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I understand the simple version answer to this question what does not make sense is if most of the matter around us is normal matter then for the blackhole to losemass from the virtual particle pairs. Doesnt more antimatter or antiparticles need to fall in then positive?

I understand the temperature or radiation one should see is one of the particles escaping the virtual pair. But unless the virtual pairs are always created with a slightly above 50% chance that the anti particle always is crossed the event horizon and the regular one on the other side how is any mass actually being lost?

Ive looked and am really surprised nobody's asking this. Conservation of energy isnt a valid answer. Because if a particle escapes why wouldn't it just vanish instead of the one in the blackhole?

My understanding is that hawking radiation has more to do with the fields in bent space time and the reference frame point then actual virtual particle pairs. Thats just the simple answer.

Also if mass is just energy and a particle and antiparticle collide they release energy How would this released energy escape the blackhole if it happened inside the event horizon. Wouldnt this just stay trapped in yhe gravity well. And if gravity separates the virtual pair and they dont annihilate then new energy has been created hasn't it?

I know im miss some things here. I just haven't found a satisfactory explanation for how the virtual pair reduces any mass inside the blackhole. Unless the two particles are quantum correlated and when one interacts with one inside the blackhole it basically teleports to the one outside the blackhole and the pair ceases to exist because it cant exist for more then a fraction of time anyways.

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    $\begingroup$ Possible duplicate of An explanation of Hawking Radiation $\endgroup$
    – StephenG - Help Ukraine
    Commented Mar 15, 2018 at 6:06
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    $\begingroup$ The whole "two virtual particles are created, and one falls in the black hole" is a rather heuristic way to talk about it. It's a nice analogy, but you shouldn't take it too seriously. $\endgroup$
    – Chris ♦
    Commented Mar 15, 2018 at 7:58
  • $\begingroup$ You seem to be under the impression that antiparticles have negative mass. They don't. However, that's not particularly relevant here since virtual particles are off-shell, so they can have nonstandard mass. And as mentioned by Chris, the virtual particle description of Hawking radiation is just a useful heuristic (like any description involving virtual particles), so don't take it too seriously. Please see math.ucr.edu/home/baez/physics/Relativity/BlackHoles/… $\endgroup$
    – PM 2Ring
    Commented Mar 15, 2018 at 13:10

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You are missing that the particle and antiparticle loops in vacuum are closed and have the quantum numbers of the vacuum and zero energy/mass. They are a mathematical construct in quantum field theory, and can affect real measurements only if energy is supplied to them, as here:

loop

An electron running along, interacts with vacuum loops of electron positrons, and the incoming energy, is shared by the group, and taken back at the edge.

Near the event horizon of the black hole there is a quantum mechanical probability of such a loop gaining energy from a particle on the horizon, (or even a graviton leaving the black hole, with small probability). The strong gravitational field can break open the loop ( in a mathematically complicated manner) pulling in one of the two ( e+ or e-) and the other, by energy and momentum conservation leaving off if the energy supplied is enough. In the case of e+e- loops, the energy supplied by the horizon particles has to be at least twice the mass of electron plus enough energy for one electron or positron to escape the black hole gravitational field. All this energy has to be supplied by the mess in the horizon, half of it will fall back, but half will leave with the free particle, diminishing the total energy momentum vector of the black hole.

Please see my answer here also.

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