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The experiment involving Ferenc Krausz combined a subfemtosecond soft-X-ray pulse with a few-cycle visible electric field. Its delay-dependent krypton photoelectron spectra tested the field-controlled response. The reported pulse duration was $650\pm 150$ attoseconds. Changing delay compared the emitted electron distributions at different reference phases; the spectral-width trace supplied a further duration-sensitive observable.^[ref]

The distinction from RABBITT is the measurement regime and reconstruction, rather than merely calling one trace a clock. The attosecond streaking momentum map is a limiting model; a finite-duration wave packet requires a quantum transition calculation. Inferring an emission delay also requires separating the illuminating pulse's spectral phase from the target response. The measurement supports time-dependent photoionization in a specified model, with no unique classical electron trajectory inside an atomic orbital implied.

An absolute hydrodynamic instability makes the response to a localized impulse grow at a fixed spatial point in a specified frame. This differs from convective hydrodynamic instability, which amplifies a moving wave packet while its response decays at fixed positions. For analytic dispersion relations, the relevant absolute frequency is associated with an accessible zero-group velocity saddle, with the spatial-branch selection checked rather than assumed for an arbitrary dispersion relation.

In one dimension the band effective mass is defined by $1/m^{\ast }(k)=E^{\prime \prime }(k)/{\hslash }^2$. Together with $v=E^{\prime }(k)/\hslash$ and $\hslash \dot{k}=F$, this gives $\dot{v}=F/m^{\ast }$. Negative band curvature produces negative effective mass for the electron wave packet; near a filled-band maximum the equivalent missing-electron description uses positively responding holes. In several dimensions the inverse mass is the Hessian ${\partial }_i{\partial }_{j}E/{\hslash }^2$.

In a single isolated energy band, a wave packet subject to a constant force satisfies $\hslash \dot{k}=F$. Its group velocity is periodic in $k$ with reciprocal-lattice period $2\pi /a$, so its position oscillates with period $2\pi \hslash /(|F|a)$ when its mean velocity over a band is zero. For $E(k)=E_0-2J\cos ka$, integrating the sinusoidal velocity gives the oscillation explicitly. This ideal description neglects scattering and interband tunnelling.

A capillary wave is a wave along a fluid interface whose restoring stress is predominantly supplied by surface tension. A curved free surface requires a pressure jump by the Young–Laplace equation; increasing its corrugation also costs surface energy. In the ideal limit, inertia carries the interface through its flat position and the resulting oscillation propagates. When gravity also contributes appreciably, use the combined gravity-capillary wave model. Within the capillary wave regime, shortening a ripple makes it travel faster. ^[ref]

The laser-diffraction measurement of capillary waves below turns this mechanism into a test: set the frequency, measure the wavelength independently with light, and ask whether the dispersion relation follows. The theory is useful both for predicting how ripple wave packets separate and for inferring surface tension from their motion.

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.

The massless Dirac equation in one spatial dimension separates into opposite chirality sectors. Its initial profiles evolve as ${\psi }_R(t,x)=f(x-t)$ and ${\psi }_L(t,x)=h(x+t)$, by the chain rule along the straight characteristics. Thus wave packets translate without distortion at the speed of light. A nonzero Dirac mass term couples the two sectors, so these separate rigid-translation formulas do not apply to a massive field.

Take a cold, unmagnetized, collisionless plasma with uniform equilibrium electron density $n_e$, fixed neutralizing ions, and a small electromagnetic wave that is a transverse wave. Cold means that thermal-motion corrections are neglected. For electron mass $m_e$, elementary charge magnitude $e$ and electric field $\mathbf{E}=\mathrm{Re}(\hat{\mathbf{E}}{e}^{-i\omega t})$, the linear Newton's second law equation is $m_e\dot{\mathbf{v}}=-e\mathbf{E}$. Thus $\hat{\mathbf{v}}=-ie\hat{\mathbf{E}}/(m_e\omega )$ and the current density is $\hat{\mathbf{J}}=i{n}_e{e}^2\hat{\mathbf{E}}/(m_e\omega )$. Writing the induced electric polarization as $\hat{\mathbf{J}}=-i\omega \hat{\mathbf{P}}$ gives a negative electric susceptibility. With vacuum permittivity ${ϵ}_0$,The plasma frequency ${\omega }_p$ controls the response. For $\omega >{\omega }_p$, the propagating branch has real refractive index $0<n<1$. For $\omega <{\omega }_p$, this homogeneous model has an evanescent wave, not an ordinary propagating wave with a negative real refractive index. In a slowly varying vertical profile, a normally incident ray approaches a turning region where ${\omega }_p(z)=\omega$; the geometrical-ray approximation fails right at that region. At oblique incidence the conserved horizontal wave vector changes the turning condition. A smooth density gradient can therefore return a skywave without a mirror or a layer of metal.

