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Theory predicts that uniform acceleration leads to experiencing thermal radiation (so called Fulling Davies Unruh radiation), associated with the appearance of an event horizon. For non uniform but unidirectional acceleration the shape of the experienced radiation changes from thermal to other spectral densities, but also is predicted to exist. But suppose the acceleration is periodic and oscillatory, i.e. no permanent horizon persists? In particular, what about the case of harmonic motion, for a full cycle, half a cycle, etc.?

Here is an even simpler related problem that makes the apparent paradox easier to see. Suppose at proper time t=0, I accelerate at constant acceleration k in the x direction for t0 seconds, presumably experiencing Unruh radiation. Then I accelerate with acceleration -k, (in the -x direction,) for 2*t0 seconds, seeing more Unruh radiation coming from the opposite direction, and then I finish with with acceleration +k for the final t0 seconds. At the end of the 4*t0 proper seconds, I'm back where I started, at rest, without any event horizon. Was the Unruh radiation I felt when reversing acceleration secretly correlated or entangled with the radiation I initially and finally saw? Otherwise, from a more macro scale, I didn't actually necessarily move much, and the acceleration event horizon was instantaneous, evanescent and fleeting, so whence arose the Unruh radiation?

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this is a really great question - i would love to see a paper or reference if this has been calculated before –  lurscher Dec 24 '10 at 17:14
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I get your original question, but I don't get the paradox... –  Bruce Connor Dec 24 '10 at 23:03
    
You say "But suppose the acceleration is periodic and oscillatory, i.e. no permanent horizon persists?" - how does that follow? And when you say oscillatory are you speaking of acceleration going from $-a$ to $a$ or from $a_0 - \delta a$ to $a_0 + \delta a$, where $a_0$ is some non-zero acceleration? The two are distinct cases. –  user346 Dec 25 '10 at 9:13
    
-a to a in your notation, k to -k in mine, i.e. the former situation. –  sigoldberg1 Dec 26 '10 at 2:52
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I'm not sure what the paradox is supposed to be exactly. So there's no permanent horizon, the horizon is not a real physical thing anyway it's just an artifact of a particular coordinate frame. –  user1631 Aug 18 '11 at 22:41
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4 Answers

I will attempt a qualitative answer to the question.

We assume that we have an observer that is performing an oscillatory motion in which we have a repetition of two constant acceleration phases. The trajectory would be the sum of parts of hyperbola, where each part would correspond to a constant acceleration phase ($k$ or $-k$) with the acceleration directed towards the $x=0$. In this motion there will be no event horizon, but there will be several apparent horizons made out of the corresponding Rindler horizons of each part of the orbit that has constant acceleration.

There is a lot of work that supports the fact that Unruh/Hawking-like radiation can be observed from apparent horizons (see here, here and here for example). So my guess would be that if the essential features that Visser describes exist, then you would see radiation. Thus in this particular case, for each accelerated branch the observer would se thermal radiation corresponding to acceleration $k$ as long as the acceleration phase is long enough to satisfy the condition for slow evolution of the apparent horizon.

I am guessing that you can apply the same reasoning for the case of harmonic motion, as long as the period of oscillation is long enough. The spectrum of the radiation would evolve in time and would probably be something like an integrated Planck spectrum over a range of temperatures which would probably be something like a power law.

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why is important that the period of oscillation is long enough? long enough compared to what? –  lurscher Dec 28 '10 at 15:18
    
Vissers argument is based on the assumption that the geometry is quasi-static, so that there is no evolution in the timescale set by the frequency of the modes that you observe. So, in order to observe a thermal spectrum of some temperature, then you should have $\frac{kT}{\hbar}\approx\omega_{peak}\gg\frac{\dot{k}}{k},$ where $\omega_{peak}$ is the frequency of the peak of the thermal spectrum. That in our case translates to something like $\frac{k}{c}\gg\frac{\dot{k}}{k}.$ –  Vagelford Dec 28 '10 at 18:46
    
-1: "Apparent horizon" has a technical meaning which is not this, so you should say "approximate horizon". Except you can't, because you build your entire answer on this confusion. An apparent horizon is a closed trapped surface. –  Ron Maimon Dec 10 '11 at 16:31
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I am still not sure of the answer, but I suspect the following paper to be highly relevant
http://arxiv.org/abs/1108.3377 . "Entanglement Dynamics between Inertial and Non-uniformly Accelerated Detectors"

David C. M. Ostapchuk, Shih-Yuin Lin, Robert B. Mann, B. L. Hu

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This is very similar to the issue discussed by Peierls in his classic book "Surprises in Theoretical Physics". He looks at the apparent violation of the equivalence principal by (1) a uniformly accelerated charge vs. (2) a charge sitting at rest on the Earth's surface. IIRC he shows that the location of the spacetime region of the radiation from the accelerated charge in (1) doesn't exist in (2). Essentially the equivalence principal is a statement about small regions, while the radiation from (1) is found in asymptotic regions.

Another way of looking at Unruh radiation is that it is "the same as" the acceleration radiation of the charged particles of the thermal detector, as seen from a non-accelerating frame. This radiation would exist even for the oscillating case. I remember seeing a paper showing that synchrotron radiation could be derived as the scattering of the Unruh radiation seen by the accelerated particles. (Actually, this is just a 2-d version of the case you were asking about.)

The Unruh effect produces a thermal radiation bath that has the physical consequence of depolarizing the charged particles circulating in a synchrotron. See http://arxiv.org/abs/hep-ph/0610391. So there is an observable physical consequence in the purely oscillatory case.

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This is the basic idea, but the point is that the radiation which is emitted for the oscillating case is monochromatic and coherent, while for the uniformly accelerated case, the transitions in the accelerated object are entirely incoherent, and the outgoing radiation is hard to understand, because it is classically all boundary terms (the derivative of the acceleration is the radiation reaction, and it vanishes). The oscillating case isn't thermal, but should be approximately thermal in some large oscillation limit. –  Ron Maimon Dec 10 '11 at 18:39
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This is very mathematical deep, and probably needs lots of delicate calculations. However since most particles oscillating in some field, retain there quantum coherence I would think that dispite the Uhruh radiation supposedly coming from infinity, the radiation in the accelerating part of the cycle would need to be entangled to cancel that of the decelerating part of the cycle leaving the particle oscillating in the field without changing the oscillation. Perphaps though the Unruh radiation represents the Time-reverse (c.f. Feynman Wheeler Absorber theory), of the potential of the particle to radiate, as all accelerated charges do (but see Fooster above for the note about a fixed particle in a gravitional field).

If the Unruh heat bath does self correlation, then it would lead to quantum decoherence of any oscillating system, and non possible non unitary evolution of the quantum state of an oscillating particle. The Unruh radiation is deemed to coming from a Rindler horizon, but during each point in the oscillation cycle, this horizon is created moves and disappears again.

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