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

[tom.stoer]

The question is rather old but I am not sure if we have a final conclusive answer (I was not able to figure it out for myself, nor have I found consensus in the literature)

1) do free-falling charges in a static gravitational field radiate?
2) do charges at rest in a static gravitational field (= accelerated charges w.r.t. to geodesic motion) radiate?
3) would a free-falling, radiating charge imply deviation from geodesic motion?
4) how would we define radiation in GR (in contrast to distortion of el.-mag fields or simple frame-dependent effects)? I mean in SR a charge with constant velocity w.r.t. to an inertial frame does not radiate but it "generates" a magnetic field, an effect which can be understood looking at Lorentz transformations solely (so I wouldn't call this radiation)
5) what are the relevant equations in GR to describe this coupling of motian in a gravitational background coupled to el.-mag. fields

 

[Meir Achuz]

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).

 

[Bill_K]

4) The 1/r coefficient in an asymptotic expansion of the field at infinity.
5) The Einstein-Maxwell equations.

 

[tom.stoer]

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).

If we are not able to answer 4) or 5) how we answer 1-3)
1) why? a free-falling charge does not feel any acceleration
2) why? a charge at rest in a gravitational field does feel acceleration (of course I am with you, static electros on earh do not radiate ;-)
free-affling charge does not feel any acceleration

4) The 1/r coefficient in an asymptotic expansion of the field at infinity.
5) The Einstein-Maxwell equations.

4) great; I was thinking about non-curl-free electric fields, but this is rather difficult in contrast to 1/r
5) I was thinking about a geodesic equation which defines the motation of a pointlike charge; this charge would then act as a delta-function source term to Maxwells equation in a gravitational background

Any references?

 

[Ben Niehoff]

Motion of test particles under the EM force goes like

$$\frac{d^2 x^\mu}{d \tau^2} + \Gamma^{\mu}_{\nu \rho} \frac{d x^\nu}{d\tau} \frac{d x^\rho}{d \tau} = \frac{q}{m} F^\mu{}_\nu \frac{d x^\nu}{d\tau}$$
where q/m is the charge-to-mass ratio.

Also, radiation becomes trickier to define in spacetimes that are not asymptotically flat...

 

[Ben Niehoff]

If we are not able to answer 4) or 5) how we answer 1-3)
1) why? a free-falling charge does not feel any acceleration
2) why? a charge at rest in a gravitational field does feel acceleration (of course I am with you, static electros on earh do not radiate ;-)
free-affling charge does not feel any acceleration

It seems to me that it is feasible that the relative acceleration between emitter and observer could be involved. We know from other contexts (e.g. Unruh effect) that merely being in an accelerated frame can change one's notion of whether one is seeing "radiation". Maybe acceleration plays some clever trick with the EM field...

 

[tom.stoer]

It seems to me that it is feasible that the relative acceleration between emitter and observer could be involved. We know from other contexts (e.g. Unruh effect) that merely being in an accelerated frame can change one's notion of whether one is seeing "radiation". Maybe acceleration plays some clever trick with the EM field...

Of course in the case 1) I would not consider any acceleration, so both charge and observer would follow geodesics ...

... but do charges follow geodesics (question 3)?

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 the usual one with F=0 (b/c I think one can safely neglect self-coupling of the charge with its own electromagnetic field)

 

[Ben Niehoff]

Yeah, I really have no idea. My impression is the same as yours, that these are difficult problems that no one has been able to work out for sure.

 

[Bill_K]

1) why? a free-falling charge does not feel any acceleration

A charge, freely falling or not, is an extended object.The stress-energy associated with it is not concentrated at a point, it extends to infinity in all directions. Even to the extent that part of it is on the other side of the Earth. Different parts of the charge experience different accelerations. In a fully accurate treatment this must be taken into account. To say that the charge is a singularity that follows a geodesic, or even a geodesic plus force term, is an approximation. Radiation produced, if any, is so slight that considerations like this are necessary.

 

[Bill_K]

Yeah, I really have no idea. My impression is the same as yours, that these are difficult problems that no one has been able to work out for sure.

Bryce DeWitt gave one of Dicke's seminars at Princeton on this topic back in the 60's. If there is anyone who's conclusions I trust, it is DeWitt. But I don't know if or where it was published.

EDIT: Probably this: (With Robert W. Brehme) “Radiation Damping in a Gravitational Field,” Annals
of Physics (N.Y.), 9, 220–259 (1960)..

 

[tom.stoer]

Bill, I certainly agree that strictly speaking this requires Einstein's equations with gravity coupled to the stress-energy tensor of the el.-mag. field plus Maxwell's equations.

