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From the illuminated side downward, the p-type gallium arsenide cap provides a low-resistance route to the front metal contact. It should remain beneath the contact grid but be removed from the optical aperture to reduce parasitic absorption. The wider-band gap p-type aluminium gallium arsenide window transmits much of the useful light and reduces front-surface recombination of carriers in the adjacent absorber. The optical-window and reduced-recombination rationale has a primary experimental basis in early heteroface devices.^[ref]

The p-type gallium arsenide emitter and n-type gallium arsenide base form the p-n junction. Absorbed photons generate electrons and electron holes; diffusion and the junction's electric field direct them toward the appropriate contacts. The n-type aluminium gallium arsenide back-surface-field layer provides a barrier to minority holes moving toward the substrate and reduces rear recombination. The n-type gallium arsenide substrate supports the epitaxial layers and conducts majority carriers to the rear contact.

The front metal must collect current with limited shading and the back metal provides the return contact. Interpreting the patterned hatched region as the optical antireflection coating, it reduces reflected light over the exposed window. Carrier selectivity, low recombination, low series resistance and low shading must work together; the equilibrium junction field alone is not a source of electrical energy.

forward bias opposes the built-in electric field and lowers the junction barrier from approximately $V_{\mathrm{bi}}$ to $V_{\mathrm{bi}}-V$. In the abrupt-junction depletion approximation,Thus the depletion region narrows and carriers cross the reduced barrier more readily. electron injection into the p side and hole injection into the n side increase strongly, allowing radiative recombination in an LED.

The injected minority-carrier concentrations, and therefore the ideal diffusion electric current, vary approximately as $\exp (qV/k_{B}T)$. This produces the steep forward branch of the electric current–voltage curve. One should not extrapolate the depletion-width formula to claim that the junction must lose its entire depletion region before conducting; appreciable electric current flows while a barrier remains, and high injection and series resistance eventually invalidate the simple formulas.

Positive dielectrophoresis acts up the gradient of the squared RMS field, towards stronger-field regions; negative dielectrophoresis acts down that gradient, towards weaker-field regions. The sign is the sign of $\Re \tilde{K}(\omega )$.

The force is not generally tangent to an electric field line. Field lines show the direction of the field itself, whereas $\nabla |{\mathbf{E}}_{\mathrm{RMS}}{|}^2$ describes how its strength changes in space. A spatially uniform field gives zero dipole-order dielectrophoretic force even when it is large. Reversing the field each half-cycle does not reverse this time-averaged force, because the relevant intensity-like quantity is quadratic.

Take a two-dimensional conducting sheet in the $xy$ plane with magnetic flux density $\mathbf{B}=B\hat{\mathbf{z}}$ and electric current density along $x$. Moving carriers feel the transverse Lorentz force $q{\mathbf{v}}_d\times \mathbf{B}$. They accumulate at opposite edges, producing a transverse electric field until its force balances the magnetic deflection:For the transverse-voltage convention $E_y/j_x$, this gives $E_y=v_{x}B$ andThe Hall effect therefore measures carrier sign as well as density in the single-carrier approximation: electrons give the opposite transverse-voltage sign from positive carriers. Some tensor conventions call $E_y/j_x$ ${\rho }_{yx}$ rather than ${\rho }_{xy}$; the physical sign is fixed by the stated lead and field directions.

The Hall voltage is the integral of this transverse electric field across the sample width. In a three-dimensional slab of thickness $t$, its magnitude is $|V_H|=IB/(n|q|t)$; for a sheet, use the sheet number density $n_s$ and omit $t$.

For a second-order response, $P_i^{(2)}({\omega }_3)={ϵ}_0{\sum }_{jk}{\chi }_{ijk}^{(2)}({\omega }_3;{\omega }_1,{\omega }_2)E_j({\omega }_1)E_k({\omega }_2)$, with the appropriate frequency conservation and negative-frequency convention for difference-frequency processes. Interchanging the two input field-frequency pairs gives intrinsic permutation symmetry of nonlinear susceptibility:For second-harmonic generation the input frequencies agree, so the last two Cartesian indices are symmetric and can be contracted into six input-pair labels. This reduces the unrestricted 27 Cartesian entries to at most 18 before crystal symmetries are imposed.

Crystal point-group transformations must leave the material tensor invariant and can force entries to vanish or become related. Under spatial inversion both electric field and polarization reverse, whereas the field product is even. A centrosymmetric bulk therefore has zero electric-dipole ${\chi }^{(2)}$. Surfaces and interfaces can break inversion symmetry, and higher-multipole processes are outside that simple prohibition.

