A collection of accepted answers on Physics Exchange paint a seemingly inconsistent picture of Unruh radiation and it's isotropy. One of the most sophisticated answers, from Lubosz Motl, to the question Is Hawking radiation really the same as Unruh radiation states that Unruh radiation is the flat-space limit of Hawking radiation. Explicitly, an observer maintaining a fixed position in Schwarzschild coordinates is necessarily accelerating; that observer experiences Unruh radiation, and if the observer is at infinity, that radiation is interpreted as Hawking radiation.

The answer to Direction of Unruh radiation invokes the idea of a "Rindler wedge" and states that "Rindler particles (are) without physical direct meaning." I find this confusing: surely if I accelerate and carry a thermometer with me, it will quickly reach thermal equilibrium, and report a non-zero temperature? It is sometimes said that spacetime itself is falling into a black hole (that is certainly the interpretation invited by the Christodolou and Rovelli work at arXiv:1411.2854) and so, in that question, it is imagined that an observer accelerating above the horizon would experience a "headwind" much as a bicyclist might, yet curiously in the form of a thermal bath. Again, this seems to contradict the notion of Hawking radiation coming "from" the event horizon (a "tailwind"). I believe that the correct resolution is that one should not think of Hawking radiation as originating from some thin shell of distance epsilon above the horizon, but rather as momentum eigenstates of indeterminate position: this is why its not directional.

One last: in Why is there a flux of radiation in the Hawking effect but not in the Unruh effect? (and other questions), the answer provided by Lubosz states that "the static observer sitting on a heavy star doesn't see any radiation!" Yet other discussion suggests that an observer riding a rocket near that very same star would perceive Unruh radiation, because there is a horizon that forms due to acceleration. If there's a horizon, then surely it matters not if one is sitting on a chair on the surface of the heavy star, or is riding a rocket just slightly above it: either way, there is a horizon!

Insofar as Hawking radiation is a flux, one could, in principle, ride it on a solar sail, as, after all, I should feel a momentum transfer from it. Which should provide an acceleration. If I deploy my solar sail just above the event horizon, is that flux sufficient to nudge me, err, um, I'm not sure what I'm asking, but, uh, how would that work right at the horizon? Is it no longer a flux, there?

The core confusion appears to have something to with the Bogoliubov transformation, which "feels like it should be" a "local" transformation applied at the (accelerated) space-like tangent space of a manifold. Why does it "feel like that"? Because this tangent space is what provides the coordinate frame for Hamiltonians to be written in: a local symplectic manifold, over there, where one is accelerating, providing the coordinates, and then using those coords to write conventional 2nd-quantized (yet covariant!) QFT in that frame. That all sounds peachy-keen, except that those local coords need to be connected to, or transported to other, distant, inertial frames, in such a way that those field excitations which "fell and disappeared" behind the horizon are no longer present, leaving behind a real particle flux (at the temperature of the horizon), i.e. the Hawking radiation.

So maybe my real question is: is there a recommended review article? One that treats the topic from a geometric view, i.e. showing how to take a Hamiltonian/symplectic frame of reference, in the accelerated frame, and transport/connect the tangent vectors and 1-forms over to an inertial frame, such that those vectors and 1-forms that "fell behind the horizon" are no longer present in the inertial frame? (Maybe I need to work with jets, and not 1-forms, to get it right? some non-commutative thingy on a 2-jet?) Bonus points for doing it on generic pseudo-Riemannian spaces (won't work on Riemannian spaces, because it seems that at least one time dimension is needed) and super-duper bonus points for working with spin structures? (why? Because spin structures are cough cough "where fermions come from", roughly speaking, when one 2nd quantizes them.)

