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Equilibrium radiation and cavity frequencies. A blackbody absorbs all incident radiation at every frequency and direction. In thermal equilibrium its emission spectrum depends only on its temperature. Radiation inside an isothermal absorbing cavity, sampled through a small aperture that hardly disturbs equilibrium, realizes black-body radiation to a good approximation. Microscopic absorption and emission establish the equilibrium distribution; a perfectly reflecting wall is the convenient idealization used to count the modes once that distribution is specified.

In a rectangular electromagnetic cavity, the wave equation and perfect conductor wall boundary conditions select component wavenumbers $k_x=\pi l/L_x$, $k_y=\pi m/L_y$, $k_z=\pi n/L_z$. Combining the vacuum dispersion relation $\omega =c|\mathbf{k}|$ and $\nu =\omega /(2\pi )$ givesFor example, electric-field components can have the spatial dependenceFor three positive indices, Gauss's law requires $k_x{A}_x+k_y{A}_y+k_z{A}_z=0$, leaving two independent polarizations of an electromagnetic wave.

Index qualification. The original paper describes the indices as positive integers. That describes the three-nonzero-index family but omits valid electromagnetic modes with exactly one zero index, which have one polarization of an electromagnetic wave. For instance, with $n=0$, $l,m\ge 1$, only $E_z$ above remains; it is normal to the top and bottom faces and vanishes at the side walls. Two zero indices give no nontrivial cavity field. The omitted one-zero-index family does not affect the leading three-dimensional mode count, but is essential to the thin-cavity limit below. Retaining the paper's strict positivity restriction for every mode would instead give a thickness gap and would not yield its requested two-dimensional power law.

Three-dimensional mode count. A lattice cell in positive $\mathbf{k}$ space has volume ${\pi }^3/V$, where $V=L_x{L}_y{L}_z$. In a continuum approximation, a shell of radius $k$ and thickness $dk$ contains its positive octant, multiplied by two polarizations:Using $k=2\pi \nu /c$ givesFor a finite cavity the exact density of states is discrete; this formula is its bulk continuum limit, appropriate when many modes contribute and wall corrections are small.

Mean photon number. A mode is a quantum harmonic oscillator with thermal excitation energies $sh\nu$, $s=0,1,2,\dots$. The zero-point energy does not enter its thermal occupation. Put $q=e^{-h\nu /(k_{B}T)}$. Its partition function and mean occupation areThis is the Bose-Einstein distribution with zero photon chemical potential: equilibrium absorption and emission do not conserve photon number.

Thermal energy. Multiply the density of states by the mean excitation energy $h\nu ⟨s⟩$ and integrate:The substitution $x=h\nu /(k_{B}T)$ isolates all the dependence on cavity dimensions and temperature:Expanding $1/(e^x-1)={\sum }_{r=1}^{\infty }{e}^{-rx}$ gives an integral ${\sum }_{r}6/r^4=6\zeta (4)={\pi }^4/15$. HenceThis is the thermal energy only, excluding the zero-point contribution. The three-dimensional continuum regime requires $k_{B}T$ large compared with the geometric level spacings in all three directions.

Spectral sketch. The spectral energy density per unit ordinary frequency is Planck's law,At small frequency it rises as $8\pi k_{B}T{\nu }^2/c^3$, while at large frequency it decays exponentially. Raising temperature increases the spectrum at every fixed positive frequency, moves the peak to larger frequency in proportion to $T$, and increases the integrated area in proportion to $T^4$. A plot per unit wavelength has a different peak position because the integration variable changes.

Thin-cavity radiation. Let $A=L_x{L}_y$. The relevant low-temperature window isNonzero thickness indices are then thermally suppressed, while many in-plane modes are populated. The $n=0$ family has one polarization of an electromagnetic wave. Its allowed in-plane wavenumbers form a lattice of cell area ${\pi }^2/A$; a quarter of a circular shell therefore containsThus the dimensional crossover of thermal cavity radiation givesThe integral is $2\zeta (3)$ by the same geometric-series argument. A model with a different number of independent polarizations of an electromagnetic wave changes the prefactor but not the $T^3$ scaling. For a cavity with fixed finite $L_x,L_y$, cooling still further eventually freezes its lowest in-plane mode as well: the strict $T\to 0$ thermal energy is exponentially suppressed. The requested area law is the intermediate quasi-two-dimensional regime, not an arbitrarily low-temperature continuum limit.

