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A reciprocal optical circular polarizer can be assembled from a linear polarizer followed by a quarter-wave plate whose axes make $45^{\circ }$ to the transmitted linear state. Forward passage converts unpolarized light into circular light of half the ideal incident intensity. Reverse passage encounters the retarder first; it leaves unpolarized light unpolarized until the linear polarizer transmits half as linear light. Thus the one-pass intensity can be equal in both directions even though the outgoing polarization differs. Use a coherency matrix, not a single Jones vector, for unpolarized input.

The optical coherency matrix records the second moments of transverse complex amplitudes. Normalize its trace to light intensity. Under a deterministic Jones matrix $M$, $C\mapsto MC{M}^{†}$. Unpolarized light has $C=(I_0/2)I_2$, so its output intensity is $(I_0/2)\mathrm{tr}(M{M}^{†})$; it cannot be represented by one fixed Jones vector.

For incident light with linear polarization, an ideal linear polarizer transmitswhere $I$ is light intensity and $\theta$ is the angle between the incident electric field's line of oscillation and the polarizer's pass axis. The input and output are compared in the same medium. The classical explanation is an orthogonal projection: for a unit vector $\mathbf{u}$ along the pass axis,The dot product reduces the transmitted wave amplitude by $\cos \theta$. For a plane electromagnetic wave in the same medium, light intensity is proportional to squared electric field wave amplitude, giving Malus's law. For example, $\theta =60^{\circ }$ transmits one quarter of the incident light intensity, although the transmitted wave amplitude is one half. ^[ref].

Start with light already in linear polarization along the horizontal $0^{\circ }$ axis and let its light intensity be $I_0$. Two ideal absorbing linear polarizers with pass axes $0^{\circ }$ and $90^{\circ }$ give $I_0{\cos }^2{0}^{\circ }{\cos }^{2}90^{\circ }=0$. Insert a $45^{\circ }$ polarizer between them: the first leaves $I_0$ unchanged, the middle transmits $I_0/2$ with linear polarization along $45^{\circ }$, and the last transmits $I_0/4$ with linear polarization along $90^{\circ }$. Each angle in Malus's law is relative to the preceding output, so the final step uses $90^{\circ }-45^{\circ }=45^{\circ }$:

For these ideal absorbing polarizers, the rejected electric field component transfers energy to the filter. In the three-filter arrangement the middle absorbs the fraction $I_0/2$, and the last absorbs $I_0/4$, leaving $I_0/4$ transmitted. Conservation of energy holds at each step. Inserting the middle polarizer changes the linear polarization arriving at the last one; it does not create light. The effect illustrates why successive orthogonal projections cannot generally be replaced by a single comparison of the first and last axes.

The initially horizontal assumption matters. For unpolarized light of light intensity $I_0$, the first ideal linear polarizer transmits $I_0/2$: averaging ${\cos }^2\theta$ over uniformly distributed incident orientations gives $1/2$. The same $0^{\circ },45^{\circ },90^{\circ }$ sequence then transmits $I_0/8$. This extra first-stage factor does not apply to the already horizontally polarized input in the figure.

With B nearest the observer, light from the reflecting target near the observer first crosses Q and then P on its way to the mirror. Unpolarized light remains unpolarized through Q, then becomes x linear after P. Mirror reflection preserves that linear orientation. On return it passes the same P without further ideal rejection and then Q, becoming circular toward the observer. Thus the reflected target remains visible.

Algebraically, with input and output at the observer side,The target here is used only as a source/reflector of ordinary unpolarized light; no physiological model is involved.

With A nearest the observer, the outward order is P then Q. The initially unpolarized light becomes x linear, then circular, before reaching the mirror. Reflection reverses its helicity. Its return through Q gives y linear, which is blocked by P:Hence the ideal reflected filter is dark and the target cannot be seen through it. This is a double-pass polarization rejection effect, not unequal one-pass transmission from its two sides. Real mirrors at oblique incidence, imperfect retardance, finite bandwidth and depolarization can leave leakage.