• Closely related: physics.stackexchange.com/q/634972, and physics.stackexchange.com/q/75750 – Chiral Anomaly Jun 6 at 22:00
• You asked for a recommended review article... Review of which part? There are lots and lots of reviews of the Unruh effect, the Hawking effect, and their relationship (I cited a handful of them in this long and unsatisfying answer), but all of the reviews that I've seen share the same limitations: they either assume that the quantum field is a free field, and/or they make specific assumptions about the quantum state. Because of those assumptions, they don't directly answer questions about how the quanta interact with other material entities. – Chiral Anomaly Jun 6 at 22:04
• Thanks, will read. My understanding of GR is from the viewpoint of conventional Riemannian geometry: frame bundles, spin connections. It's highly geometric; such texts never use the word "acceleration" cause everything is coordinate-free. I know many approaches to QFT, again, the word "acceleration" never appears. For the present case, it feels like some Hamiltonian formalism applied in a geometric setting would work best; in a Hamiltonian setting its easy to say "these are my creation annihilation operators." and that horizons would cut some of them. Particle interactions don't seem relevant. – Linas Jun 7 at 1:36

I should say as a disclaimer that I am pretty confused by Unruh radiation myself, but I think I can answer part of your question.

The idea that the radiation comes from near the horizon, and can be blocked by a thermal insulator below your detector, isn't inconsistent with isotropy. If you imagine Unruh radiation as ideal gas particles that are thrown upward from near the horizon with some continuous distribution of velocities, then in the Rindler case, every particle that crosses a surface at a fixed height in the upward direction will (with probability 1) later cross it in the downward direction, and with the same kinetic energy. So the temperature that you see at a point will at least have an up/down symmetry, and plausibly could be isotropic. Nevertheless, if you imagine the particles to be interacting, they can be blocked on the way up, and by your skin.

Black hole horizons differ from Rindler horizons in that the escape velocity outside the horizon is less than $$c$$. Radiation that falls back down through a surface will still have the same energy as on the way up, but some of it never falls back, so there is a net upward energy flux (and pressure, if you consider the particles to be interacting).

All that said, it clearly can't be the case that Unruh radiation comes from event horizons, or requires them at all. If you move inertially until some time, then accelerate uniformly for a while (longer than $$c/a$$), then move inertially again, you clearly must see Unruh radiation while accelerating (if Unruh radiation exists in the real world at all, and doesn't violate causality), but there is no event horizon. So it must be sufficient for an apparent horizon (somehow defined) to exist, but now you have an apparent problem, since the difference between temporary acceleration in Minkowski space and, say, standing on the surface of a gravitating sphere seems like it should be one of degree and not kind. But there's a convincing argument that you should see precisely zero Unruh/Hawking radiation in the latter case, namely that heating your detector costs energy which must come from somewhere, and there's nowhere it can come from that is independent of the composition of the sphere. I don't know how to resolve that paradox, so I can't answer the rest of your questions.

• According to Einstein's equivalence principle, there should be no difference whatsoever between standing on the surface of a gravitating sphere, and accelerating in Minkowski space. This principle has so far held as a golden, unassailable rule (cf. Eotvos experiment) but must surely fail for QGR to work. A pop-sci description of why is here: Clarke, Stuart (19 January 2013). "Sacrificing Einstein: the keystone of relativity that just has to go" (PDF). New Scientist. 217 (2900): 32–36. – Linas Jun 8 at 13:52
• @Linas I don't believe that QG has to violate the equivalence principle. Even classically, an accelerating charge in Minkowski space radiates but a charge on a gravitating sphere doesn't. The difference has to do with global properties of the field and with how energy is defined. Unruh radiation is nonlocal in a similar way (characteristic wavelength comparable to $c^2/a$) so I imagine there's a similar resolution to the paradox I mentioned. – benrg Jun 9 at 3:03

An observer staying fixed in the spacetime of a black hole is different from an observer in an accelerated frame of reference. Locally they look pretty much the same but the difference is that in the black hole spacetime the observer will see a huge tidal force which makes the radiation higher (and real!) than that in the Unruh effect. That is why both are not the same, and the Unruh effect is said to be Hawking radiation in the flat spacetime limit.

