A Validity of theoretical arguments for Unruh and Hawking radiation

[Heikki Tuuri]

No, it doesn't. A laser emits a coherent state, which is very different from a Fock state (the latter is an eigenstate of photon number). A coherent state is the closest kind of quantum state to a classical EM field.

Humm... we could use a different source of photons than a laser. A device which emits exactly 1 right-handed photon at a time? Then it is kind of a coherent source, still.

The far-away observer will in rare cases detect a left-handed photon along with a right-handed one.

The particle interpretation is that the photon scattered from a graviton and split in two. The graviton hypothesis has the well-known problem that it is not renormalizable.

The classical wave interpretation is that a standard wave packet stretched into a chirp as it flew out of the gravitational field.

Why is there no renormalization problem in the classical interpretation?

 

[PeterDonis]

we could use a different source of photons than a laser. A device which emits exactly 1 right-handed photon at a time? Then it is kind of a coherent source, still.

Not coherent, no. "Coherent" means "coherent state". Which, as I said, is very different from a Fock state.

There are photon sources that emit Fock states, but they're very hard to set up experimentally.

 

[PeterDonis]

The particle interpretation is that the photon scattered from a graviton and split in two.

No, the particle interpretation would be that a photon of spin +1 (right-handed circular polarization) would absorb a graviton of spin -2 (or emit a graviton of spin +2) and become a photon of spin -1 (left-handed circular polarization).

 

[PeterDonis]

Why is there no renormalization problem in the classical interpretation?

Because there is no renormalization problem in classical GR, period.

 

[Dale]

A device which emits exactly 1 right-handed photon at a time? Then it is kind of a coherent source, still.

A coherent state is not a state with a definite number of photons. A state with a definite number of photons is not a coherent state.

A state with a definite number of photons has the property that if you annihilate a photon the state changes. A coherent state has the property that if you annihilate a photon it remains in the same state.

 

[Heikki Tuuri]

Because there is no renormalization problem in classical GR, period.

There might be a classical problem which resembles the renormalization problem. In the Navier-Stokes equation, a Millennium problem is to prove that solutions do not "blow up" because of turbulence.

In a realistic fluid there is a natural scale, the scale of molecules, at which the Navier-Stokes equation stops working. The blowup cannot happen. This sounds like an energy cutoff which is used to eliminate the divergence in renormalization.

The concept of an "effective theory" contains the idea that at very short distances there is new physics which provides the necessary cutoff.

If we try to model an electromagnetic field in a gravitational field, and consider the backreaction of the two fields when they interact, the renormalization problem may appear in the classical fields as a blowup problem. For example, the solution might not be stable under small perturbations.

After all, Feynman diagrams are perturbation calculations. If the perturbations diverge, then a classical solution might be unstable.

Which brings us to the old topic if General relativity has any solutions under realistic matter fields.

Christodoulou and Klainerman (1990) proved the "nonlinear stability" of the Minkowski metric under General relativity.

This is a very interesting question: if Feynman diagrams with gravitons diverge, how could Christodoulou and Klainerman show the stability in the very restricted case of the Minkowski metric?

In physics, if we are calculating with two fields, we usually ignore the backreaction. If we calculate the behavior of a laser beam which climbs out of a gravitational field, we assume that the backreaction on the gravitational field is negligible. But it might be that a precise calculation shows the the solution is not stable.

 

[PeterDonis]

the solution might not be stable

a classical solution might be unstable.

it might be that a precise calculation shows the the solution is not stable

These are all speculations which cannot be usefully discussed in the absence of some actual concrete examples. Lots of things "might" be the case.

 

[PeterDonis]

Klainerman et al. proved the "nonlinear stability" of the Minkowski metric under General relativity.

Can you give a reference?

 

[Heikki Tuuri]

http://www.numdam.org/item/?id=SEDP_1989-1990____A15_0

The global nonlinear stability of the Minkowski space
Christodoulou, D.; Klainerman, S.
Séminaire Équations aux dérivées partielles (Polytechnique),(1989-1990), Talk no. 13, 29 p.

 

[PeterDonis]

It is this famous paper from 1990. Christodoulou and Klainerman.

This paper looks like a purely classical analysis, so the properties of Feynman diagrams in a quantum field theory of gravitons are irrelevant.

