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

Only the $1/r$ fields transport a nonzero outgoing power to infinity. Retain those terms from the Hertzian dipole fields:The brackets mean evaluation at the retarded time,for a point source at the origin in vacuum. They are not a time average. The radial Poynting vector in the radiation zone isIntegrating over a sphere usesThereforeNear-field terms either give inverse powers of $r$ after the surface integral or describe reversible exchange with the near field. The emitted power labelled by source time $t_{\mathrm{src}}$ is ${\mu }_0\ddot{p}(t_{\mathrm{src}}{)}^2/(6\pi c)$; the enclosing surface receives it one light-travel time later. For a changing dipole orientation, the general form is $P={\mu }_0|\ddot{\mathbf{p}}{|}^2/(6\pi c)$.

The prescribed surface charge density is that of a uniformly polarized sphere and has zero total charge because $\int \cos \theta d\Omega =0$. Its electric dipole moment at $t=0$ isThe transverse components cancel by axisymmetry, while the $z$ component isThusThe angular surface distribution is a pure degree-one spherical harmonic. Hence its static external electric potential is exactly that of this dipole, with no higher electric multipole moments; it is not an assertion that the potential inside the finite sphere equals a singular point-dipole potential.

For right-handed uniform rotation about $+\hat{\mathbf{x}}$, the rigidly attached dipole isConsequently $|\ddot{\mathbf{p}}{|}^2=p_0^2{\omega }^4$ is constant. The electric dipole radiation formula givesThere is no extra factor of one-half: the squared sine and cosine contributions of the two perpendicular oscillating components sum to one. This is the leading electric-dipole approximation for radiation, requiring $\omega R/c\ll 1$ and observation in the radiation zone. The stated permanent surface distribution is interpreted as the actual charge producing the given external dipole; adding another dielectric screening factor would contradict the dipole moment the question asks to establish. The relative permittivity enters the induced-polarization problem in part (d), not this prescribed total dipole moment.

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.

The given static induced surface charge and the dipole integral from part (b) giveHere $ϵ$ is the dimensionless relative permittivity, as required by the stated $ϵ\pm 1,2$ expression; ${\alpha }_E$ is the SI electric polarizability. Take ${\mathbf{E}}_{\mathrm{inc}}(t)=E_0\cos \omega t\hat{\mathbf{e}}$ and $k=\omega /c=2\pi /\lambda$. In the Rayleigh scattering regime the field is approximately uniform across the sphere. Its induced dipole radiates mean powerThe incident time-averaged light intensity in vacuum is $I_{\mathrm{inc}}={ϵ}_{0}c{E}_0^2/2$. Dividing power by intensity gives the total cross section,The units are ${\mathrm{m}}^2$; the comparison is with the geometric projected area $\pi R^2$, not with a length:The inequality holds for an ordinary nonabsorbing dielectric, with finite contrast factor and $kR\ll 1$. It explains why a sufficiently small transparent sphere scatters weakly and why its scattered power has the characteristic ${\lambda }^{-4}$ dependence. Near a material resonance where $ϵ+2$ is small, this simple inequality and static-polarizability approximation need reconsideration, including loss and radiative corrections. For the stated low-frequency ordinary dielectric the supplied polarizability is the intended approximation.