(Note: I'm only considering flat spacetime in this question.) The Unruh effect is that the quantum state that looks like the vacuum in an inertial frame looks like a thermal bath of particles in a frame with proper acceleration $a$, with an effective temperature $T = \frac{\hbar a}{2 \pi c k_B}$. Presumably the accelerated observer not only sees Unruh radiation, but also interacts with it. Indeed, if the radiation is hot enough, then the accelerated observer will necessarily burn up. But I'm confused what this burnup process would look like back in the inertial frame, in which there are no particles except for the accelerated observer herself. For example, it seems to me that in the inertial frame, one might observe someone accelerating more and more, and then suddenly "spontaneously combusting." What would account for this "combustion" in the inertial frame?

Some people I've talked to have pointed out that "burning up" is not a very precisely defined concept at the scale of particle physics - fair enough. Here's a (hopefully) completely precise example problem (but feel free to ignore it if you think that the paragraph above already lays out the problem precisely enough to answer). Suppose you were to subject a single proton to a static, uniform electric field of strength $5 \times 10^{24}$ V/m.* Such a field would impart a force of $800$ kN on the proton and accelerate it at a proper acceleration of $5 \times 10^{32}$ m/s$^2$. The proton would then observe Unruh radiation at an effective Hagedorn temperature of $2 \times 10^{12}$ K, so it would deconfine and its constituent quarks would fly apart. But in the inertial frame, the temperature is still zero and there is no "real" thermal bath for the proton to interact with - it would just look like the proton disintegrated just because of the electric field. From the inertial frame, what would the deconfinement mechanism be?

Also see my related question Can you ride Hawking radiation away from a black hole?.

*Such a strong electric field would not be achievable in the real world, because it would induce vacuum polarization and charged virtual particle pair-creation that would reduce the field strength. But I think that if one were to consider a simplified model with only photons, gluons, and up and down quarks, then it would be theoretically possible.


3 Answers 3


In its own, non inertial, frame the charged particles in the body absorb radiation from what they perceive as a thermal state --- and so the body heats up.

From the point of view of an inertial-frame observer the charged particles in the body emit radiation and the recoil from this process makes the paritcles move with respect to one another --- and so the body heats up. Emission or absorbtion is a frame-dependent notion.

  • $\begingroup$ If the constituent charged particles are all identical and feel the same electric field, then presumably their radiation reactions would also be identical - so why would they move with respect to each other and gain thermal energy? $\endgroup$
    – tparker
    Feb 7, 2017 at 21:43
  • $\begingroup$ When a bunch of $\pm$ charged particles are accelerated together they would emit coherent classical radiation. The quantum process is more random. Now this guy emits a photon, then that one. Actually the emission even of classical of radiation from a uniformly accelerated particle is a bit complicated. There is a large literature on the subject. $\endgroup$
    – mike stone
    Feb 7, 2017 at 22:00

Leonard Susskind discusses a similar issue to this in the book An Introduction to Black Holes, Information, and the String Theory Revolution, in the context of explaining black hole complementarity.

Physical processes can look very different to different observers when there is an extreme difference in reference frames, especially if one of them is inertial and the other non-inertial. Just as it may be possible for a distant observer to see someone burn up in a wall of fire at the horizon of a black hole while a free-falling observer sails past the horizon without experiencing anything unusual, it may be possible that an observer in an inertial frame sees an ordinary proton while one in a highly accelerated non-inertial frame sees a quark-gluon plasma.

Even without a difference in reference frames, there are always different descriptions of the same physics when you consider different time scales--or equivalently--different energy scales. A Wilsonian understanding of the renormalization group in quantum field theory implies that there are effective field theories which yield the same measurable predictions for nearly all practical purposes as the more exact field theories defined at a higher energy scale (shorter time scale). For example, in some cases, there are "duality cascades" where electric charges take on the role of magnetic charges and vice versa. As you zoom in to shorter and shorter time scales and higher spacial resolution, the physics can start to look very different.

An example Susskind gives is the stability of the proton. Even though it is stable on a time-scale of at least $10^{34}$ years, boosting to a non-inertial frame where it takes that amount of time for less than an ordinary second to pass could cause an observer to see it decay into a positron or other particles. From the point of view of an inertial observer, the particles resulting from the "decay" are just temporary virtual particles contributing to the proton propagator, they aren't physically measurable. But to the non-inertial observer the proton may be highly unstable.


Some people I've talked to have pointed out that "burning up" is not a very precisely defined concept at the scale of particle physics

Perhaps, but I think there is a way of making this a bit more specific in a very straightforward manner that doesn't require the proton example. Regardless of what "burn up" means, it involves interacting with the particles in the Unruh thermal bath. Deep down, our question is how an inertial observer describes what a Rindler observer calls a "particle absorption". If an accelerated observer detects a particle, how does an inertial observer interprets this experiment? Notice that a thermometer is, in a certain sense, just a fancy particle detector.

This was studied by Unruh and Wald in Phys. Rev. D. 29 1047 (1984). Shortly, the detection of a Rindler particle (i.e., a particle seen by an accelerated observer) is interpreted by an inertial observer as the emission of a Minkowski particle. This can be roughly interpreted (pictorially) using the virtual particle illustration of Hawking radiation. The inertial observer interprets Rindler particles as vacuum fluctuations, which pictorially are particle-antiparticle pairs. Once a particle is detected, its pair is released as a real particle. Notice this is a pictorial interpretation meant to give some intuition, but shouldn't be taken too seriously—we are working in a quantum theory of fields, not of particles.

About protons

First and foremost, I think I should point out that the Hagedorn temperature might mean something completely different in this context. Being in thermal equilibrium at an accelerated frame or at an inertial frame are quite different things. For example, suppose a box filled with a gas in thermal equilibrium. If it is inertial, the gas is uniformly distributed. If it is accelerated, the gas is being pushed against one of the walls due to an inertial force. Hence, these two situations are not equivalent. Similarly, not necessarily the Hagedorn temperature in an accelerated frame means the same thing as in an inertial frame.

Furthermore, accelerated protons can decay to neutrons and leptons. In fact, this has been used by Matsas and Vanzella (arXiv: gr-qc/0205078 and references therein) as an argument that the Unruh effect is mandatory, or the calculation in an accelerated frame would not match the calculation in an inertial frame. I'm not sure of what is the proton's lifetime under the acceleration you are proposing, but this might be a related issue that either works to solve the problem (through a sequence of different decays) or makes it way more complicated.

Yet another issue is the Schwinger effect. Shortly, strong electromagnetic fields can pull real particles out of the vacuum, and create real particle-antiparticle pairs. According to Wikipedia, the order of magnitude of the electric field expected to create these effects seems to be of about $10^{18} \mathrm{V}/\mathrm{m}$. Hence, this should be extremely relevant in the situation you are proposing. Since this is a QED effect, it should apply (in principle at least) to any charged fermions, and hence might make the situation way more complicated. As with the proton decay, perhaps it helps solving the issue (maybe the quarks created by the Schwinger effect can hadronize into an antiproton and annihilate your initial proton), or perhaps it just makes the situation way more difficult to study.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.