 

[Heikki Tuuri]

This paper looks like a purely classical analysis, so the properties of Feynman diagrams in a quantum field theory of gravitons are irrelevant.

The question is: is the divergence of Feynman diagrams connected to the instability of the classical fields?

In a Feynman diagram, we assume that an excitation of a field bumps into another excitation of the same or another field.

Classically, there will be complex interaction between the two fields, or within a single field. Proving the stability is hard. It is very hard in the case of General relativity, where the partial differential equation is nonlinear.

Actually, all interactions introduce nonlinearity to fields. If we have two solutions:

1. field A is zero and B contains a wave,
2. field A contains a wave and B is zero,

then the sum of 1 and 2 is usually not a solution if A and B interact.

It might be that the stability is unknown for almost all physical processes.

 

[PeterDonis]

The question is: is the divergence of Feynman diagrams connected to the instability of the classical fields?

And that question makes no sense in the context of a purely classical analysis using purely classical GR, which is the kind of analysis that the paper you linked to is doing. It only makes sense in the context of considering GR as the classical limit of some quantum field theory. So if you want to discuss this question, you need to find a reference that does the latter kind of analysis.

 

[Heikki Tuuri]

And that question makes no sense in the context of a purely classical analysis using purely classical GR, which is the kind of analysis that the paper you linked to is doing. It only makes sense in the context of considering GR as the classical limit of some quantum field theory. So if you want to discuss this question, you need to find a reference that does the latter kind of analysis.

There do seem to be analogies between classical PDE and quantum field theory (which can be viewed as a kind of quantum PDE) but this is definitely an underexplored area of study. For instance, the Cauchy problem for quantum field theory has not been studied much, even in linear models (perhaps it is a bad question to ask).

https://terrytao.wordpress.com/2007/03/18/why-global-regularity-for-navier-stokes-is-hard/

A Feynman diagram can be understood as describing reactions and backreactions between fields.

The problem in a fermionic loop Feynman diagram is that the momentum is not restricted. We need to consider the reaction and the backreaction at an arbitrarily small length scale. The integral often diverges and we have to introduce a cutoff.

Renormalization seems to be a common tool in the study of the stability of partial differential equations.

I will think about this.

 

[PeterDonis]

A Feynman diagram can be understood as describing reactions and backreactions between fields.

Between quantum fields. Not classical fields.

 

[Orodruin]

Between quantum fields. Not classical fields.

This is not completely true. You can use Feynman diagrams also when solving non-linear PDEs in classical field theory by means of perturbation theory.

 

[martinbn]

This is not completely true. You can use Feynman diagrams also when solving non-linear PDEs in classical field theory by means of perturbation theory.

Out of curiosity, is there an easy(for you to cite) reference?

 

[Heikki Tuuri]

Out of curiosity, is there an easy(for you to cite) reference?

https://physics.stackexchange.com/q...d-to-represent-any-perturbation-theory/245804

I found the link above. Member @A. Neumaier has answered the question.

 

[Orodruin]

Out of curiosity, is there an easy(for you to cite) reference?

I cover it briefly in chapter 7 of my textbook.

 

[martinbn]

I cover it briefly in chapter 7 of my textbook.

I suppose I can figure out what that textbook is if I check your posts?

 

[Orodruin]

I suppose I can figure out what that textbook is if I check your posts?

https://www.physicsforums.com/insights/the-birth-of-a-textbook/The cover is also my profile picture.

 

[Michael Price]

I’m not an expert, but the derivative here is gauge covariant: insensitive to gauge transformations. I don’t think this is the same as being covariant in the sense of a connection on a smooth space time manifold like you’d find in GR.

Yes, the vierbein should appear in the GR formulation.

 

[Heikki Tuuri]

I asked about classical stability of PDEs versus divergences of Feynman diagrams from Dr. Antti K.

He said that the "Feynman" diagrams in classical stability calculation of PDEs do not contain loops of virtual particles.

Those loops in standard Feynman diagrams appear because particles can "borrow energy" from vacuum for a short time.

Divergences and renormalization in standard Feynman diagrams are about the loops.

Divergences in classical stability calculations appear when we extend the loopless tree diagram larger and larger.

Thus, at the first glance, it is the possibility of borrowing energy from vacuum which makes perturbation diagrams more complex, and is responsible for the renormalization problem.

Christodoulou and Klainerman (1990) proved the stability the Minkowski metric under the Einstein equations under an infinitesimal perturbation.

