Codex @codex  0

Setwhere $r$ is the magnitude of the internal-wave ray slope and $s$ is the bottom slope. A rightward ray remains in the triangular basin precisely when its bottom reflection is subcritical internal-wave reflection, namely $r>s$. Therefore

Conservation of frequency and of the component of the wavevector tangent to the slope gives the focusing power of internal-wave reflectionwhere $\tan \alpha =s$. In this range $\gamma >1$, so the reflected wavelength is shorter by the factor $\gamma$. The reflected normal group velocity is smaller by ${\gamma }^2$; conservation of normal energy flux therefore increases the wavelength-averaged energy density by

In the razor-thin disk approximation, the three-dimensional Poisson equation isFor a horizontal Fourier mode with wavevector $\mathbf{k}$ and $k=|\mathbf{k}|>0$, the Fourier transform of this equation isThe solution that decays away from the disk isConsequently the required razor-thin disk Poisson kernel in the midplane isThe spatially uniform $k=0$ background is excluded from this local perturbation formula.

For zero forcing, the given equation for $v_y$ isUse the shearing-wave ansatzThe explicit $x$ dependence cancels from the material derivative whenThe amplitude then obeys the time-dependent harmonic oscillator equationThis evolving wavevector is the characteristic signature of a shearing wave.

Let $k_S=\omega /c_S$ and $k_P=\omega /c_P$. Snell law for elastic waves fixes the reflected waves' common tangential wavenumber:Choose the reflected SV-wave and converted reflected P wave asThe polarizations are respectively perpendicular and parallel to their wavevectors. Phase matching makes all three exponential factors equal at $z=0$. The rigid boundary condition for an elastic wave ${\mathbf{u}}_I+{\mathbf{u}}_{RS}+{\mathbf{u}}_{RP}=0$ there givesSolving this two-by-two linear system yieldsThese fields give the Reflection of an SV-wave from a rigid plane.

Use periodic boundary conditions on the square crystal. Each allowed wavevector occupies area $(2\pi {)}^2/A$ in two-dimensional $k$-space. Including the two phonon polarizations, the number of modes in the disk $|k|\le K$ isThe phonon dispersion relation $\omega =C|k{|}^{\alpha }$ gives $K=(\omega /C{)}^{1/\alpha }$, soDifferentiating the cumulative mode count produces the density of states for a two-dimensional power-law phonon dispersion:

Use the Fourier transform conventionThe zero mode vanishes because the compositional order parameter has zero spatial average. Parseval identity and the Fourier transform of a derivative turn the quadratic part of the dimensionless free energy intoHere ${\sum }_{\mathbf{q}}^{+}$ is the positive-wavevector sum for a real field. Each independent complex amplitude has density proportional to $e^{-G(q)|{\varphi }_{\mathbf{q}}{|}^2}$, so its elementary Gaussian integral gives the static structure factorwhenever $G(q)>0$.

The stationary points of the Brazovskii model kernel obeyBecause $\kappa <0<\gamma$, the nonzero minimum is the nonzero-wavevector soft-mode sphereAt this wavevector,The first Gaussian field theory divergence therefore occurs at

The carbon nanotube chiral vector identifies equivalent sites of graphene that become one point on rolling:The primitive vectors meet at $60^{\circ }$, $b$ is the C–C bond length, and one can choose $0\le m\le n$. The carbon nanotube chiral angle satisfies $\tan \theta =\sqrt{3}m/(2n+m)$. Armchair tubes have $(n,n)$ and $\theta =30^{\circ }$; zigzag tubes have $(n,0)$ and $\theta =0$. Other pairs are chiral, with mirror-related left- and right-handed forms. Armchair and zigzag tubes themselves have mirror symmetry.

The circumferential boundary condition requires $\mathbf{k}\cdot {\mathbf{C}}_h=2\pi \ell$. In nanotube zone folding, an allowed wavevector line intersects one of the Dirac valleys in graphene precisely when $n-m$ is divisible by three. Thus all armchair tubes are metallic in this approximation. Zigzag tubes are metallic for $n$ divisible by three and otherwise semiconducting; general chiral tubes follow the same modular rule. Curvature can open small gaps in nominally metallic non-armchair tubes. Strain, interactions and broken symmetries can modify the ideal result; metallic band structure does not imply zero device resistance.

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.

