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GRDixon
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An isolated charge, accelerated by a constant force, theoretically radiates (Larmor). Does the same charge, held at rest in a gravitational field, constantly radiate?
DaleSpam said:How do you intend to detect the radiation?
GRDixon said:An isolated charge, accelerated by a constant force, theoretically radiates (Larmor). Does the same charge, held at rest in a gravitational field, constantly radiate?
Count Iblis said:I think this paper is better:
http://arxiv.org/abs/gr-qc/9303025
Put simply, if you're going to apply the equivalence principle, you have to do it correctly, and thus also look at the proper boundary conditions/initial conditions on the fields.
If the charge is held at rest, meaning mgh = constant, then where would the radiated energy come from? Wouldn't conservation of energy apply?GRDixon said:An isolated charge, accelerated by a constant force, theoretically radiates (Larmor). Does the same charge, held at rest in a gravitational field, constantly radiate?
Bob S said:If the charge is held at rest, meaning mgh = constant, then where would the radiated energy come from? Wouldn't conservation of energy apply?
Bob S
GRDixon said:Those were questions I had in mind when I submitted the thread. The majority consensus seems to be that the EP doesn't apply to electric charge. I'll give some thought to an analogous situation that doesn't involve electric charge. Thanks to all for responding, and for the links.
Well, I don't know about using heat to measure it. I was thinking about measuring it with an antenna. My guess is that a charge accelerating past a stationary antenna will induce the same signal in the antenna as an antenna falling past a stationary charge.GRDixon said:If you refer to the radiation emitted (or not emitted) in the gravitational field, I envisioned placing the charge in a blackened container and monitoring to see if the container's temperature increases (up to a limit).
Very interesting comment. My guess is that a charge moving past a stationary antenna at constant velocity will induce the same signal in the antenna as an antenna falling past a stationary charge at constant velocity. Both are Faraday induction of an electric field.DaleSpam said:Well, I don't know about using heat to measure it. I was thinking about measuring it with an antenna. My guess is that a charge accelerating past a stationary antenna will induce the same signal in the antenna as an antenna falling past a stationary charge.
Well, that is not exactly what I was saying, but that may be correct also. I really don't know.Bob S said:My guess is that a charge accelerating past a stationary antenna will radiate the same signal in the antenna as an antenna accelerating past a stationary charge.
In the absence of gravity case you would probably say that the energy comes from the work done by the rocket, and in the gravity case you would probably say that the energy comes from the lost gravitational potential energy of the antenna. But again, I don't have a good feel for it.Bob S said:So how does the charge then radiate energy?
kev said:Is it just coincidence that the article http://www.maxwellsociety.net/Charge%20and%20the%20Equivalence%20Principle.html" that you linked to in the other thread on the same subject, was authored by someone called G.R.Dixon?
bcrowell said:I like Chiao's thought experiment a lot better than the ones involving linear acceleration. I think it's conceptually simpler. For one thing, there's been some debate over the significance of horizons in the linear case. Boulware says that the radiation disappears behind the event horizon of the accelerated observer, where it can't be observed, but Parrott argues that this is wrong.
There are different ways of defining "local." After all, you can determine the Riemann tensor by "local" measurements, but the e.p. is specifically supposed to be talking about things that are "more local" than that. The e.p. basically says you can do a local linear approximation to the structure of spacetime, just like the freshman calculus idea of approximating a function with a line. But that doesn't mean that the second derivative isn't also a "local" quantity.pellman said:This business about local vs non-local.. the equivalence principle is local, Maxwell's equations are local (curved spacetime or flat), so what is there that is nonlocal?
I think the basic answer is: accelerate relative to what Isaac Newton would have considered to be a good, God-fearing inertial frame. This seems to give the right answer for all the cases I have in mind (charge lying on a table, charge in an accelerating elevator, charge orbiting the earth). The other rule of thumb that seems to work is just to check conservation of energy. All of these answers are completely inconsistent with how we usually think about the e.p., but I think they just go to show that the e.p. has fuzzy boundaries.pellman said:If we are asking, "Does a charged particle radiate when it accelerates?" we have to specify: accelerate relative to what?