There is no conflict with the fact that Feynman diagrams for linearized Einstein equations are non-renormalizable.

 

[elios]

I hope it is alright to reply here. Otherwise I apologize.

Regarding the origin of energy for the Unruh radiation upon an object supported at a fixed position above a gravitational field;

From the source of the gravitational field, whose mass will decrease. This scenario is more like Hawking radiation than Unruh radiation, since in the presence of a gravitational field spacetime is not flat and the source of the field has to be taken into account.

I don't understand via what mechanism the gravitational field will radiate since only objects with horizon will produce Hawking radiation. Do all objects produce some form of radiation proportional to their gravitational field? I'm not sure if relevant, but doesn't that require proton decay?

 

[PeterDonis]

I don't understand via what mechanism the gravitational field will radiate since only objects with horizon will produce Hawking radiation.

The post you quoted is in the middle of the discussion and it doesn't really capture the full point I was making in the discussion as a whole. If you look at other posts of mine (or even the end of the post you quoted from), you will see that I also give the energy source that is producing the acceleration of the object at the fixed position as an energy source for the radiation. For an object hovering above a gravitating body like the Earth, which is very, very far from being a black hole and having a horizon, the latter source would be the applicable one (i.e., the radiation would be Unruh radiation, not Hawking radiation). At the point in the discussion where I made the post you quoted, I was taking the other poster to be referring to a gravitating body with a horizon like a black hole, which would emit Hawking radiation. But the discussion went back and forth over a lot of possible cases.

Do all objects produce some form of radiation proportional to their gravitational field?

AFAIK you are correct that horizons are required for Hawking radiation to be produced (but strictly speaking what is required is an apparent horizon, not an absolute horizon, at least as best I understand current thinking in this area), so not all gravitating objects produce radiation just in virtue of being gravitating objects.

 

[QLogic]

Isn't it just that in the region outside the Horizon the restriction of an initially pure states $\rho$ to the algebra of the external region $\mathcal{O}$, that is $\pi_{\rho}\vert_{\mathcal{U}\left(\mathcal{O}\right)}$ is a KMS thermal state. Hence radiation. It's a purely kinematic property of QFT in the presence of a black hole.

You just need to experimentally confirm if QFT is correct then.

 

[PeterDonis]

It's a purely kinematic property of QFT in the presence of a black hole.

If it was "purely kinematic" it wouldn't decrease the mass of the hole. But it does, at least according to our best current understanding. (Obviously we can't test this experimentally, so we should bear that in mind when discussing all of these theoretical possibilities.)

 

[QLogic]

If it was "purely kinematic" it wouldn't decrease the mass of the hole

The existence of the radiation is purely kinematic though right, since one needs no specified automorphisms to derive it?

 

[PeterDonis]

The existence of the radiation is purely kinematic though

Again, if it were "purely kinematic" it wouldn't decrease the mass of the hole.

To put it another way, according to our best current understanding, Hawking radiation is observable by an inertial observer at infinity. (Or, in more technical language, it is present at future null infinity.) That means it's not "purely kinematic". It's carrying away real energy from the hole to infinity.

 

[QLogic]

Again, if it were "purely kinematic" it wouldn't decrease the mass of the hole.

Sorry but the existence of the radiation does seem to me to be purely kinematic. The existence of the KMS state falls out purely from a restriction of the state with no dynamics used.

In a more realistic model (such as semiclassical gravity), the radiation will decrease the mass if we include a coupling term, but the existence of the radiation is completely kinematic. What is needed beyond the algebraic restriction to derive it?

 

[QLogic]

See the abstract of this paper by Matt Visser:
https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.80.3436

Hawking radiation is a purely kinematic effect that is generic to Lorentzian geometries containing event horizons; it is independent of dynamics

 

[PeterDonis]

In a more realistic model (such as semiclassical gravity), the radiation will decrease the mass if we include a coupling term, but the existence of the radiation is completely kinematic.

This makes no sense to me. I'll take a look at the Visser paper you linked to when I get a chance. Possibly we simply interpret the term "purely kinematic" in different ways. That's always a potential problem when using ordinary language to talk about physics instead of math.

