Do Black Holes become visible at relativistic speeds?

[tionis]

Ok, so if I were to travel towards a black hole at close to c, would the event horizon become visible?

Let us assume there are no stars, CMB, or any other luminous body in the universe.

 

[Orodruin]

No.

The event horizon does not consist of anything and regardless how fast you move relative to it it does not change the fact that no light escapes the black hole even if it did.

 

[tionis]

Orudruin, but what about the firewall of particles hovering on the surface of the BH? Won't they become visible at some point if we move fast enough?

 

[PeterDonis]

what about the firewall of particles hovering on the surface of the BH?

The firewall is speculation at this point; there is an ongoing debate on this, and I'm not sure the firewall speculation is actually doing very well in the debate at this point.

That said, I'm not sure what difference the speed at which you move towards the hole makes for "visibility" of anything near (but outside) the horizon. You'll obviously see what's near the horizon when you get near the horizon yourself; you'll do that faster if you accelerate towards the horizon than if you just free-fall, yes. And if you only have detectors that can detect, for example, visible light, moving faster towards the horizon will blueshift what you're trying to detect, which may make it easier for your detector to detect it. But it doesn't seem like these kinds of effects are what you're talking about.

 

[tionis]

Hi Peter.

Yes, the question is whether I can make the black hole itself become visible by virtue of my speeding towards it at a phenomenal speed (close to c). I don't care what is near the horizon, that is why I got rid of any possible thing that might shift into visible view like the CMB or any cosmic dust, starlight, etc. I only care about the BH becoming visible in an empty space devoid of any source of radiation. Does that make sense lol?

 

[PeterDonis]

I don't care what is near the horizon

But the firewall is "near the horizon" in the sense you mean; it's just outside the horizon (in a sort of "membrane" one Planck length above the horizon, at least in the original formulation of the idea).

I only care about the BH becoming visible in an empty space devoid of any source of radiation. Does that make sense lol?

Orodruin gave the correct answer to that: the BH itself can't "become visible" because it isn't a "thing" you can see in the first place. But I still don't understand why moving towards the hole very fast would make any difference, even if there was something potentially visible.

 

[tionis]

But the firewall is "near the horizon" in the sense you mean; it's just outside the horizon (in a sort of "membrane" one Planck length above the horizon, at least in the original formulation of the idea).

So the firewall would indeed become visible?

Orodruin gave the correct answer to that: the BH itself can't "become visible" because it isn't a "thing" you can see in the first place. But I still don't understand why moving towards the hole very fast would make any difference, even if there was something potentially visible.

I don't understand this. A black hole is an object in space with a temperature, right? If I understand correctly, that temperature can go up or down with relative speed, no?

 

[PAllen]

So the firewall would indeed become visible?
I don't understand this. A black hole is an object in space with a temperature, right? If I understand correctly, that temperature can go up or down with relative speed, no?

Let's try to separate concerns.

1) A classical BH, no quantum effects at all (presumably doesn't exist in our quantum universe).

A) Eternal classical BH: Then, no visibility however fast you approach it.

B) A BH that formed from a collapse. Theoretically, it will always be emitting some low frequency radiation from just outside the horizon 'left over' from the collapsing matter. Approaching at close to c will blue shift this near horizon emission of weak, low frequency, near horizon radiation into visible light. The older the BH, the closer to c you would need to approach it to 'see' it in this specific sense.

2) First quantum correction: A BH has temperature and emits Hawking radiation. Temperature and Hawking radiation go up as a BH ages. In the sense of seeing Hawking radiation as visible light, just wait long enough (very very long). Also, approach at near c will make Hawking radiation detectable as visible light. You would have to worry about not being killed by blue shifted CMB which would become gamma rays at the speed needed to 'see' Hawking radiation for a stellar mass BH, but let's put such issues aside. Note that Hawking radiation is normally assumed to originate outside the Horizon.

3) Firewall: as Peter noted, this is a debated speculation, not considered theoretically established like Hawking radiation. While I have read parts of some of the papers debating this, I have no firm idea about any luminance due to the firewall. This not something the debate focuses on. However, I have also seen no hint that an observer some distance away is expected to see anything other than Hawking radiation. If this is true, then (3) is the same as (2) for purposes of rapid approach to BH, until you get incredibly close to the horizon.

 

[PeterDonis]

So the firewall would indeed become visible?

To the extent that it emits radiation outward, yes. But the radiation it emits is just Hawking radiation, and Hawking radiation is not necessarily visible to all observers. See below.

