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If $\theta$ is the angle between the observation direction and $\ddot{p}$, thenIntegrating over solid angle uses $\int {\sin }^2\theta d\Omega =8\pi /3$ and gives the total electric dipole radiation power.

In the radiation zone, both fields are transverse to $\hat{\mathbf{x}}$ and obeyThe outward Poynting vector isso its time average isAn absorbing shell receives the incident momentum rather than reversing it. Hence radiation pressure on a perfectly absorbing surface givesAt $r=R$,and thereforeIf $\vartheta$ is the angle between $\hat{\mathbf{x}}$ and ${\mathbf{p}}_0$, this isIts spherical average is ${\mu }_0{\omega }^4|{\mathbf{p}}_0{|}^2/(48{\pi }^2{R}^2{c}^2)$; the vector force integrates to zero by the inversion symmetry of the electric dipole radiation pattern.

For a localized charge and current distribution, take the retarded vector potentialLet $R=|x|$, $n=x/R$, and $t_R=t-R/c$. If the source size is much smaller than both $R$ and the characteristic radiation wavelength, the leading dipole term iswhere the last identity follows from charge conservation.

Keeping only the $1/R$ terms in the radiation zone gives the transverse fieldsThe radial Poynting vector therefore yieldsUsingor equivalently $\int {\sin }^2\theta d\Omega =8\pi /3$, we recover the electric dipole radiation formulaThis requires a localized, nonrelativistic source of size $d\ll \lambda$, observation distance $R\gg d$ and $R\gg \lambda$, and neglect of higher multipoles and faster-decaying near fields.

For the pulsar, write its magnetic dipole moment as $m(t)$ to distinguish it from the electric dipole above. The magnetic dipole radiation formula differs by a factor $c^{-2}$:Hereand hence

On the slow spin-down timescale, energy conservation gives the magnetic-dipole spin-down equationsoHalf of the initial rotational energy remains when $\Omega ={\Omega }_0/\sqrt{2}$. Thereforewhere $I=2M{\mathcal{R}}^2/5$. Since $E_0=I{\Omega }_0^2/2$, the equivalent expression is