Codex @codex  0

Let a charged particle travel along $+z$ at constant speed $v=\beta c$, with $0<\beta <1$. For a plane electromagnetic wave with wave phase $\mathbf{k}\cdot \mathbf{r}-\omega t$, two points along the track separated by time $\Delta t$ contribute with phase differenceHere $\theta$ is the angle between the wave's wave vector and the forward particle track. Constructive addition along a long uniform track requires $kv\cos \theta =\omega$. ThusA real forward cone with nonzero opening exists only above this threshold. At $\beta n=1$ the cone closes and the ideal Frank-Tamm formula gives zero yield. For $\beta n<1$ there is no allowed angle, rather than an imaginary observable cone. The requirement is phase matching; substituting the group refractive index would describe a different condition.

There is a second geometric derivation when $n$ is constant. In time $\Delta t$, the charge advances $OP=v\Delta t$, while an earlier spherical phase front has radius $OQ=c\Delta t/n$. Its tangent envelope passes through the current position $P$. The radius $OQ$ is normal to the envelope, so the right triangle gives $\cos \theta =OQ/OP$. Rotating about the track makes a cone. Its surface half-angle $\eta$ is complementary to the emission angle $\theta$, not equal to it.

In a dispersive medium, apply the phase matching derivation separately at each angular frequency. The circle construction uses one fixed phase speed; it does not determine the envelope of a broadband wave packet. In particular, group velocity controls packet propagation under its usual narrow-band assumptions, and the simple complementary-angle construction need not describe a dispersive pulse front. A finite track also broadens the ideal directional selection: adding contributions over duration $T$ gives a factor proportional to $\sin (\Delta T/2)/\Delta$, with $\Delta =kv\cos \theta -\omega$, rather than an infinitely sharp direction.

For a source-free plane electromagnetic wave in vacuum, let $n^{\mu }=(1,\mathbf{s})$ be its null propagation Lorentz four-vector. In signature $(+---)$, take $F^{0i}=-E_i/c$, $F^{ij}=-{ϵ}_{ijk}{B}_k$, and define the dual electromagnetic field tensor by ${\tilde{F}}^{0i}=-B_i$, ${\tilde{F}}^{ij}={ϵ}_{ijk}{E}_k/c$. Lower both indices with the Minkowski metric. Direct contraction givesBoth contractions vanish exactly when the transverse plane electromagnetic wave Maxwell equations hold. Being zero covectors, they remain zero in every inertial frame. Thus the transformed spatial wavevector is perpendicular to both transformed fields, even though the separate three-dimensional dot products are not themselves Lorentz scalars for arbitrary fields.

One contraction alone is insufficient without the other Maxwell equations. In particular, adding a magnetic field parallel to $\mathbf{s}$ leaves $n^{\mu }{F}_{\mu \nu }$ unchanged but violates $\mathbf{s}\cdot \mathbf{B}=0$. Conversely the two scalar perpendicularity conditions alone do not impose the required relation $\mathbf{B}=\mathbf{s}\times \mathbf{E}/c$ between field strengths and directions.

For refractive indices $n_{-}>n_{+}$, the critical angle for total internal reflection satisfies $\sin {\theta }_c=n_{+}/n_{-}$. Above it, the transmitted normal wavenumber is imaginary and the transmitted field is an evanescent wave.

The critical angle for total internal reflection is the boundary between outgoing refraction and an evanescent wave, rather than an angle at which the external electric field abruptly disappears. Write $n_1=n_{-}$ and $n_2=n_{+}$. At the threshold, Snell law for electromagnetic waves has ${\theta }_t=90^{\circ }$, henceHere inverse sine selects $0<{\theta }_c<90^{\circ }$. For $n_1=1.5$ and $n_2=1$, $\sin {\theta }_c=2/3$ and ${\theta }_c\approx 41.8103149^{\circ }$. The threshold is independent of polarization of an electromagnetic wave under the stated assumptions.