In a homogeneous linear medium, write $\mathbf{D}=ϵ\mathbf{E}$ and $\mathbf{B}=\mu \mathbf{H}$. Without free charge or current, Maxwell's equations becomeTaking the curl of Faraday's law, using $\nabla \times (\nabla \times \mathbf{E})=\nabla (\nabla \cdot \mathbf{E})-{\nabla }^2\mathbf{E}$, givesTheir plane-wave solutions have ${\omega }^2=k^2/(\mu ϵ)$ and speed $v=1/\sqrt{\mu ϵ}$; in vacuum this is $c=1/\sqrt{{\mu }_0{ϵ}_0}$. For a wave $\mathrm{Re}[{\mathbf{E}}_0{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}]$, the divergence equations give $\mathbf{k}\cdot {\mathbf{E}}_0=\mathbf{k}\cdot {\mathbf{B}}_0=0$, while Faraday's law gives ${\mathbf{B}}_0=\mathbf{k}\times {\mathbf{E}}_0/\omega$. Thus the fields and propagation direction form an orthogonal triad, with $B_0=E_0/v$.

The two independent transverse electric components allow linear, circular and elliptical polarization of an electromagnetic wave. Their relative amplitudes and phase determine the polarization state. In a lossless nondispersive medium the energy density and energy flux areA travelling plane wave has equal electric and magnetic energy contributions and time-averaged intensity $⟨S⟩=E_0^2/(2Z)$, where $Z=\sqrt{\mu /ϵ}$ is its wave impedance and $E_0$ denotes peak electric-field amplitude. Reflection and refraction follow by imposing the electromagnetic boundary conditions. Frequency-dependent constitutive response introduces dispersion; conductivity introduces attenuation, so the simple real constant-speed result has a specified domain of validity.

For the overlapping monochromatic wave components, let $d$ be the thickness and ${\lambda }_0$ the vacuum wavelength. Define $\Delta n=n_s-n_f>0$ using the slow and fast refractive indices. With complex amplitude convention $e^{i(kz-\omega t)}$, the accumulated slow-minus-fast phase difference isThis is the optical path difference $\Delta nd$ expressed as a wave phase. It is a signed accumulation convention; with the opposite time convention the sign reverses. The outgoing polarization of an electromagnetic wave depends on $\delta$ modulo $2\pi$. Here the positive slow-minus-fast $\Delta n$ differs from the signed principal-index $n_e-n_o$ used in the parent birefringence entry when $n_e<n_o$.

For an original worked example, stipulate $\Delta n=0.0100$ at ${\lambda }_0=600\mathrm{nm}$ and a uniform thickness $d=15.0\mu \mathrm{m}$. These are design assumptions, not measured material data. The optical path difference is $150\mathrm{nm}$, soIt is a quarter-wave plate. At the same wavelength and refractive indices, doubling $d$ to $30.0\mu \mathrm{m}$ gives $\delta =\pi$, a half-wave plate. A $1.0\mu \mathrm{m}$ thickness increase changes $\delta$ by $\pi /30$, or $6^{\circ }$; retaining the same birefringence at $500\mathrm{nm}$ would instead give $\delta =3\pi /5$ for the $15.0\mu \mathrm{m}$ plate. Real refractive indices also vary with wavelength, so this last calculation explicitly neglects optical dispersion.

For these equal transmitted wave amplitudes, orthogonal projection onto a polarizer crossed with the initial $45^{\circ }$ linear polarization gives a simple independent readout. After removing a common wave phase, the component complex amplitudes are proportional to $(1,e^{i\delta })/\sqrt{2}$; orthogonal projection onto $(1,-1)/\sqrt{2}$ gives $(1-e^{i\delta })/2$. Squaring its modulus yields $I_{\perp }/I_{\mathrm{in}}={\sin }^2(\delta /2)$: zero at $d=0$, one half at $15\mu \mathrm{m}$, and one at $30\mu \mathrm{m}$. This assumes ideal polarizers, negligible specular reflection and wave absorption, overlapping components and stable relative wave phase. It measures a periodic function of phase retardation, so one brightness measurement cannot determine the thickness uniquely.