For a homogeneous propagating section, direct differentiation of the dispersion relation givesA phase velocity greater than $c$ does not send information faster than $c$. Here a narrow wave packet has group velocity $v_g$, while the full causal response must still respect special relativity. Even this stripped-down plasma is strongly dispersive near its plasma frequency. In the actual ionosphere, collisions produce absorption, and the magnetic field of Earth changes the response with direction and wave polarization; a single real scalar $n$ cannot capture all of that behavior.

A material particle carries a quantum phase whose wave vector and angular frequency obey $\mathbf{p}=\hslash \mathbf{k}$ and $E=\hslash \omega$. Interference and diffraction test this relation; a localized particle requires a wave packet.

The de Broglie wavelength is the distance over which a matter wave accumulates one full spatial quantum phase cycle in its propagation direction:Here $h$ is the Planck constant, $\mathbf{p}$ is the momentum, and $\mathbf{k}$ is the angular wavevector. The scalar $p$ in $\lambda =h/p$ means the positive momentum magnitude, not a signed component. Reversing a momentum reverses the wavevector without making the wavelength negative. At zero momentum there is no finite spatial phase period: the limiting wavelength diverges.

This relation makes microscopic diffraction accessible with particles: increasing the momentum shortens the wavelength until it is comparable with the spacings in a crystal scattering problem. The electron accelerating-voltage wavelength gives a practical way to tune this scale, while the Davisson–Germer experiment tests it by comparing electron diffraction directions with independently known crystal geometry. The same relation applies to neutral particles in a neutron interferometer; electric charge is not a requirement for a matter wave.

The wavelength belongs to the wavefunction, not to a sinusoidal path of a particle. A free-particle momentum convention makes the distinction especially clear: a plane wave has a rotating complex quantum phase but a constant probability density. A localized wave packet contains a range of momenta and therefore generally has a range of wavelengths, rather than one exact spatial period.

For constant $A,\rho ,K$ on all space and time, the active waveform transformation ${\psi }_b(z,t)=\psi (bz,bt)$ leaves the action of $\mathcal{L}=A(\rho {\dot{\psi }}^2-K{\psi }^{\prime 2})/2$ invariant: the density gains $b^2$ and the integration measure gains $b^{-2}$. With canonical momentum $\Pi =A\rho \dot{\psi }$, its Noether current can be chosen asThe integrated charge $D=\int j_D^{0}dz$ is conserved if its boundary flux vanishes and the weighted integral exists. For a rightgoing wave packet $\psi =f(z-ct)$, $c=\sqrt{K/\rho }$, it becomes $D=-A\rho c\int \xi [f^{\prime }(\xi ){]}^{2}d\xi$. A fixed finite interval generally breaks this scale symmetry unless its boundaries and boundary data are transformed too.

A dispersion relation specifies which angular frequencies and wavevectors a linear wave model admits. Fix the plane wave conventionHere $a$ is a constant complex amplitude, $\mathbf{k}$ is the wavevector, and $\omega$ is the angular frequency. Substitution into a constant-coefficient linear partial differential equation replaces ${\partial }_t$ by $-i\omega$ and $\nabla$ by $i\mathbf{k}$, turning the wave problem into an algebraic condition. A system can have several roots; $\omega =\omega (\mathbf{k})$ means one chosen branch, rather than a promise that the full dispersion relation is single-valued. For real $\mathbf{k}$, $\mathrm{Im}\omega >0$ means temporal growth in this convention, while $\mathrm{Im}\omega <0$ means damping. The real-branch discussion of phase velocity and group velocity below assumes negligible loss.

