Past exam of the optics and electrodynamics course of the University of Cambridge / 2022 / Section B / 5 / c / Solution

Let $\mathcal{I}$ be the sphere's mechanical moment of inertia about its rotation axis, and let ${\omega }_i=\omega (0)$. For an undriven rigid sphere the rotational energy is $E_{\mathrm{rot}}=\mathcal{I}{\omega }^2/2$. In the weak-radiation-loss approximation, conservation of energy givesIntegrating $d({\omega }^{-2})/dt=2K/\mathcal{I}$ givesEquivalently, the characteristic spin-down time isThe radiative spin-down of a rotating electric dipole therefore has angular frequency that decays as $t^{-1/2}$ at late times within this model, rather than exponentially. The original paper gives no mass or mass-density profile, so $\mathcal{I}$ cannot be determined from $R$ and the relative permittivity. If a homogeneous solid sphere of mass $M$ is additionally assumed, insert $\mathcal{I}=2M{R}^2/5$; uniform dielectric response alone does not imply this mechanical mass distribution.

The approximation is important. If the dipole angle is $\varphi (t)$ with $\dot{\varphi }=\omega (t)$, direct differentiation gives $|\ddot{\mathbf{p}}{|}^2=p_0^2({\omega }^4+{\dot{\omega }}^2)$. Using the constant-frequency radiation law instantaneously neglects the second term and assumes $|\dot{\omega }|\ll {\omega }^2$. This gives the standard leading slow spin-down solution above. An exact nonadiabatic radiation reaction evolution would require the mechanical and near-field dynamics beyond the information supplied, rather than treating this leading law as exact at arbitrarily rapid damping.