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From what I've read, Hawking radiation is produced far outside the event horizon. The radiation is produced by the quantum field of the black hole outside the event horizon. As more of these radiations are emitted, the quantum field's energy decreases, hence the black hole's mass decreases.

My problem with this explanation is, it appears that the energy of the quantum field inside the event horizon could somehow ‘flow’ out of it, to fill the lost energy of the quantum field outside. Even if the energy in the quantum field inside the event horizon couldn't ‘move’ to the quantum field outside, wouldn't the energy of quantum field inside the event horizon stays same, while the outside of event horizon will be devoid of energy?

If such assumptions are wrong, how can Hawking radiation reduce a black hole's mass, without resorting to virtual particles or negative energy?

Qmechanic
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Prido1024
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    What is "the energy in the quantum field"? How are you localizing it? Why does it need to "flow" anywhere? – ACuriousMind Feb 08 '22 at 16:16
  • Based on my understanding, a field like a gravitational one have energy, and in a black hole it's localized because the curvature is big there, and it probably ‘flows’ to ‘balance’ the field. – Prido1024 Feb 08 '22 at 16:32
  • It might be worth pointing out that the quantum field involved in the Hawking effect is usually taken to be some non-gravitational quantum field. In other words, you usually put a quantum field $\phi$ (often a scalar field) on top of the classical geometry of a black hole. The field involved in the Hawking calculation does not need to be the gravitational field (although, afaik, you can derive Hawking radiation for gravitons just like you do for a scalar field) – Níckolas Alves Feb 08 '22 at 18:39

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Hawking radiation depends on a flow of negative energy. There is a negative energy density in the surroundings of the black hole. By analyzing the energy flux at infinity, one notices that the spacetime is losing energy as time goes by. However, no energy can escape the black hole, as you noticed, so what happens is precisely that you get a flux of negative energy flowing into the black hole. Of course, there is no need to resort to "particles", since everything can be defined in terms of fields.

To understand how this is physically viable, let me quote Hawking's seminal paper:

The above discussion shows that the particle creation is really a global process and is not localised in the collapse: an observer falling through the event horizon would not see and infinite number of particles coming out from the collapsing body. Because it is a non-local process, it is probably not reasonable to expect to be able to form a local energy-momentum tensor to describe the back-reaction of the particle creation on the metric. Rather, the negative energy density needed to account for the decrease in the area of the horizon, should be thought of as arising from the indeterminacy of order of $M^{-4}$ of the local energy density at the horizon. Equivalently, one can think of the area decrease as resulting from the fact that quantum fluctuations of the metric will cause the position and the very concept of event horizon to be somewhat indeterminate.

Also,

A renormalised [energy-momentum] operator which was regular at the horizon would have to violate the weak energy condition by having negative energy density. This negative energy density is not observable locally.

Notice that, in a quantum setting, negative energy densities are a prediction, not an issue. For an example, see, e.g., Sec. 1.2 of 1208.5399 [gr-qc] and/or Problem 6 of Chapter 14 of Wald's General Relativity. In short, if $\hat{T}_{ab}$ denotes the normal-ordered Klein–Gordon energy-momentum tensor and $t^a$ is some timelike vector, one can find a state $|\psi\rangle$ such that $\langle \psi | \hat{T}_{ab} | \psi \rangle t^a t^b < 0$ in a region of spacetime. Hence, one can have negative energy densities as a consequence of quantum effects.

I should point out that Hawking also writes

Thus it will give positive energy flux out across the event horizon or, equivalently, a negative energy flux in across the event horizon.

So in principle, both descriptions are somewhat equivalent. Of course, positive energy flowing out of the black hole is a problematic interpretation since it is acausal, while there is nothing wrong with negative energy flowing in, since quantum theory allows for negative energy densities.

