Does a free falling charge radiate ?

[greswd]

It appears paradoxical because to an observer falling with the charge it is as though the charge is at rest and therefore should not radiate.

Also, if we place a charge on a table, shouldn't it radiate as there is a normal reaction force from the table?

 

[Bill_K]

Frequently asked question. See the links to previous discussions at the bottom of the page. Yes it radiates, because a charge is not a point object. The field surrounding it is extended, and even though the center may be following a geodesic, the other parts of the field are not.

 

[TrickyDicky]

a charge is not a point object.

I''m confused, what happened to test charges? and aren't electrons(isn't an electron a charge?) point particles according to QFT-QED?

 

[Bill_K]

The electron itself is a pointlike object. But any charged object is surrounded by a Coulomb field which contains stress-energy and extends to infinity. For a charge falling in the Earth's gravitational field, for example - part of the EM field is on the other side of the planet!

In addition to radiating electromagnetic waves, a falling charge gets distorted by the varying gravitational field as it goes along, and radiates gravitational waves too.

 

[TrickyDicky]

The electron itself is a pointlike object. But any charged object is surrounded by a Coulomb field which contains stress-energy and extends to infinity. For a charge falling in the Earth's gravitational field, for example - part of the EM field is on the other side of the planet!

That's a nice explanation, only problem is that the contradiction between the first sentence and the rest of it is too evident to let it pass. Oh well, that rising damp again, let's pretend it's not there.

 

[Nugatory]

That's a nice explanation, only problem is that the contradiction between the first sentence and the rest of it is too evident to let it pass. Oh well, that rising damp again, let's pretend it's not there.

I must confess that I don't see any contradiction between saying that an electron is a point-like object and saying that its electrical field extends out to infinity.

Indeed, I remember watching Edward Purcell standing in front of a blackboard, describing the physics in exactly those terms, just so that we couldthen consider what would happen when the point particle was instantaneously displaced by a small amount...

 

[WannabeNewton]

Indeed, I remember watching Edward Purcell standing in front of a blackboard, describing the physics in exactly those terms, just so that we couldthen consider what would happen when the point particle was instantaneously displaced by a small amount...

You got to meet Purcell? I'm so jealous...T_T but yes this is also how he talks about it in his text.

 

[WannabeNewton]

By the way, Rindler's text has a small discussion on this caveat but it pretty much just reiterates the point made by Bill already.

 

[TopiRinkinen]

I should disagree. The Coulomb field external to electron does not carry electrical charge.
If you calculate the Gauss integral for the electric field of the electron, such that the electron is not enclosed by integrating surface, the integral is zero Coulombs.

On the other hand, if the accelerating electric field generates EM radiation, then it could be the the explanation.

BR, -Topi

The electron itself is a pointlike object. But any charged object is surrounded by a Coulomb field which contains stress-energy and extends to infinity. For a charge falling in the Earth's gravitational field, for example - part of the EM field is on the other side of the planet!

In addition to radiating electromagnetic waves, a falling charge gets distorted by the varying gravitational field as it goes along, and radiates gravitational waves too.

 

[Bill_K]

The Earth itself is a "freely falling particle", following a geodesic in its orbit around the sun. If the Earth had a net charge (and it may well have!), the problem lies entirely within the classical physics of Newton and Maxwell, and the circular motion of this charge would necessarily produce an outgoing EM wave.

 

[Bill_K]

I should disagree.

The bottom line is, for the reasons I gave, a freely falling charge does not exactly follow a geodesic. Due to its extended size there will be additional forces acting on it, that depend not just on the local gravitational field, but everywhere.

 

[TrickyDicky]

The Coulomb gauge is not Lorentz covariant, why would you use it in a QED context?

 

[Jano L.]

This is notorious question. Part of the reason for disagreements is that the situation is often not specified well enough, leaving the contributors to let their imagination fill in the details.

I propose to focus to the original question which is stated almost well enough:

Also, if we place a charge on a table, shouldn't it radiate as there is a normal reaction force from the table?

Well, the experience says that it will not radiate, besides the thermal radiation and scattered radiation. One can isolate charged object and have it on table indefinitely without any time-dependent fields connected due to force of gravity.

This is the answer just for the situation proposed in the question above. It does not say anything about any other scenario, like what free falling observer sees. That is a different question.

 

[tom.stoer]

How should we define 'radiation' in curved and non-stationary spacetimes?
- non-geodesics trajectories
- non-conservation of energy along a trajectory
- 1/r behaviour in the Coulomb potential for large r
- ...

