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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.

Interpreting the cone as Cherenkov radiation, its angle gives the speed, not directly the mass:Combining this with the relativistic kinetic energy $T=(\gamma -1)m{c}^2$ gives the Cherenkov mass inferenceThe material must be transparent at the observed frequency, $n$ is its phase refractive index there, and $n\cos \theta >1$. For $T>0$ this requires $0<\theta <\arccos (1/n)$ for a massive particle above threshold. Radiation of an unspecified other origin would not determine the mass from these data alone.

Synchrotron radiation is radiation from relativistic charges undergoing transverse acceleration, for example electrons bent by a magnetic field. Relativistic aberration concentrates the radiation within an angle of order $1/\gamma$ about the instantaneous velocity. An observer sees short pulses as this cone sweeps past; synchrotron pulse compression produces a broad spectrum extending far above the orbital frequency.

For an ultrarelativistic circular trajectory of curvature radius $\rho$, the characteristic scales are^[ref] The synchrotron critical frequency sets the spectral turnover; it is not a sharp upper cutoff. The transverse-acceleration Lienard formula givesThe strong energy dependence is a major loss mechanism in electron storage rings. Radiation observed in the orbital plane is linearly polarized in that plane; away from it the two components generally produce elliptical polarization, with opposite handedness on opposite sides. Accelerator light sources exploit its brightness, broad tunable spectrum and pulsed timing. At low speeds the relativistic beaming and high-frequency extension disappear, approaching ordinary cyclotron radiation.^[ref]

A photonic crystal has a spatially periodic dielectric response on the scale of the optical wavelength. For a nonmagnetic, lossless, nondispersive dielectric, the source-free Maxwell equations can be written as the eigenproblemThe Bloch theorem gives ${\mathbf{H}}_{n\mathbf{k}}=e^{i\mathbf{k}\cdot \mathbf{r}}{\mathbf{u}}_{n\mathbf{k}}(\mathbf{r})$ with periodic $\mathbf{u}$. The allowed frequencies form photonic band structures. Coupling of counterpropagating waves near a Brillouin zone boundary opens a stop band. A one-dimensional dielectric multilayer is a simple example; two-dimensional hole arrays and three-dimensional dielectric networks allow more extensive control.

A complete photonic band gap excludes bulk propagating modes for every direction and polarization in a frequency interval. Periodicity alone does not guarantee such a gap: dimensionality, symmetry and refractive-index contrast matter. A missing or altered element can support a localized defect state in the gap, making a cavity; a line defect can guide light. Finite crystals have leakage and wavelength-dependent transmission rather than an infinitely sharp exclusion. Applications include selective mirrors, optical filters, small resonators and integrated waveguides.^[ref]

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.

Let $I_0$ be the intensity from either fibre alone at the observation point, assuming equal output intensities, matching polarizations and negligible drift. The screen samples the relative phase $\Phi (X)$ between the two outputs. For an ideal monochromatic laser, first-order optical coherence is complete andThe interference fringes are approximately equally spaced in the small-angle far field, with a slowly varying diffraction envelope. If the output separation is $a$ and the screen distance is $D$, the interference fringe spacing is approximately ${\lambda }_{0}D/a$. Equal phases at the two input ends do not ensure a central bright fringe: the extra fibre contributes carrier phase $2\pi n_{\mathrm{eff}}(l_2-l_1)/{\lambda }_0$ in addition to the free-space optical path difference.

For a finite linewidth, the time-averaged interference signal becomesHere $g^{(1)}$ may be defined for the slowly varying field envelope, with the carrier phase already included in $\Phi$. A realistic laser therefore gives reduced interferometric visibility; for $V=1/2$, its extrema are $I_0$ and $3I_0$.

