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A recent paper "Gravitational Pair Production and Black Hole Evaporation" (discussed in short here) says that any spacetime curvature would produce Hawking radiation, no need for event horizon.

If this is correct then a question arises - how negative energy would take energy from observable object?

Fermion cannot be partially annihilated and full annihilation is way more energetic then Hawking photon. So the only source of energy would be object's heat.

But would that mean that object with temperature near absolute zero would stop radiating? There is nothing about temperature in equation.

Does it mean that the paper is wrong? Or is there any mechanism to provide Hawking energy from object's mass?

P.S. "Black holes are white hot" Hawkins:

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Vashu
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  • From the introduction in link you give I see a serious misunderstanding: "According to quantum mechanics, a vacuum state is populated by virtual particle pairs undergoing spontaneous creation and annihilation processes. These quantum fluctuations can turn into real particle pairs in the presence of a background field. " They do not turn into real particles just in the presence of passive fields. One needs energy input from real particles , four vectors on mass shell. In a black hole horizon the four vector comes from the black hole, that is why it is diminishing. – anna v Jun 07 '23 at 06:03
  • @annav Even ignoring that, I'm hesitant to accept these findings solely based on the reliance on the virtual particle heuristic explanation of Hawking radiation. It seems like an interesting result, but I need to see it reproduced from calculations on tidal disruptions of quantum fields. – Logan J. Fisher Jun 07 '23 at 06:07
  • @annav How about Schwinger effect? – Vashu Jun 07 '23 at 07:06
  • @Vashu The electric field is created by real four vectors somehow, feeding energy to it, read https://en.wikipedia.org/wiki/Schwinger_effect#Experimental_prospects – anna v Jun 07 '23 at 08:08
  • Authors mention that there is no need for event horizon for detecting Hawking radiation but it doesn't eliminate other kinds of horizon. See the article by M. Visser, reference [14] in that paper – KP99 Jun 07 '23 at 08:38

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This study proposes that local tidal forces may be enough to separate virtual particle pairs without the need for any horizon, resulting in Hawking-like radiation. In other words, the energy is purely derived from the stress-energy tensor - while thermal energy contributes to that, it's not a necessary condition.

It's worth noting that I am hesitant to accept this perspective built entirely on the virtual pairs explanation of Hawking radiation, which is fundamentally only a heuristic approximation of the disruption of quantum fields resulting in fluctuations that appear to far-away observers as emitted radiation. I won't concretely say it's wrong, but I personally am not prepared to adopt this idea until a proper QFT in curved spacetime approach is shown to produce similar results.

In any case, I'll be attending a conference also attended by Wondrak next month, and I'm sure they'll be presenting a poster on this. If I learn more there, I'll update this answer.

Logan J. Fisher
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  • "the energy is purely derived from the stress-energy tensor" We have a single proton - it has stress-energy. Suppose it managed to radiate hawking photon, how would proton state change if the law of the energy conservatio is not broken? – Vashu Jun 07 '23 at 07:09
  • @Vashu I don't have an answer for that. Possibly there's a constraint that the proton would have to "fully" decay, or perhaps there's simply a bottom limit and you get left with stable remnants like protons, or maybe the proposal is just fundamentally flawed. I really couldn't say. – Logan J. Fisher Jun 07 '23 at 07:22
  • I am hesitant to accept this perspective built entirely on the virtual pairs explanation of Hawking radiation I would argue that the paper does not really assign to virtual pairs ontological significance, the ontology there is pure QFT. Rather, particle pairs here arise as artifacts of approximation used (since particle action is an ingredient in effective action calculation). – A.V.S. Jun 09 '23 at 19:44
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That is one of the issues I personally see with that paper. Logan J. Fisher already mentioned the issue with taking particles too literally, but also there is the matter that since they use path integral methods, they are implicitly selecting a vacuum with respect to which they are computing expectation values. Furthermore, your choice of vacua carries information about whether the spacetime has or not a horizon. They worked with Euclidean path integrals in Schwarzschild, which will typically give you the Hartle—Hawking vacuum, which only makes sense if you assume the presence of an eternal black hole. Hence, I believe they are assuming the presence of a horizon implicitly, by means of a path integral.

I discussed more about the relation between states and the presence of a horizon in this answer. My master's thesis also includes extensive discussions on how path integrals depend on a choice of state and how that is related to boundary conditions and the presence of a horizon.

Shortly, I believe the paper overlooked a few subtleties and is, in fact, assuming the presence of a horizon implicitly.

Níckolas Alves
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The paper is just plain wrong and not in a few subtleties but in its basic assumption of relevance of obtained local effective action for black hole radiance.

Since the paper puts a lot of emphasis on analogy of Hawking radiation with Schwinger pair production let us take this analogy a step further by considering pair production in the background electrostatic field of a point charge. Just like the paper in question does for gravitational field, we can evaluate the imaginary part of effective action (a well known expression due to Heisenberg and Euler) for Coulomb field and obtain some nonzero “rate”. If we are to believe the authors, then this would indicate that background EM field of a charge (for example a heavy nucleus) would spontaineously produce electron–positron pairs. This is, of course, wrong. The Coulomb field of a charge $Z$ (in units of elementary charge) is stable against pair production for $Z<137$. For larger charge values situation is more complicated, for example, nuclei of finite size can have negative energy bound electron states for $150<Z<173$ while ultra-heavy nuclei with $Z>173$ would spontanously emit positrons. For details and links to original literature see section 6.5 of the review:

The important point here is that “imaginary part of effective action” calculations fail for such situations. This is not surprising, since the result of Heiseberg–Euler is derived for a uniform field, while Coulomb field is strongly non-uniform and the larger its value the more it varies. Similarly, the effective lagrangian derived by Wondrak et al. is based on the assumption that derivatives of curvature tensor do not contribute significantly to functional integral. But this assumption fails for Schwartzschild spacetime in basically the same way as for Coulomb field: the larger the curvature tensor, the larger are its covariant derivatives, so higher terms of covariant expansion cannot be discarded.

In conclusion, I think that it is safe to assume that this paper did not open a “new avenue to black hole evaporation” and that it is better to analyze Hawking radiation via any of the already established methods. And for static spacetimes, event horizon is required for Hawking radiation.

A.V.S.
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