Níckolas Alves
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  • Thx for answer. I had some vague idea that a classical negative $T_{00}$ can result in instability of the Einstein field equation. Is that right and if so then do you know how quantum field theory avoids it? – Andrew Steane Feb 08 '22 at 18:18
  • @AndrewSteane I've never heard of any instability issues, or at least not in these terms. As far as I know, one often imposes $T_{00} \geq 0$ and other similar energy conditions on the grounds of trying to obtain a physically reasonable stress tensor and avoiding describing matter that doesn't exist (this proves to be more subtle once quantum effects are taken into account). Without imposing such "physicality" conditions, any Lorentzian metric is a solution to the EFE (https://physics.stackexchange.com/q/483259/168783) – Níckolas Alves Feb 08 '22 at 18:33
  • A nice example is that of the Alcubierre drive, a proposal for hyperfast travel within GR, which was derived by first prescribing the metric, and then computing the stress tensor. It is an interesting solution with $T_{00} < 0$, but problematic in the sense that apart from quantum effects, we don't know of any sorts of matter that could be arranged to yield the stress tensor necessary for the solution to describe something that can actually happen – Níckolas Alves Feb 08 '22 at 18:35
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    Nice references to big names, but the truth is that there is no “negative energy” in nature. Any math quoting “negative energy” is an interpretation that involves additional assumptions and references, without which it quickly becomes meaningless and deceptive. The bottom line is, once you admit the actual existence of physically “negative energy”, you part with physics and reality and lose a delicate logical chain that keeps your thinking rigorous. Energy is a Fourier conjugate of time in any branch of physics, so real energy is always positive simply because time is irreversible. – safesphere Feb 09 '22 at 17:23
  • @safesphere There are some remnants of positivity, such as the Quantum Energy Inequalities mentioned in the arXiv link I provided. Nevertheless, notice that within QFT one can quite easily find a state in which the expectation value of the energy density of the stress tensor is negative. Of course, within semiclassical gravity what is relevant to study backreaction effects on the metric is precisely the expectation value of the stress tensor, so it is reasonable to talk of negative energies in this specific sense. – Níckolas Alves Feb 09 '22 at 19:13
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The answer to your question is the gravitational field of the black hole, which already exists before the (for example neutron star) collapses into a black hole and, moreover, this gravitational field extends to infinity (so outside the horizon).

Since the gravitational field already exists for the collapsing star, when the collapse "happens", the event horizon (which is not a physical thing, rather, a boundary) happens to be so that the gravitational field extends outside of it too.

A black hole, however, can have an electric charge, which means there is an electric field around it. This is not a paradox because a static electric field is different from electromagnetic radiation. Similarly, a black hole has a mass, so it has a gravitational field around it. This is not a paradox either because a gravitational field is different from gravitational radiation.

How does gravity escape a black hole?

The fact that the gravitational field extends outside the horizon, is no contradiction. Nothing, no information or particles need to travel from inside, because the gravitational field of the collapsing star (for example a neutron star) already existed outside the boundary (that we later call event horizon), and so the energy of this gravitational field can (and in your example does) influence the quantum fields outside the horizon.

This is what is used in the Hawking radiation arguments, where the virtual loop at the event horizon has one particle interacting with the gravitational field and energy is supplied by it so that the other becomes real and exits as real.

Can virtual particles be 'boosted' into becoming real particles by fields other than gravity?

Now there are many interpretations of Hawking radiation, in this one, the extreme energy of the gravitational field gives this "boost" to the quantum fields, helping fluctuations to manifest as (seen from a far away observer) as photons coming from the (outside region of the) black hole. This radiation is interpreted as Hawking radiation, and the reason it decreases the mass of the black hole is because it takes energy from the gravitational field of the black hole. No need for particles to go in or out the horizon or anything to be influenced from inside the horizon (of course there are many equivalently good interpretations, including particles going in and out but these all originate from outside the horizon), but just energy taken from the gravitational field of the black hole, decreasing its mass, which corresponds to the fact that in mainstream physics, the mass of the black hole is derived from its gravitational field's energy (ADM, Komar, Bondi).

So again, the answer is the gravitational field (and its energy) of the black hole which already exists outside the horizon. Please note that there is a way to view this intuitively (if I understand correctly that you are assuming an external view of the black hole) where all the energy of the black hole is outside the horizon (because it never crosses in a fininte amount of time), and it all evaporates before the black hole could fully form for a far away observer.

