Validity of theoretical arguments for Unruh and Hawking radiation

[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.

 

[QLogic]

But you get Hawking radiation from using QFT on a curved background.

 

[martinbn]

I am confused about the meaning of "the collapse stops". Can someone clarify?

 

[QLogic]

The star undergoes a process initially similar to collapse into a black hole, but the process doesn't reach the stage of the forming of a horizon.

 

[martinbn]

The star undergoes a process initially similar to collapse into a black hole, but the process doesn't reach the stage of the forming of a horizon.

And what does this have to do with whether black holes radiate or not?

 

[QLogic]

And what does this have to do with whether black holes radiate or not?

That's what I was wondering above. Basically if the collapse stops there is no black hole to radiate and since Hawking radiation is radiation from a black hole it means there is no Hawking radiation.

As I mentioned above if you have a black hole it will automatically radiate since Hawking radiation is a kinematic effect. So the only way to stop a black hole from radiating in QFT on curved spacetimes is to have no black hole.

 

[Elias1960]

So the only way to stop a black hole from radiating in QFT on curved spacetimes is to have no black hole.

Yes. This is not questioned: QFT on a curved background predicts that a BH resulting from a collapse radiates.

But the domain of applicability of QFT on curved background is limited, and the limits can be seen in the theory itself. Namely, it is not a consistent theory, because it has no back-reaction of the quantum fields on the classical background. And the rough estimate for when we need more, full QG, is when we reach effects of order of Planck distance, Planck time or Planck energy. The trans-Plackian problem is that the semiclassical QFT derivation depends on semiclassical QFT remaining valid deep in the trans-Planckian domain. This assumption is nonsensical.

The question what one names Hawking raditation is irrelevant. There is a mass M, if it is of an actually collapsing star or a BH is irrelevant, because what we can measure outside is anyway the same. Hawking radiation is simply thermal radiation with a particular temperature depending on this mass M. If we see it coming from the direction of a BH candidate, and trace back the corresponding classical light ray, it ends (starts) from the collapsing surface before horizon creation, which comes from a trans-Planckian distance from the horizon, has gone through trans-Planckian time dilation (this is when a Planck time on the surface translates into more than the time after BB for the outside observer) an redshifted down from a trans-Planckian energy. This is what makes the whole thing trans-Planckian. But if it is really a BH or not yet observation cannot decide.

 

[QLogic]

There is a mass M, if it is of an actually collapsing star or a BH is irrelevant, because what we can measure outside is anyway the same. Hawking radiation is simply thermal radiation with a particular temperature depending on this mass M. If we see it coming from the direction of a BH candidate, and trace back the corresponding classical light ray, it ends (starts) from the collapsing surface before horizon creation

This isn't my understanding of Hawking radiation. Hawking radiation is usually defined as radiation that results from the state restricted to observables outside the horizon being a mixed state of KMS form.

If you have a star, i.e. no horizon, the mixed state has quite a different form that lacks Hawking radiation's completely thermal profile. Also Hawking radiation does not originate from the surface of the star since it's a fully kinematic effect in the post-horizon formation spacetime.

 

[Elias1960]

This isn't my understanding of Hawking radiation. Hawking radiation is usually defined as radiation that results from the state restricted to observables outside the horizon being a mixed state of KMS form.

Maybe some people consider it that way, but IMHO it makes no sense.

You need an initial state, the star before the collapse, with the usual ~ Minkowski vacuum. Then you have a collapse and the vacuum state changes together with the metric. The result differs from the vacuum at that time, the difference is the radiation. As long as the collapse continues, the change of the vacuum state continues, thus, new radiation appears.

If you have a star, i.e. no horizon, the mixed state has quite a different form that lacks Hawking radiation's completely thermal profile. Also Hawking radiation does not originate from the surface of the star since it's a fully kinematic effect in the post-horizon formation spacetime.

In Hawking's derivation, the Hawking modes are those modes which go through the star during the collapse and, because of the change which happens while they are inside, appear different than they have started, thus, differ from the vacuum modes. To name this a "fully kinematic effect in the post-horizon formation spacetime" is at best misleading (but I would guess simply nonsense). You need the element of change, without it there is no Hawking radiation. Which is what the paper

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

proves for a Hawking-like situation, but which is a quite general principle which follow from the way Hawking radiation is derived - it is caused by the difference between the time-evolved in-vacuum and the out-vacuum. But without change, the vacuum is stable and therefore there is no difference.

 

[QLogic]

Maybe some people consider it that way, but IMHO it makes no sense.

