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There are a lot of questions on this site about photon speed, none of them answer my question:

Why and how is the speed of light in vacuum constant, i.e., independent of reference frame?

If refraction slows down light, isn't it possible to hold light still?

Photons are massless, elementary particles as per the SM, they always move at speed c in vacuum, when measured locally.

And, photons do not have a rest frame.

I have read this question:

Is the Schwarzschild horizon lightlike?

where Yukterez says:

If an observer on a timelike path emitts a radially outwards directed photon when he crosses the horizon, it will stay there forever.

Can a photon have little to no energy and/or speed?

where annav says:

No, a photon can never move more slowly than the speed of light in vacuum, because as an elementary particle it is either in vacuum or interacting with another elementary particle or field.

Now there are other questions and answer on this site, but those are talking about light in a medium, slowing down.

I am talking about photons in vacuum, when emitted outwards from the EH. And when measured locally, the speed of photons is always c in vacuum. But photons do not have a rest frame, and to all observers, the photon's speed seems to be c, regardless of the motion of the light source or the observer as per SR.

https://en.wikipedia.org/wiki/Special_relativity

enter image description here

After the comments, it needs to be clarified, which frame will observe the photon frozen, the emitter, or the external observer?

Question:

  1. Which one is correct, can a photon be frozen on the EH, or not?
Árpád Szendrei
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  • The instantaneous speed of a photon at a given event $E$ is always $c$ in the coordinate system induced by the frame of any inertial observer located at $E$. There are also (of course) coordinate systems in which the speed is not $c$. You seem to already know all this, so what is your question? – WillO Feb 26 '20 at 23:27
  • @WillO "If an observer on a timelike path emitts a radially outwards directed photon when he crosses the horizon, it will stay there forever." What does this mean? To which observer will it seem frozen? The emitter or an external? – Árpád Szendrei Feb 27 '20 at 02:41
  • Comments are not for extended discussion; this conversation has been moved to chat. – ACuriousMind Feb 27 '20 at 22:03
  • The light cone of any event is light-like. Is the photon emitted at a given event frozen on the light cone? – MBN Mar 02 '20 at 07:53

2 Answers2

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The event horizon is light like, so yes light may be “frozen” on the event horizon in whatever way it makes sense to say that something is frozen on a light like surface.

If you are freely falling through the EH then in your local inertial frame the EH is moving past you at c. The light stays fixed on the EH as they both move past you together at c. That is what it means that the EH is light like. It moves at c in any local inertial frame. Although it has a constant $r$ Schwarzschild coordinate it is light like as described above.

Dale
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  • thank you, can you please tell me which observer will observe the photon to be frozen, the emitter or the external observer? – Árpád Szendrei Feb 27 '20 at 19:47
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    an observer might not even notice crossing the horizon as far as I know. Though, if it is possible, to somehow detect it (the event of being at the EH) and emit a photon from the horizon outwards, would the emitting observer observe the photon frozen or moving at speed c? – Árpád Szendrei Feb 27 '20 at 21:52
  • In the edited picture, will the emitter observer see the reflection of the photon? Can the external observer observe radiation pressure on the mirror? – Árpád Szendrei Feb 27 '20 at 23:10
  • this is exactly the question. both cannot happen. Just one of them, and that is the question which one happens, does it remain at the EH or does it hit the mirror? – Árpád Szendrei Feb 28 '20 at 04:26
  • what do you mean different scenarios? you emit light outwards from the EH. it can either leave the Eh or not. which one? – Árpád Szendrei Feb 28 '20 at 04:34
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It's a matter of semantics. But I don't like the term "frozen" because what does that mean in the context of a photon which moves with c locally, as any observer could measure in principle while crossing the horizon. It's sufficient to say that photon "stays" at r = 2M.

timm
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  • thank you, you are saying that the photons is staying at r=2M, but at the same time you say that any observer (when crossing the EH) would measure the local speed of the photon to be c? So the photon would stay at r=2M for the external observer only? – Árpád Szendrei Feb 27 '20 at 19:04
  • Staying at r=2M means staying at a lightlike surface. This doesn’t depend on any observer but could be measured by an observer who is just crossing this surface, the horizon. This is nothing special, because an observer measures the speed of light locally frame independed always with c, above, at and below the horizon. – timm Feb 28 '20 at 17:29