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Assume an air-filled or evacuated lossless electromagnetic waveguide, and a pulse bandwidth narrow enough to use its carrier's local group velocity. The longitudinal dispersion relation isIt givesHere $f_c/f=20/40=1/2$, soThe phase velocity exceeds $c$, but the pulse envelope travels at the subluminal group velocity; a moving constant-phase surface does not by itself transport information. Their product is $v_{\mathrm{p}}{v}_{\mathrm{g}}=c^2$. A broad pulse can distort because this waveguide is dispersive.

Assume a hollow, perfectly conducting rectangular electromagnetic waveguide filled with air approximated by vacuum. Let its wider side be $a=0.040\mathrm{m}$ and its narrower side $b=0.020\mathrm{m}$. The longitudinal propagation constant of a mode isso the waveguide cutoff frequency isFor a mode with transverse electric polarization, one of $m,n$ may vanish, but they cannot both vanish. For a mode with transverse magnetic polarization, both must be positive. A single hollow conducting guide has no transverse electromagnetic mode with zero cutoff.

The lowest cutoff therefore belongs to the ${\mathrm{TE}}_{10}$ mode, varying across the wider side:Strictly, a traveling mode transporting energy down the guide requires $f>f_c$. At exactly $f_c$, $k_z=0$ and the axial group velocity vanishes. Below cutoff the field is evanescent. A different dielectric filling would replace $c$ by the wave speed in that material.

Electromagnetic waves drive mobile charges in a conducting medium. Their current changes both the phase and attenuation of the wave. A description in terms of a real, frequency-independent refractive index is therefore generally insufficient: the material response and the propagation constant are complex and frequency dependent.

Use phasors proportional to $e^{i(kz-\omega t)}$, with $k=k^{\prime }+i{k}^{\prime \prime }$ and $k^{\prime \prime }\ge 0$ for a passive medium. In a homogeneous, isotropic, linear material, let ${ϵ}_b$ represent the permittivity associated with bound-charge electric polarization, $\mu$ the magnetic permeability and $\sigma (\omega )$ the mobile-charge conductivity. The macroscopic Maxwell equations giveFor a transverse plane wave, taking another curl yieldsThis division of the response avoids counting mobile electrons both in ${ϵ}_b$ and in $\sigma$. Define the complex refractive index $\tilde{n}=ck/\omega =n+i\kappa$. For a nonmagnetic medium, ${\tilde{n}}^2={ϵ}_{\mathrm{eff}}/{ϵ}_0$. Its real part determines the phase variation, while the positive optical extinction coefficient $\kappa$ gives amplitude attenuation $e^{-\omega \kappa z/c}$ and intensity attenuation $e^{-2\omega \kappa z/c}$. The amplitude penetration length is

A simple microscopic explanation comes from the Drude model. Let electrons of charge $-e$, mass $m$ and number density $n_e$ move against an effectively stationary positive background. Collisions relax their mean velocity at rate $\nu$:ConsequentlyWith a vacuum-like bound-charge background, this givesIts imaginary part is positive for positive $\omega ,\nu$, consistent with dissipation under the chosen phasor convention. Real materials can require an effective electron mass, a nontrivial bound-charge response, or additional absorption associated with the electronic band structure.

A plasma is a conducting ionized medium with mobile electrons and ions, approximately neutral on large scales because of plasma screening. For the electron-frequency transverse waves considered here, the heavy ions can often be treated as stationary. In a dilute weakly collisional plasma, $\nu \ll \omega$, soFor $\omega >{\omega }_p$, $k$ is real in this collisionless model and the wave propagates. Its phase velocity is $v_{\mathrm{ph}}=c/n>c$, while its group velocity is $v_g=cn<c$; their product is $c^2$. A phase crest exceeding $c$ does not imply superluminal information transport. For $\omega <{\omega }_p$, the refractive-index square is negative andThe transmitted field is evanescent rather than a traveling bulk wave. A sufficiently thick collisionless plasma reflects such radiation without Ohmic absorption. Collisions make the response complex and permit absorption. This accounts for a plasma cutoff and, qualitatively, the reflection of some radio frequencies by ionized atmospheric layers. The simple model describes transverse waves; plasmas also support longitudinal collective modes.

In a metal, positive ions usually form a crystal lattice. Mobile conduction electrons occupy states described by the electronic band structure and normally form a degenerate Fermi gas. The conduction-electron density is usually much higher than in a dilute gas plasma. Thus its plasma frequency is commonly in the optical or ultraviolet range, rather than the lower radio-frequency values possible in dilute plasmas. Electron scattering from the lattice, defects and other excitations gives a finite relaxation rate. The distinction is not that every plasma is collisionless and every metal collisional: the controlling ratio for either material is $\omega /\nu$, together with its carrier density and bound-charge response.