 

[TrickyDicky]

What I'd like to get right is if from Bill's reply we must infer that point charges don't exist.

 

[Bill_K]

How should we define 'radiation' in curved and non-stationary spacetimes?
- non-geodesics trajectories
- non-conservation of energy along a trajectory
- 1/r behaviour in the [STRIKE]Coulomb potential[/STRIKE] field for large r

The last one is correct. Radiation is defined as the presence of a 1/r field at future null infinity, in asymptotically flat coordinates. For GR this issue was resolved back in the 60's by Bondi, Newman, Penrose, et al, where "field" means Riemann tensor.

 

[tom.stoer]

Bill, we had this discussion a couple of times and my answer was always that this definition does not work in not-asymptotically flat spacetimes, therefore one should look for a local definition; this is a general idea in GR: replace global definitions by local ones, look at all horizon discussions where one tries to get rid of null-infinity

(of course "field" Is correct and "potential" was nonsense, I am sorry for the confusion)

So why not using a comparison of trajectories of a non-charged and a charged particle? Of course this does not answer the radiation question directly, but it turns it round: we do no longer ask whether free falling particles radiate, but whether charged particles are in free fall according to the equivalence principle.

In parallel we should address the question whether (why) charged particles which are not in free fall do or don't radiate, i.e. particles which are stationary in a gravitational field, e.g. at fixed radius in a lab on the earth.

 

[atyy]

In parallel we should address the question whether (why) charged particles which are not in free fall do or don't radiate, i.e. particles which are stationary in a gravitational field, e.g. at fixed radius in a lab on the earth.

http://arxiv.org/abs/physics/9910019 suggested this answer. I believe the paper is serious, except for the claim that the principle of equivalence is validated, since I'm sure they know it's not applicable to this situation.

 

[TrickyDicky]

Bill, we had this discussion a couple of times and my answer was always that this definition does not work in not-asymptotically flat spacetimes,

This was my concern from the start of this thread, why mix notions from static time-independent scenarios (Coulomb fields, 1/r...) in which there would seem no radiation is even possible in principle with a question that requires moving charges-time dependent scenario?
IMHO it can only contribute to confuse even more the issue and the OP.

 

[Bill_K]

Bill, we had this discussion a couple of times and my answer was always that this definition does not work in not-asymptotically flat spacetimes, therefore one should look for a local definition

Tom, the understanding of radiation is rather firmly established - it's a global phenomenon, not a local one, in which a bounded system irreversibly loses energy to infinity. In the framework of GR one could perhaps extend the analysis from asymptotically flat spacetimes to de Sitter, or some other open cosmology.

But the radiation concept is not limited to GR. It's a basic feature of electromagnetism, as well as mechanical systems, such as elastic media. There are several reasons we treat it asymptotically.

One is simplicity - radiation exhibits common features at infinity which are far simpler than the details of what is happening at the source. For example, radiation may be conveniently described in terms of time-varying multipole moments, and knowing only these you know the energy loss.

A second reason is that some source motions transfer energy without producing radiation. Energy may be transferred from one part of the source to another "inductively", e.g. a pair of orbiting planets in Newtonian gravity, which constantly exchange energy and momentum through the inductive zone, which is 1/r² rather than 1/r. Likewise many electromagnetic examples.

Or even such a simple system as a pair of pendulums, coupled to each other, and also coupled to an infinitely long spring. You imagine there could be a local definition of radiation? As one of the two pendulums loses amplitude, it would be impossible to tell from its (local) motion alone whether the energy is being transferred (temporarily) to the other, or (permanently) lost to infinity. The answer must necessarily involve an analysis of the entire system, not just the one pendulum. That is, it must be global.

 

[Bill_K]

Don't believe everything you read on arXiv! The paper cited by atyy claims that a motionless particle radiates, the necessary energy coming about because, although the charge itself remains stationary, its surrounding electric field sags somewhat due to gravity, and goes on indefinitely sagging more and more!

Indeed, depending on how the charge is supported its field may sag, but an equilibrium is eventually reached where the distortion of the field is enough to resist more sagging. So there is not an indefinite source of energy to feed the radiation.

 

[tom.stoer]

Bill, regarding radiation you are right; the only question is whether this is the appriate question; I think that the main is geodesic motion = free fall.

 

[atyy]

Don't believe everything you read on arXiv! The paper cited by atyy claims that a motionless particle radiates, the necessary energy coming about because, although the charge itself remains stationary, its surrounding electric field sags somewhat due to gravity, and goes on indefinitely sagging more and more!

