Codex @codex  0

Write $\psi =kx-\omega t$ and $v_p=\omega /k$. The electric field isIt has linear polarization, along the fixed line $E_z=6E_y$, at angle ${\tan }^{-1}6\simeq 80.54^{\circ }$ from $+y$ toward $+z$. For the associated plane electromagnetic wave, Faraday's law gives $\mathbf{B}=v_p^{-1}\hat{\mathbf{x}}\times \mathbf{E}$. HenceIn vacuum $v_p=c$. Both fields are transverse, and $\mathbf{E}\times \mathbf{B}$ points along $+x$.

Let $\psi =kx-\omega t$ and use bold $\mathbf{B}$ for the magnetic flux density, to distinguish it from the scalar amplitude $B$ in $E_z$. For a plane electromagnetic wave travelling along $+x$, Faraday's law gives the wave-associated fieldHenceFor an explicit component check, $\nabla \times \mathbf{E}=(0,kB\sin \psi ,kA\cos \psi )$, while differentiating the displayed field gives $-\partial \mathbf{B}/\partial t$ equal to that vector. The field is transverse and also satisfies $\nabla \cdot \mathbf{B}=0$.

In vacuum, Ampère-Maxwell equation compatibility requires $\omega /k=c$, so each factor $k/\omega$ is $1/c$. More generally it is the inverse phase velocity on a suitable homogeneous-medium plane-wave branch. A time-independent background field could be added without changing the electric wave's Faraday equation; the requested propagating-wave field takes that independent background to be zero. Finally, $\mathbf{E}\times \mathbf{B}=(k/\omega )(E_y^2+E_z^2)\hat{\mathbf{x}}$ confirms the propagation direction and the relative signs.

This is a plane electromagnetic wave with linear polarization. $\mathbf{E}$ is the electric field; $E_0$ its peak amplitude; $x$ the position along propagation; $t$ the time; $k=2\pi /\lambda$ the wavenumber; and $\omega =2\pi f$ the angular frequency. Here $\lambda$ is the wavelength, $f$ the frequency, and $\hat{\mathbf{y}}$ the unit vector in the direction of the electric field. Constant phase $kx-\omega t$ moves along $+x$ with phase velocity $\omega /k$.

Put $\theta =kx-\omega t$. The differential Faraday's law of induction givesThe accompanying oscillating magnetic field, excluding any independent static background, is therefore $\mathbf{B}=(k/\omega )E_0\cos \theta \hat{\mathbf{z}}$. To determine the propagation speed, also use the source-free Ampere-Maxwell law:ThusThe electric field and magnetic field oscillate in phase, perpendicular to one another and to propagation; $\mathbf{E}\times \mathbf{B}$ points along $+x$.

For a plane electromagnetic wave with linear polarization, write $\mathbf{E}=E_0\hat{\mathbf{x}}\cos (kz-\omega t)$ and $\mathbf{H}=E_0\hat{\mathbf{y}}\cos (kz-\omega t)/({\mu }_{0}c)$. The Poynting vector is instantaneous energy flux:The light intensity is its cycle average, not its peak:An incident electromagnetic wave transports longitudinal momentum $\mathcal{E}/c$ per energy $\mathcal{E}$. Perfect reflection back along the incident direction reverses that momentum, transferring $2\mathcal{E}/c$ to the mirror. Dividing by illuminated area and time gives the mean radiation pressureHere $\mathbf{S}$ denotes the incident travelling wave. The average net Poynting vector of the incident-plus-reflected standing wave is zero, although its radiation pressure is nonzero.

In SI units, the Maxwell equations in vacuum with prescribed charge density $\rho$ and current density $\mathbf{J}$ areWhen $\rho =0$ and $\mathbf{J}=0$, take the curl of the third equation and substitute the fourth. Using the curl of the curl identity and $\nabla \cdot \mathbf{E}=0$ gives the electromagnetic wave equation:For example, $\mathbf{E}=E_0\hat{\mathbf{x}}\cos (kz-\omega t)$, $\mathbf{B}=(E_0/c)\hat{\mathbf{y}}\cos (kz-\omega t)$ is a plane electromagnetic wave provided $\omega =ck$. The electric field, magnetic field and propagation direction are mutually perpendicular.

