I do understand that nothing can escape a BH (not particles, no information), and this question is not about why you cannot escape. This question is not about Hawking radiation, and I do understand that there are different explanations about Hawking radiation, and one of them is tunneling.

I have read this question:

Quantum tunneling through the event horizon (EH): Is the EH a potential barrier for Quantum tunneling?

Why can't you escape a black hole?

where Anixx says:

It is actually possible to get out of the horizon. To get out one only have to reach speed higher than c. Note that in all cases of tunelling (say at nucleus decay) the emitted particle reaches higher than c speed.

Quantum tunnelling near the speed of light

where annav says:

The impotant thing to keep in mind is that the energy level is the same inside and outside the barrier, thus the particle must have the same momentum i.e. velocity all through.

So one of them says that tunneling particles are able to escape a black hole, and they reach a speed faster then the speed of light. In this case, nuclear decay just inside the horizon would cause particles to escape the BH.

The other one says during tunneling, the velocity of the particle must be the same inside and outside the barrier, so it must be less then c (if it has rest mass), but it cannot be faster then c.

Now I believe there are two main contradictions here:

  1. particles with rest mass traveling at speed c or any particle traveling faster then c is a violation of SR (causality)

What everyone else said, but note that this STILL violates causality if you use general relativity to create one of these "warp drive" scenarios--the "warp drive" can always be restricted to an arbitrarily small region of spacetime, and then special relativity will be true over the rest of spacetime, and the problems will still arise.

How does faster than light travel violate causality?

  1. actually even tunneling particles cannot escape a black hole, because Hawking radiation is explained by negative energy particles entering the BH, thus reducing its mass (evaporating)


Anyhow, since energy is associated with the time coordinate and momentum with the spatial coordinates, the energy and radial momentum of a particle also swap "roles" when crossing the horizon. The negative energy of the particle becomes negative momentum and the positive momentum of the particle becomes positive energy.

Black holes and positive/negative-energy particles

So in reality, no real particles can tunnel out of a black hole, and even those (tunelling) do not exceed the speed of light.


  1. Can you escape a BH by tunneling faster then the speed of light?
  • 1
    $\begingroup$ Leaving the speed question aside for a moment (and, in fact, leaving aside the more important question of whether it's even sensible to say that tunneling has a speed in the first place), tunneling out of the event horizon would require that part of the wavefunction is inside the horizon and part of it is outside. What makes you think this configuration is possible? $\endgroup$ – probably_someone Nov 24 '19 at 23:20
  • $\begingroup$ Also, doesn't anna v's answer to the original question, which says in particular that the "probability of tunneling is zero by mathematical arguments", answer this question? $\endgroup$ – probably_someone Nov 24 '19 at 23:21
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    $\begingroup$ It doesn't make sense to talk about "tunneling faster than the speed of light." The kinetic energy of a particle is negative in the classically forbidden region, so its velocity, if it had one, would be imaginary. $\endgroup$ – user4552 Nov 24 '19 at 23:24
  • $\begingroup$ @probably_someone I am on your side, I believe Hawking radiation is not tunneling, but negative energy particles entering (and reducing the mass of) the BH. I would like to understand why he (Anixx) says that you can get out by tunneling and tunneling means exceeding the speed of light. $\endgroup$ – Árpád Szendrei Nov 24 '19 at 23:25
  • $\begingroup$ why the downvote? $\endgroup$ – Árpád Szendrei Nov 24 '19 at 23:26

In this recent analysis, it is shown that the logic leading to statements of group velocity showing velocities larger than c, is misidentifying what "group velocity " means.

In the simple quantum mechanical diagram for single particles ( the experiments that gave the misuse of group velocity were not single particle experiments) that I gave in my answer, reproduced here :


it is the total wave function for the whole system , which it breaks down to $Ψ_{inc}$ and $Ψ_{exit}$ ( plus the part with no label within the barrier)that is normalized to 1, to be a probability density when one takes the $Ψ^*Ψ$ .

So the speed in this case has no meaning, because no velocity can be defined for the (x,y,z,t) points represented by the probabilities given by $Ψ^*Ψ$ .


  • $\begingroup$ I read this in the references of the link experts.illinois.edu/en/publications/… "The peak of the photon wave packet appears ", so even though they are using single photons, they are talking about the accumulation to be talking of wave-packet. a single photon is a point when measured. ( see the single photon double slit) $\endgroup$ – anna v Nov 25 '19 at 6:38

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