Black hole horizon and Hawking evaporation

In summary: Hawking radiation which evaporates the collapsing star in a finite time.This is not correct. If the horizon was eternal then the radiation would never be able to escape. The radiation is a consequence of the Hawking radiation, not the other way around. Thank you for your input.In summary, the distant observer does not agree with the opinion that the horizon forms in the future light cone of an external observer. Hawking radiation occurs at the horizon for a falling observer, but not for a distant observer.
  • #1
Marilyn67
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TL;DR Summary
Apparent contradiction between the formation of a horizon and the evaporation of black holes
Hello,

I take the example of two observers :

- A distant observer
- A falling observer

For the distant observer, the formation of the horizon is not part of his future cone of light, we agree.

For the falling observer, the consensus says it is crossing the horizon.

First question: the evaporation of the black hole occurs in a very long time, but finished for the distant observer.
How can the falling observer cross a horizon that has not happened, since the black hole has already evaporated? (for him the "film" is simply "accelerated").

Second question: The consensus is that the Hawking radiation occurs at the horizon. How can a distant observer perceive radiation coming from a horizon that does not yet exist (as if the radiation came from an event that occurs in the future?)

In advance, thank you for your answers.

see you soon

Marilyn
 
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  • #2
Marilyn67 said:
For the distant observer, the formation of the horizon is not part of his future cone of light, we agree.
I don’t agree. In the Oppenheimer Snyder metric there certainly are observers for whom the formation of the horizon is in their future light cone. This set of observers includes both observers that eventually cross and those that do not cross the horizon.

Personally, I do have issues with Hawking radiation. I do not know of a valid metric for an evaporating black hole. The outgoing Vadiya metric is the closest, but it is a white hole not a black hole. So I am unsure what metric to use to actually calculate the correct answer to questions like this.
 
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  • #3
Marilyn67 said:
For the distant observer, the formation of the horizon is not part of his future cone of light, we agree.
Horizon formation is never in an external observer's past light cone. It can be in the future light cone.
Marilyn67 said:
How can the falling observer cross a horizon that has not happened, since the black hole has already evaporated?
I don't think an evaporating black hole has an event horizon, strictly speaking. It has an apparent horizon, something that's indistinguishable from an event horizon on timescales much less than the evaporation time of the black hole, but not an actual event horizon.
Marilyn67 said:
The consensus is that the Hawking radiation occurs at the horizon.
I don't think that's correct. Hawking radiation comes from the region just above the horizon.
 
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  • #4
Hello Ibix,

Thank you for your quick and concise response.
I think we agree on all but one point.
You say :

Ibix said:
Horizon formation is never in an external observer's past light cone. It can be in the future light cone.
When I describe a distant observer, I mean a distant observer who stays that way (ad infinitum).

In this case, isn't talking about a past light cone or a future light cone the same thing?
The point is, for a distant observer (and who stays that way), the horizon of the black hole never forms, right ?

From what I understand, the (real) horizon never forms for anyone (distant or falling observer).

The model of General Relativity (with horizon) becomes incompatible with the Hawking radiation which evaporates the collapsing star in a finite time.

We agree ? If not, where is the error ?

Thank you very much,
See you soon,

Marilyn,
 
  • #5
Marilyn67 said:
In this case, isn't talking about a past light cone or a future light cone the same thing?
The point is, for a distant observer (and who stays that way), the horizon of the black hole never forms, right ?
No. The formation of the event horizon is a specific event in the manifold. That event has a past light cone, and that past light cone goes out to infinity. For every event inside the past light cone, including those on the distant observers worldline, the formation of the event horizon is in the future light cone.

Marilyn67 said:
From what I understand, the (real) horizon never forms for anyone (distant or falling observer).
This is not correct. Perhaps you are thinking of the Schwarzschild spacetime instead of the Oppenheimer Snyder spacetime. In the Schwarzschild spacetime the horizon is eternal and static so there is no formation. So if you are talking about the formation of an event horizon then you cannot use a spacetime where it always existed.
 
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  • #6
Marilyn67 said:
When I describe a distant observer, I mean a distant observer who stays that way (ad infinitum).