"fully kinematic effect in the post-horizon formation spacetime" is at best misleading (but I would guess simply nonsense)

I've given two papers, one from an expert on QFT in curved spacetime where the fact that it is a kinematic effect is not only mentioned but described as "well known". I can give several other sources that say this. One of the papers even explains exactly how it is a kinematic effect. What is actually wrong with what Visser is saying?

As long as the collapse continues, the change of the vacuum state continues, thus, new radiation appears.

It's not thermal until a horizon has formed. Prior to the formation of the horizon the vacuum has a statistical profile that differs from the Minkowski profile but it's not yet thermal so it usually isn't called radiation.

proves for a Hawking-like situation, but which is a quite general principle which follow from the way Hawking radiation is derived - it is caused by the difference between the time-evolved in-vacuum and the out-vacuum. But without change, the vacuum is stable and therefore there is no difference

Of course. The radiation is still a property of the state itself though, not of the dynamics. Hawking radiation is a property of the statistics inherent in the vacuum. You need a horizon for states to have those statistics and the horizon itself can only be created by the dynamics, but the radiation is not produced by the collapse dynamics that form the horizon. It's simply a property of the state.

I will say as a side note that I am the one using standard terminology here. It's difficult to have a discussion if you don't stick to standard terminology and in standard terminology Hawking radiation is kinematic. It's a bit strange to have to argue for textbook terminology.

 

[Elias1960]

I've given two papers, one from an expert on QFT in curved spacetime where the fact that it is a kinematic effect is not only mentioned but described as "well known". I can give several other sources that say this. One of the papers even explains exactly how it is a kinematic effect. What is actually wrong with what Visser is saying?

That it ignores the trans-Planckian problem.

It's not thermal until a horizon has formed. Prior to the formation of the horizon the vacuum has a statistical profile that differs from the Minkowski profile but it's not yet thermal so it usually isn't called [Hawking?] radiation.

If you use Schwarzschild time, the horizon never forms. So, the radiation will be never thermal. In fact, the difference is completely negligible already in the Planckian region.

The radiation is still a property of the state itself though, not of the dynamics. Hawking radiation is a property of the statistics inherent in the vacuum. You need a horizon for states to have those statistics ...

A radiation which stops immediately after the dynamics stops and therefore the state remains unchanged, and you need a horizon to have it even if in Schwarzschild coordinates you have no horizon at any finite time. Sorry, I continue to consider such claims as misleading at best.

I will say as a side note that I am the one using standard terminology here. It's difficult to have a discussion if you don't stick to standard terminology and in standard terminology Hawking radiation is kinematic. It's a bit strange to have to argue for textbook terminology.

I have no doubt that this is textbook terminology. There is no doubt that I'm questioning the mainstream approach, given that (better as far as) it ignores the trans-Planckian problem. Any textbook which introduces Hawking radiation but does not discuss the trans-Planckian problem is misleading at best. In democratic physics, you can easily find the 51% votes to prove me wrong.

 

[QLogic]

That it ignores the trans-Planckian problem.

That it's a statistical property of a state restricted to observables outside the horizon has nothing to do with the Trans-Planckian problem.

Let me put it this way. Hawking radiation is due to detection events outside the horizon having a thermal profile. The radiation doesn't emanate from the black hole. It's not driven by the dynamics. It requires a horizon but it is not dynamical. It's just a correlation property of the vacuum not driven by dynamics.

A radiation which stops immediately after the dynamics stops

It doesn't stop. Prior to the hole's formation there is no Hawking radiation. After the horizon forms there is. The radiation emitted during collapse discussed in Padmanabhan's paper is not Hawking radiation.

If the collapse stops it means there never is Hawking radiation.

 

[PeterDonis]

That it's a statistical property of a state restricted to observables outside the horizon has nothing to do with the Trans-Planckian problem.

Mathematically, it doesn't, no. But physics is not mathematics.

I have read the Visser paper you linked to; basically, by "kinematic property" he means "you can derive it from the math of generic classical Lorentzian spacetimes with horizons, without having to make any use of the particular dynamical law that determines the spacetime geometry". That's math, not physics. Math can't emit actual radiation that actual detectors detect. You would need to make a physical argument that the math in Visser's paper is physically relevant for the case under discussion. So far you have not made any such argument in this thread, nor does Visser in his paper.

 

[Elias1960]

That it's a statistical property of a state restricted to observables outside the horizon ...
It doesn't stop. Prior to the hole's formation there is no Hawking radiation. After the horizon forms there is.

