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If $\theta$ is the angle between the observation direction and $\ddot{p}$, thenIntegrating over solid angle uses $\int {\sin }^2\theta d\Omega =8\pi /3$ and gives the total electric dipole radiation power.

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.

A uniform electrically short current element of length $\ell$ with sinusoidal peak current $I_0$ has electric dipole moment amplitude $p_0=I_0\ell /\omega$. The mean electric dipole radiation power is ${\omega }^4{p}_0^2/(12\pi {ϵ}_0{c}^3)$, so fixing current makes it proportional to ${\ell }^2{\omega }^2$, rather than ${\ell }^2{\omega }^4$. Current, charge excursion and dipole moment are different quantities to hold fixed when comparing antennas. The approximation requires $\ell \ll \lambda$ and does not impose the nonuniform standing-current profile of a long antenna.

Multipole radiation expands radiation from localized time-dependent sources into electric and magnetic multipoles. Electric dipole radiation is the leading electric term, while magnetic dipole radiation and higher electric multipoles contribute when the leading term vanishes or higher precision is needed.

The leading electric dipole radiation fields areThe radiation electric field is transverse, $\mathbf{n}\cdot {\mathbf{E}}_{\mathrm{rad}}=0$. The vector triple product therefore gives the leading Poynting vectorIf $\theta$ is the angle between ${\mathbf{p}}^{\prime \prime }(t-r/c)$ and $\mathbf{n}$, thenUsing ${\mu }_0{ϵ}_0{c}^2=1$, the final result isThe leading Poynting flux is outward, vanishes along the instantaneous retarded ${\mathbf{p}}^{\prime \prime }$ axis, and is maximal in its perpendicular plane. If ${\mathbf{p}}^{\prime \prime }=0$, the radiation term is zero and no angle needs to be assigned. The large-$r$ expansion assumes bounded retarded electric dipole moment and time derivatives; it may also be read at fixed retarded time.

Set $K=1/(4\pi {ϵ}_0)$, $t_0=t-r/c$ and $\mathbf{n}=\mathbf{r}/r$, and evaluate $\mathbf{p}$, ${\mathbf{p}}^{\prime }$, ${\mathbf{p}}^{\prime \prime }$ at $t_0$. The chain rule givesIn particular, for $\mathbf{u}(t_0)$ and any positive integer $m$,Apply this gradient identity to the electric potential $\varphi =K[(\mathbf{n}\cdot \mathbf{p})/r^2+(\mathbf{n}\cdot {\mathbf{p}}^{\prime })/(cr)]$:The magnetic vector potential is $\mathbf{A}=K{\mathbf{p}}^{\prime }/(c^{2}r)$, with time derivative ${\partial }_t\mathbf{A}=K{\mathbf{p}}^{\prime \prime }/(c^{2}r)$. Thus the electric field $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ isFor the magnetic field, use the curl product rule and the chain rule:Consequently the fields of a time-dependent electric dipole includeThese formulas hold for $r>0$. The $r^{-3}$ electric field is the retarded electric dipole field, the $r^{-2}$ terms are induction contributions, and the $r^{-1}$ terms describe electric dipole radiation. Retaining the dependence of $t_0$ on position in every spatial derivative is what produces both the induction and radiation contributions.

In the radiation zone, both fields are transverse to $\hat{\mathbf{x}}$ and obeyThe outward Poynting vector isso its time average isAn absorbing shell receives the incident momentum rather than reversing it. Hence radiation pressure on a perfectly absorbing surface givesAt $r=R$,and thereforeIf $\vartheta$ is the angle between $\hat{\mathbf{x}}$ and ${\mathbf{p}}_0$, this isIts spherical average is ${\mu }_0{\omega }^4|{\mathbf{p}}_0{|}^2/(48{\pi }^2{R}^2{c}^2)$; the vector force integrates to zero by the inversion symmetry of the electric dipole radiation pattern.

For a localized charge and current distribution, take the retarded vector potentialLet $R=|x|$, $n=x/R$, and $t_R=t-R/c$. If the source size is much smaller than both $R$ and the characteristic radiation wavelength, the leading dipole term iswhere the last identity follows from charge conservation.