At sufficiently low frequency, $\omega \ll \nu$, the metal's conductivity is approximately real, ${\sigma }_0=n_e{e}^2/(m\nu )$. In a good conductor the conduction current dominates the displacement current, ${\sigma }_0\gg \omega {ϵ}_b$. The wave equation then givesFor example, a tangential electric field varies asThe current $\mathbf{J}={\sigma }_0\mathbf{E}$ is therefore concentrated within a few skin depths of the surface: this is the skin effect. As frequency increases within this regime, $\delta \propto {\omega }^{-1/2}$. In a wire whose radius greatly exceeds $\delta$, only a thin outer region carries appreciable alternating current, increasing the effective resistance. In a conducting shield, the same penetration suppresses the field transmitted through a sufficiently thick wall.

The Poynting theorem identifies the loss as Joule heating. The cycle-averaged heating density for a complex conductivity isThe wave impedance is $Z=\mu \omega /k$. For incidence from vacuum at normal incidence, the electric reflection amplitude is $r=(Z-Z_0)/(Z+Z_0)$, where $Z_0=\sqrt{{\mu }_0/{ϵ}_0}$. A good conductor has small $|Z|$ and therefore strong reflection, while the part of the incident power entering it is dissipated over the skin depth.

At high enough frequency that $\omega \gg \nu$, a metal's free-electron response also approaches the collisionless plasma form. The low-frequency formula $\delta \propto {\omega }^{-1/2}$ must not simply be extrapolated into that regime. Real optical behavior additionally depends on the electronic band structure and transitions between bands. Both plasmas and metals are therefore described by the same electromagnetic response framework, but their relevant densities, scattering rates and frequency regimes differ.

For the propagating mode of an air-filled or evacuated rectangular electromagnetic waveguide, the longitudinal dispersion relation isThe phase velocity and group velocity along the guide are consequentlyWith waveguide cutoff frequency $f_c=10\mathrm{GHz}$ and carrier frequency $f=30\mathrm{GHz}$, $\sqrt{1-(f_c/f{)}^2}=\sqrt{8}/3$. ThusThe pulse envelope travels at the group velocity for a sufficiently narrow range of frequencies. The phase velocity exceeding the speed of light does not describe superluminal signal transport. The check $v_p{v}_g=c^2$ follows directly from the dispersion relation. A different uniform filling medium would require replacing $c$ by its electromagnetic propagation speed; the stated numerical answer uses the usual empty-guide assumption.

The displacement is along $x$, transverse to propagation in the membrane's $yz$ plane. Write $A=|A|e^{i\varphi }$; then $d=|A|\cos (\omega t-k_{y}y-k_{z}z+\varphi )$. Here $|A|$ is the displacement amplitude, $\varphi$ the initial phase, $\omega$ the angular frequency and $\mathbf{k}$ the wavevector normal to constant-phase lines. Its magnitude is $2\pi /\lambda$, and its direction specifies phase propagation for the chosen sign convention.

For equal-phase two-wave membrane modes, two equal complex amplitudes with opposite $k_y$ sum toThe edge condition gives $\cos (k_{y}a/2)=0$, hence $k_y=(2j+1)\pi /a$, $j=0,1,\dots$, for this even-in-$y$ family. Opposite relative phase gives the sine family; together they yield all Dirichlet transverse modes $k_y=n\pi /a$, $n\ge 1$. For the equal-phase family,The lowest mode has $k_y=\pi /a$. ThereforeIt approaches $c$ for $\lambda \to 0$ and diverges as $\lambda \to 2a^{-}$; there is no real travelling-mode $k_z$ beyond the cutoff. This is an axial phase velocity, not the group velocity: $v_g=c\sqrt{1-(\lambda /2a{)}^2}<c$. Here $c$ denotes the membrane wave speed, not the speed of light.

For the evanescent membrane response below cutoff, below $f_c=c/(2a)$, put $k_z=-i\kappa$ for decay along positive $z$, with $\kappa =\sqrt{(\pi /a{)}^2-(\omega /c{)}^2}$. A localized central drive creates an evanescent, sinusoidally oscillating deformation decaying away in both axial directions. Far enough from the drive the leading profile is proportional to $\cos (\pi y/a)e^{-\kappa |z-z_0|}\cos (\omega t+{\varphi }_0)$. Higher evanescent modes contribute near the point source. The pattern has no steadily advancing axial phase front or far-field propagating wave; decreasing frequency increases its spatial localization.

