Regarding the Hawking radiation

[Matterwave]

Hello guys.

From my understanding of the Hawking radiation, it is a consequence of QFT in curved space time. The fact that a uniformly accelerated observer sees the Unruh radiation means that by the equivalence principle, a stationary observer at a distance from a black hole will see Hawking radiation. Is this correct?

If my understanding is correct, what would a freely falling observer see? Would an observer in a free fall orbit around a black hole no longer detect a Hawking radiation, just as a freely floating observer in free space detects no Unruh radiation? How can this be reconciled with the fact that the black hole will evaporate in finite time? Or is my understanding incorrect?

Thanks.

 

[Simon Bridge]

Basically: is hawking radiation the same as the unruh effect?
https://www.physicsforums.com/showthread.php?t=584942

 

[Matterwave]

I don't...really see a direct answer to my question in there though. One paper in the thread says there is a detailed violation of the strong equivalence principle...but that's all I could get out of that. Perhaps it's above my understanding level.

I'd just like to know if a freely falling observer would still see Hawking radiation, and if so, is not the Equivalence principle strongly violated, and if not, how do we reconcile this observation with a finite time of black hole evaporation?

 

[Simon Bridge]

What sort of reference frame is a free-falling observer in?

 

[craigi]

Hello guys.

From my understanding of the Hawking radiation, it is a consequence of QFT in curved space time. The fact that a uniformly accelerated observer sees the Unruh radiation means that by the equivalence principle, a stationary observer at a distance from a black hole will see Hawking radiation. Is this correct?

If my understanding is correct, what would a freely falling observer see? Would an observer in a free fall orbit around a black hole no longer detect a Hawking radiation, just as a freely floating observer in free space detects no Unruh radiation? How can this be reconciled with the fact that the black hole will evaporate in finite time? Or is my understanding incorrect?

Thanks.

This is still being debated. One proposed answer is a firewall at the absolute horizon and another is that the infalling observer sees Hawking radiation from the apparent horizon and nothing unusal at the absolute horizon. Both are controversial. It seems that it's a case where either GR or QFT fail and we need a theory of quantum gravity to get an answer.

 

[Matterwave]

What sort of reference frame is a free-falling observer in?

That would be an inertial reference frame. I don't really know where you're going with this?

@craigi: So the answer is "theoretically...we don't know"?

 

[Demystifier]

Hello guys.

From my understanding of the Hawking radiation, it is a consequence of QFT in curved space time. The fact that a uniformly accelerated observer sees the Unruh radiation means that by the equivalence principle, a stationary observer at a distance from a black hole will see Hawking radiation. Is this correct?

If my understanding is correct, what would a freely falling observer see? Would an observer in a free fall orbit around a black hole no longer detect a Hawking radiation, just as a freely floating observer in free space detects no Unruh radiation? How can this be reconciled with the fact that the black hole will evaporate in finite time? Or is my understanding incorrect?

Thanks.

The standard answer (see e.g. the book by Birrell and Davies) is the following:

Even though the presence or absence of particles depends on the observer, the presence or absence of energy-momentum tensor does not depend on the observer.

In particular, in the case of Hawking radiation, all observers agree that there is a negative energy-momentum that flows into the black hole. It is this negative energy-momentum falling into the black hole that makes the evaporation objective and observer-independent.

By contrast, in the case of Unruh effect, all observers agree that the energy-momentum tensor is zero.

 

[Matterwave]

The standard answer (see e.g. the book by Birrell and Davies) is the following:

Even though the presence or absence of particles depends on the observer, the presence or absence of energy-momentum tensor does not depend on the observer.

In particular, in the case of Hawking radiation, all observers agree that there is a negative energy-momentum that flows into the black hole. It is this negative energy-momentum falling into the black hole that makes the evaporation objective and observer-independent.

By contrast, in the case of Unruh effect, all observers agree that the energy-momentum tensor is zero.

This does indeed seem to be the only covariant way of thinking about it. But from a practical standpoint, how does one measure the stress energy tensor if not through detection of particles and their energies?

