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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.

A charge $q$ undergoing simple harmonic motion of angular frequency $\omega$ and slowly varying amplitude $A$ radiates the averaged powerIts mechanical energy $E=m{A}^2{\omega }^2/2$ therefore has exponential decay rate $\gamma ={\mu }_0{q}^2{\omega }^2/(6\pi mc)$ and half-life