In March 2012, Joseph Polchinski claimed that the following three statements cannot all be true [1]: (i) Hawking radiation is in a pure state, (ii) the information carried by the radiation is emitted from the region near the horizon, with low energy effective field theory valid beyond some microscopic distance from the horizon, (iii) the infalling observer encounters nothing unusual at the horizon. Joseph Polchinski argue that the most conservative resolution is that [1]: the infalling observer burns up at the horizon. In Polchinski's account, quantum effects would turn the event horizon into a seething maelstrom of particles. Anyone who fell into it would hit a wall of fire and be burned to a crisp in an instant. As pointed out by physics community such firewalls would violate a foundational tenet of contemporary physics known as the equivalence principle [2], it states in part that an observer falling in a gravitational field --- even the powerful one inside a black hole --- will see exactly the same phenomena as an accelerated observer floating in empty space [3].
[1] Almheiri, A., Marolf, D., Polchinski, J.,Sully, J. http://arxiv. org/abs/1207.3123 (2012). [2] Ginzburg V.L.,Eroshenko, Yu.N.,Once again about the equivalence principle, 1996 Phys.-Usp. 39 1275 [3] Merali, Z., 4 April 2013,Vol 496,Nature,23
With a complete theory combining general relativity with quantum theory missing, it is vain to say who is wrong and who isn't. It is all but speculation! And in any case I am doubtful anyone in this forum will have a definite answer because of exactly that.
The only thing that can be said for sure is: whatever the physics of a radiating/evaporating black hole is, it should -- for logic's and consistency's sake -- somehow merge seamlessly with some semiclassical analysis, which in turn should merge seamlessly with the classical laws of physics.
With regards to this, an infalling object/observer undergoes in principle a scattering process in a curved spacetime. Classically, the only effect (in the observer's proper time) is being either crossing the horizon and vanish in the singularity, or not to cross the horizon if possible and being transferred far far away into the universe's future.
With an evaporating black hole, whatever the exact physics near the horizon is, it should have this classical effect as a limiting case, and the limiting case being large, supermassive black holes, small spacetime curvature.
No, in fact, he isn't-because there's no reason to expect low energy effective field theory to be an appropriate description. So (i) and (iii) are perfectly valid statements and it's (ii) that's of limited validity.
Incidentally, the event horizon of a black hole isn't a space-like surface; so the statements about ``vicinity of the event horizon'' or ``near (or far from) the event horizon'' aren't meaningful. It's a light-like surface. This leads to considerable confusion.
An observer in free fall in a black hole spacetime isn't equivalent to an accelerating observer in flat spacetime, because the black hole spacetime has singularities.
The freely falling (non-extremal) observer in a (non-extremal) black hole spacetime has a space-like singularity in his/her future, to be experienced in finite proper time; an accelerating observer in flat spacetime doesn't.
>the black hole spacetime has singularities
But that is only coordinate singularities.
SN: .. the black hole spacetime has singularities
JF: But that is only coordinate singularities.
The event horizon is a coordinate singularity only, the central singularity seems to be physical.
Regarding the original question, point (ii) is wrong as Stam said, see this blog article by Sabine Hossenfelder for more detail:
http://backreaction.blogspot.co.uk/2015/12/hawking-radiation-is-not-produced-at.html
If we assume, that small black holes do not exist, then evaporation of black holes caused by Hawking radiation cannot occur and there is no information carried away from the black hole.
In large black holes generated by gravitational collapse, the Hawking radiation generated at the event horizon has no chance to reach the outer space, because the escape speed is larger or equal to the speed of light. Therefore large black holes, which exceed this escape speed limit, will never evaporate. The remaining question is:
Can medium size or rotating black holes be generated by gravitational collapse, which do not exceed the escape speed limit?
GD:..The event horizon is a coordinate singularity only, the central singularity seems to be physical.
Ok.But what about "the infalling observer burns up at the horizon"? [1].
If (ii) is wrong, the radiation is only evident far from the horizon. At the horizon, there is just the usual local quantum vacuum.
No (besides the fact that ``at the event horizon'' doesn't make sense! Nor does ``far from the event horizon''), in a curved spacetime the vacuum isn't unique-that's the whole point. However, since it's not known how the metric adjusts to the presence of the non-trivial particle profile, any statements about the evolution aren't solid. It's not known how to compute the transition amplitude from a given spacetime geometry, with some distribution of Hawking quanta, to another one.