In this case, isn't talking about a past light cone or a future light cone the same thing?
For an eternal black hole like the Schwarzschild solution the direction of time doesn't make much difference, no. But you were talking about the formation and evaporation of a black hole. This is a different model, and there is clearly a unique sense of future and past light cones, because there's a star there before the black hole. The horizon formation event is part of the final collapse of the star, which can be in my causal future but not in my causal past until the hole evaporates.
Marilyn67 said:
The model of General Relativity (with horizon) becomes incompatible with the Hawking radiation which evaporates the collapsing star in a finite time.
A model of an eternal black hole is incompatible with a model of a black hole that forms and evaporates, certainly. But GR can certainly describe the former (for example the Oppenheimer-Snyder collapse model mentioned by @Dale), and can describe the latter with a semi-quantum extension without which you can't describe Hawking radiation at all. So I think it's wrong to say that GR is incompatible with finite-lifetime black holes. GR is more than just the Schwarzschild black hole.
 
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  • #7
Dale said:
Personally, I do have issues with Hawking radiation. I do not know of a valid metric for an evaporating black hole. The outgoing Vadiya metric is the closest, but it is a white hole not a black hole. So I am unsure what metric to use to actually calculate the correct answer to questions like this.

Ibix said:
I don't think an evaporating black hole has an event horizon, strictly speaking. It has an apparent horizon, something that's indistinguishable from an event horizon on timescales much less than the evaporation time of the black hole, but not an actual event horizon.

I must react. I am really happy to read such a reply, which I definitely agree. A few years ago standard answer for Marylins question was usually strong rejection with some mantra about obvious difference between freefalling and Schwarzschild observer (which by the way says nothing to this problem).
The question is pretty common and seems naive (and as naive is often answered) , but it has some logic.
Of course this naiv arguing breaks the (probably) fact that Hawking radiation do not occurs exactly at the horizon, but when you go deeper to problem of evaporating BH it seems that you really need different solution (metric). This is my personal belief. I will be not so surprised if finally this simple and first view approach has correct result, of course the result will have much more complex and deeper reasons.
 
  • #8
Marilyn67 said:
Summary:: Apparent contradiction between the formation of a horizon and the evaporation of black holes

See this Insights article:

https://www.physicsforums.com/insights/black-holes-really-exist/

It discusses the scenario you describe and why there is no actual contradiction.

Marilyn67 said:
For the distant observer, the formation of the horizon is not part of his future cone of light, we agree.

No, we don't. The horizon is in the future light cone of a distant observer.

In a spacetime with a black hole that does not evaporate, the horizon is not in the past light cone of distant observers, so they cannot see the horizon, but that does not prevent observers from falling into it.

In a spacetime with a black hole that does evaporate, the horizon (but not the region inside it) eventually does enter the past light cone of distant observers; but there is a very long period of time before that happens when it is still perfectly possible for those distant observers to fall into the hole.
 
  • #9
Dale said:
I do not know of a valid metric for an evaporating black hole. The outgoing Vadiya metric is the closest, but it is a white hole not a black hole.

I think a spacetime that includes a region similar to the Oppenheimer-Snyder collapse, joined to its future with an outgoing Vaidya region, in turn joined to its future with a Minkowski region inside the last "shell" of outgoing radiation, would work. The fact that the outgoing Vaidya region in this heuristic model is only a region, not the entire spacetime, avoids the white hole issue you refer to.
 
  • #10
Ibix said:
I don't think an evaporating black hole has an event horizon, strictly speaking.

There are models of evaporating black holes that do have one; the simplest is the one Hawking originally used. But it's still an open question whether those models actually describe things in our real universe.
 
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  • #11
PeterDonis said:
The fact that the outgoing Vaidya region in this heuristic model is only a region, not the entire spacetime, avoids the white hole issue you refer to.
I am not convinced that it does, but I have never seen a formal derivation one way or another. I can’t see how the “stitching” you describe changes the fact that the outgoing Vaidya metric is a white hole. It produces an event horizon which allows matter to cross inward during the OS portion and then suddenly does not allow matter to cross inward during the outgoing V portion. I can’t visualize how that could work (which is of course not a proof, but just a limitation of mine)

I have wondered if the ingoing V metric would work using a negative radiation rate but this is beyond my capability to evaluate.
 
