= Bound-charge response
Put $\omega_0^2=k/m$, where $k$ is the binding spring constant rather than an optical <wavenumber>. The steady <complex amplitude> solution of the forced <Lorentz oscillator model> is
$$
\mathbf x_0=\frac{q\mathbf E_0}{k-m\omega^2}
=\frac{q\mathbf E_0}{m(\omega_0^2-\omega^2)},\qquad
\alpha(\omega)=\frac{q^2}{m(\omega_0^2-\omega^2)}.
$$
The acceleration <complex amplitude> is $-\omega^2\mathbf x_0$. Hence
$$
\langle P\rangle=\frac{q^2\omega^4|\mathbf x_0|^2}{12\pi\epsilon_0c^3},\qquad
\boxed{\sigma(\omega)=\frac{q^4\omega^4}{6\pi\epsilon_0^2c^4(k-m\omega^2)^2}
=\sigma_T\frac{\omega^4}{(\omega_0^2-\omega^2)^2}.}
$$
This is the <scattering cross-section of a harmonically bound charge>. The displacement and induced <electric dipole moment> are parallel to the incident <linear polarization>, and integrating their dipole radiation over all directions gives the total <cross section> above.
For $\omega\ll\omega_0$, binding dominates and
$$
\boxed{\sigma\simeq\sigma_T(\omega/\omega_0)^4,}
$$
recovering <Rayleigh scattering> with static <electric polarizability> $q^2/k$. For $\omega\gg\omega_0$, inertia dominates: $\mathbf x_0\simeq-q\mathbf E_0/(m\omega^2)$, so its second time derivative becomes frequency independent and
$$
\boxed{\sigma\simeq\sigma_T.}
$$
This is a classical <Thomson scattering> regime only while $\hbar\omega\ll mc^2$. Exactly at $\omega=\omega_0$, an undamped driven oscillator has no finite steady-state amplitude; the divergent formula must not be used. Finite <damping> and <radiation reaction> regulate the physical resonant response. Coherent contributions from many charges in an extended medium can also require a structure calculation rather than multiplying this single-charge result by a particle count.
Back to article page
