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From what I understand about Hawking radiation, a virtual particle comes into existence very near the event horizon of a black hole. This causes one part of the particle to fall into the black hole, and the other can radiate outwards. When the anti-particle falls in, it will collide with a particle and annihilate, making the black hole lose mass.

Does that mean that it's only anti-particles that are being captured by the black hole? Wouldn't we expect the non-anti-particle to fall in just as often, which would make the black hole stay about the same mass? Is there any reason that an anti-particle from a pair of virtual particles would be more likely to descend into the black hole?

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  • $\begingroup$ One should point out, again, that the particle/antiparticle analogy is simply wrong. A black hole will overwhelmingly emit uncharged photons, not charged particles (which would leave it in a charged state). This still requires a balance of angular momentum, but whether that exchange is really handled by a second real photon is (in my opinion) questionable. None of the current theoretical explanations for Hawking radiation is likely to be correct, since there simply is no established theory of quantum mechanics at the event horizon. What we have is QFT shoehorned on top of a classical theory. $\endgroup$
    – CuriousOne
    Commented Sep 30, 2014 at 5:03

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