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I may have some specifics wrong here. If so, don't focus on those. Just focus on the general thrust of my question.

I "understand" (cough) that particle/anti-particle pairs form spontaneously in space. I understand that they can form near the event horizon of a black hole, and that one particle can fall in, where as the other particle can barely escape. I understand that an anti-particle will annihilate with a particle. What I don't understand is why only the anti-particles of these virtual particle pairs fall into the black hole, while the other ones just manage to escape. Shouldn't both particle and anti-particle have equal chance to be the one to fall in, or just manage to escape?

It seems that there should be an equal chance of either the particle, or the anti-particle, would be captured while the other "ejected." So it seems that the black hole should be somewhat steady-state as far as mass change with respect to virtual particles.

Explain?

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Your premise is wrong. Particle type doesn't matter whether mass is added or removed. –  this Apr 21 at 13:23
    
I thought that the anti-particle was annihilating with "normal" mass inside the black hole? No? –  user3355020 Apr 21 at 14:21

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I thought that the anti-particle was annihilating with "normal" mass inside the black hole? No?

No. First, both particles and anti-particles have "normal" mass (should they have mass in the first place) and "normal" (positive) energy. The distinction between them is either a matter of convention or a question of which type is more common in the universe. Furthermore, for typical-massed blacked holes, the bulk of Hawking radiation would be made of photons, which properly speaking do not even have anti-particles, though one could also say that they are their own anti-particle.

Shouldn't both particle and anti-particle have equal chance to be the one to fall in, or just manage to escape?

Yes, and uncharged ones do. A smaller black hole would radiate both neutrinos and anti-neutrinos, assuming all neutrinos are massive (otherwise, all black holes would do it already), and a sufficiently small (and thus sufficiently hot) one would radiate both electrons and positrons. Very roughly, a black hole will radiate non-negligible amounts of massive particles when the temperature of the black hole is on the order of the particle mass or greater, in natural units.

It seems that there should be an equal chance of either the particle, or the anti-particle, would be captured while the other "ejected."

Correct, with a minor exception that if a hot black hole has electric charge, it is more likely to radiate particles of the same sign of charge.

So it seems that the black hole should be somewhat steady-state as far as mass change with respect to virtual particles.

If either a particle or an anti-particle falls into a black hole, its mass will go up. It doesn't matter. Fundamentally, the "reason" for Hawking radiation is that the vacuum state in quantum field theory is a state of lowest energy, but different observers can disagree about which state is the vacuum. Thus, since particles are fluctuation on top of the vacuum, they can disagree about whether or not there are particles.

I don't think there is a good way to repair the "antiparticle falls in" story except some roundabout appeal to energy conservation: if the escaping particle is real and have positive energy, the one that fell in must have negative energy, and would therefore decrease the mass of the black hole. Unfortunately, that only shows what must happen for the situation to be consistent, not that it does actually happen.

Although with some knowledge of general relativity, one can motivate this slightly further--e.g., for the Schwarzschild black hole, there is energy conservation given by a Killing vector field, which goes from timelike to spacelike at the horizon--so what an external observer considers time/energy would be space/momentum inside the black hole, and momentum is allowed to be negative.

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I do not understand very well your answer, but most of all your first one: are you saying that an electron and a positron do NOT annihilate each other? –  Py-ser Apr 22 at 5:40
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@Py-ser: I'm saying that whereas an electron an a positron may annihilate each other, that has nothing to do with why a black hole decreases mass through Hawking radiation. A hypothetical particle/antiparticle annihilation inside the black hole would do nothing to the mass, since both have positive energy. You're treating antiparticles as something special in regards to this process, but this is a mistake. A better (though slightly handwavy) view is that whichever particle falls in, it has negative energy rel. to an observer at infinity. That's completely different from m/am annihilation. –  Stan Liou Apr 22 at 14:26

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