# Evaporating BH and infalling observers - I am puzzed

Here is an article Count Iblis recommended me:
http://arxiv.org/abs/gr-qc/0609024

Page 11

The infalling observer never crosses an event horizon, not because it takes an infinite time, but because there is no event horizon to cross. As the infalling observer gets closer to the collapsing wall, the wall shrinks due to radiation back-reaction, evaporating before an event horizon can form. The evaporation appears mysterious to the infalling observer since his detectors don’t register any emission from the collapsing wall. Yet he reconciles the absence of radiation with the evaporation as being due to a limitation of the frequency range of his detectors

However, even without the radiation, an infalling observer is not supposed to cross it - that is what I was thinking:

http://en.wikipedia.org/wiki/Event_horizon#Interacting_with_an_event_horizon

In practice, all event horizons appear to be some distance away from any observer, and objects sent towards an event horizon never appear to cross it from the sending observer's point of view (as the horizon-crossing event's light cone never intersects the observer's world line).

Observers who fall into the hole are moving with respect to the distant observer, and so perceive the horizon as being in a different location, seeming to recede in front of them so that they never contact it.

Also, if (in his frame) the radiation is so low-energy that the wavelength can not be detected, then how the evaporation can be so fast? I suspec that somewhere in the article there is an error - they mixed frames of distant and infallinf observers....

JesseM
A calculation like this is probably based on semiclassical quantum gravity, in which case the results would be open to question since we don't have a complete theory of quantum gravity and thus wouldn't know if the approximations are good (and having just read https://www.amazon.com/dp/0316016403/?tag=pfamazon01-20 I think a lot of physicists in quantum gravity would dispute the conclusion that an infalling observer doesn't reach the horizon before it evaporates)...anyway, if this is a question that touches on quantum gravity issues, maybe it should be in the "Beyond the Standard Model" forum?

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Yes, it is a good idea.

But how an observer can cross HIS horizon - in HIS frame? I can not imagine such spacetime diagram...

George Jones
Staff Emeritus
Gold Member
Here another post on the Vachaspati, Stojkovic, Krauss paper,

However, even without the radiation, an infalling observer is not supposed to cross it - that is what I was thinking:

http://en.wikipedia.org/wiki/Event_horizon#Interacting_with_an_event_horizon

Also, if (in his frame) the radiation is so low-energy that the wavelength can not be detected, then how the evaporation can be so fast? I suspec that somewhere in the article there is an error - they mixed frames of distant and infallinf observers....

Wikipedia is wrong. In particular,
Wikepedia said:
Observers who fall into the hole are moving with respect to the distant observer, and so perceive the horizon as being in a different location, seeming to recede in front of them so that they never contact it.

is wrong.

The (absolute) event horizon[/edit] is a particular set of spacetime events, and, for an infalling observer, one of the events on the event horizon is also an event on the observer's worldline.

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Wait, when I look here: http://www.valdostamuseum.org/hamsmith/DFblackIn.gif [Broken]

I see a point where an observer crosses a horizon, but from an observer's point of view, nothing strange happens at that moment.

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George Jones
Staff Emeritus
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Wait, when I look here: http://www.valdostamuseum.org/hamsmith/DFblackIn.gif [Broken]

I see a point where an observer crosses a horizon, but from an observer's point of view, nothing strange happens at that moment.

Exactly.

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Wait, but it is exactly what Wiki is explaining!
When observer crosses a 'horizon' (calculated based on the distant observers data - or from a 'bird's view' - the picture is a sort of 'bird's view') for him there is no horizon at that point! But for an observer there is still ANOTHER horizon at ANOTHER place - obviously, particles entered the BH much earlier can not interact with him. The closer he approaches the singularity, the closer is the horizon to him.

George Jones
Staff Emeritus
Gold Member
Wait, but it is exactly what Wiki is explaining!
When observer crosses a 'horizon' (calculated based on the distant observers data - or from a 'bird's view' - the picture is a sort of 'bird's view') for him there is no horizon at that point! But for an observer there is still ANOTHER horizon at ANOTHER place - obviously, particles entered the BH much earlier can not interact with him. The closer he approaches the singularity, the closer is the horizon to him.

But this isn't the event horizon of the black hole.

Hm, then how it is called?
Does it emit Hawking radiation? I guess yes, because if a virtual pair falls inside then it can be torn apart.