For propagation along z, write $\mathbf{E}=\mathrm{Re}\{(E_x\hat{\mathbf{x}}+E_y\hat{\mathbf{y}})e^{i(kz-\omega t)}\}$. A Jones vector contains the complex transverse components. Equal phase modulo $\pi$ gives linear polarization; equal magnitudes with phase difference $\pm \pi /2$ give circular polarization. A common overall phase has no effect on the polarization ellipse. A deterministic optical element acts by a Jones matrix, with successive matrices ordered from right to left along the light path.

A circular polarizer uses a linear polarizer transmitting x and a quarter-wave plate whose principal axes are at $\pm 45^{\circ }$. Up to an overall phase, convenient matrices in fixed physical x/y axes areThe plate is lossless, $Q{Q}^{†}=I_2$, and $Q(1,0{)}^T=(1,i{)}^T/\sqrt{2}$ is circular in the supplied convention. Put P on side A and Q on side B: forward passage is $J_{A\to B}=QP$.

Unpolarized light has no single Jones vector. Represent its intensity by the coherency matrix $C_{\mathrm{in}}=(I_0/2)I_2$ and propagate it as $C_{\mathrm{out}}=J{C}_{\mathrm{in}}{J}^{†}$. Forward transmission givesThus the transmitted light is circular with intensity $I_0/2$. The initial polarizer removes one incoherent linear component; the plate changes phase without restoring the lost intensity.

Reverse passage encounters Q before P and has $J_{B\to A}=PQ$ in the same fixed transverse axes. It givesThe output is linearly polarized along x and again has intensity $I_0/2$. A retarder alone leaves an unpolarized coherency matrix unchanged. Consequently an unpolarized scene appears darker from either side, although the transmitted polarization differs. No polarization-dependent response of the observer is needed to explain this reduction.

For the mirror calculations below, use an ideal near-normal isotropic reflector, $M=-I_2$ up to a common phase. In fixed physical transverse axes reflection leaves the polarization ellipse unchanged; it reverses circular helicity because the propagation direction reverses. A reciprocal plate has the same relative phase delay on the return traversal in this fixed-axis convention. In particular $Q^2=i\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$: two passes exchange x and y linear states.

For two-element polarization identification, assume ideal elements and use an intensity detector. For each source, test both element orders while rotating their relative axes. The circular source followed by the quarter-wave plate and then the polarizer is the combination that gives full intensity modulation. The plate converts circular light to linear light, and the downstream polarizer then obeys Malus's law.

If the polarizer comes first, it transmits half the circular source's intensity and the downstream lossless plate cannot change that intensity. Unpolarized light also gives constant half-intensity in either order: a wave plate leaves it unpolarized, while a polarizer followed by a wave plate has no final analyzer. Thus the modulating source is circular, its first element is the quarter-wave plate, and its second is the polarizer; the remaining source is unpolarized. Testing both orders avoids needing to know either source's original brightness.

Let ${\mathbf{u}}_{\theta }=(\cos \theta ,\sin \theta {)}^{\mathsf{T}}$ be the transmission direction. An ideal linear polarizer projects the incident electric field onto this direction, so its Jones matrix isSince ${\mathbf{u}}_{\beta }^{\mathsf{T}}{\mathbf{u}}_{\alpha }=\cos (\alpha -\beta )$, the amplitude transmitted by a sequence of polarizers is the product of the projections between successive axes. Here the four directions are $0,\omega t,-2\omega t,0$. After the first polarizer the unpolarized light input has intensity $I/2$ and is in linear polarization along $x$. The transmission through a sequence of linear polarizers therefore follows from successive applications of Malus's lawThe relative angle between the rotating polarizers is $3\omega t$, not $\omega t$. The first factor $1/2$ is an ensemble average over incident polarization: unpolarized light cannot be represented by a single deterministic Jones vector. Equivalently, its coherency matrix is $(I/2)I_2$; the total matrix $P(0)P(-2\omega t)P(\omega t)P(0)$ gives the same output intensity by the trace rule. The quasistatic optical-element approximation assumes the mechanical rotation is slow compared with the optical oscillation.

