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


by tom.stoer
Tags: charges, falling, field, free, gravitational, radiation, rest, static
tom.stoer
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#37
Feb13-12, 12:05 AM
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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

[tex]\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}[/tex]
reduces to

[tex]\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[/tex]
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)
universal_101
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#38
Feb13-12, 12:22 AM
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Quote Quote by clem View Post
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).
tom.stoer
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#39
Feb13-12, 12:37 AM
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Quote Quote by universal_101 View Post
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 Quote by universal_101 View Post
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.
universal_101
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#40
Feb13-12, 01:08 AM
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Quote Quote by tom.stoer View Post
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.
tom.stoer
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#41
Feb13-12, 01:16 AM
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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)
PAllen
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#42
Feb13-12, 01:49 AM
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Quote Quote by tom.stoer View Post
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?
tom.stoer
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#43
Feb13-12, 02:00 AM
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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.
strangerep
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#44
Feb13-12, 02:27 AM
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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.
TrickyDicky
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#45
Feb13-12, 03:45 AM
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Quote Quote by PAllen View Post
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 Quote by atyy View Post
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 Quote by PAllen View Post
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 Quote by universal_101 View Post
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 Quote by tom.stoer View Post
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 Quote by tom.stoer View Post
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.
tom.stoer
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#46
Feb14-12, 12:51 AM
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I was thinking about this post and #24 and I have to say that I don't understand your objections.

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

Quote Quote by TrickyDicky View Post
... 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 Quote by TrickyDicky View Post
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 Quote by TrickyDicky View Post
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 Quote by TrickyDicky View Post
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 Quote by TrickyDicky View Post
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.
lugita15
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#47
Feb14-12, 01:07 AM
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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.
tom.stoer
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#48
Feb14-12, 01:29 AM
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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)
TrickyDicky
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#49
Feb14-12, 05:20 AM
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Quote Quote by tom.stoer View Post


??? 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.
tom.stoer
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#50
Feb14-12, 06:00 AM
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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
Dickfore
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#51
Feb14-12, 06:23 AM
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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?
ApplePion
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#52
Jul22-12, 09:02 AM
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"I think one can safely neglect self-coupling of the charge with its own electromagnetic field)"

In some places that is a good approximation, but this is definitely a place where you should not use such an approximation.

When a particle radiates energy, energy goes into the electromagnetic field. Unless there is a radiation reaction force on the particle, the field gets energy for "free", and energy will not be conserved.
bcrowell
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#53
Jul22-12, 10:50 AM
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ApplePion, you're posting in a thread that's been inactive since February.

There is a huge literature on this subject, dating back to the 60's. Here are some references:

C. Morette-DeWitt and B.S. DeWitt, "Falling Charges," Physics, 1,3-20 (1964), http://www.scribd.com/doc/100745033/Dewitt-1964

http://arxiv.org/abs/gr-qc/9303025 -- Parrott, 1993

http://arxiv.org/abs/physics/9910019 -- Harpaz and Soker, 1999

http://arxiv.org/abs/0905.2391 -- Gralla, Harte, and Wald, 2009

http://arxiv.org/abs/0806.0464 -- Grøn and Næss, 2008

http://arxiv.org/abs/gr-qc/0006037

http://arxiv.org/abs/gr-qc/9909035

Some expositions:

http://www.itp.uni-hannover.de/~giul...Talks/Lyle.pdf

http://www.mathpages.com/home/kmath528/kmath528.htm
ApplePion
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#54
Jul22-12, 10:52 AM
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"ApplePion, you're posting in a thread that's been inactive since February."

So pretty much are you!

Best wishes.


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