= Solution
Define $p(t)=\int x'\rho(t,x')d^3x'$. Expanding the retarded potential for source size $d\ll r$ and speed $v\ll c$, with radiation zone $r\gg d,c/\omega$, and using charge conservation gives $B=\mu_0\ddot p(t-r/c)\times\hat x/(4\pi rc)$. Then $S=\mu_0|\ddot p\times\hat x|^2\hat x/(16\pi^2cr^2)$ and $dP/d\Omega=\mu_0(|\ddot p|^2-(\hat x\cdot\ddot p)^2)/(16\pi^2c)$. For circular motion, $P=\mu_0q^2\omega^4R^2/(6\pi c)$ and $\langle dP/d\Omega\rangle=\mu_0q^2\omega^4R^2(1+\cos^2\theta)/(32\pi^2c)$. The stated axial and equatorial polarizations follow directly. Relativistically, perpendicular <acceleration> gives $P=\mu_0q^2\gamma^4\omega^4R^2/(6\pi c)$.
Back to article page