Planck's law gives the spectral radiance of an ideal blackbody in thermal equilibrium at absolute temperature $T>0$. It predicts how much power is carried in each small frequency or wavelength interval, per projected area and per solid angle. The whole curve is determined by $T$: no material-specific parameter enters this ideal Planck function.

Writing $\nu$ for frequency in hertz and $\lambda$ for wavelength in metres, the two forms of the Planck function areHere $h$ is the Planck constant, $k_B$ the Boltzmann constant, and $c$ the speed of light in vacuum. Their exact values in the International System of Units are $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$, and $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$. ^[ref]

The units matter: $B_{\nu }$ is measured in $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\mathrm{H}{\mathrm{z}}^{-1}$, whereas $B_{\lambda }$ is measured in $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}{\mathrm{m}}^{-1}$. Both are spectral radiance, rather than radiant exitance or energy density. To report $B_{\lambda }$ per micrometre, multiply its per-metre value by $10^{-6}$; a micrometre-wide interval is $10^{-6}$ metres wide.

For a computed example, at $T=6000\mathrm{K}$ and $\lambda =0.500\mu \mathrm{m}$, the exponent is $hc/(\lambda k_{B}T)\approx 4.79592$. Substitution into Planck's law gives $B_{\lambda }\approx 3.17569\times 10^{13}\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}{\mathrm{m}}^{-1}$, or $3.17569\times 10^7\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$ per micrometre. At the same wavelength, the illustrative temperatures $3000\mathrm{K}$ and $4500\mathrm{K}$ give $2.60268\times 10^5$ and $6.37793\times 10^6$ in the latter units. These are ideal predictions for stipulated temperatures.

The key change of variables is $\nu =c/\lambda$. Corresponding positive intervals carry the same power, so $B_{\lambda }d\lambda =B_{\nu }|d\nu |$ andEqual-width wavelength intervals have unequal widths in frequency; the factor $c/{\lambda }^2$ compensates for this. Merely replacing the horizontal-axis labels of $B_{\nu }$ by $c/\nu$ does not produce $B_{\lambda }$. This explains the frequency and wavelength peaks of Planck's law.

A short derivation uses the Boltzmann distribution for a single electromagnetic mode with thermal energies $nh\nu$, $n=0,1,\dots$. Put $q=e^{-h\nu /(k_{B}T)}$. The geometric series gives $Z={\sum }_{n\ge 0}{q}^n=(1-q{)}^{-1}$ and mean thermal energy $h\nu q/(1-q)$. The electromagnetic mode density in a box, including both independent polarization of an electromagnetic wave states of a photon, is $8\pi {\nu }^2/c^3$ per volume per unit frequency. Multiplication gives spectral energy density $u_{\nu }=8\pi h{\nu }^3/[c^3(e^{h\nu /(k_{B}T)}-1)]$; isotropy gives $B_{\nu }=c{u}_{\nu }/(4\pi )$. This counts thermal excitation energy, excluding zero-point energy. ^[ref]

Planck's law connects quantum mechanics to a measurable radiation curve. Its small-$h\nu /(k_{B}T)$ limit is the Rayleigh-Jeans law, $B_{\nu }\approx 2k_{B}T{\nu }^2/c^2$; its large-$h\nu /(k_{B}T)$ limit is the Wien approximation, $B_{\nu }\approx (2h{\nu }^3/c^2)e^{-h\nu /(k_{B}T)}$. The suppression at high frequency makes the total energy density finite. Integrating the curve gives the Stefan–Boltzmann law, while locating its peak gives Wien's displacement law. Thus one formula predicts both the spectral shape and the total emission.

Polarization of waves describes the direction and time-dependent geometry of a transverse wave's vector during its oscillation at a fixed position. In three spatial dimensions, a transverse wave can have two independent directions of oscillation, both perpendicular to its direction of propagation. For polarization of an electromagnetic wave, the relevant vector is the transverse electric field; for a mechanical transverse wave, it can be the transverse displacement. A purely longitudinal wave has no choice between these transverse directions, and the orientation of a static electric field is not polarization of waves.

Choose transverse axes $x,y$ in a Cartesian coordinate system. A monochromatic wave with a fixed phase difference has the real electric fieldHere $A_x,A_y$ are wave amplitudes, $\omega$ is the angular frequency, and $\delta$ is the phase difference of the $y$ oscillation relative to the $x$ oscillation. The vector's tip traces its path during the wave period $T=2\pi /\omega$. Both plotted coordinates are real electric field components; neither axis represents the imaginary part of a complex amplitude.