Use base dimensions $\mathsf{M},\mathsf{L},\mathsf{T}$ for mass, length and time. The proposed dimensional analysis givessoEquating the independent dimensional powers requiresConsequentlyThe mass density drops out because there is no other mass-bearing quantity with which to combine it under the stated assumptions.

Dimensional analysis fixes the powers but leaves a dimensionless coefficient undetermined. For a linear deep-water gravity wave, the dispersion relation is ${\omega }^2=gk$, giving phase velocity $\omega /k=g/\omega$ and group velocity $d\omega /dk=g/(2\omega )$. Both obey the inferred scaling. The phrase wave speed alone does not distinguish these two velocities or fix the coefficient.

The vacuum wave equation gives ${\omega }^2=c^2[k_g^2+(\pi /a{)}^2]$. The cutoff frequency is $f_c=c/(2a)=1.499\mathrm{GHz}$. In a lossless empty guide, energy travels at group velocity,The larger phase velocity $\omega /k_g=c^2/v_E$ is not the speed of energy transport.

Let $K=(k^2+l^2{)}^{1/2}$ and $c=\sqrt{g^{\prime }{h}_0}$. The three branches areFor nonzero $K$ and $f$, the zero-frequency mode is stationary geostrophic flow, with $f\mathbf{k}\times \mathbf{u}=-g^{\prime }\nabla \eta$ and zero horizontal divergence. It carries the linear potential-vorticity disturbance. There is no Rossby-wave propagation on a constant-$f$ plane; a varying Coriolis parameter would be needed for that.

The other two branches are oppositely phased inertia-gravity waves. Their phase-speed magnitude and radial group velocity are $|\omega |/K=\sqrt{c^2+f^2/K^2}$ and $|\partial \omega /\partial K|=c^{2}K/\sqrt{f^2+c^2{K}^2}$. When $K{L}_R\gg 1$, gravity dominates: $|\omega |\simeq cK$ and the waves are approximately nondispersive, with both speeds approaching $c$. When $K{L}_R\ll 1$, rotation dominates: $|\omega |\simeq |f|+c^2{K}^2/(2|f|)$, the frequency approaches the inertial frequency and the group velocity tends to zero. The large long-wave phase speed is not a large signal speed.

Without rotation the propagating branches become nondispersive gravity waves, $\omega =\pm cK$, and the stationary branch is a nondivergent vortical velocity disturbance rather than a finite-pressure geostrophic balance. At $K=0$, the modes reduce to constant interface height and spatially uniform inertial oscillations. The wave branches have zero linear potential-vorticity perturbation; it is the stationary branch that retains the initial potential vorticity.

Put $K=(k^2+m^2{)}^{1/2}$ and $P=p^{\prime }/{\rho }_0$. Acting on the Fourier mode, $D_0=-i(\hat{\omega }-Uk)=-i\omega$. Continuity and horizontal momentum give $u=-mw/k$ and $P=\omega u/k$. The buoyancy equation gives $g{\rho }^{\prime }/{\rho }_0=i{N}^{2}w/\omega$. Substitution into vertical momentum yieldsThe two signs describe the two frequency branches; when the intrinsic frequency is required positive, $\omega =N|k|/K$. For a stationary disturbance $\hat{\omega }=0$, so $\omega =-Uk>0$. With $U>0$ this requires $k<0$ and selects the minus sign, $\hat{\omega }=Uk-Nk/K$. The stationary dispersion condition isDifferentiate the unconstrained dispersion before imposing stationarity. The stationary mountain-wave group velocity is obtained from the ground-frame dispersion; before imposing stationarity it isOn the stationary locus this becomesDifferentiating after substituting $K=N/U$ as though all neighbouring modes were stationary would incorrectly erase the frequency gradients.

For upward energy propagation from the lower boundary, choose $m<0$, since $k<0$ and ${\hat{c}}_{gz}>0$. It requires real nonzero $m$, henceThe downstream ray inclination above horizontal isIt is fixed for a given monochromatic forcing, with the two signs of $m$ giving the upward/downward rays.

Frame qualification for the requested phase angle: the normal intrinsic phase velocity is ${\mathbf{c}}_p=\omega (k,m)/K^2$. Because $\omega =-Uk$, the boxed stationary group velocity is exactly $-{\mathbf{c}}_p$: its angle to intrinsic phase propagation is $\pi$, and it is perpendicular to the stationary phase lines. By contrast the intrinsic group velocity ${\mathbf{c}}_g=\widehat{{\mathbf{c}}_g}-U\hat{\mathbf{x}}$ satisfies ${\mathbf{c}}_g\cdot (k,m)=0$, so its angle to intrinsic phase propagation is $\pi /2$. In the ground frame the phase velocity of a strictly stationary pattern is zero and has no nonzero propagation direction. For example $k=-0.6N/U$, $m=-0.8N/U$ gives $\widehat{{\mathbf{c}}_g}=(0.36U,0.48U)$ and ${\mathbf{c}}_p=(-0.36U,-0.48U)$. These distinctions prevent applying the usual zero-flow perpendicularity rule to two velocities expressed in different frames.

