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This question is motivated by similar one. If an accelerated point charge $q$ radiates with power $W$ then I assume the same particle with charge $-q$ will radiate with the same rate $W$. Now what if we make a dipole with these two charges and accelerate it with the same acceleration? What will be the radiation power?

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up vote 3 down vote accepted

According to this paper, New approach to the classical radiation fields of moving dipoles, the answer is:

$P = \dfrac{18d^2a^4}{35c^7} + \dfrac{2d^2\dot a^2}{15c^5}$

Here, $d$ is the fixed electric dipole moment and the acceleration, velocity, and dipole moment are along the $z$ axis.

From the paper:

This formula may be considered as the dipole analogue of the Larmor formula of the point charge.

And the Larmor formula for a point charge is:

$P = \dfrac{2q^2a^2}{3c^3}$

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The link doesn't work for me – Larry Harson Jul 23 '13 at 20:34
@LarryHarson, thanks for the heads up. Funny, I tried the link while I was editing and it worked but anyhow, it's fixed now. – Alfred Centauri Jul 23 '13 at 20:37
@AlfredCentauri, Thanks, nice answer. This result looks a bit puzzling to me! – richard Jul 24 '13 at 7:49

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