I don't understand this. A black hole is an object in space with a temperature, right? If I understand correctly, that temperature can go up or down with relative speed, no?

A black hole has a temperature, but it's not an "object in space with a temperature". A black hole is a geometric configuration of spacetime. Moving radially inward in this spacetime does not really move you "toward" the black hole--or more precisely, toward the horizon; that's because the horizon is really in the future, so the "direction" it is in, with respect to you, is not really best thought of as a spatial direction.

It's true that, when quantum effects are included, the geometric configuration of spacetime that is a black hole produces radiation--Hawking radiation--that travels outward, yes. And the temperature that's quoted for a black hole, and which corresponds to the spectrum of the Hawking radiation, is "temperature at infinity", yes; so if you're at a finite radius, you will see a larger temperature of the Hawking radiation than an observer at infinity.

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; but that's only a heuristic guess, based on the fact that the ingoing and outgoing radial directions are not symmetric in a black hole spacetime. If my guess is correct, then moving faster towards the hole would *not* make the Hawking radiation more visible.

Now, as to where the firewall speculation fits into all this: the standard view about Hawking radiation, as above, is that, if you freely fall through the event horizon of a black hole, you will not see any Hawking radiation. However, not everyone is comfortable with this view, and some people have built a speculative model on the fact that an observer who is accelerating outward in order to "hover" at an altitude just above a black hole's horizon will see a hot "membrane" of Hawking radiation (because, as above, they are seeing the radiation highly blueshifted from its temperature at infinity). This speculative model postulates that quantum effects (which must therefore be different than we currently understand them to be) actually cause this hot membrane to be detectable by observers freely falling through the horizon--so "detectable", in fact, that the membrane destroys any such observer and re-emits the individual particles that compose the observer as Hawking radiation. The term "firewall" comes from this latter property.

If the "firewall" model is correct, it would seem to imply that Hawking radiation would indeed be visible to all observers, so that moving radially inward very fast would indeed cause you to more easily detect the Hawking radiation (unlike in the standard model I gave above). However, as I said, the "firewall" model is just speculation at this point, and AFAIK it is not very likely to pan out.

 

[PeterDonis]

A BH that formed from a collapse. Theoretically, it will always be emitting some low frequency radiation from just outside the horizon 'left over' from the collapsing matter. Approaching at close to c will blue shift this near horizon emission of weak, low frequency, near horizon radiation into visible light. The older the BH, the closer to c you would need to approach it to 'see' it in this specific sense.

This is a good point; I hadn't considered this possibility, but you're right, any real BH will have formed from a collapsing object and will have near-horizon radiation of this kind that can be detected.

approach at near c will make Hawking radiation detectable as visible light.

I'm not sure this is true; as I noted in my previous post, I've never actually seen an analysis of this case, and the fact that an observer free-falling into the hole sees no Hawking radiation (at least on the standard view) seems to indicate otherwise.

 

[PAllen]

I'm not sure this is true; as I noted in my previous post, I've never actually seen an analysis of this case, and the fact that an observer free-falling into the hole sees no Hawking radiation (at least on the standard view) seems to indicate otherwise.

I had forgotten about this result (that is closely related to complementarity ideas as understood before the firewall proposal). However, my intuition for an observer accelerating toward the BH is different from yours. I suspect they would see radiation. However, I am not prepared to justify this mathematically, and I am unable to find published analysis of this case. So, who knows?

 

[Haelfix]

2) First quantum correction: A BH has temperature and emits Hawking radiation. Temperature and Hawking radiation go up as a BH ages. In the sense of seeing Hawking radiation as visible light, just wait long enough (very very long). Also, approach at near c will make Hawking radiation detectable as visible light. You would have to worry about not being killed by blue shifted CMB which would become gamma rays at the speed needed to 'see' Hawking radiation for a stellar mass BH, but let's put such issues aside. Note that Hawking radiation is normally assumed to originate outside the Horizon.

3) Firewall: as Peter noted, this is a debated speculation, not considered theoretically established like Hawking radiation. While I have read parts of some of the papers debating this, I have no firm idea about any luminance due to the firewall. This not something the debate focuses on. However, I have also seen no hint that an observer some distance away is expected to see anything other than Hawking radiation. If this is true, then (3) is the same as (2) for purposes of rapid approach to BH, until you get incredibly close to the horizon.

