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If 'quantum foam'-generated particles are made of matter and antimatter in equal amounts, why don't the matter particles that fall into the black hole add to the black hole's mass just as much as the antimatter particles subtract from it?

How, then, can the black hole evaporate due to Hawking radiation?

P.S.: Someone else asked a very similar question, but it was never answered....

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  • $\begingroup$ Related: physics.stackexchange.com/q/251385/123208 & physics.stackexchange.com/a/30601/123208 & links therein. $\endgroup$
    – PM 2Ring
    Jun 21, 2021 at 6:37
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    $\begingroup$ Can you link the unanswered similar question? Someone might want to close it as a duplicate of this, now that this has answers. $\endgroup$ Jun 21, 2021 at 13:14
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    $\begingroup$ Pop-sciencey answer: because the energy that initially created the pair comes from somewhere and the only thing nearby is the black hole. It only gets half of that back. $\endgroup$ Jun 21, 2021 at 22:44
  • $\begingroup$ @safesphere I called it pop-sciencey because of how thoroughly I handwaved the particles getting energy out of black hole in the first place, basically amounting to saying "because it's there." $\endgroup$ Jun 23, 2021 at 21:00
  • $\begingroup$ @safesphere Yes, that's the part I glazed over. You gave a proper answer, and I gave a pop-science answer because I skipped all of that. You should probably build this up and post it as a full answer where it won't get so easily missed. $\endgroup$ Jul 14, 2021 at 19:49

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The pop science picture of Hawking radiation as particles popping into existence near the event horizon is grossly oversimplified and in many respects misleading. See https://www.forbes.com/sites/startswithabang/2020/07/09/yes-stephen-hawking-lied-to-us-all-about-how-black-holes-decay for details.

In fact Hawking radiation is, like any black body radiation, mostly composed of photons (of large wavelength, on the order of the Schwarzschild radius) and not arbitrary particles and anti-particles. It's emitted due to the curvature of spacetime near the black hole, and comes not just from the event horizon but from a whole volume of space around the black hole. In fact the photons have such large wavelength that the uncertainty in their positions are greater than the size of the black hole, and so they may be thought of as "tunneling" out of the black hole in some sense, as mentioned in the highest upvoted answer here.

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    $\begingroup$ I don't think this answers the question. Classically, an electron would fall into an atomic nucleus; quantum mechanically, it can't be localized in such a small area; but it doesn't escape to infinity either. It just hangs around near the nucleus (with exponentially decaying chance to be found farther away). Why doesn't Hawking radiation do that? $\endgroup$
    – benrg
    Jun 21, 2021 at 2:28
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    $\begingroup$ Electrons not falling into the nucleus is a very different phenomenon, that's due to quantization of allowed energy levels. Hawking radiation is more like quantum tunneling through a barrier. In both cases words are not really adequate to describe the reality of the situation, and one should not take the analogies too strictly. The OP's particle analogy is particularly flawed, and resulted in confusion. $\endgroup$
    – Eric Smith
    Jun 21, 2021 at 11:31
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    $\begingroup$ "composed mostly of photons" assumes that there are no other quantum fields with extremely low rest mass. Given that known quantum fields only account for about 5% of the energy of our present day universe, this is a questionable assertion. $\endgroup$
    – John Doty
    Jun 21, 2021 at 13:39
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    $\begingroup$ @JohnDoty Interesting suggestion! Note that, in general, the emitted particles must also interact via low-mass fields to participate in low-temperature blackbody radiation. Consider that there is a thermal "cosmic neutrino background," but neutrino emission is a negligible part of how your stove cools off. Whether a black hole could couple thermally to low-mass particles which aren't accessible to weakly gravitating systems is an interesting question, but it's speculative enough that leaving it unmentioned in this answer seems reasonable. $\endgroup$
    – rob
    Jun 21, 2021 at 17:38
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    $\begingroup$ @rob Since the electromagnetic coupling constant plays no role in the theory, it would seem, if the theory is correct, that the black hole can couple to any quantum field. $\endgroup$
    – John Doty
    Jun 21, 2021 at 21:16
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Disclaimer: not an expert; I understand this at the pop science level only.

Near the Black Hole, there is at first nothing, and then there is a matter-antimatter pair. That violates conservation of energy and is "not possible" except in quantum mechanics, which has this equation:

$\Delta E \Delta t \geq \hbar/2$

Which can be interpreted as energy can be "borrowed" (i.e. created) for a certain amount of time, but the more energy is "borrowed", the less the amount of time one can borrow it for. When the matter-antimatter pair pops up out of nothing, it needs to annihilate quickly.

Hawking radiation arises when one of the two particles falls into the black hole before it can annihilate. Both the matter and antimatter particle have positive energy (since they both have positive mass), so what actually looks like it happened is that a new particle is created out of thin air. For this to happen, the required energy must come out of the black hole, hence it loses mass. This applies regardless of whether the emitted particle is a matter particle or an antimatter particle.