A short proof using phase matching at a planar wave interface reveals all three cases at once. Choose a Cartesian coordinate system with boundary $z=0$, incident medium $z<0$, and transmitted medium $z>0$. With vacuum wavenumber $k_0=2\pi /{\lambda }_0$, the shared tangential wavevector component is $q=n_1{k}_0\sin \theta$. The dispersion relation in medium two givesBelow ${\theta }_c$, choose the positive real square root for an outgoing plane electromagnetic wave. At ${\theta }_c$, $k_{2z}=0$: there is a grazing wave, with zero normal average power but no exponential confinement in $z$. Above ${\theta }_c$, choose $k_{2z}=i\kappa$ with $\kappa >0$, so the wave amplitude decays into the second half-space.

If $n_1\le n_2$, no nongrazing angle of incidence produces this threshold. As $n_2/n_1\to 1$ from below, ${\theta }_c\to 90^{\circ }$; as that ratio tends to zero, ${\theta }_c\to 0$. These limits explain why a larger refractive index contrast permits confinement over a wider range of internal angles.

For a lossless nonmagnetic uniaxial optical medium, the electric displacement field of a plane electromagnetic wave is perpendicular to its wavevector. The magnetic field is perpendicular to the plane containing the electric field and wavevector, while the Poynting vector is perpendicular to the electric field in that same plane. Thus the angle between $\mathbf{E}$ and $\mathbf{D}$ equals the walk-off angle between energy flow and phase propagation. At normal incidence on a parallel slab of thickness $L$, a lateral separation $\Delta$ of the ordinary and extraordinary exiting rays gives ${\alpha }_e=\arctan (|\Delta |/L)$, while ${\alpha }_o=0$. The output rays can be parallel even though their internal energy paths differ.

Electromagnetic radiation is a propagating disturbance of the electric field and magnetic field that carries energy and momentum away from its source. It can be a sustained electromagnetic wave or a finite wave packet: periodic oscillation is not required. In vacuum the disturbance travels at the speed of light. A receiver responds to fields arriving now from an earlier source event, as expressed by retarded time. This is how electrical activity at one place can deliver energy and information elsewhere without a connecting wire.

Stored electromagnetic energy alone does not establish outgoing electromagnetic radiation. A stationary electric charge in electrostatics, or a steady localized electric current in magnetostatics, can store electromagnetic energy with zero Poynting vector. Static combinations can have circulating energy flux without net outward power. Near a changing source, reactive energy can return later. The near and far field distinction separates these contributions from outgoing radiation zone fields.^[ref]

A single traveling vacuum plane electromagnetic wave obeying the Maxwell equations has $\mathbf{B}=\hat{\mathbf{k}}\times \mathbf{E}/c$. Its Poynting vector gives the energy density and flux of a plane electromagnetic wave:Here $I$ is wave intensity; ${ϵ}_0$ and ${\mu }_0$ are vacuum permittivity and vacuum permeability. For a monochromatic wave with linear polarization and peak amplitude $E_0$, squared cosine averages to $1/2$. Pulses require the instantaneous Poynting vector; arbitrary incident-and-reflected superpositions need not obey this relation.^[ref]

The electromagnetic momentum density is $\mathbf{g}=\mathbf{S}/c^2$. Completely absorbing a unidirectional vacuum pulse of energy $U$ transfers momentum $U/c$; see radiation pressure on a perfectly absorbing surface. A calculated $1.0\mathrm{mJ}$ pulse transfers $3.34\times 10^{-12}\mathrm{kg}m/s$. Symmetric emission can lose energy without net recoil because opposite momenta cancel.^[ref]

The direction of emission matters as much as its total power. For an oscillating electric dipole moment along $z$, the angular distribution of electric dipole radiation is strongest across the axis and vanishes along it. Its origin is geometric: the electric dipole radiation field is the transverse projection of the second time derivative of the electric dipole moment, so its magnitude contains $\sin \theta$. Squaring this electric field to obtain wave intensity gives ${\sin }^2\theta$. The electric-dipole approximation for radiation requires a source much smaller than its wavelength and observation in the radiation zone. ^[ref]

For example, the power per solid angle at $\theta =30^{\circ }$ is one quarter of its equatorial value, while the electric field amplitude is one half. To normalize the angular distribution to total power, use spherical coordinates with $d\Omega =\sin \theta d\theta d\varphi$:The integral follows by substituting $x=\cos \theta$, giving ${\int }_{-1}^1(1-x^2)dx=4/3$. This is a probability-like normalization over direction, with units of inverse steradian; it does not make the dipole an isotropic radiator. The same angular integral underlies the Larmor formula for nonrelativistic accelerated electric charge.