Keeping only the $1/R$ terms in the radiation zone gives the transverse fieldsThe radial Poynting vector therefore yieldsUsingor equivalently $\int {\sin }^2\theta d\Omega =8\pi /3$, we recover the electric dipole radiation formulaThis requires a localized, nonrelativistic source of size $d\ll \lambda$, observation distance $R\gg d$ and $R\gg \lambda$, and neglect of higher multipoles and faster-decaying near fields.

For the pulsar, write its magnetic dipole moment as $m(t)$ to distinguish it from the electric dipole above. The magnetic dipole radiation formula differs by a factor $c^{-2}$:Hereand hence

On the slow spin-down timescale, energy conservation gives the magnetic-dipole spin-down equationsoHalf of the initial rotational energy remains when $\Omega ={\Omega }_0/\sqrt{2}$. Thereforewhere $I=2M{\mathcal{R}}^2/5$. Since $E_0=I{\Omega }_0^2/2$, the equivalent expression is

Spherical-particle frequency scaling. For a dielectric sphere of radius $a\ll \lambda$, the quasistatic electric polarizability isThe electric dipole radiation formula and $I_0={ϵ}_{0}c|E_0{|}^2/2$ giveThus Rayleigh scattering scales as ${\omega }^4$, or ${\lambda }^{-4}$, when the polarizability varies slowly with frequency. Material resonances and absorption can change this simple scaling. For the incident $z$ polarization, the dipole is along $z$: intensity is proportional to ${\sin }^2\chi$, where $\chi$ is the angle from $z$. There is no radiation on the dipole axis and a maximum in the $xy$ plane. In a direction $\hat{\mathbf{n}}$ the scattered electric field is along $\hat{\mathbf{z}}-(\hat{\mathbf{z}}\cdot \hat{\mathbf{n}})\hat{\mathbf{n}}$, hence linearly polarized wherever nonzero.^[ref]

Needle model used in parts (a)–(d). Let its axis be $\mathbf{u}=(u_x,u_y,u_z)=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$. A very slender, high-permittivity needle has dominant longitudinal response, so use the rank-one needle polarizability $\mathbf{\alpha }\simeq {\alpha }_{\parallel }\mathbf{u}{\mathbf{u}}^{\mathsf{T}}$. Its transverse response is neglected at leading order. For observation along $y$, define $\mathbf{v}=u_x\hat{\mathbf{x}}+u_z\hat{\mathbf{z}}$, $s=|\mathbf{v}{|}^2=1-u_y^2$ andFor a unit incident Jones vector $\mathbf{e}$, the common far-field relation is ${\mathbf{E}}_s\propto \mathbf{v}(\mathbf{u}\cdot \mathbf{e})$, and $I_s=K{I}_{0}s|\mathbf{u}\cdot \mathbf{e}{|}^2$. Here $I_s$ is irradiance at distance $R$; multiplying by $R^2$ gives power per solid angle. No absolute value of ${\alpha }_{\parallel }$ is given or needed for the requested relative intensities.

Diffuse random cloud, with the incident polarization of (a). The single-particle amplitudes through $x$ and $z$ analysers are proportional to $u_x{u}_z$ and $u_z^2$. The isotropic orientation fourth moment gives the needle-scattering analyzer momentsFor example, $⟨u_z^4⟩=\frac{1}{2}{\int }_{-1}^1{\mu }^{4}d\mu =1/5$; symmetry and $u_x^2+u_y^2+u_z^2=1$ give $2⟨u_x^2{u}_z^2⟩=1/3-1/5$. Since the particles scatter mutually incoherently, intensities add. For $N$ illuminated particles,The cross correlation vanishes by reflection symmetry. The degree of linear polarization is $(I_z-I_x)/(I_z+I_x)=\boxed{1/2}$: random anisotropic scatterers produce partially depolarized light, even though every individual needle produces a single linear polarization.

Continuous homogeneous dielectric. One must sum amplitudes coherently and include their spatial phases, rather than retain the preceding sum of intensities. Random orientations give $⟨\mathbf{u}{\mathbf{u}}^{\mathsf{T}}⟩=\mathbf{I}/3$, so the averaged response is isotropic and its induced polarization is along $z$. Its coherent radiation toward $y$, when allowed by the sample geometry, is therefore purely $z$ polarized, with $I_x=0$. The independent-needle prefactor is no longer a quantitative dense-medium model; local-field and collective effects determine the effective permittivity.

