From what I understand so far, when one of virtual particles crosses the event horizon and the other does not, they can not annihilate each other. The latter wanders off into the universe (btw, is it still virtual at this point, and what does 'virtual' mean at this point, if so?), while the other gets consumed by the black hole. I don't see how this event contributes to evaporation of the black hole (since the particles do not originate from the black hole). Shouldn't the consumed particle actually add-up to the black hole mass?

The closest question to mine is Does Hawking radiation in fact bring mass into the universe?, but I don't find the answers satisfactory.

I.e. "the escaped virtual particle is 'boosted' by black hole's gravitational field into becoming a real particle", rather adds to the question then answer it.

EDIT: I am humbled by the knowledge presented in the answers and feel incompetent to mark any as the most suitable. I hope this is okay.

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    $\begingroup$ Forget about virtual particle pairs, that's more like a metaphor. The thing is, black holes emit radiation, regardless of how they do that. Radiation carries energy, which must come from somewhere, there's no free lunch here. But energy equals mass. It all comes out of the black hole's "bank account" of mass, because that's the only thing nearby. $\endgroup$ Commented Apr 25, 2019 at 9:32
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    $\begingroup$ @Marko36 It's not that virtual particles are a metaphor in general (although in a sense they are, all particles, virtual or otherwise are just a way of viewing some aspects of the underlying fields), but they are not really a very good explanation of Hawking Radiation. This, however, doesn't answer your question. I look forward to seeing an answer. $\endgroup$ Commented Apr 25, 2019 at 10:04
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    $\begingroup$ You might like to check out math.ucr.edu/home/baez/physics/Quantum/virtual_particles.html and physics.stackexchange.com/questions/185110/… and other related questions on Physics about virtual particles. $\endgroup$
    – PM 2Ring
    Commented Apr 25, 2019 at 12:53
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    $\begingroup$ This question on physics (and its accepted answer come the closest I have yet found to addressing this question) but they still don't answer it completely. physics.stackexchange.com/questions/251385/…. $\endgroup$ Commented Apr 25, 2019 at 14:43
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    $\begingroup$ @Marko36 "virtual particles" are internal lines in Feynman diagrams and should not be reified more than that. QFT never assigns a state to a "virtual particle". They are entirely metaphorical. See the answer by Arnold Neumaier (physics.stackexchange.com/a/252183) and his links. $\endgroup$ Commented Apr 26, 2019 at 17:24

5 Answers 5


I'm going to give you an intuitive answer. Keep in mind, this is not the "actual" answer, as the Hawking radiation is quite a bit more complex than the typical pop-sci explanation with virtual particles. But some intuitive justification is possible nevertheless.

I don't see how this event contributes to evaporation of the black hole (, since the particles do not originate from the black hole).

You're missing a key point here.

When the pair was generated, those were virtual particles. After one side of the pair was absorbed by the black hole, and the other side was released, the released part is a real particle. Huge difference there - virtual vs real.

Virtual particles don't really exist the same way that you and me exist. They seem to exist for a very short time; the more energetic they are, the shorter the interval of their virtual "existence", per the Heisenberg equation. In many ways they are just a mathematical trick.

Think of the vacuum, where no real particles exist. Before, it's just vacuum. Right now, a virtual pair flickers briefly, then it's gone. In the future, it's vacuum again.

What was the energy before? Zero. What is the energy in the future? Zero. What's the energy during the flicker? Well, it basically zero, within the limits permitted by Heisenberg's equations. Bottom line is, virtual particles come and go, and they do not contribute to the energy balance of some empty chunk of space.

(I am ignoring here the concept of vacuum energy, for the sake of an intuitive explanation.)

But let's say one of the virtual particles gets trapped by the black hole, so it cannot annihilate with its counterpart. The other particle flies off in the opposite direction and escapes the black hole. What's worse, this is now a real particle - we've exceeded the duration permitted by the Heisenberg equations, so the one that escapes is not virtual anymore.

How did that particle become real?

This is a big issue, because virtual particles don't require an energy budget to briefly exist, while real particles do carry energy forever. Something prevented the virtual pair from annihilating itself, and boosted one of the components to the status of real particle. The virtual pair has zero energy. The real particle that gets away has non-zero energy. That energy has to come from somewhere.

It comes from the black hole. The black hole gives up some of its mass / energy (same thing) to boost one particle from virtual to real. The other particle is captured - but being virtual anyway, it doesn't really matter.

What this intuitive explanation doesn't say is how the boost actually happens. I dunno, magic. Somehow one of the virtual particles gets a chunk of energy from the black hole and becomes real.