To be honest I don't think Unruh radiation is real radiation. If it heats a thermometer up in an accelerated frame and the frame is suddenly stopped what happens to the thermometer? Will it return to its value for the non-accelerating frame? From where comes the energy that makes the radiation real? A virtual pair of particles can only be made real if tidal forces are present, not if the same acceleration acts on both. so the claim that for an accelerated observer the perceived vacuum contains real particles because creation operators are present isn't true. Particles can't be created. How? There is now metric that allows this. Only around a BH this metric is present.

• At first, you seem to suggest I should discard the principle of equivalence in some way. On closer reading, you suggest that Hawking radiation is entirely due to tidal components!? – Linas Jun 6 at 21:52
• @Linas No we are not, but the equivalence principle holds only locally. For spacetime points there is no difference, but for different spacetime points there exists a tidal force that is not observed in an accelerated frame in flat spacetime. So HR is due to both gravity and tidal forces. – user303670 Jun 6 at 21:52
• Even for a single space-time point, the stress-energy tensor will, in general, have tidal components. I don't know how to make use of that, to convert vacuum-state quanta into physical quanta. – Linas Jun 6 at 22:06
• @Linas Do you mean of diagonal components of the stress energy tensor, due to rotation? In the case of a stationary BH there are none of these (only for a rotating one, giving rise to frame drag). – user303670 Jun 6 at 22:12
• @Linas Hawking won the Noble prize for the conversion you have in mind! HR can't be produced in an accelerating frame in flat spacetime. – user303670 Jun 6 at 22:14

I think there a confusion here between equilibrium Hawking radiation, in which the black hole is in (unstable) equilibrium with the a heat bath at the Hawking temperature, and the non-equilibrium situation in which the black hole is radiating into asymptoticaly cold space. Unruh radiation is more closely related to the former than the latter.

Having read a couple of existing answers, I will throw in some items not mentioned yet, but this answer will be another incomplete one.

Regarding the Unruh effect first of all, and the sense in which it is real, it is good to notice that insight into these phenomena can be obtained by considering an inertial reference frame, and looking at correlation functions involving the field and an accelerated body (e.g. particle, detector, atom) which can interact with the field. By "field" here I mean any quantum field, e.g. scalar or electromagnetic, which can be taken to be in its ground state in the initial conditions. One thus finds that the accelerated body has non-zero quantum amplitude to become excited and subsequently to emit back into the field (e.g. emit photons in the electromagnetic case). So there is a transfer of energy; the source of this energy is ultimately whatever is providing the force to accelerate the body.

The second piece of information I have to offer is a standard result but worth mentioning. It is a combination of the following facts:

(1.) The Schwarzschild metric (which arguably should be named "Schwarzschild-Droste") is equal to the Rindler metric in the vicinity of any given event, up to some approximation.

(2.) The red shift between any given place and $$r \rightarrow \infty$$ is finite in the Schw case but shifts all the way to zero frequency in the Rindler case.

(3.) The quantum effects at the heart of either Hawking or Unruh effects are non-local, in the sense of requiring an analysis over an extended region not just at a point. In particular, Hawking radiation is produced from an extended region around a Schwarzschild black hole, not just from the horizon. To get an impression of this, note for example that in a simple "tunneling" picture, a particle can tunnel from inside the horizon straight out to places well outside the horizon. The significant region extends, as I understand it, out to $$r - r_{\rm s} \sim O( r_{\rm s} )$$ where $$r_{\rm s}$$ is the Schwarzschild radius.

(4.) Having said that, in the limit as the location approaches the horizon, it is sufficient, to calculate the Hawking temperature, to restrict the analysis to a smaller range of radii, and it then turns out that the approximation involved in (1) becomes more accurate. One can in this way obtain the Hawking temperature by taking the Unruh temperature in the limit as one approaches the horizon (for an observer held fixed in the Schwarzschild coordinates) and then applying the Schwarzschild (N.B. not the Rindler) red shift as the radiation propagates out to the asymptotic region.