For a Lorentzian line shape of frequency full width at half maximum $\Delta \nu$, temporal optical coherence givesThe relevant index is the group index. Using the intended nondispersive approximation $n_g\simeq n_{\mathrm{eff}}=2$,This delayed-fibre linewidth measurement assumes the loss of contrast comes from linewidth. Unequal output intensities, polarization mismatch or mechanical phase fluctuations also lower interferometric visibility. If dispersion matters, the stated phase index alone does not specify the necessary group index.

A phased array uses fixed antennas driven coherently, with adjustable electronic delays or phases. Changing those relative phases changes the direction of constructive interference.

For $N$ identical elements at ${\mathbf{r}}_j=jd\hat{\mathbf{x}}$, take time dependence $e^{-i\omega t}$ and feed phase $j\psi$. In a far-field direction at angle $\theta$ from the array normal, the array factor isMultiply $|F{|}^2$ by the single-element antenna beam pattern to obtain the total angular intensity. All elements add in phase in the selected direction ifThus electronic control of $\psi$ steers the beam to ${\theta }_0$ without moving an element. Reciprocity gives the corresponding directionality in reception.

The aperture $Nd$ controls angular resolution: the nearest nulls obey $|\sin \theta -\sin {\theta }_0|=\lambda /(Nd)$, giving a main-beam width of order $\lambda /(Nd\cos {\theta }_0)$ away from endfire. Additional equally strong maxima can occur when $\chi =2\pi m$; these are grating lobes. Choosing $d<\lambda /2$ avoids them for all steering directions, with $d=\lambda /2$ the usual limiting spacing. Amplitude tapering reduces side lobes at the cost of a wider main beam. For a broad frequency band, true time delays preserve steering more accurately than frequency-independent phase shifts.

A charge moving uniformly through a transparent dielectric can emit Cherenkov radiation if its speed exceeds the phase velocity of light in that medium:It need not exceed the vacuum speed $c$. The passing charge polarizes the medium; the induced response produces radiation that adds coherently along selected directions. Equivalently, phase matching along the trajectory requires $\mathbf{k}\cdot \mathbf{v}=\omega$. Since $|\mathbf{k}|=n(\omega )\omega /c$,Here ${\theta }_C$ is the angle of the emitted light rays to the forward particle velocity: the rays form a cone around the track. With dispersion, both the threshold and cone angle depend on frequency; only transparent frequency bands satisfying the threshold contribute escaping radiation.

In the wavefront construction, a disturbance emitted a time $t$ earlier has radius $ct/n$, while the particle has moved $vt$. Tangents from the present particle position to these wavefronts make a backward-opening wavefront cone of half-angle $\alpha$, where $\sin \alpha =1/(\beta n)$. The light rays are normal to this cone. Consequently $\alpha +{\theta }_C=\pi /2$: the Cherenkov photon-cone and wavefront-cone angles are complementary, rather than identical.

Interpret the stated ${\tau }_1=1\mathrm{ns}$ as the coherence-decay time, or the mean interval between fully phase-randomizing collisions in the simple impact model of pressure broadening. Then each line has correlation $e^{-|\tau |/{\tau }_1}{e}^{-i{\omega }_j\tau }$. The collisional phase relaxation and Lorentzian linewidth transform givesThe intrinsic spectrum is the sum of two such Lorentzian line shapes, still centred $5.0\mathrm{GHz}$ apart. With equal weights and balanced arms, its normalized visibility isFor the intensity normalization of the first source plot, the corresponding model interferogram isThe constant background remains $4$; only the oscillatory contrast decays. The beating zeros and recurrence spacing remain fixed, but successive visibility revivals shrink, reaching a factor $e^{-0.2}\simeq 0.819$ at $|\tau |=200\mathrm{ps}$ and $e^{-1}$ at $1\mathrm{ns}$. Long-delay fringes disappear toward the background.

The sketch chooses a carrier frequency for legibility; the envelope and linewidth follow from the inferred spacing and specified coherence time. A finite experimental delay window additionally convolves these intrinsic Lorentzians with the instrumental line profile. If ${\tau }_1$ were instead a population lifetime, the lifetime-limited field decay rate would be $1/(2{\tau }_1)$ and the corresponding linewidth would be half the boxed value. The source does not define that time convention explicitly; pressure-collision coherence decay is the interpretation used here.

