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.


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    $\begingroup$ Your premise is wrong. Particle type doesn't matter whether mass is added or removed. $\endgroup$
    – this
    Commented Apr 21, 2014 at 13:23
  • $\begingroup$ I thought that the anti-particle was annihilating with "normal" mass inside the black hole? No? $\endgroup$ Commented Apr 21, 2014 at 14:21
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    $\begingroup$ I've wondered this for ages. $\endgroup$ Commented Mar 31, 2017 at 20:40

4 Answers 4


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.

  • $\begingroup$ 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? $\endgroup$
    – Py-ser
    Commented Apr 22, 2014 at 5:40
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    $\begingroup$ @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. $\endgroup$
    – Stan Liou
    Commented Apr 22, 2014 at 14:26
  • $\begingroup$ Thanks @StanLiou, so you've clarified the misunderstanding that both me and the OP made. So both matter and anti-matter both have positive mass/energy, of which is borrowed very temporarily from quantum fluctuations. So where does the -ve energy come from when one of the particles falls in? $\endgroup$ Commented Mar 31, 2017 at 20:47
  • $\begingroup$ @ChrisWalsh I would assume that it comes from the decrease in mass of the black hole. But one thing that is still unclear in my mind - is this answer saying that a (small enough) black hole radiates matter and anti-matter evenly, or is it preferential one way or another? $\endgroup$
    – Michael
    Commented Jun 22, 2018 at 16:52

First, I'd like to point out and commend @user83692435's reply which came first and is correct. Expanding on it:

The image of a virtual particle/anti-particle pair being created and then one of the pair being swallowed by the event horizon leaving the other as real is an analogy which provides a picture of what is happening, but is definitely not correct. Popularizers continue to use it because what is really going on is extremely complex and not easily explainable in words. (And I won't try!) But here's a link to a technical paper on the subject.

But perhaps the most telling point against the simple explanation is that Hawking radiation doesn't come from the event horizon which the analogy requires, but from the space outside it!

A second point against the Hawking analogy is that the event horizon is enormously deep in the black hole's potential well. For a particle or photon to escape the BH (which Hawking radiation must) it must be created with enough additional energy to escape the BH -- and a BH can be thought of an an object with an escape velocity greater than the speed of light. Wimpy little virtual particles which lost their partner to the BH would never make it out.

If you'd like to dig a bit deeper, I recommend Sabine Hossenfelder's blog Backreaction which has a long post with many links to further information. Backreaction is one of the best of the frontier physics popularizing blogs going these days, in significant part because Hossenfelder is an active researcher and a good writer.


This explanation of Hawking radiation as virtual particles forming and one particle falling into black hole is an incomplete one. Stephen Hawking originally imagined a path from distant past to distant future and and null geodesic(path of light) travelling it. A black hole is formed in the path of the geodesic just before it passes the place where black hole is formed. It is the last geodesic to do so.
The vacuum is not empty. It consists of some vibrations because of uncertainty principle. This vacuum field is made up of many frequency modes. They keep creating virtual particle anti-particle pair annihilating each other. The anti particle can be thought a vibration in the quantum field having negative frequency, ie, travelling back in time. The black hole formed nudges some frequencies of the geodesic that passed. So the geodesic creates it's fields from remaining frequencies. And as anti particle can be thought of as particle with frequency travelling backwards in time its frequency is always lost to black hole and the field creates a virtual particle from remaining frequency modes.


You have slightly misremembered a common (though bad) way of describing Hawking radiation. Physics popularizers sometimes describe it as a pair of particles being created, one of which is matter and the other of which is negative matter. Or one of which is antimatter and the other of which is negative antimatter. So your proton escapes and your negative-matter-proton is absorbed. Or your antiproton escapes and your negative-antimatter-antiproton is absorbed. The negative matter (or negative antimatter) shrinks the black hole.

Although this is a common way to describe things for non-physicists, it's a bad way to describe it. It's confusing because it suggests the exact question that you raised: why doesn't the negative matter fly outward, and shrink the first star or planet it hits? Also, negative matter has never been detected. There is no particular reason to think it's a useful construct for understanding black holes. (Though if it did exist, you might be able to use it to stabilize wormholes, which could be very useful).

It's better to describe Hawking radiation as the other answers here have done, without resorting to virtual negative particles.

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    $\begingroup$ "Negative-matter" - what's that then? $\endgroup$ Commented Nov 22, 2016 at 14:45
  • $\begingroup$ @adrianmcmenamin - all known matter has positive mass. Negative matter would have negative mass and theoretically negatively curve spacetime. It's highly theoretical...like, completely speculative level theoretical. Basically just a "what if?" type thing. $\endgroup$ Commented Feb 9, 2021 at 22:57

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