# Thread: Black hole evaporation depending on the observer?

1. Hi everyone,

From the discussion we had regarding the Unruh effect, we discussed briefly the fact that an in-falling observer would not see the Hawking radiation ( identical apparently to an inertial observer who wouldn't see the Unruh radiation, not completely clear... ).

Anyways the question is. For the infalling observer, the black hole would not evaporate so would not lose mass.

Let's consider 2 observers. The observer A is stationnary relative to the black hole, so is accelerating to keep its position. The observer B is extremely far from the black hole and free falling towards it.

When observer B passes A during his fall, B would still see the black hole with the original mass, but A would see it with less mass due to the evaporation that happened during the time that it took for B to reach A.

Isn't there a problem with that?

Nic.

2.

3. and why the black hole would not lose mass?

4. Isn't there a problem with that?
Yes, there is indeed a problem with that
The problem arises quite simple from using the wrong solution to the Einstein equations. Most of the time when people use the term "black hole" they in fact mean Schwarzschild black hole, as described by the Schwarzschild metric, which is a vacuum solution to the field equations. This means we are talking about a static and stationary black hole in otherwise completely empty space, where the mass M is constant and hence the same for all observers. However, for the purpose of discussion Hawking radiation we do in fact refer to black holes which, for a stationary observer somewhere far away, radiate a thermal bath of particles; and a space-time filled with radiation of any kind is most certainly not empty, so we cannot use the Schwarzschild solution to describe such objects. The solution that models radiating black holes is called the Vaidya-Bonnet metric, and the geometry of that space-time is a fair bit more complicated than Schwarzschild geometry, so I will not attempt to give a complete description here. The part that is relevant for us is that in the Vaidya-Bonnet metric, the total mass of the black hole becomes a function of time, and not just a constant; and since a stationary far-away observer and a freely falling observer do not share the same concept of time, they will also disagree as to the mass of the black hole at any given instant ( this is analogues to the difference in coordinate time and proper time in Schwarzschild coordinates ). In Schwarzschild space-times that would be a problem as you correctly pointed out since the mass M is supposed to be a constant, but the same is not true in Vaidya-Bonnet space-time, so the apparent paradox immediately vanishes.

Black holes are very counter-intuitive things, and most especially so once you get quantum field theories involved as well !

5. Originally Posted by Markus Hanke
Isn't there a problem with that?
Yes, there is indeed a problem with that
The problem arises quite simple from using the wrong solution to the Einstein equations. Most of the time when people use the term "black hole" they in fact mean Schwarzschild black hole, as described by the Schwarzschild metric, which is a vacuum solution to the field equations. This means we are talking about a static and stationary black hole in otherwise completely empty space, where the mass M is constant and hence the same for all observers. However, for the purpose of discussion Hawking radiation we do in fact refer to black holes which, for a stationary observer somewhere far away, radiate a thermal bath of particles; and a space-time filled with radiation of any kind is most certainly not empty, so we cannot use the Schwarzschild solution to describe such objects. The solution that models radiating black holes is called the Vaidya-Bonnet metric, and the geometry of that space-time is a fair bit more complicated than Schwarzschild geometry, so I will not attempt to give a complete description here. The part that is relevant for us is that in the Vaidya-Bonnet metric, the total mass of the black hole becomes a function of time, and not just a constant; and since a stationary far-away observer and a freely falling observer do not share the same concept of time, they will also disagree as to the mass of the black hole at any given instant ( this is analogues to the difference in coordinate time and proper time in Schwarzschild coordinates ). In Schwarzschild space-times that would be a problem as you correctly pointed out since the mass M is supposed to be a constant, but the same is not true in Vaidya-Bonnet space-time, so the apparent paradox immediately vanishes.

Black holes are very counter-intuitive things, and most especially so once you get quantum field theories involved as well !
Ok, they wouldn't see the mass evolve the same way over time, but is it true that an observer in free fall towards the black hole would not see any Hawking radiation at all? ( so would not see it lose mass at all )

Physicist mentionned that in this post although he was not sure.

