- **General Physics**
(*http://www.physicsforums.com/forumdisplay.php?f=111*)

- - **Radiation's equivalence principle**
(*http://www.physicsforums.com/showthread.php?t=141667*)

Radiation's equivalence principleWhy is that an accelerated charge radiates in a zero-valued
gravitational field, whereas a charge in free-fall in a non-zero gravitational field does not? Doesn't this contradict Einstein's Equivalence Principle? Are we in fact already assured, via measurement, of the free-fall case's outcome? |

Re: Radiation's equivalence principleIn article <SQSFf.10817$1n4.2693@newsread2.news.pas.earthlink.net>,
"Lord Snooty" <bonzo@dog.com> writes: > Why is that an accelerated charge radiates in a zero-valued > gravitational field, whereas a charge in free-fall in a non-zero > gravitational field does not? Doesn't this contradict Einstein's > Equivalence Principle? Are we in fact already assured, via measurement, > of the free-fall case's outcome? For the definitive account, read http://arxiv.org/abs/gr-qc/9303025 . |

Re: Radiation's equivalence principleIn <SQSFf.10817$1n4.2693@newsread2.news.pas.earthlink.net> Lord Snooty
wrote: > Why is that an accelerated charge radiates in a zero-valued > gravitational field, whereas a charge in free-fall in a non-zero > gravitational field does not? Doesn't this contradict Einstein's > Equivalence Principle? Are we in fact already assured, via measurement, > of the free-fall case's outcome? > Good question. The difference is that a charge falling in a gravitational field (which is accelerating relative to the position of the gravitating body) is in free-fall. "Free-fall" means that is not accelerating in its own frame it, it does not "feel" inertial forces. A charge that is accelerating in a zero-valued field must, even in its own frame, experience the energy difference from whatever is accelerating it, and react accordingly. To accelerate a charge in flat space requires another source of force (energy). Now, here is another thing to ponder. Why then does a charge that is 'stationary' in a gravitational field, say if it were sitting atop a table and in which case it is indeed "feeling" the accelerating forces, not radiate? Marc |

Re: Radiation's equivalence principleIn article <SQSFf.10817$1n4.2693@newsread2.news.pas.earthlink.net>,
Lord Snooty <bonzo@dog.com> wrote: >Why is that an accelerated charge radiates in a zero-valued >gravitational field, whereas a charge in free-fall in a non-zero >gravitational field does not? Doesn't this contradict Einstein's >Equivalence Principle? Are we in fact already assured, via measurement, >of the free-fall case's outcome? To take the last question first, I think the answer has got to be no. For gravitational accelerations of the amount we get around here in the solar system, the typical wavelength of radiation is a light-year, and the amount of energy radiated is really tiny. But the question is a very interesting one from a purely theoretical point of view. It gets rediscovered here on sci.physics.research fairly often. I suggest you dig through back issues of the newsgroup (say on groups.google.com), searching for the words "charged particle" and "equivalence principle". You'll find me, among others, going on at some length about it. The biggest chunk of the answer, I think, is that the equivalence principle is local -- that is, it makes sense only in small neighborhoods. Most of the time, when someone says that something is radiating, they're referring to a measurement that can only be made far away from the something. At small distances, you can't tell the thing's radiating, and at large distances, you can't use the equivalence principle. So everything works out OK. That's not the whole story. You can think of reasonable radiation criteria that can be defined in a small neighborhood. So it's quite a subtle problem! -Ted -- [E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.] |

