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I have some trouble asking this question, so I will try a roundabout approach to explain what I mean.

If (outside of Earth atmosphere) one looks at the sun from different distances, the light coming from any direction intersecting the Sun's surface has a BB spectrum which is always the same, that of the surface temperature of the Sun. Looking in any other direction it is a BB spectrum at 2.7°K. So the spectrum at any place is "thermal by parts". Far from the Sun the high temperature spectrum is within a smaller solid angle than closer but however close one gets (while still just outside the Sun) the "high temperature" thermal spectrum never concerns more than a 2$\pi$ solid angle.

Now consider a black hole emitting Hawking's radiation. Let us assume it is small enough its Hawking temperature is higher than 2.7°K. So if I understand correctly, if I am stationary outside the black hole (not free falling into it, which means I am effectively in an accelerated frame) I will see BB radiation at the Hawking temperature, but only in the solid angle where the line of sight intersects the black hole horizon. For other directions, I only get 2.7°K. Or am I already mistaken ? The closer I am the larger this solid angle. There is the added complication that when I get closer the spectrum is blue-shifted, but anyway the solid angle where I see BB temperature will never be more than 2$\pi$. Indeed if I cross the horizon, then I cannot be stationary anymore, and the logic of Hawking's radiation cannot apply in the same way.

So here is my question. The Hawking radiation is often compared to the Unruh effect, because in an accelerated frame there is also a horizon, and the expression of the Unruh temperature and Hawking temperature are the same.

But from what I read, it always seems that the Unruh temperature concerns a BB radiation in all directions, over a full 4$\pi$ solid angle. Unless I am mistaken.

So why is the Unruh temperature BB in the full 4$\pi$ solid angle rather than just in a "half-space" 2$\pi$ solid angle like the Hawking radiation at the limit where I get very close to the horizon ?

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    $\begingroup$ The Hawking radiation is often compared to the Unruh effect, because in an accelerated frame there is also a horizon” - But no tidal forces. Accordingly, the Hawking radiation has two modes. One is similar to Unruh, the other is different. $\endgroup$
    – safesphere
    Oct 13, 2023 at 14:42

2 Answers 2

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In the Unruh case the thermal radiation is in equilibrium and appears isotropic to the accelerated observer. The argument by Gibbons and Perry using the imaginary-time-periodic Euclidean section of the Schwarzschild metric shows that the Black hole can only be in (actually unstable) equilibrium with an isotropic radiation bath that, far away from the B-hole, is at $T_{\rm Hawking}$, so the two equilibrium cases are quite similar.

The usually-described case of outgoing-only Hawking radiation is not in equilibrium because the far-away universe is at $T=2.7K$ and so is somewhat different from Unruh. One can deduce the latter case from the first however just as one can deduce Stefan-Boltzmann radiation power of a hot body from the equilibrium black-body energy density.

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  • $\begingroup$ What is the other equilibrium case? Accelerated observer from the Unruh effect is not in equilibrium with thermal radiation. $\endgroup$ Oct 12, 2023 at 21:37
  • $\begingroup$ @ Jan Lalinsky Why do you say that? The detector carried by each observer reaches an (acceleration dependent) temperature. The euclidean spacetime is periodic for each Rindler Killing curve with $\beta=1/T_{\rm Unruh}$ and so its $n$-point functions are equilibrium corrrelators. $\endgroup$
    – mike stone
    Oct 12, 2023 at 23:48
  • $\begingroup$ I'm trying to understand what you mean by "the two equilibrium cases are quite similar". In the case of a black hole Hawking radiation, it is clear that equilibrium may or may not be established between the black hole and the sources of radiation coming from the outside (CMB). In case of the uniformly accelerated observer, as I understand it (maybe I'm wrong), the apparent black body radiation observed has no source, and also is not necessarily in equilibrium with any other source of radiation. It's just there, observable with instruments, regardless of whether those have the same temperature. $\endgroup$ Oct 13, 2023 at 0:51
  • $\begingroup$ @Jan Lalinsky I just mean that an eternally accelerating detector reaches equilibrium with its environment at the local value of the Unruh temperature. $\endgroup$
    – mike stone
    Oct 13, 2023 at 1:00
  • $\begingroup$ Is that a necessary assumption for derivation of the Unruh effect? Or is it just a possibility? I don't think it is a necessity. $\endgroup$ Oct 13, 2023 at 8:42
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however close one gets (while still just outside the Sun) the "high temperature" thermal spectrum never concerns more than a 2π solid angle.

If Sun was a solid body with impenetrable surface, then yes. But we believe there is no such thing there - instead, we believe most of the radiation is coming from a photosphere, an opaque layer of dilute, but hot ionized gas. In theory, one could submerge a measuring device into Sun's photosphere gas deep enough, where very similar thermal radiation would be coming from all directions. Not exactly isotropic though, as there has to be net flux of radiation energy in the radial direction.

So if I understand correctly, if I am stationary outside the black hole (not free falling into it, which means I am effectively in an accelerated frame) I will see BB radiation at the Hawking temperature, but only in the solid angle where the line of sight intersects the black hole horizon. For other directions, I only get 2.7°K.

Almost, but because of the gravitational lensing, and gravitational frequency shifting (blue for some and red for other directions), radiation intensity per unit solid angle won't be the same for all angles, thus cannot be assigned single temperature - if the radiation has black body spectrum at all, the temperature will have to vary with position in the sky. For radiation coming from outside, I would expect the overhead direction to be the most bright (blueshift+lensing). For the Hawking radiation, I'm not sure, but radiation from the black hole apparent edge could be stronger than from the center, due to gravitational red-shift.

but anyway the solid angle where I see BB temperature will never be more than $2\pi$

Because of the gravitational lensing, this is not exactly right. According to a memoir by Riazuelo (2018), Fig. 20, the outside world appears to cover less than $2\pi$ of the sky when the observer is under the photon sphere, but still above the event horizon. So the Hawking radiation could be coming from greater solid angle than $2\pi$.

https://arxiv.org/pdf/1511.06025.pdf

Regarding the Unruh radiation being isotropic and Hawking radiation not, I don't know enough to answer confidently, but it seems to me they are more different than similar. Also see this answer by @cesaruliana

https://physics.stackexchange.com/a/512410/31895

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