Equivalence principle implies uniformly accelerated charge doesn't radiate?by johne1618 Tags: accelerated, charge, equivalence, implies, principle, radiate, uniformly 

#19
Jan413, 12:19 AM

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Here you are referring to Schwinger's book to say they do not radiate?? 



#20
Jan413, 12:30 AM

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#21
Jan413, 12:40 AM

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To me it shows how nontrivial this problem is that we are dipping into quantum mechanics in order to explain it. 



#22
Jan413, 12:47 AM

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Later in their series they reach the same technical conclusion regarding the observability of the radiation as the Almeida and Saa article that the_emi_guy (#6) and I (#5) posted. My view is that there is no paradox in the first place, since it arises from an abuse of the equivalence principle (which applies locally in curved spacetime). Nonetheless, the question is interesting as a classical special relativistic electrodynamics problem. 



#23
Jan413, 12:51 AM

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#24
Jan413, 01:17 AM

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#25
Jan413, 06:24 AM

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#26
Jan413, 07:20 AM

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#27
Jan413, 09:18 AM

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In GR, when we write the equivalence principle, roughly speaking we have to use the form of the laws of physics in which only first derivatives or lower appear. This is because higher derivatives are more "nonlocal" in the sense that each time you take a derivative, you take a difference of values at different points. Technically, the derivatives all exist at a point after taking the limit, but this is the idea behind higher derivatives being considered "nonlocal". 



#28
Jan413, 09:22 AM

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I think better example is the fall of a charged sphere due to gravity in the vicinity of the Earth surface, where the gravitational field is almost uniform. Consider two frames: 1) In the frame of the sphere, there is no gravitational force. The principle of equivalence and experience leads us to believe that both the motion of the sphere is uniform and its fields are given by the standard retarded solution of the Maxwell equations. That is, the sphere is at rest and the EM field in its vicinity is static. 2) In the frame of the Earth, there is gravitational force. The sphere accelerates, but this motion is referred to noninertial frame of reference. Therefore there is no easy way to find the EM fields of the sphere. The only clue we have is to go back to the situation 1), find the field there and get back to 2) by coordinate transformation. Now, since the field in 1) is frozen into the sphere, it seems very probable that the "free fall transformation" will make it into EM field which is comoving with the sphere. Of course, the field will be timedependent, as the charge moves. But I think we should not call it radiation. This is because in ordinary usage of that word, one has in mind some sort of propagation of phase or signal from the source to infinity, and in this case, there is such propagation. If the fall happens in the z direction, in the planes xy no propagation happens. In the direction z, there is propagation of the sphere, but this is motion with slow velocity, not an EM wave. 



#29
Jan413, 10:57 PM

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#30
Jan413, 11:54 PM

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So how does one apply the EP to charged particles? In the Lagrangian for all known charged particles, there are at most first derivatives coupled to the metric. So one can apply the EP to the Lagrangian by asserting that the form of the Lagrangian in curved spacetime is the "generally covariant" form of the Lagrangian in flat spacetime. This is known as the "minimal coupling" prescription. There's a discussion about similar issues in section 24.7 of Blandford and Thorne http://www.pma.caltech.edu/Courses/p...2006/text.html. 



#31
Jan513, 09:02 AM

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#32
Jan513, 11:07 PM

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I think the dipole is important because one could always create an electric dipole by separating + and  charges in a neutral body. Many molecules (eg. water) have an electric dipole moment. If a dipole in freefall radiates, then a great deal of matter in freefall must radiate. Mind you, there is likely a quantum mechanical threshold for radiation for an electric dipole ie. quantum harmonic oscillator. AM 



#33
Jan613, 09:35 AM

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But at some level, one might think the nonneutrality of water at small scales does come into play. I don't know the answer for what one might expect then. For example, is it a limiting factor in our ability to do extremely precise equivalence principle tests? 



#34
Jan613, 12:54 PM

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