Again, this is not the actual process. The actual process is more complex. This is just a pop-sci fairy tale.

EDIT: To hit closer to home, Hawking radiation is more like a close relative to the Unruh effect. Say an inertial observer sees empty space here in this chunk of volume. An accelerating observer would not see empty space in the same volume, but instead would see blackbody radiation. That's the Unruh effect.

Well, gravity and acceleration are the same thing, per general relativity. So the strong gravity near a black hole is equivalent to strong acceleration. Something similar to the Unruh effect must happen there. That's the Hawking radiation.


EDIT2: The other answers currently on this page provide useful alternative points, so check them out too.

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    $\begingroup$ This "pop-sci fairytale", as you called it yourself, is quite a pleasant read, I even laughed. Thanks. But it is this "I dunno magic" I am after: how does the virtual particle get it's real state (besides magic) and how does this contribute to BH evaporation, having that nothing can escape the black hole.. $\endgroup$
    – marko-36
    Commented Apr 25, 2019 at 11:21
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    $\begingroup$ @Marko Also see math.ucr.edu/home/baez/physics/Relativity/BlackHoles/… $\endgroup$
    – PM 2Ring
    Commented Apr 25, 2019 at 12:49
  • $\begingroup$ Do you get particles and anti-particles in equal proportion? $\endgroup$ Commented Apr 26, 2019 at 12:11
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    $\begingroup$ I'm struggling with one part of the intuition - why is the particle which escapes converted to "real" while the one which is captured (and thus doesn't annihilate any more than the escaping one) remains "virtual"? I think the answer needs to address what is meant by real vs virtual. My intuition tells me that the escaping particle and the captured particle are just as real as each other, and thus the black hole may have lost mass/energy allowing one to escape, but it has gained the mass/energy of the captured one, resulting in an overall loss of zero. $\endgroup$
    – JBentley
    Commented Apr 26, 2019 at 16:52
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    $\begingroup$ Indeed the "Heisenberg/virtual-particle" explanation is false. Uhruh radiation is a better analogy given the Bogoliubov transformations' use.. But a simplified, accurate explanation of why mass decreases is still missing from this site. $\endgroup$ Commented Sep 13, 2020 at 2:32

These lecture notes address the issues to some degree, especially on slides 33-35.

Because in the strongly warped spacetime near the horizon, virtual particles made from vacuum fluctuations turn out to have negative energy density.

Energy density = energy per unit volume.

These particles indeed have positive mass -- look at the one that escaped! -- but their mass is distributed very strangely over spacetime. (Quantum-mechanically speaking, particles have nonzero volume; this is an aspect of the wave-particle duality.)

Matter with negative energy density is generally called exotic matter

and, a bit later:

Quantum mechanical vacuum fluctuations in flat spacetime - far from any strong gravitational field - always have zero net energy density; they can never be exotic.

However, in warped spacetime, vacuum fluctuations are in general exotic: their net energy density is negative, according to a distant observer measuring the energy density by observation of the deflection of light by the ensemble of fluctuations. The stronger the curvature, the more negative the energy density looks.

This is the best explanation I have seen so far.

  • $\begingroup$ This is great ! $\endgroup$ Commented Apr 25, 2019 at 21:46
  • $\begingroup$ I like the explanation in this lecture as well. Which is why I linked it in my answer hours ago... $\endgroup$
    – user27815
    Commented Apr 25, 2019 at 21:53
  • $\begingroup$ @jakub_d. Oops sorry. Would you like me to delete mine? $\endgroup$ Commented Apr 25, 2019 at 22:46
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    $\begingroup$ This is a complex phenomenon. I think the question would benefit from having a diversity of answers. $\endgroup$ Commented Apr 25, 2019 at 23:11
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    $\begingroup$ I don't mind, but if people keep copy-pasting different bits of the lecture, soon we will have the whole thing on here. :) I would encourage interested people to go read the lecture notes, they're really interesting. $\endgroup$
    – user27815
    Commented Apr 26, 2019 at 9:21

Heisenberg's principle allows you to temporarily violate energy conservation laws (e.g. create pairs of particles out of nothing) as long as you repay everything in time. The larger the particle-antiparticle pair, the quicker it has to be repaid. Converting a virtual pair to a real pair can be seen as generating a bit of negative-energy "exotic matter" (whatever that is) to represent the unpaid debt. Its energy is equal in size to the pair with the opposite sign. This then falls into the black hole along with one of the particles, decreasing the mass of the black hole overall.