There is also coherent continuum phase matching: a uniform bulk driven by a plane wave has no arbitrary outgoing $90^{\circ }$ plane-wave component. In an ideal infinite homogeneous medium there is no diffuse side scattering at all; propagation, refraction and reflection are determined by the medium and its interfaces. A finite sample's coherent side signal depends on its shape and boundaries. Thus the old $1:3$ diffuse ratio is replaced by a coherent isotropic-medium response, not by a universal nonzero $N^2$ side-scattering intensity.

Treat a single weakly driven electron or an independent bound charge, so that the incident electric field is approximately uniform across its displacement. The classical calculation uses electric dipole radiation and excludes both a resonant divergence and substantial quantum recoil.

Work in the dilute, single-scattering limit and add the diffuse scattered light intensities of randomly positioned particles. The incident unpolarized light has equal, mutually uncorrelated $x$ and $y$ electric components. Each particle therefore has an induced electric dipole moment along its permitted axis, with the same mean squared magnitude proportional to $|\alpha {|}^2$.

For an observer in direction $\hat{\mathbf{n}}=(\sin \psi ,0,\cos \psi )$, choose transverse unit directions ${\mathbf{e}}_{\parallel }=(\cos \psi ,0,-\sin \psi )$ in the scattering plane and ${\mathbf{e}}_{\perp }=\hat{\mathbf{y}}$ perpendicular to it. The electric dipole radiation field is proportional to the transverse projection of the electric dipole. Hence an $x$ electric dipole contributes a wave amplitude factor $\cos \psi$ along ${\mathbf{e}}_{\parallel }$, while a $y$ electric dipole contributes its full wave amplitude along ${\mathbf{e}}_{\perp }$.

Let $C>0$ collect the common incident light intensity, electric polarizability, distance and frequency factors. ThenThe two components are uncorrelated, so the coherency matrix is diagonal in this basis. The degree of polarization isA signed degree of linear polarization, defined positive perpendicular to the scattering plane, has the same expression without the absolute value. The dominant direction is $\hat{\mathbf{y}}$ when $N_2>N_1{\cos }^2\psi$, and ${\mathbf{e}}_{\parallel }$ when the inequality reverses. If the denominator vanishes there is no scattered signal and its degree of polarization is undefined. A coherent exactly forward component, including the transmitted incident beam, is not part of this diffuse-intensity calculation; random particle positions do not by themselves destroy forward optical coherence.

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.

Viewed perpendicular to the magnetic field, the normal Zeeman effect has a $\pi$ component with linear polarization parallel to the magnetic field and two $\sigma$ components polarized perpendicular to it. Along the magnetic field, $\pi$ vanishes and the $\sigma$ pair has opposite circular polarizations. Pieter Zeeman described these geometry-dependent patterns in his lecture.^[ref]

The geometry can be derived from electric dipole radiation. Let the observation unit vector be $\mathbf{n}=(\sin \theta ,0,\cos \theta )$, with $\theta$ measured from the magnetic field. Radiation retains only the transverse part of the oscillating electric dipole moment $\mathbf{d}$:For $\pi$, take $\mathbf{d}\parallel \hat{\mathbf{z}}$; its angular light intensity factor is ${\sin }^2\theta$, so it vanishes on axis. For ${\sigma }^{\pm }$, use normalized complex dipoles $(\hat{\mathbf{x}}\pm i\hat{\mathbf{y}})/\sqrt{2}$. Their angular factors are $(1+{\cos }^2\theta )/2$. Projection gives circles on axis, lines at $\theta =\pi /2$, and generally elliptical polarization between. These factors assume equal dipole magnitudes; actual line light intensities also depend on initial quantum state populations, transition strengths and collection geometry. No absolute handedness name is needed: it depends on the viewing and phase conventions.

A transverse polarizer can separate $\pi$ from $\sigma$ even before the spectral line components are fully resolved. Along the magnetic field, a quarter-wave plate followed by a polarizer converts the two opposite circular polarizations into distinguishable linear settings. This explains why polarization analysis carried information beyond a small increase in line width.

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.