Both Thomson scattering and Rayleigh scattering can be understood as radiation from a dipole driven by the incident electric field. Their different frequency dependences come from the response of a free charge compared with a bound charge.

For a free particle of charge $q$ and mass $m$, a harmonic field gives $m\ddot{\mathbf{x}}=q\mathbf{E}$, hence $\mathbf{p}=q\mathbf{x}=-q^2\mathbf{E}/(m{\omega }^2)$. The radiated field contains two time derivatives of $\mathbf{p}$, cancelling the response's ${\omega }^{-2}$. The Thomson cross section is consequently independent of frequency in the classical low-recoil regime:For incident linear polarization with unit polarization $\mathbf{ϵ}$, summing over outgoing polarizations gives $d\sigma /d\Omega =r_q^2[1-|\hat{\mathbf{n}}\cdot \mathbf{ϵ}{|}^2]$. Radiation vanishes along the acceleration and is strongest perpendicular to it. Averaging an incident unpolarized light beam giveswhere $\Theta$ is the scattering angle. Scattering through $90^{\circ }$ is fully linearly polarized perpendicular to the scattering plane for this ideal isotropic dipole model. The electron case dominates over equally charged heavier particles because the cross section scales as $m^{-2}$. recoil corrections become necessary when photon energy is no longer small compared with $m{c}^2$.

For a bound charge with natural frequency ${\omega }_0$ and damping rate ${\gamma }_d$,Far below resonance, ${\alpha }_{\mathrm{e}}$ is approximately constant, soThis is Rayleigh scattering. It requires a scatterer small compared with the wavelength and, for the simple expression, an isotropic linear dipole response. Atoms, molecules and small dielectric particles provide examples. The stronger scattering of shorter visible wavelengths explains blue skylight; absorption and the nonuniform solar spectrum prevent treating that simple scaling as a complete color calculation.^[ref]

For isotropic electric scatterers the normalized angular and polarization patterns are the same as in Thomson scattering. The frequency scaling distinguishes their limits: a free charge has a displacement proportional to ${\omega }^{-2}$, whereas a tightly bound charge has an approximately frequency-independent displacement. Near resonance, far outside the small-particle limit, or with anisotropic polarizability, the simple Rayleigh formula must be modified. In particular, the electric-only pattern must not silently be applied to a perfectly conducting sphere whose magnetic dipole response is also retained.

A scattering measurement should subtract stray-light and detector backgrounds, control scatterer density and optical thickness, and calibrate detector response and incident power when comparing wavelengths. A polarization analyzer at $90^{\circ }$ tests the dipole pattern; multiple scattering or unresolved nonisotropic contributions can reduce its polarization.

Polarization under a boost. Lorentz invariance of normalized plane-wave polarization preserves the polarization ellipse's shape, handedness and degree of polarization when its transverse basis is carried to the aberrated ray. The propagation direction changes; equality of a fixed laboratory Cartesian pair of field components is not the appropriate comparison.

Rotate the $yz$ axes so that a general incident vacuum plane wave has unit direction $\hat{\mathbf{n}}=(\cos \vartheta ,\sin \vartheta ,0)$. Choose the transverse basis ${\mathbf{e}}_p=(-\sin \vartheta ,\cos \vartheta ,0)$, ${\mathbf{e}}_s=(0,0,1)$, and arbitrary complex amplitudes $E_p,E_s$. ThenPutting $\beta =v/c$, the given transformation givesThe four-wavevector transforms with Doppler factor $\mathcal{D}=\gamma (1-\beta \cos \vartheta )>0$, givingIn the corresponding basis ${\mathbf{e}}_p^{\prime }=(-\sin {\vartheta }^{\prime },\cos {\vartheta }^{\prime },0)$, ${\mathbf{e}}_s^{\prime }={\mathbf{e}}_s$, the transformed field isBoth components of the Jones vector acquire the same positive real multiplier. Their amplitude ratio and relative wave phase are unchanged, proving preservation of arbitrary linear, circular or elliptical polarization. All Stokes parameters scale by ${\mathcal{D}}^2$, so their normalized values and the degree of polarization are unchanged. An unpolarized light ensemble remains unpolarized under this single-ray boost. A mixture of different ray directions requires separate transformations and is outside this statement.

