Does a free falling charge radiate ?

1. Feb 16, 2013

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?

2. Feb 16, 2013

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.

3. Feb 16, 2013

TrickyDicky

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

4. Feb 16, 2013

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.

5. Feb 16, 2013

TrickyDicky

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. :uhh:

6. Feb 16, 2013

Staff: Mentor

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....

7. Feb 16, 2013

WannabeNewton

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

8. Feb 16, 2013

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.

9. Feb 17, 2013

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

10. Feb 17, 2013

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.

11. Feb 17, 2013

Bill_K

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.

12. Feb 17, 2013

TrickyDicky

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

13. Feb 17, 2013

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:

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.

14. Feb 18, 2013

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
- ...

15. Feb 18, 2013

TrickyDicky

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

16. Feb 18, 2013

Bill_K

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.

17. Feb 18, 2013

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.

18. Feb 18, 2013

atyy

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.

19. Feb 18, 2013

TrickyDicky

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.

20. Feb 18, 2013

Bill_K

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.