Consider a fixed, homogeneous, nonabsorbing, nonmagnetic sphere of radius $a$ in vacuum. Let $m$ be its relative refractive index, a fixed real number; in vacuum this equals the sphere's refractive index. Treat $m$ as independent of wavelength over the range being compared, neglecting optical dispersion. Air can be approximated by vacuum here because its refractive index is close to one. These assumptions describe a toy particle, rather than a literal model of an individual air molecule.

The Rayleigh scattering cross-section $\sigma$ is an effective area: the total scattered power is $P_{\mathrm{sca}}=I_{\mathrm{inc}}\sigma$, where $I_{\mathrm{inc}}$ is the incident light intensity. For $ka\ll 1$ its leading approximation isThe units are area, since $k^4{a}^6$ has units ${\mathrm{m}}^{-4}{\mathrm{m}}^6={\mathrm{m}}^2$. This Rayleigh scattering cross-section measures scattering, not conversion of the incident electromagnetic radiation into heat. The definition follows ^[ref].

To see the mechanism, the nearly uniform applied electric field produces an electric dipole moment with peak magnitudewhere $E_0$ is the incident electric field peak wave amplitude and ${ϵ}_0$ is vacuum permittivity. This is the sphere's response in electrostatics, with its permittivity divided by vacuum permittivity equal to $m^2$ under the nonmagnetic assumption. ^[ref]. The averaged electric dipole radiation power is $P_{\mathrm{sca}}={\omega }^4{p}_0^2/(12\pi {ϵ}_0{c}^3)$, with angular frequency $\omega =ck$ and speed of light $c$. Dividing by $I_{\mathrm{inc}}={ϵ}_{0}c{E}_0^2/2$ gives the displayed Rayleigh scattering cross-section. Thus doubling radius multiplies the electric dipole moment by eight and the scattered power by $64$, provided the larger sphere still satisfies $ka\ll 1$. ^[ref].

For a charge $q$ of mass $m$ bound by spring constant $k=m{\omega }_0^2$, the undamped equation $m\ddot{x}+kx=q{E}_0{e}^{-i\omega t}$ has complex amplitude $x_0=q{E}_0/[m({\omega }_0^2-{\omega }^2)]$. The induced electric dipole moment is $p_0=q{x}_0$. Averaged electric dipole radiation power, divided by incident light intensity, givesThe Rayleigh scattering limit $\omega \ll {\omega }_0$ gives $\sigma \sim {\sigma }_T(\omega /{\omega }_0{)}^4$, proportional to ${\lambda }^{-4}$. For ${\omega }_0\ll \omega \ll m{c}^2/\hslash$, the charge behaves as free and the Thomson cross-section is recovered. At resonance the undamped steady-state formula is invalid; a finite damping mechanism or radiation reaction must be included. The formula describes independent weakly driven scatterers, not arbitrary coherent scattering from an extended medium.

Assume one harmonic normal mode at thermal equilibrium, weak spontaneous excitation far from electronic resonances, the same collection solid angle and wave polarization, and matched response after instrumental correction. The Boltzmann distribution givesHere $k_B$ is the Boltzmann constant, $T>0$ is temperature, and $P_v$ is the probability of vibrational quantum number $v$. Thus $P_1/P_0=r$. Linear coupling to $Q$ has squared oscillator matrix elements proportional to $v+1$ for Stokes Raman scattering and $v$ for Anti-Stokes Raman scattering. Summing over initial levels gives strengths proportional to $\overline{v}+1$ and $\overline{v}$ respectively; their ratio is $r$, even when levels above $v=1$ are occupied.

Let $I_{\mathrm{S}}$ and $I_{\mathrm{AS}}$ denote line-integrated radiant intensities, meaning power per collection solid angle. The off-resonant electric dipole radiation law adds the fourth power of emitted frequency:The photon rate divides each line's power by its own $h{\nu }_s$. Consequently, the ideal response-corrected anti-Stokes/Stokes count-rate ratio has the third power of the same frequency ratio. Peak heights alone do not give line-integrated radiant intensities if linewidths differ. Uncorrected detector counts can also include unequal detector response and unequal optical filter transmission. Resonant enhancement, overlapping spectral lines, nonthermal populations and local optical heating can invalidate a temperature inferred from this simple formula.