Can any object pass into the event horizon of a black hole and then escape?

With the exception of Hawking radiation - which (as I understand) is the capture of one member of a virtual pair of photons that appears at the event horizon by the black hole with the "escape" of the other photon into the surroundings - I find no conjectures that any particle of mass could cross a black hole's event horizon and then pass back across the event horizon - essentially "escaping" from the black hole in the sense that it avoids falling into the singularity. I am not claiming others have not addressed this conjecture , just that I have not found any in my literature search.

Certainly there are many published explanations that say "no", but consider the following scenario:

Although it appears that an electrically charged black hole may not ever occur naturally, there are two general-relativity solutions that do allow for the possibility of an electrically charged black hole (the Reissner-Nordstrom and the Kerr-Newman black holes).

If a charged particle's velocity trajectory intersected the event horizon of a charged black hole at the limit of a tangent trajectory (an infinitesimal fraction of a degree below the tangent and therefore crossing the event horizon);could there be a case where the magnitude of the electric charge of the black hole and the opposite magnitude of the charge of the particle of mass combined with a sufficiently high magnitude of velocity of the particle would allow the particle to then recross the event horizon into normal space?

Essentially, could the coulombic repulsive force between the particle and the singularity diminish the net force of the singularity acting on the mass of the particle sufficiently for this hypothesized escape trajectory to be possible?

The answer is no. But you are not too far from something that is possible to extract energy, charge and mass from a BH, without extracting any actual particles. Two reasons, besides the obvious one that no particle escapes the horizon

First, why can't what you say happen: The repulsive electric Coulomb like field from the BH is already being accounted in whether the charged particle can get close enough to the horizon. A BH with a certain mass and angular momentum J can only exist up to a certain amount of charge. It's called an extremal (Kerr Newman) BH. If a charge tries to join it, even at the speed of light, it won't be able to. The BH is already at that point holding as much charge as possible to remain a BH, any more and the self repulsive would repel its own mass. It's been tried in simulations and numerical solutions. The same is true for a maximum J, any more and the 'centrifugal-like effect would force matter out.

How is that relevant? A charged BH has its own mechanism for repelling charge approaching it. If it turns out it can pass the horizon, it's because the BH was not extremal, and could support more charge. Once inside it stays there.

But there is a second reason, phenomena and factor that elucidates even more, and that's close to what you were saying, but without going inside the horizon. The Penrose process. Penrose did it first for J. He found that if you have a particle rotating around the BH in the ergosphere (if I remember right, I'm going a little fuzzy here but the effect is real, and I have below a reference and it's references you can get more on), and it's J is aligNed with (again, not sure if anti-aligned) the J for the BH, it split into a particle and a virtual particle, with the virtual,particle falling inside the BH horizon, and the real particle can escape from the ergosphere (which is outside the horizon) and carry with it more J and actually more energy than what it came in with. The virtual particle goes in with negative mass and opposite J, and the net effect is That the BH looses some mass and J, and gives it to the outgoing particle. It's called extracting energy from the BH. The BH gets a lower mass, and J, but looses no real particle. The same is true by extracting charge,having a charged particle going in, it can extract charge from the BH, and mass.

All the laws of BH thermodynamics (such as total entropy increases, with BH entropy given by its horizon area) still hold. The process can extract large percentage of the energy of the BH, and could be the basis for energetic jets. It does require a particle getting close to the BH, and some of the right directions of the motion. The effect has not been specifically observed yet.

See the wiki article below about the Penrose process for J. Also true for charge, see Ref. 5 in the wiki article. There's been many other papers, there's no controversy, it's accepted. See https://en.m.wikipedia.org/wiki/Penrose_process

• "If a charge tries to join it, even at the speed of light, it won't be able to." Actually it can, but the kinetic energy of the charge would need be very high (it would have to be "force-fed" and would be repelled until close to the horizon). All this kinetic energy adds to the mass of the hole keeping it from becoming super extremal. Commented May 31, 2022 at 0:10

Hawking radiation is not escaping the black hole. Instead, the effect of this apparent "escaping" is due to two random particles appearing, one extremely close to the event horizon of the black hole, the other just slightly further away. The particles decouple as the one next to the event horizon falls into the black hole, while the other particle floats away.

• This virtual particle pair explanation accounting for Hawking radiation does not work, Hawking radiation causes a black hole to lose mass. For this to happen the ingoing particle would have to have some sort of negative mass or energy while the escaping particle has positive or real mass or energy. The problem with this explanation is it that it is equally likely on average that the ingoing particle has positive mass and the outgoing particle has negative mass and the net loss of mass from the black hole is zero.
– KDP
Commented Apr 29 at 9:45