 

[atyy]

These seem relevant. Not sure whether the answers agree.

http://arxiv.org/abs/0805.1876
Taking the Temperature of a Black Hole
Erling J. Brynjolfsson, Larus Thorlacius
We use the global embedding of a black hole spacetime into a higher dimensional flat spacetime to define a local temperature for observers in free fall outside a static black hole. The local free-fall temperature remains finite at the event horizon and in asymptotically flat spacetime it approaches the Hawking temperature at spatial infinity. Freely falling observers outside an AdS black hole do not see any high-temperature thermal radiation even if the Hawking temperature of such black holes can be arbitrarily high.

http://arxiv.org/abs/0806.0628
Hawking radiation as seen by an infalling observer
Eric Greenwood, Dejan Stojkovic
(Submitted on 3 Jun 2008 (v1), last revised 8 Sep 2009 (this version, v2))
We investigate an important question of Hawking-like radiation as seen by an infalling observer during gravitational collapse. Using the functional Schrodinger formalism we are able to probe the time dependent regime which is out of the reach of the standard approximations like the Bogolyubov method. We calculate the occupation number of particles whose frequencies are measured in the proper time of an infalling observer in two crucially different space-time foliations: Schwarzschild and Eddington-Finkelstein. We demonstrate that the distribution in Schwarzschild reference frame is not quite thermal, though it becomes thermal once the horizon is crossed. We approximately fit the temperature and find that the local temperature increases as the horizon is approached, and diverges exactly at the horizon. In Eddington-Finkelstein reference frame the temperature at the horizon is finite, since the observer in that frame is not accelerated. These results are in agreement with what is generically expected in the absence of backreaction. We also discuss some subtleties related to the physical interpretation of the infinite local temperature in Schwarzschild reference frame.

http://arxiv.org/abs/1101.4382
Hawking radiation as perceived by different observers
Luis C. Barbado, Carlos Barceló, Luis J. Garay
(Submitted on 23 Jan 2011 (v1), last revised 20 Jul 2011 (this version, v2))
We use a method recently introduced in Barcelo et al (2011 Phys. Rev. D 83 41501), to analyse Hawking radiation in a Schwarzschild black hole as perceived by different observers in the system. The method is based on the introduction of an 'effective temperature' function that varies with time. First we introduce a non-stationary vacuum state for a quantum scalar field, which interpolates between the Boulware vacuum state at early times and the Unruh vacuum state at late times. In this way we mimic the process of switching on Hawking radiation in realistic collapse scenarios. Then, we analyse this vacuum state from the perspective of static observers at different radial positions, observers undergoing a free-fall trajectory from infinity and observers standing at rest at a radial distance and then released to fall freely towards the horizon. The physical image that emerges from these analyses is rather rich and compelling. Among many other results, we find that generic freely-falling observers do not perceive vacuum when crossing the horizon, but an effective temperature a few times larger than the one that they perceived when it started to free-fall. We explain this phenomenon as due to a diverging Doppler effect at horizon crossing.

 

[Haelfix]

The answer to what is seen by a freely falling observer is simply not known at the present time. There does indeed seem to be a paradox at the horizon of black holes when quantum mechanics is taken into account. The previous best idea and way out of this impasse was known as black hole complementarity (where two different observers see different things but are never able to communicate the disparity). Unfortunately this solution was shown to be flawed or at least more subtle than previously imagined.

Does a free falling observer see a Hawking flux or not? One way or the other, whatever answer you pick seems to lead to general physical paradoxes.

Generally speaking, the belief is that there will be a measurement, similar to the Unruh calculation of a nearly thermal spectrum for a near horizon accelerating observer. The local observer would need to go out to scri plus (infinity) to check to see the difference between that measurement and normal Unruh radiation.

 

[atyy]

The answer to what is seen by a freely falling observer is simply not known at the present time. There does indeed seem to be a paradox at the horizon of black holes when quantum mechanics is taken into account. The previous best idea and way out of this impasse was known as black hole complementarity (where two different observers see different things but are never able to communicate the disparity). Unfortunately this solution was shown to be flawed or at least more subtle than previously imagined.

Does a free falling observer see a Hawking flux or not? One way or the other, whatever answer you pick seems to lead to general physical paradoxes.

Generally speaking, the belief is that there will be a measurement, similar to the Unruh calculation of a nearly thermal spectrum for a near horizon accelerating observer. The local observer would need to go out to scri plus (infinity) to check to see the difference between that measurement and normal Unruh radiation.

I guess you are referring to the firewall question? But how about a non-evaporating Schwarzschild black hole? I think one still gets Hawking radiation in that case, if one calculates semiclassically. In that approximation, does one get any consistent answer for the free falling observer?

 

[Haelfix]

I'm not quite sure what you mean by a non evaporating bh with Hawking radiation. Do you mean during the phase where a bh is actively absorbing exterior thermal radiation?

It was thought, for a long time, that a freely falling observer would see nothing special at the horizon of a large enough black hole. That there was a coordinate system and a frame whereby such an observer would agree with classical reasoning and pass through unmolested.

We have no idea if that's correct or not anymore unfortunately.

 

[atyy]

I'm not quite sure what you mean by a non evaporating bh with Hawking radiation. Do you mean during the phase where a bh is actively absorbing exterior thermal radiation?

It was thought, for a long time, that a freely falling observer would see nothing special at the horizon of a large enough black hole. That there was a coordinate system and a frame whereby such an observer would agree with classical reasoning and pass through unmolested.