Re: Radiation's equivalence principleIn article <dscjn4$jed$2@online.de>,
Phillip Helbig---remove CLOTHES to reply <helbig@astro.multiCLOTHESvax.de> wrote: >In article <SQSFf.10817$1n4.2693@newsread2.news.pas.earthlink.net>, >"Lord Snooty" <bonzo@dog.com> writes: > >> Why is that an accelerated charge radiates in a zero-valued >> gravitational field, whereas a charge in free-fall in a non-zero >> gravitational field does not? Doesn't this contradict Einstein's >> Equivalence Principle? Are we in fact already assured, via measurement, >> of the free-fall case's outcome? > >For the definitive account, read http://arxiv.org/abs/gr-qc/9303025 . I'll say what I always say when this comes up: 1. I heartily second Phillip's recommendation: Everyone who's at all interested in understanding this issue should read this paper. It lays out the relevant issues very clearly and makes arguments that are well worth considering. BUT 2. I don't think that the author has justified his main conclusion, which is that "purely local experiments" can distinguish a charge in uniform acceleration from a charge at rest in a uniform gravitational field. The purely local experiment he has in mind is a measurement of the force required to support the charge in its motion. If it's at rest in a uniform gravitational field, the force should be equal to the charge's weight. If it's accelerating (he argues), it should be more than that, because of the radiation reaction force. The problem is that he hasn't shown that the radiation reaction force for a uniformly accelerating charge is nonzero. The standard theory of radiation reaction forces on classical point charges, the Lorentz-Dirac equation, says that the force is zero in this situation. The Lorentz-Dirac theory is famously pathological, but it's not clear to me that the pathologies are relevant in this situation. Anyway, since there's not really any better competing theory, there's no good way to justify the claim that the force is nonzero. -Ted PS in case anyone's wondering: It may seem obvious that the radiation reaction force on a uniformly accelerated charge must be nonzero -- after all, the charge is sending electromagnetic waves off to infinity, and the energy's got to come from somewhere! But I claim it's not the least bit obvious. Start with a charge at rest in the lab, gently nudge it into accelerated motion, let it uniformly accelerate for a while, and then gently reduce the acceleration to zero, leaving it in uniform motion relative to the lab. Lorentz-Dirac theory says that there's a radiation reaction force only during the two "nudges" at the beginning and the end -- during the uniform acceleration, the total force required is the same as it would be for an uncharged particle of the same mass. And if you calculate the work done by the external force during the whole time, it works out right: The work equals the final energy (mc^2 + KE of the charge + radiated energy) minus the initial energy (just mc^2). Granted, Lorentz-Dirac theory is sick, but this is the sort of situation in which it's expected to be approximately correct (acceleration for only a finite period, all nudges smooth and gentle). Anyway, it's an existence proof that you can conserve energy even if the reaction force is zero during periods of uniform acceleration. -- [E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.] |

Re: Radiation's equivalence principleebunn@lfa221051.richmond.edu wrote:
> In article <dscjn4$jed$2@online.de>, > Phillip Helbig---remove CLOTHES to reply <helbig@astro.multiCLOTHESvax.de> wrote: >> In article <SQSFf.10817$1n4.2693@newsread2.news.pas.earthlink.net>, >> "Lord Snooty" <bonzo@dog.com> writes: >>> Why is that an accelerated charge radiates in a zero-valued >>> gravitational field, whereas a charge in free-fall in a non-zero >>> gravitational field does not? Doesn't this contradict Einstein's >>> Equivalence Principle? Are we in fact already assured, via measurement, >>> of the free-fall case's outcome? >> For the definitive account, read http://arxiv.org/abs/gr-qc/9303025 . > > 2. I don't think that the author has justified his main conclusion, > which is that "purely local experiments" can distinguish a charge in > uniform acceleration from a charge at rest in a uniform gravitational > field. I agree. Here's a simple gedanken that I think shows that no local experiment can do so, given the equivalence principle: Consider a small rocket with a bare charge suspended inside, and an observer who can measure the EM field at rest with respect to the rocket, anywhere within it; all other sources of the EM field are negligible. With the rocket far from any massive objects and thrusting with constant proper acceleration, at any given position within the rocket the EM field will be constant in time[#], so this observer cannot possibly _measure_ any radiation[@]. With the rocket at rest on a (large) massive planet, the equivalence principle implies the same result. So for both cases this observer must necessarily conclude there is no radiation[@], and therefore no radiation reaction and no observable difference between the two cases. [#] Consider the instantaneously-comoving inertial frame (ICIF) of the observer. In this frame the EM field is due to the retarded position of the charge, so the charge is slightly offset in position and has a small velocity relative to the observer in this ICIF. At any time this relationship remains the same (even though the ICIF is different) because the _proper_ acceleration is constant. Since the observer always measures the field in the then-current ICIF, the observer measures the field to be constant in time (even though it is not purely Coulomb). [@] Note that ExB is nonzero, but the absence of any time variation in either E or B implies no radiation. The corresponding transport of energy and momentum is of course constant in time and is offset by the mechanism suspending the charge. Note this depends on the rocket being small, and the acceleration being uniform. As I have remarked elsewhere, "radiation" can be a rather slippery concept; but the time variation of the EM field (or lack thereof) is definitive, and in this case gives a clear conclusion. I believe the above reference's error is in not basing the conclusion on _measurements_ made locally. Tom Roberts tjroberts@lucent.com |

