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.

Write a nonzero vacuum plane electromagnetic wave with wavevector $\mathbf{k}=(\omega /c)\mathbf{s}$, where $\mathbf{s}$ is a unit propagation direction. Its propagation Lorentz four-vector is $K^{\mu }=(\omega /c,\mathbf{k})$, or, up to a scalar normalization in any one frame, $n^{\mu }=(1,\mathbf{s})$. It is null because $K_{\mu }{K}^{\mu }=0$.

Choose coordinates $x^{\mu }=(ct,\mathbf{x})$ and Minkowski metric ${\eta }_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$. With the supplied three-dimensional boost formulas, a consistent convention assigns the printed array to upper-index components,The index conventions for the electromagnetic field tensor then require $F_{0i}=+E_i/c$ and $F_{ij}=F^{ij}$ when both indices are lowered. The printed lower-index label therefore cannot be kept together with its electric signs and a forward propagation vector in this convention.

This is not merely an arbitrary overall sign. For a valid forward wave with $\mathbf{s}=\hat{\mathbf{z}}$, $\mathbf{E}=E_0\hat{\mathbf{x}}$ and $\mathbf{B}=(E_0/c)\hat{\mathbf{y}}$, multiplying the printed array, unchanged, by $n^{\mu }=(1,0,0,1)$ gives an $x$ component $-2E_0/c$, rather than zero. One consistent repair is to retain the printed array as $F^{\mu \nu }$ and contract with $n_{\mu }=(1,-\mathbf{s})$. Equivalently, lower both tensor indices before forming the contraction asked for in the question.

Using that corrected lower-index tensor givesTo establish both electric and magnetic transversality, include the dual electromagnetic field tensor. Choose ${\tilde{F}}^{0i}=-B_i$ and ${\tilde{F}}^{ij}={ϵ}_{ijk}{E}_k/c$; lowering both indices then givesThe source-free plane-wave Maxwell equations are precisely the pair of covariant conditionsTheir temporal components give $\mathbf{k}\cdot \mathbf{E}=0$ and $\mathbf{k}\cdot \mathbf{B}=0$. Their spatial components give the associated amplitude and orientation relations, $\mathbf{E}/c=-\mathbf{s}\times \mathbf{B}$ and $\mathbf{B}=\mathbf{s}\times \mathbf{E}/c$.

To see the Lorentz covariance explicitly, let $V_{\nu }=K^{\mu }{F}_{\mu \nu }$. It is a covector, so if $V_{\nu }=0$ in one inertial frame, its Lorentz transformation is also zero in every other inertial frame. The same is true of $K^{\mu }{\tilde{F}}_{\mu \nu }$. Since the transformed wave still has $K^{\prime 0}>0$ and $|{\mathbf{k}}^{\prime }|={\omega }^{\prime }/c$, their temporal components implyHere $K^{\mu }$, rather than a vector renormalized to have temporal component one in every frame, is the actual Lorentz four-vector; multiplying a zero contraction by a nonzero normalization does not affect the argument.

Finally, the two scalar perpendicularity statements alone are not equivalent to one tensor contraction. They do not enforce the amplitude relation between $\mathbf{E}$ and $\mathbf{B}$. Conversely $n^{\mu }{F}_{\mu \nu }=0$ alone cannot exclude a longitudinal magnetic field: adding $B_{\parallel }\mathbf{s}$ leaves its cross product unchanged. For a genuine source-free plane electromagnetic wave, the remaining Maxwell equations supply that missing constraint, and both tensor contractions vanish. This qualification and the corrected index placement are necessary for the intended invariance argument.

