Does Hawking radiation take place at the apparent horizon?

[wywong]

When the apparent horizon differs from the event horizon, as in the case of an observer falling into a black hole, does Hawking radiation take place at the former, the latter, or both?

 

[rootone]

The infalling observer doesn't witness any horizon, although they might witness that are they becoming spaghettified.
The event horizon is something which is determined to be where it is according to a distant observer, perhaps standing where the traveler started.

The idea of Hawking radiation is that at this event horizon, some 'virtual particles' will fall into the BH, while others escape.
That which escapes is the Hawking radiation

 

[PeterDonis]

When the apparent horizon differs from the event horizon, as in the case of an observer falling into a black hole

I assume you mean an object falling in with significant enough mass to change the hole's mass, so that the horizons change during the process, correct?

does Hawking radiation take place at the former, the latter, or both?

I don't think the answer to this is known for sure. Hawking's original derivation gave the answer that it's the event horizon; however, other treatments have made it seem like it's the apparent horizon, and recently I believe Hawking has come to that view. I don't know that we'll know for sure until we have a confirmed theory of quantum gravity.

 

[wywong]

I assume you mean an object falling in with significant enough mass to change the hole's mass, so that the horizons change during the process, correct?

Sorry to have confused you. I actually mean the horizon perceived visually by the falling observer, which always seems some distance below the observer. It appears that "apparent horizon" is not the correct term. Is "visual horizon" a better term?

What puzzles me most is how the observer perceives the Hawking radiation as he crosses the event horizon. Will he still see Hawking radiation coming from below?

Hawking's original derivation gave the answer that it's the event horizon; however, other treatments have made it seem like it's the apparent horizon... I don't know that we'll know for sure until we have a confirmed theory of quantum gravity.

Thanks for enlightening me.

 

[rootone]

Sorry to have confused you. I actually mean the horizon perceived visually by the falling observer, which always seems some distance below the observer. It appears that "apparent horizon" is not the correct term. Is "visual horizon" a better term?.

As far as everything I have read goes, that's incorrect, but if I have been misled I'll happily stand corrected.
The generally accepted idea is that the infalling observer does not see an apparent or visual horizon which is always below him, he sees nothing at all below him.
He passes right though the place where the event horizon was deemed to be before he started his journey. then continues deeper into the hole.
In his own frame of reference he shouldn't see any significant increase in radiation when crossing the event horizon.

Edit: Although it does seem that Hawking is himself now not so sure.

 

[jartsa]

When the apparent horizon differs from the event horizon, as in the case of an observer falling into a black hole, does Hawking radiation take place at the former, the latter, or both?

Below the event horizon, you can not block any Hawking radiation going to infinity.
Above the event horizon, you can block Hawking radiation going to infinity.
Those two facts suggest that Hawking radiation going to infinity comes from the event horizon.

 

[PeterDonis]

I actually mean the horizon perceived visually by the falling observer, which always seems some distance below the observer.

Ok, yes, this is not the apparent horizon, at least not in the usual terminology. The apparent horizon is the surface at which radially outgoing light no longer moves outward locally, but stays at the same radius.

I don't know that there is a standard term for what you're describing. Possibly that's because, at least classically, the horizon does not emit any light, so it can't be seen. But if it did emit light, and some of that light went inward instead of outward, then light that went inward at various angles would be bent by the spacetime curvature inside the horizon so that it came at the infalling observer from the front. So there would appear to be a "horizon" in front of the observer even though he had already passed through it.

Will he still see Hawking radiation coming from below?

I'm not sure. The usual heuristic argument for why there is Hawking radiation is that you have pairs of virtual particles spontaneously appearing from the vacuum near the horizon, and before they can annihilate each other, the tidal gravity of the hole separates them, so that one falls down the hole while the other escapes to infinity. In this picture, Hawking radiation would only be emitted outward; the particle that falls inward would have negative energy and would not be detectable inside the hole (the only effect would be a slight decrease in the hole's mass). However, I don't know if the detailed math actually has that restriction.