It was thought for years that an observer passing through the event horizon of a large black hole would NOT see anything special there, even though in principle one might think that he/she would see Hawking radiation (let us ignore the fact that the wavelength is on the order of the size of the Bh to begin with). This apparent paradox was accepted to be resolved b/c of arguments from BH complementarity, Holography and semiclassial calculations in the context of AdS/CFT that seemed to give credence to that idea). However once the AMPS paper was released, they showed in a sort of no go proof (within the framework of semiclassical gravity) that complementarity was NOT quite enough, therefore the infalling observer does see highly excited modes in complete contradiction to the standard GR picture, the equivalence principle and so on.

Its worth pointing out that the authors of the AMPS paper were actually trying to resolve exactly what you guys were looking for. Eg trying to build a concrete model of the local physics which tracked exactly what would happen to the local infalling observer as he/she was approaching the (stretched) horizon. This program failed, which led to formalizing the extent of the failure as a no go theorem.

As of this current moment, no one has really found a consistent way out of this mess, which is why it is an active and controversial subject.

 

[timmdeeg]

B) A BH that formed from a collapse. Theoretically, it will always be emitting some low frequency radiation from just outside the horizon 'left over' from the collapsing matter. Approaching at close to c will blue shift this near horizon emission of weak, low frequency, near horizon radiation into visible light. The older the BH, the closer to c you would need to approach it to 'see' it in this specific sense.

Isn't this the only case considering what happens in finite time?
From the "frozen stars" point of view I would expect that an arbitrary observer will see their light in his finite-time future but never the formation of the horizon and thus never any Hawking radiation.

 

[PAllen]

Isn't this the only case considering what happens in finite time?
From the "frozen stars" point of view I would expect that an arbitrary observer will see their light in his finite-time future but never the formation of the horizon and thus never any Hawking radiation.

In this paragraph I was explicitly discussing only pure classical picture, not Hawking radiation. A collapse leading to a BH, viewed from outside, at any finite time, detects states closer and closer to horizon formation, with signal ever dimmer and redder. In practice, it becomes blacker than anything else in the universe in a quite short time. However, in priniciple, at any distance (even far away) from the BH there is some infinitesimally dim, redshifted radiation from just before horizon formation. If you are heading toward the BH sufficiently close to C, this radiation (by both beaming and blueshifting) should become detectable. It is in this specific sense (as I carefully worded it) that you could see a BH that had become invisible to any detector, become visible again, on approach close enough to c.

 

[timmdeeg]

If you are heading toward the BH sufficiently close to C, this radiation (by both beaming and blueshifting) should become detectable. It is in this specific sense (as I carefully worded it) that you could see a BH that had become invisible to any detector, become visible again, on approach close enough to c.

Yes, understand. Its my fault, using the phrase "see their light" I meant detectable (at least in principle) electromagnetic radiation. But that wasn't the question, sorry for that.

 

[PAllen]

Isn't this the only case considering what happens in finite time?
From the "frozen stars" point of view I would expect that an arbitrary observer will see their light in his finite-time future but never the formation of the horizon and thus never any Hawking radiation.

I realize there is another question here. A static outside observer DOES detect Hawking radiation from the collapsed body after finite time. This is the key, undisputed feature of Hawking radiation - that despite not being able to get any classical signal from (or inside) the horizon, Hawking radiation is received within finite time. Evaporation occurs within within finite time. Coincident with final evaporation (according to many models), you would see light from the horizon, verifying that it had actually formed.

 

[timmdeeg]

A static outside observer DOES detect Hawking radiation from the collapsed body after finite time. This is the key, undisputed feature of Hawking radiation - that despite not being able to get any classical signal from (or inside) the horizon, Hawking radiation is received within finite time. Evaporation occurs within within finite time. Coincident with final evaporation (according to many models), you would see light from the horizon, verifying that it had actually formed.

Thank you, that's convincing. I'm just still a bit curios, having in mind that the formation of a black hole happens in infinite coordinate time. Would it make sense to argue that the "last photon" emitted just before the horizon was formed arrives in finite time? Or, how is the "frozen star" puzzle usually resolved otherwise?

 

[WannabeNewton]

Null geodesics emitted from the worldline of an observer on the collapsing star before the worldline intersects the event horizon always reach the distant observer. A null geodesic emitted when the worldline intersects the event horizon will remain tangent to the event horizon as the horizon is their generator. You can see this all easily by looking at the local light cones in a spacetime diagram in advanced Eddington Finkelstein coordinates for the schwarzschild spacetime outside the star.

 

[timmdeeg]

Null geodesics emitted from the worldline of an observer on the collapsing star before the worldline intersects the event horizon always reach the distant observer.