This is a second answer from me, in which I explicitly quote and respond to things asked in the question.

"Explicitly, an observer maintaining a fixed position in Schwarzschild coordinates is necessarily accelerating; that observer experiences Unruh radiation, and if the observer is at infinity, that radiation is interpreted as Hawking radiation."

No, that's wrong. An observer maintaining a fixed position in Schwarzschild coordinates at $$r \rightarrow \infty$$ receives Hawking radiation at the Hawking temperature if there is a horizon at $$r_{\rm s}$$ and the whole setup has existed long enough for Hawking's calculation to apply. Such an observer has zero proper acceleration so the Unruh temperature associated with their proper acceleration is zero; i.e. they experience no Unruh radiation.

"Can Hawking radiation be blocked by a thermal blanket (or a lead brick) or is it experienced as a bath?"

At positions far from the black hole the Hawking radiation is just like radiation from a star; it can be blocked. Near the horizon ($$r - r_{\rm s} \sim O(r_{\rm s})$$), though, it is not so clear. I think the source of the radiation is throughout this region but I am not sure of the details.

"A freely falling observer experiences no Unruh radiation"

this is correct but it is an observation that can only be applied rigorously in flat spacetime. In a curved spacetime this observation can serve as a rough guide for distance and times scales small compared to the local Gaussian curvature of spacetime. This means it does not contribute much to an understanding of a freely falling observer near a Schwarzschild horion.

"Yet, near a BH, there is still an event horizon for the freely falling observer, and thus the observer would necessarily observe Hawking radiation (but not Unruh radiation). Is that correct?"

Yes. As I understand it, the observer in radial free fall registers a temperature of order of the Hawking temperature. Such an observer does not have long to measure such a temperature however (not much proper time before they pass the horizon). The reason why the equivalence principle is not holding is that the physical effect in question (i.e. the quantum tunneling or whatever you wish to call it) involves distances of the order of the Schwarzschild radius, and on this distance scale the observer's frame is not inertial.

"Rindler particles (are) without physical direct meaning."

This is a half-truth. In the Unruh effect the excitation of the detector is real enough. It gets warm. But you are not obliged to say that this is because it absorbed "particles".

"the static observer sitting on a heavy star doesn't see any radiation!"

This seems plausible at first, but on further thought I am not sure, because this observer has non-zero 4-acceleration. Should one expect the Unruh effect for such an observer? On a flat spacetime model one would say yes, but perhaps it is not correct to neglect the spacetime curvature: I don't know.

"Yet other discussion suggests that an observer riding a rocket near that very same star would perceive Unruh radiation, because there is a horizon that forms due to acceleration."

Again I am not sure, but I would not rule out that this case might be different from the static observer sitting on a star because, once again, the effect concerns distance scales where spacetime curvature cannot be neglected.

Finally, to return to the core question: does Unruh radiation heat a body internally not just from the surface and can you shield it? I don't know, but I think the method of calculation where one considers an accelerating detector in Minkowski vacuum suggests the Unruh effect occurs throughout the bulk of a body and so cannot be shielded. However one should note that if you add a mirror to the overall setup then this changes the quantum vacuum state. Calculations have been done to some extent but I have not done them.

I have found the following helpful:

Marlan O. Scully, Anatoly A. Svidzinsky, and William Unruh Phys. Rev. Research 1, 033115 – Published 20 November 2019

Enhancing Acceleration Radiation from Ground-State Atoms via Cavity Quantum Electrodynamics Marlan O. Scully, Vitaly V. Kocharovsky, Alexey Belyanin, Edward Fry, and Federico Capasso Phys. Rev. Lett. 91, 243004 – Published 10 December 2003

https://arxiv.org/abs/1906.01729