Let $\mathbf{u}$ be the boost velocity, and decompose the fields into components parallel and perpendicular to it. The parallel components are unchanged. The perpendicular components obeyAll vectors inside these expressions are perpendicular to $\mathbf{u}$. Thus $|\mathbf{u}\times {\mathbf{B}}_{\perp }{|}^2=u^2{B}_{\perp }^2$, and similarly for ${\mathbf{E}}_{\perp }$. In the difference of squares, the mixed terms cancel becauseIt follows thatAdding the unchanged parallel components provesFor the mixed dot product, expand directly. The terms ${\mathbf{E}}_{\perp }\cdot (\mathbf{u}\times {\mathbf{E}}_{\perp })$ and $(\mathbf{u}\times {\mathbf{B}}_{\perp })\cdot {\mathbf{B}}_{\perp }$ vanish. The remaining cross-product identity isHenceParallel-perpendicular mixed dot products are zero in both frames, soRotations also preserve these two expressions, completing the proof for proper Lorentz transformations. The second expression is a pseudoscalar under spatial parity, but is invariant under the boosts at issue here.

These results also follow from tensor contractions. In the convention used below, $F_{\mu \nu }{F}^{\mu \nu }=2(B^2-E^2/c^2)$ and $F_{\mu \nu }{\tilde{F}}^{\mu \nu }=-4\mathbf{E}\cdot \mathbf{B}/c$. The numerical sign of the latter depends on the chosen orientation of the dual electromagnetic field tensor, while its invariance under proper Lorentz transformations does not.

A distinct classical Thomson scattering window requires two separated scales:To remove that window, the binding frequency must be comparable with or greater than the Compton scattering recoil scale. Thus the formal condition for bound-electron scattering without a Thomson frequency window isUsing the stipulated approximate electron mass $m=10^{-30}\mathrm{kg}$, $c=2.998\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$ and $\hslash =1.055\times 10^{-34}\mathrm{J}\mathrm{s}$ givesThe dimensions are $[m^3{c}^4/{\hslash }^2]=\mathrm{kg}{\mathrm{s}}^{-2}=\mathrm{N}{\mathrm{m}}^{-1}$. The inequality denotes a scale comparison, not a uniquely specified spring constant or a sharp transition.

There is also a physical limitation to the question's premise. Compton scattering does not begin discontinuously at $\hslash \omega =m{c}^2$; recoil corrections grow as photon energy approaches the rest energy. If $\hslash {\omega }_0\gtrsim m{c}^2$, the assumed nonrelativistic binding model is itself no longer reliable. The calculation therefore locates where no well-separated classical Thomson scattering range exists; it cannot establish a literal direct transition of a real strongly bound system using this oscillator equation alone.

Put ${\omega }_0^2=k/m$, where $k$ is the binding spring constant rather than an optical wavenumber. The steady complex amplitude solution of the forced Lorentz oscillator model isThe acceleration complex amplitude is $-{\omega }^2{\mathbf{x}}_0$. HenceThis is the scattering cross-section of a harmonically bound charge. The displacement and induced electric dipole moment are parallel to the incident linear polarization, and integrating their dipole radiation over all directions gives the total cross section above.

For $\omega \ll {\omega }_0$, binding dominates andrecovering Rayleigh scattering with static electric polarizability $q^2/k$. For $\omega \gg {\omega }_0$, inertia dominates: ${\mathbf{x}}_0\simeq -q{\mathbf{E}}_0/(m{\omega }^2)$, so its second time derivative becomes frequency independent andThis is a classical Thomson scattering regime only while $\hslash \omega \ll m{c}^2$. Exactly at $\omega ={\omega }_0$, an undamped driven oscillator has no finite steady-state amplitude; the divergent formula must not be used. Finite damping and radiation reaction regulate the physical resonant response. Coherent contributions from many charges in an extended medium can also require a structure calculation rather than multiplying this single-charge result by a particle count.

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.

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.

Denote the orbit radius by $b$ here to distinguish it from observation distance; it is the quantity called $R$ in this part of the paper. WriteFor observation at spherical coordinates $(\theta ,\varphi )$, measured from the orbit normal,The acceleration magnitude is constant, $|\mathbf{a}|=b{\omega }^2$. Therefore the radiation pattern of a circularly moving nonrelativistic charge isTime averaging removes the azimuthal dependence. The axial intensity is twice the equatorial intensity, and no direction has zero mean intensity: the instantaneous dipole axis continually rotates. This is also the sum of the two orthogonal linear-oscillation patterns with their relative phase quadrature. At equal radius and linear-oscillation amplitude, the total mean power is twice the linear result. The calculation requires $b\omega \ll c$; relativistic synchrotron radiation is outside this approximation.