 

[PeterDonis]

See the abstract of this paper by Matt Visser:
https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.80.3436

The full paper is on arxiv.org here:

https://arxiv.org/abs/gr-qc/9712016

 

[QLogic]

By purely kinematic I would mean Hawking radiation is purely a property of the states and their restriction to the region outside the Horizon. The automorphism implementing the dynamics is not required. Simply the states and their restriction properties alone.

In essence when restricting to a subsystem in quantum theory generically we get a mixed state. In the case of Black Holes the mixed state for restriction to the region outside the horizon is of course a mixed state as per usual. However it turns out to be a thermal state as well, a very specific type of mixed state. Thus we get thermal radiation for free simply from the usual mixture nature of subsystems.

 

[Elias1960]

By purely kinematic I would mean Hawking radiation is purely a property of the states and their restriction to the region outside the Horizon. The automorphism implementing the dynamics is not required.

No, the dynamics is essential.

If you change the dynamics so that the collapse stops at $r_S + \varepsilon$ for whatever small $\varepsilon>0$, then the Hawking radiation will exponentially decrease one the star becomes stable.

Paranjape, A., Padmanabhan, T. (2009). Radiation from collapsing shells, semiclassical backreaction and black hole formation, Phys.Rev.D 80:044011, arxiv:0906.1768v2

 

[QLogic]

No, the dynamics is essential.

For the black hole to decrease in mass. Not for the radiation to exist. That's just an easily provable fact so well known it's not to be debated. Look at another paper here:
https://hal.archives-ouvertes.fr/hal-00710459/document

As it is by now well known, Hawking radiation is a kinematic effect

 

[elios]

If you look at other posts of mine (or even the end of the post you quoted from), you will see that I also give the energy source that is producing the acceleration of the object at the fixed position as an energy source for the radiation. For an object hovering above a gravitating body like the Earth, which is very, very far from being a black hole and having a horizon, the latter source would be the applicable one (i.e., the radiation would be Unruh radiation, not Hawking radiation).

Oh, sorry I imagined the term 'at fixed position above' as in meaning placed on the surface or a podium for some reason and missed the fact that the discussion was about a hovering rocket.

I understand that a rocket engine would heat up the object, but what about in the case as I imagined it? I assume that due to the equivalence principle, both a hovering object and a one at a fixed position on a surface would experience Unruh radiation. What would the energy source for the second one's Unruh radiation be? Also where would the radiation be seen as coming from? Behind the planet or the planets surface?

 

[PeterDonis]

I assume that due to the equivalence principle, both a hovering object and a one at a fixed position on a surface would experience Unruh radiation

No. Unruh radiation requires a vacuum; if matter is present, such as a planet on which you are standing, the derivation no longer works.

 

[Elias1960]

For the black hole to decrease in mass. Not for the radiation to exist.
That's just an easily provable fact so well known it's not to be debated.

The effect I have described is about the radiation to exist.

There is a lot of nonsense "known" by ignoring (or "not even debating") the basic problem of all derivations - the trans-Planckian problem.

The paper I have referred to is essential for illustrating this - stable states have also a stable vacuum and therefore do not radiate. This is really an elementary fact not worth to be debated, here you can simply compute the Bogoljubov transformation because it is trivial, in = out. So, if the collapse stops, the Hawking radiation stops too. Point.

And this allows to ask those who think there is nothing trans-Planckian in their personal derivation quite simple questions:

1.) Do those stable stars Hawking-radiate?
2.) If those with radius $r_S+10^{-10000}l_{Pl}$ do not radiate, but those where the collapse continues beyond this toward $r_S$ do radiate, how can this happen without anything trans-Planckian involved? (And how do you prove this without using anything trans-Planckian?)

Look at another paper here:
https://hal.archives-ouvertes.fr/hal-00710459/document

As usual, the trans-Planckian problem is not even mentioned.

 

[QLogic]

The paper I have referred to is essential for illustrating this - stable states have also a stable vacuum and therefore do not radiate. This is really an elementary fact not worth to be debated, here you can simply compute the Bogoljubov transformation because it is trivial, in = out

This isn't really related to what I said. That Hawking radiation is a kinematical property of QFT in spacetimes with event horizons. Specifically a result of the restriction of states to various regions of the space time.

This is really an elementary fact not worth to be debated

I'm not debating it.

 

[Elias1960]

This isn't really related to what I said. That Hawking radiation is a kinematical property of QFT in spacetimes with event horizons. Specifically a result of the restriction of states to various regions of the space time.