bcrowell said:A good example of this is Parrott's criticism of Boulware's idea about how the radiation gets dumped behind a horizon where you can't see it. He says that the distinction between the near-field region and the radiation region is only an approximate one. He argues that any analytic function can be determined everywhere from knowledge of all its derivatives at a point, so an observer in the "near field" region can actually extrapolate to find the radiation field. IMO this is kind of over-reaching, because by the same argument you could do local measurements of all the derivatives of the metric (well, modulo gauge) and determine the metric everywhere in space.
atyy said:But the argument from derivatives is not so different from Chiao's argument that radiation can only be detected non-locally. Basically Parrott's, Gron and Naess, Chiao's etc resolutions are either EP does not hold/does not apply because it's intrinsically non-local/non-free-falling etc. I suppose there are subtle differences between "does not hold" and "does not apply", but I'd say those are all quite different from the resolution of say Almeida and Saa, or Peierl's (one of his surprises in theoretical physics) which says the EP does apply and hold, and the horizons save it.
bcrowell said:To me, the take-home thought from all this is that when we try to understand how gravity works, we understand it in terms of ideas like the e.p. and Mach's principle, but that those ideas are not crisp, well-defined mathematical statements. In general, all you can say is that our universe is highly e.p.-ish and fairly non-Mach-ish (in the sense that the Brans-Dicke [itex]\omega[/itex] parameter, which measures non-Mach-ish-ness, is constrained by the latest measurements to be at least 40000).
bcrowell said:If the equivalence principle applies regardless of charge, then these two particles must go on orbiting amicably, side by side. But then we have a violation of conservation of energy, since the charged particle, which is accelerating, will radiate electromagnetic waves (with very low frequency and amplitude). It seems as though the particle's orbit must decay.
bcrowell said:The resolution of the paradox, as demonstrated by detailed calculations by Gron and Naess http://arxiv.org/abs/0806.0464 is interesting because it exemplifies the local nature of the equivalence principle. When a charged particle moves through a gravitational field, in general it is possible for the particle to experience a reaction from its own electromagnetic fields. This might seem impossible, since an observer in a frame momentarily at rest with respect to the particle sees the radiation fly off in all directions at the speed of light. But there are in fact several different mechanisms by which a charged particle can be reunited with its long-lost electromagnetic offspring.
bcrowell said:An example (not directly related to Chiao's scenario) is the following.
Bring a laser very close to a black hole, but not so close that it has strayed inside the event horizon at rH. It turns out that at r=(3/2)RH, a ray of light can have a circular orbit around the black hole. Since this is greater than RH, we can, at least in theory, hold the laser stationary at this value of r using a powerful rocket engine. If we point the laser in the azimuthal direction, its own beam will come back and hit it.
Since matter can experience a back-reaction from its own electromagnetic radiation, it becomes plausible how the paradox can be resolved. The equivalence principle holds locally, i.e., within a small patch of space and time. If Chiao's charged and neutral particle are released side by side, then they will obey the equivalence principle for at least a certain amount of time --- and "for at least a certain amount of time" is all we should expect, since the principle is local. But after a while, the charged particle will start to experience a back-reaction from its own radiated electromagnetic fields. Since Chiao's particles are orbiting the earth, and the Earth is not a black hole, the mechanism clearly can't be as simple as the one described above, but Gron and Naess show that there are similar mechanisms that can apply here, e.g., scattering of light waves by the nonuniform gravitational field.
[later...]
The equivalence principle says that electromagnetic waves have gravitational mass as well as inertial mass, so it seems clear that the same must hold for static fields. In Chiao's paradox (p. 26), the orbiting charged particle has an electric field that extends out to infinity. When we measure the mass of a charged particle such as an electron, there is no way to separate the mass of this field from a more localized contribution. The electric field "falls" through the gravitational field, and the equivalence principle, which is local, cannot guarantee that all parts of the field rotate uniformly about the earth, even in distant parts of the universe. The electric field pattern becomes distorted, and this distortion causes a radiation reaction which back-reacts on the particle, causing its orbit to decay.
bcrowell said:So if the question is whether the equivalence principle applies to charged particles or not, I think the answer is not a yes/no answer.
However, a well-known paradox now arises when we ask the following
question: Is it the falling charged object, or is it the stationary charged object
at rest on the ground, that radiates electromagnetic waves?