For constant $N^2$, substitution of the vertical cosine and horizontal plane wave givesThe top rigid-lid condition is satisfied by $\cos (mz)$. To select a discrete vertical mode one must also impose the corresponding no-flow, adiabatic condition at the flat bottom, ${\psi }_z^{\prime }(-H)=0$. It requires $mH=n\pi$. The mode $n=0$ is barotropic; the gravest nonzero baroclinic mode has $m=\pi /H$. HenceThe lid condition alone is insufficient: $\cos [\pi z/(2H)]$ also has zero derivative at the lid but violates the bottom condition. Constant depth motivates the additional physical boundary, which must be used to obtain the printed denominator.

Let $a_d=\pi |f_0|/(HN)=L_d^{-1}$ and set $l=0$. The zonal phase velocity and group velocity areThe phase curve is even in $k$, with limiting value $U-\beta /a_d^2$ at zero and approach to $U$ from below at large $|k|$. The group curve is also even: it starts at the same long-wave limit, equals $U$ at $|k|=a_d$, reaches $U+\beta /(8a_d^2)$ at $|k|=\sqrt{3}{a}_d$, and approaches $U$ from above at large $|k|$. The normalized curves appear in the bottom panels of the preceding figure. For $k=0$, the phase speed means the limit $\omega /k$; there is no zonal crest propagation to define directly at exactly zero wavenumber.

For $U=0$ and fixed $\omega >0$, completing the square in the dispersion relation givesThese Rossby-wave isofrequency circles lie in the $k\le 0$ half-plane and touch the origin, where the dispersion relation is undefined. Their radii decrease as frequency increases. The zero-frequency line is $k=0,l\ne 0$.

The group velocity is normal to the constant-frequency curves, toward increasing frequency:At $l=0$, the group velocity is eastward; close to $k=0$ it is westward. For the positive-frequency circles its arrows point inward toward the circle centre. Its meridional component is southward in $k<0,l>0$ and northward in $k<0,l<0$.

For $k\ne 0$, the zonal phase velocity, meaning propagation at fixed $y$, isIt is westward everywhere it is defined. The zonal group velocity is eastward for $|k|>|l|$, westward for $|k|<|l|$, and zero on $|k|=|l|$. There is no region of eastward zonal phase velocity. At $k=0$, $\omega /k$ is undefined, though its continuous limiting expression is westward and $c_{gx}=-\beta /l^2$. Using instead the normal-to-wavecrest phase velocity vector gives ${\mathbf{c}}_p=\omega (k,l)/(k^2+l^2)$; its zonal component is also westward for $k\ne 0$, but is not the same scalar as $\omega /k$.

For a nonzero zonal wavenumber $k$, stationarity requiresA real propagating solution therefore requires a westerly $U>0$. For $0<|k|<K_s$ there are two meridional wavenumbers:The source's assertion of two stationary solutions needs these qualifications: at $|k|=K_s$ they coincide, and at $|k|>K_s$ there is no real meridionally propagating solution. For example $k=2\sqrt{\beta /U}$ gives $l^2=-3\beta /U$. Modes with $k=0$ are a separate zonally invariant degeneracy and do not have a defined zonal phase velocity.

Adding $U$ to the zonal group velocity and imposing stationarity givesThus a localized midlatitude disturbance radiates stationary Rossby wave activity downstream, along one poleward and one equatorward branch. For $k>0$ in the Northern Hemisphere, $l_{+}$ gives the northeastward branch and $l_{-}$ the southeastward branch. With strictly constant $U,\beta$, these wave rays are straight. Changing latitude cannot bend them within that constant-coefficient model.

To discuss a planet, extend it locally to slowly varying $\beta (\varphi )=2\Omega \cos \varphi /a_p$ and possibly $U(\varphi )$. The conserved zonal wavenumber in a locally zonally uniform medium then hasFor constant $U$ and northward propagation in the Northern Hemisphere, decreasing $\beta$ reduces $l$ until $l=0$: the wave ray becomes zonal at a turning latitude and can refract equatorward. The equatorward branch initially encounters larger allowed $K_s$ and becomes more meridional. This local midlatitude description eventually needs replacement near the equator or wherever the background varies rapidly.