Kleinman symmetry is a stronger approximate permutation symmetry used sufficiently far from resonances when dispersion can be neglected over the participating frequencies. It is not justified merely because two input frequencies are equal; resonant or appreciably dispersive media need the full frequency-dependent relations. These distinctions determine which polarization and propagation geometries have nonzero effective coupling, in addition to whether they satisfy phase matching.^[ref]

The electric field of the reflected beam is perpendicular to the vertical plane of incidence, hence parallel to the water surface. At Brewster's angle the reflected p-polarization amplitude vanishes; only the s-polarization component remains. For incidence from air, approximated by refractive index one,Incidence is measured from the surface normal, whereas solar elevation is measured from the horizontal. The Brewster solar elevation is thereforeThis assumes an ideally flat, transparent, nonmagnetic interface; ripples, absorption and diffuse light would spoil complete polarization.

An accelerating point charge has a transverse radiation field in addition to its near field. At nonrelativistic speed and large distance $R$,The acceleration is evaluated at retarded time. Radiation fields decay as $1/R$, so their outward energy flux integrated over a sphere is independent of radius; the near electric field instead decays as $1/R^2$. The radiation vanishes along the acceleration and is strongest perpendicular to it.^[ref]

Using the Poynting vector givesThe integral uses $\int {\sin }^2\vartheta d\Omega =8\pi /3$. This is the Larmor formula. A harmonically accelerated charge with peak acceleration $a_0$ has mean power $q^2{a}_0^2/(12\pi {ϵ}_0{c}^3)$; external work supplies this radiated energy in steady operation. Uniform inertial motion does not radiate.

At relativistic speed the angular pattern becomes directionally concentrated and retarded-source factors matter. The total instantaneous radiated power generalizes to $q^2{\gamma }^6[a^2-|\mathbf{\beta }\times \mathbf{a}{|}^2]/(6\pi {ϵ}_0{c}^3)$, with $\mathbf{\beta }=\mathbf{v}/c$. The displayed simple ${\sin }^2\vartheta$ pattern is restricted to the nonrelativistic limit.

The electric potential and magnetic vector potential represent the fields through $\mathbf{B}=\nabla \times \mathbf{A}$ and $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$. They are not unique: a differentiable function $\chi$ giveswithout changing either field. Gauge fixing restricts this freedom to suit the sources, boundaries or symmetry; it does not alter observables.

The Lorenz gauge $\nabla \cdot \mathbf{A}+c^{-2}{\partial }_t\varphi =0$ is Lorentz covariant. Maxwell's equations then reduce to sourced wave equations,Retarded solutions implement causal radiation conditions. This gauge is useful for radiation and relativistic calculations; residual transformations obey the homogeneous wave equation.^[ref]

The Coulomb gauge $\nabla \cdot \mathbf{A}=0$ instead gives ${\nabla }^2\varphi =-\rho /{ϵ}_0$. Its electric potential is an instantaneous Poisson solution, but the accompanying magnetic vector potential makes the full physical electric field causal. It is useful in separating longitudinal electric fields from transverse radiation. In a source-free vacuum region, suitable radiation boundary conditions permit radiation gauge $\varphi =0$, $\nabla \cdot \mathbf{A}=0$, leaving two transverse plane-wave polarizations. These extra conditions are not a gauge choice applicable unchanged to arbitrary nonzero charge distributions.

A transverse static electric field can induce linear birefringence through the electro-optic Kerr effect. Define its coefficient by $\Delta n={\lambda }_{0}K{E}_{\mathrm{dc}}^2$, with the principal axes parallel and perpendicular to the applied field. A length $d_K$ then gives ${\delta }_K=2\pi K{d}_K{E}_{\mathrm{dc}}^2$.

For input polarized at $45^{\circ }$ to those axes, switching from zero retardation to $|{\delta }_K|=\pi /2$ switches linear to circular polarization. Other input angles generally become elliptical, and axis-aligned input stays linear. The required field magnitude for the lowest-order setting isA Kerr cell is suitable as a switchable quarter-wave plate. The field can be changed electronically without mechanically rotating the axes; actual switching time depends on the material response and driving circuit. This is induced linear birefringence, not polarization-selective absorption.^[ref]

In $S$, adding the stationary opposite line cancels charge density and electric field but contributes no current. HenceThe neutral current filament in a boosted frame is not neutral. Transform its line-integrated four-current:The same result follows by adding the original resting line $({\lambda }^{\prime },0)$ to the new negative line, whose primed density and current are $(-\gamma \lambda ,+\gamma \lambda u)$. In particular ${\lambda }^{\prime }-\gamma \lambda =-\gamma \lambda u^2/c^2$ because $\lambda =\gamma {\lambda }^{\prime }$.

