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For a source of characteristic size $a$ and angular frequency $\omega$, the electric-dipole approximation requiresso the source is much smaller than its radiation wavelength. Observing the pure $1/R$ radiation field additionally requires the radiation zone, $R\gg c/\omega$ and $R\gg a$.

Ignoring magnetization, electric polarization charge transport gives bound current ${\mathbf{J}}_{\mathrm{b}}={\partial }_t\mathbf{P}$. The assumed weak dielectric contrast therefore gives ${\mathbf{J}}_{\mathrm{b}}\simeq -i\omega {ϵ}_0({ϵ}_r-1){\mathbf{E}}_0{e}^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}$ inside the sphere. This is the first Born approximation: the incident field supplies the electric polarization to leading order in ${ϵ}_r-1$.

For $\mathbf{x}=r\mathbf{n}$, the far-distance expansions are $|\mathbf{x}-{\mathbf{x}}^{\prime }|=r-\mathbf{n}\cdot {\mathbf{x}}^{\prime }+O(a^2/r)$ and $|\mathbf{x}-{\mathbf{x}}^{\prime }{|}^{-1}\simeq r^{-1}$. Substitute the retarded time into the phase in the Lorenz gauge vector potential. The phase becomes $kr-\omega t+(\mathbf{k}-k\mathbf{n})\cdot {\mathbf{x}}^{\prime }$, so, using ${\mu }_0{ϵ}_0=c^{-2}$,The assumptions $r\gg a$ and $k{a}^2/r\ll 2\pi$ control amplitude and phase errors respectively. In the radiation zone the supplied electric and magnetic fields satisfy ${\mathbf{B}}_{\mathrm{scatt}}=c^{-1}\mathbf{n}\times {\mathbf{E}}_{\mathrm{scatt}}$. Complex phasors therefore give the time-averaged radial Poynting flux $|{\mathbf{E}}_{\mathrm{scatt}}{|}^2/(2{\mu }_{0}c)$. The incident flux is $|{\mathbf{E}}_0{|}^2/(2{\mu }_{0}c)$, soOrient spherical coordinates along $\mathbf{q}$ to evaluate the form factor:ConsequentlyAt $q=0$ the bracket is interpreted by its limit $a^3/3$, not a singularity. The transverse factor accounts for polarization of an electromagnetic wave; for complex polarization of an electromagnetic wave its squared norm means the Hermitian norm.

The outgoing solution obtained from the retarded fundamental solution of the wave equation isIn the radiation zone, put $r=|\mathbf{x}|$, $\mathbf{n}=\mathbf{x}/r$, and retain the leading $1/r$ term at retarded time $u=t-r$. The trace term disappears after applying the transverse-traceless projectorsoTwice using stress-energy conservation, ${\partial }_a{T}^{ab}=0$, and integrating by parts givesThe projector removes the trace, so in terms of the mass quadrupole momentthe far field isThe gravitational-wave energy flux isFor a trace-free symmetric tensor $A_{ij}$, the isotropic tensor integral over the observation direction givesConsequently the standard quadrupole formula, in the units $G=c=1$ used by the paper, isThus the printed coefficient $4\pi /5$ is inconsistent with the stated definition of $Q_{ij}$ and the standard wave-energy normalization; it appears to be a typographical error.

For the planet, choose its circular orbit in the $xy$-plane and write $\omega =2\pi /\tau$. With the star treated as fixed,and henceDirect differentiation givessoRestoring units multiplies this by $G/c^5$. If the printed $4\pi /5$ coefficient is followed literally, the answer is instead $4\pi$ times larger,For two bodies of comparable mass, $M$ is replaced by the reduced mass and $R$ by their separation.

The electric-dipole approximation for radiation requires the source size to be much smaller than the radiation wavelength:The displayed $1/r$ fields also describe the radiation zone, so the observation point must satisfy $r\gg a$ and $r\gg c/\omega$.

Write $t_r=t-r/c$. The given vector potential and the time derivative of the electric dipole moment giveWhen taking the curl, differentiating $1/r$ produces a near-field term of order $r^{-2}$, while differentiating the retarded time produces the leading radiation term because $\nabla t_r=-\hat{\mathbf{x}}/c$. Hence

The assumed electric field $\mathbf{E}=-c\hat{\mathbf{x}}\times \mathbf{B}$ and the transversality $\hat{\mathbf{x}}\cdot \mathbf{B}=0$ make the Poynting vectorOn a sphere of radius $R$, the power through surface element $R^{2}d\Omega$ is thereforeChoose the polar axis along $\ddot{\mathbf{p}}$ and use the angular distribution of electric dipole radiation, for which $|\hat{\mathbf{x}}\times \ddot{\mathbf{p}}{|}^2=|\ddot{\mathbf{p}}{|}^2{\sin }^2\theta$. Since $\int {\sin }^2\theta d\Omega =8\pi /3$,

For the charged particle in simple harmonic motion,ThusHere the angular frequency obeys $\omega =2\pi /T$, and the average uses $⟨{\sin }^2⟩=1/2$. The dipole condition is

Allow the amplitude to vary slowly as radiation removes energy. The mechanical energy isso the radiation damping of a harmonically oscillating charge givesThis is exponential decay with rate $\gamma ={\mu }_0{q}^2{\omega }^2/(6\pi mc)$. Its half-life isFinally, the largest particle speed is $v_{\max }=A\omega$. Therefore the dipole condition $A\omega /c\ll 1$ itself implies $v_{\max }\ll c$, precisely the nonrelativistic limit required by the mechanical energy formula.

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.

Assume a spatially compact source of size $d\ll r=|\mathbf{x}|$, internal speeds much smaller than one, and wavelength much larger than $d$. In the radiation zone,soDefine the mass quadrupole momentTwice using ${\partial }_{\mu }{T}^{\mu \nu }=0$, discarding boundary terms, givesHence

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