Indeed, depending on how the charge is supported its field may sag, but an equilibrium is eventually reached where the distortion of the field is enough to resist more sagging. So there is not an indefinite source of energy to feed the radiation.

Thanks for the warning! I'd long been wondering whether that article was correct, and hoped to get your opinion!

 

[Demystifier]

Even though I agree one shouldn't trust everything on arXiv, someone might want to see my opinion on that stuff, written in an arXiv paper:
http://arxiv.org/abs/gr-qc/9909035

For those who do not have time to read all this, let me just write down the main conclusions (for classical theory):
1. Radiation does not depend on the observer.
2. A charge radiates if and only if it does not move along a geodesic.

 

[Bill_K]

2. A charge radiates if and only if it does not move along a geodesic.

I gave what I thought was a valid counterexample of that. The Earth moves along a geodesic in its orbit about the sun. It surely radiates gravitational waves, and if it should carry a slight nonzero charge it will radiate electromagnetic waves also.

 

[TrickyDicky]

I gave what I thought was a valid counterexample of that. The Earth moves along a geodesic in its orbit about the sun. It surely radiates gravitational waves, and if it should carry a slight nonzero charge it will radiate electromagnetic waves also.

If it radiates gravitational waves it cannot be moving along a geodesic.

 

[TrickyDicky]

More specifically if you consider the Earth an idealized test particle orbiting the sun in a perfect geodesic it is obvious that it cannot radiate gravitational waves (nor EM ones following the correct point 2. by Demystifier).

If you consider it a real body with real mass and net charge, you run into problems with GR (two body problem etc) and if you were to claim that there is a point about the Earth's core that can be considered to be following an exact geodesic you stumble on the difficult problem that in GR there is no defined center of gravity as there is in Newtonian mechanics, so you can't really claim the Earth as a whole is following an exact geodesic path unless you follow the usual idealization of the Earth as a test body.

 

[Demystifier]

I agree with TrickyDicky. If Earth was a point particle, it would indeed move along a geodesic and would not radiate gravitational waves. But it is an extended object consisting of many particles between which other (non-gravitational) forces act, so that individual particles do not move along a geodesic.

In my arXiv paper I explain that a charged particle in curved spacetime also does NOT move along a geodesic, and therefore radiates.

 

[atyy]

I agree with TrickyDicky. If Earth was a point particle, it would indeed move along a geodesic and would not radiate gravitational waves. But it is an extended object consisting of many particles between which other (non-gravitational) forces act, so that individual particles do not move along a geodesic.

In my arXiv paper I explain that a charged particle in curved spacetime also does NOT move along a geodesic, and therefore radiates.

http://relativity.livingreviews.org/Articles/lrr-2011-7/fulltext.html (after Eq 19.84) discusses conditions under which one can use two complementary pictures (approximate, but very good) in which a point mass radiates and moves on a geodesic.

"It should be noted that Eq. (19.84) is formally equivalent to the statement that the point particle moves on a geodesic in a spacetime ... This elegant interpretation of the MiSaTaQuWa equations was proposed in 2003 by Steven Detweiler and Bernard F. Whiting [53]. Quinn and Wald [151] have shown that under some conditions, the total work done by the gravitational self-force is equal to the energy radiated (in gravitational waves) by the particle."

 

[TrickyDicky]

http://relativity.livingreviews.org/Articles/lrr-2011-7/fulltext.html (after Eq 19.84) discusses conditions under which one can use two complementary pictures (approximate, but very good) in which a point mass radiates and moves on a geodesic.

"It should be noted that Eq. (19.84) is formally equivalent to the statement that the point particle moves on a geodesic in a spacetime ... This elegant interpretation of the MiSaTaQuWa equations was proposed in 2003 by Steven Detweiler and Bernard F. Whiting [53]. Quinn and Wald [151] have shown that under some conditions, the total work done by the gravitational self-force is equal to the energy radiated (in gravitational waves) by the particle."

We had heated debates about this in the past, a distinction was considered important between the linearized gravity approximation and full non-linear GR, your example belongs to the former case.
In any case gravitational radiation has certain idiosyncratic properties for instance relating to the SET and energy of gravity that have generated lots of threads here and that make it a world of its own. If anyone wants to claim that objects following geodesics(and thus subjected only to the gravitational interaction) radiate just like the ones that don't , it's fine (but it kind of makes the distinction between geodesic motion and non geodesic motion as based in whether one can measure the acceleration with an accelerometer moot).
But EM radiation stress energy doesn't have the problems of gravitational SET(or absence of), and there is nothing in that article that allows a test particle with charge following a geodesic to radiate.