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## Radiation of free falling charges and charges at rest in a static gravitational field

 Quote by atyy Won't the charge flow through the ground wire be different?
I don't know. If it were, then the energy for the flow has to come from somewhere, suggesting there would still be some back reaction on the charge in the free falling cage. But I don't know how to answer the question:

For a free falling 'ideal' Faraday cage, with a free falling charge in it, will grounding wire(s) (assumed to not interfere in any way with free fall) develop a current flow?
 Blog Entries: 1 Recognitions: Gold Member Science Advisor Actually, let's make this really simple. Suppose you have a spherical charge inside a conducting spherical shell of exact opposite charge. In an inertial frame in empty space far from gravity, there is no E or M field at all outside the sphere. What happens (theoretically) if this object is free falling in gravitational field? The fact that neutral matter of all different internal charge distributions obeys EP, suggests the above configuration would not radiate if falling in gravity. If this is so, it supports the model that violation of EP for a falling charge is due to distant field interaction.
 Recognitions: Science Advisor I would neither use the spherical charge (b/c then we would have an external electric field which modifies the equations of motion) nor the Faraday cage (b/c we don't need it in a universe w/o any external electric field - and we cannot measure the electromagnetic far-field generated by the charge). We take a spaceship in free fall; we place a proton and a neutron at exactly the same point (at rest w.r.t. the spaceship); we switch of the strong interaction between them and look what happens. We are not interested in the difference due to external fields. The modified geodesic equation $$\frac{d^2 x^\mu}{d \tau^2} + \Gamma^{\mu}_{\nu \rho} \frac{d x^\nu}{d\tau} \frac{d x^\rho}{d \tau} = \frac{q}{m} F^\mu{}_\nu \frac{d x^\nu}{d\tau}$$ reduces to $$\frac{d^2 x^\mu}{d \tau^2} + \Gamma^{\mu}_{\nu \rho} \frac{d x^\nu}{d\tau} \frac{d x^\rho}{d \tau} = 0$$ as long as we neglect backreaction of the electromagnetic field generated by the charged particle itself. Now we can safely separate two effects: a) a deviation of the worldline of the proton from the geodesic motion b) the electromagnetic far-field generated by the proton which we can observe in different (free falling) inertial frames Any deviation from geodesic motion can be observed locally w/o ever referring to the electromagnetic field. If we observe a deviation, then we can try to study the effect for the far-field. If we don't observe a deviation, then every effect in the far-field is due to different inertial (inertial) frames. Of course a deviation from geodesic motion would indicate a violation of the equivalence principle. And b/c we start with a geodesic equation for the charge (neglecting backreaction) it should be clear where we have to look for a correction of the equation of motion. Hope this makes clear what I have in mind. (I have to admit that I haven't checked the references; I'll do that asap)

 Quote by clem 1) Yes. 2) No. 3) Yes, but it could depend on your definition of 'geodesic motion'. I don't know if anyone has an answer yet to 4) and 5).
But I think that if the free-falling charge particles radiate then they are no longer free-falling!!

Because we have to account for the energy dissipated due to radiation, now this can only mean that object is not free-falling but experiencing a net force(radiation reaction force).

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 Quote by universal_101 But I think that if the free-falling charge particles radiate then they are no longer free-falling!!
Yes; I think so, too; but perhaps it's the other way round: if the particle is not free-falling, then this indicates that it radiates (w/o looking at the far-field).

 Quote by universal_101 Because we have to account for the energy dissipated due to radiation
The problem is that we can neither measure nor define this energy; b/c it's a Coulomb field the energy diverges both a r=0 and at r→∞; b/c we may have arbitrarily curved (expanding) spacetime we cannot define E = ∫d³x T00 even for well-behaved Tab; w/o a timelike Killing vector ka field (e.g. in an expanding universe) we cannot define the 4-vector Ja = Tab kb and again we do not have a reasonable definition of the el.-mag. field energy E = ∫ d³x J0

That's why I would like to get rid of the energy and the far-field.