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  • #12
PeterDonis said:
There are models of evaporating black holes that do have one; the simplest is the one Hawking originally used. But it's still an open question whether those models actually describe things in our real universe.
Ok - I misunderstood something. Having tracked down a Penrose diagram I see you are correct. There's a part of spacetime where you must hit the singularity before evaporation happens. So whether there's actually an event horizon or not depends on what actually happens where GR says there's a singularity?
 
  • #13
Dale said:
I can’t see how the “stitching” you describe changes the fact that the outgoing Vaidya metric is a white hole.

Because you're only using a portion of the full spacetime described by the outgoing Vaidya metric. The portion being used does not include the white hole region of the outgoing Vaidya metric. It's no different from the fact that the Oppenheimer-Snyder model you refer to does not have a white hole issue, even though the maximally extended Schwarzschild spacetime includes a white hole, because the portion of maximally extended Schwarzschild spacetime used in the Oppenheimer-Snyder model does not include the white hole region.

Dale said:
It produces an event horizon which allows matter to cross inward during the OS portion and then suddenly does not allow matter to cross inward during the outgoing V portion.

See Figure 2 of this paper:

https://www.groundai.com/project/a-linear-mass-vaidya-metric-at-the-end-of-black-hole-evaporation/1

This is the kind of model I was describing. The region marked "III" is the outgoing Vaidya region. Note that that region is only a portion of the full outgoing Vaidya spacetime, which is shown in Figure 1 (the region marked "I" in that figure, if I'm understanding it correctly, is the white hole region).
 
  • #14
PeterDonis said:
The portion being used does not include the white hole region of the outgoing Vaidya metric.
I think you may be misunderstanding the outgoing V spacetime (or I am). This has noting to do with the maximally extended Schwarzschild spacetime where there are two horizons. To my knowledge the outgoing V spacetime has only one horizon and it is a white hole horizon.

As you say, this is similar to the OS spacetime which has only one horizon and it is a black hole horizon.

I need to read that paper.
 
  • #15
Dale said:
To my knowledge the outgoing V spacetime has only one horizon and it is a white hole horizon.

I know. But that doesn't mean you can't put a portion of the outgoing Vaidya spacetime, one that doesn't include the white hole region, into a model that includes portions of other spacetimes as well. The model I described, and which is shown in Figure 2 of the paper I referenced, does that: it uses only the portion of outgoing Vaidya spacetime that is outside the white hole horizon of that spacetime, and joins it to a portion of Schwarzschild spacetime that includes the Schwarzschild exterior region and the Schwarzschild black hole region. The black hole horizon in that model is in the Schwarzschild region, not the outgoing Vaidya region.
 
  • #16
PeterDonis said:
it uses only the portion of outgoing Vaidya spacetime that is outside the white hole horizon of that spacetime
Yes, I see that in the paper, but that then does not represent exactly the portion of primary interest: the horizon during evaporation.

In this paper it looks like the V spacetime is used to transition from Schwarzschild to Minkowski, but not for the event horizon except possibly right at the end in Figure 6.

I think the portion of primary interest is what they describe as “adiabatically evaporating Schwarzschild” which I have never seen before.
 
  • #17
Dale said:
In this paper it looks like the V spacetime is used to transition from Schwarzschild to Minkowski, but not for the event horizon except possibly right at the end in Figure 6.

Yes, that's my understanding.

Dale said:
I think the portion of primary interest is what they describe as “adiabatically evaporating Schwarzschild” which I have never seen before.

I'm not entirely sure what that means. It might be that that portion is also technically an outgoing Vaidya region, just with a different mass function.

I think part of the problem might be that the model in that figure is leaving out the collapse part--the Oppenheimer-Snyder type region where the black hole is initially formed. In the figure it looks like the Stretched Horizon (SH) line goes all the way back to past timelike infinity, which obviously can't be the case for an actual collapsing object.
 
