Do Black Holes become visible at relativistic speeds?

  • Thread starter tionis
  • Start date

timmdeeg

Gold Member
926
60
This statement: "event horizon of the collapsing star is formed in the infinite future "
is not accurate. Accurate is:

No event at or inside event horizon is ever in the causal past of an eternally external observer.

See the difference? "Not in causal past" is not at all the same as "infinite future". The horizon formation CAN be treated as in the past, present, or future of an external observer (just not the causal past) depending on how they choose to foliate the region of specetime that is neither in their past nor future light cone. Note, in particular, there is a precise moment when the horizon formation is no longer in the future light cone of an external observer. This makes it fair game to be treated as 'now' or even 'past', but not causal past.
So, one has to take the causal structure of spacetime into consideration. Good insight, thanks for clarifying.

WannabeNewton, thanks for your hint.
 

timmdeeg

Gold Member
926
60
When you include quantum effects, so that the black hole emits Hawking radiation and eventually evaporates, it's no longer clear exactly how to even construct standard Schwarzschild coordinates; and no matter how you do, it can't take infinite coordinate time for the hole to form, because at some finite coordinate time, an observer far away will see the hole's final evaporation, and the light signals that carry the image of that evaporation will also carry images of the hole forming.
Yes, that's indeed a very interesting side aspect, thanks for mentioning.
 

timmdeeg

Gold Member
926
60
However, all of this assumes that you are stationary with respect to the black hole--more precisely, it assumes that you are accelerating in order to maintain altitude, not freely falling inward. At least on the standard view of Hawking radiation (by "standard" I mean not including speculations like firewalls--see below), it is only detectable by observers accelerating outward, not by observers freely falling through the hole's horizon. I'm not sure anyone has ever really analyzed the case of an observer accelerating *inward* towards the hole, instead of outward in order to maintain altitude above the hole. My guess is that such an observer would also not detect any Hawking radiation;
It seems widely accepted that the thermal bath experienced locally by a near-horizon observer may be interpreted as Unruh radiation. In my understanding from this point of view no more radiation should be detectable if he stops accelerating as he then is freely falling.
Now you are mentioning the 'accelerating inward' case. I wonder why shouldn't one expect the same Unruh radiation as in the 'accelerating outward' case, provided the local acceleration is the same? But if correct, do we still talk about Hawking radiation then? I can't imagine that. Kindly correct my amateurish reasoning in case ... .
 

PAllen

Science Advisor
7,719
1,074
Back to the original thread topic, Peter and I discussed the visibility of Hawking radiation to an observer rapidly approaching a BH. The traditional view (pre-firewall) was presented, by among others, Verlinde in papers from the 1990s. This is the view Peter mentioned: infaller's see no Hawking radiation. On reviewing my own notes on this, I see that I had not so much forgotten this view as pushed it out of mind as no longer convincing to me. The following paper, especially, led me to doubt its truth. If this and similar ones are right, then there is every expectation that approaching a BH at extremely near c will make it's Hawking radiation increasingly visible:

http://arxiv.org/abs/1101.4382
 
26,555
7,091
It seems widely accepted that the thermal bath experienced locally by a near-horizon observer may be interpreted as Unruh radiation. In my understanding from this point of view no more radiation should be detectable if he stops accelerating as he then is freely falling.
Yes.

Now you are mentioning the 'accelerating inward' case. I wonder why shouldn't one expect the same Unruh radiation as in the 'accelerating outward' case, provided the local acceleration is the same?
Because the derivation of Hawking radiation as equivalent to Unruh radiation depends on the fact that, mathematically, the black hole's event horizon is equivalent (modulo some technicalities that don't affect this discussion) to a Rindler horizon for static observers--that is, observers who are accelerating outward in order to remain at a constant altitude above the horizon. But this equivalence *only* holds for those static observers--it does *not* hold for observers with other accelerations. (Which means it doesn't even hold, strictly speaking, for observers who are accelerating outward either less or more than a static observer.)

In other words, in a black hole spacetime, the "Unruh radiation" associated with observers with a particular acceleration--static observers--has special properties because those observers' worldlines are orbits of the time translation symmetry of the spacetime. In flat Minkowski spacetime, *any* observer with constant acceleration is following an orbit of a time translation symmetry of the spacetime; so all accelerating observers are "equivalent" in Minkowski spacetime in a way that they are *not* in Schwarzschild spacetime.

I emphasize that this is all heuristic; as I said before, it doesn't seem like anyone has actually analyzed the case of an observer accelerating inward toward a black hole in detail (and PAllen appeared to agree, at least until his latest post linking to a paper that analyzes, if not an inward accelerating observer, at least observers free-falling inward, in a way that leads to a different conclusion than my heuristic one :redface:).
 

