Comment Re:Gravitational tides will kill you (Score 1) 412
Instead, it can take a few swings around the black hole in a rapidly decaying orbit, until it slingshots out on a hyperbolic path. The smaller the black hole gets, the more definite the position is for every matter/antimatter particle pair, and by Heisenberg's uncertainty principle applied to position-momentum, this makes it easier for one of the two particles to escape.
Erm, that's not how orbital/slingshot mechanics work. In fact, a mass-bearing particle (from a virtual pair) could never escape a stationary black hole, because it wouldn't have enough energy to do so, and normal orbital mechanics wouldn't increase its energy.
Instead, you'd need a rotating black hole with an ergosphere. This is a weird area where space-time is dragged along the black hole faster than the speed of light relative to outer space. Here it is possible to extract energy from the black hole with what is called the Penrose process, and thus the electrons/positrons may gain enough speed to escape.
I was simplifying things for the audience, and wasn't even remotely about to bring up frame dragging. That said, I'm an interested layman who's never taken a physics class touching GR so correct me if I'm totally off-base, but I'm pretty sure that Hawking's conclusion was initially solved for Schwarzschild black holes, and if so the ergosphere around Kerr black holes clearly doesn't come into it. The region between the photon sphere and the event horizon has no stable orbits, but wouldn't there still be some trajectories that would send a particle out past the photon sphere? (On a more circuitous path than a straight line, I mean.) And the Wikipedia article for the photon sphere says that "[a]ny orbit that crosses [the photon sphere] from the inside escapes to infinity".
I will admit to never having plugged numbers into a tensor equation in my life, so I could be totally bullshitting here.