George Jones
Staff Emeritus
Gold Member
Hm, then how it is called?
Does it emit Hawking radiation? I guess yes, because if a virtual pair falls inside then it can be torn apart.

From the Wikipedia article,
The definition of "event horizon" given by Hawking & Ellis,[1] Misner, Thorne & Wheeler,[2] and Wald[3] differs from the one presented here. Their definition rules out the cosmological and particle horizons presented below (as well as the apparent horizon). However, modern usage has brought those ideas under the umbrella of the term "event horizon".[4] To make the distinction clearer, some authors refer to their more specific notion of a horizon as an "absolute horizon". In the context of black holes, event horizon almost always refers to the absolute horizon, as distinct from the apparent horizon.

Thank you, it was really helpful. Now I can answer my own question: based on the equivalence principle, apparent horizon also emits radiation.

BTW I think Hawking is wrong here (oh my God, what am I saying?)

Hawking has argued [9] that the infalling observer does not detect significant Hawking adiation since the emission is dominantly at low frequencies compared to 1/RS, while the infalling observer can only have local detectors of size less than RS. Thus the infalling observer would appear to see event horizon formation in a finite time, with no significant radiation emanating from the black hole.

For example, 21cm stellar hydrogen line can be emiited and hence be absorbed by tiny atoms, which are obviously much smaller then 21cm

perhaps, i am simply too dim, but i find hawking's technical writing to be completely incomprehensible. his chapters in "The Nature of Space and Time" are so confoundingly obtuse, i get lost within a few paragraphs, whereas the chapters by Penrose are succint, clear, and meaningful. It makes me wonder about hawking - it reminds me of feynman's comment that "if you are not able to express an idea clearly to a layperson, you dont understand it well enough yourself..." (not a direct quote)

i have much more confidence in the works of MTW and Wald.

JesseM
Yes, it is a good idea.

But how an observer can cross HIS horizon - in HIS frame? I can not imagine such spacetime diagram...
What do you mean by "HIS" horizon? The event horizon of a black hole is an objective thing, it isn't frame-dependent...it refers to the boundary of the region where it becomes impossible for light to escape. Of course, there is nothing about this definition that would suggest an infalling observer would notice anything weird or different at the moment they crossed the horizon, the definition just says that if the observer sets off a spherical flash of light before crossing the horizon some of the light will eventually get far away from the black hole, while if the observer sets off a flash inside the horizon all of the light will eventually hit the singularity (at least in the case of a nonrotating black hole, for a rotating black hole some trajectories avoid the singularity even though they never get back to the original region of spacetime they came from outside the horizon). But at the moment the observer sets off the flash it looks just the same to him regardless of whether he's inside the horizon or outside.

By HIS horizon I meant HIS APPARENT horizon,
AH is frame-dependent.

JesseM
By HIS horizon I meant HIS APPARENT horizon,
AH is frame-dependent.
It looks like the definition of the "apparent horizon" of a black hole is given on p. 320 of Hawking and Ellis' The Large Scale Structure of Space-Time, it's the "outer boundary of a connected component of the trapped region", with the "trapped region" defined at the top of the page--when you say this is frame-dependent, do you mean to suggest that the set of events in spacetime that lie on the apparent horizon would depend on your choice of coordinate system?

Also, why are you interested in the apparent horizon rather than the absolute horizon? When physicists discuss the event horizon of a black hole they are more often talking about the latter I think, and I'd bet the quote from the paper in your OP about there being "no event horizon to cross" probably was too unless they said otherwise (or unless someone here can follow enough of the math to know for sure).

By the way, on the subject of what actually happens to observers and light falling into evaporating black holes, this paper shows some different proposed Penrose-Carter diagrams for evaporating black holes that correspond to different theories about this. And on the subject of different definitions of the horizon, here is a review paper comparing and contrasting various definitions (along with a discussion of the conservation of energy problem in GR), and this paper talks about some new ways of defining it that are more local than previous definitions. Finally, here is a good paper on the classical theory of black holes, and here is one on the current state of the astrophysical evidence for them.