PAllen

Science Advisor
7,719
1,074
Back to the original thread topic, Peter and I discussed the visibility of Hawking radiation to an observer rapidly approaching a BH. The traditional view (pre-firewall) was presented, by among others, Verlinde in papers from the 1990s. This is the view Peter mentioned: infaller's see no Hawking radiation. On reviewing my own notes on this, I see that I had not so much forgotten this view as pushed it out of mind as no longer convincing to me. The following paper, especially, led me to doubt its truth. If this and similar ones are right, then there is every expectation that approaching a BH at extremely near c will make it's Hawking radiation increasingly visible:

http://arxiv.org/abs/1101.4382
I think a key logical argument from this paper is asking about a static observer starting free fall (rather than the simplest case typically picked: free fall from infinity). Does the Hawking radiation immediately disappear? That seems implausible, and this paper derived that it is not so. So what happens over time for free fall after having been static? This paper argues that the observed Hawking radiation increases to limiting value at horizon crossing.
 

timmdeeg

Gold Member
926
60
Because the derivation of Hawking radiation as equivalent to Unruh radiation depends on the fact that, mathematically, the black hole's event horizon is equivalent (modulo some technicalities that don't affect this discussion) to a Rindler horizon for static observers--that is, observers who are accelerating outward in order to remain at a constant altitude above the horizon. But this equivalence *only* holds for those static observers--it does *not* hold for observers with other accelerations. (Which means it doesn't even hold, strictly speaking, for observers who are accelerating outward either less or more than a static observer.)
Ok, that's a very good explanation, thanks for spending your time.
 
26,555
7,091
Does the Hawking radiation immediately disappear? That seems implausible
But the same thing happens in the case of Unruh radiation; if you shut off your rocket engine and stop accelerating, the Unruh radiation disappears immediately.

I haven't read the paper through yet, so perhaps they address this.
 

WannabeNewton

Science Advisor
5,774
529
That seems implausible...
Why exactly is it implausible? In flat space-time it's just a result of the Bogoliubov transformation for creation/annihilation operators of e.g. a Klein-Gordon field on the Minkowski background that if you have no particles with respect to one vacuum you will in general have particles with respect to another vacuum. The act of shutting off the rockets is the same as instantaneously transforming from one vacuum to another. The difference in particle fluctuations in vacuum is after all what leads to the Planck distribution when the rockets are not off. Why would it be any different in curved space-time?

Like Peter I too have not read the paper yet so hopefully it's addressed in the paper.
 

PAllen

Science Advisor
7,719
1,074
Here is a paper referencing the one I gave earlier, validating its main results with different methodology and assumptions, and extending it to the case of circular orbits, with the interesting conclusion that a detector in a circular orbit around a BH detects more Hawking radiation than the static detector at the same radial position.

http://arxiv.org/abs/1304.2858
 
Last edited:

Haelfix

Science Advisor
1,948
209
To really have a concrete picture as to what happens to Alice when she falls into the blackhole, we really need a fully fleshed out model of the quantum mechanics of the near horizon degrees of freedom. What modes are being excited, how fast is the thermalization and what sort of measuring device are we using (and what local operators are we measuring).

There are some papers that try to treat this very hard question (like the one that is linked), but I believe the consensus is that they are still hopelessly naive without a real model of quantum gravity.

Again, if you claim to see a calculation of a hot horizon (eg Hawking modes, Fuzzballs, bouncing stars, Firewalls etc) the trivial thought experiment is to pick a very massive black hole (so that curvature invariants along the horizon are arbitrarily tiny) and argue based on effective field theory what is known as the adiabiatic principle/no drama hypothesis, which seems to suggest (naively) that a local observer should see departures of the equivalence principle at most up to statements that include functions of these curvature invariants, together with suppression factors that contain powers of the Planck Mass.

It is very hard to save locality and the classical theory of GR in those circumstances (eg we are talking about arbitrarily large modifications of GR at arbitrarily long distances) and although it is not unheard of for effective field theory reasoning to fail, the magnitude of the failure here would be quite unheard off.
 

PAllen

Science Advisor
7,719
1,074
Perhaps they are naive, but so is the confident claim that someone accelerating toward a BH (or even free falling) detects no Hawking radiation.
 

Haelfix

Science Advisor
1,948
209
Don't get me wrong, I am making no claims here. I have no idea what the correct statement is. It does however seem like no matter what physics we choose, that we are forced to give up something cherished, which is why there has been so much work on this subject in the past several years.

I actually believe that the hot horizon camp seem to have the strongest overall intellectual case as it currently stands, even though I suspect that one day it will go away somehow.
 

PAllen

Science Advisor
7,719
1,074
I don't really see the papers I linked as being in the 'firewall camp', or proposing a hot horizon (more like a warm horizon for most in fallers, with a limiting special infaller that sees no Hawking radiation). They analyze within the semi-classical framework in which Hawking radiation is traditionally derived. Whether they correspond to what would be observed in our universe (or what would be predicted by a complete theory of quantum gravity) is obviously not known.