Transforming the total fields givesFor positive $\lambda$, $\mathbf{u}\times \mathbf{B}$ points inward, explaining the electric-field sign. The primed Gauss and Ampère fields calculated from ${\lambda }_{\mathrm{tot}}^{\prime }$ and $I_{\mathrm{tot}}^{\prime }$ reproduce these expressions. Both a moving charge distribution and a moving current must be transformed; boosting just one constituent would give an inconsistent total.

An ideal current-carrying wire consists of opposite charge populations, one associated with the stationary lattice and one with drifting carriers. They may cancel in the lattice frame while carrying a net current, as in part (c), up to exchanging the charge signs and current direction. In a carrier-rest description their densities no longer cancel because the two populations undergo different length contractions. The wire then has both electric fields and magnetic fields.

A neutral steady current in one frame generally becomes a charged steady current in another. This illustrates how electric fields and magnetic fields are parts of one relativistic field rather than separately invariant forces. Real resistive wires also have a driving longitudinal electric field and geometry-dependent surface charges; the infinite uniform filament model isolates the transverse fields of a neutral current and does not describe every field of a complete circuit.

Spatial coherence compares the electric fields at different positions at a common time; temporal coherence compares the field with itself at a delay. Experimentally, they control whether separated apertures or unequal optical paths can form stable interference fringes. They are correlation properties, not a requirement that every emitted wave have a deterministic phase.

The source diameter is $\alpha =40/206265=1.94\times 10^{-4}\mathrm{rad}$. At a representative visible wavelength $\lambda =550\mathrm{nm}$,For a uniformly bright circular disc, the first visibility zero is at $1.22\lambda /\alpha =3.5\mathrm{mm}$; the coefficient depends on the chosen coherence criterion and brightness distribution. Atmospheric phase fluctuations can further spoil measured coherence.

For an explicitly assumed broad visible band from $400$ to $700\mathrm{nm}$, the frequency span is $\Delta \nu =c(1/{\lambda }_{\min }-1/{\lambda }_{\max })=3.21\times 10^{14}\mathrm{Hz}$. ConsequentlyThus broadband planetary light has millimetre-scale transverse and micrometre-scale longitudinal coherence. The receiver band is not supplied: a narrow filter would increase the temporal coherence length, so this is a stated broadband estimate, not a unique source-only constant.

A Hertzian dipole is an ideal localized oscillating source much smaller than its radiation wavelength. Its electric version is an infinitesimal current element; its magnetic version is a small current loop with moment $\mathbf{m}=IA\mathbf{n}$. Such elements model the far-field radiation and radiation resistance of electrically small antennas. The approximation requires source dimensions much smaller than $c/\omega$.

The supplied magnetic field has a quasistatic $[m]/r^3$ term, an induction $[\dot{m}]/(c{r}^2)$ term and a transverse radiation $[\ddot{m}]/(c^{2}r)$ term. The electric field has induction and radiation terms. Brackets mean evaluation at the retarded time $t_r=t-r/c$, not a time average. Near terms mainly exchange stored energy; only the $r^{-1}$ fields give a nonvanishing outgoing power integral as $r\to \infty$.

For a fixed dipole axis the radiation fields satisfy $E_{\varphi }=-c{B}_{\theta }$. The radial Poynting flux is thereforeUsing $\int {\sin }^2\theta d\Omega =8\pi /3$ yieldsFor a general changing dipole direction, replace the squared scalar acceleration by $|\ddot{\mathbf{m}}{|}^2$.

Radiation resistance is the equivalent input resistance whose mean power equals radiated power, not ohmic heating. For a fixed small loop carrying $I(t)=I_0\cos \omega t$, $m=AI$, soThis uses peak current $I_0$; with rms current the power is $I_{\mathrm{rms}}^2{R}_{\mathrm{rad}}$.