Detection requires the incident electromagnetic field to do something observable. An antenna develops an electrical response; a thermopile converts absorbed energy into a thermal and then electrical signal; a photodetector can respond through photon absorption. Hertz spark transmission and detection experiment makes this conversion visible as a tiny spark. Detector orientation, sensitivity, and surrounding reflections all matter: failure to see a signal does not prove that no electromagnetic radiation reached the receiver.

An electromagnetic wave is a propagating disturbance of the electric field and magnetic field obeying the Maxwell equations. In source-free vacuum its speed isthe speed of light. An electromagnetic wave can travel through vacuum: a conducting medium is unnecessary. A single traveling plane electromagnetic wave has mutually perpendicular electric field, magnetic field, and propagation direction. This statement applies to that wave; arbitrary near and far field configurations need not share these directions or the relation $|\mathbf{B}|=|\mathbf{E}|/c$.

The visualization compares perpendicular components on a common scale: $c\mathbf{B}$ has the same units as $\mathbf{E}$. The curves are field values along space, rather than paths traced by particles. Their common phase moves toward $+z$, while the Poynting vector records the associated energy flux.

The electromagnetic wave equation gives a propagation law, while the full Maxwell equations additionally enforce the field directions and relative amplitudes. This is why solving an independent scalar wave equation for each component is insufficient to construct an electromagnetic wave. ^[ref].

For a single traveling vacuum plane electromagnetic wave, $\mathbf{B}=\hat{\mathbf{k}}\times \mathbf{E}/c$ gives equal electric and magnetic contributions to electromagnetic energy:For the fields with linear polarization along $x$ and magnetic orientation along $y$ specified above,Angle brackets mean an average over one period at fixed position. The factor $1/2$ comes from averaging squared cosine; $E_0$ is the peak amplitude, whereas the root mean square field is $E_{\mathrm{rms}}=E_0/\sqrt{2}$. The equal energy contributions oscillate together, rather than trading energy as in an RLC circuit. ^[ref].

For $E_0=100V/m$, using $c=299792458m/s$ and ${ϵ}_0\simeq 8.854\times 10^{-12}F/m$,A $0.010{\mathrm{m}}^2$ surface normal to $+z$ therefore has average crossing power $0.133\mathrm{W}$. This is incident power; the amount absorbed depends on the surface. Doubling the field amplitude quadruples the average flux.

Define the amplitude evanescent penetration depth as the distance at which the electric field wave amplitude falls to $1/e$ of its surface value:The squared-field envelope decays twice as fast:The same normalized exponential decay describes the tangential average Poynting vector for the TE polarization above. An intensity penetration depth convention often means $d_{1/e}$; it must not be confused with amplitude depth $\delta$ or interpreted as outgoing normal light intensity.

For the stipulated $n_1=1.5$, $n_2=1$, ${\lambda }_0=633\mathrm{nm}$ and $\theta =50^{\circ }$,At $z=200\mathrm{nm}$ the normalized wave amplitude is approximately $0.325099$, and its squared envelope is approximately $0.105689$. At $z=\delta$, these become $e^{-1}$ and $e^{-2}$, respectively. The quantity inside the square root is dimensionless, so $\kappa$ has inverse-length units and both depths have length units.

The sharp angular dependence has a simple explanation. For $\theta ={\theta }_c+\epsilon$ with $\epsilon >0$ small and measured in radians, a first-order Taylor expansion givesThus $\delta \to \infty$ as $\theta \downarrow {\theta }_c$: exponential confinement disappears continuously. With fixed refractive indices, $\delta$ decreases as the angle of incidence increases and approaches ${\lambda }_0/(2\pi \sqrt{n_1^2-n_2^2})\approx 90.1091\mathrm{nm}$ as $\theta \to 90^{\circ }$. This is a limit of the normalized depth; exactly grazing incidence supplies zero incident normal power. Real beam widths, angular spreads and optical dispersion matter when applying these ideal plane electromagnetic wave predictions near the threshold.

The Fresnel equations give reflection and transmission amplitudes for both transverse-electric and transverse-magnetic polarization of an electromagnetic wave at a planar interface. They follow by applying interface conditions for electromagnetic fields to incident, reflected and refracted plane electromagnetic waves.