Take $u>0$ to mean that the mirror recedes along the incident plane electromagnetic wave. Put $\beta =u/c$, with $|\beta |<1$. The moving-mirror Doppler shift can be obtained directly by matching wave phase on the moving surface. Write the incident wave phase as $k_{i}z-{\omega }_{i}t$ and the reflected wave phase as $-k_{r}z-{\omega }_{r}t$, where $k_i={\omega }_i/c$ and $k_r={\omega }_r/c$ are positive wavenumbers. At the mirror, $z=z_0+ut$, the two phases must have the same time dependence; their constant offset incorporates the reflection phase shift. ThusConsequently the observed reflected angular frequency and ordinary frequency areThis is a redshift for a receding mirror and a blueshift for an approaching mirror. At small velocity, ${\omega }_r\simeq {\omega }_i(1-2u/c)$.

Equivalently, the relativistic Doppler effect gives ${\omega }_i^{\prime }=\gamma (1-\beta ){\omega }_i$ in the mirror rest inertial frame. Reflection there reverses propagation and preserves frequency. Transforming the backward-going reflected wave to the laboratory gives another factor $\gamma (1-\beta )$, so ${\omega }_r={\gamma }^2(1-\beta {)}^2{\omega }_i={\omega }_i(1-\beta )/(1+\beta )$. A single Doppler factor would miss the effect of the return path.

The coherence length from a finite spectral band is most safely estimated by converting its endpoints to frequency, because the quoted band is broad:The reciprocal-bandwidth optical coherence time and corresponding longitudinal optical coherence length areThe given temperature is consistent with the spectral peak: Wien's displacement law gives ${\lambda }_{\mathrm{peak}}=2.898\times 10^{-3}\mathrm{m}\mathrm{K}/5778\mathrm{K}\simeq 502\mathrm{nm}$. Using the common narrow-band estimate ${\lambda }_{\mathrm{peak}}^2/\Delta \lambda$, with $\Delta \lambda =600\mathrm{nm}$, gives about $0.42\mu \mathrm{m}$. Its closeness here does not justify the narrow-band approximation for arbitrary broad spectra.

The $0.40\mu \mathrm{m}$ result is a bandwidth-scale estimate. A flat spectral density per unit frequency would make it the first-zero optical coherence length; a truncated Planck law spectrum is not flat and would have a different correlation function. Neither the stated endpoints nor the word “coherence” specifies an exact visibility threshold. Propagation to Earth sets the angular size and transverse scale, but does not by itself narrow the solar optical spectrum or lengthen this longitudinal scale.

In Rayleigh scattering, a small bound system develops an induced electric dipole moment $\mathbf{p}(t)=\mathrm{Re}[\alpha (\omega ){\mathbf{E}}_0{e}^{-i\omega t}]$. Well below its binding resonances, the electric polarizability is approximately its static value $\alpha (0)$, independent of frequency. The mean radiated power and incident light intensity areDividing gives the total scattering cross sectionThis Rayleigh scattering wavelength dependence comes from the two time derivatives of an oscillating electric dipole moment, whose radiated field is proportional to ${\omega }^2{p}_0$. It requires the same scatterer with approximately constant low-frequency electric polarizability; resonances and strong optical dispersion invalidate the simple fourth-power law.

Thomson scattering describes the low-recoil scattering of electromagnetic radiation by a free charge. For charge $q$ and mass $m$, the weak incident field gives peak acceleration $|q|E_0/m$. The mean Larmor formula power is thereforeThere is no binding force to introduce a characteristic frequency. Dividing by the same incident light intensity gives the Thomson cross-section:For an electron, the length inside parentheses is the classical electron radius. The cross section has units ${\mathrm{m}}^2$ and is independent of wavelength in this classical limit. The driven velocity must remain small compared with $c$, and $\hslash \omega \ll m{c}^2$ must make Compton recoil kinematics negligible.

