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For example, the displacement R(r,t) of the single charge q is given by

$$R = R_0\cos(\omega t)$$

In the view of multipole expansion, what is the monopole moment and the dipole moment for this configuration? Why the monopole does not contribute to the final EM field?

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An electric monopole is just a charge, isn't it? Yes, oscillating charges radiate. Any accelerating charge radiates. – Colin K Sep 28 '12 at 4:01
Its important to note that the directivity of this oscillation is the worst. You will have a field, but you won't be able to do much with only one. – Mikhail Sep 28 '12 at 7:57
Your electric monopole has a dipole moment $d(t)=q\cdot R(t)$. – Vladimir Kalitvianski Sep 28 '12 at 13:24
Yes, there seems to be a dipole moment, but just hard to understand in that there is no actual opposite charge. – Tao Sheng Sep 28 '12 at 20:48
FWIW, here's an applet that aids in picturing the field of a moving point charge:… – Alfred Centauri Sep 28 '12 at 21:41

There is a famous nonradiating configuration, which is not exactly what you asked, but it is very interesting, since it is also true in GR. If you have a sphere of fixed total charge whose radius oscillates arbitrarily, there is no electromagnetic radiation emitted. The solution is just given by Gauss's law outside the sphere, and you can verify that this satisfies Maxwell's equations even in the presence of everywhere equal radial currents, since the magnetic fields produced by these currents cancel out.

The deep reason that this doesn't radiate is that photons are spin 1, so they can't be emitted by an always spherically symmetric source distribution. The analogous Birkhoff theorem in GR tells you that a radially oscillating massive sphere doesn't radiate, although there it's because gravitons are spin 2.

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Wouldn't that be true ONLY if the charge on the sphere were completely continuous? In our world, it would have to be made of individual electrons and therefore cannot be completely spherically symmetric. This configuration would greatly suppress radiation, but would not eliminate it entirely. – FrankH Sep 28 '12 at 11:39
Thanks, but GR is far beyond my ability to comprehend. I just need to find a succinct and elegant derivation for the E-field and M-field produced by this sinusoidal oscillation charge. – Tao Sheng Sep 28 '12 at 18:42
@TaoSheng: The far-field is given by the dipole approximation for slow oscillations, which is all anyone ever uses--- you can get the field for an oscillating dipole in many sources, it's from the field far away from the dipole and you don't need to muck around with the detailed shape of the charge. – Ron Maimon Sep 28 '12 at 23:56
@FrankH: Yes, sure, but this is classical EM. – Ron Maimon Sep 28 '12 at 23:58
this doesn't answer the question and shouldn't be up voted. – Physiks lover Sep 29 '12 at 20:18

I believe that to get radiation you need a "jerk," that is a non-zero third time-derivative of the displacement. Hence, this will radiate.

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Accelerating charges radiate. You don't need jerk. – Colin K Sep 28 '12 at 4:02
This doesn't deserve a downvote +1--- the time derivative of the acceleration is the back-reaction in Dirac's point limit, and this led to the question of whether an eternally accelerating charge radiates, with back and forth going on for 50 years, until the question petered out with no completely agreed upon answer. I am not sure the question is well defined in classical mechanics, but it is definitely true that radiation reaction goes as the jerk. – Ron Maimon Sep 28 '12 at 7:32
@ColinK, "Does a uniformly accelerating charge radiate?" – Alfred Centauri Sep 28 '12 at 21:40

An electrical monopole is just charged particle. Any charged particle that accelerates will definitely radiate an electromagnetic wave. A sinusoidal oscillation you gave as an example will produce radiation at the frequency of the sinusoidal acceleration. But any acceleration of any kind will produce some kind of electromagnetic wave.

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Not any acceleration will result in radiation. please see the Nonradiation condition <>; for more details. This one will radiate. But I need find a rigid derivation to convince myself. Thanks anyway. – Tao Sheng Sep 28 '12 at 18:37
@Tao Sheng - dead link. – Rob Jeffries Nov 26 '14 at 22:39

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