Wave interference is the reinforcement or cancellation that occurs when overlapping wave amplitudes add by the superposition principle. For two monochromatic waves of the same frequency, mutual optical coherence means that their relative wave phase is stable during the observation. Equal wave phases give constructive interference; a half-cycle phase difference gives the destructive interference intensity minimum. Wave amplitudes add before their squares determine wave intensity. ^[ref].

Consider two overlapping beams in a homogeneous medium, with the same polarization of an electromagnetic wave and individual local time-averaged light intensities $I_1,I_2$. Their phase difference $\delta$ is measured in radians. In the scalar detection convention, the combined local time-averaged wave intensity isThe last term records wave interference; adding $I_1+I_2$ alone misses it. The formula describes irradiance when both fields propagate along the same direction, or when the same detector geometry factor is included in all three intensities. For crossing beams it describes the squared electric field response, proportional to scalar wave intensity; a specified component of the Poynting vector also requires the propagation geometry. The ideal crossing plane waves below make that distinction explicit.

For equal individual light intensities $I_1=I_2=I_0>0$, the cosine half-angle identity givesThe constructive interference intensity is $4I_0$ at $\delta =0$ modulo $2\pi$, and the destructive interference intensity is zero at $\delta =\pi$ modulo $2\pi$. At phase quadrature, $\delta =\pi /2$, it is $2I_0$. With $I_1=1$ and $I_2=0.25$ in the same arbitrary wave intensity unit, the maximum is $2.25$, the minimum is $0.25$, and the phase quadrature value is $1.25$: unequal wave amplitudes cannot cancel completely.

Both curves use the same wave intensity unit and a horizontal angle in radians. The dashed levels are the averages over a full phase difference cycle. Equal beams have interferometric visibility one; the unequal example has interferometric visibility $0.8$.

A spatially varying phase difference turns the same law into interference fringes. Take two ideal plane electromagnetic waves with common wavelength $\lambda$, equal individual light intensities $I_0$, and propagation directions at angles $+\theta$ and $-\theta$ from the $z$ axis in the $xz$ plane, with $0<\theta <\pi /2$. Both have linear polarization along $y$, perpendicular to both propagation directions. Choose equal initial wave phases; their wavevectors and phase difference areHere $k$ is the wavenumber and $\omega$ the angular frequency. The common $z$ and time terms cancel, so the interference fringes stay at fixed $x$ for every $z$. Their interference fringe spacing isSet the length unit to $\lambda$, so that $\lambda =1$, and take $\theta =30^{\circ }=\pi /6$: then $\Lambda =1$. The constructive interference intensity bands lie at integer $x/\lambda$; the destructive interference intensity bands lie at half-integer $x/\lambda$.

The grid covers $-3\le x/\lambda \le 3$ and $0\le z/\lambda \le 4$; both axes are dimensionless lengths in wavelength units. The colour scale is $I/I_0$ from zero to four, using the same single-beam normalization as the equal-beam curve above. This is an ideal plane wave superposition, with no slits or point sources and no near-field approximation. For these $y$-polarized plane electromagnetic waves in vacuum, the longitudinal time-averaged Poynting vector obeys $⟨S_z⟩=I\cos \theta$; the plotted scalar $I$ therefore differs from the irradiance through a plane normal to $z$ by that common factor. This relation follows by adding the magnetic fields as well as the electric fields.

The bright peaks do not create energy. Averaging over one interference fringe spacing, or equivalently over one full phase difference cycle, givesbecause the integral of the cosine is zero, with sine as its antiderivative. Equal beams thus average to $2I_0$, despite bright bands at $4I_0$. For the crossing-wave geometry, the average longitudinal energy flux is correspondingly $2I_0\cos \theta$. Bright and dark regions redistribute the incoming energy; conservation of energy concerns the full energy flux, rather than the height of one peak.

Wave interference is useful because phase differences convert small changes in optical path length into measurable changes of wave intensity, as in a Mach-Zehnder interferometer. Its beauty is that the same superposition principle produces both complete cancellation and fourfold brightening; adding more coherent contributions also produces the narrow peaks of a diffraction grating. Feynman's treatment connects this many-source pattern directly to addition of complex amplitudes. ^[ref].