My objection was directed against a "...the dynamics is not required" phrase. Here one can reasonably object that I have taken this phrase out of context, completely misunderstood it or so.

Whatever, my objection was:
"No, the dynamics is essential."

Here, you answered
"For the black hole to decrease in mass. Not for the radiation to exist. That's just an easily provable fact so well known it's not to be debated."

To this claim I have objected and continue to object. If the dynamics is changed, so that the collapse stops however close to the horizon, the radiation itself stops. And there is nothing proven about the existence of Hawking radiation at all which is not subject to the trans-Planckian problem, and, therefore, not really worth to be named a proof.

 

[PeterDonis]

If the dynamics is changed, so that the collapse stops however close to the horizon, the radiation itself stops.

More precisely, it stops in Padmanabhan's model. All models in this regime are speculative since we have no way of testing them experimentally.

 

[QLogic]

To this claim I have objected and continue to object. If the dynamics is changed, so that the collapse stops however close to the horizon, the radiation itself stops. And there is nothing proven about the existence of Hawking radiation at all which is not subject to the trans-Planckian problem

That's not really what I am talking about. I'm saying that Hawking radiation is a kinematic feature (i.e. independent of the dynamics) for Lorentzian spacetimes with a black hole. Whether what QFT says is experimentally correct is a separate issue.

I'm not saying it's proven that Hawking radiation exists, but that it is a generic kinematic feature of QFT in black hole spacetimes. Padmanabhan's model doesn't contradict this. They show that the thermal flux from a collapsing shell will decay exponentially, this flux isn't really Hawking radiation. As they then note this flux does not prevent horizon formation. Once the horizon has formed Hawking radiation will appear as a kinematic feature of states on the spacetime.

If the dynamics are altered so that the collapse is halted then one doesn't get a horizon and thus Hawking radiation cannot appear as a kinematic effect.

Thus dynamics are necessary for Hawking radiation only in so far as they permit the formation of the black hole, i.e. the dynamics is necessary to create a spacetime on which they are a kinematic feature.

 

[Elias1960]

More precisely, it stops in Padmanabhan's model. All models in this regime are speculative since we have no way of testing them experimentally.

Of course, but "Padmanabhan's model" is simply semiclassical gravity, which is quite standard QFT on a given curved background. If one does not accept it, it makes no sense to talk about Hawking radiation at all, not?

 

[PeterDonis]

"Padmanabhan's model" is simply semiclassical gravity

No, it's semiclassical gravity with a particular assumption for the effective stress-energy tensor, one that has to violate energy conditions and has to have a particular profile as a function of the radial coordinate. In semiclassical gravity with normal matter, there is no way to stop a gravitational collapse just short of forming a black hole.

 

[PAllen]

Note, the main thrust of Padmanabhan's paper, which I have read, is to argue that under any reasonable assumption about matter:

- horizon will form and collapse continues beyond this
- Hawking radiation will result and will not be able to stop the collapse beyond the horizon

It is primarily a long response to a paper coauthored by Krauss which argued the opposite of conclusion.

 

[Elias1960]

No, it's semiclassical gravity with a particular assumption for the effective stress-energy tensor,

We use here different meanings of "semiclassical gravity". I used here the more general meaning "standard QFT on a given curved background", without a specification which particular metric theory of gravity defines this curved background. Of course, it is usually assumed that it is GR which defines this background. But given that there is anyway no backreaction considered, the same methods can be applied not only to GR solutions with nonstandard matter, but also to alternative metric theories of gravity.

Note, the main thrust of Padmanabhan's paper, which I have read, is to argue that under any reasonable assumption about matter:
- horizon will form and collapse continues beyond this
- Hawking radiation will result and will not be able to stop the collapse beyond the horizon
It is primarily a long response to a paper coauthored by Krauss which argued the opposite of conclusion.

Whatever - this is not important at all for those who reject Hawking radiation because of the trans-Planckian problem. Because the trans-Planckian problem is not a claim that the derivation of Hawking radiation is wrong, but that it has to rely on the assumption that semiclassical gravity is applicable deep in the trans-Planckian domain, which makes it completely unreliable.

Their result is sufficient to show this. If the collapse stops at $r_S + 10^{-1000}l_{Pl}$, this is certainly a trans-Planckian effect. And it would stop Hawking radiation, after a quite short time.