Orographic forcing by a mountain range or persistent spatially localized heating can supply such a disturbance. Its steady forcing selects the ground-stationary part of the response. The corresponding local total wavelength isFor fixed $U$ it increases toward the pole as $\beta$ decreases. Locally the zonal integer scale has upper bound $n_{\max }\simeq a_p\cos \varphi K_s=\sqrt{2\Omega a_p{\cos }^3\varphi /U}$, so fewer planetary-scale zonal waves fit around higher-latitude circles. This is not an unconditional trend in actual weather: changing $U$, its curvature, stratification and the forcing spectrum can change the dominant Rossby wave wavelengths.

Coastal Kelvin-wave radiation of long Rossby waves requires a common phase convention to match the two waves. The supplied Rossby wave formula conventionally multiplies $e^{i(k_{R}x+{\ell }_{R}y-{\omega }_{R}t)}$, while the coastal expression uses $e^{i({\omega }_{K}t-{\ell }_{K}y)}$. Since physical fields are real, matching may use the complex-conjugate representative; in either convention the boundary forcing fixes the absolute frequency and alongshore wavenumber. Thus $|{\omega }_R|=|{\omega }_K|$ and $|{\ell }_R|=|{\ell }_K|$.

For $k_R^2+{\ell }_R^2\ll L_d^{-2}$, the shallow-water Rossby-wave dispersion relation reduces toChoosing $k_R>0$ recovers the printed magnitude relation. Both the zonal phase velocity and zonal group velocity are approximately $-\beta L_d^2$, so the long Rossby waves carry the coastal signal westwards into the basin.

The coastal condition $|{\ell }_K|L_d\ll 1$ alone does not guarantee a long zonal wavelength. Since $|{\omega }_K|=c|{\ell }_K|$, matching requires $|k_R|=c|{\ell }_K|/(\beta L_d^2)$ and hence the stronger consistency conditionFor example, at $10^{\circ }$ a coastal wavenumber with $|{\ell }_K|L_d=0.05$ gives $|k_R|L_d\simeq 0.56$, which is not asymptotically small even though the meridional wavelength is long. Thus the source's proposed excitation is possible at sufficiently low frequency; it is not an automatic consequence of its meridional long-wave assumption.

At $10^{\circ }$ N or S the beta plane parameter isTaking a representative Pacific basin width $W\simeq 1.5\times 10^7$ m givesA width of $1.4$–$1.7\times 10^7$ m gives roughly $2.0$–$2.4$ years. This is propagation near $10^{\circ }$ latitude, not an equatorial $f=0$ calculation. Slow Rossby wave adjustment supplies a basin-scale delay relevant to El Niño–Southern Oscillation: changing thermocline depth can affect later sea surface temperature and coupled wind forcing. Reflection and coupling may help produce interannual memory, but this travel-time estimate alone does not establish a climate oscillation or its period.

Take the two-dimensional electron gas in the $xy$ plane, the Hall bar along $x$, and $\mathbf{B}=B\hat{\mathbf{z}}$ with $B>0$ perpendicular to it. An electron has charge $-e$. The confining electron potential energy $U(y)$ is approximately flat inside a wide strip and rises at either boundary; smooth quenched disorder adds hills and valleys. For a slowly varying edge, each spin-resolved Landau level approximately followsHere $s=\pm 1$, and $Y$ is the guiding centre. For $\mathbf{A}=(-By,0,0)$, $Y={\ell }_B^2{k}_x$ and the edge group velocity isThus electrons move along $+x$ at the upper edge and $-x$ at the lower edge. The Lorentz force gives counterclockwise bulk cyclotron motion in this convention. At a hard boundary its interrupted circles become directed skipping orbits; a smooth boundary gives the corresponding guiding-centre drift. Smooth bulk quenched disorder generally gives closed drift contours around localized hills or valleys, with extended trajectories appearing near a broadened level's critical energy. Realistic electrostatic screening can produce alternating compressible quantum Hall edge strips and incompressible quantum Hall edge strips, so the unscreened bending curves are an effective sketch rather than an exact self-consistent edge profile.

In the original diagram, contacts are ordered around the electron propagation direction: upper-left $1\to 2\to 3$, then lower-right $4\to 5\to 6$. Let $I_1=I$, $I_4=-I$ and the other leads be ideal equilibrating voltage probes. Define $R_H=(V_2-V_6)/I$ and $R_{xx}=(V_2-V_3)/I$. These choices fix the voltage sign; reversing the field reverses chirality and the Hall polarity. For spin-resolved plateau index $N$,At fixed sheet carrier density and increasing field, $N$ decreases, giving successively higher Hall plateaus; longitudinal peaks occur between them. Unresolved spin branches can suppress odd plateaus. The sketch uses resistance magnitudes, and every integer step is not promised in an arbitrary sample.