Árpád Szendrei
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  • But since the energy escapes from the gravitational field beyond the horizon, while the energies within the gravitational field inside the horizon can't get out of the horizon (just like everything else), wouldn't the gravitational field be left with lack of energy outside the horizon and unchanging energy inside it? – Prido1024 Feb 09 '22 at 00:57
  • @Prido1024 the total mass (energy) of the black hole would still decrease because it includes the total gravitational field energy (which extends outside too). The rule is that changes inside the horizon cannot influence the outside world, so changes in the gravitational field that originate from inside the horizon should not influence the outside, but there is no need for that at all. Since these changes (Hawking radiation) originate from outside the horizon, there is no need for any influence from inside the horizon. The changes originate from outside, and can influence the inside though. – Árpád Szendrei Feb 09 '22 at 03:53
  • @Prido1024 so you are asking about the "distribution" of the loss of energy (mass) of the gravitational field, and you are saying that the energy is only lost outside, and not inside. But that does not need to be the case, because if energy is lost outside it can influence the inside, and can equalize (distribute) the changes throughout the whole gravitational field (including inside and outside). – Árpád Szendrei Feb 09 '22 at 03:55
  • How can the energy is equalized/distributed throughout the whole gravitational field, outside and inside, if energy inside the horizon supposedly can't get out of it? – Prido1024 Feb 09 '22 at 04:14
  • @Prido1024 when the energy of the black hole decreases (that changes the curvature itself too), that changes the mass, changing the event horizon too (the event horizon is determined by the energy/mass too). Decreasing event horizon means certain part of the gravitational field that was inside, gets (happens to be) now outside. Again, no information or energy got outside, nothing can influence the outside world from inside. – Árpád Szendrei Feb 09 '22 at 04:19
  • So as a black hole decreases in curvature, the space itself ‘creeps’ out of the horizon, carrying in it the energy formerly within the horizon (because energy can't get out of an event horizon, but space can). Am I getting it right? – Prido1024 Feb 09 '22 at 04:55
  • @Prido1024 correct, the energy of the gravitational field itself is (connected to) the curvature of spacetime. – Árpád Szendrei Feb 09 '22 at 05:05
  • Your answer is essentially correct (so +1), but it doesn’t directly answer the posted question. The OP is asking how the outside process can reduce the energy inside. You explain that the energy is already outside. This is correct. The part you are missing is that there is no energy inside an evaporating black hole in the first place. All it’s energy is outside, because everything evaporates before anything crosses the horizon. See my comment under the question. This is similar to your questions on merging black holes - they merge (and then evaporate) before anything crosses any horizon. – safesphere Feb 09 '22 at 17:12
  • @safesphere thank you I edited. I understand, my only question was, what about the energy (mass) of the collapsed star (like a neutron star) that was already inside when the collapse happens (so it was already inside the region that we later called event horizon). – Árpád Szendrei Feb 09 '22 at 18:46
  • The radial direction inside the horizon is the direction in time, not in space. The “inside” is not a “place closer to the center”, but the future time at the same place - the spatial radius of the horizon is already zero. Since the “inside” is the future, nothing is there yet. Your question is equivalent to asking, “What about the star matter that was already in tomorrow when the collapse happened today?” Nothing is yet in tomorrow, as far as we are concerned today. So no energy of the star is inside the horizon before the star completely evaporates. I can explain this in chat, if you want. – safesphere Feb 10 '22 at 17:58
  • @safesphere yes please I am interested. – Árpád Szendrei Feb 10 '22 at 18:36
  • Sure, I'll ping you next week :) – safesphere Feb 11 '22 at 06:49
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When stuff falls in through the horizon it gets entangled with the virtual particles (quantum bubbles, disconnected closed one particle propagators) in the vacuum. The entanglement last long enough for the virtual particles to get promoted (by the strong curvature of spacetime) to real positive and negative energy particles. From the perspective of the infalling matter it takes the blink of an eye, while for an outside observer it takes long years.

The infalling negative energy particles reduce the mass, while the outgoing positive ones carry inside information because of the entanglement. Particles and antiparticles annihilate and radiate away as Hawking radiation. Virtual photon bubbles get promoted too into positive and negative energy parts and radiate directly outward and inward.

MatterGauge
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