Another question would be, what would an observer accelerating towards a black hole with the thrust of his propulsion, see? If an observer accelerating away ( fixed relative to the black hole ) sees the Hawking radiation, and if the free falling observer doesn't see the Hawking radiation, an observer accelerating towards it should see the contrary, that it, it should see it get larger and absorb some radiation. That doesn't seem to make sense...

6. Originally Posted by Quantum immortal
and why the black hole would not lose mass?
That is what was mentionned in another thread. That's what I would like to clarify.

7. Originally Posted by Nic321
but is it true that an observer in free fall towards the black hole would not see any Hawking radiation at all? ( so would not see it lose mass at all )
To answer this authoritatively I would first have to acquire a thorough understanding of quantum field theories in curved space-times, which I haven't done yet. My personal opinion based purely on a phenomenological understanding of Vaidya-Bonnet space-time without taking QFTs into account would be that the freely falling observer should not see any Hawking radiation, but he will disagree with the external stationary observer on where the event horizon is located at each instance in time. The other subtle issue is that aforementioned external observer will also notice the presence of an apparent horizon in addition to an event horizon, which the freely falling observer does not see. Furthermore, the two observers do not share the same vacuum ground state in their respective frames ( in terms of QFTs ), so they disagree on the energy content of the vacuum they are in.

This is all quite complicated, and, having done a very quick survey of the relevant literature, it would appear that even the experts have not reached a full consensus as to all the details of this scenario. The only way to really answer what happens in the free-fall frame would be to perform an explicit calculation that accounts for all relevant quantum effects at the horizon, which I don't think is even possible based on current theories. There are too many question marks as to the quantum structure of space-time at the horizon, in the absence of a full model of quantum gravity.

an observer accelerating towards it should see the contrary, that it, it should see it get larger and absorb some radiation.
I don't really think so, though I cannot be completely certain, not having studied this in detail. Unruh radiation is a function of the magnitude of proper acceleration, so it should be the same whether you accelerate towards the black hole or away from the black hole. It simply manifests as a thermal bath in the vacuum within the accelerating frame.

8. Ok, ignore what I said before - I have done some further research into this, and it turns out that I was wrong. In fact, Hawking radiation does exist even for a freely falling observer, and the source of Hawking radiation isn't the event horizon at all, but rather the past ( or "illusory" ) horizon, which is the redshifted surface of collapsing matter that originally formed the black hole. That past horizon always remains below the in-falling observer ( unlike the event horizon ), and is only met at the central singularity. For both far-away as well as in-falling observer the temperature ( i.e. radiation flux ) is a function of proper time, and hence they don't agree on the black hole mass at a given instant, but they do agree on the fact that it evaporates in a finite time.

Here is the source : http://ae100prg.mff.cuni.cz/pdf_proc...s/Hamilton.pdf

The upshot is that Unruh radiation and Hawking radiation are physically distinct effects, and not the same thing at all. This kind of makes sense, because unlike the Rindler horizon in an accelerated frame, the past horizon of a black hole doesn't vanish simply by putting yourself into free fall, it is still always below you, so of course there must still be radiation present. Personally I always thought that they are just two names for the same effect, but that is not the case. Learned something new here

As the paper above states, there seems to be much confusion in the literature about this; I have come across numerous sources that explicitly state the freely falling observers do not see Hawking radiation. This cannot be right though because of the above mentioned very simple reason; a far more detailed mathematical treatment of this is found here :

http://edoc.ub.uni-muenchen.de/6024/1/Deeg_Dorothea.pdf

9. Originally Posted by Markus Hanke
Ok, ignore what I said before - I have done some further research into this, and it turns out that I was wrong. In fact, Hawking radiation does exist even for a freely falling observer, and the source of Hawking radiation isn't the event horizon at all, but rather the past ( or "illusory" ) horizon, which is the redshifted surface of collapsing matter that originally formed the black hole. That past horizon always remains below the in-falling observer ( unlike the event horizon ), and is only met at the central singularity. For both far-away as well as in-falling observer the temperature ( i.e. radiation flux ) is a function of proper time, and hence they don't agree on the black hole mass at a given instant, but they do agree on the fact that it evaporates in a finite time.