 Quote by tom.stoer Yes; I think so, too; but perhaps it's the other way round: if the particle is not free-falling, then this indicates that it radiates (w/o looking at the far-field). The problem is that we can neither measure nor define this energy; b/c it's a Coulomb field the energy diverges both a r=0 and at r→∞; b/c we may have arbitrarily curved (expanding) spacetime we cannot define E = ∫d³x T00 even for well-behaved Tab; that means that w/o a timelike Killing vector field we cannot define the energy of the el.-mag. field
I believe that the above problem arises because we don't have a theory for radiation in classical electrodynamics itself. In short, we don't know how to deal with particles radiating due to motion in Electromagnetic fields.
 Recognitions: Science Advisor The problem may be due to a missing understanding of the self-energy. But even w/o this problem we would not be able to define energy (as a volume integral) in arbitrary spacetimes; this has nothing to do with the self-energy problem but simply with the geometry of spacetime. (I changed my post #39)

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 Quote by tom.stoer I would neither use the spherical charge (b/c then we would have an external electric field which modifies the equations of motion) nor the Faraday cage (b/c we don't need it in a universe w/o any external electric field - and we cannot measure the electromagnetic far-field generated by the charge).
I don't follow the objection to the spherical shell. I am proposing treating the charge and shell as one object free falling as a together. Inside the shell you have pure coulomb force; outside you have no E or M field at all (at least in an inertial frame far from gravity). What external field are you talking about?
 Recognitions: Science Advisor First there is an electric field inside the shell, so the particle at the center feels an additional force (this applies to the shell as well). Second the shell is not pointlike, so it does not follow a single geodesic but a family of geodesics and is subject to deformation. I do not say that the problem is uninteresting or wrong, but it's much more complicated.
 Recognitions: Science Advisor Tom, Regarding (approximate) motion of test particles in GR (possibly spinning, possibly charged), R.P. Kerr worked out a reasonable amount of this. R. P. Kerr, "The lorentz-covariant approximation method in general relativity.", [Series of three papers]: Il Nuovo Cimento (1959) Volume 13, Number 3, 469-491 Volume 13, Number 3, 492-502, Volume 13, Number 4, 673-689 It's an improved (i.e., lorentz-covariant) version of the old Einstein-Infeld-Hoffman method which showed (by solving the field equations) that test particles do indeed follow geodesics, (and hence this need not be postulated as a separate hypothesis). The Lorentz-covariant method reveals extra terms if the test particle has intrinsic spin. The final paper above extends the treatment to a charged test particle in an Einstein-Maxwell context.

 Quote by PAllen I wonder about this scenario. Suppose you had a free falling box that was a Faraday cage. Within it, you dropped a small uncharged body and small charge body. Would they diverge? Assume the Faraday cage can completely cancel EM field outside the box. Now you have no distant interaction.
 Quote by atyy OK, perhaps that is a local question, I'm not sure. But if it is, then I suppose it is more accurate to say that the EP does not apply because a charged particle cannot fall freely - it is acted on not only by the gravitational field, but also by the electromagnetic field. More generally though, it would still be non-local, because of the electromagnetic field.
 Quote by PAllen The fact that neutral matter of all different internal charge distributions obeys EP, suggests the above configuration would not radiate if falling in gravity. If this is so, it supports the model that violation of EP for a falling charge is due to distant field interaction.
 Quote by universal_101 But I think that if the free-falling charge particles radiate then they are no longer free-falling!! Because we have to account for the energy dissipated due to radiation, now this can only mean that object is not free-falling but experiencing a net force(radiation reaction force).
 Quote by tom.stoer Yes; I think so, too; but perhaps it's the other way round: if the particle is not free-falling, then this indicates that it radiates (w/o looking at the far-field).
It is comforting to see others share the concerns I raised in #24 about the problems of considering a charge as free-falling and how to distinguish it from a neutral particle in that case. After giving it some more thought I think these doubts come from thinking in classical (meaning pre-GR) terms about radiation and energy. According to GR as it is understood at least from the 50s-60s the answer is the one clem and Bill_K gave, and test charges follow geodesics and radiate, the hard part is to fully understand it intuitively, but thinking in 4-dimensions is not easy.
It is also discouraging that there seems not to exist a practical way to decide this experimentally.
 Quote by tom.stoer The problem may be due to a missing understanding of the self-energy. But even w/o this problem we would not be able to define energy (as a volume integral) in arbitrary spacetimes; this has nothing to do with the self-energy problem but simply with the geometry of spacetime. (I changed my post #39)
Maybe the self-energy problem has something to do with the geometry of spacetime after all.