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stevebd1
Gold Member
Wiki states that the apparent horizon is inside the absolute horizon (EH) but when looking at the tortoise coordinate (r*) in Eddington-Finkelstein coordinates, I'm almost under the impression that the apparent horizon is outside the EH, synonymous in some way with v=t+r*=0 for ingoing radial null coordinates, sometimes referred to as 'the surface of last influence'-

http://www.sron.nl/~jheise/lectures/kruskal.pdf" [Broken] (page 66)

'..Figure 8.9 Surface of last influence. Spherical gravitational collapse is shown here in ingoing E-F coordinates. For each external particle or external observer there is a moment of the ”birth of the black hole”. The set of such moments form the ”surface of last influence”. Before passing this surface external observers can in principle still shine a flashlight onto the contracting star and receive the bounced light or he can collect a few baryons from the surface. After passing surface of last influence observers cannot interact (matter cannot be influenced) and can consider the object a black hole..'

the above implies there are trapped surfaces between the surface of last influence and the event horizon, possibly giving rise to an apparent horizon.

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JesseM
Wiki states that the apparent horizon is inside the absolute horizon (EH) but when looking at the tortoise coordinate (r*) in Eddington-Finkelstein coordinates, I'm almost under the impression that the apparent horizon is outside the EH, synonymous in some way with v=t+r*=0 for ingoing radial null coordinates, sometimes referred to as 'the surface of last influence'-
Kip Thorne's book Black Holes and Time Warps has a diagram on p. 418 of both the absolute and apparent horizon in the case of a black hole that grows because it swallows some additional matter, he definitely shows the apparent horizon as being inside the absolute horizon during the period when the two horizons disagree. What makes you think the apparent horizon, defined in terms of "trapped surfaces", has anything to do with the concept of the "surface of last influence"? I don't really understand the technical definition of a "trapped surface" in GR so I'm not saying you're definitely wrong, just that you don't really explain your reasoning for equating the two.
stevebd1 said:
http://www.sron.nl/~jheise/lectures/kruskal.pdf" [Broken] (page 66)

'..Figure 8.9 Surface of last influence. Spherical gravitational collapse is shown here in ingoing E-F coordinates. For each external particle or external observer there is a moment of the ”birth of the black hole”. The set of such moments form the ”surface of last influence”. Before passing this surface external observers can in principle still shine a flashlight onto the contracting star and receive the bounced light or he can collect a few baryons from the surface. After passing surface of last influence observers cannot interact (matter cannot be influenced) and can consider the object a black hole..'

the above implies there are trapped surfaces between the surface of last influence and the event horizon, possibly giving rise to an apparent horizon.
Why do you think the quote implies that? The quote doesn't even mention the notion of trapped surfaces...

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stevebd1
Gold Member
If Black Holes and Time warps shows the apparent horizon inside the absolute horizon then that's pretty conclusive but I'll shed some light on why I thought the apparent horizon had anything to do with the surface of last influence.

I initially thought that the quote I provided was in some way describing this effect from wiki-

'..An apparent horizon is a surface defined in general relativity as the boundary between light rays which are directed outwards and moving outwards, and those which are directed outwards but moving inwards..' I now realise that's not the case.

In mass inflation, there is the mass-energy of the radiation influx (δm) which is associated with the inverse power law decrease of late-time fields, (sometimes referred to as Price's power law damping of radiative tails) which are radiated then backscattered into the BH at the potential barrier (http://arxiv.org/PS_cache/gr-qc/pdf/9411/9411050v1.pdf" [Broken], page 2, last para) the potential barrier appearing synonymous with the surface of last influence, giving the impression that this information, while outside the event horizon, is 'trapped'.

Also, the surface of last influence is also synonymous with the ingoing null coordinates (v) becoming negative. On that note, in the case of v=t+r*, what units are used for t? In http://www.phys.ufl.edu/~det/6607/public_html/grNotesMetrics.pdf" [Broken], (page 10) time seems to be counting down from infinity to zero, parallel (if not exactly) with the coordinate radius.

Source where r0 appears to be the surface of last influence-
http://www.damtp.cam.ac.uk/user/sg452/black.pdf [Broken] diagrams page 8-9

r0 as the potential barrier-
http://relativity.livingreviews.org/open?pubNo=lrr-1999-2&key=Chandra83 [Broken] eq 30

Would it at least be reasonable to say that the surface of last influence and the potential barrier are the same thing?

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stevebd1
Gold Member
I now get it that the surface of last influence is a temporary phase that exists just as a star collapses into a black hole, anything on the inside of the surface of last influence as the black hole forms, regardless of its velocity, will be dragged into the black hole, the surface of last influence eventually becoming the event horizon (http://www.sron.nl/~jheise/lectures/kruskal.pdf" [Broken] where BH radial perturbation equations are similar to equations used in QM.

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