I've heard "light cannot escape a black hole" explained several ways. One is that if a photon inside the event horizon tries to escape a black hole it loses energy to gravity. As it loses energy its wavelength gets longer and longer until its energy is zero.

Where does that energy go and how is it transported?


This is just ordinary potential energy from first semester physics -- when the photon is close to the black hole, it's deep in the potential well. As it goes away from the black hole, it picks up gravitational potential energy, so therefore, it must lose kinetic energy. For a photon, the kinetic energy is given by the Planck formula $E = hf$, so the photon redshifts until its frequency is zero.

  • $\begingroup$ Near a black there are large number of photons that came from the black hole and can't be detected, because they only have potential energy. Do you and the guy in the video really claim this? $\endgroup$ – stuffu Apr 23 '16 at 21:39
  • $\begingroup$ @stuffu: I know this is late, but no, that's not the claim I'm making at all. $\endgroup$ – Jerry Schirmer Nov 27 '16 at 20:04

Since due gravitational time dilation something takes forever to fall inside a black hole from the perspective of an outside observer, there is nothing in it yet that could come out.

The light which is emitted from outside the horizon does reach an outside observer, but since it hasn't yet fallen in, it technically doesn't escape from inside the black hole because it was always on the outside.

The energy stays conserved, just the power of the light gets reduced, but therefore the duration of the received signal is longer. Since energy = power x time, there is no energy problem.

If you don't throw photons but stones, the sum of kinetic and potential energy also stay conserved: if you throw it up and it comes back or someome catches it and throws it back down with the same impulse as he reveiced it, the impact when it comes back will have the same kinetic energy as the launch. The same goes for photons if you replace the stonethrower with a mirror.

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    $\begingroup$ I understand that the energy remains inside the event horizon. What happens to it inside the event horizon? Or is this a nonsense / not known to physics question? $\endgroup$ – Schwern Apr 19 '16 at 19:14
  • $\begingroup$ As I said, all the energy that gets emitted outside the horizon reaches the outside observer in a finite time. Everything that gets emitted inside the horizon does not have a corresponding time on the outside for its own finite proper time, since it already took forever to fall in. So if something from inside should reach you it would first have to be emitted, but it can't, since in your system it is still outside the horizon. $\endgroup$ – Yukterez Apr 19 '16 at 19:16

A photon that escapes from black hole's neighborhood does work on the black hole. The photon causes the black hole not to be in the photon's gravity well after the photon has escaped. In other words the photon increases the potential energy of the black hole.

The following paragraph may not be science, I just want to say something sane, as opposed to "photon that tries to escape loses energy":

A photon that tries to escape, tries to do all the things that a photon that escapes does: it tries to escape, it tries to redshift, it tries to do work on the black hole, it does not do those things, it tries to.

What does a upwards directed photon exactly at the event horizon do? Well, it stays at the event horizon, so no change of potential energy of photon happens, and no redshift of photon happens.


protected by Qmechanic Apr 20 '16 at 6:06

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