Mentor

## photons for accelerating charges and frames

I can't figure out the correct explanation for this gedanken.

1) We have a room in deep space which is shielded from external EM fields. The walls are lined with photon detectors. Inside is a charged ball attached to an insulating rod. The other end of the rod is on an axis and the rod is rotated so that the charged ball moves in uniform circular motion.

So far so good. I understand that the charged ball is continually accelerating and therefore is continually emitting photons which are detected by the detectors on the wall. Now, what if we change the experiment:

2) Same as 1) except that the detectors are mounted around the ball on the end of the rod.

Now, in the detector's frame the charge is stationary so there is no acceleration and therefore no emission of photons.

I know that different frames can disagree on distances, times, frequencies, energies etc., but I thought that all frames agreed on the existence of particles. What am I missing? Are the photons in the inertial frame somehow redshifted into non-existence in the rotating frame or is there some other explanation?

-Thanks
Dale
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 Quote by DaleSpam I know that different frames can disagree on distances, times, frequencies, energies etc., but I thought that all frames agreed on the existence of particles. Dale
This turns out not to be the case. The "unruh effect" is the best documented example, illustrating how the vacuum state for a non-accelerated observer appears to contain particles when viewed by an accelerated observer.

See for instance

http://en.wikipedia.org/wiki/Unruh_effect
 Mentor OK, so this Unruh effect is where an inertial observer sees a vacuum of absolute zero temperature and therefore no photons. At the same time an accelerating observer sees the same vacuum with some temperature and therefore detects photons in some black-body spectrum. So you are saying that my analysis of the experiment is correct, the inertial detectors detect photons and the rotating detector does not, so the two frames disagree on the existence of certain photons. Is there any consistent rule for which photons they will agree on and which they will disagree on? I mean, there is so much in relativity that is described by two observers in different frames emitting light pulses and measuring. The fact that they can disagree about wether or not photons were even emitted seems to make such gedanken's much more difficult. -Dale PS On the link they gave a formula relating temperature to acceleration. Is there any reason why it can't be used to claim that the cosmic background temperature indicates that the universe is accelerating at about 6.7E20 m/s^2

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## photons for accelerating charges and frames

 Quote by DaleSpam PS On the link they gave a formula relating temperature to acceleration. Is there any reason why it can't be used to claim that the cosmic background temperature indicates that the universe is accelerating at about 6.7E20 m/s^2
I know I am not accelerating at 6e20 m/s^2 because that would turn me into strawberry jam.

Therfore I must find some other explanation than the unruh effect to explain the CMB background radiation that I see.
 a good example is when you hover over a black hole, if you were falling into it you wouldnt see anything, but since you are accelerating compared to the outside you see light, depending on how close you are to the black hole
 Hi DaleSpam, Try reducing your thought experiment to just the magnetic field. The charged particle moving relative to an observer creates a magnetic field (seen by the observer). If however, the observer "rides along" with the charged particle so that there is no relative motion, the observer will experience no magnetic field. Clearly, the existence of the magnetic field is not absolute, it depends on the reference frame. Therefore so do photons since they consist of changing magnetic and electric fields. Regarding the photons' redshift: If an EM detector rode along with the rotating charged particle, it would measure just an electric field while the EM detectors on the walls would still “see” photons. However, if you increased the experiment’s apparatus radius to infinity so that the revolving charged particle’s acceleration approached zero, then the photons seen by the wall’s EM detectors would indeed redshift “out-of-existence”. Cheers, Redhat

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 Quote by Redhat Try reducing your thought experiment to just the magnetic field. The charged particle moving relative to an observer creates a magnetic field (seen by the observer). If however, the observer "rides along" with the charged particle so that there is no relative motion, the observer will experience no magnetic field. Clearly, the existence of the magnetic field is not absolute, it depends on the reference frame. Therefore so do photons since they consist of changing magnetic and electric fields.
Thanks Redhat! I had continued thinking about this principle quite a bit since I started the thread and basically also came to the conclusion that it is the fields that you need to consider, not just the acceleration of a charge. My specific thoughts went something like this:

In a circular loop of superconducting material you can set up a current that will not lose any measurable amount of energy, so it must not radiate any photons. However, the individual electrons in the circular current are in uniform circular motion and therefore are constantly accelerating. So how can one charge undergoing centripetal acceleration radiate while millions do not radiate? My conclusion was that the EM fields are time-invariant for the superconducting loop, but time varying for the individual charge. So the fact that a charge is accelerating is insufficient, only the time variation of the EM fields is important for determining if photons are radiated.

-Dale
 Hi Dale, Are you sure that electrons in a superconducting loop don’t radiate photons? There are no ohmic losses due to resistance but I can’t think of a reason why the electrons wouldn’t still emit synchrotron radiation. How much energy loss the superconducting loop’s photons produce is the question. In a synchrotron, the electrons are moving at nearly the speed of light so their centripetal acceleration is very high. A quick calculation shows that an electron moving at nearly (90%) the speed of light while emitting a 1 nm X-ray from synchrotron radiation loses about 0.25% of its kinetic energy. Not really that much of a loss. In a superconducting loop, the electrons are moving within a crystalline lattice so their speed is much slower and consequently their centripetal acceleration is much lower. The frequency (or energy) of the "synchrotron" photons is proportional to the electron's centripetal acceleration. I’m just guessing here but I would predict that (even though quite small) there are indeed energy losses due to synchrotron radiation in a superconducting loop and if given enough time would stop the current within it. Redhat

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 Quote by Redhat Hi Dale, Are you sure that electrons in a superconducting loop don’t radiate photons? There are no ohmic losses due to resistance but I can’t think of a reason why the electrons wouldn’t still emit synchrotron radiation. How much energy loss the superconducting loop’s photons produce is the question. In a synchrotron, the electrons are moving at nearly the speed of light so their centripetal acceleration is very high. A quick calculation shows that an electron moving at nearly (90%) the speed of light while emitting a 1 nm X-ray from synchrotron radiation loses about 0.25% of its kinetic energy. Not really that much of a loss. In a superconducting loop, the electrons are moving within a crystalline lattice so their speed is much slower and consequently their centripetal acceleration is much lower. The frequency (or energy) of the "synchrotron" photons is proportional to the electron's centripetal acceleration. I’m just guessing here but I would predict that (even though quite small) there are indeed energy losses due to synchrotron radiation in a superconducting loop and if given enough time would stop the current within it. Redhat
The problem here is that once you have a supercurrent, you can no longer treat this as having "individual electrons". The whole glob is now the "good quantum number". You don't have these "particles" going around in circles aorund the loop, but rather a condensate that has no well-defined position, analogous to why electrons in atomic orbitals don't radiate.

So this is not the same as what you get in a synchrotron.

Zz.
 Interesting point!

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 Quote by Redhat Are you sure that electrons in a superconducting loop don’t radiate photons? ... I’m just guessing here but I would predict that (even though quite small) there are indeed energy losses due to synchrotron radiation in a superconducting loop and if given enough time would stop the current within it.
Well, I work with superconducting magnets all the time (MRI), and if you keep them cryogenic there has never yet been any detectable energy loss. You are right that it is possible that we just haven't been observing superconductivity long enough to notice the energy loss that is there, so the words "yet" and "detectable" are definitely caviats. But I would think that someone would have looked specifically at detecting this kind of radiation.

 Quote by ZapperZ The problem here is that once you have a supercurrent, you can no longer treat this as having "individual electrons". The whole glob is now the "good quantum number". You don't have these "particles" going around in circles aorund the loop, but rather a condensate that has no well-defined position, analogous to why electrons in atomic orbitals don't radiate.
Woah, cool! Thanks a lot for the info. That makes sense. Well, it makes as much sense as anything else in QM anyway

-Dale

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