We have no idea if that's correct or not anymore unfortunately.

By a non-evaporating black hole with Hawking radiation, I'm thinking of something like what's mentioned in Mathur's http://arxiv.org/abs/0909.1038. There he uses the Schwarzschild metric (Eq 12), and shows that one gets Hawking radiation (section 4). I think http://arxiv.org/abs/0704.1814 mentions something similar, because they say (p6) "Note that there is a natural limit where the geometry (R_S) is kept fixed, M → ∞ but G → 0 so that S_BH → ∞. In this limit there is still a black hole together with its horizon and singularity, and it emits Hawking radiation with temperature T_H ∼ R_S⁻¹, but t_ev → ∞ so the black hole never evaporates." They also say (p11) "Notice however that this breaking down is not merely a kinematical effect due to the presence of the horizon, after all EFT is perfectly good for computing Hawking radiation. In fact in the limit described at the beginning of this section, when we keep the geometry fixed and we decouple dynamical gravity, there still is an horizon but effective field theory now gives the right answer for arbitrarily long time scale: the black hole doesn’t evaporate and information is entangled with states behind the horizon."

I understand that if the black hole does evaporate, one runs into conceptual problems like the sub-additivity that Mathur pointed out, and the AMPS firewall. But since one can get Hawking radiation in the limit of a non-evaporating black hole, can one also calculate a reasonable answer for the free-falling observer in that limit?

 

[Haelfix]

Its a good question, I don't know. I don't really understand the limits that Nima et al are taking and how physical or not the question is. Mathur's paper was an important one, and one that is frequently cited and referred too, b/c it contains a proof that small corrections in the Hawking radiation cannot unitarize the physics.

This is somewhat controversial still (see Raju and Papodimas's talk at strings 2014), but a priori it seems quite robust (and seemed to me at the time).

As far as I know, the controversy over what happens at the horizon is pretty much ongoing and involves the horizon at any stage of a black holes lifetime (Polchinski and Marolf believe that there is a hot horizon even before the Page time for instance).

 

[Demystifier]

But from a practical standpoint, how does one measure the stress energy tensor if not through detection of particles and their energies?

Perhaps by measuring the gravitational field sourced by the energy-momentum?

 

[Matterwave]

Perhaps by measuring the gravitational field sourced by the energy-momentum?

So are you saying that even though in different frames I might see different number of particles or some frames where I detect no particles at all, I still will measure an invariant S-E tensor due to gravitational effects...of the radiative matter?

If this is the only way to measure the SE tensor, then practically speaking, it doesn't seem likely that any of these findings can be experimentally verified...since the Hawking radiation from a stellar mass black hole is tiny and the corresponding gravitational effects must be even tinier.

In addition, I'm not sure how one calculates the gravitational effect due to free streaming radiation. This in itself seems to be a difficult problem.

 

[craigi]

If this is the only way to measure the SE tensor, then practically speaking, it doesn't seem likely that any of these findings can be experimentally verified...since the Hawking radiation from a stellar mass black hole is tiny and the corresponding gravitational effects must be even tinier.

Indeed. Measurement of Hawking radiation from stellar mass black holes isn't viable.

Unruh discovered that supersonic fluid flow can exhibit a direct analog to Hawking radiation in the form of phonons and argues that Hawking radiation has actually now been experimentally confirmed.

Nevertheless, it seems unlikely that Hawking is due a Nobel prize.

 

[Demystifier]

So are you saying that even though in different frames I might see different number of particles or some frames where I detect no particles at all, I still will measure an invariant S-E tensor due to gravitational effects...of the radiative matter?

Yes, that's exactly what I am saying.

If it looks paradoxical, perhaps the following analogy might help. Consider a Stern-Gerlach device (i.e., a device which measures spin in a given direction) which clicks only when the spin is up. Consider a spin-1/2 particle which in any given direction has two posible projections od spin: spin-up and spin-down. The corresponding states, traditionally called |+> and |->, can be renamed as |1> and |0>, respectively. This is a useful notation because the presence of click corresponds to the state |1>, while the absence of click corresponds to the state |0>.

Now suppose that you have prepared your system in the state |0>. Does it mean that your detector cannot click? No! You can make it click by rotating the detector, i.e. by changing its orientation. For the rotated detector the natural basis is |1'>, |0'>, which differs from the original basis |1>, |0>. In particular, your state |0> is a superposition of |1'> and |0'>, so there is a finite probabilty that this superposition will collapse to |1'>, i.e., that the detector will click.

So whether the detector can click depends on the observer, or more precisely, on how the observer orientates his detector. But energy of the observed object (in this case, of the spin-1/2 particle) does not depend on it.