Re: Radiation's equivalence principleIn article <1139856844.342067.266560@g44g2000cwa.googlegroups.com>,
Igor Khavkine <igor.kh@gmail.com> wrote: >Actually, the study of the motion of a point particle coupled to a >classical field has recently been a fairly active research topic. For >example, there is a review article by Eric Poisson published in Living >Reviews in Relativity: > >The Motion of Point Particles in Curved Spacetime >by Eric Poisson >http://relativity.livingreviews.org/...es/lrr-2004-6/ > >And it seems that, with a slight modification, the Lorentz-Dirac >equation is no longer pathological. The modification is described in > >An introduction to the Lorentz-Dirac equation >by Eric Poisson >http://arxiv.org/abs/gr-qc/9912045 Thanks for this information! I've been going around for years telling people that there is no satisfactory classical theory of point charges, but this makes it look like I was wrong. I seem to remember reading about the order-reduction trick Poisson describes and thinking that it wasn't a satisfactory way to remove the pathology in the Lorentz-Dirac equation for some reason. But I can't see what that reason might be at the moment -- it looks convincing to me. Given this information, I'll now strengthen my claim that Parrott is wrong in his about charged particles and the equivalence principle. I used to tell anyone who'd listen that he hasn't adequately shown that the reaction force on a uniformly accelerating charge is nonzero. Instead, I'll now say that, in the only sensible theory to apply in this situation, the force is definitely zero. >These are fairly recent developments. I wonder when they'll make it >into a newer edition of Jackson or some texbook on electrodynamics. Actually, Jackson does give the order-reduced version of the Lorentz-Dirac equation, although without a particularly careful or rigorous derivation like Poisson's. The nonrelativistic limit is in the main text, and the full equation is in the problems. I suspect it wasn't there in the edition I used in grad school (the one with the reddish cover), or else I think I'd probably have known about it. I could be wrong, though. -Ted -- [E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.] |

[Spam?] Re: Radiation's equivalence principleIn article <HPbHf.54823$PL5.41279@newssvr11.news.prodigy.com>,
Tom Roberts <tjroberts@lucent.com> wrote: >Consider a small rocket with a bare charge suspended inside, and an >observer who can measure the EM field at rest with respect to the >rocket, anywhere within it; all other sources of the EM field are >negligible. With the rocket far from any massive objects and thrusting >with constant proper acceleration, at any given position within the >rocket the EM field will be constant in time[#], so this observer cannot >possibly _measure_ any radiation[@]. With the rocket at rest on a >(large) massive planet, the equivalence principle implies the same >result. So for both cases this observer must necessarily conclude there >is no radiation[@], and therefore no radiation reaction and no >observable difference between the two cases. This is a correct argument showing that the equivalence principle implies no radiation reaction. The point of Parrott's paper is to question whether the equivalence principle is valid in this situation. In other words, he agrees with your argument, up to the last step. He then combines it with the assumption that the radiation reaction force is nonzero, and concludes that the equivalence principle doesn't apply. In other words, to use a word I may not have used since high-school geometry, he does the contrapositive of what you do (not-Q ==> not-P instead of P ==> Q). Personally, I'm with you. I just want to make sure Parrott's argument gets a fair shake. -Ted -- [E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.] |

