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Picture a light wave orthogonal to and just above the event horizon of a black hole, fired directly away from the black hole.

If the black hole is of sufficient mass, the light would be pulled back towards the black hole and would eventually reverse directions.

My two questions are:

  1. Is this assumption correct?
  2. If so, how would the light be able to reverse directions to travel back towards the black hole without reaching a speed of 0?
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  • $\begingroup$ @PM2Ring Just above the event horizon $\endgroup$ Sep 8, 2021 at 0:27
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    $\begingroup$ It does not reverse direction $\endgroup$
    – user65081
    Sep 8, 2021 at 0:32
  • $\begingroup$ @Wolphramjonny could you give some context? Does it not reverse direction solely because it is beyond the event horizon? Does it slow down or turn in another direction at all? $\endgroup$ Sep 8, 2021 at 0:39

2 Answers 2

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If the black hole is of sufficient mass, the light would be pulled back towards the black hole

If the light was emitted from outside the event horizon, then the black hole is necessarily not of sufficient mass. Once the photon is emitted, it moves at the local speed of light. It is too late to add mass to the black hole to change its trajectory. The light when it is far from the black hole will be observed to have less energy than an observer at the event would have seen.

the light would be pulled back towards the black hole and would eventually reverse directions.

No, the light never "reverses" directions the way a ball might when thrown upward. Spacetime near the black hole (inside the event horizon) is sufficiently curved that all directions point into the black hole. There is no direction that you can point the light that leads to outside.

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  • $\begingroup$ Great answer. Thank you! $\endgroup$ Sep 8, 2021 at 0:47
  • $\begingroup$ Just checking...that's really "all timelike directions", correct? Or do all spacelike directions get eaten up too if you're close enough to the singularity? $\endgroup$ Sep 8, 2021 at 2:33
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    $\begingroup$ Spatial directions for a light beam, which would allow you to enumerate all possible lightlike paths with that event as an origin. Outside the horizon, some spatial directions intersect the singularity. Inside the horizon they all do. $\endgroup$
    – BowlOfRed
    Sep 8, 2021 at 4:11
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    $\begingroup$ What I'd like to know is, what if the photon start exactly at the event horizon? $\endgroup$
    – j4nd3r53n
    Sep 8, 2021 at 9:07
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    $\begingroup$ @j4nd3r53n : A photon is not a point, so what does "exactly at" mean? $\endgroup$ Sep 8, 2021 at 17:33
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No, your assumption is wrong.

All vanilla black holes (no charge, no spin) have an event horizon. If you are outside the event horizon, you are far enough away that light pointed directly away can escape to infinity. As in all gravitational wells, it will be red shifted. The light that reaches infinity will be less energetic that it would have been without the gravitational well. Light emitted very close to the event horizon is strongly red shifted.

Time passes slowly in a gravitational well. The deeper in, the more slowly. So it takes longer to escape than you might expect.

Inside the event horizon, everything, light included, travels on a trajectory that ends at the singularity.

Outside the black hole, there is a direction perpendicular to the event horizon toward the outside. An excited atom that decays can send a photon in that direction. It will travel infinitely far away. Photons sent in other directions travel on "straight" geodesics. But in curved space, "straight" bends toward the event horizon. The trajectory may well pass through the event horizon and end on the singularity.

However, if the excited atom falls past the event horizon, all directions lead to the singularity. Light pointed in what would have been the outward direction does not travel outward. Inside the event horizon, space is curved enough that intuition is not helpful. Perhaps you might think of the photon swimming upstream, but not fast enough. But of course there are problems with that thought.

Exactly at the event horizon, light would never go anywhere. Of course, you can't put light exactly at a point. It would either be a little inside and get sucked in. Or a little outside and an extremely red shifted remnant would escape.

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  • $\begingroup$ Thank you for the great answer! $\endgroup$ Sep 8, 2021 at 0:48
  • $\begingroup$ "The light that reaches infinity will be less energetic that it was" - The energy of an ascending photon doesn't change when measured in the same coordinate system. If a photon arrives to me "redshifted", then in my coordinates it was emitted already "redshifted" and never changed its energy or color. So the redshift here refers to the fact that energy is generally frame dependent, but not to the energy of a photon changing in flight. $\endgroup$
    – safesphere
    Sep 8, 2021 at 6:06
  • $\begingroup$ "Time passes slowly in a gravitational well. The deeper in, the more slowly. So it takes longer to escape than you might expect. - It takes only a fraction of a second for light to escape far enough from any physically meaningful distance, even from a Planck length. $\endgroup$
    – safesphere
    Sep 8, 2021 at 6:25
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    $\begingroup$ "Inside the event horizon [...] Light pointed outward does not travel outward." - How do you point "outward" (or "inward") inside the horizon? These are directions in time, not in space. How do you point a flashlight to yesterday? $\endgroup$
    – safesphere
    Sep 8, 2021 at 6:32
  • $\begingroup$ @safesphere - Your objections are correct. I was talking too loosely. I have made some edits, but I don't know that it improves matters. I will say that a fraction of a second can be significant. GPS would not work in Earth's gravity without relativistic corrections. $\endgroup$
    – mmesser314
    Sep 8, 2021 at 14:50

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