 

[PAllen]

Here is a nice modern reference on Hawking radiation, which takes the view that the apparent horizon is the key factor:

http://arxiv.org/abs/hep-th/0106111

I have seen papers arguing that you can detect Hawking radiation inside the horizon, but I don't think this is the consensus view, and I am not presently interested in trying to locate those papers.

 

[jartsa]

What puzzles me most is how the observer perceives the Hawking radiation as he crosses the event horizon. Will he still see Hawking radiation coming from below?

Oh, an observer free falling through event horizon. Well he does not observe any horizons or any radiation coming from any horizons.

Observer hovering near the event horizon feels a large acceleration, therefore he must observe Unruh-radiation. The black hole is the only possible energy source for that radiation.

If the observer starts free falling, acceleration felt by the observer becomes zero, therefore Unruh-radiation absorbed by the observer must become zero.

Mathematically: Observer hovering at distance D from event horizon absorbs X % of all the radiation emitted by the black hole. When the observer starts free falling, he starts absorbing X% / K of all the radiation emitted by the black hole. As D approaches zero, K approaches infinity.

 

[PeterDonis]

Observer hovering near the event horizon feels a large acceleration, therefore he must observe Unruh-radiation. The black hole is the only possible energy source for that radiation.

No, there is another obvious source: whatever it is that is producing the acceleration of the observer. Unruh radiation occurs in flat spacetime, where there is no black hole, so if Hawking radiation is really just Unruh radiation, there is no need for a black hole as an energy source.

The actual derivation of Hawking radiation, while it has similarities to the derivation of Unruh radiation, does not work out quite the same physically. It involves (heuristically) scattering of virtual wave packets off of the spacetime curvature of the black hole, so a quantum field state that is a particular vacuum state in the far past becomes a thermal state in the far future. This would not occur in flat spacetime, so it can't be the same as Unruh radiation.

The reason that, at least in the original derivation of Hawking radiation that I just referred to, an observer free-falling into the hole cannot detect any radiation, is that the "incoming" vacuum state, the one the quantum field starts out in in the far past, happens to evolve in just the right way to remain a vacuum state for infalling observers. However, it's not clear to me that it is a vacuum state for inertial observers in other states of motion (for example, for an observer who is moving inertially at escape velocity just above the horizon). Such observers might still be able to observe Hawking radiation, i.e., they might not see the quantum field as being in a vacuum state. (In flat spacetime, they would, since the vacuum should be Lorentz invariant; but near the horizon of a black hole, there is an obvious anisotropy due to the hole's gravity.)

Mathematically: Observer hovering at distance D from event horizon absorbs X % of all the radiation emitted by the black hole. When the observer starts free falling, he starts absorbing X% / K of all the radiation emitted by the black hole. As D approaches zero, K approaches infinity.

Can you show math for this? Or give a reference?

 

[wywong]

As far as everything I have read goes, that's incorrect, but if I have been misled I'll happily stand corrected.
The generally accepted idea is that the infalling observer does not see an apparent or visual horizon which is always below him, he sees nothing at all below him.
He passes right though the place where the event horizon was deemed to be before he started his journey. then continues deeper into the hole.
In his own frame of reference he shouldn't see any significant increase in radiation when crossing the event horizon.

Edit: Although it does seem that Hawking is himself now not so sure.

According to http://jila.colorado.edu/~ajsh/insidebh/penrose.html

The Penrose diagram shows that the horizon is really two distinct entities, the Horizon, and the Antihorizon.The Antihorizon might reasonably called the illusory horizon. ...The diagram shows that when you look at a black hole from the outside, you are looking at its Antihorizon, or illusory horizon. When you fall through the horizon, you fall not through the Antihorizon, but rather through the Horizon, or true horizon. The Horizon becomes visible only after you have fallen through it. The Antihorizon continues to remain ahead of you, and you never fall through it.

According to http:///http://en.wikipedia.org/wiki/Event_horizon

Observers crossing a black hole event horizon can calculate the moment they have crossed it, but will not actually see or feel anything special happen at that moment. In terms of visual appearance, observers who fall into the hole perceive the black region constituting the horizon as lying at some apparent distance below them, and never experience crossing this visual horizon.

