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In case of a black hole, the direction of the Hawking radiation is from the horizon to the observer. The corresponding effect in the Rindler spacetime is the Unruh radiation.

Intuitively, a rapidly accelerating observer should face the Unruh radiation coming from the direction of motion, like a wind felt by a biker. The faster I accelerate forward, the stronger "wind" hits me in the face.

Yet, if the Unruh radiation comes from the horizon like the Hawking radiation does, it would hit the observer in the back. The faster I accelerate forward, the stronger "wind" hits me in the back. This seems counter-intuitive.

From which direction does the Unruh radiation hit an accelerating observer? In the face from the direction of motion or in the back from the Rindler horizon?

Hiroyashu
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  • There is no radiation: it is a thermal bath...that is different from the Hawking radiation... – Valter Moretti Aug 08 '20 at 11:36
  • @ValterMoretti "the Unruh radiation may be considered a flat space (large black hole) limit of the Hawking radiation" - https://physics.stackexchange.com/questions/259338#259342 and "The equivalence principle of GR states that the gravitational forces observed locally can be formulated in terms of fictitious forces experienced in a non-inertial reference frame. Therefore, if a black hole releases a thermal emission of particles, then there should be an analogous phenomenon for accelerated observers - this is the Unruh effect" - https://www.math.ucla.edu/~laurenst/Resources/undergrad_thesis.pdf – Hiroyashu Aug 09 '20 at 07:11
  • Nope, the Unruh effect corresponds to the fact that Minkowski vacuum in Rindler static coordinates is a thermal state with respect to the Rindeler Killing time: a thermal bath (KMS condition). Instead, the Hawking radiation is not an equilibrium state and we have a radiation emitted from the Killing horizon. A more precise analogy would be with the Hartle-Hawking state in the Kruskal spacetime. – Valter Moretti Aug 09 '20 at 07:46
  • In the Kruskal manifold there are at least two quantum states with thermal properties. One is the HH state which is an equilibrium state, the other is the Unruh state which describes the Hawking radiation. Both have the same temperature. Minkowski vacuum which has thermal properties with respect to the Rindler time is the analog of HH state. The equivalence principle should be applied that way. It being an equilibrium state, a thermal bath, there is no flux of radiation. – Valter Moretti Aug 09 '20 at 07:51
  • A discussion on the various quantum states around black holes can be found in the introduction of this very long paper of mine and coworkers I wrote several years ago https://arxiv.org/abs/0907.1034. Advances in Theoretical and Mathematical Physics, Volume 15, Number 2, 2011 – Valter Moretti Aug 09 '20 at 08:11
  • @ValterMoretti Are you saying that Unruh photons hit the accelerating observer equally from all directions and with an equal average energy (not redshifted in any particular direction)? – Hiroyashu Aug 09 '20 at 08:15
  • I do not know, I never analysed these facts in detail, I am just saying that there is no preferred direction in the thermal bath experienced by the accelerated observer in Minkowski vacuum: no emitted radiation, since the state describes a thermal equilibrium differently from what happens with the Hawking radiation where the radiation exits the Killing orizon and nothing enters it. – Valter Moretti Aug 09 '20 at 08:19
  • @ValterMoretti Can a "thermal bath" with "no preferred direction" consist of photons that don't hit the observer equally from all directions? Also, can you please move your comments to an answer for me to accept? Thank you! – Hiroyashu Aug 09 '20 at 08:30
  • The geometrical setup is a delicate matter here: the rest space of the Rindler observer appears to be homogeneous, but the norm of the Killing time used to define the thermal equilibrium depends on the spatial non-Cartesian coordinate usually denoted by $\rho$. Therefore to answer your question one should fix adapted definitions. – Valter Moretti Aug 09 '20 at 08:34
  • @ValterMoretti Please consider posting an answer, even if incomplete. Your comments and links have been helpful enough. Thank you! – Hiroyashu Aug 09 '20 at 08:41
  • Ok @Hiroyashu done! – Valter Moretti Aug 09 '20 at 09:19
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    Good question Hiroyashu. – John Duffield Aug 09 '20 at 09:58