For nonzero wave amplitudes, $\delta =0$ or $\pi$ modulo $2\pi$ gives linear polarization: the vector moves back and forth along a fixed line. Equal nonzero wave amplitudes and $\delta =\pm \pi /2$ modulo $2\pi$ give circular polarization: the tip traces a circle of constant radius. Unequal nonzero wave amplitudes with $\delta =\pm \pi /2$ give a noncircular ellipse whose axes coincide with $x,y$. More generally, nonzero wave amplitudes and $\sin \delta \ne 0$ give elliptical polarization, with circular polarization as a special case. If exactly one wave amplitude vanishes, the nonzero wave has linear polarization for every $\delta$; if both vanish, there is no nonzero wave to assign a polarization of waves. ^[ref].

For example, take $\omega =2\pi {\mathrm{s}}^{-1}$, so the wave period is $T=1\mathrm{s}$. The choices $(A_x,A_y,\delta )=(2,1,0)$, $(1,1,\pi /2)$ and $(2,1,\pi /2)$, with electric field wave amplitudes in $V/m$, produce respectively the line segment $E_y=E_x/2$, the circle $E_x^2+E_y^2=1$ and the ellipse $E_x^2/4+E_y^2=1$, when coordinates are measured in $V/m$. At $t=T/4$, their tips are respectively $(0,0)$, $(0,-1)$ and $(0,-1)V/m$. The last two share a point without having the same path.

Resolving the electric field into two transverse components makes polarization of waves useful: the same frequency and wave amplitudes can describe different paths when their phase difference changes. An ideal linear polarizer selects the component along its pass axis, letting its orientation probe the direction of linear polarization through Malus's law. A fixed pair of wave amplitudes and a fixed phase difference describe one stable path; unpolarized light requires a statistical description rather than one fixed ellipse.

A linear polarizer transmits one linear polarization and suppresses the orthogonal one. Two ideal crossed polarizers block light unless an intervening medium changes its polarization of an electromagnetic wave. A twisted nematic field effect uses this selection to convert field-driven reorientation into brightness modulation.

A microscope that uses controlled incident and analysed polarization of an electromagnetic wave to study specimens. Crossed polarizers reveal orientation-dependent phase retardation in a birefringent specimen.

For the field-free, spin-free linear rigid rotor in one fixed internal quantum state, the electric-dipole interaction is $-\mathbf{d}\cdot \mathbf{E}(t)$. A nonzero permanent electric dipole moment along the molecular axis allows pure rotational photon absorption with $\Delta J=+1$ and photon emission with $\Delta J=-1$. The matrix element between initial and final quantum states must be nonzero: the existence of an energy gap alone is insufficient. Relative to a chosen $z$ axis, polarization of an electromagnetic wave along $z$ selects $\Delta M=0$; transverse polarization of an electromagnetic wave permits $\Delta M=\pm 1$.^[ref]

Hydrogen chloride and carbon monoxide are examples with a permanent electric dipole moment. By symmetry, molecular hydrogen and carbon dioxide have no permanent electric dipole moment in the ideal symmetric internal state. They still rotate and have rotational energy eigenvalues, but the simple permanent-electric dipole spectrum described here is absent. This restriction concerns this interaction and these assumptions, not every possible way of detecting molecular rotation.

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.

For incident unpolarized light and the isotropic electric dipole response of Rayleigh scattering by a small sphere, let $\theta$ be the angle between the incident beam's propagation direction and the outgoing direction. The differential scattering cross-section and its relative angular pattern arewhere $d\Omega$ is an element of solid angle. Averaging over the two transverse directions of incident polarization of an electromagnetic wave gives this pattern. Forward scattering at $0^{\circ }$ and backward scattering at $180^{\circ }$ are equally strong: $F=1$. Sideways scattering at $90^{\circ }$ is weakest, but nonzero: $F=1/2$. ^[ref].

The normalization is consistent becauseThus integrating $d\sigma /d\Omega$ gives the total Rayleigh scattering cross-section $\sigma$. The plotted $F$ has maximum one; it is not itself normalized to unit solid angle integral.