But in classical electrodynamics (incl. special relativity) one can answer these questions by using pointlike charges acting as sources for the el.-mag. field. My idea was to try something similar - except for taking spacetime curvature into account. I agree that this may not be fully correct, but
- this is the same approx. as in Maxwell's theory
- this is the same approx. as for the geodesic equation in GR

Which terms would be neglected?
- I think only backreaction of the el.-mag field on spacetime curvature

So as a first approx. this should be sufficient.

 

[tom.stoer]

I found this comment in one blog

"John Wheeler once asked a group of relativity theorists to vote on whether the falling charge radiates or not; their responses were split almost precisely down the middle."

 

[Bill_K]

Electromagnetic radiation is caused by time-varying electromagnetic fields. In classical electromagnetism we can throw the time variation over onto the charges, and it's possible and highly convenient to talk about the radiation being caused by moving charges. In general relativity we can no longer do this and must deal directly with the fields.

Here's an example. Take a Coulomb field, which of course is spherically symmetric. Suppose we could somehow squash the field pole to pole so that it becomes oblate. Although the charge has not moved, the field now has a quadrupole moment. Now alternately squash and release, and we have a time varying quadrupole field. And by squashing we are doing work on the field, which shows up as quadrupole electromagnetic radiation.

How could the field get squashed? Suppose the charge is immersed in a plane gravitational wave. The charge itself won't accelerate, but the time-varying quadrupole gravitational field will see the stress-energy in the Coulomb field and cause a time-varying quadrupole oscillation, as above. The gravitational wave does work on the electromagnetic field, which shows up as electromagnetic radiation.

The moral is that in general relativity, charges often find themselves in inhomogeneous surroundings. The Coulomb field is a nonlocal, spreadout thing, and what happens to it depends not only on what happens to the singularity, e.g. whether the singularity is accelerating or not, but also what is happening to the outskirts of the field. Shaking the outskirts can produce electromagnetic waves as well. And so to answer the question "does the charge radiate", it's a function not only of the spacetime curvature at the singularity but the spacetime curvature everywhere, or at least in a vicinity.

You can't write an equation of motion for the singularity and expect that to answer the question. The charge has internal degrees of freedom which feel the effect of gravity and react to it, and these must be included in the answer.

 

[tom.stoer]

OK; what you are saying is that the approximation of point-like charges I am proposing is not allowed in GR.

Here are my questions from the beginning

1) do free-falling charges in a static gravitational field radiate?
2) do charges at rest in a static gravitational field (= accelerated charges w.r.t. to geodesic motion) radiate?
3) would a free-falling, radiating charge imply deviation from geodesic motion?

slightly rephrased:

1) does a 'free-falling' electron in the gravitational field of the Earth radiate?
2) does an electron at rest here on the surface of the Earth radiate?
3) does a 'free-falling' electron near the Earth follow a geodesic?

 

[TrickyDicky]

OK; what you are saying is that the approximation of point-like charges I am proposing is not allowed in GR.

Here are my questions from the beginning

1) do free-falling charges in a static gravitational field radiate?
2) do charges at rest in a static gravitational field (= accelerated charges w.r.t. to geodesic motion) radiate?
3) would a free-falling, radiating charge imply deviation from geodesic motion?

slightly rephrased:

1) does a 'free-falling' electron in the gravitational field of the Earth radiate?
2) does an electron at rest here on the surface of the Earth radiate?
3) does a 'free-falling' electron near the Earth follow a geodesic?

1) I'm not sure if a charge can be truly "free-falling", meaning that its being a charge seems to imply it is subject to EM forces and therefore wouldn't be free falling.
2)do electrons that radiate spontaneously in matter at rest count?
3)I tend to equate free-falling path with geodesic motion and as I said I find difficult to consider a charge as free-falling, but I could be mistaken.

 

[Bill_K]

2) is the easy one. For a charge sitting still in the lab, the situation is totally a static one. Hence simply by energy conservation there is no way it can radiate, or we would have invented a perpetual motion machine of the first kind.

1) For a freely falling charge, consider its electric field asymptotically far from the Earth, where spacetime may be treated as flat. In this region the field is Q/|R|² where R = r₀ - r(t). In other words it is a Coulomb field that varies as the charge producing it moves around. It looks for all the world like the field produced by a charge moving around in Minkowski space. You can worry about effects of the Schwarzschild geometry, and whether r(t) needs to be modified, but it does change. It is impossible for a falling charge to fall without producing a change in its distant field. And so it radiates.

3) I'd be surprised if it followed a geodesic, but it's possible I guess.