This would be a contradiction with relativistic causality or relativistic symmetry.

Once it is a property detectable by observables outside the horizon, it has to be detectable before horizon formation, because it has to be detectable in Schwarzschild time too (assuming the effect does not depend on coordinates). But Schwarzschild time covers only the part before horizon formation, and cannot be causally influenced by anything happening after horizon formation according to Einstein causality. It would follow that there cannot be any Hawking radiation in the whole region of spacetime covered by Schwarzschild coordinates. So, in the form you present it here, it is self-contradictory, and makes no sense.

 

[PeterDonis]

Schwarzschild time covers only the part before horizon formation

This is only true for an "eternal" black hole that never evaporates. If the hole evaporates, it is no longer true that "horizon formation" occurs at $t = + \infty$ in Schwarzschild coordinates. In fact, the usual definition of Schwarzschild coordinate time doesn't even work in such a spacetime.

I think a better way of phrasing your underlying (valid) point would be that @QLogic needs to be a lot more specific about exactly what he means by "before" and "after" horizon formation. Most ways of specifying what those terms mean are highly coordinate-dependent. There are ways of doing it in an invariant manner, but it's not easy, and it's not altogether clear whether such an invariant definition has all of the properties he is implicitly assuming.

 

[PAllen]

A further point is simply that is false that Schwarzschild time only covers events before horizon formation, even in a classical non evaporating collapse. Instead, there is a clear demarcation of exterior events into which horizon formation is the causal future, versus not the causal future, thus possibly present. For the whole region in which horizon formation is not causal future, it is plausible to consider the horizon as currently existing.

 

[Elias1960]

This is only true for an "eternal" black hole that never evaporates. If the hole evaporates, it is no longer true that "horizon formation" occurs at $t = + \infty$ in Schwarzschild coordinates.
In fact, the usual definition of Schwarzschild coordinate time doesn't even work in such a spacetime.

What happens during the evaporation process is already part of a different theory, it is not covered by standard semiclassical QFT, because one has, somehow, to incorporate backreaction. How to do this without specifying preferred coordinates is completely unclear.

I think a better way of phrasing your underlying (valid) point would be that @QLogic needs to be a lot more specific about exactly what he means by "before" and "after" horizon formation. Most ways of specifying what those terms mean are highly coordinate-dependent. There are ways of doing it in an invariant manner, but it's not easy, and it's not altogether clear whether such an invariant definition has all of the properties he is implicitly assuming.

Ok, one can formulate it this way. But I think my remark remains correct, given that I specify the time coordinate I use, so that I can use "before" and "after" horizon formation in their usual meaning as referring to the particual system of coordinates used. Those who don't specify the coordinates should live with the fact that their statements can be rejected as false if applied to some systems of coordinates they are false.

A further point is simply that is false that Schwarzschild time only covers events before horizon formation, even in a classical non evaporating collapse. Instead, there is a clear demarcation of exterior events into which horizon formation is the causal future, versus not the causal future, thus possibly present. For the whole region in which horizon formation is not causal future, it is plausible to consider the horizon as currently existing.

One can, indeed, use the possibility of a system of coordinates with a time so that the horizon already exists as a reasonable way to give "before horizon formation" and "after horizon formation" a coordinate-independent meaning.
But this does not make my claim about the Schwarzschild time coordinate false. .

 

[PAllen]

...

One can, indeed, use the possibility of a system of coordinates with a time so that the horizon already exists as a reasonable way to give "before horizon formation" and "after horizon formation" a coordinate-independent meaning.
But this does not make my claim about the Schwarzschild time coordinate false..

But then who cares about Schwarzschild time coordinate? Physics s about coordinate independent statements.

 

[Elias1960]

But then who cares about Schwarzschild time coordinate? Physics s about coordinate independent statements.

Then one would better avoid talking about "before/after horizon creation" too.

It is, moreover, far from sure that the theory we need to describe back-effects will not have preferred coordinates.

 

[PAllen]

Then one would better avoid talking about "before/after horizon creation" too.

It is, moreover, far from sure that the theory we need to describe back-effects will not have preferred coordinates.

The statement that the horizon is not in the causal future of some exterior event is a coordinate independent statement.

 

[PeterDonis]

I can use "before" and "after" horizon formation in their usual meaning as referring to the particual system of coordinates used.

No, you can't, because standard Schwarzschild coordinates don't cover the horizon, and don't cover the event of horizon formation. So talking about "before" and "after" horizon formation is meaningless in Schwarzschild coordinates.