For a single electric dipole driven by incident linear polarization, the angular distribution of electric dipole radiation instead has a ${\sin }^2\alpha$ factor, where $\alpha$ is measured from the electric dipole moment direction, parallel to the incident electric field. That direction is perpendicular to the incident beam. Confusing $\alpha$ with the beam angle $\theta$ would incorrectly predict zero forward scattering. Averaging the dipole patterns over incident unpolarized light yields the displayed $F(\theta )$.

For a uniform linear medium with isotropy and local constitutive response, Maxwell equations give $k^2={\omega }^2\epsilon (\omega )\mu (\omega )$. Here $\epsilon$ is permittivity and $\mu$ is magnetic permeability. Since $c^{-2}={\epsilon }_0{\mu }_0$, the relative permittivity ${\epsilon }_r=\epsilon /{\epsilon }_0$ and the relative magnetic permeability ${\mu }_r=\mu /{\mu }_0$ obeyIf both responses are real and positive with negligible loss, and the ordinary positive-index branch is intended, then $n=+\sqrt{{\epsilon }_r{\mu }_r}$. If additionally ${\mu }_r\simeq 1$, then $n\simeq \sqrt{{\epsilon }_r}$. All these responses must be evaluated at the electromagnetic wave's frequency; a static relative permittivity cannot simply be substituted at an optical frequency. ^[ref]

With loss, use a complex index $\tilde{n}=n^{\prime }+i\kappa$ and explicitly fix the phase convention for complex amplitudes. For $E\propto \exp [i(kz-\omega t)]$ with $k=\tilde{n}\omega /c$, its wave amplitude has factor $\exp (-\kappa \omega z/c)$. A forward attenuating branch has $\kappa \ge 0$, while its wave phase is controlled by $n^{\prime }$. The complex relation ${\tilde{n}}^2={\epsilon }_r{\mu }_r$ alone leaves a choice of sign: the physical solution also needs causality and the appropriate radiation or decay condition. In negative-index media the physically selected real part can be negative. The positive square root is therefore not a universal rule for complex or negative-index media. Direction and polarization of an electromagnetic wave also matter in a plane wave in an anisotropic dielectric, for which one scalar index need not describe every propagation branch.

For the spin-free linear rigid rotor in thermal equilibrium at temperature $T>0$, assume all integer $J\ge 0$ are allowed and ignore nuclear-spin restrictions. Each $J$ level has $2J+1$ quantum states sharing one energy eigenvalue, an instance of degenerate energy levels, so the Boltzmann distribution givesHere $Q_{\mathrm{rot}}$ is a rotational partition function, $h$ the Planck constant, $k_{\mathrm{B}}$ the Boltzmann constant, and $B$ a frequency in $\mathrm{Hz}$. Thus $hB/(k_{\mathrm{B}}T)$ is dimensionless. The $2J+1$ factor counts magnetic quantum numbers; it does not mean that a whole level is one quantum state. For the stipulated $B=5\mathrm{GHz}$, the most populated levels are $J=2$ at $3\mathrm{K}$ and $J=7$ at $30\mathrm{K}$.

Population alone is insufficient to predict intensity. In weak, unsaturated absorption, the integrated absorption coefficient includes the transition frequency, the squared electric dipole moment matrix element, and the difference of lower and upper substate populations because stimulated emission returns energy to the field. A pulsed FTMW spectroscopy signal additionally depends on excitation, quantum coherence, cavity tuning, and receiver response.^[ref]

For an isotropic sample with polarization of an electromagnetic wave along $z$, write $d_z=d\cos \theta$. Normalized spherical harmonics giveThe last equality follows from ${\sum }_{M=-J}^J{M}^2=J(J+1)(2J+1)/3$. Combining this sum with the per-substate Boltzmann distribution and subtracting stimulated emission gives relative integrated absorption areasThe common factor includes $d^2/3$ and cancels upon normalization. This formula describes integrated weak absorption in a thermal, spin-free linear rigid rotor, not arbitrary peak heights. The example's strongest integrated lines originate at $J=3$ and $J=13$ for $3$ and $30\mathrm{K}$, respectively: they differ from the most populated levels. A missing line can reflect population, coupling, coverage, or sensitivity; it does not prove that the corresponding quantum state is absent.

