Does a free falling charge radiate ?

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.

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.

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.

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

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

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

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

http://relativity.livingreviews.org/.../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

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.

Bill_K

That noise you just heard was my jaw hitting the floor. Needless to say (I hope it is needless to say!) it is a basic fact in general relativity that two massive particles orbiting each other do radiate gravitational waves.
This is reversing cause and effect. The backreaction from the emitted radiation may well make the particle deviate from a geodesic, but that is a result of the radiation, not a cause of it.

Demystifier

That's true. But our real issue here is electromagnetic radiation of a test charge in a fixed gravitational background.

It is a sort of a chicken-or-egg dilemma. There is a self-reaction involved here, so I think it is both a cause and a result.

A.T.

So what if the charged object is forced to move on a geodesic, by a rail or something. Can it radiate if the rail goes around a big mass?

Bill_K

Yes, the lesson of the 60's tells us that the gravitational radiation emitted by a source depends only on its mass multipole moments and not on the details of its composition or the nature of the forces holding it together, or whether the internal fields are weak or strong. In lowest order the radiated power is proportional to the third time derivative of the quadrupole moment. And these are completely well-defined quantities.

Just as we learned that in basic electromagnetism it's the second time derivative of the electric dipole moment that comes into play. And GR has not repealed this fact.