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  • #18
PeterDonis said:
I'm not entirely sure what that means. It might be that that portion is also technically an outgoing Vaidya region, just with a different mass function.
Me neither, and a cursory reading of the text didn’t provide details. I will need to read it more in depth to see if there are additional hints
 
  • #19
Thanks @PeterDonis for your response, as well as for the very extensive article link you wrote.
I understood that my question is simplistic and that intuition, the most "rational" explanation is not necessarily the right one.
Your knowledge is superior to my knowledge, and it's completely beyond me.
I can't get a clear picture of the chronology of events.
I must reread your article several times.
I understood that the answer is not yet clear and that it depends on the models.

Cordially
Marilyn
 
  • #20
Marilyn67 said:
I understood that the answer is not yet clear

The answer to whether black holes actually evaporate in our actual universe (and even whether the objects we call "black holes" are actually black holes in the sense of having true event horizons) is not yet clear, yes.

But, as I said in the article, if it turns out that our current ideas about black holes and black hole evaporation need modification, it won't be because there is any inherent contradiction between black hole formation and black hole evaporation. We have consistent models that contain both; the open question is whether those models are actually realized in our universe.
 
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  • #21
PeterDonis said:
it won't be because there is any inherent contradiction between black hole formation and black hole evaporation. We have consistent models that contain both; the open question is whether those models are actually realized in our universe.

Yes, I understood the meaning of your article correctly.
It's these consistent models that I don't yet understand very well.

Thanks also for your second link !

I have to read it all with a clear head !
Cordially
Marilyn
 
  • #22
Dale said:
Personally, I do have issues with Hawking radiation. I do not know of a valid metric for an evaporating black hole. The outgoing Vadiya metric is the closest, but it is a white hole not a black hole. So I am unsure what metric to use to actually calculate the correct answer to questions like this.
Strictly speaking you are saying that there isn't a known explicit solution. Not that there are no solutions. Most solutions are not possible to write in closed form, but they do exist.
 
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  • #23
Hello @PeterDonis,

I just found a link to an article published by Stephen Hawking in the journal "Nature" in 2014.
His conclusions are the same as yours :

https://www.sciencesetavenir.fr/fondamental/stephen-hawking-les-trous-noirs-n-existent-pas_23069

D'une manière générale, précise Hawking, on ne peut rendre compte parfaitement de ces astres curieux tant que nous n'avons pas élaboré une théorie de la gravité unifiée. C'est à dire capable de concilier les lois de la physique quantique des échelles subatomiques avec la relativité générale qui rend compte de l'astronomie…

It's in French (sorry, I am of French-speaking origin) but if we translate :

Generally speaking, Hawking specifies, we cannot fully account for these curious stars until we have developed a unified theory of gravity. That is to say capable of reconciling the laws of quantum physics of subatomic scales with general relativity which accounts for astronomy…

Consistent models are a real problem for me !

As a distant observer, and who stays that way indefinitely, I don't "see" the horizon forming before the black hole evaporates, it's impossible.

In my opinion, these "models" are highly speculated:

S.H himself says:

En effet, la description des trous noirs par la physique classique, telle qu'on l'enseigne aux cours des premières années universitaires aux étudiants, n'est pas exacte.

If we translate :

Indeed, the description of black holes by classical physics, as it is taught during the first years of university to students, is not exact.

Cordially
Marilyn
 
  • #24
martinbn said:
Strictly speaking you are saying that there isn't a known explicit solution. Not that there are no solutions. Most solutions are not possible to write in closed form, but they do exist.
If you have a published numerical solution I would certainly accept that also. I don’t know anywhere that has such a solution either. Do you?
 
  • #25
Marilyn67 said:
As a distant observer, and who stays that way indefinitely, I don't "see" the horizon forming before the black hole evaporates, it's impossible.
I don’t believe that claim can be asserted either. Without a valid metric you cannot assert this.

With the OS metric the formation of the EH is a specific event. I am skeptical that feature would change with an evaporating model. And without such a model you have no basis to make this claim one way or the other
 
  • #26
Marilyn67 said:
In my opinion, these "models" are highly speculated

Hawking is not saying what you think he is saying. He is not saying the models that are actually used in black hole physics, for example to study quasars, or to model the black hole at the center of our galaxy, or to model black hole mergers in order to analyze data from LIGO, are speculative or inexact.