2 Answers2

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An interesting feature of the Unruh effect at any one point is that it is isotropic. So the intuition that the radiation comes from the horizon and "hits you in the back" is wrong. One way to interpret this is to say that the radiation comes from the horizon, rises up high, and then falls back down again, in such a way that the net result at any one point is isotropic. I note that Valter Moretti's good answer prefers to avoid the word "radiation" for technical reasons, but I think it remains an acceptable word in this context, as a way of discussing energy-momentum transfers between an observer and the field. The point here is that flux at the observer is isotropic, and I believe a detector which absorbs or reflects the radiation will undergo Brownian motion consistent with isotropic fluctuating illumination.

The temperature of the radiation is not homogeneous; it gets smaller as you move away from the horizon. The radiation arriving at any given height $x_1$ from other heights $x_2$ gets just the right Doppler shift to make it all arrive at $x_1$ with the same temperature and flux independent of what height $x_2$ it came from.

This feature of the Unruh effect is different from Hawking radiation. In the case of Hawking radiation, once you are far from the black hole, the radiation approaches you from the black hole and not the other way. For observers near the black hole horizon (within a Schwarzschild radius or two) the situation is more complicated.

Andrew Steane
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    This answer appears to directly contradict what Lubosz is saying here. An observer attempting to maintain a constant position in Schwarschild coordinates would need to be accelerating. At infinity, the observer experiences that Unruh radiation as Hawking radiation, coming from the direction of the black hole. That implies (to me) that, at closer distance, the observer would still see radiation coming from the direction of the BH, thus "hitting the observer in the back". – Linas Jun 06 '21 at 18:30
  • @Linas I'm not quite sure how you are using the term "at infinity" here, since the proper acceleration of an observer at rest in Schw coords goes to zero as $r \rightarrow \infty$. Such an observer has no Unruh effect, but does receive the (red shifted) Hawking radiation from the black hole if there is one. – Andrew Steane Jun 06 '21 at 19:00
  • @Linas In Schw black hole It is near the horizon that one can use an Unruh calculation to estimate the Hawking radiation, and the estimate becomes precise in the limit as one approaches the horizon. After allowing for red shift as the radiation propagates outwards, one then finds what is observed at infinity for the Schw case. – Andrew Steane Jun 06 '21 at 19:04
  • I reformulated all of my confusions into this question -- @Andrew Steane – Linas Jun 06 '21 at 21:05
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To answer the question, a comparison is convenient with the phenomenology of the Hawking radiation to make evident some crucial differences.

First of all, what actually happens for an accelerated observer moving in the Minkowski invariant state is that this state appears as a thermal bath with respect to the Rindler Killing time. A thermal bath is different from a radiation state: the former is an equilibrium state, the latter is not.

Strictly speaking the phenomenology is different from the one of an observer in the spacetime of a large black hole observing the Hawking radiation exiting from the horizon. There a net flux of particles exiting the horizon exists (with thermal properties with respect to the Schwarzshild Killing time) and nothing enters it. That is quite different from an equilibrium state, in fact the black hole eventually evaporates!

This situation, in the Kruskal extension of the Schwarzshild spacetime, is described by the so-called Unruh state of the background quantum field.

Conversely, what describes a thermal bath at the Hawking temperature is the so-called Hartle-Hawking state. Here the flux of particles entering the horizon is equal to the one of particles exiting the horizon.

What happens in the Rindler wedge for an accelerated observer in the Minkowski vacuum is an approximation of the phenomenology of Hartle-Hawking state (in accordance with the equivalence principle) and not of the Unruh state.

A crucial difference, distinguishing black hole phenomenology from Rindler phenomenology, is however that the particles of HH and U state around a black hole are standard particles. In the sense that, far from the black hole where the spacetime becomes flat, they are described by modes of standard QFT in flat spacetime.

Conversely, particles used to describe the thermal bath for the accelerated observer are Rindler particles without physical direct meaning. Their existence is furthermore confined to the Rindler wedge, so that their physical relevance is disputable. This does not automatically means that the abovementioned thermal properties do not exist since different theoretical descriptions of extended thermal states of a quantum field are at our disposal, in particular, the one relying on the KMS identity.