The Maxwell equations are linear, so the superposition principle applies to fields. However, the Poynting vector is quadratic:One may add the average wave intensities only when the cross terms vanish under the actual detector average. For co-propagating plane electromagnetic waves of parallel linear polarization, averaging over optical cycles but potentially resolving the difference frequency givesIf the detector integrates for $T$, the last term acquires $\mathrm{sinc}[({\omega }_1-{\omega }_2)T/2]$, with sinc function convention $\mathrm{sinc}x=\sin x/x$, evaluated at the midpoint phase. Thus $T\gg 1/|f_1-f_2|$ suppresses beats, and the infinite-time average adds the wave intensities exactly for distinct fixed frequencies. Random mutually uncorrelated wave phases also remove the cross terms in an ensemble average; orthogonal linear polarizations remove them for co-propagating waves in polarization-insensitive detection. Merely having different frequencies does not remove a resolved beat signal: two nearly equal frequencies with fixed relative wave phase are a concrete counterexample. For different propagation directions, the relevant addition is of average Poynting vectors, or their components through the chosen detector surface.

For a sinusoidal plane electromagnetic wave in vacuum, write the maximum electric field as $E_0$. The magnetic field maximum is $B_0=E_0/c$, so the cycle-averaged Poynting vector magnitude, or light intensity, isEach photon carries energy $hf$, where $h$ is the Planck constant. Therefore the photon number flux density, namely the number per unit normal area per unit time, isMultiplying by an illuminated normal area gives the photon arrival rate. Each absorbed photon transfers momentum $hf/c$ at normal incidence, giving the radiation pressure on a perfectly absorbing surfaceThese are time-averaged quantities. At oblique angle of incidence $\theta$, the normal radiation pressure is $I{\cos }^2\theta /c$; the simple displayed expression assumes normal incidence.

An incident electromagnetic wave generally produces reflected and transmitted waves because the two media have different electromagnetic response. Consider a flat interface between homogeneous, linear, lossless media with permittivities ${ϵ}_j$ and magnetic permeabilities ${\mu }_j$. Their wave speeds and wave impedances areFor each plane electromagnetic wave, $\mathbf{H}=\hat{\mathbf{k}}\times \mathbf{E}/Z_j$. With no free surface charge or current, the interface conditions for electromagnetic fields require continuous tangential $\mathbf{E},\mathbf{H}$ and normal $\mathbf{D},\mathbf{B}$. Equality of the phase at every point on the boundary gives equal angular frequencies and tangential wavevectors. Hence the reflection angle equals the incidence angle and Snell's law iswhere $n_j=c/v_j$ and each angle is measured from the normal.

At normal incidence use the same tangential axis for all electric amplitudes. The two independent boundary equations areSolving gives the amplitude Fresnel equationsFor nonmagnetic media, these become $r=(n_1-n_2)/(n_1+n_2)$ and $t=2n_1/(n_1+n_2)$. Reflection into a medium of larger refractive index therefore reverses the electric amplitude's sign at normal incidence.

The normal Poynting vector determines the reflected and transmitted fractions of incident power. At normal incidence,It is the power fractions, rather than the squared electric amplitudes alone, that add to unity.

At oblique incidence separate transverse electric polarization ($s$, electric field perpendicular to the incidence plane) from transverse magnetic polarization ($p$, electric field in that plane). For nonmagnetic media, writing $c_i=\cos {\theta }_i$ and $c_t=\cos {\theta }_t$, the Fresnel equations areHere choose a common $s$ direction and, for each ray, the $p$ direction as $\hat{\mathbf{s}}\times \hat{\mathbf{k}}$. This makes the reflected $p$ basis reverse its tangential direction at normal incidence, explaining why the displayed $r_p$ then has the opposite sign to $r_s$. Physical reflectance is independent of this sign convention. For either polarization with propagating transmission,For $p$ polarization, zero reflection occurs at the Brewster angle $\tan {\theta }_B=n_2/n_1$. If $n_1>n_2$, incidence above the critical angle for total internal reflection $\sin {\theta }_c=n_2/n_1$ produces an evanescent field in total internal reflection and $\mathcal{R}=1$ for an isolated lossless interface. A lossy second medium requires complex response coefficients and an absorption term in the energy balance.

In vacuum, Maxwell's equations in SI units areThe displacement current term makes the last equation consistent with the charge continuity equation. In a source-free region, take the curl of Faraday's law and use $\nabla \times (\nabla \times \mathbf{E})=\nabla (\nabla \cdot \mathbf{E})-{\nabla }^2\mathbf{E}$. This givesA plane electromagnetic wave therefore propagates at $c=1/\sqrt{{\mu }_0{ϵ}_0}$. The divergence equations make both fields transverse; the curl equations giveThe electric fields and magnetic fields are in phase and perpendicular to one another and to the propagation direction. Their energy densities are equal:For electric-field peak amplitude $E_a$, the time-averaged intensity is $⟨S⟩={ϵ}_{0}c{E}_a^2/2$. Mutually coupled changing electric fields and magnetic fields transport energy through vacuum, even where no charges or conduction currents are present.

