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For a localized charge distribution in an external field varying little across its extent, expand the external scalar potential about a reference point $\mathbf{R}$. The constant term couples to total charge; the first spatially varying term gives the electric-dipole interaction $-\mathbf{d}\cdot \mathbf{E}(\mathbf{R},t)$, where $\mathbf{d}$ is the electric dipole moment. Higher spatial multipoles are neglected.

For a passing charge at minimum distance $b$, the atomic extent must be small compared with $b$. For a plane electromagnetic wave, the analogous long-wavelength condition is $ka\ll 1$, as in the electric-dipole approximation for radiation. A short interaction time and a spatially uniform field across the atom are different assumptions; neither implies the other.

Electromagnetic radiation is a propagating disturbance of the electric field and magnetic field that carries energy and momentum away from its source. It can be a sustained electromagnetic wave or a finite wave packet: periodic oscillation is not required. In vacuum the disturbance travels at the speed of light. A receiver responds to fields arriving now from an earlier source event, as expressed by retarded time. This is how electrical activity at one place can deliver energy and information elsewhere without a connecting wire.

Stored electromagnetic energy alone does not establish outgoing electromagnetic radiation. A stationary electric charge in electrostatics, or a steady localized electric current in magnetostatics, can store electromagnetic energy with zero Poynting vector. Static combinations can have circulating energy flux without net outward power. Near a changing source, reactive energy can return later. The near and far field distinction separates these contributions from outgoing radiation zone fields.^[ref]

A single traveling vacuum plane electromagnetic wave obeying the Maxwell equations has $\mathbf{B}=\hat{\mathbf{k}}\times \mathbf{E}/c$. Its Poynting vector gives the energy density and flux of a plane electromagnetic wave:Here $I$ is wave intensity; ${ϵ}_0$ and ${\mu }_0$ are vacuum permittivity and vacuum permeability. For a monochromatic wave with linear polarization and peak amplitude $E_0$, squared cosine averages to $1/2$. Pulses require the instantaneous Poynting vector; arbitrary incident-and-reflected superpositions need not obey this relation.^[ref]

The electromagnetic momentum density is $\mathbf{g}=\mathbf{S}/c^2$. Completely absorbing a unidirectional vacuum pulse of energy $U$ transfers momentum $U/c$; see radiation pressure on a perfectly absorbing surface. A calculated $1.0\mathrm{mJ}$ pulse transfers $3.34\times 10^{-12}\mathrm{kg}m/s$. Symmetric emission can lose energy without net recoil because opposite momenta cancel.^[ref]

The direction of emission matters as much as its total power. For an oscillating electric dipole moment along $z$, the angular distribution of electric dipole radiation is strongest across the axis and vanishes along it. Its origin is geometric: the electric dipole radiation field is the transverse projection of the second time derivative of the electric dipole moment, so its magnitude contains $\sin \theta$. Squaring this electric field to obtain wave intensity gives ${\sin }^2\theta$. The electric-dipole approximation for radiation requires a source much smaller than its wavelength and observation in the radiation zone. ^[ref]

For example, the power per solid angle at $\theta =30^{\circ }$ is one quarter of its equatorial value, while the electric field amplitude is one half. To normalize the angular distribution to total power, use spherical coordinates with $d\Omega =\sin \theta d\theta d\varphi$:The integral follows by substituting $x=\cos \theta$, giving ${\int }_{-1}^1(1-x^2)dx=4/3$. This is a probability-like normalization over direction, with units of inverse steradian; it does not make the dipole an isotropic radiator. The same angular integral underlies the Larmor formula for nonrelativistic accelerated electric charge.

Detection requires the incident electromagnetic field to do something observable. An antenna develops an electrical response; a thermopile converts absorbed energy into a thermal and then electrical signal; a photodetector can respond through photon absorption. Hertz spark transmission and detection experiment makes this conversion visible as a tiny spark. Detector orientation, sensitivity, and surrounding reflections all matter: failure to see a signal does not prove that no electromagnetic radiation reached the receiver.

A Hertzian dipole is an ideal electrically short oscillating electric dipole used to model radiation from a localized current element. In vacuum its far-field electric dipole radiation has ${\sin }^2\theta$ angular power distribution and total power $P={\mu }_0|\ddot{\mathbf{p}}{|}^2/(6\pi c)$, evaluated at the retarded time. The electric-dipole approximation for radiation requires a source much smaller than its radiation wavelength.

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.

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.

A permanent electric dipole moment of magnitude $p_0$ rotating perpendicular to a fixed axis emits $P=K{\omega }^4$ in the electric-dipole approximation for radiation, with $K={\mu }_0{p}_0^2/(6\pi c)$. For fixed mechanical moment of inertia $\mathcal{I}$, weak radiative loss gives $\mathcal{I}\omega \dot{\omega }=-K{\omega }^4$ and hence the displayed spin-down law. It assumes $|\dot{\omega }|\ll {\omega }^2$: the full dipole acceleration has squared magnitude $p_0^2({\omega }^4+{\dot{\omega }}^2)$, whose second term is beyond this leading adiabatic approximation.

Rayleigh scattering is elastic scattering of electromagnetic radiation by particles much smaller than the incident wavelength. In the stationary particle approximation, the scattered electromagnetic radiation keeps the incident frequency: its direction changes, rather than its color. The incident electric field induces an oscillating electric dipole moment, which produces electric dipole radiation. This is a useful example of how a microscopic response can explain a familiar large-scale sight, why the sky is blue. ^[ref].

The smallness condition for Rayleigh scattering by a small sphere of radius $a$ is $ka\ll 1$, where the wavenumber is $k=2\pi /\lambda$ and $\lambda$ is the surrounding medium's wavelength. For ordinary materials with modest refractive index, the incident electric field then varies little across the particle, making the electric-dipole approximation for radiation useful. For very large refractive index, also check the internal size $|m|ka\ll 1$ before applying the simple electric dipole model. Once these conditions fail, use Mie scattering for a sphere instead of extrapolating the Rayleigh scattering wavelength dependence.

Historically, the 1871 paper ^[ref] connected the color and polarization of an electromagnetic wave of scattered skylight with particles smaller than the wavelength. Its dimensional analysis anticipates scattered light intensity proportional to particle volume squared and to ${\lambda }^{-4}$. The original mechanical picture is historical; the modern explanation uses induced electric dipole radiation.