# Electromagnetic Radiation of Charged particles

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?

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:

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

• The link doesn't work for me Commented Jul 23, 2013 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. Commented Jul 23, 2013 at 20:37
• @AlfredCentauri, Thanks, nice answer. This result looks a bit puzzling to me! Commented Jul 24, 2013 at 7:49