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Does a free falling charge radiate ?

by greswd
Tags: charge, falling, free, radiate
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TrickyDicky
#19
Feb18-13, 09:32 AM
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Quote Quote by tom.stoer View Post
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
#20
Feb18-13, 10:42 AM
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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
#21
Feb18-13, 10:50 AM
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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
#22
Feb18-13, 11:01 AM
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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
#23
Feb18-13, 11:04 AM
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Quote Quote by Bill_K View Post
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
#24
Feb19-13, 02:02 AM
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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
#25
Feb19-13, 05:32 AM
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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
#26
Feb19-13, 05:42 AM
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Quote Quote by Bill_K View Post
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
#27
Feb19-13, 06:04 AM
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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
#28
Feb19-13, 06:34 AM
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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
#29
Feb19-13, 06:58 AM
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Quote Quote by Demystifier View Post
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/.../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
#30
Feb19-13, 07:54 AM
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Quote Quote by atyy View Post
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."
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
#31
Feb19-13, 08:04 AM
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If Earth was a point particle, it would indeed move along a geodesic and would not radiate gravitational waves.
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.
a charged particle in curved spacetime also does NOT move along a geodesic, and therefore radiates.
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
#32
Feb19-13, 09:11 AM
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Quote Quote by Bill_K View Post
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.
That's true. But our real issue here is electromagnetic radiation of a test charge in a fixed gravitational background.

Quote Quote by Bill_K View Post
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.
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.
#33
Feb19-13, 09:21 AM
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Quote Quote by Bill_K View Post
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.
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
#34
Feb19-13, 09:38 AM
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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.
pervect
#35
Feb19-13, 07:01 PM
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Quote Quote by Bill_K View Post
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.
I suspect this is the most common view, but I've seen enough papers with contradictory views to suggest that one needs to make sure that there is agreement about the definition of "radiating" before one tries to answer the question.

I think it'd be handy to have the Cliff Notes version (a short description of the proposed definition and the conditions required for radiation) but I don't think I've seenone.
pervect
#36
Feb19-13, 07:42 PM
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Quote Quote by Demystifier View Post
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.
You write:

On the other hand, if the charge accelerates,
then, even in the small neighborhood, Eqs. (11) no longer look like the Maxwell equations in Minkowski spacetime. This gives rise to a more complicated solution, which includes the terms proportional to r−1.
If we apply this to to an accelerating particle in Minkowskii coordinates, I don't quite understand how you conclude that it radiates.

Whatever the solution is, it must be static in those coordinates, because the space-time is static.

You also write

Now we turn back to the attempt to give an operational definition of radiation at large distances. In our opinion, the only reason why radiating fields deserve special attention in physics, is the fact that they fall off much slower than other fields, so their effect is much
stronger at large distances. Actually, the distinction between “radiating” and “nonradiating” fields is quite artificial; there is only one field, which can be written as a sum of components
that fall off differently at large distances. If one knows the distance of the charge that produced the electromagnetic field Fμ
ext and measures the intensity of its effects described
by (12), then one can determine whether this effect is “large” or “small”, i.e., whether the charge radiates or not.
Can you demonstrate, explicitly, such an effect ("slow falloff) in Rindler coordinates?

To insure coordinate independence, I'd like to see an argument for radiation that applies whichever coordinate system is used. Saying that "fermi coordinates are preferred because they are more physical" is sort of a cop-out. (I'm not sure that you actually said such a thing, I'm tempted to think it after a brief reading of your paper though.)


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