The satellite loop needs its changing orientation as well as its induced current. Put the spin axis along z, the external field along x, and let $\mathbf{n}(t)=(\cos \omega t,\sin \omega t,0)$. With $A=\pi b^2$ and resistance $R=10^{-4}\Omega$, the resistance-only approximation givesWrite $m_0=B{A}^2\omega /R$. The full magnetic moment isIt has a static component and a rotating component at $2\omega$. Consequently $|\ddot{\mathbf{m}}{|}^2=4m_0^2{\omega }^4$, andHolding the dipole axis fixed while inserting the sinusoidal current would instead give $B^2{A}^4{\omega }^6/(12\pi {ϵ}_0{c}^5{R}^2)$, eight times smaller. That is the fixed-axis-loop result; it does not describe the stated rotating loop. This is frequency doubling of an induction-driven rotating loop; the changing vector direction is essential.

The resistance-only model is conditional: loop inductance is not supplied. If self-inductance L matters, solve $L\dot{I}+RI=BA\omega \sin \omega t$. In the steady state $I_0=BA\omega /\sqrt{R^2+{\omega }^2{L}^2}$. Its phase changes the static moment direction but not the twice-frequency component's magnitude, so the power becomesThe first expression is the requested resistive idealization and an upper estimate within this passive RL correction. Radiation reaction is negligible compared with the stipulated ohmic resistance. The radiated energy comes from rotational kinetic energy: the induced magnetic torque brakes the loop. Most lost rotational energy is Joule heating; a small part escapes as radiation. Maintaining a constant spin would require external mechanical power.

For scale only, take $B=50\mu \mathrm{T}$, a representative near-Earth field rather than a supplied orbital measurement. The terrestrial surface range reported by NOAA is roughly $25$–$65\mu \mathrm{T}$; the actual orbital value depends on position.^[ref] With $b=1\mathrm{m}$ and $\omega =2\pi (500{\mathrm{s}}^{-1})$, the resistance-only model givesInductance would reduce this power. The kilohertz signal is exceptionally weak and a metre-scale receiving antenna is electrically tiny. Ionospheric absorption/reflection and strong natural or instrumental low-frequency backgrounds make ordinary ground reception impractical. This is not a theorem that the ionosphere blocks every low-frequency mode: field-guided propagation and condition-dependent leakage are real, as the VIPER mission's measurements were designed to test.^[ref]

At a separation $r=100\mathrm{km}$, $k_{\mathrm{rad}}r\simeq 2.1$, so the observer is not securely in the far field. The radiation term alone has at most the characteristic magnetic amplitudein this optimistic resistance-only vacuum model. Near and induction fields must be retained for a full prediction; a simple far-field inverse-square power estimate is not an exact received signal here. Another satellite avoids the whole ground ionospheric path, but plasma propagation, receiving sensitivity, bandwidth, orientation, integration time and spacecraft noise remain unspecified. Detection is not forbidden in principle, but routine practical detection is implausible at this scale; a definitive instrument-level yes/no does not follow from the supplied data.

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.

antenna quantities. radiation resistance represents emitted power by $P_{\mathrm{rad}}=I_{\mathrm{rms}}^2{R}_{\mathrm{rad}}$ at the feed. It is not Joule heating in the antenna; a separate loss resistance accounts for that. antenna gain is the directional far-field intensity relative to an isotropic radiator supplied with the same input power:It includes radiation efficiency and equals antenna directivity only for a lossless antenna. antenna effective area is defined by the power available to a matched load, $P_{\mathrm{av}}=A_{\mathrm{eff}}{S}_{\mathrm{inc}}$, for incidence from a specified direction and matched polarization. It need not equal a literal geometric collecting area.

Gain–area relation. To derive the antenna gain-effective-area relation, take a passive reciprocal antenna. Its electromagnetic reciprocity makes the receive pattern proportional to the transmit pattern; write $A_{\mathrm{eff}}(\Omega )=KG(\Omega )$. The normalization follows from thermal equilibrium. In the Rayleigh-Jeans law limit, blackbody radiation radiance per unit frequency per unit solid angle is $2k_{B}T/{\lambda }^2$ for two polarizations. A single antenna polarization couples to one half, so the available incident noise power isThe open-circuit Johnson voltage spectrum of the antenna resistance is $4k_{B}T{R}_{\mathrm{rad}}d\nu$. A matched load receives half the open-circuit voltage, giving available power $k_{B}Td\nu$. In a lossless antenna at thermal equilibrium these powers agree; hence $\int A_{\mathrm{eff}}d\Omega ={\lambda }^2$. Since $\int Gd\Omega =4\pi$,The same efficiency factor multiplies receiving area and gain if reciprocal dissipative losses are included consistently. Impedance mismatch or polarization mismatch reduces delivered power further. The factor of two for thermal polarization modes and the factor of four between open-circuit voltage variance and matched-load power are separate effects.