Reflection from the moving slab. Take the upper interface as $z=0$, the material below it, and the incident ray along $-\hat{\mathbf{z}}$. This fixes the sign left implicit by “along the $z$-axis”; choosing the opposite surface convention changes no result below. Treat $n$ as the real refractive index in the slab's rest frame, with unit relative magnetic permeability. In the laboratory,Transform to the slab rest frame:The incidence is now oblique, in the $xz$ plane, at angle $i$ to the surface normal:The incoming ensemble remains unpolarized light. In this frame the ordinary stationary-interface Fresnel equations give the amplitude coefficients:For generic oblique incidence, $R_s=|r_s{|}^2$ and $R_p=|r_p{|}^2$ differ. Equal incident $s$ and $p$ intensities therefore produce unequal reflected intensities: the reflection is partially polarized. Its degree of linear polarization isThe inverse boost preserves this degree, so the same conclusion holds in the laboratory even though the laboratory incident ray is normal to the surface. The moving material cannot be treated as an isotropic stationary dielectric in that frame.

The reflected rest-frame wavevector is $(\omega /c)(\gamma ,-\gamma \beta ,0,+1)$. Boosting back gives $(\omega /c)(1,0,0,+1)$: the reflected laboratory ray is normal to the surface and has the incident frequency. There is no contradiction with the oblique rest-frame geometry. Translation tangential to a uniform interface does not itself introduce a moving normal mirror.

Complete linear polarization. For Brewster reflection from a tangentially moving dielectric, at Brewster's angle, $r_p=0$ and $\tan i_B=n$. Combining this with the aberration relation givesOnly the $s$ component is reflected. The plane of incidence in the slab frame is $xz$, so the reflected electric field is along $y$, perpendicular to the slab velocity. Boosting this $s$ field back keeps its electric direction along $y$. For example, the reflected rest-frame $s$ field has $B_z^{\prime }=-\beta E_y^{\prime }/c$; hence $E_y=\gamma (E_y^{\prime }+v{B}_z^{\prime })=E_y^{\prime }/\gamma$ and $E_x=E_z=0$.

This exact extinction assumes a transparent nonmagnetic dielectric with $n\ne 1$. At $v=0$ the reflected intensities are equal and the reflected beam is unpolarized. If $n=1$, both reflection coefficients vanish and there is no reflected beam whose polarization can be assigned. With dispersion, evaluate $n$ at the shifted frequency ${\omega }^{\prime }=\gamma \omega$; the Brewster-speed equation is then implicit. absorption generally prevents a real incidence angle from making the complex $r_p$ exactly zero.

At an air–glass interface, resolve the incident unpolarized light into components perpendicular ($s$) and parallel ($p$) to the plane of incidence. The Fresnel equations give the parallel reflection coefficientAt Brewster's angle, $r_p=0$. Together with Snell's law, this gives ${\theta }_i+{\theta }_t=\pi /2$ and $\tan {\theta }_i=n_2/n_1$. Thus, taking the incident medium to be air,The remaining reflected component has linear polarization perpendicular to the plane of incidence. This assumes a transparent, nonmagnetic glass with negligible absorption.

Molecules in clean air are much smaller than visible wavelengths. Their induced electric dipole moments undergo Rayleigh scattering. Away from molecular resonances, the induced dipole amplitude is approximately $p=\alpha E$, with only weak wavelength variation in the polarizability $\alpha$. The radiation-zone electric field is proportional to $\ddot{p}$, so the scattered intensity satisfiesThis is the Rayleigh scattering wavelength dependence. For equal incident intensities, $450\mathrm{nm}$ light scatters about $(650/450{)}^4=4.35$ times as strongly as $650\mathrm{nm}$ light. Thus light scattered out of the direct solar beam and into our line of sight is enriched in blue wavelengths.