A short proof outline uses local real fields $E_j(t)=A_j\cos (\omega t+{\varphi }_j)$ with parallel linear polarizations. Write $I=C⟨(E_1+E_2{)}^2{⟩}_t$ and $I_j=C{A}_j^2/2$ for the same positive conversion factor $C$. Averaging over many wave periods while the phase difference stays fixed uses $⟨{\cos }^2(\omega t+{\varphi }_j){⟩}_t=1/2$ and $⟨\cos (\omega t+{\varphi }_1)\cos (\omega t+{\varphi }_2){⟩}_t=\cos \delta /2$, giving the stated law. Equivalently, add the complex amplitudes and take the squared complex modulus. ^[ref].

For different fixed linear polarizations in a common transverse plane, the cross term acquires the dot product of their real unit vectors, equal to $\cos \alpha$ for relative angle $\alpha$. Orthogonal linear polarizations remove the cross term in polarization-insensitive scalar detection; different orientations generally only reduce it. A common polarizer can select overlapping components and recover wave interference. Likewise, rapid random variation of the phase difference removes the observed cross term only when the detector's time average has $⟨e^{i\delta (t)}{⟩}_t=0$. Slowly varying wave phases can instead move resolved interference fringes, and biased or incompletely sampled wave phases can leave residual contrast. Finite optical coherence therefore depends on the observation time as well as the fields.

For a vacuum plane electromagnetic wave, a Lorentz boost multiplies both transverse Jones vector amplitudes by the same positive Doppler factor when the basis is carried to the aberrated ray. Their ratio and phase difference therefore remain unchanged. All Stokes parameters gain the same squared factor, leaving normalized parameters, ellipse shape, handedness and degree of polarization unchanged. This does not mean the propagation direction or the components in an arbitrarily fixed pair of spatial axes remain unchanged.

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

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

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

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

In vacuum the Maxwell equations areSubstitution of the complex plane electromagnetic wave amplitudes into these equations givesThus the electric field must be transverse to $k$; taking the positive-frequency branch gives $\omega =c|k|$.

For $E_0=E(1,i,0)$, $k=k\hat{z}$, and $\phi =kz-\omega t$, the real fields areThe electric vector rotates with constant magnitude, so this is circular polarization. Its Poynting vector isThe Poynting vector is the electromagnetic energy flux: its direction is the direction of propagation and its magnitude is the power crossing unit area.

In a vacuum with no charge or current, the Maxwell equations areFor the proposed plane electromagnetic wave, $\nabla \cdot B=0$ givesThe Ampère-Maxwell equation givesand substituting this into Faraday's law gives the dispersion relationThus both fields have transverse polarization relative to $k$, and the corresponding real electric field is

For incidence in the positive $x$-direction, choose the incident fieldsThe perfect conductor requires the tangential electric field to vanish at $x=0$. The reflected wave therefore hasThis is normal reflection of an electromagnetic wave from a perfect conductor. The magnetic field in $x\le 0$ isso its tangential value at the surface is

Define the dual tensor byFor metric signature $(-,+,+,+)$ and ${ϵ}^{0123}=+1$, two independent quadratic electromagnetic field invariants areandChanging orientation or the sign convention for $F$ can change the displayed overall signs but not their vanishing.

For a vacuum plane electromagnetic wave with wavevector $\mathbf{k}$,Hencefor every transverse polarization. Both Lorentz scalars therefore vanish.

Substitution of the sinusoidal plane waves into Faraday's law givesThe source-free Ampère-Maxwell equation, with $\mathbf{H}=\mathbf{B}/\mu$ and $D_i={\epsilon }_{ij}{E}_j$, givesEliminating $\mathbf{b}=(\mathbf{k}\times \mathbf{e})/\omega$ yields the wave equation for a plane wave in an anisotropic dielectric:

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 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.

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.

The phase $kz-\omega t$ describes propagation along $+z$. With $\psi =kz-\omega t$, the vacuum plane electromagnetic wave hasThus $B_x=-3A\sin \psi /c$, $B_y=2A\sin \psi /c$, and $B_z=0$. The field amplitudes are $E_0=\sqrt{13}|A|$ and $B_0=\sqrt{13}|A|/c$, givingThe instantaneous magnitude ratio also equals $c$ wherever the fields are nonzero; at a node both vanish together. Their Poynting vector points along $+z$, checking the relative component signs.