That's why it's not known how quantum effects resolve the black hole singularity-which is the only thing that matters. The event horizon simply separates the observers, that have the singularity in their future from those that don't. A firewall would imply that, in fact, no quantum observers have the singularity in their future-but those that would, in the classical limit, don't contribute to any process. The problem is that it isn't at all obvious how to make the distinction precise.
in my view the more tangible question is about the validity of the equivalence principle. should QG show GR (and therefore SR) limits then the Rindler wedge is a good framework for analysis. the radiation is emanating from striped pairs 'outside the horizon' and it's temperature is measured at infinity where the BH is a classical object that adheres to Newtonian limits anyway. generally speaking and as can be expected from astrophysics, the horizon can be hot unless the BH undergoes a phase transition. what bothers the philosophers is loss of information and the violation of the GSL of thermodynamics through breakdown of quantum entanglement (QE). there are empirical means by which one should be able to attempt at least to observe the effect of gravity on QE. see for example -- https://www.researchgate.net/post/How_does_gravity_affect_quantum_entanglement
G.D.:The event horizon is a coordinate singularity only, the central singularity seems to be physical.
No at all. The event horizon singularity also seems to be physical.
The reason for this is that the spatial metric (97.15) in contrast to the space-time metric (97.14), actually has a singularity at r = rg .
see Landau. Lifshitz. The Classical. Theory of Fields. Third Revised English Edition. Course of Theoretical Physics. Volume 2. p.287
http://www.elegio.it/mc2/LandauLifshitz_TheClassicalTheoryOfFields_text.pdf
The singularity of the coordinate system at the horizon can be eliminated by a general coordinate transformation-it would be better to actually study the book, not just quote it.
This transformation is known, cf. http://phya.snu.ac.kr/php/subject_list/Notice/data/1401260740.pdf
Loss of information isn't a philosophical issue, it's a physical one, since the observers that belong to the interior of the black hole, that's the result of gravitational collapse-and, therefore, experience the singularity in finite proper time-become causally disconnected from those that don't (these belong to the exterior).
S.N.:The singularity of the coordinate system at the horizon can be eliminated by...
No at all.
The singularity of the coordinate system at the horizon cannot be eliminated by a general coordinate transformation!
The “soft” singularity at r = 2m of the standard metric (see eq. (1) of sect.2) is “hidden” in the differentials of the new coordinates with respect to the standard ones.see eq. (12) of sect.4)
https://arxiv.org/pdf/0906.0168.pdf
Such transformantion with singularity in the differentials, as mentioned above, of course beyond rigorous Riemannian geometry.
Note that. If you remove to the cat a tail that you have no the same cat but in fact you have another cat without a tail.
The singularity at r=2m isn't hidden, any more or less than the singularities of any coordinate system, that isn't flat, is hidden, by a local change of coordinates. But since a local change of coordinates can be found, in which the surface r=2m isn't singular, this is what matters. The future singularity at r=0 can't be removed, for freely falling and, certain, accelerating observers; it can be removed for certain accelerating observers.
S.N. :by a local change of coordinates...
There is no such singular change of coordinates in rigorous Riemannian geometry. Only diffeomorphism as change of coordinates is possible. Clearly why it so or not?
The change of coordinates isn't singular, at r=2m-and that's trivial to check. The fact that it's observer dependent doesn't matter-that all observers, whatever their motion, can define coordinates that are regular on light-like surfaces, by a diffeormorphism, is what matters. Claiming that this isn't true doesn't make sense, given the explicit expressions for the change of coordinates. What matters is their regularity on the light-like surface, that's at r=2m, when using a particular coordinate system, not anywhere else.
Freely falling observers can't define coordinates that are regular at r=0, in the same coordinate system; nor can certain accelerating observers. More precisely, such observers can't define coordinates that are free of a space-like singularity at all points of their worldline.
These are the observers that ``fall in the black hole''. And the fact that the singularity is space-like is what implies that they will encounter it in finite proper time.
Certain other accelerating observers can define coordinates that are regular everywhere along their worldline -these are those that ``don't fall in the black hole''-which means that they don't experience the singularity in finite proper time.
The non-trivial question is how to distinguish the three kinds of observers.
https://en.wikipedia.org/wiki/Kruskal%E2%80%93Szekeres_coordinates
dT/dr=dX/dr=infinity at r=2m. Terefore Kruskal–Szekeres transform is not
diffeomorphism as it should be in rigorous Riemannian geometry.
By using Kruskal–Szekeres transform we know nothing about Schwarzschild horizon at r=2m.
A black hole must establish thermal equilibrium with the CMB of nearly flat space to remain stable. Gravitational blue shifting occurs to incoming radiation, and gravitational red shifting occurs for outgoing radiation. To accomplish this the region very close to the black hole must be extremely hot.
