According to what I've read on special relativity, $c$ is the speed limit for every object in the universe, and according to Einstein, an object's speed through the three spatial dimensions plus its speed through the fourth temporal dimension always sums to $c$.

I once watched a video demonstrating what it would be like to fall into a black hole. At one point the author stated that if a photon was emitted directly away from the singularity and at a distance equal to the Schwarzschild radius, the photon would hover there for eternity.

My question is based on these assumptions so please let me know if either is incorrect.

With its motion through the spatial dimensions halted, will the photon in question not experience time at roughly the same rate we do? Will it decay?


First, a couple things:

  • photons do not "experience" time in general, precisely because they always travel at $c$.
  • because of the above point, photons do not decay, either.

To better visualize what's happening, consider that the event horizon is a place where spacetime itself is "falling" into the black hole at the speed of light. So, if you emit a photon precisely as you pass the event horizon, the photon's physical motion through space would be exactly counteracted by the spacetime curvature at the horizon, and it would effectively "hover" at that location.

However, the way that this photon would be observed is very different for different observers:

The in-falling observer, who emits the photon just as she passes the event horizon, will believe that the photon is traveling away from her (at $c$, as usual).

However, a distant observer (far from the black hole) will never see the photon at all. Or rather, to the distant observer, the photon will appear to be infinitely redshifted.

An observer who falls into the black hole after the first in-falling observer still has a chance to observe the photon (not redshifted).

But more practically speaking, this kind of "equilibrium" would be highly unstable (the photon would not be able to "hover" for very long). This is because the black hole's Schwarzschild radius is always changing slightly, whether it's due to the black hole absorbing CMB radiation, or emitting Hawking radiation. So, the photon will eventually either escape, or be dragged deeper into the black hole.

  • $\begingroup$ Is it necessarily true that massless particles don't decay? That seems wrong to me, though I have to invoke non-free states like gluons to come up with an obvious counterexample. $\endgroup$ – Jerry Schirmer Oct 31 '13 at 21:13
  • $\begingroup$ @JerrySchirmer: I don't think gluons can be used as counter example - the decaying ones should be off mass-shell... $\endgroup$ – Christoph Oct 31 '13 at 21:19
  • $\begingroup$ @Christoph: Yeah, and they're not physical states anyway. $\endgroup$ – Jerry Schirmer Oct 31 '13 at 21:44
  • $\begingroup$ @Jerry I think it is necessarily true, because if a massless particle (on shell) were to decay, the products would necessarily have a rest frame, in which conservation of momentum would have been violated. The only exception AFAIK is collinear branching, but I wouldn't consider that to be a decay process. $\endgroup$ – David Z Oct 31 '13 at 21:59
  • $\begingroup$ @JerrySchirmer: for what it's worth, the fact that massless particles cannot decay makes sense from a classical (relativistic) point of view: quantum-mechanical decay is random and has a finite probability to happen in a certain time interval; now, think about proper time for massless particles... $\endgroup$ – Christoph Oct 31 '13 at 22:00

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