For Fermi-energy oscillations in a two-dimensional electron gas, first distinguish fixed density from fixed reservoir chemical potential. At negligible temperature the former is determined by state counting. Each resolved Landau level contains $eB/h$ states per unit area, soIgnoring spin and broadening for the counting illustration, if $N<\nu <N+1$ then $N$ levels are full and level $N$ is partly full: $\mu =E_N$. At fixed $n_s$, increasing $B$ increases both level energy and degeneracy. The partly filled level rises in energy until the lower levels can contain all electrons; $\mu$ then drops to the next lower level. The resulting sawtooth is periodic in $1/B$, with $\Delta (1/B)=e/(n_{s}h)$ per resolved branch. At an exactly integer filling in an ideal gap, any chemical potential within the gap gives the same zero-temperature particle number; the plotted discontinuity denotes its limiting values.

At fixed $B$ while $n_s$ increases, the ideal chemical potential stays at a partly occupied level and steps upward after each additional density $eB/h$. It therefore has a staircase superposed on the rising zero-field trend, not a repeated downward sawtooth. Landau-level broadening and finite temperature smooth both ideal graphs. With an externally pinned chemical potential, density changes instead; the fixed-density sawtooth cannot simultaneously be imposed. In a realistic disordered sample, the Fermi energy sweeps localized states between transport-carrying level centres. The following schematic omits spin for clarity and does not attempt an interaction-corrected density of states.

For $N$ transmitted spin-resolved quantum Hall edge states, the Landauer-Büttiker formula reduces, in that contact order, toZero probe currents imply $V_2=V_3=V_1$ and $V_5=V_6=V_4$. Hence $I=(N{e}^2/h)(V_1-V_4)$ and the boxed resistance results follow. The spin-resolved conductance quantum arises because the one-dimensional density of states times velocity is $1/h$ per unit energy; a reservoir imbalance supplies $e^2\Delta V/h$ per channel. Its value does not depend on edge slope or sample width.

On a plateau, opposite-moving edge channels are separated by a wide bulk with a mobility gap. Ordinary elastic scattering on one edge has no counterpropagating channel into which to reflect; closed bulk orbits do not connect the contacts. Changing $B$ or $n_s$ can alter the occupancy of localized states without changing the number of transmitting edge channels. This accounts for the finite plateau width, which an entirely clean set of sharp levels would not explain. Accuracy requires negligible backscattering between opposite edges, low enough current to avoid breakdown, equilibrating contacts and negligible probe loading. The finite-width plateau and the precision of its quantized value are related but distinct observations.

An antidot is a depleted potential hill, not a potential well. Its closed edge circulates oppositely to the outer boundary. In a constriction, tunnelling through an antidot orbit can transfer an electron from a forward edge into the returning edge: resonant antidot backscattering produces conductance dips. For $N$ otherwise open spin-resolved channels and one active resonance,At symmetric coupling a single resonance can remove $e^2/h$; asymmetry or thermal averaging reduces its depth. If the background constriction is closed and the resonant route instead supplies forward transmission, peaks occur. The lead connectivity must therefore accompany the conductance sketch. In the fixed-area single-particle limit, successive alignments change enclosed flux by ${\Phi }_0=h/e$, giving the Aharonov-Bohm antidot periodThe effective orbit area, not merely the etched island area, enters. Charging, screening and spin can modify this simple sequence; two coupled spin branches can yield a near-$h/(2e)$ pattern without changing the elementary flux quantum.^[ref]

In Landau gauge for a uniform magnetic field $\mathbf{A}=(-By,0,0)$, the guiding centre is $Y=\hslash k_x/(eB)$. The group velocity isThus the upper edge in the original trajectory diagram carries electrons to the right and the lower edge to the left. Reversing the magnetic field reverses both arrows. Semiclassically these are skipping trajectories; quantum mechanically the guiding centres label propagating quantum Hall edge states. The larger orbital-plus-spin energy crosses the Fermi energy farther toward the interior, because less confinement energy is needed. Opposite spins in one orbital branch are therefore spatially separated.

In a clean bulk with the Fermi energy between Landau levels, there are no extended bulk states at that energy. Occupied bulk cyclotron orbits lie below it; one must not draw them as extra conducting Fermi-level channels. Smooth disorder can instead produce localized bulk contours without adding a direct source-to-drain channel.