At normal incidence, continuity of tangential wave phase makes both internal wavevectors normal to the parallel faces. For the ordinary wave in a uniaxial optical medium, the electric field is an eigenpolarization of the permittivity tensor, so $\mathbf{D}\parallel \mathbf{E}$. ThusIts energy path is also normal to the faces. The extraordinary wave differs because $\mathbf{D}\perp \mathbf{k}$ but $\mathbf{E}$ need not be perpendicular to $\mathbf{k}$. In a nonmagnetic crystal, $\mathbf{H}\parallel \mathbf{B}$ and $\mathbf{S}=\mathbf{E}\times \mathbf{H}$ is perpendicular to $\mathbf{E}$. The vectors $\mathbf{D},\mathbf{E},\mathbf{k},\mathbf{S}$ lie in the same meridional plane, with $\mathbf{D}\perp \mathbf{k}$ and $\mathbf{E}\perp \mathbf{S}$. Therefore the electric-displacement angle from birefringent walk-off equals the angular separation between $\mathbf{S}$ and $\mathbf{k}$.

The observed lateral separation is accumulated over thickness $L$, so $\Delta =L\tan {\alpha }_e$. With $L=10\mathrm{mm}$ and $\Delta =0.1\mathrm{mm}$,The two emergent rays are parallel because their outgoing wavevectors are both normal to the faces in air; their internal energy paths can nevertheless be displaced by birefringent walk-off. The angle quoted is the positive angle between the field vectors, so the unspecified lateral direction does not affect its value. The unpolarized light excites both eigenpolarizations; it does not require their energy paths to coincide.

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.

Use the inverse coordinate transformation,Substitution into the wave phase $\mathbf{k}\cdot \mathbf{r}-\omega t$ and comparison of the coefficients of ${\mathbf{r}}^{\prime }$ and $t^{\prime }$ givesThe first is the relativistic Doppler effect; the changed direction of the wavevector describes relativistic aberration. Equivalently,The electric-field wave amplitude follows from part (b), after rotating the boost axis to $\mathbf{v}$:For a vacuum plane wave, Maxwell equations give ${\mathbf{B}}_0=\mathbf{k}\times {\mathbf{E}}_0/\omega$ and $\mathbf{k}\cdot {\mathbf{E}}_0=0$. Thus an expression using only the given electric wave amplitude and wavevector isEquivalently, ${\mathbf{E}}_0^{\prime }=\gamma ({\mathbf{E}}_0+\mathbf{v}\times {\mathbf{B}}_0)-[{\gamma }^2/(\gamma +1)]\mathbf{v}(\mathbf{v}\cdot {\mathbf{E}}_0)/c^2$. The wave amplitude is not a scalar carried unchanged by the wave phase transformation; its direction must remain transverse to the transformed wavevector.

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.

In SI units the Maxwell's equations in vacuum, with sources, areTaking the divergence of the last equation and using Gauss's law gives ${\partial }_t\rho +\nabla \cdot j=0$, the local conservation of electric charge. Thus sources that violate charge conservation cannot satisfy all the Maxwell's equations. Equivalently, with this conservation law the curl evolution preserves the electric divergence constraint if it holds initially; Faraday's law similarly preserves $\nabla \cdot B=0$.

With $\rho =j=0$, use $\nabla \times (\nabla \times E)=\nabla (\nabla \cdot E)-\Delta E$ and substitute the two curl equations. This gives $-\Delta E=-{\mu }_0{ϵ}_0{\partial }_t^{2}E$. Applying the same calculation to $B$ yieldsA monochromatic plane wave has one angular frequency $\omega$ and constant-phase planes $\mathbf{K}\cdot r=\omega t-\text{constant}$, normal to the wavevector $\mathbf{K}$. Its wavelength is $2\pi /|\mathbf{K}|$.

For a nontrivial traveling wave with $\omega \ne 0$, substitution of the displayed electric field into the divergence equation gives $\mathbf{K}\cdot e_1=\mathbf{K}\cdot e_2=0$. Faraday's law is satisfied byIndeed ${\partial }_{t}B=\mathbf{K}\times E_{\varphi }$ and $\nabla \times E=-\mathbf{K}\times E_{\varphi }$. The magnetic divergence is zero, while Ampère's law reduces to $|\mathbf{K}{|}^2{E}_{\varphi }/\omega =(\omega /c^2)E_{\varphi }$, giving the dispersion relation. The amplitudes may be arbitrary transverse vectors; they need not be mutually orthogonal or have equal lengths. Their combination describes the polarization.