For the prescribed displacement along $z$, the source acceleration is $\mathbf{a}(t_r)=-a{\omega }^2\cos (\omega t_r)\hat{\mathbf{z}}$. If $\theta$ is the angle from the $z$ axis, $|\mathbf{a}{|}^2-(\mathbf{n}\cdot \mathbf{a}{)}^2=a^2{\omega }^4{\cos }^2(\omega t_r){\sin }^2\theta$. Averaging ${\cos }^2$ over one wave period givesThe angular distribution of electric dipole radiation has zero intensity on the oscillation axis and maximum intensity in the equatorial plane, with no azimuthal dependence. Its total power isThe condition $a\omega \ll c$ is the required nonrelativistic condition. The plotted radial coordinate below is angular power, rather than field wave amplitude.

Since $\mathbf{E}\cdot \mathbf{n}=0$, the Poynting vector isThe area on a distant sphere corresponding to solid angle $d\Omega$ is $r^{2}d\Omega$. ConsequentlyThe average is over the source motion at retarded time. At leading nonrelativistic order, differences between source-time and reception-time averaging are beyond the approximation. The units are power per solid angle, $\mathrm{W}\mathrm{s}{\mathrm{r}}^{-1}$.

The acceleration covariance in nonrelativistic radiation is another way to organize this result: with $C_{ij}=⟨a_i{a}_j⟩$, the angular factor is $\mathrm{tr}C-{\mathbf{n}}^{T}C\mathbf{n}$. Integrating over all observation directions uses $\int [|\mathbf{a}{|}^2-(\mathbf{n}\cdot \mathbf{a}{)}^2]d\Omega =(8\pi /3)|\mathbf{a}{|}^2$ and gives the averaged Larmor formula, $⟨P⟩=q^2⟨a^2⟩/(6\pi {ϵ}_0{c}^3)$.

The consistent Liénard–Wiechert potential obeys $\mathbf{A}=\mathbf{u}\phi /c^2$. To first order in $\mathbf{\beta }=\mathbf{u}/c$, its large-distance form isTerms varying as $1/r^2$ give near fields and are negligible for the radiated power at large $r$. Keep the first-order velocity term in $\phi$: although small, its retarded-time derivative contributes a leading acceleration field. To the required order, $\partial t_r/\partial t\simeq 1$ and $\nabla t_r\simeq -\mathbf{n}/c$. Holding the observation direction fixed at leading order givesThus the outgoing electric field is transverse:Here $\mathbf{a}=d\mathbf{u}/d{t}_r$. The magnetic field follows from $\nabla \times \mathbf{A}$, or from the outgoing plane electromagnetic wave relation:The minus sign is essential. Both fields fall as $1/r$, are mutually orthogonal and obey $|\mathbf{E}|=c|\mathbf{B}|$. The corresponding Liénard-Wiechert field also contains nonradiative terms, omitted in this leading far-field calculation.

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.

Convert the distances to SI units: $r=1.5\times 10^{11}\mathrm{m}$ and $D=1.4\times 10^9\mathrm{m}$. The angular solar diameter isTwo receivers separated by a transverse baseline $b$ see a spread of relative wave phases of order $2\pi b\Theta /\lambda$. Averaging over incoherent source directions therefore reduces interferometric visibility when $b$ reaches the transverse optical coherence length scale $\lambda /\Theta$. At $\lambda =500\mathrm{nm}$,If “coherence length” specifically means the first zero of uniform-disk stellar visibility, a uniformly bright circular solar disk gives $g^{(1)}(b)=2J_1(x)/x$, with $x=\pi b\Theta /\lambda$. The first zero then lies at $\boxed{b_0=1.21967\lambda /\Theta \simeq 65\mu \mathrm{m}}$. This more specific number assumes a uniform disk and a first-zero criterion; the characteristic estimate is approximately $50$–$60\mu \mathrm{m}$. Solar brightness variation across the disk changes the exact contrast curve.