A modulation-doped GaAs-AlGaAs electron gas is confined on the GaAs side of the interface. The larger-gap AlGaAs supplies a conduction-band barrier; charge transfer from remote donors and the resulting electric field bend the bands into an approximately triangular well. transverse electronic mode wavefunctions are confined normal to the interface but free in its plane. electrons settle into the well until their electrochemical potential agrees with the donor/contact environment. An undoped spacer separates mobile electrons from ionized donors, reducing impurity scattering. Equilibrium has one Fermi level across the structure; the conduction-band edge is not flat. Surface depletion and the cap also affect the profile. The original drawing illustrates one occupied confinement transverse electronic mode; a single spatial electron gas need not in general imply only one occupied vertical transverse electronic mode.

molecular beam epitaxy can grow a clean GaAs buffer/channel, an undoped AlGaAs spacer, a Si-doped AlGaAs supply region and a protective GaAs cap on a suitable substrate. Control composition, interface quality and background impurities during growth; placing dopants directly in the channel defeats the principal mobility advantage. Define a mesa or isolation region, fabricate low-resistance ohmic contacts to the buried gas, and pattern fine metal split gates by electron beam lithography and the lift-off process. Negative gate bias depletes the underlying gas, leaving a narrow channel between the gates. Check contact resistance, leakage, carrier density and mobility; operation with ballistic electron transport requires channel length shorter than the relevant scattering lengths and smooth confinement on the wavelength scale. The lithographic gap does not equal the electrical width because of lateral depletion.

For a long uniform channel supporting ballistic electron transport, transverse confinement creates thresholds $E_n$ with continuous longitudinal propagation above each threshold. One directed spin branch of a transverse electronic mode has states per length $dk/(2\pi )$ and group velocity $v={\hslash }^{-1}dE/dk$. Hence $vdk/(2\pi )=dE/h$: velocity cancels the one-dimensional density of states. Including two unresolved spin states,For low temperature and small bias, a fully open transverse electronic mode has $T_n(E_F)=1$, givingThe assumptions are ideal reservoirs/contacts, negligible reflection and interactions that would invalidate the independent-subband description, and unresolved spin splitting. A resolved spin branch contributes $e^2/h$. The finite two-terminal electrical resistance is a contact/mode-injection effect, despite the absence of dissipative scattering inside the ideal channel.

For voltage partition in a ballistic constriction, a short saddle has strong local electric-field and electrostatic screening variation near its entrances and barrier, rather than the uniform Ohmic voltage gradient of a diffusive wire. Right-moving incident states inherit the source distribution and left-moving states the drain distribution; there is no unique local equilibrium chemical potential inside an ideal ballistic saddle. The electrostatic potential must be determined self-consistently. In a symmetric rigid-barrier approximation, the barrier follows the mean lead potential and half the electrochemical bias lies on each side. This is a stated approximation, not a universal spatial voltage profile.

For a saddle $U=U_0-m^{\ast }{\omega }_x^2{x}^2/2+m^{\ast }{\omega }_y^2{y}^2/2$, transverse electronic mode thresholds are $E_n=U_0+\hslash {\omega }_y(n+1/2)$. The smooth-barrier transmission is $T_n(E)=[1+\exp (-2\pi (E-E_n)/(\hslash {\omega }_x)){]}^{-1}$, approaching a step for a sharp threshold on the bias/thermal scale.^[ref]

Let ${\mu }_{S,D}=E_F\pm eV/2$ and assume no additional bias dependence of $T_n$. At zero temperature,When the source and drain sample $N+1$ and $N$ thresholds, the differential electrical conductance is $(N+1/2)2e^2/h$. These half plateaus arise from the two ends of the bias window, not fractional carriers. The dc ratio $I/V$ need not share their plateau values. Heating, spin splitting and interaction-induced features must be distinguished experimentally.

In bias spectroscopy of one-dimensional subbands, plot differential electrical conductance or its gate derivative against gate bias and source-drain bias. Threshold branches occur where $E_n(V_g)={\mu }_S$ or ${\mu }_D$. When one bias-window edge meets $E_n$ and the other $E_{n+1}$,Thus the intersections of adjacent threshold branches calibrate the one-dimensional spacing directly in energy. A gate lever arm $\alpha =-e^{-1}\partial E_n/\partial V_g$ can then convert gate intervals to energy; it should be measured rather than guessed. Asymmetric bias partition, a moving barrier or a gate-dependent spacing modifies slopes and plateau weights, requiring the corresponding self-consistent bias model.

For group velocity from a period-dependent phase velocity, start with $k=\omega /c_n(\omega )$ rather than equating group velocity to phase velocity. Differentiating,For the stated inverse-frequency term, $d{c}_n/d\omega =-a/{\omega }^2$, soThe inequality assumes positive frequency and physical positive phase velocity, with $a>0$. The symbol $a$ here has units of acceleration and is unrelated to the plate thickness in Question 1.

