed of light relative to X at any distance when passing the event horizon, as the event horizon accelerates toward you to c the moment you reach it, which is implicitly impossible. this means no two particles can be at an angle that intersects any point inside the event horizon - they must have an outbound or parallel trajectory. Same goes for everything that fell toward the black hole. – Russell King Sep 27 '18 at 23:03
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    The event horizon only exists at a particular distance from Y's point of view. When you're traveling alongside X, the horizon relative to you will be different from the horizon seen by Y. The horizon will continue to be "in front of you" as you continue falling, all the way until you hit the singularity. – Dmitry Brant Sep 28 '18 at 00:11

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The event horizon is an ideal surface, surrounding the black hole, which disconnects the exterior region of spacetime from the interior region. An observer far away from the horizon will never measure a radially free falling object to cross the horizon, as the latter will appear to progressively slow down when approaching the horizon until to freeze on it.

However that is not the experience of the free falling object, which instead will cross the horizon and will hit the physical singularity at the center of the black hole in a finite amount of its time (proper time).

Moreover, a free falling object can cross the horizon even if it has an angular momentum, provided its energy is enough to overcome the effective potential. This is a new feature of general relativity vs. Newtonian gravity, where the centrifugal barrier can avoid an object to fall to the gravitational mass.

Michele Grosso
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