 Solution

Define $p(t)=\int x^{\prime }\rho (t,x^{\prime })d^3{x}^{\prime }$. 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 $S={\mu }_0|\ddot{p}\times \hat{x}{|}^2\hat{x}/(16{\pi }^{2}c{r}^2)$ and $dP/d\Omega ={\mu }_0(|\ddot{p}{|}^2-(\hat{x}\cdot \ddot{p}{)}^2)/(16{\pi }^{2}c)$. For circular motion, $P={\mu }_0{q}^2{\omega }^4{R}^2/(6\pi c)$ and $⟨dP/d\Omega ⟩={\mu }_0{q}^2{\omega }^4{R}^2(1+{\cos }^2\theta )/(32{\pi }^{2}c)$. The stated axial and equatorial polarizations follow directly. Relativistically, perpendicular acceleration gives $P={\mu }_0{q}^2{\gamma }^4{\omega }^4{R}^2/(6\pi c)$.