Use the same fixed laboratory $x,y$ axes for the incident and returning Jones vectors, with $x$ along the fast axis. Remove a common phase factor so that the one-pass Jones matrix is $J=\mathrm{diag}(1,i)$. The reverse pass through this reciprocal plate accumulates the same relative phase retardation, not its inverse. The mirror has scalar reflection coefficient $r_m$ in these fixed axes. ThereforeThe effective retardance is $\pi$: the complete return acts as a half-wave plate. For initial linear polarization at angle $\alpha$ to the fast axis,In particular, the $45^{\circ }$ input used for linear-to-circular conversion returns in linear polarization perpendicular to its original line: $(1,1)/\sqrt{2}\mapsto (1,-1)/\sqrt{2}$. An input along either plate axis returns along that same axis.

The wording does not specify the initial polarization of waves. Thus “orthogonal linear return” requires the usual $45^{\circ }$ linearly polarized input; it is not a universal answer for arbitrary polarized input. The Jones matrix above supplies the general answer. A non-normal metallic reflection may introduce different complex reflection coefficients for the two components and would require additional geometry and optical data. Descriptions of the intermediate circular polarization must also account for the reversed propagation direction when assigning handedness.

Birefringence gives two orthogonal linear eigenpolarizations with different refractive indices, $n_f$ and $n_s$. They accumulate different wave phases through the same physical thickness $d$. Their relative phase retardation isHere ${\lambda }_0$ is the vacuum wavelength. For the thinnest positive-retardance plate, $d={\lambda }_0/[4(n_s-n_f)]$. The plate uses a difference in phase velocity, rather than differential wave absorption. Equal wave amplitudes along the two axes, obtained from input linear polarization at $45^{\circ }$ to them, acquire phase quadrature and become circular polarization. Reversing that conversion gives linear polarization from the corresponding circular input. A different input angle generally gives elliptical polarization on one pass.

Keep wire 1's electrons moving at $+v$, but let wire 2's electrons move at $-v$, so $I_2=-I_1$. The positive charges remain lab-stationary and still feel zero external electrostatic force in the lab. The appropriate rest frame for wire 2's electrons now moves at $-v$.

In this frame, wire 1's total line charge isIt produces an electric field pointing toward wire 1. The stationary test electron is negatively charged, so its electric force points away from wire 1. Transforming back again divides the individual transverse force by $\gamma$, and multiplying by $\lambda /e$ givesThe changed sign arises from the changed transformed charge imbalance, without needing a magnetic-force calculation. More generally the signed result is $F_x/\ell =-{\mu }_0{I}_1{I}_2/(2\pi d)$ for the chosen outward radial direction.

Put wire 1 on the $z$ axis and wire 2 at $x=d$, with positive $x$ pointing away from wire 1. Let each wire have stationary positive linear charge density $+\lambda$ and moving electron line density $-\lambda$ in the lab. For equal currents, both electron populations have drift velocity $+v\hat{\mathbf{z}}$, so the signed conventional current is $I=-\lambda v$.

First consider wire 2's positive charges in their own rest frame, which is the lab. Wire 1 has zero total line charge there, so its external electrostatic field is zero. The positive-charge force per length is therefore $\boxed{f_{+}=0}$.

Next use the electron rest frame $S^{\prime }$ moving at $+v\hat{\mathbf{z}}$. Wire 1's ion line density becomes ${\lambda }_{+}^{\prime }=\gamma \lambda$. Its electron density becomes ${\lambda }_{-}^{\prime }=-\lambda /\gamma$: the electron rest density is lower than its lab density. The transformed total isEquivalently this follows from the four-current transformation ${\lambda }_1^{\prime }=\gamma ({\lambda }_{1,\mathrm{net}}-v{I}_1/c^2)$, with ${\lambda }_{1,\mathrm{net}}=0$ and $I_1=-\lambda v$. By the electrostatic field of a line charge,A wire-2 electron is at rest in $S^{\prime }$ and feels only the electric force, $F_{e,x}^{\prime }=-e{\lambda }_1^{\prime }/(2\pi {ϵ}_{0}d)$, toward wire 1.