In any case, if these papers are taken as true, then my post #8 remains an accurate summary. Rapidly approaching a BH formed from collapse would increase its visibility due both to a classical effect (blue shift of leaking, nearly trapped, light) and a quantum effect (blue shift of Hawking radiation).

[edit: A further observation is that the papers I referenced are not really discussing anything about distinguishability of the horizon - thus Haelfix general argument does not apply. They discuss phenomena visible at a distance from the BH and the limiting behavior (temperature) on approach to the horizon. For example, observation of Hawking radiation in circular orbits of a BH is certainly not a horizon phenomenon (though, of course, the origin of Hawking radiation is related to the existence of a horizon.]
 
Last edited:

tionis

Gold Member
458
67
In any case, if these papers are taken as true, then my post #8 remains an accurate summary. Rapidly approaching a BH formed from collapse would increase its visibility due both to a classical effect (blue shift of leaking, nearly trapped, light) and a quantum effect (blue shift of Hawking radiation).
True. I was curious about this and someone in academia was kind enough to workout some of the details:

According to the standard picture of black hole thermodynamics, a black hole has a temperature (due to Hawking radiation). When one travels towards a black hole with a speed very close to speed of light, one is traveling with a very large "Lorentz factor" (e.g. 0.999994 c corresponds to Lorentz factor 300; and c corresponds to Lorentz factor infinity). All the light seen by you would be boosted by a factor of order the Lorentz factor. So is the black hole temperature. For a stellar-size black hole, the temperature is extremely low, say 10^{-8} kelvin. Suppose there are no stars, CMB and other luminous objects, yes, when you travel very close to speed of light, say, your Lorentz factor is 5x10^11, then the black hole temperature can be 5000 K -- this is essentially a star. Then you can "see" the black hole. Notice that the required speed is really close to c: 0.9999999999999999999999992 c!
^^That's hot!
 

Haelfix

Science Advisor
1,948
209
[edit: A further observation is that the papers I referenced are not really discussing anything about distinguishability of the horizon - thus Haelfix general argument does not apply. They discuss phenomena visible at a distance from the BH and the limiting behavior (temperature) on approach to the horizon. For example, observation of Hawking radiation in circular orbits of a BH is certainly not a horizon phenomenon (though, of course, the origin of Hawking radiation is related to the existence of a horizon.]
I finally had a chance to read 1101.4382, and you are correct. The paper is really dealing with a semi realistic collapsing geometry, rather than the case of an eternal Schwarschild black hole (so Hawking radiation switches on at some finite proper time). In other words, we are not in the Unruh vacuum.

Now, of course there is going to be a blue shift for noninertial observers, which they compute explicitly (and looks correct to me, as i've seen similar expressions albeit with less generality in eg Wald and other papers).

There is a bit of a funny claim/calculation very close to horizon crossing, and i'm not sure I agree with the paper on that point. The real issue is how far you can trust quantum field theory in curved spacetime past a certain limit. Here we are dealing with modes that diverge as something like e^k(u)T, so a detector with a finite response frequency v will have to resolve trans Planckian modes for T > 1/K(u). The authors are evidently aware of this point, as its related to their discussion on the validity of the adiabatic limit.

But anyway, I don't think this paper really challenges anything about the standard lore (except perhaps for worldlines that are very/very close to the horizon, where there is a claimed jump in the Unruh temperature) but this doesn't really run against the equivalence principle in any way.

edit: Thinking about this a bit more, i'm convinced the paper is correct, however a near horizon observer is not really measuring what I would call the Hawking effect perse. Instead he/she is just measuring a perfectly classical effect of a time varying gravitational field, NOT the subtle features of the quantum vacuum that traditionally gives rise to the Hawking effect. Analysis of that point, really requires putting in a detector (along with the deep subleties about quantum measurement in curved space) and carefully subtracting off the classical effects.
 
Last edited:

tionis

Gold Member
458
67
however a near horizon observer is not really measuring what I would call the Hawking effect perse. Instead he/she is just measuring a perfectly classical effect of a time varying gravitational field, NOT the subtle features of the quantum vacuum that traditionally gives rise to the Hawking effect. Analysis of that point, really requires putting in a detector (along with the deep subleties about quantum measurement in curved space) and carefully subtracting off the classical effects.
From the author...

Dear Haelfix: The vacuum of the quantum field gets altered due to the presence of horizon structure in the spacetime. It is like changing the boundary conditions. The observer who is falling in sees two effects one due to time varying gravitational field as he falls and the spacetime has horizon structure. Both of these alter the quantum vacuum and hence give rise to radiation. The effects cannot be resolved as such but one can show for example that the temperature seen by a freely falling observer from far off when he is near the horizon is 4 times the usual hawking temperature seen by the observer who is at rest very far away from the black hole.
 

Related Threads for: Do Black Holes become visible at relativistic speeds?

Replies
34
Views
5K
Replies
18
Views
12K
Replies
2
Views
3K
Replies
4
Views
921
Replies
20
Views
2K
Replies
2
Views
2K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top