From my understanding, instead of seeing total darkness below, the observer can see a lot from below when he is inside the EH (event horizon). If an object crosses the event horizon a short while before the infalling observer, he will still be able to see the object when he catches up with the photons from the object inside the event horizon; the faint stars on the other side of the BH are visible outside the shrinking illusory horizon before and after he crosses the EH.

 

[rootone]

Well I'm not going to argue with Penrose about mathematical models.
Logically, (well Intuitively), it seems to me that you cannot see anything deeper into the black hole than you are yourself,
Since the deeper thing has experienced even more gravity than you have it is therefore going in the same direction but faster than you.

Doesn't this also contradict the notion of 'spaghettification' ?
In principle the observer sees his feet moving faster into the BH than his head is, and within a short time they become too distant to be observable.

 

[PeterDonis]

According to http://jila.colorado.edu/~ajsh/insidebh/penrose.html

This is a description of an idealized "eternal" black hole that has always existed (the technical name is "maximally extended Schwarzschild spacetime"). It is not a description of a real black hole that forms through gravitational collapse of a massive object. Such a real hole does not have an antihorizon. (Neither does a quantum hole that can evaporate through Hawking radiation.)

 

[PeterDonis]

it seems to me that you cannot see anything deeper into the black hole than you are yourself

No, that's not really true. What is true is that you won't see the light from anything deeper than yourself, until you have fallen to where the light is.

Also, this comment doesn't really apply anyway to the question of whether you can see the horizon from inside the hole, since once you're inside the hole the horizon is above you.

Doesn't this also contradict the notion of 'spaghettification' ?

No.

 

[rootone]

No, that's not really true. What is true is that you won't see the light from anything deeper than yourself, until you have fallen to where the light is.

But by the time you get there, wouldn't the light itself have gone still deeper into the well?

 

[wywong]

But by the time you get there, wouldn't the light itself have gone still deeper into the well?

If a boat sailing eastwards towards a boat sailing westwards, they will meet in a time given by (starting distance)/(relative speed). The fact that the latter is actually also sailing eastwards (due to the Earth's rotation) doesn't prevent the former from catching up with it.

 

[PeterDonis]

by the time you get there, wouldn't the light itself have gone still deeper into the well?

Yes, but heuristically, the light will be decreasing its $r$ coordinate more slowly than you, so you can still catch up with it. (Note that this assumes the light was emitted in an outward radial direction.)

 

[jartsa]

Can you show math for this? Or give a reference?

This paper seems to say that the closer an observer goes to the event horizon of a black hole, the more Hawking radiation will look like Unruh-radiation to him.
[PLAIN]http://arxiv.org/abs/1102.5564[/PLAIN]
http://arxiv.org/abs/1102.5564

So if an observer in a lab hovering near an event horizon jumps into the air, he observes almost no Hawking radiation during the jump, because Unruh-radiation would completely disappear during a jump, I mean during the free fall part of the jump.

 

[PeterDonis]

This paper seems to say that the closer an observer goes to the event horizon of a black hole, the more Hawking radiation will look like Unruh-radiation to him.

Sort of. It says that, as you get closer to the horizon, the difference between the radiation detected by an observer hovering at rest in Schwarzschild spacetime, and the radiation that would be detected by an observer in flat spacetime with the same proper acceleration, goes to zero.

if an observer in a lab hovering near an event horizon jumps into the air, he observes almost no Hawking radiation during the jump, because Unruh-radiation would completely disappear during a jump, I mean during the free fall part of the jump.

The paper you linked to doesn't say anything about free-fall observers, as far as I can see.

More generally, the issue here is that what any observer observes, free-fall or accelerated, depends on the state of the quantum field. In flat spacetime, there is a unique notion of a "vacuum state" of the quantum field, that looks like a vacuum to all inertial observers. (It does not look like a vacuum to accelerated observers, which is where Unruh radiation comes from.) However, in curved spacetime, there are different possible "vacuum" states, and AFAIK a given state that looks like a vacuum to one inertial observer may not look like a vacuum even to other inertial observers. In particular, the state that is assumed in the derivation of Hawking radiation looks like a vacuum to inertial observers free-falling into the hole, but I'm not sure if it looks like a vacuum to inertial observers who "jump" upward from a hovering platform close to the horizon.