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?
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.
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.
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.
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.