 

[atyy]

I'm pretty sure they're settled along the lines given by clem and Bill K, 4) is the only one I'm not sure of, but it should be something like what Bill K says.

Rindler gives a common sense proof;)
http://books.google.com/books?id=LkEhsgmP4vEC&dq=rindler&source=gbs_navlinks_s (p22)

Also
http://relativity.livingreviews.org/Articles/lrr-2011-7/
http://arxiv.org/abs/quant-ph/0601193
http://arxiv.org/abs/0806.0464

 

[TrickyDicky]

One example of a different take on this:
http://aflb.ensmp.fr/AFLB-301/aflb301m196.pdf

 

[lugita15]

I think that the consensus view, inasmuch as there is one, is that appropriately defined, radiation is frame-dependent. The definition of radiation is that the EM-fields have a 1/r term. If a charged particle is accelerating with respect to you, you will detect a 1/r EM-field. Thus a free-fall observer detects radiation from charges sitting on the earth, and an observer sitting on the Earth detects radiation from freely-falling charges. But neither observer detects radiation from charges at rest with respect to them; they only detect a 1/r^2 dependence of the EM- field.

 

[atyy]

The basic idea is that the equivalence principle "does not apply" to charged particles and radiation. The EP only applies to the "local laws" of physics, roughly those which don't involve derivatives higher than second. The EP applies to charged particles in the sense that the actions are minimally coupled. However, radiation is not directly seen from the minimally coupled action of a charged particle, but involves a second "non-local" derivative.

In addition to the discussions of the EP in the references in post #17, Blandford and Thorne's http://www.pma.caltech.edu/Courses/ph136/yr2011/1024.1.K.pdf (section 24.7) may also be useful: "What is the minimum amount of nonlocality that can produce curvature-coupling modifications in physical laws? As a rough rule of thumb, the minimum amount is double gradients: Because the connection coefficients vanish at the origin of a local Lorentz frame ..."

If we are not able to answer 4) or 5) how we answer 1-3)

I think Bill K's answer for 4) is about right, but here's one way to answer 1-3) without 4-5): consider two particles of equal mass and initial velocity and position, but one uncharged and the other charged - will they fall in the same way? In other words, we first answer a variant of 3).

From Poisson et al's http://relativity.livingreviews.org/Articles/lrr-2011-7/ (section 24): "In the scalar and electromagnetic cases, the picture of a particle interacting with a free radiation field removes any tension between the nongeodesic motion of the charge and the principle of equivalence."

 

[tom.stoer]

I think that the consensus view, inasmuch as there is one, is that appropriately defined, radiation is frame-dependent. The definition of radiation is that the EM-fields have a 1/r term. If a charged particle is accelerating with respect to you, you will detect a 1/r EM-field. Thus a free-fall observer detects radiation from charges sitting on the earth, and an observer sitting on the Earth detects radiation from freely-falling charges. But neither observer detects radiation from charges at rest with respect to them; they only detect a 1/r^2 dependence of the EM- field.

OK, that seems to make sense b/c it resolves both problems 1) and 2) with the same kind of reasoning.

@All: Thanks for the refereces

 

[TrickyDicky]

I still would like to understand how a free falling "charge", that I guess it really is universally free falling in the sense of only subject to gravity, is different from a free falling neutral particle.

I think that the consensus view, inasmuch as there is one, is that appropriately defined, radiation is frame-dependent.

Here I'm also confused, certainly the region of the radiation spectrum an observer detects is frame-dependent as Doppler shifts dictate, but I thought the existence or inexistance of radiation was an invariant at least classically if only because of energy conservation reasons. Maybe this is different in GR.

 

[tom.stoer]

Perhaps we should consider a different characterization of "radiation" which is frame-independent and which does neither depend on the asymptotic behavour of the el.-mag. field nor on asymptotics of space-time.

In classical electrodynamics the radiation of charges is equivalent to it's loss of kinetic energy. The charge does no longer follow a geodesic e.g. due to the effects of an external magnetic field; but synchrotron radiation may be a kind of secondary effect, the primary one would be the deviation from geodesic motion.

Therefore my proposal is to study simply the deviation from geodesic motion. If the charge follows a geodesic then a least locally (w.r.t. to an observer co-free-falling with the charge) the charge does not lose anything like energy b/c otherwise its motion would deviate from a geodesic.

So the central question is whether - in the absence of external el.-mag fields - charges follow geodesics

The answer is "yes" - as long as we neglect back-reaction.

