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It occurred to me in passing that the Lorentz contraction of a black hole from the perspective of an ultra-relativistic (Lorentz factor larger than about 10^16) particle could reduce the thickness of a black hole to less than the DeBroglie wavelength of the particle.

It would seem to me that under those conditions the particle would have a non-insignificant probability of tunnelling right through the black hole rather than being adsorbed by it.

Is this so? If not, why?

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4 Answers 4

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The important thing is the cross-sectional area of the horizon, and this is independent of Lorentz transformation, since the $y$ and $z$ coordinates are not changed.

Additionally, you can calculate that light will be captured by the horizon with non-zero cross section, and the geodesics ultra-relativistic particles will asymptote to the geodesics of null particles.

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    $\begingroup$ I'm not getting why the cross-sectional area of the horizon matters. We are talking about the probability of the particle tunnelling through a thin potential barrier. The size of the barrier in the directions perpendicular to the direction of motion would seem to be irrelevant. To put it another way - if I were to replace the black hole with a normal matter wall, how would that differ for the question of tunnelling a ultra-relativistic particle through it? $\endgroup$
    – Snowhare
    Jan 29, 2013 at 17:26
  • $\begingroup$ @BenjaminFranz: thinking about it as a potential barrier is a bad way to think about it. In general relativity, the only important thing is the geodesics. You will get messed up if you take the classical picture of energy too seriously. And irrespective of that, the limit to null particles circumvents that issue. $\endgroup$
    – Zo the Relativist
    Jan 29, 2013 at 17:29
  • $\begingroup$ And above I should say, rather than the "classical picture of energy", "the idea of the energy of a single particle" $\endgroup$
    – Zo the Relativist
    Jan 29, 2013 at 17:39
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    $\begingroup$ I'm not sure we aren't talking past each other. My question is about the quantum mechanical possibility of the particle quantum tunnelling 'past' the barrier posed by black hole without ever actually 'entering' it (ala the Klein Paradox). I'm happy either way - I just want a clear explanation of why it can or cannot. $\endgroup$
    – Snowhare
    Jan 30, 2013 at 1:32
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    $\begingroup$ This answer (v1) argues purely classically within GR (where the word classical here is used in the sense $\hbar=0$, which was also covered recently here), and seemingly fail to address OP's main quantum tunneling question. $\endgroup$
    – Qmechanic ♦
    Jan 30, 2013 at 15:47
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Quantum tunnelling occurs when a particles does not have enough energy to break through a potential barrier (classically). The wave function that describes the probability distribution of the position cannot instantly drop to zero at the boundary so there is a small chance the particle can exist on the other side. A black hole does not present a potential barrier the particle would just travel into the black hole normally.

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Actually I've been thinking about this and its not as simple as I thought.

The most complete description we have of black hole in GR is the Kerr–Newman metric so we should use that description. This describes a rotating charged black hole.

A charged particle could be repelled by the Kerr-Newman black hole so in this case it would present a potential barrier and indeed the particle could tunnel through it.

So it depends on the properties of the particle.

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Light (apart form high energy particles) can't even tunnel through a thin piece of metal let along a black hole!

If you packed enough energy into your particle, perhaps it would release gravitational waves on the way in, so some of the incoming energy would escape.

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