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    $\begingroup$ I think this answer is quite good. In my understanding the gravitational field is so strong that is separates the virtual particle pair, making them real and losing therefore the energy equivalent to two particles. One of that it gets back while the other escapes, so one particle mass is lost in the budget. However, all this is pure hand waving arguments because Hawking radiation is complicated math that in principle cannot be put in a mental picture. $\endgroup$ Jun 21, 2021 at 6:24
  • $\begingroup$ How do your matter and antimatter particle pairs generate photons? $\endgroup$
    – PM 2Ring
    Jun 21, 2021 at 20:54
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In order to have the creation of a particle, as far as we know in mainstream physics, one needs energy, and energy is provided by an interaction. Interactions must obey special relativity and quantum mechanics, assuming that the horizon is very far removed from the singularity.

So here is a horizon with a lot of charged and neutral particles falling in, with large energies, but let us ignore them. For the black hole to lose mass the interaction has to be with the gravitational field of the black hole, so one has to assume quantum gravity.

As an example, assume a graviton from the black hole interacts with a vacuum loop of electrons at the horizon. What can happen?

a) The loop radiates a photon, and if the direction of the photon is favorable it can escape the gravitational field, taking away part of the graviton energy, and thus diminishing the mass of the black hole.

b) the energy is enough for the loop to break up, and a particle falls in while the antiparticle escapes (or vice versa).

These are rough images for describing what has been calculated by Hawking in a complicated manner. The fact is that the energy is provided from the gravitational field of the black hole, and thus its mass diminishes with this radiation.

My answer here may help.

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  • $\begingroup$ Are you suggesting that an anti-particle is repelled by a gravitational field? Otherwise, how does the it escape from the field of the black hole? $\endgroup$
    – R.W. Bird
    Jun 21, 2021 at 14:43
  • $\begingroup$ @R.W.Bird (vice versa) was meant to mean or the antiparticle is eaten and the particle leaves..It escapes if it has enough kinetic energy and direction against the gravitational field, it is a matter of probabilities. $\endgroup$
    – anna v
    Jun 21, 2021 at 15:36
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The matter/anti-matter distinction is often mentioned here, but you do not need it to understand Hawking radiation because Hawking radiation is mostly made of photons (which are their own antiparticles). If we consider a pair of virtual photons in the vacuum with no energy input, then if one has energy $E$ then the other must have energy $-E$, in some local inertial frame. In consequence the quantum amplitude for this process gives zero overall probability once the integral over all paths is carried out in flat spacetime (you will have to take this bit on trust). (To be precise, there is zero probability for such a pair of virtual photons to propagate away to infinity, and zero contribution to any other process from any loop they may form.)

Near a horizon, on the other hand, among the paths in which the negative energy photon passes into the horizon there are some in which that photon now has an energy-momentum 4-vector with temporal part positive. In consequence the quantum amplitude does not vanish when the integral is carried out. In this case the net effect is that there is a photon of energy $E$ moving away from the black hole, and a change in the rest energy of the black hole by $-E$. Therefore its mass goes down.

So far I considered cases where both photons escape (zero probability) or where the photon with positive energy escapes and the other one does not (non-zero probability). The other two cases are: both enter black hole (non-zero probability; net result no change in the black hole rest energy), or the positive energy virtual photon enters the horizon and the negative energy one does not. This last case contributes a zero net probability. Basically to get a non-zero overall quantum amplitude you need energy-momentum 4-vectors with positive temporal components in the limit $t \rightarrow \infty$.

[The above picture is based on field theory on a classical spacetime background. In this picture we don't need gravitons; we let classical G.R. account for gravity.]

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  • $\begingroup$ What about the paths in which the positive energy photon passes into the horizon? A variant of the original question is if these would then increase the mass of the black hole, counteracting the negative energy photons. Why is this not the case? Do they not pass into the horizon in the first place, or what is the difference? $\endgroup$ Jun 21, 2021 at 16:10
  • $\begingroup$ @StephanMatthiesen thanks; I added a paragraph to deal with this $\endgroup$ Jun 21, 2021 at 16:19
  • $\begingroup$ why does "the positive energy virtual photon enters the horizon and the negative energy one does not" contribute net zero? it means a -E escaped and an E stayed? $\endgroup$ Jun 21, 2021 at 18:09
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    $\begingroup$ @AwokeKnowing it is to do with the way the motion under gravity conserves the covariant temporal component of the energy-momentum 4-vector along a geodesic. For worldlines passing the horizon there is a sign change, for other worldlines there is not. It is this sign change which allows an integral which would otherwise be zero to be non-zero. $\endgroup$ Jun 21, 2021 at 18:31
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There are basically three ways to describe why the black hole does not gain the same amount of energy it loses:

  1. it is not a particle-antiparticle pair, rather, the far away observer's view is different of the vacuum, and the particles there that are created in entangled pairs will have a opposite energy content (from the far away observer's view, the zero energy is different then an observer at the horizon, so the energy scale is shifted). The one of the pair that escapes is the one with the energy, the other one that falls into the black hole has negative energy content and reduces the black hole's total energy

When a pair of virtual particles are produced there isn't a negative energy particle and a positive energy particle. Instead the pair form an entangled system where it's impossible to distinguish between them. This entangled system can interact with the black hole and split, and the interaction guarantees that the emerging particle will be the positive one. NB "positive" and "negative" doesn't mean "particle" and "anti-particle" (for what it does mean see below), and the black hole will radiate equal numbers of particles and anti-particles. When you quantise a field you get positive frequency and negative frequency parts. You can sort of think of these as representing particles and anti-particles. How the positive and negative frequencies are defined depends on your choice of vacuum, and in quantum field theory the vacuum is unambiguously defined. The problem is that in a curved spacetime, like the region near a black hole, the vacuum changes. That means observers far from the black hole see the vacuum as different from observers near the black hole, and the two observers see different numbers of particles (and antiparticles). A vaccum near the event horizon looks like excess particles to observers far away, and this is the source of the radiation.

Black holes and positive/negative-energy particles

  1. the creation of the photon that the far away observer sees needs energy to be taken away from the gravitational field of the black hole (outside the event horizon, because the gravitational field extends there too), and so the net energy of the black hole decreases

The dry facts are that two real particles (e.g., two photons, or an electron and a positron) are created from the energy in the very strong gravitational field near the horizon of the black hole - from a classical external gravitational field (if gravitation is treated classically), or possibly from two gravitons (in effective quantum gravity at lone loop), not from the vacuum. [Strong external fields with energies significantly above the pair creation energy threshold necessarily create the corresponding particle pairs. See the postscript below for more details.] The particle pair creation reduces the gravitational energy by the energy (including the rest mass energy equivalent) of the two particles. One particle escapes, the other is absorbed by the black hole. The net result (black hole energy - 2 particle energies + 1 particle energy) is a loss of mass corresponding to the energy of the escaping particle.

An explanation of Hawking Radiation

  1. the wavelength of this radiation (or the energy of the photons is so low) is so long, that it might exceed the radius of the black hole itself, so asking where they are originating from is ambiguous (because they are QM entities), so some might suggest they just originate from inside the black hole, but the fact is, they take away energy from the black hole, reducing its total energy

This is possible because the Hawking radiation wavelength is of the same order as the BH diameter.

From where (in space-time) does Hawking radiation originate?

These are all equivalently satisfying answers, but I rather would say that all three phenomena (quantum vacuum, gravitational field, wavelength) are at the same time at work.

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First off, one should get out of the way that there are no "virtual particles popping in and out of the vacuum" - this is a bad picture of quantum field theory whose reference is uncertain; it could refer to a result involving the Reeh-Schlieder theorem, or it could refer to how that vacuum states possess fuzzy field values at each point in space. Hence, given this is wrong, there can be no "taking in one particle and spitting out another". In fact, the positions of involved particles are so quantum-mechanically fuzzy that the fuzziness is bigger than the black hole event horizon by several times. Also, "quantum foam" is an altogether different concept, which belongs to still-hypothetical fully-quantum mechanical models (which actually go beyond Hawking's original idea) of spacetime, referring to quantum fuzziness or indeterminacy in the geometry of spacetime itself.

What, then, is going on? The answer is that if an agent is initially informed that a black hole is present with a given mass and space around it is vacuum, then it will project a future history of that black hole involving a steadily-growing quantum amplitude that there are radiation particles outside. Likewise, the expected value of the black hole mass will be dropping. This process is very complicated to describe precisely in terms of mathematics, but it is (at least I believe) what is going on from a conceptual or intuitive point of view(*).

Broadly, such is the same thing as what happens with an unstable atomic nucleus. An agent which has initially been informed that the nucleus exists and is of a given unstable type, will then project a future history with rising quantum amplitude that the nucleus is in some once- or even multiply-decayed form, and that there are fragments or radiation particles present outside. And in either case, if the agent then queries the system at this later time for its decay state, there is likewise greater probability that such decay will be recorded as having taken place.

To put it another way, Hawking radiation is nothing more or less than the radioactive decay of a black hole. The black hole is coupled to nongravitational quantum fields, and these provide channels by which it can lose energy. But the precise shape of the decay curves is different between the two situations - we get a simple decay to stability for the radioactive nucleus, while for the black hole the decay becomes more and more furious as its mass is depleted, meaning more and more and more particles are sent forth. There is a singularity here as the black hole mass goes to zero - this singularity represents a limit of the theories used to derive the black hole decay, namely semiclassical gravity.


(*) I should note I am taking this via the subjective/informational conception of quantum states; one may of course argue for an alternative interpretation. Something has to be used, after all.

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