Here is the source : http://ae100prg.mff.cuni.cz/pdf_proc...s/Hamilton.pdf

The upshot is that Unruh radiation and Hawking radiation are physically distinct effects, and not the same thing at all. This kind of makes sense, because unlike the Rindler horizon in an accelerated frame, the past horizon of a black hole doesn't vanish simply by putting yourself into free fall, it is still always below you, so of course there must still be radiation present. Personally I always thought that they are just two names for the same effect, but that is not the case. Learned something new here
That article is very interesting but is it an accepted theory about Hawking radiation? It sounds more like a new theory from reading it.

Concerning the fact that the Unruh and Hawking radiations are physically different effects, I am not sure that the paper says that. It seems to me that it says the contrary in fact. Look at chapter 3:

At its most fundamental level, Hawking [6] or Unruh [7, 8] radiation arises when an observer watches an emitter that is accelerating relative to the observer. When waves that are pure negative frequency (positive energy) in the emitter’s frame are propagatedtotheobserver,theaccelerationcausesthewa vestoappeartobeamixof negativeandpositivefrequenciesintheobserver’sframe .Inparticular,theemitter’s vacuum (“in” vacuum) is not the same as the observer’s vacuum (“out” vacuum). A classic calculation (e.g. [9, 10]) shows that if the acceleration is approximately constantoverseveralaccelerationtimescales,thentheo bserverwillseetheemitter’s vacuum as a thermal state with temperature proportional to the acceleration. An observer watching a black hole sees Hawking radiation because matter that collapsed to the black hole long ago appears classically frozen at the illusory hori- zon, apparently accelerating away from the observer, redshifting and dimming into the indeﬁnite future. When an infaller free-falls through the true horizon, they do
If I understand correctly, it is purely an observational effect due to the acceleration. The acceleration does in sort that the observer doesn't see the same frequencies coming from the vacuum/matter of the accelerating object.

In the case of the Hawking radiation, the observer sees a radiation because the matter is "apparently accelerating away from the observer".

But what I don't understand is the relationship between the "matter" that is observed and the "vaccuum" of that matter. I believe that the Hawking radiation does depend on the mass ( the less massive a black hole is the faster it evaporates ), but in the case of Unruh, there doesn't seem to be a mass involved, it is only a vacuum??

As the paper above states, there seems to be much confusion in the literature about this; I have come across numerous sources that explicitly state the freely falling observers do not see Hawking radiation. This cannot be right though because of the above mentioned very simple reason; a far more detailed mathematical treatment of this is found here :

http://edoc.ub.uni-muenchen.de/6024/1/Deeg_Dorothea.pdf
I have concentrated on the first pdf, I am going to check this one as soon as I can.

Another thing apart from this, isn't there a sort of "anti-Hawking radiation", which is the radiation of the anti-particles with negative energy that go behind the horizon. Wouldn't those have to be taken into account? It doesn't come from the illusory horizon of the infalling observer, but it should still be able to reach him.

Finally, I am not sure but the illusory horizon sounds like a deformed cosmological horizon due to the acceleration between the matter that fell into the black hole and the infalling observer.

It is complicated but interesting.

10. Ok, I read the other document. Ouch... I understood very little , but the conclusion seems to suggest that indeed the infalling observer will see the Hawking radiation both in Schwarzschild space-time and in Vaidya space-time.

Frankly I don't really see the link between what is discussed in the first paper regarding the illusory horizon and the second paper.

Well, that's kind of tough...