In summary, there is no experienced radiation of Rindler particles for an accelerating observer but a thermal equilibrium state of those particles takes place. A quantitative description of the effect of the action of this bath on physical devices is more difficult. In particular, the geometrical setup is a delicate matter here: the rest space of the Rindler observer appears to be homogeneous, but the norm of the Killing time used to define the thermal equilibrium depends on the spatial non-Cartesian coordinate usually denoted by $\rho$. Therefore to quantitatively answer your question one should fix suitably adapted definitions.

Valter Moretti
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    Interesting stuff Valter. For myself, I note Rugh and Zinkernagel’s 2002 paper on the quantum vacuum and the cosmological constant problem (http://philsci-archive.pitt.edu/398/) . They point out that photons do not scatter on the vacuum fluctuations of QED, saying if they did, “astronomy based on the observation of electromagnetic light from distant astrophysical objects would be impossible”. They end up saying the QED vacuum energy concept “might be an artefact of the formalism with no physical existence independent of material systems”. I wonder if the same is true for Unruh radiation. – John Duffield Aug 09 '20 at 10:08
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    Dear John, thank you for this ref. I gave up with qft in curved space time some yet ago (after 20 years). Presently I am dealing with pure mathematics on the one hand and experimental physics on the other hand (entanglement, quantum information) and, say, foundations of quantum theories. But I think I will find some time to read that paper since it seems quite interesting. – Valter Moretti Aug 09 '20 at 10:17
  • Very interesting your blog, and your last post! – Valter Moretti Aug 09 '20 at 10:19
  • Could you comment on the word "homogeneous" near the end? I think Rindler metric is not homogeneous, nor is Unruh bath, because the temperature depends on distance from horizon. The effects at any point are isotropic though. – Andrew Steane Aug 09 '20 at 10:29
  • Best not to mention it Valter. When I think of QFT and "the strong curvature regime " I think of Wheeler's geon, and then I take note of all the papers I see when I google on electromagnetic geometry. – John Duffield Aug 09 '20 at 10:31
  • @Andrew Strange I was referring to the spatial part $ds^2 =- \rho^2 d\tau^2 + d\rho^2 + dy^2 + dz^2$. You see that the spatial part in the said coordinates is the Euclidean flat metric. Non homogeneity arises from the temporal part (the norm of the Killing time), which depends on $\rho$. – Valter Moretti Aug 09 '20 at 10:37
  • Regarding the temperature, we have to distinguish between the one, uniform, referred to the Killing time,and the thermodynamical one measured with respect to the proper time, which is not uniform as you correctly point out. The difference is the well known Tolman factor. However I agree, the bath is not uniform due to the nature of Rindler Killing time. – Valter Moretti Aug 09 '20 at 10:39
  • @Andrew Steane, Sorry, the automatic corrector of my phone changed your family name to "Strange". It was not my intention to do it! I apologise. – Valter Moretti Aug 09 '20 at 10:44
  • Tolman factor https://en.m.wikipedia.org/wiki/Ehrenfest%E2%80%93Tolman_effect – Valter Moretti Aug 09 '20 at 10:49
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    No problem! And thanks for the answer. – Andrew Steane Aug 09 '20 at 10:49
  • @ValterMoretti "a thermal bath at the Hawking temperature is the so-called Hartle-Hawking state" - Does this mean that in addition to the Hawking radiation, there is also a thermal bath near a black hole? Thanks for the answer! – Hiroyashu Aug 10 '20 at 07:49
  • The H H state is quite unphysical as it needs the whole Kruskal spacetime to be defined, so also the white hole. The Unruh state only requires the black hole and the external region. Around a real black hole there is no thermal bath, only radiation exiting the horizon. – Valter Moretti Aug 10 '20 at 14:13
  • @JohnDuffield re the reality of the vacuum state: apparently the "Sokolov–Ternov effect" can be obtained by assuming it's an artifact of Unruh radiation. – Linas Jun 06 '21 at 18:02