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 Quote by TrickyDicky It is comforting to see others share the concerns I raised in #24 about the problems of considering a charge as free-falling and how to distinguish it from a neutral particle in that case.
Neglecting backreaction there seems to be no distinction.

 Quote by TrickyDicky According to GR ... test charges follow geodesics ...
Yes, neglecting backreaction they follow geodesics.

 Quote by TrickyDicky ... and radiate.
??? Whether they radiate or not (defined via a 1/r term) is not only a matter of the worldline of the charge itself but depends crucially on the oberver's frame and on global aspects of spacetime. So I can't they 'yes', I have to say 'it depends'.

 Quote by TrickyDicky It is also discouraging that there seems not to exist a practical way to decide this experimentally.
Yes, unfortunately.

In #24 you are writing
 Quote by TrickyDicky I can see the logic of your reasoning, but there seems to be a circular element to all this, at least the way I perceive it. When it is demanded the absence of ext. EM fields which is what I'd understand neglecting back-reaction means, how exactly is one still dealing with a charge in the context of curved spacetime.
There should not be any circular argument. You determine the worldline, then you fix a reference frame and calculate the 1/r term w.r.t. this reference frame.

[perhaps there's the possibility to calculate some kind of energy loss which could be defined w.r.t. a small spherical shell, but I doubt that is is possible in general]

 Quote by TrickyDicky Is it correct to equate in the context of curved spacetime geodesic motion with free-falling?
Of course yes (in the absence of other external fields). This is related to the fundamental principles of GR.

 Quote by TrickyDicky If yes, I would say backreaction can't be neglected and a charge cannot be considered free falling by definition.
To be clear about that: backreaction means an effect of the el.-mag. field created by the charge in its own worldline. This goes beyond GR, of course.

Your reasoning here is a bit strange. It's not that the question if free-fall corresponds to geodesic motion determines whether one can neglect backreaction, but that backreaction may cause deviations from free-fall i.e. from geodesic motion.
 Blog Entries: 1 Regarding experimental confirmation, it should be noted that the idea of radiation from charges accelerating with respect to you is not limited to classical physics; it remains in quantum field theory on curved spacetime, where it gives rise to so-called Unruh radiation: people on earth should be able to detect blackbody radiation from charges in outer space that are more or less in geodesic motion. (I take the point many in this thread have raised that back-reaction may lead to slight deviations from free-fall.) I think to date there has only been one experiment claiming to detect the Unruh radiation, and even that is disputed.
 Recognitions: Science Advisor But Unruh radiation is not the effect of a free-falling, not-so-free-falling or accelerated charge, but of an accelerated observer. Inertial observers will not detect Unruh radiation; non-inertial obversevers will detect Unruh radiation, but this is caused by vacuum in the absence of charges. So I don't understand how this is related. (I agree that backreaction could be studied using QFT in curved spacetime)

 Quote by tom.stoer ??? Whether they radiate or not (defined via a 1/r term) is not only a matter of the worldline of the charge itself but depends crucially on the oberver's frame and on global aspects of spacetime. So I can't they 'yes', I have to say it depends.
But is proper acceleration of the charge observer dependent?
I find it difficult to understand.
 Recognitions: Science Advisor no, proper acceleration is of course not observer-dependent; but if a charge is free-falling it is by definition not accelerated (w.r.t. a local inertial observer), nevertheless we discuss radiation; for a distant observer proper acceleration of the charge becomes meaningless - and in addition the radiation he observes may be due to his own motion i.e. observer dependent that's why I want to get rid of the observer and the far-field with its 1/r behavior at all
 Someone had pointed out that the static field of a point charge may become inhomogeneous and/or time-dependent in a grav field. I remember that a curved metric plays an analogous role to a(n inhomogeneous) dielectric constant. This means that the speed of propagation of em-waves is less than c. But, then, there arises the possibility of wakefields being created by charged particles traveling with speed higher than the corresponding phase velocity? Is my conclusion correct?