Seismic surface-wave dispersion. Rayleigh waves combine vertical and longitudinal motion in the vertical plane of propagation; Love waves have horizontal transverse motion and can be guided by a low-shear-speed surface layer. A homogeneous elastic half-space supports a Rayleigh wave with a frequency-independent speed in the ideal elastic limit, but depth-dependent Earth structure makes the surface-wave eigenfunctions and their propagation speeds depend on frequency. This structural dispersion can occur even when each constituent material is itself nondispersive.

Distinguish phase velocity $c=\omega /k$ from group velocity $U=d\omega /dk$, which controls a narrow-band envelope in the appropriate weakly attenuating limit. Longer periods generally have deeper sensitivity, while shorter periods emphasize shallower structure. In a profile whose shear speed increases with depth this often makes long-period waves faster, but low-velocity zones, anisotropy and higher modes can change that trend. Frequency dependence need not be monotonic in every Earth model.

Measure phase delays or wave-packet arrival times across known source–receiver paths for many periods, accounting for source phase, cycle ambiguity, multipathing and mode identification. Compute predicted dispersion from trial elastic profiles and compare with observations. Linearization expresses the sensitivity schematically aswhere $\alpha ,\beta$ are compressional and shear speeds. A regularized inverse problem then estimates depth structure; many crossing paths additionally provide surface-wave tomography. Joint Love wave and Rayleigh wave measurements help distinguish shear-wave polarizations and constrain radial seismic anisotropy. Finite bandwidth, depth averaging and correlated sensitivities mean that one dispersion curve does not uniquely determine a detailed Earth profile.

Seismic ray theory, the eikonal equation and Fermat's principle. In a smoothly varying isotropic medium, insert a high-frequency ansatz $u=A(\mathbf{x})e^{i\omega [T(\mathbf{x})-t]}$ into the scalar wave equation. The leading powers of frequency giveThe travel-time surfaces are wavefronts and rays are their normal trajectories, with slowness vector $\mathbf{p}=\nabla T$. The next-order equation transports amplitude, including geometrical spreading; attenuation is an additional material effect.

Fermat's principle says that the ray makes $\int ds/c$ stationary under path variations with fixed endpoints. Stationarity, rather than always the global minimum, permits multiple arrivals. Its Euler–Lagrange equation, with arclength $s$ and unit tangent $\hat{\mathbf{t}}$, isFor horizontal layering, the horizontal slowness $p_h=\sin i/c$ is conserved, giving Snell's law at interfaces and smooth refraction between them. Here $i$ is measured from vertical. A turning point satisfies $p_{h}c(z)=1$; increasing speed with depth bends a downgoing ray back upwards. Travel-time tomography uses these ray integrals to infer spatial slowness variations.

The approximation requires wavelength to be small compared with structural variation scales. Caustics, turning-point amplitudes, diffraction, scattering and finite-frequency sensitivity require wave-based corrections. Thus seismic ray theory gives an economical geometric description and useful travel times, while surface-wave dispersion retains essential interference and boundary-mode physics.

For $k>0$, phase velocity is the speed of a fixed phase, $c_p=\omega /k$. The group velocity $c_g=d\omega /dk$ transports a narrow-band wave envelope. With $q=kH$,As $H\to 0$ with $k$ fixed, $\tanh (kH)/(kH)\to 1$. ThereforeBoth velocities tend to the shallow-water speed $\sqrt{gH}$ at leading order.

The material surface is $F(x,y,t)=y-\eta (x,t)=0$. Its material derivative vanishes, giving the exact kinematic conditionFor irrotational flow $u=\nabla \varphi$, the Euler equations and $(u\cdot \nabla )u=\nabla (|u{|}^2/2)$ integrate spatially to ${\varphi }_t+|\nabla \varphi {|}^2/2+gy+p/\rho =C(t)$. Adding a suitable function of time to $\varphi$ changes no velocity and makes the right side $p_a/\rho$. ThusTake $p_a$ as the air-side pressure and use the given capillary pressure jump on the fluid side. At first order in wave amplitude, the product ${\varphi }_x{\eta }_x$ and kinetic-energy term are quadratic, the curvature denominator is one to the required order, and replacing $y=\eta$ by $y=0$ changes first-order quantities only at second order. ConsequentlyFor a harmonic wave $\eta =A\cos (kx-\omega t)$, the harmonic potential decaying with depth isIt satisfies the kinematic condition. Substituting into the dynamic condition givesThe phase speed is $c=\omega /k$ and the group velocity of a narrow wave packet is $c_g=d\omega /dk$. For the positive-frequency branch and $\sigma =T/\rho$,For positive surface tension this ratio exceeds one exactly when $\sigma k^2>g$, proving