On the one hand, a freely-falling observer, who is co-moving with the freely
falling neutral and charged objects, sees these two objects as if they were freely
floating in space. The falling charged object would therefore appear to him
not to be accelerating, so that he would conclude that it is not this charge
which radiates. Rather, when he looks downwards at the charged object which
is at rest on the ground, he sees a charge which is accelerating upwards with
an acceleration −g towards him. He would therefore conclude that it is the
charged object at rest on the ground, and not the falling charge, that is radiating
electromagnetic radiation.
On the other hand, an observer on the ground would come to the opposite
conclusion. She sees the falling charge accelerating downwards with an acceleration
g towards her, whereas the charged object at rest on the ground does not
appear to her to be undergoing any acceleration. She would therefore conclude
that it is the falling charge which radiates electromagnetic radiation, and not the charge which is resting on the ground. Which conclusion is the correct one?
Hi, kev -- No need to apologize; I explicitly asked for comments, so constructive criticism is most welcome :-)kev said:I hope you do not think I am being too critical of your book which I think is generally well written and presented and you can not be blamed for being a bit sketchy on this issue, as most of the literature on this isssue is also sketchy or downright contradictory.
kev said:You seem to be assuming that an orbiting charged particle must be accelerating, but that is not necessarily the case.
That wasn't what I intended. Maybe the edit above will clarify that. The back-reaction is a mechanism that allows it to *lose* energy.kev said:Here you seem to be saying that the orbiting charged particle can radiate energy and later recover its energy.
Hmm...from the point of view of nonrelativistic mechanics, we expect that given an initial-value problem, there is only one possible motion that simultaneously satisfies all the conservation laws. This comes up when I teach freshman engineering physics, because I use an unusual order of topics that does conservation laws first, and then Newton's laws. For the most part uniqueness holds pretty straightforwardly, but you do get some pathological cases that are harder to rule out. For instance, if you release a rock above the ground, you can satisfy conservation of energy and momentum in all inertial frames of reference if the rock simply hovers. I think the question of radiation by an orbiting particle is similar. It could satisfy all the conservation laws while failing to radiate, and yet it would fail to satisfy the field equations.kev said:Why not just say the charged particle radiates energy and as a result of the loss of energy, its orbit decays?
The answer is clearly no in the lab frame of reference, based on conservation of energy. It's conceivable that in other frames of reference, the answer is different, because of horizons.kev said:Actually the question posed by the OP can be paraphrased as "Does a charged particle that is stationary in a gravitational field (GF) radiate or not? Yes or no?" and you seem to have avoided that question.
As far as I can tell, all the papers agree that a charge sitting on my desk does not radiate. I think they also all agree that if you see a charge whiz by you, being accelerated by a rocket engine, it does radiate. These are both 100% solid conclusions based on classical electromagnetism, and they're consistent with conservation of energy. I think the real disagreement comes when you talk about an observer aboard the rocket.kev said:However, you are not alone, as most of the papers linked to in this thread avoid answering that direct question too.
kev said:Q1) Does a charged particle that is stationary in a GF radiate, from the point of view of an observer that is also stationary in the GF?
Q2) Does a charged particle that is stationary in a GF radiate, from the point of view of an observer that free falling in the GF?
But Gron and Naess do explicitly solve it.kev said:Unfortunately he does not explicitly solve the paradox in his paper and digresses off into a new design for a gravitational wave detector.
This possible frame-dependence of the radiative character of a wave is sort of what I had in mind in my answer to Q2 above. However, I don't think the Unruh effect is the right way to view this, because that's a purely quantum-mechanical effect (and far too weak to observe in any practical experiment anyone has been able to devise), whereas Q2 is posed as a classical problem.kev said:Nevertheless, one interesting possibiltiy is to declare both points of view as valid and not mutually exclusive. In other words radiation may be relative. What appears to be radiating to one observer, is not necessarily radiating according to another observer. If this seems too radical, consider the Unruh effect. An accelerating observer in flat space sees radiation in the form of photons and massive particles coming from an event horizon behind him, while an observer in the same flat space does not see or detect any of the radiation particles. That might seem paradoxical in itself, but the Unruh effect is considered to be a serious possibility by many physicists.
bcrowell said:my book, http://www.lightandmatter.com/html_books/genrel/ch01/ch01.html#Section1.5
atyy said:How did you get the photo of the butterfly on your forehead?
bcrowell said:
kev said:...
Does a charged particle ... radiate, from the point of view of an observer that ...?
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