From a distance the Hawking Temperature of a black body radiator applies. For a traveler the apparent temperature increases during approach to the event horizon as the radius of the event horizon appears to shrink. The traveler gets incinerated by the blue shifted radiation before reaching the standard radius of the event horizon. It happens quickly for the traveler, but appears slow to a distant observer.
Blue shift (or red shift) isn't relevant for the topic under discussion. A black hole isn't in thermal equilibrium with a bath at 2.7 K, but at the Hawking temperature-which for astrophysical black holes is a hundredth of a microkelvin. That's why it's not possible, in practice, to detect Hawking radiation from astrophysical black holes. The topic represents a consistency check, only.
A freely falling observer is inertial and doesn't detect any radiation-that's what the terms mean, since such an observer isn't accelerating in the spacetime. The only thing is that such an observer experiences the black hole singularity in finite proper time.
Accelerating observers exist, that don't experience the singularity in finite proper time-these could detect, of course, radiation, but it would be at Hawking temperature.
Accelerating observers exist, that will experience the singularity in finite proper time; and, also, detect Hawking radiation, since they're accelerating.
Once more, the event horizon is light-like surface, so there's no notion of a distance from it. Nor is there any notion of the radius of the event horizon, for that reason. For a non-extremal black hole, the singularity is a space-like surface, that extends over the whole black hole spacetime; and the only relevant notion of distance is the proper time from it, along the worldline of an observer.
S.N.:Accelerating observers exist, that don't experience the singularity in finite proper time-these could detect, of course, radiation, but it would be at Hawking temperature.
Ok. But string theory predict that Unruh account is wrong and revision needed.
No it doesn't. It just can't provide, for the moment, the framework for calculating all effects of Hawking radiation, since it's not known, yet, how to compute transition amplitudes with arbitrary number of Hawking quanta in the initial and final states, nor what are the intermediate string states, in that case.
The Unruh effect is perfectly well understood at scales where string effects aren't relevant.
S.N.:No it doesn't. It just can't provide,
Ok.But what about firewalls?
Once more-read the previous messages. Firewalls aren't inconsistent with Unruh radiation-they imply that Hawking radiation is Unruh radiation.
It is widely believed that firewalls violated the Einstein equivalence principle.
Belief doesn't have anything to do with this-the statement's wrong, since firewalls don't imply that matter couples to gravity through properties other than its energy-momentum tensor; nor do they imply that fields other than the metric are required to describe spacetime.
They do imply that freely falling observers can't be defined and that all accelerating observers are equivalent in black hole spacetimes-which does pose the problem of how freely falling observers are eliminated, since this can't be a local process.
S.N.:The singularity of the coordinate system at the horizon can be eliminated by a general coordinate transformation...
The singularity of the coordinate system at the horizon can be eliminated also by using well known isotropic coordinates. However Schwarzschild spacetime in these coordinates, essentially different in comparison with Schwarzschild spacetime in kruskal-szekeres coordinates. For example metric tensor is allowed to become degenerate at horizon in isotropic coordinates but in kruskal-szekeres is not degenerate.
The metric tensor isn't a physical quantity-it changes under general coordinate transformations. So it doesn't make sense to focus on it, any more than it is to focus on the properties of any gauge-variant quantity-it's the gauge invariant quantities that can be constructed from it that are relevant.
S.N:The metric tensor isn't a physical quantity-
Yes of course. But canonical Riemannian Geometry break down for these metrics (see attached picture). Note that: if the fundamental tensor is allowed to be degenerate, there are some obstructions in constructing the geometric objects normally associated to the fundamental tensor. Also,local orthonormal frames and coframes no longer exist, as well as the metric connection and its curvature operator.
The vacuum energy density of free scalar quantum field phgr in a Rindler distributional space-time with distributional Levi-Cività connection is considered. It has been widely believed that, except in very extreme situations, the influence of acceleration on quantum fields should amount to just small, sub-dominant contributions. Here we argue that this belief is wrong by showing that in a Rindler distributional background space-time with distributional Levi-Cività connection the vacuum energy of free quantum fields is forced, by the very same background distributional space-time such the Rindler distributional background space-time, to become dominant over any classical energy density component. This semiclassical gravity effect finds its roots in the singular behavior of quantum fields on a Rindler distributional space-times with distributional Levi-Cività connection. In particular we obtain that the vacuum fluctuations 〈phgr2〉 has a singular behavior on a Rindler horizon. Therefore sufficiently strongly accelerated observer burns up near the Rindler horizon. Thus Polchinski's account doesn't violate of the Einstein equivalence principle.
Conference Paper Was Polchinski wrong? Colombeau distributional Rindler space...
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