The dipole mechanism also determines the Rayleigh scattering angular pattern. For unpolarized light, its angular factor is $1+{\cos }^2\vartheta$; near $90^{\circ }$ scattering, the scattered light has strong linear polarization. The sky's perceived colour depends additionally on the solar spectrum, atmospheric transmission and the eye's spectral response, so the ${\lambda }^{-4}$ law alone does not predict a violet sky. Over a long near-horizontal path the direct beam loses more short-wavelength light, explaining red sunsets. Large aerosol particles instead produce Mie scattering, which is less strongly wavelength-selective and can make haze look whitish.

At normal incidence, continuity of tangential wave phase makes both internal wavevectors normal to the parallel faces. For the ordinary wave in a uniaxial optical medium, the electric field is an eigenpolarization of the permittivity tensor, so $\mathbf{D}\parallel \mathbf{E}$. ThusIts energy path is also normal to the faces. The extraordinary wave differs because $\mathbf{D}\perp \mathbf{k}$ but $\mathbf{E}$ need not be perpendicular to $\mathbf{k}$. In a nonmagnetic crystal, $\mathbf{H}\parallel \mathbf{B}$ and $\mathbf{S}=\mathbf{E}\times \mathbf{H}$ is perpendicular to $\mathbf{E}$. The vectors $\mathbf{D},\mathbf{E},\mathbf{k},\mathbf{S}$ lie in the same meridional plane, with $\mathbf{D}\perp \mathbf{k}$ and $\mathbf{E}\perp \mathbf{S}$. Therefore the electric-displacement angle from birefringent walk-off equals the angular separation between $\mathbf{S}$ and $\mathbf{k}$.

The observed lateral separation is accumulated over thickness $L$, so $\Delta =L\tan {\alpha }_e$. With $L=10\mathrm{mm}$ and $\Delta =0.1\mathrm{mm}$,The two emergent rays are parallel because their outgoing wavevectors are both normal to the faces in air; their internal energy paths can nevertheless be displaced by birefringent walk-off. The angle quoted is the positive angle between the field vectors, so the unspecified lateral direction does not affect its value. The unpolarized light excites both eigenpolarizations; it does not require their energy paths to coincide.

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.

Choose the first polarizer's pass axis as ${\mathbf{e}}_1$ and write the second as $\mathbf{a}=({\mathbf{e}}_1+{\mathbf{e}}_2)/\sqrt{2}$. The positive signed rotation takes ${\mathbf{e}}_1$ toward $\mathbf{a}$ by the short $\pi /4$ angle. Let $B_{\parallel }$ be signed according to $\theta ={\rm Y}L{B}_{\parallel }$. The Jones matrices of ideal polarizers are their orthogonal projection matrices,Removing the unimportant common propagation phase, the lossless glass block has the rotation matrixThe elements act from right to left, soWith the common optical phase retained, multiply this by $e^{i\bar{\varphi }}$. Extra scalar attenuation would also multiply the result, but none is supplied. A single Jones vector does not represent unpolarized light; the device's Jones matrix is nevertheless well defined for each incident component and can be applied to their statistical mixture.

Let $I_0$ be the incident light intensity. The first ideal polarizer transmits $I_0/2$ and produces linear polarization along ${\mathbf{e}}_1$. Faraday rotation turns that direction through $\theta ={\rm Y}L{B}_{\parallel }$, and Malus's law at the second polarizer givesEquivalently, use the input coherency matrix $C_{\mathrm{in}}=(I_0/2)I_2$ and $I_f=\mathrm{tr}(J{C}_{\mathrm{in}}{J}^{†})$. This produces the same one-half factor required by unpolarized light.

