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.

  • $\begingroup$ Much of the question is the same as this: physics.stackexchange.com/q/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. $\endgroup$ – Alan Rominger Mar 28 '12 at 21:13
  • $\begingroup$ @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. $\endgroup$ – Nathaniel Mar 29 '12 at 8:42

This paper discusses these issues in a fairly comprehensible way. Faraway observers (like your observer A) see thermal Hawking radiation with an effective temperature given by the Hawking temperature $$T_H := \frac{\hbar c^3}{8 \pi G M k_B},$$ where $M$ is the black hole's mass.

If an observer on a string is very slowly lowered toward the black hole (so that her $dr/d\tau$ is very small), then the effective temperature increases without bound and diverges at the horizon, so your observer B inevitably gets burned up. (You can think of her as needing unboundedly large acceleration to stay out of the hole, so that she sees a huge Unruh radation. More realistically, of course, the string would break first.) See Figure 1 of the paper, which refers to the temperature observed by a strongly accelerated observer at constant $r$ as the "fiducial temperature" $T_\text{FID}$.

If an observer free-falls into the black hole, the effective temperature that she observes (which the paper calls the "free-falling at rest temperature" $T_\text{FFAR}$) gradually increases from $T_H$ very far away to $2 T_H$ at the horizon (your observer C), also plotted in Figure 1. You might think that her thermometer hitting $2 T_H$ would give her a local probe of exactly when she crosses the horizon, which would violate the equivalence principle. But this is not the case, for a subtle reason given below Fig. 1:

We note that our method gives a physically reasonable answer for $T_\text{FFAR}$ at all values of $r ≥ 2m$. However, the precise numerical value $T_\text{FFAR} = 2TH$ at the event horizon has limited operational meaning. First of all, as was discussed in the introduction, the local free-fall temperature is not a precise notion. On top of this, a freely-falling observer passing through the horizon has only a proper time of order $m$ left before running into the curvature singularity at $r = 0$, and since the characteristic wavelength of thermal radiation at $T ∼ 2T_H$ is also of order $m$, the observer cannot measure temperature to better than $O(1/m)$ precision near the horizon. Although our result for the free-fall temperature is thus only qualitative in the region near the event horizon, it does confirm the expectation expressed in early work of Unruh [13] that an infalling observer will not run into highly energetic particles at the horizon.

The case of observer C illustrates an important subtlety regarding Hawking radiation. It's often stated that Hawking radiation is just the Unruh radiation seen by an observer near the horizon accelerating away from it in order to keep from falling in. But this is not quite correct, as explained in this answer, because Unruh radiation is a flat-spacetime effect and spacetime is curved near an event horizon. If they were truly equivalent, then a freefalling observer would not observe any Hawking radiation, but in fact observer C does. But for a very large black hole, the curvature at the horizon (as measured, say, by the Kretschmann scalar $$R_{\mu \nu \rho \sigma} R^{\mu \nu \rho \sigma} \big|_{r = 2GM} = \frac{48 (G M)^2}{(2 G M)^6} = \frac{3}{4 (GM)^4}$$ ) becomes arbitrarily small. So near the horizon of a very large black hole, Hawking radiation and Unruh radiation become essentially the same thing, and indeed a freefalling observer sees negligible Hawking radiation (at an arbitrarily low temperature $2 T_H \propto 1/M$).

Edit: I should clarify that the temperature you observe as you free-fall through the horizon depends on your speed, or more precisely, how far away you were from the horizon when you were released from rest. $T_\text{FFAR}$ is the temperature that you observe if you are both inertial ("free-fall") and have proper velocity $dr/d\tau = 0$ ("at rest"). The paper https://arxiv.org/abs/1608.02532 explains this in more detail, distinguishes between the observed temperatures of the outgoing and incoming radiation, and elaborates on the distinction between the Hawking and Unruh effects. Thanks to Akoben for pointing this out.

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    $\begingroup$ This is really helpful and fully answers the question, so I'm accepting it, thank you! $\endgroup$ – Nathaniel Feb 7 '17 at 3:14
  • $\begingroup$ @Nathaniel I added a paragraph expanding on my answer $\endgroup$ – tparker Feb 7 '17 at 7:47
  • $\begingroup$ Your second reference (§4.2) gives a different value of $T_{FFAR}(r=r_s)$ than the first ($2T_H$). Which is correct? $\endgroup$ – alexchandel Feb 5 at 18:44

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.

  • $\begingroup$ 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. $\endgroup$ – Nathaniel Mar 31 '12 at 10:00
  • $\begingroup$ @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. $\endgroup$ – Ron Maimon Mar 31 '12 at 17:08
  • $\begingroup$ 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. ... $\endgroup$ – Nathaniel Apr 1 '12 at 10:37
  • $\begingroup$ ... 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.) $\endgroup$ – Nathaniel Apr 1 '12 at 10:41
  • $\begingroup$ ... 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. $\endgroup$ – Nathaniel Apr 1 '12 at 10:51

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.


Hot means B sees the temperature (and acceleration) go to infinity as the horizon is approached. Normal means the hawking temperature (what the temperature would be stationary and far away from the hole). The reconciliation is the Unrah effect. In one frame of reference, B is accelerating and C is not accelerating. B sees a hot horizon due to the Unrah effect. In another frame, B is stationary and C's acceleration goes to infinity at the horizon. For C, both the Unrah effect and Hawking radiation grow to infinity, but these effects cancel out. A proper semiclassical quantum field theory calculation would probably show C's Unrah radiation acting out of phase with the Hawking radiation she receives to cancel it out. It's as if we shine two flashlights on C but the lights almost perfectly destructively interfere and C is left in the dark. Both frames are valid to use.

  • $\begingroup$ Could you clarify what you mean by "hot" and "normal" temperatures? By "hot" do you mean "unboundedly hot as one approaches the horizon," as Ron Maimon's answer would indicate? By "normal" do you mean the Hawking temperature $1/(8 \pi G)$? How do you reconcile your answer with Ron Maimon's claim that observer $C$ does not observer any Hawking radiation at all as she crosses the horizon? $\endgroup$ – tparker Feb 6 '17 at 4:42
  • $\begingroup$ @tparker: Yes, hot means B sees the temperature (and acceleration) go to infinity as the horizon is approached. $\endgroup$ – Kevin Kostlan Feb 6 '17 at 22:13
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    $\begingroup$ @tparker: I tried to clarify it better. $\endgroup$ – Kevin Kostlan Feb 6 '17 at 22:24
  • $\begingroup$ According to this paper, a free-falling observer measures a temperature of $2 T_H$ as she crosses the horizon. But this does not provide a local probe for identifying the horizon (which is forbidden by the equivalence principle), because ... $\endgroup$ – tparker Feb 7 '17 at 1:21
  • $\begingroup$ ... "a freely-falling observer passing through the horizon has only a proper time of order $m$ left before running into the curvature singularity at $r = 0$, and since the characteristic wavelength of thermal radiation at $T \sim 2 T_H$ is also of order $m$, the observer cannot measure temperature to better than $O(1/m)$ precision near the horizon ... Our result for the free-fall temperature is thus only qualitative in the region near the event horizon." $\endgroup$ – tparker Feb 7 '17 at 1:22

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