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

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

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

The transmitting antenna is matched to its input transmission line, so its real terminal electrical resistance is $Z_i$. The identical receiving antenna has the same electrical impedance at the operating frequency. Its available received power for a matched load is the incident energy flux times its antenna effective area:The real receiving load is $Z_t$, so its voltage-wave reflection coefficient relative to $Z_i$ is ${\Gamma }_L=(Z_t-Z_i)/(Z_t+Z_i)$. The fraction of available power delivered to the output line isConsequently, neglecting path losses, mismatch of polarization of an electromagnetic wave and line losses,This is the Friis transmission equation with a receiving impedance-mismatch factor. For fixed $P_i,G_a,d,Z_i,Z_t$, it scales as ${\lambda }^2$. That scaling is conditional on holding antenna gain fixed: a fixed physical aperture can instead have $G_a\propto {\lambda }^{-2}$, and any wavelength dependence of the impedances also enters the expression. The matched-output check $Z_t=Z_i$ gives $P_t=P_i\mathcal{F}$.

A uniform Lorentz boost preserves the plane wave form, while changing both its frequency and its vector wave amplitude. Decompose vectors parallel and perpendicular to $\mathbf{v}$, so $\mathbf{k}={\mathbf{k}}_{\parallel }+{\mathbf{k}}_{\perp }$. In the observer's primed coordinates,The transformed wave amplitude is specified in part (d). The unprimed “rest frame” in this question is the original coordinate frame, not a rest frame of a vacuum light wave, which has no inertial rest frame.

The frequency and longitudinal wavevector formulas are exactly the boost of a contravariant Lorentz four-vector:where the matrix is understood on the time component and the signed scalar component along the boost direction. Hence the four-wavevector obeysFor the metric signature $(+,-,-,-)$, lowering the position index gives $R_{\alpha }=(ct,-\mathbf{r})$, soIt is a dimensionless Lorentz scalar, equal to minus the wave phase used in the stated complex wave. Invariance follows by contracting the mutually inverse position and wavevector transformations. Reversing metric signature reverses this displayed scalar's sign, but not its invariance or any observable phase difference. In vacuum, $K_{\alpha }{K}^{\alpha }=0$ is an additional null-vector check; the wave remains lightlike in every inertial frame.

Yes: the same physical resonant cavity and its field remain resonant in every inertial description. A Lorentz transformation changes the mirror worldlines and the field together. The boundary phase matching and the constructive-interference condition are statements about the same reflection events, and the contracted wave phase is a Lorentz scalar. A boosted resonant solution therefore still satisfies the moving-mirror boundary conditions.

It would be incorrect merely to insert a Lorentz-contracted length into the formula for two stationary mirrors while leaving both propagation frequencies unchanged. For example, let the mirrors be separated by $L$ in their rest frame and let ${\omega }_{0}L/c=m\pi$, including the usual equal-mirror reflection-phase convention. For an observer boosted at $+v$ along the cavity axis, the cavity moves at $-v$ and its two travelling components have frequenciesThey need not have one common laboratory frequency. The simultaneous mirror separation is $L^{\prime }=L/\gamma$, while the forward and return flight times are $L^{\prime }/(c+v)$ and $L^{\prime }/(c-v)$. The round-trip interval between events on the same mirror is therefore $2\gamma L/c$.

Along that moving mirror the wave phase rate of either wave component is ${\omega }^{\prime }-k^{\prime }{v}_{\mathrm{mirror}}={\omega }_0/\gamma$. Its advance over the round trip isThus the resonance condition is unchanged even though the observed travelling frequencies, lengths and flight times all change. For a boost in a general direction the beam directions also undergo relativistic aberration; the event-by-event phase-invariance argument still applies. A separate stationary cavity placed in the new frame, or an external laser whose frequency is not transformed with the original setup, is a different physical system.