With Cartesian Minkowski spacetime coordinates $x^{\mu }=(ct,x,y,z)$, introduce the four-current $J^{\mu }=(c\rho ,\mathbf{J})$ and electromagnetic four-potential ${\mathcal{A}}^{\mu }=(\varphi /c,\mathbf{A})$. The antisymmetric electromagnetic field tensor $F^{\mu \nu }={\partial }^{\mu }{\mathcal{A}}^{\nu }-{\partial }^{\nu }{\mathcal{A}}^{\mu }$ has components $F^{0i}=-E_i/c$ and $F^{ij}=-{ϵ}^{ijk}{B}_k$. Maxwell equations becomeThe time and space components reproduce Gauss's law and the Ampère-Maxwell equation; the dual equation gives absence of magnetic monopoles and Faraday's law of induction. Antisymmetry gives charge conservation ${\partial }_{\nu }{J}^{\nu }=0$. The relativistic Lorentz force is $d{p}^{\mu }/d\tau =q{F}^{\mu \nu }{u}_{\nu }$. Every displayed equation transforms tensorially under a Lorentz transformation; the electric field and magnetic field are different components of the same object.

Two independent algebraic electromagnetic field invariants areThe second is a pseudoscalar: it is invariant under proper Lorentz transformations but changes sign under spatial reflection. For perpendicular fields, the sign of the first distinguishes a frame with purely magnetic field from one with purely electric field. Both invariants vanish for a plane electromagnetic wave with $\mathbf{E}\perp \mathbf{B}$ and $E=cB$, although the field itself is nonzero.

Use Cartesian coordinates $x^a=(t,x,y,z)$ with $c=1$ and ${\eta }_{ab}=\mathrm{diag}(1,-1,-1,-1)$. Spatial dot and cross products below use the Euclidean three-metric. The alternating tensor is given the orientation ${ϵ}^{0123}=+1$.

Vector and covector transformation. The vector $A=A^b{\partial }_b$ is independent of coordinates, and the chain rule gives ${\partial }_b=(\partial x^{\prime a}/\partial x^b){\partial }_a^{\prime }$. Likewise the one-form $A_{b}d{x}^b$ is independent of coordinates. Comparing coefficients provesTheir contraction $A_a{B}^a$ is unchanged because the forward and inverse Jacobians cancel. Here lowering an index uses the metric, not an identification of the numerical upper and lower components.

Why antisymmetrization matters. Write $J^c{}_a=\partial x^c/\partial x^{\prime a}$. Differentiating the covector law givesThe last term obstructs the general tensor transformation law. It is symmetric in $a,b$, so it cancels from ${\partial }_a^{\prime }{A}_b^{\prime }-{\partial }_b^{\prime }{A}_a^{\prime }$. ThereforeEquivalently $F=dA$ is the exterior derivative of the electromagnetic four-potential one-form. The statement that ${\partial }_a{A}_b$ is not a tensor needs a qualification: under affine Cartesian Lorentz transformations it does transform tensorially, because the second derivatives vanish. The failure concerns arbitrary nonlinear coordinate changes, including curvilinear coordinates in flat spacetime. This distinction is precisely the difference between general covariance and covariance under the restricted inertial-frame transformations.^[ref]

Electric and magnetic components. Choose $A^a=(\Phi ,\mathbf{A})$, so $A_a=(\Phi ,-\mathbf{A})$, $\mathbf{E}=-\nabla \Phi -{\partial }_t\mathbf{A}$ and $\mathbf{B}=\nabla \times \mathbf{A}$. ThenThus the lower-index field tensor isThis is a derived component representation of the field, with the source's sign convention.

Raised indices and the first invariant. Raising both indices changes the time-space signs and leaves the spatial block unchanged:In the scalar contraction, each independent antisymmetric pair is counted twice, givingUnder a Lorentz transformation each inverse transformation from a lower index cancels its matching transformation from an upper index, leaving the same scalar. In particular a plane electromagnetic wave has $E=B$ in these units and makes this invariant vanish, while a pure electric field makes it negative.