A single-mode optical fiber supports only its fundamental spatial guided normal mode at the operating wavelength, with possible distinct states of polarization of an electromagnetic wave. The designation must include wavelength: MIT's fibre laboratory explicitly notes that a fibre sold for single-mode operation can support several normal modes at its helium–neon laser wavelength.^[ref]

For an illustrative $n=2\to 1s$ emission calculation, retain only the first-order shifts above and neglect the quadratic shift of the hydrogen atom ground state. The unperturbed Lyman-alpha line frequency is ${\nu }_0=(E_2-E_1)/h$. The possible offsets areThe outer pair originates in the $p_0$ parts of $|{\psi }_{\pm }⟩$ and has transition electric dipole moments along the applied electric field, conventionally called $\pi$ components. The central pair originates in $p_{\pm 1}$ and has transverse rotating transition electric dipole moments, conventionally called ${\sigma }^{\pm }$ components. Because electromagnetic radiation is transverse, the $\pi$ components disappear when viewed along the electric field. Viewed from the side, the $\pi$ and $\sigma$ components have polarization of an electromagnetic wave respectively parallel and perpendicular to the applied electric field. Line intensities additionally require the initial quantum state populations and the collection geometry.

In 1913 Johannes Stark reported that an applied electric field resolved the hydrogen $H_{\beta }$ and $H_{\gamma }$ spectral lines into five components at his instrument's spectral resolving power, viewed perpendicular to the electric field. Three inner components had polarization of an electromagnetic wave perpendicular to the electric field; two outer components had it parallel.^[ref]

In his lecture of 3 June 1920 for the 1919 prize, Johannes Stark describes canal rays emerging through a perforated cathode into the electric field between it and a supplementary electrode. A spectrograph recorded the emitted light. His earlier Doppler effect comparison used a collimator perpendicular to the beam and then along approaching rays. Later beam/electric field reversal changed component intensities.^[ref]

The historical account distinguishes observation perpendicular to the electric field, with five resolved components, from observation parallel to it, with three. These are viewing directions, distinct from the direction of the polarization of an electromagnetic wave.^[ref]

The experimental distinctions matter. Changing the viewing direction tests a Doppler effect contribution; changing the applied electric field tests the Stark effect; analyzing polarization of an electromagnetic wave separates components that may overlap in an unpolarized record. A comparison with the unperturbed spectral line fixes the origin of a displacement. These are different controls, and the earlier Doppler effect comparison should not be described as a field-off control of the later discovery. A spectrograph's finite spectral resolving power can merge several components into one. In particular, a count of visible spectral lines does not by itself count all energy eigenstates.

For a modern idealization of the gap, an electric potential difference $\Delta V$ over separation $d$ gives $|\mathcal{E}|\simeq |\Delta V|/d$ only if edge effects and the charge density in the gas are negligible. This is a calibration model, not a historical value assigned to Stark's apparatus. Similarly, a canal ray beam need not be a pure beam of neutral excited hydrogen atoms; the quantum mechanics model below isolates one neutral hydrogen atom from the source's preparation and collision processes.

The Verdet constant is the coefficient relating Faraday rotation to the magnetic flux density path integral. Its SI units are $\mathrm{rad}{\mathrm{T}}^{-1}{\mathrm{m}}^{-1}$. It depends on material, wavelength and temperature; a sign requires the stated viewing convention.^[ref]

The name commemorates Émile Verdet. For nonuniform material or temperature, use $\theta =\int V(\ell )B_{\parallel }(\ell )d\ell$ when the local scalar description remains valid. A single central magnetic field reading cannot generally replace this weighted integral. Strong additional linear birefringence or appreciable circular dichroism requires following the full polarization of an electromagnetic wave, rather than treating the output as merely a rotated line.

A waveplate uses birefringence to change polarization of an electromagnetic wave through phase retardation. Its fast axis has the smaller refractive index, hence the larger phase velocity; its slow axis has the larger index. Consider normal incidence with the optic axis in the plate face, so both components follow the same straight path. A linear polarization at $45^{\circ }$ to these axes excites equal wave amplitudes.

A complementary experiment uses two sheet polarizers with a cellophane sheet between them. First cross the polarizers and remove the sheet: the field is dark. Insert and rotate the sheet about the beam: light returns at intermediate orientations, but extinction recurs at two perpendicular orientations. The sheet has changed the polarization of an electromagnetic wave, without requiring a resolved double image. With white light, different wavelengths can produce different transmitted colors.^[ref]. Brightness alone does not establish circular polarization: an ideal rotating linear polarizer also gives orientation-independent light intensity for unpolarized light.