# Intensity of Hawking radiation for different observers relative to a black hole

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Consider three observers in different states of motion relative to a black hole:

Observer A is far away from the black hole and stationary relative to it;

Observer B is suspended some distance above the event horizon on a rope, so that her position remains constant with respect to the horizon;

Observer C is the same distance from the horizon as B (from the perspective of A), but is freefalling into it.

All of these observers should observe Hawking radiation in some form. I am interested in how the spectra and intensity of the three observations relate to one another.

My previous understanding (which might be wrong, because I don't know how to do the calculation) was that if you calculate the radiation that B observes, and then calculate how much it would be red shifted as it leaves the gravity well, you arrive at the spectrum and intensity of the Hawking radiation observed by A. I want to understand how the radiation experienced by C relates to that observed by the other two.

The radiation fields observed by B and C are presumably different. B is being accelerated by the tension in the rope, and is thus subject to something like the Unruh effect. C is in freefall and therefore shouldn't observe Unruh photons - but from C's point of view there is still a horizon ahead, so presumably she should still be able to detect Hawking radiation emanating from it. So I would guess that C observes thermal radiation at a lower intensity than B, and probably also at a lower temperature (but I'm not so sure about that).

So my question is, am I correct in my understanding of how A and B's spectra relate to one another, and has anyone done (or would anyone be willing to do) the calculation that would tell us what C observes? References to papers that discuss this would be particularly helpful.

This post imported from StackExchange Physics at 2014-04-08 05:12 (UCT), posted by SE-user Nathaniel
Much of the question is the same as this: physics.stackexchange.com/questions/10811/… But you asked more specifically about moving vs. stationary observers. This is encompassed in the "apparent" horizon definition, but it wasn't directly addressed in that question, just saying.

This post imported from StackExchange Physics at 2014-04-08 05:12 (UCT), posted by SE-user AlanSE
@AlanSE I would say this is more of a follow-on from that question. I know that all horizons emit thermal radiation (apparent or otherwise - there isn't really much of a difference). In this question I'm asking for a calculation of the temperature of the observed radiation in particular situations.

This post imported from StackExchange Physics at 2014-04-08 05:12 (UCT), posted by SE-user Nathaniel

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The Hawking radiation seen by A and B are related as you say, by the redshift factor of the black hole's gravity field, which is the square root of the time-time component of the metric tensor. This is determined by the full Hawking radiating equilibrium state, which is the path-integral in the Euclidean continued geometry, whose period is everywhere the same in the imaginary time variable, and is constant at large distances, but goes to zero near the horizon, corresponding to a diverging temperature there.

For observer C, as the observer gets close to the black hole, so that the distance to the horizon becomes smaller than the black hole radius, the Hawking radiation becomes invisible, and the observer crosses the BH without any awareness that anything has happened.

The reason this is not paradoxical is because when the suspended observer B is close to the horizon, B is accelerating very fast, and the apparent temperature B sees can be interpreted by B to be the local Unruh temperature corresponding to B's acceleration. The Hawking temperature interpretation is only when you extend this near-horizon Unruh profile to infinity using the redshift factor, which is what the stable imaginary-time Hawking solution describes.

This post imported from StackExchange Physics at 2014-04-08 05:12 (UCT), posted by SE-user Ron Maimon
answered Mar 29, 2012 by (7,535 points)
So you're saying that as C approaches the horizon the radiation she observes reduces in temperature and intensity, approaching zero asymptotically as she approaches the horizon? Note that I'm interested in what C sees before she crosses the horizon. Any real black hole will evaporate before C crosses the horizon, so I don't consider the question of what she sees after she passes it to be a physical one. I'd still be very interested in references to papers that discuss this - I'm unlikely to follow a full GR calculation, but I'm really after a mathematical expression of the spectrum C sees.

This post imported from StackExchange Physics at 2014-04-08 05:12 (UCT), posted by SE-user Nathaniel
@Nathaniel: You are the second person to claim that nothing crosses the horizon! The evaporation is irrelevant. The observer will cross at a finite proper time. It's holography, and it's consistent with finite lifetime. It's understood today as "black hole complementarity". There is no "freezing on the horizon". As C crosses the horizon, C sees nothing, and also well before. There is no spectrum. Far away, C sees the same as A.