Normalize the radar beam. Interpret the azimuth as signed about the beam centre at $\varphi =0$, so both sides of the narrow lobe are included. Put $u=\theta -\pi /2$. To leading order in the small angular width, $\sin \theta \simeq 1$ and the angular integrals can be extended over the real line:The Gaussian antenna beam solid angle gives the peak on the positive $x$-axis consequently hasHere $N$ is power per solid angle, equal to $r^2{S}_r$ in the far field. The leading neglected solid-angle correction is relative order ${\alpha }^2$.

Return using the supplied electric-dipole response. The electric-dipole radar cross section of a small conducting sphere determines this return. At target range $D$, the incident intensity is $S_t=C/D^2$. Write its root mean square electric field as $E_t$, with $S_t={ϵ}_{0}c{E}_t^2$. The induced electric dipole is $p_{\mathrm{rms}}=4\pi {ϵ}_0{a}^3{E}_t$. A dipole's angular radiation pattern, normalized by $\int {\sin }^2\chi d\Omega =8\pi /3$, gives $d{P}_{\mathrm{sc}}/d\Omega =(3P_{\mathrm{sc}}/8\pi ){\sin }^2\chi$. The backward direction is perpendicular to the induced dipole, so the supplied total dipole power givesThe received intensity is this angular power divided by $D^2$. Using the monostatic radar equation, or directly multiplying by the receive area, givesThe receiver threshold is a voltage across the matched load, so the detectable power is $V_m^2/R$, without another matching factor:This is the electric-dipole result following from the response data provided. It assumes $ka\ll 1$, far-field propagation on each leg, alignment at the beam maximum and no attenuation or additional receiver losses.

Perfect-conductor qualification. perfect-conductor sphere dipole interference arises because the sphere also excludes the oscillating magnetic field. Solving the magnetic boundary problem gives $\mathbf{m}=-2\pi a^3{\mathbf{H}}_t$ in addition to the electric dipole. The magnetic polarizability of a perfectly conducting sphere follows from its induced surface currents.^[ref] In the backward direction the magnetic contribution has half the electric scattering amplitude and the same sign, so the returned intensity is $9/4$ of the electric-only value. Equivalently, the monostatic radar cross section is ${\sigma }_{\mathrm{back}}=9\pi k^4{a}^6$, rather than $4\pi k^4{a}^6$. Including both leading dipoles therefore givesBoth results have been distinguished because the supplied auxiliary data describe only the electric dipole, while the literal perfect-conductor boundary conditions also generate the magnetic one. The backward interference is a leading-order effect, not a higher-order correction in $ka$.^[ref]

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.

Birefringence is polarization-dependent propagation in an optically anisotropic medium. Its permittivity tensor relates $\mathbf{D}$ to $\mathbf{E}$. In a lossless uniaxial dielectric, take the optic axis along $z$ and the principal relative permittivities as $(n_o^2,n_o^2,n_e^2)$. For a wave vector making angle $\vartheta$ with that axis, the ordinary mode has index $n_o$, while the extraordinary phase-index angular dependence isThe ordinary electric field is perpendicular to the plane containing the wave vector and optic axis. The extraordinary electric displacement field lies in that plane and is perpendicular to the wave vector, but its electric field need not be. Its energy-flow direction can therefore differ from the wave-normal direction: birefringent walk-off. Along the optic axis the two indices coincide. Refraction generally splits an incident beam into two differently polarized rays, producing double images, as in calcite.

A waveplate instead uses the relative phase of two eigenpolarizations. With the axes chosen for the propagation direction, the retardance is $\Delta =2\pi L(n_2-n_1)/{\lambda }_0$. A quarter-wave plate gives $|\Delta |=\pi /2$ modulo $2\pi$ and converts equal-amplitude orthogonal components between linear and circular polarization. A half-wave plate gives $|\Delta |=\pi$ and rotates a linear polarization. This is a phase effect; unequal absorption is a separate property.^[ref]

Using the needle definitions in the question-level solution, $\mathbf{u}\cdot \mathbf{e}=u_z=\cos \theta$. The scattered field is parallel to $\mathbf{v}=u_x\hat{\mathbf{x}}+u_z\hat{\mathbf{z}}$, andThe light is linearly polarized along the needle's projection into the $xz$ observation plane. Its Jones vector in the $(x,z)$ basis is $(u_x,u_z{)}^{\mathsf{T}}/\sqrt{1-u_y^2}$ whenever the intensity is nonzero. A needle perpendicular to the incident electric field, or pointing directly at the observer, has zero leading-order signal.

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.