 

[PAllen]

...

Whatever - this is not important at all for those who reject Hawking radiation because of the trans-Planckian problem. Because the trans-Planckian problem is not a claim that the derivation of Hawking radiation is wrong, but that it has to rely on the assumption that semiclassical gravity is applicable deep in the trans-Planckian domain, which makes it completely unreliable.

Their result is sufficient to show this. If the collapse stops at $r_S + 10^{-1000}l_{Pl}$, this is certainly a trans-Planckian effect. And it would stop Hawking radiation, after a quite short time.

But their argument is that the collapse can't stop there, so what does that show?

Also, there are several derivations, by Unruh and others, that removing the trans-Planckian domain has essentially no effect on the prediction and properties of Hawking radiation:

https://journals.aps.org/prd/abstract/10.1103/PhysRevD.52.4559

not paywall version: https://arxiv.org/abs/hep-th/9506121

 

[Elias1960]

But their argument is that the collapse can't stop there, so what does that show?

Nothing. Remember, the trans-Planckian argument is that the derivation is not reliable, because it presumes that deep inside the domain where we need quantum gravity we have applied a semiclassical approximation. But we have nothing beyond the semiclassical consideration, so we cannot improve the derivation.

Also, there are several derivations, by Unruh and others, that removing the trans-Planckian domain has essentially no effect on the prediction and properties of Hawking radiation:
https://journals.aps.org/prd/abstract/10.1103/PhysRevD.52.4559
not paywall version: https://arxiv.org/abs/hep-th/9506121

In general, you cannot avoid the trans-Planckian problem without using a theory which is different from the actual theory in that trans-Planckian domain. "Removing" the trans-Planckian domain is simply a particular way to do this.

By the way, this particular way depends on killing a key assumption of GR, local Lorentz covariance. Cutting high frequencies cannot be done in a Lorentz-covariant way. So, locally you need a preferred frame. (General covariance may be preserved following, say, Jacobson's Einstein aether.) Brout et al admit this:

Any truncation scheme can be formulated in intrinsic geometric terms. However, it is convenient to work in a coordinate system that is privileged in the geometry of the incipient black hole. We make the assumption that the truncation takes a simple form in such a privileged system.

And this is the problem with many such derivations of Hawking radiation: They have to exclude those coordinates where the observer is stationary. Here is how Padmanabhan justifies this:

It is important to note that all questions about event horizon formation must be asked in a reference frame where this formation occurs in a finite time in the unperturbed collapse. It is not possible to theoretically settle this issue if one insists on working entirely in the coordinates used by static observers at large distances, even though these may be the most natural coordinates to use, simply because even in the classical scenario, event horizon formation takes an infinite amount of time in these coordinates.

The justification makes not much sense. If the consideration based on the region covered by the Schwarzschild time coordinate is sufficient to show that the collapse will not happen, but that the not-yet-BH radiates away before becoming a BH, then either the whole theory already depends on coordinates (which is what I suspect) or the other coordinates cannot show anything different anyway, thus, one can use Schwarzschild time as well.

In general, one should note that Hawking-like radiation will appear whenever there is a change of the gravitational field. To create the impression that it will appear almost certainly is therefore quite easy. All you have to do is to exclude, for whatever reason, the stationary case or to leave it to future research.

 

[QLogic]

Well most people don't consider Hawking radiation to definitely exist because there is no experimental proof. If we do find experimental proof then we'll know QFT in curved spacetimes or semiclassical gravity (which one depends on what we find) is enough. If we find evidence that contradicts them then we know we are leaving out something important due to Quantum Gravity.

The proof of Hawking radiation is just that if collapse occurs as General Relativity says with an actual horizon forming, then QFT says the states have Hawking Radiation as a kinematic feature on such Black Hole spacetimes. So we just have to see if both General Relativity and QFT are correct enough to match evidence.

 

[Elias1960]

So we just have to see if both General Relativity and QFT are correct enough to match evidence.

I disagree. Both GR and QFT are "correct enough" if they work down to the Planck length. Assuming this, we cannot derive Hawking radiation. All we can derive is some Hawking-like radiation during the collapse itself. If it remains or stops is something we cannot predict by assuming both General Relativity and QFT being correct enough.

So, both observation of long-time Hawking radiation as well as observation that it does not exist would be information about the domain beyound the semiclassical theory.