The horizon of the black hole gets in the way of recombining some virtual pairs, so these conversions virtual->real will happen.

I found this lecture with the same idea (more detailed and less butchered): http://teacher.pas.rochester.edu/Ast102/LectureNotes/Lecture19/Lecture19.pdf

  • $\begingroup$ Pop-sci theorizing here, but is "exotic matter" actually necessary to balance this equation? If a virtual matter-antimatter pair self-instantiates and the anti-matter particle falls into the black hole while a regular-matter-particle escapes, shouldn't it annihilate a "particle's" (yes, I know black holes are weird™ and black hole particles aren't really a thing, that's why it's in quotes) worth of matter from the black hole, or has the black hole essentially forgot that it's constituent mass used to be composed of matter? $\endgroup$
    – Sidney
    Commented Apr 25, 2019 at 17:38
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    $\begingroup$ @Sidney antimatter still has positive mass, when a positron meets an electron, we call it annihilation, but the result is not zero, it is two gamma photons of equivalent relativistic mass. $\endgroup$
    – user27815
    Commented Apr 25, 2019 at 17:43
  • $\begingroup$ This answer triggered a thought I couldn't help but share. The concept of an Alcubierre warp drive requires said 'exotic matter', so is it theoretically possible to take advantage of this to jump-start a warp drive? Or is this not the same type of exotic matter / energy? $\endgroup$ Commented Apr 25, 2019 at 21:10
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    $\begingroup$ @KellyS.French If you want to harvest some negative energy density stuff, you would be better off playing with flying mirrors and conductive plates (see Casimir effect). It won't work either, but at least you don't need a black hole for it. :) $\endgroup$
    – user27815
    Commented Apr 25, 2019 at 21:47
  • $\begingroup$ Why this should happen as such that the negative whatever it is fall inside? By this picture one would expect an average null effect. $\endgroup$
    – Alchimista
    Commented Apr 26, 2019 at 7:19

I don't know if the experts will agree with this description, but here is how I understand it:

Both space and the event horizon are in constant quantum fluctuation. Essentially, the event horizon has tiny ripples. At points where the event horizon ripples up (above the average radius of the black hole), it has an above average amount of local energy. The intense gravity rapidly pulls that local bump back down, the falling bump sends that local energy concentration back across the rest of the event horizon.

Now let's consider possible virtual particle-pairs near the hole. If a stationary virtual particle-pair appears just above the event horizon, it will either recombine and vanish or the entire thing gets pulled into the hole and vanishes in zero. We need a virtual particle-pair that has an apparent motion away from the black hole, at nearly the speed of light. If that that virtual particle-pair is going fast enough to completely escape, they recombine and vanish. Zero net effect. We need a virtual particle-pair that is moving away from the black hole at nearly the speed of light, and we need a ripple in horizon which only catches one virtual particle. I believe the ripple must be under extreme downwards acceleration for it to pull away from the second virtual particle, to avoid catching both. And here's the key part: The energy-debt between the particle pair intensely pulls them towards each other. The trapped particle is being pulled upwards, effectively pulling upwards on the horizon that trapped it. This slows the fall of horizon-ripple, diminishing the energy that the falling ripple returns to the rest of the black hole.

The energy required to pull the two virtual particles apart equals the combined energy of the two non-virtual particles. So the falling ripple loses energy equal to two particles, and the hole eats one particle. Everything balances out with the one escaped particle.

I believe it works the same, regardless of whether the virtual particles are photons or a matter-antimatter pair.


Here's an analogy to Quantum Mechanics. A particle in QM can tunnel through an impossible barrier, which is how elements heavier than lead can have some of their neutrons "tunnel" out of the nucleus escaping the bonds of the Strong Force.

enter image description here

A small Black Hole is like a quantum barrier that a particle can be tunnel through to escape. The smaller the barrier (Event Horizon), the more likely it is to be able to tunnel out. So a micro black hole with a mass of 228 tons and a Event Horizon of 3.4 x 10^-7 femtometers (literally less than 1 millionth the size of a proton) won't hold on to it's particles for very long and at all. In fact it will explode in a burst of Hawking Radiation after exactly 1 second.

A larger Earth Mass black hole with a radius of a whole centimeter, will last much longer: 8 x 10^50 years because it's much less likely for a particle to tunnel through a whole centimeter to get to freedom.

Source: Quantum tunneling from three-dimensional black holes: https://arxiv.org/abs/1306.6380

Source: Hawking Radiation modelled as a Quantum Effect: http://cscanada.net/index.php/ans/article/view/j.ans.1715787020120502.1817


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