Maxima satisfy $\theta =\pi /4+m\pi$, with $m\in \mathbb{Z}$. For $L=1\mathrm{m}$ and ${\rm Y}=9\mathrm{rad}{\mathrm{T}}^{-1}{\mathrm{m}}^{-1}$, the smallest field magnitude isIts direction is the axial direction that rotates the polarization passed by $P_1$ toward the pass axis of $P_2$ through the shorter $45^{\circ }$. In the right-handed ray-basis convention where $\rho =+{\rm Y}{B}_{\parallel }$ and the depicted second axis is $+45^{\circ }$, this is $B_{\parallel }>0$, along $S\to O$. Reversing that field at the same magnitude gives $\theta =-\pi /4$ and zero transmission. A signed Verdet constant is meaningful only together with its rotation convention, so the polarization-alignment description fixes the physical direction without relying on ambiguous clockwise labels in a perspective diagram.

Keep the magnetic field fixed in the laboratory and exchange source and observer. The incident unpolarized light now meets $P_2$ first. In fixed transverse laboratory axes the backward glass map is still $R(\theta )$, not $R(-\theta )$: this is optical nonreciprocity of the Faraday effect. The order of the two polarizers is reversed, soThe reverse light intensity is thereforeFor the optimal forward setting $\theta =\pi /4$, it vanishes: the $45^{\circ }$ input selected by $P_2$ rotates to $90^{\circ }$ and is blocked by $P_1$. At a general field, $I_r+I_f=I_0/2$. The result assumes equal incident light intensities in the two experiments. Treating the glass as a reciprocal natural optical rotator would predict cancellation of its laboratory rotation on reversal and would give the wrong answer here.

The far fields obey $E_{\theta }=c{B}_{\varphi }$. The radial component of their Poynting vector is $E_{\theta }{B}_{\varphi }/{\mu }_0$. Averaging ${\ddot{p}}^2=p_0^2{\omega }^4{\cos }^2\omega (t-r/c)$ and integrating $\int {\sin }^2\theta d\Omega =8\pi /3$ givesFor a small isotropically polarizable particle, $p_0=\alpha E_0$ and incident intensity is $I={ϵ}_{0}c{E}_0^2/2$. Thus the total Rayleigh scattering cross-section isEach incident linear polarization gives the same integrated power; averaging unpolarized light does not add another factor of two. The scalar real polarizability approximation neglects absorption and molecular anisotropy.

For Rayleigh attenuation from refractivity in dilute air, polarization density is $\mathbf{P}=N\alpha \mathbf{E}$. Consequently ${ϵ}_r=1+N\alpha /{ϵ}_0$ and, with ${\mu }_r\simeq 1$ and $n-1\ll 1$,The ideal-gas number density at the supplied pressure and temperature is $N=P/(k_{B}T)=2.47\times 10^{25}{\mathrm{m}}^{-3}$, yielding $\alpha =2.15\times 10^{-40}\mathrm{C}{\mathrm{m}}^2{\mathrm{V}}^{-1}$. For an explicit representative visible wavelength $\lambda =550\mathrm{nm}$,Direct-beam intensity follows $I(x)=I(0)e^{-x/L_{\mathrm{att}}}$. Equivalently $L_{\mathrm{att}}=3N{\lambda }^4/[32{\pi }^3(n-1{)}^2]$. Its strong wavelength dependence gives roughly $21$–$200\mathrm{km}$ across $400$–$700\mathrm{nm}$ if refractivity is held fixed. This is a uniform-density dry-air scattering estimate, not absorption or transmission through a vertically stratified atmosphere; aerosols and molecular depolarization corrections are outside the given model.

At a transparent nonmagnetic dielectric interface, the Fresnel reflection coefficients differ for s and p linear polarization. At Brewster incidence the reflected p amplitude vanishes and reflected unpolarized light becomes s polarized. For a horizontal reflecting surface this is horizontal glare. Away from Brewster incidence it is generally partial polarization. Refraction and internal reflection in droplets also select the two components differently, explaining polarization of rainbow light without a biological mechanism.

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.