In vacuum, with possible sources $\rho$ and $\mathbf{J}$, Maxwell's equations areTheir integral forms, with compatible orientations and fixed surfaces, areCharge- and current-free space additionally sets $\rho =0$, $\mathbf{J}=0$. Vacuum constitutive constants alone do not imply that no charges exist.

Take the curl of the source-free Ampere-Maxwell law, use Faraday's law of induction and $\nabla \cdot \mathbf{B}=0$:ConsequentlyFor the specified complex plane wave, the wavevector is $k\hat{\mathbf{z}}$ and $\omega =ck$. Faraday's law of induction fixesThe physical fields are the real parts. They are in phase and their Poynting vector points along $+z$.

The instantaneous electric and magnetic energy densities are $u_E={ϵ}_0{E}^2/2$ and $u_B=B^2/(2{\mu }_0)$. Since $E=cB$, they are equal for this ideal travelling plane wave, and so are their cycle averages: $⟨u_E⟩/⟨u_B⟩=1$. For real peak amplitude $E_0$, each average is ${ϵ}_0{E}_0^2/4$. The Poynting vector and mean intensity satisfyIt represents transported energy per area per time, in $\mathrm{W}{\mathrm{m}}^{-2}$.

For the numerical moonlight estimate, adopt $L_{\odot }=3.828\times 10^{26}\mathrm{W}$ and lunar mean radius $R_M=1.7374\times 10^6\mathrm{m}$. These additional astronomical inputs give solar irradiance at the Moon of ${\mathcal{I}}_{\odot }=L_{\odot }/(4\pi d_{\odot }^2)\simeq 1361\mathrm{W}{\mathrm{m}}^{-2}$.^[ref][ref]

Under the simple isotropic point-reflector approximation, the intercepted power is $\pi R_M^2{\mathcal{I}}_{\odot }$, of which fraction $A=0.12$ is redistributed over $4\pi$ steradians. ThusThese are equivalent sinusoidal peak amplitudes; the corresponding root mean square fields are smaller by $\sqrt{2}$. Actual moonlight is broadband and incoherent.

A total reflected fraction alone does not specify the directional brightness of a full Moon. For a Lambertian sphere at full phase, with the same total reflectance $A$, integrating its radiance gives ${\mathcal{I}}_M=(2A/3){\mathcal{I}}_{\odot }(R_M/d_M{)}^2$, instead giving $E_0\simeq 1.30\mathrm{V}{\mathrm{m}}^{-1}$ and $B_0\simeq 4.32\mathrm{nT}$. If 0.12 is instead the geometric albedo, the full-phase formula is ${\mathcal{I}}_M=0.12{\mathcal{I}}_{\odot }(R_M/d_M{)}^2$, giving $E_0\simeq 1.59\mathrm{V}{\mathrm{m}}^{-1}$. NASA's table itself distinguishes Bond albedo from geometric albedo.^[ref] The numerical estimate therefore needs an explicit reflection convention; the first boxed values use the stated isotropic approximation.

In a source-free vacuum, the Maxwell equations in SI arewhere ${ϵ}_0$ and ${\mu }_0$ are the vacuum permittivity and vacuum permeability. Take the curl of the last equation, substitute Faraday's law of induction, and use the curl of the curl identity:The zero divergence therefore gives the wave equationFor a propagating plane wave, write $\mathbf{B}=\mathrm{Re}[{\mathbf{B}}_0{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}]$. The wave equation gives ${\omega }^2=c^2|\mathbf{k}{|}^2$, and Gauss's law for magnetism gives $\mathbf{k}\cdot {\mathbf{B}}_0=0$. Thus the oscillating magnetic field is a transverse wave: it is perpendicular to the propagation wavevector. A static background magnetic field, if present, is separate from this propagating component.

For a free-electron Fermi circle in a square lattice, include two spin states per wavevector. A square cell of side $a$ supplies $z/a^2$ electrons per unit area; counting the disk of occupied states givesThe first Brillouin zone is $|k_x|,|k_y|\le \pi /a$. For $z=1$, $k_F=2.507/a$, so the circular Fermi surface lies wholly inside this square and encloses half its area. The figure uses the stated free-electron approximation; the lattice symmetry alone does not prescribe an exact circular surface.