This post imported from StackExchange Physics at 2014-04-08 05:12 (UCT), posted by SE-user Ron Maimon
I understand well enough that for a non-radiating black hole of infinite duration, C would pass the event horizon in finite proper time. But from an outside point of view, C is frozen on the horizon, forever, and if we can see her clock its hands will asymptotically approach the time at which she crosses the horizon from her point of view. For a real, radiating black hole she will not remain frozen forver (from an outside point of view), but merely until the hole evaporates, and then A can catch up with her and compare notes. ...

This post imported from StackExchange Physics at 2014-04-08 05:12 (UCT), posted by SE-user Nathaniel
... Susskind proposes black hole complimentarity on the basis that once an observer crosses an event horizon, she can never compare notes with an outside observer, and hence there's no paradox if they make contradictory observations. But in this scenario A and C do compare notes, so black hole complimentarity can't be invoked. (Unless, I suppose, you want to say there are "two Cs", C1 who passes the event horizon in a finite time and hits the singularity, and C2 who is observed by A and eventually escapes. In that case I'm interested in what C2 will report observing after she escapes.)

This post imported from StackExchange Physics at 2014-04-08 05:12 (UCT), posted by SE-user Nathaniel
... but all of this is a side point. You say that C observes zero radiation, with no spectrum, a finite distance away from the horizon, but of course sees the same as A when far away. Do you really mean there's a point before the event horizon beyond which she observes literally zero radiation, or just that the radiation she observes rapidly approaches zero very rapidly as she falls in? There must be a formula for the temperature and intensity of the radiation as a function of C's velocity and distance from the horizon - I'm after some insight into what form that equation has.

This post imported from StackExchange Physics at 2014-04-08 05:12 (UCT), posted by SE-user Nathaniel
An analogy that might make this clearer: consider observers A and C, both initially at x=1 with velocity=0 in flat Einstiein-Minkowski space. A accelerates so as to experience a Rindler horizon that crosses (x,t)=(0,0), while C remains moving inertially. C crosses this horizon (at (1,1)) in finite proper time, but from A's point of view it simply never happens. If A stops accelerating the horizon disappears and they can meet again, but at no point was C ever beyond the horizon from A's point of view. Why should a black hole horizon behave differently from a Rindler one in this respect?

This post imported from StackExchange Physics at 2014-04-08 05:12 (UCT), posted by SE-user Nathaniel
@Nathaniel: yes, of course, profile exponentially going to zero in the ratio of distance to the horizon over the area of the black hole. The precise formula is complicated, involving a full mode analysis in infalling coordinates of the emission, and nobody has the heart to calculate it, but it should be possible. Your example with Rindler horizons is instructive--- as A slows down the acceleration (adiabatically) C moves farther and farther inward, because the location of the horion changes. The black hole is exactly the same--- the local properties of the horizon are Rindler.

This post imported from StackExchange Physics at 2014-04-08 05:12 (UCT), posted by SE-user Ron Maimon
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The short answer: B sees a hot horizon. A and C see a normal temperature horizon (until C gets close to the singularity). C sees the horizon appear to stay ahead of her even after she enters the hole! C may see an infinite temperature increase as she nears the singularity, but at the horizon it will be the same order of magnitude of A.

There are two interpretations of whats going on: 1. The event horizon (or a surface very near it) is really hot, but C's acceleration (with respect to nearby stationary observers) produces a Unrah effect that cancels this out (along with the blueshift due to inward motion), saving her from getting incinerated.
2. Photons are produced all over the place, at a low energy. The wavelength of the photons is on the order of the horizon radius, which makes their location of origen "fuzzy". B gets hot due to the Unruh effect as she cranks up her rockets, but C is in free-fall so notices no Unruh effect.

These two interpretations are equality valid, much like reference frames in special relativity. They superficially disagree but predict the same thing for what all the observers see in their own proper time.

This post imported from StackExchange Physics at 2014-04-08 05:12 (UCT), posted by SE-user Kevin Kostlan
answered Jul 21, 2013 by (0 points)

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