I don't really trust the "last photon" argument. But if it is true that the event horizon of the collapsing star is formed in the infinite future why then does the Hawking radiation arrive in finite time? It seems nobody doubts the latter.

 

[PAllen]

I don't really trust the "last photon" argument. But if it is true that the event horizon of the collapsing star is formed in the infinite future why then does the Hawking radiation arrive in finite time? It seems nobody doubts the latter.

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.

 

[WannabeNewton]

I don't really trust the "last photon" argument. But if it is true that the event horizon of the collapsing star is formed in the infinite future why then does the Hawking radiation arrive in finite time? It seems nobody doubts the latter.

Who said the event horizon is formed in the infinite future? A black hole is a region of spacetime which does not lie in the past of future null infinity. This does not imply that it lies in future timelike infinity.

 

[PeterDonis]

TI'm just still a bit curios, having in mind that the formation of a black hole happens in infinite coordinate time.

That's only true for a classical black hole, with no quantum effects included. 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.

In other words, the hole forming in infinite coordinate time assumes a particular spacetime structure; but including quantum effects changes the spacetime structure to a different one.

 

[tionis]

A) Eternal classical BH: Then, no visibility however fast you approach it.

Peter and PAllen, awesome explanations. I'm really having trouble understanding this ^ and unfortunately, I lack the technical vocabulary to get my point across, but let me try lol.

Now, I'm not sure how classical or quantum my scenario is. I really don't. But it's my understanding that classical relativity deals with this issue of red and blue shifting without invoking any quantum effects, though I could be mistaken. The only reason I brought up firewalls and stuff is because I thought they were part of the localized spacetime phenomenon we call a black hole, but on further reading your posts, I have come to the realization that although very close to the event horizon (a Planck length or so), I don't believe these quantum effects satisfy my scenario.

I would like my black hole to be in its purest classical state possible, i.e. devoid of any outside quantum effects no matter how close they are to the BH itself. I'm only interested in the event horizon and everything that follows after it. So, from a relativity point of view and without any other contribution from anything else but what is already present in the black hole itself, would I be able to shift that BH into view simply by going fast? You said ''no,'' but I don't understand why.

 

[PeterDonis]

I would like my black hole to be in its purest classical state possible, i.e. devoid of any outside quantum effects no matter how close they are to the BH itself. I'm only interested in the event horizon and everything that follows after it. So, from a relativity point of view and without any other contribution from anything else but what is already present in the black hole itself, would I be able to shift that BH into view simply by going fast? You said ''no,'' but I don't understand why.

Because there's nothing to see. A purely classical black hole emits nothing, and contains nothing--it's a vacuum. So no matter how fast you go towards it, you can't blueshift anything into view because there's nothing there.

 

[PAllen]

Peter and PAllen, awesome explanations. I'm really having trouble understanding this ^ and unfortunately, I lack the technical vocabulary to get my point across, but let me try lol.

Now, I'm not sure how classical or quantum my scenario is. I really don't. But it's my understanding that classical relativity deals with this issue of red and blue shifting without invoking any quantum effects, though I could be mistaken.

Correct, red and blue shifting are classical phenomena.

The only reason I brought up firewalls and stuff is because I thought they were part of the localized spacetime phenomenon we call a black hole, but on further reading your posts, I have come to the realization that although very close to the event horizon (a Planck length or so), I don't believe these quantum effects satisfy my scenario.

Hawking radiation is derived from Quantum Field Theory + curved spacetime. It is not part of the classical theory of BH at all. Firewalls are a much more recent speculation. Both of these should be left out of a classical discussion. Temperature of a BH is also strictly a QFT phenomenon (or a QG phenomenon - quantum gravity).

I would like my black hole to be in its purest classical state possible, i.e. devoid of any outside quantum effects no matter how close they are to the BH itself. I'm only interested in the event horizon and everything that follows after it. So, from a relativity point of view and without any other contribution from anything else but what is already present in the black hole itself, would I be able to shift that BH into view simply by going fast? You said ''no,'' but I don't understand why.

No. There is no radiation of any kind associated with the event horizon itself (classically). For an eternal BH (one that existed forever, rather than forming from a collapse), there also no near horizon emissions left over from a collapse. Thus, there is nothing at all to blue shift on approach.

 

[timmdeeg]

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]

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]

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]

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

 

[PeterDonis]

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 ).

 

[PAllen]

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]

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.

 

[PeterDonis]

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]

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]

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