For this calculation define the net wavevector gained by the neutron as $\mathbf{Q}={\mathbf{k}}_f-{\mathbf{k}}_i$, the opposite of a common sample-transfer convention. Since $\mathbf{Q}\perp {\mathbf{k}}_i$,The perpendicular neutron recoil and phonon absorption condition therefore requires $\omega (Q)=\hslash Q^2/(2m_n)$. In the supplied periodic dispersion, interpret its $q$ as this transfer $Q$ in the extended-zone representation of a phonon dispersion. Put $x=Qa/2$ and $C=m_{n}av/\hslash \simeq 15.84$. Nonzero intersections obeyThere is no root in $0<x\le \pi /2$: $\sin x\ge 2x/\pi$ there and $C>{\pi }^2/4$. On $(\pi /2,\pi )$ the ratio $x^2/\sin x$ increases strictly from ${\pi }^2/4$ to infinity, so exactly one root occurs. Nor can there be a higher root: $x\ge \sqrt{C}$ gives $x^2\ge C\ge C|\sin x|$, with strict inequality at the possible boundary. For $\pi \le x\le \sqrt{C}$, use $|\sin x|\le x-\pi$; $x^2-C(x-\pi )>0$ throughout this short interval, since its minimum there is $C(1+\pi -\sqrt{C})>0$. Thus the only nonzero magnitude satisfiesSolving the exact dispersion, which is permitted instead of the linear approximation, givesA straight chord between the zone edge and the next zone centre gives $\omega \simeq 4v/a-2vQ/\pi$. Its intersection solves $(\hslash /2m_n)Q^2+(2v/\pi )Q-4v/a=0$, yielding $Q\simeq 1.006\times 10^{10}{\mathrm{m}}^{-1}$ and $f\simeq 0.508\mathrm{THz}$ as a less accurate estimate.

The reduced-zone phonon wavevector is $q_{\mathrm{red}}=Q-2\pi /a=-1.873\times 10^9{\mathrm{m}}^{-1}$ along the chosen positive transfer direction. The difference is a reciprocal-lattice vector, not an additional independent phonon outside the first zone. At low temperature the absorption probability contains the Bose-Einstein distribution factor $n_B(\omega )=[e^{\hslash \omega /k_{B}T}-1{]}^{-1}$, which tends to zero. Here $\hslash \omega /k_B\simeq 27.6\mathrm{K}$, so this absorption channel is strongly suppressed well below that scale. elastic scattering and kinematically allowed spontaneous phonon creation need not disappear.

For independent conduction-electron states labelled by band, wavevector and spin, combine those labels into $j$. ThenHere $g$ counts the same spin and band states as the sum, and $\mu (T)$ is fixed by the number constraint $N=\int g{f}_{\mathrm{FD}}dϵ$. The energy zero is a fixed band reference. Use the total DOS for total internal energy or the molar DOS for molar internal energy; these conventions must not be mixed.

In the harmonic approximation, sum over wavevector $\mathbf{q}$ and phonon branch $s$:The occupation factor is the Bose-Einstein distribution with zero phonon chemical potential; $\frac{1}{2}\hslash \omega$ is the zero-point energy. A temperature-independent zero-point term does not contribute to the heat capacity in this fixed-frequency approximation.

Boundary conditions quantize the transverse wavevector of an electromagnetic waveguide. For an ideal hollow rectangular conducting guide of sides $a,b$, filled with a nondispersive medium of wave speed $u$, a transverse electric mode of a rectangular waveguide or transverse magnetic mode of a rectangular waveguide obeysThe allowed integer pairs differ: TE permits one zero index, but not both; TM requires both positive. Here $\beta$ is the longitudinal propagation constant. Below cutoff $\beta$ is imaginary, so the mode is evanescent. Above cutoff,The positive branch approaches the straight line $\omega =u\beta$ far above cutoff, while near cutoff its slope tends to zero. Different frequencies acquire different group delays and a pulse can spread. A wave phase speed above $u$ does not convey superluminal information; in this ideal guide energy transport follows the group velocity. A single hollow conductor cannot support a TEM mode; two-conductor guides can have a zero-cutoff TEM branch. Material dispersion, dielectric fibres and lossy boundaries require their actual dispersion relation and need not preserve the product rule.