The printed dual convention. The PDF explicitly defines ${}^{\ast }{F}^{ab}={ϵ}^{abcd}{F}_{cd}$, with no $1/2$. Each independent pair therefore contributes twice: for example ${}^{\ast }{F}^{01}=F_{23}-F_{32}=-2B_x$. The complete result isThe time-space pairs contribute $-4\mathbf{E}\cdot \mathbf{B}$ and the spatial pairs contribute another $-4\mathbf{E}\cdot \mathbf{B}$. The conventional Hodge dual includes $1/2$ and would instead give the contraction $-4\mathbf{E}\cdot \mathbf{B}$; it must not silently replace this paper's definition. With fixed orientation the latter contraction is a pseudoscalar: invariant under proper Lorentz transformations but odd under orientation reversal if compared using a fixed alternating symbol.

Energy density and flux. Substitution into the given mixed stress-energy tensor yieldsFor the first formula, $F^{0c}{F}_{0c}=-E^2$ and $F_{cd}{F}^{cd}/4=(B^2-E^2)/2$. For the second, $F^{0j}{F}_{ij}=(\mathbf{E}\times \mathbf{B}{)}_i$. The minus sign on $T^0{}_i$ arises from its lowered spatial index; it does not reverse the physical energy flow. Raising that index gives $T^{0i}=S_i$, and symmetry gives $T^{i0}=S_i$, with $\mathbf{S}=\mathbf{E}\times \mathbf{B}/{\mu }_0$ the Poynting vector in $c=1$ units. Thus $T^{00}$ is energy density and $T^{i0}$ represents energy flux. In SI coordinates $x^0=ct$, restore $T^{0i}=S_i/c$ and $T^{00}=({ϵ}_0{E}_{\mathrm{SI}}^2+B^2/{\mu }_0)/2$, with ${ϵ}_0{\mu }_0{c}^2=1$. This keeps the mixed-index signs and the units distinct.

Let $s=k^{\mu }{x}_{\mu }$. The Lorenz gauge condition on the plane electromagnetic wave givesThe vacuum wave equation gives $k_{\mu }{k}^{\mu }=0$ for a nonzero amplitude; equivalently $k^0=\omega /c$ and $|\mathbf{k}|=\omega /c$. Use the gauge transformation $A^{\prime \mu }=A^{\mu }+{\partial }^{\mu }\chi$ withBecause $\square \chi =0$, this is a residual Lorenz gauge symmetry and $k_{\mu }{ϵ}^{\prime \mu }=0$. For a propagating plane electromagnetic wave, $k^0\ne 0$, so take $\lambda =-{ϵ}^0/k^0$. ThenThe second equality follows from $-k^0{ϵ}^{\prime 0}+\mathbf{k}\cdot {\mathbf{ϵ}}^{\prime }=0$. The electric field and magnetic field are unchanged by this gauge transformation. Thus Lorenz gauge permits a representative in radiation gauge; Lorenz gauge alone does not require the spatial amplitude of every representative to be transverse.

At the surface of a perfect conductor, the tangential electric field and normal magnetic field vanish. An incident plane electromagnetic wave is therefore accompanied by a reflected wave whose tangential electric field cancels the incident field at the surface.

A monochromatic vacuum plane electromagnetic wave has phase $\mathbf{k}\cdot \mathbf{r}-\omega t$, where $\mathbf{k}$ is the wavevector, $\omega >0$ the angular frequency, and $\omega =c|\mathbf{k}|$. Its field amplitudes obey $\mathbf{k}\cdot {\mathbf{E}}_0=\mathbf{k}\cdot {\mathbf{B}}_0=0$ and ${\mathbf{B}}_0=\mathbf{k}\times {\mathbf{E}}_0/\omega$.

For the linear polarization and orientation in the visualization, choose a right-handed Cartesian coordinate system and putConstant phase means $z=ct+\text{constant}$, so propagation is along $+z$. The cross product $\hat{\mathbf{x}}\times \hat{\mathbf{y}}=\hat{\mathbf{z}}$ fixes the positive Poynting vector. Both fields reverse together every half-period and are in phase. Reversing the propagation direction with the same electric orientation reverses the magnetic orientation.

These fields are spatially uniform across every transverse plane. An infinite plane electromagnetic wave is an idealization with infinite total electromagnetic energy; it is useful as a local approximation and as a building block for superposition. For frequency $f=100\mathrm{MHz}$, $\omega =2\pi f$ and wavelength $\lambda =c/f\simeq 3.00\mathrm{m}$, independently of $E_0$.

