Can you shield yourself from Unruh particles? In an old answer to a different question, Ron Maimon says yes:

You should think of the radiation as coming from the horizon--- if you place a refrigerated barrier between you and the horizon, you won't see any radiation past the barrier (at least not until it heats up). The reason is that the temperature of the barrier at the end furthest from the horizon forms the boundary condition for the Rindler Hamiltonian [...] and if it has a very long period in imaginary time, so does all the spacetime further along the Rindler x coordinate [...].

The barrier in his setup accelerates with you. What if it's inertial? For example, suppose there's a barrier suspended below you by a rope which you then cut. Does it instantly become transparent when it starts to fall toward the horizon? It seems that it must – it can't absorb the particles without continuing to heat up, and an inertially moving object in a vacuum can't heat up. But it's hard to believe that Unruh particles are similar enough to ordinary radiation that they seem to travel from the horizon to you, but different enough that they choose whether to interact with a solid object or completely ignore it based on the second derivative of its position.

Another possibility is that Ron Maimon is wrong and the particles don't travel from the horizon, but that seems no better. It would seem to imply, for instance, that if you're near a glowing black hole, you can't block the glow by closing your eyes, in stark contrast to ordinary blackbody radiation which seems to closely resemble Hawking radiation otherwise.

What really happens, and why isn't it as crazy as it looks?


2 Answers 2


For Unruh radiation, the typical wavelength is comparable to the distance to the horizon, because both quantities are determined by the acceleration. Contrast this with Hawking radiation, for which the typical wavelength is determined by the size of the black hole, even if the distance from horizon to observer is much greater than the size of the black hole.

I don't know what Ron Maimon meant by "You should think of the radiation as coming from the horizon," but notice his comment below that answer:

...the typical wavelength is about the distance to the horizon, so it is hard to establish direction of motion on the radiation...

A more general way to say this is that the typical wavelength is determined by the acceleration. For a quantitative example, consider an object with an acceleration of $a = 100$ km/s$^2$. That's an intense (bone-crushing!) acceleration by everyday standards, but the corresponding Unruh "radiation" still has a typical wavelength of $\sim c^2/a\sim 10^{9}$ km.

Can it be shielded? Well, introducing any additional material into the scenario will change the conditions and therefore possibly change the phenomena. The important point is that Unruh "radiation" can equally well be regarded as a local effect, as local as anything with such an enormous wavelength can possibly be. Introducing new material whose distance from the object is less than the wavelength can certainly have an effect regardless of the direction of the "radiation," so it's not clear to me that there is any paradox.

  • 1
    $\begingroup$ We do not know whether real black holes have true event horizons (this depends on features of quantum gravity currently unknown), but for purposes of astrophysics we could act as if event horizons do exist. Similarly, while observer/detector accelerating for a finite time may not have one a true Rindler horizon, effectively we can assume that it has. Reason: detector (e.g. Unruh-de Witt one) do not have long memory. $\endgroup$
    – A.V.S.
    Aug 13, 2020 at 20:43
  • $\begingroup$ @A.V.S. Good point. Whether or not the horizon/acceleration is temporary is not really important. The important thing is that the typical wavelength is comparable to the distance to the horizon (or to where the horizon would be), in contrast to Hawking radiation where the typical wavelength is determined by the size of the black hole. I edited the answer. $\endgroup$ Aug 13, 2020 at 23:59
  • $\begingroup$ Thanks. I understand what you're saying but I'm still confused by my rope-cutting example. I think I'd be happy with the claim that the particles interact nondirectionally or just fail to notice the barrier because of wavelength. What I don't understand is that they seemingly are blocked by it (directional interaction), then suddenly switch to ignoring it completely, even though it's the same size and in the same place and they have the same wavelengths as before. (I'm assuming of course that the barrier is much larger than the wavelength in the directions perpendicular to the acceleration.) $\endgroup$
    – benrg
    Aug 14, 2020 at 3:35
  • $\begingroup$ @benrg The barrier can't block anything unless it interacts with the field, and if we introduce a material that interacts with the field, then we've changed the problem. The original Unruh-radiation calculation is no longer valid. We need to do a whole new calculation with the material in place, and I don't know if that's ever been done. Just setting up the calculation would probably be enlightening. $\endgroup$ Aug 14, 2020 at 13:43
  • $\begingroup$ @benrg For perspective, remember that (1) Unruh radiation is in all directions, (2) the observer stops experiencing Unruh radiation as soon as the acceleration stops, (3) even classically, an accelerating charge radiates but a non-accelerating charge doesn't. In view of all this, I'm not sure the idea of a simple "barrier" is viable. We can't stop low-frequency radio waves by closing our eyes, and if the frequency is high enough that we could block them that way, we wouldn't survive: the distance from our retina to the horizon would be less than the diameter of our eyeball! $\endgroup$ Aug 14, 2020 at 13:43

A good way to calculate the Unruh effect is to do the calculation in an inertial frame. In that case it becomes a way of discussing the interaction of an accelerating detector with the ordinary Minkowski vacuum.

If a mirror is fixed in an inertial frame then it only makes a modest change to the vacuum state of the field so I think it does not destroy the Unruh effect for detectors accelerating away from the mirror. If the mirror is accelerating with you, on the other hand, then it makes a substantial change to the vacuum state and then I don't know the outcome but I suspect the Unruh effect may then vanish.

This paper is helpful in this context:

Quantum optics approach to radiation from atoms falling into a black hole Marlan O. Scully, Stephen Fulling, David M. Lee, Don N. Page, Wolfgang P. Schleich, and Anatoly A. Svidzinsky PNAS August 7, 2018 115 (32) 8131-8136; https://doi.org/10.1073/pnas.1807703115


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