Apply the supplied transverse relativistic force transformation with the electron's lab velocity $\mathbf{u}=\mathbf{v}$. Its denominator is $\gamma (1-v^2/c^2)=1/\gamma$, so $F_{e,x}^{\prime }=\gamma F_{e,x}$, or $F_{e,x}=F_{e,x}^{\prime }/\gamma$. There are $\lambda /e$ electrons per unit lab length. ThusAdding $f_{+}=0$ gives exactly the attractive magnetic result. This electrostatic-frame derivation of parallel-wire forces uses electrostatics in each test carrier's own rest frame; no magnetic force acts on that stationary test carrier there. The other wire can still carry current in that frame. Transform the force per electron first and then count electrons per lab length: transforming a force per unit length as if it were a particle force would introduce an incorrect extra contraction factor.

A long straight wire carrying current $I_1$ produces field magnitude $B_1={\mu }_0|I_1|/(2\pi d)$ at a parallel wire a distance $d$ away. The magnetic force on a current-carrying wire is $d\mathbf{F}=I_{2}d\mathbf{\ell }\times {\mathbf{B}}_1$. Hence for equal currents,The two forces have equal magnitude and opposite directions. Net electrical neutrality removes a lab-frame external electrostatic force between ideal uniform wires, but does not remove their magnetic interaction.

Let an object be at rest in $S_0$, with proper charge density ${\rho }_0$ and no rest-frame current. In a frame moving at $\mathbf{v}$ relative to it, the object moves with velocity $-\mathbf{v}$. A simultaneous volume element is Lorentz-contracted along the motion, $dV=d{V}_0/\gamma$, while its charge is invariant. ThereforeThe object's four-velocity in this frame is $U^{\mu }=\gamma (c,-\mathbf{v})$, soA scalar proper density times a Lorentz four-vector transforms as a Lorentz four-vector, demonstrating the claim. With the alternative convention that $\mathbf{v}$ denotes the object's velocity in the observer frame, the spatial component is $+\gamma {\rho }_0\mathbf{v}$ instead. Summing the four-currents of different carrier species preserves this transformation law even when the combined current has no rest frame, as for a lab-neutral current-carrying wire.

A Lorentz four-vector is a set of four components transforming as $A^{\prime \mu }={\Lambda }^{\mu }{}_{\nu }{A}^{\nu }$ under a Lorentz transformation. Examples include the four-position and four-velocity. A Lorentz invariant is a quantity unchanged by that transformation. With metric signature $(+---)$, the scalar product $A_{\mu }{B}^{\mu }=A^0{B}^0-\mathbf{A}\cdot \mathbf{B}$ is invariant. The components of a four-vector are generally not individually invariant.

The proton's total electric charge is taken to be a scalar: it does not acquire a factor $\gamma$ when the proton moves. Experimental evidence comes from sensitive charge neutrality tests of matter. Hydrogen molecules and helium atoms remain neutral despite different binding and internal electron and nuclear motions. If constituent charge scaled with kinetic energy as relativistic mass does, the same cancellation of fixed numbers of proton and electron charges would generally fail in different bound systems.

In a gas-efflux test of charge neutrality, gas escapes from an electrically isolated vessel connected to an electrometer. Any net charge per escaping molecule would leave a measurable opposite vessel charge, after residual ions and charged contaminants are removed. The hydrogen/helium efflux experiment supports velocity-independent additive charge. ^[ref]

This is evidence for charge invariance, not merely conservation of electric charge: the latter concerns constancy during a process, while invariance compares descriptions or internal motions without rescaling charge by a boost factor. Net-matter neutrality constrains combinations of constituent charges; it should not be described as an exact proof covering every possible velocity-dependent microscopic model.