A free nonrelativistic charge driven by a weak plane electromagnetic wave has peak acceleration $|q|E_0/m$. Its time-averaged Larmor formula power is $q^4{E}_0^2/(12\pi {ϵ}_0{m}^2{c}^3)$. Divide by incident light intensity ${ϵ}_{0}c{E}_0^2/2$ to obtain the displayed cross section, independent of frequency. For an electron, it is $8\pi r_e^2/3$. The approximation assumes negligible Compton recoil kinematics and small driven velocity compared with the speed of light.

Total internal reflection occurs when a wave travels from higher to lower refractive index above the critical angle for total internal reflection, leaving no propagating transmitted plane electromagnetic wave normal to the interface.

At an infinite planar interface, total internal reflection returns all incident normal power in the reflected plane electromagnetic wave. Assume homogeneous dielectrics with isotropy, linearity, no dielectric loss, magnetic permeability ${\mu }_0$, and positive real refractive indices $n_1>n_2$. Above the critical angle for total internal reflection, both TE polarization and TM polarization have unit optical reflectance despite the external evanescent wave.^[ref]

This makes total internal reflection useful for guiding light in an optical fiber: suitable internal incidence angles keep energy inside successive boundaries. Its more surprising feature is that perfect optical reflectance coexists with an external electric field. The evanescent field in total internal reflection can interact with a nearby object, as the frustrated total internal reflection experiment demonstrates.

For an original idealized example, choose $n_1=1.5$, $n_2=1$ at vacuum wavelength ${\lambda }_0=633\mathrm{nm}$. At angle of incidence $40^{\circ }$, Snell law for electromagnetic waves gives an angle of refraction $\arcsin (1.5\sin 40^{\circ })\approx 74.6186^{\circ }$, so a transmitted light ray still emerges. At ${\theta }_c\approx 41.8103^{\circ }$, its direction becomes tangent to the boundary. At $50^{\circ }$, the proposed sine of a real angle of refraction would be $1.5\sin 50^{\circ }\approx 1.14907>1$: the transmitted wavevector must instead have an imaginary normal component. These are predictions for the stipulated media, not measured properties of a particular block.

A real semicircular-block experiment isolates this threshold. The University of Maryland laboratory instructions place a laser, motor-driven rotation stage and photodetector together; an aperture limits the detector voltage. The semicircular block is centered on the rotation axis, with the photodetector aligned along its flat face. The instructions identify a strong grazing-beam signal and a weaker signal just before the critical angle for total internal reflection. The readout is detector voltage versus stage angle. Retracing the incident laser beam identifies normal incidence for alignment.^[ref]

The semicircular shape supplies a geometric control: a light ray passing radially through the curved face enters at normal incidence, avoiding additional entrance refraction. The weaker signal near the flat face locates a limiting angle of refraction; a vanishing remote-detector signal alone would not establish that the external electric field vanishes. Beam width and alignment limit angular resolution; no measured threshold or decay depth is supplied.^[ref]

A monochromatic wave in vacuum has frequency $\nu$ and wavelength $\lambda$ related by $\nu \lambda =c$, where $c$ is the speed of light. During one oscillation period $T=1/\nu$, a constant-wave phase surface travels one wavelength, so $\lambda =cT$. Equivalently, the plane electromagnetic wave relation $\omega =ck$ with angular frequency $\omega =2\pi \nu$ and wavenumber $k=2\pi /\lambda$ gives the same result.

As a calculated example, a vacuum radio wave with $\nu =300\mathrm{MHz}=3.00\times 10^8\mathrm{Hz}$ has $\lambda =0.999308\mathrm{m}$ using $c=299792458m/s$. Doubling frequency halves wavelength, independently of wave amplitude and wave intensity. The frequency spectrum of a pulse contains multiple frequencies; the relation applies to each vacuum component, rather than assigning one exact wavelength to the whole pulse. In a homogeneous medium replace $c$ by the appropriate phase velocity; the vacuum equation does not assert that all media have the same wavelength at a given frequency.

For a plane electromagnetic wave in a homogeneous, linear, lossless, isotropic medium with permittivity $ϵ$ and magnetic permeability $\mu$, the wave impedance relates the field amplitudes by $Z=E/H=\sqrt{\mu /ϵ}$. Its units are ohms. Impedance differences determine the normal-incidence amplitude Fresnel equations, $r=(Z_2-Z_1)/(Z_2+Z_1)$ and $t=2Z_2/(Z_2+Z_1)$, with a common tangential electric-field axis.