Re: Radiation's equivalence principle<ebunn@lfa221051.richmond.edu> wrote in message
news:dsig3n$jv$1@bigbang.richmond.edu... > In article <dscjn4$jed$2@online.de>, > Phillip Helbig---remove CLOTHES to reply <helbig@astro.multiCLOTHESvax.de> wrote: > >In article <SQSFf.10817$1n4.2693@newsread2.news.pas.earthlink.net>, > >"Lord Snooty" <bonzo@dog.com> writes: > > > >> Why is that an accelerated charge radiates in a zero-valued > >> gravitational field, whereas a charge in free-fall in a non-zero > >> gravitational field does not? Doesn't this contradict Einstein's > >> Equivalence Principle? Are we in fact already assured, via measurement, > >> of the free-fall case's outcome? > > > >For the definitive account, read http://arxiv.org/abs/gr-qc/9303025 . > > I'll say what I always say when this comes up: > > 1. I heartily second Phillip's recommendation: Everyone who's at all > interested in understanding this issue should read this paper. It > lays out the relevant issues very clearly and makes arguments that are > well worth considering. > > BUT > ... The issue of radiation from a *uniformly* accelerating charge is clearly problematical, considering disagreements and the rather casual throwing around of claims that the EP may just not work for charges (er, wouldn't that mean complete revision of GR and replacement with...what?) However, I ask an even more challenging question: what if we had an oscillating charged body (not necessarily a "particle," but perhaps an insulated object) undergoing natural SHM inside a diameter tunnel through the earth. This body is in free fall, and "feels" like it is floating in space. There are tidal fields, but they are symmetrical per up-down distinctions. The charge must radiate in some external sense, because the local existence of changing EM fields (say, on the earth's surface) must propagate to distant locations. That implies energy transport to those distant locations. However, there is no reasonable basis for such a charge to feel the reactive self-force: F_rad = (2/3)k(q^2)(da/dt)/c^3. An accelerometer attached to the charge measures no acceleration, nor change in acceleration. Furthermore, because the actual coordinate value da/dt = -(4/3)G(rho)(dr/dt), any effect involving say tidal fields must be velocity sensitive! Hence, either: (1.) The charge behaves as expected in terms of simple application of the EP. It radiates but is not impeded by a self force. Thus it radiates energy for free. (2.) The charge is subject to the self-force that would apply in a non-gravitating system (i.e., as if it were in an antenna free of gravitational fields.) Energy is conserved, but we must (?) adjust GR to violate (?) the EP and explain how the charge can "sense" its actual coordinate acceleration changes. I wouldn't know which will apply, although I expect (1.) to be considered distasteful to many. Option (2.) if true may have some application to peculiar recent observations and commentary on charged particles near or falling into black holes (sadly I can't find the reference/s - can you?) It has odd consequences, such as a breaking force on charges moving at high speed through the matter of the universe (from equivalence to free-fall inside a planet) and acceleration of fast-moving charges approaching the earth etc. - and aren't we wondering about how some charged particles hitting our atmosphere can have so much kinetic energy? |

Re: Radiation's equivalence principleLord Snooty wrote:
> Why is that an accelerated charge radiates in a zero-valued > gravitational field, whereas a charge in free-fall in a non-zero > gravitational field does not? Doesn't this contradict Einstein's > Equivalence Principle? Actually, it *follows* from the Equivalence Principle. The "accelerating" in "accelerating charge", in all cases, is that taken with respect to a free fall frame since the Equivalence Principle states that free fall is to be taken as the curved spacetime analogue of inertial motion. |

All times are GMT -5. The time now is 12:52 PM. |

Powered by vBulletin Copyright ©2000 - 2014, Jelsoft Enterprises Ltd.

© 2014 Physics Forums