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My understanding of Hawking Radiation is that it is a pair of virtual particles that pop into existence at the edge of an even horizon. The anti-particle falls into the event horizon thus slightly reducing the holes mass.

My question is this: The anti particle will fall in 50% of the time... Wouldn't that mean that 50% of the time, the regular particle falls in... thus adding the same amount of mass that was just deducted?

In some cases, couldn't a black hole increase in mass if just a tiny bit more regular particles fall in that anti- particles?

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    $\begingroup$ Anti-particles don't have negative mass. I've linked questions that discuss more accurate explanations of Hawking radiation than the virtual-pairs analogy as duplicates. $\endgroup$ – ACuriousMind Sep 27 '20 at 0:46
  • $\begingroup$ I think he means that if the antiparticle (positron) falls in and meets an electron it will annihilate creating energy - he maybe missing the point that mass/energy are equivalent. $\endgroup$ – joseph h Sep 27 '20 at 1:08
  • $\begingroup$ @Drjh - No... I am trying to wrap my non scientific head around something that is not logical, but probably makes sense, mathematically. How does a particle entering an event horizon bleed energy from a singularity? If it's because of an anti-particle falling in, then the BH should never lose mass as a regular particle has the same chance of being captured. $\endgroup$ – Rick Sep 27 '20 at 14:40