11. Originally Posted by Nic321
Concerning the fact that the Unruh and Hawking radiations are physically different effects, I am not sure that the paper says that.
They have to be physically distinct effects, because Unruh radiation is a function of proper acceleration, which, for a freely falling observer, is exactly zero. Nonetheless, a freely falling observer sees Hawking radiation, so the two are not the same. If you compare a stationary observer in a gravitational field to a uniformly accelerating observer with an equivalent proper acceleration, you will find that Hawking temperature ( due to gravity ) is hotter as compared to Unruh temperature ( due to uniform acceleration ) - the equivalence principle is hence violated :

http://arxiv.org/pdf/1102.5564v2.pdf

However, the two approach asymptotically at the event horizon, where the temperatures would be the same.

the conclusion seems to suggest that indeed the infalling observer will see the Hawking radiation both in Schwarzschild space-time and in Vaidya space-time
Yup

Frankly I don't really see the link between what is discussed in the first paper regarding the illusory horizon and the second paper.
The link is the conclusion that freely falling observers do indeed see Hawking radiation. The first paper gives a conceptual explanation, whereas the second one performs the explicit calculation.

12. It is beginning to make a bit more sense. Your last article is a bit easier to understand for me.

It is interesting that the equivalence principle is violated far from the horizon, but wouldn't that suggest that GR has to be modified in low gravitational field? In particular when objects of similar masses are attracting each other, because the termal bath that each object is in would add to its energy, so this extra energy would have to be taken into account in the calculating the curvature of space-time, etc? On another hand, in the case where a light object falls into a heavy black hole the effect would be insignificant I guess.

Do the ideas of firewall at the horizon have something to do with this phenomenon? They also relate to a problem of combining GR with QM as far as I know.

I understand a bit better the issue of different vacua in different space-times now, but how do they physically configure the detector to detect particles from each different vacuum? That's what they try to do, isn't it?

13. Originally Posted by Nic321
but wouldn't that suggest that GR has to be modified in low gravitational field?
Both Hawking radiation and Unruh radiation are quantum effects, whereas GR is a purely classical theory of gravity. Marrying the two will most definitely mean that some of the classic principles - such as the equivalence principle - will need to be abandoned or generalised.

In particular when objects of similar masses are attracting each other, because the termal bath that each object is in would add to its energy, so this extra energy would have to be taken into account in the calculating the curvature of space-time, etc?
Yes, Vaidya geometry ( but not Schwarzschild geometry ) accounts for the radiation flux.

Do the ideas of firewall at the horizon have something to do with this phenomenon?
Yes, it would have something to do with it, but I haven't studied the "firewall" idea so I can't really comment much.

how do they physically configure the detector to detect particles from each different vacuum?
I'm not really sure what you mean by that.

14. I'm not really sure what you mean by that.
They say that the detector will detect a radiation for a given space-time with respect to a given vacuum. For instance the detector will not detect a radiation in Rindler s-t with respect to Rindler vacuum, but will detect a radiation in Rindler s-t with respect to Minkowski vacuum.

Also, they say with more detail:
It is important to note that G+ is deﬁned with respect to some vacuum state |0> and picking a diﬀerent vacuum state can lead to diﬀerent F(E)’s. Some vacuum choices can lead to the detector being excited while other choices of vacuum leave the detector in the ground state. It is this subtle issue of the choice of vacuum that prevents simple violations of the equivalence principle.
What do they physically change in the detector so that it either detects a radiation or not? It has to be configured to pick up the radiation with respect to a given vacuum, no?

15. Originally Posted by Nic321
What do they physically change in the detector so that it either detects a radiation or not? It has to be configured to pick up the radiation with respect to a given vacuum, no?
By "vacuum" they mean the state of motion of the detector, specifically the presence or absence of proper acceleration in the frame of the detector. A Rindler vacuum means that there is uniform acceleration in the detector's frame, a Minkowski vacuum means the detector is in free fall, or stationary very far away from the black hole. Basically, the vacuum ground state the detector "sees" is a function of proper acceleration - a result of quantum field theory.

16. Ah ok, I understand.

Thank you Markus for all these explanations. That pretty much answers the original question, as far as I can understand at least.

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