radiation due to a harmonic source

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Consider a distribution of charge and current that oscillates harmonically in time with angular frequency \omega\,:

\rho(\mathbf{x},t) = \rho(\mathbf{x})e^{-i\omega t}\,,
\mathbf{J}(\mathbf{x},t) = \mathbf{J}(\mathbf{x})e^{-i\omega t}\,.

Here it is understood that in all quantities that depend linearly on these two functions, we only take the real component. Furthermore, suppose that these sources are localized near \mathbf{x} = 0\, and have spatial extent a\,. We would like to determine the form of the fields in the radiation zone x \gg a\,. Recall the forms of the scalar and vector potentials in the Lorenz gauge

\Phi(\mathbf{x},t) = \frac{1}{4\pi\epsilon_0} \int\!dt'\,\int\!d^3x'\, \frac{\rho(\mathbf{x}',t')}{|\mathbf{x}-\mathbf{x}'|}\delta\left(t' - (t-\tfrac{|\mathbf{x}-\mathbf{x}'|}{c}) \right)\,,
\mathbf{A}(\mathbf{x},t) = \frac{\mu_0}{4\pi} \int\!dt'\, \int\!d^3x'\, \frac{\mathbf{J}(\mathbf{x}',t')}{|\mathbf{x}-\mathbf{x}'|} \delta\left(t' - (t-\tfrac{|\mathbf{x}-\mathbf{x}'|}{c}) \right)\,.

Including the time dependence we get

\Phi(\mathbf{x},t) = \frac{1}{4\pi\epsilon_0} e^{-i\omega t} \int\!d^3x'\, \frac{\rho(\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|}e^{i k |\mathbf{x}-\mathbf{x}'|}\,,
\mathbf{A}(\mathbf{x},t) = \frac{\mu_0}{4\pi} e^{-i\omega t} \int\!d^3x'\, \frac{\mathbf{J}(\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|} e^{i k |\mathbf{x}-\mathbf{x}'|}\,,

where k = \tfrac{\omega}{c}\,.

Radiation zone

If x \gg a\, then the integral's dominant contribution is due to \tfrac{x'}{x}\ll 1\,, and

\left|\mathbf{x}-\mathbf{x}'\right| = x\left(1 - 2\frac{\mathbf{x}\cdot\mathbf{x}'}{x^2} + \frac{x'^2}{x^2} \right)^\frac{1}{2} \approx x - \hat{\mathbf{x}}\cdot\mathbf{x}'\,. Then
\Phi(\mathbf{x},t) = \frac{1}{4\pi\epsilon_0} \frac{1}{x} e^{i(k x - \omega t)} \int\!d^3x'\, \rho(\mathbf{x}') e^{i \mathbf{k} \cdot \mathbf{x'}}\,,
\mathbf{A}(\mathbf{x},t) = \frac{\mu_0}{4\pi} \frac{1}{x} e^{i (k x -\omega t)} \int\!d^3x'\, \mathbf{J}(\mathbf{x}') e^{i \mathbf{k} \cdot \mathbf{x}'} \,,

where \mathbf{k} \equiv k \hat\mathbf{x}\, and we only expand \tfrac{1}{|\mathbf{x}-\mathbf{x}'|}\, to lowest order as the next term would be of order \tfrac{1}{x^2}\,.

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