That means that we separate the problem in
1) dynamics of the charge in spacetime
2) charge acting as source of an el.-mag field
3) geometry of spacetime and other free-falling observers

In 3) we mix effects of the el.-mag. field, spacetime and observers. That's why I am looking for a local definiton (which is always preferable - think about the concept of isolated horizons of black holes which tries to get rid of timelike or lightlike infinity to characterize a "local" phenomenon).

 

[TrickyDicky]

Perhaps we should consider a different characterization of "radiation" which is frame-independent and which does neither depend on the asymptotic behavour of the el.-mag. field nor on asymptotics of space-time.

In classical electrodynamics the radiation of charges is equivalent to it's loss of kinetic energy. The charge does no longer follow a geodesic e.g. due to the effects of an external magnetic field; but synchrotron radiation may be a kind of secondary effect, the primary one would be the deviation from geodesic motion.

Therefore my proposal is to study simply the deviation from geodesic motion. If the charge follows a geodesic then a least locally (w.r.t. to an observer co-free-falling with the charge) the charge does not lose anything like energy b/c otherwise its motion would deviate from a geodesic.

So the central question is whether - in the absence of external el.-mag fields - charges follow geodesics

The answer is "yes" - as long as we neglect back-reaction.

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. I know in the context of electrodynamics it is a habitual situation to neglect it in test charges but then in those situation generally gravity is not considered, they are ideally set in a flat space because gravitational considerations have a negligible effect.
Is it correct to equate in the context of curved spacetime (that is, not neglecting gravity) geodesic motion with free-falling?
If yes, I would say backreaction can't be neglected and a charge cannot be considered free falling by definition.
If not, I'm not completely sure how a neutral particle in free fall can be distinguished from a "free-fall" charge with immunity to EM fields.

 

[ofirg]

I am still a bit confused regarding the geodesic motion of charged particles. On one hand, the charged particle experiences acceleration (at least in the Newtonian sense) when moving in a gravitational field. This acceleration should cause it to radiate and the back reaction should cause its trajectory to deviate from the geodesic motion a neutral particle would make. On the other side, it is basic to general relativity that in the absence of other forces the trajectory of a particle with given initial conditions should not depend on the nature of that particle. Because of that, an observer falling with the neutral and charged particle can locally apply the laws special relativity (is locally blind to the gravitational field) and find they will not be pulled apart from each other.

Does this reasoning fall down because the electric field of the charged particle is infinite and the notion of a local inertial frame isn't useful anymore?

Thanks.

 

[atyy]

Because of that, an observer falling with the neutral and charged particle can locally apply the laws special relativity (is locally blind to the gravitational field) and find they will not be pulled apart from each other.

"Pulled apart from each other" - is that a question that can be defined locally (roughly, using first derivatives or lower)?

If a question cannot be defined locally, such as the existence of radiation (look at Bill K's answers - he uses the field at infinity), then the EP and local inertial frames exist, but don't apply.

 

[tom.stoer]

"Pulled apart from each other" - is that a question that can be defined locally (roughly, using first derivatives or lower)?

It's enough to have a deviation in the worldlines for identical initial conditions.

My idea is based on the fact that a) global spacetime should not be used to describe local effects and that b) of course we know that charged particles DO follow geodesics w/o or w/ immeasurably small deviations.

 

[atyy]

It's enough to have a deviation in the worldlines for identical initial conditions.

My idea is based on the fact that a) global spacetime should not be used to describe local effects and that b) of course we know that charged particles DO follow geodesics w/o or w/ immeasurably small deviations.

Yes, but if the deviation is defined using a second derivative, then it is "non-local", with regard to the EP.

 

[tom.stoer]

Yes, but if the deviation is defined using a second derivative, then it is "non-local", with regard to the EP.

What's the problem?

You have a particle following a geodesic, and a second particle (with identical initial conditions) not following a geodesic. That's enough to show that for the second there's some additional effect beyond spacetime curvature.

 

[atyy]

What's the problem?

You have a particle following a geodesic, and a second particle (with identical initial conditions) not following a geodesic. That's enough to show that for the second there's some additional effect beyond spacetime curvature.

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.

 

[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.

 

[atyy]

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.

My guess is yes, because a Faraday cage only works for static fields.

 

[PAllen]

My guess is yes, because a Faraday cage only works for static fields.

Actually, that's not true. They can block many dynamic fields as well, the range of wavelengths depending on the design. So I am proposing an idealized form in which all EM field lines are canceled within the thickness of the cage.

 

[atyy]

Actually, that's not true. They can block many dynamic fields as well, the range of wavelengths depending on the design. So I am proposing an idealized form in which all EM field lines are canceled within the thickness of the cage.

Won't the charge flow through the ground wire be different?

 

[PAllen]

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?