He is only saying that the highly idealized and simplified models that are taught in introductory university courses in GR are not exact (he is not saying they are speculative, just that they are highly idealized and simplified and so are not exact).

Big difference.
 
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  • #27
PeterDonis said:
He is only saying that the highly idealized and simplified models that are taught in introductory university courses in GR are not exact (he is not saying they are speculative
And that is true of nearly everything taught in introductory courses. So it is not a big criticism at all
 
  • #28
Hello,

PeterDonis said:
He is only saying that the highly idealized and simplified models that are taught in introductory university courses in GR are not exact (he is not saying they are speculative, just that they are highly idealized and simplified and so are not exact).
Yes, I know, I was the one who said that these idealized models were speculative. The term is abusive.
I don't understand these models, and I don't have your level, I recognize it.
I have read and re-read your article.

My problem with these models is that I don't understand, as a distant observer (and who stays that way indefinitely), how we can witness the formation of a real horizon, with an infinite redshift (we observes a star undergoing asymptotic collapse) before a finite time corresponding to the evaporation of the black hole.

How can a model remove this contradiction ? Where is the error in this current description?
Dale said:
And that is true of nearly everything taught in introductory courses. So it is not a big criticism at all
Yes I agree
 
  • #29
Marilyn67 said:
My problem with these models is that I don't understand, as a distant observer (and who stays that way indefinitely), how we can witness the formation of a real horizon

In a spacetime with an evaporating black hole, the distant observer sees the formation of the horizon at the same time as he sees the final evaporation of the black hole. That finite time (which will be very, very long--something like ##10^{67}## years for a black hole of roughly the mass of the Sun, and the time increases as the cube of the mass) takes the place of the "infinite time" for the distant observer in the non-evaporating black hole model.
 
  • #30
Hello @PeterDonis,

PeterDonis said:
In a spacetime with an evaporating black hole, the distant observer sees the formation of the horizon at the same time as he sees the final evaporation of the black hole. That finite time (which will be very, very long--something like ##10^{67}## years for a black hole of roughly the mass of the Sun, and the time increases as the cube of the mass) takes the place of the "infinite time" for the distant observer in the non-evaporating black hole model.
I thank you for your message.
This question has kept me awake for years and you are the first to give me a relevant answer! Thank you !

Most of the speakers with a sufficient culture on the French forums bypass the question and never answer it.
You are the first to give me a relevant answer.
Thank you very much.

You have both the level of study and the pedagogy to explain what escapes us.

I will finally be able to better understand these models with this contradiction that you have just resolved and which seems to me to be completely consistent.

Thank you very much,

Marilyn
 
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  • #31
Marilyn67 said:
Thank you very much

You're welcome! Glad I could help.
 

1. What is a black hole horizon?

A black hole horizon is the boundary surrounding a black hole where the escape velocity exceeds the speed of light. This means that anything, including light, that crosses the horizon cannot escape the gravitational pull of the black hole.

2. How does a black hole's horizon form?

A black hole's horizon forms when a massive star runs out of fuel and collapses under its own gravity. As the star collapses, its mass becomes concentrated in a small area, creating a singularity at the center. The horizon forms around this singularity, marking the point of no return for anything that falls into the black hole.

3. What is Hawking evaporation?

Hawking evaporation is a process proposed by physicist Stephen Hawking in which black holes emit radiation and gradually lose mass. This occurs due to quantum effects near the black hole's horizon, where particles and antiparticles are constantly being created and annihilated. If one of these particles is created just outside the horizon, it can escape and carry away some of the black hole's energy, causing it to shrink over time.

4. Can a black hole's horizon ever disappear?

According to current theories, a black hole's horizon cannot disappear. As the black hole loses mass through Hawking evaporation, its horizon will shrink, but it will never completely disappear. However, the black hole can eventually become so small that its horizon is no longer detectable.

5. How does Hawking evaporation affect the lifespan of a black hole?

Hawking evaporation causes black holes to gradually lose mass and shrink over time. The rate of evaporation increases as the black hole's mass decreases, so smaller black holes evaporate faster. This means that smaller black holes have a shorter lifespan compared to larger ones. A black hole with the mass of our sun would take about 10^67 years to evaporate, while a black hole with the mass of a mountain would evaporate in less than a second.

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