For the one-dimensional wave equation $u_{tt}=c^2{u}_{xx}$, the dispersion relation is ${\omega }^2=c^2{k}^2$. The branch $\omega =ck$ carries right-moving travelling waves for $c>0$; both its phase velocity and group velocity are $c$. One may instead use the positive-angular frequency branch $\omega =c|k|$, with right-moving components at $k>0$ and left-moving components at $k<0$. A real wave pairs each $(k,\omega )$ with $(-k,-\omega )$; that conjugate pair represents the same travelling pattern. Branch and sign choices therefore matter when interpreting a dispersion diagram.

A particularly tangible dispersive example is a small-amplitude surface gravity wave on still water of uniform depth $h$. Neglect surface tension and viscosity, assume incompressible flow and irrotational flow, and impose a flat impermeable bottom. With $k>0$ and positive angular frequency,Here $g$ is gravitational acceleration and $\tanh$ is the hyperbolic tangent. Deep water means $kh\gg 1$: $\omega \simeq \sqrt{gk}$ and longer wavelengths have larger phase speed. Shallow water means $kh\ll 1$: $\omega \simeq \sqrt{gh}k$, an approximately nondispersive wave. Thus depth relative to wavelength, rather than depth alone, determines whether crests and wave packets travel together.^[ref]

The dispersion diagram below also includes the ideal deep-water capillary wave limit $\omega =\sqrt{\gamma /\rho }{k}^{3/2}$, where $\gamma$ is surface tension and $\rho$ is mass density. This limit requires the surface tension restoring term to dominate gravity; it reverses the trend of the surface gravity wave, giving larger phase speed at shorter wavelength. The combined gravity-capillary wave has ${\omega }^2=gk+(\gamma /\rho )k^3$.^[ref]

The curves use dimensionless normalizations chosen separately for each model; they are analytical diagrams, not experimental data. Their beauty is that a single dispersion relation encodes several observable behaviours: its value gives the angular frequency, its origin-to-point slope gives the one-dimensional phase velocity, its tangent slope gives the group velocity, and variation of that tangent slope predicts distortion of a wave packet. A straight line through the origin describes a nondispersive wave; an affine branch $\omega =ck+b$ with $b\ne 0$ has constant group velocity but phase velocity $c+b/k$, so those properties must be distinguished.

A coherent or fluctuating swirling region of a velocity field. Eddies transport momentum, heat and material constituents, and their averaged fluxes can modify a mean current. They need not be nonlinear: a linear wave packet can also have nonzero averaged quadratic fluxes.

For a semiclassical wave packet in an isolated energy band, $v_i={\hslash }^{-1}{\partial }_{k_i}\epsilon$ and $\hslash {\dot{k}}_j=F_j$ imply ${\dot{v}}_i={\sum }_j(m^{\ast -1}{)}_{ij}{F}_j$. The inverse effective-mass tensor is the Hessian matrix of the dispersion divided by ${\hslash }^2$. Along a unit vector $\mathbf{e}$, $1/m_{\mathbf{e}}^{\ast }={\sum }_{ij}{e}_i(m^{\ast -1}{)}_{ij}{e}_j$. The Laplacian is its trace: ${\hslash }^{-2}{\nabla }_k^2\epsilon =\mathrm{tr}(m^{\ast -1})$. Thus in $d$ isotropic dimensions the scalar mass is $d{\hslash }^2/({\nabla }_k^2\epsilon )$, rather than ${\hslash }^2/({\nabla }_k^2\epsilon )$.

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.

A nonzero probability density inside a barrier does not by itself imply transport. In one dimension, the probability current of a wavefunction is $J=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }^{\prime })$. A single evanescent wave $C{e}^{-\kappa x}$ has ${\psi }^{\ast }{\psi }^{\prime }=-\kappa |C{|}^2{e}^{-2\kappa x}$, a real quantity, hence $J=0$. This is what happens on the inaccessible side of a potential barrier extending to infinity with $E<V_0$.^[ref]

For the finite-barrier wavefunction $A{e}^{\kappa x}+B{e}^{-\kappa x}$, the cross terms instead giveThe relative quantum phase of the two probability amplitudes permits a nonzero, spatially constant probability current, even though neither exponential function separately carries it. The probability continuity equation gives ${\partial }_{x}J=0$ for this stationary state. Matching the wavefunction and its derivative also makes $J$ continuous at both faces, soHere $R=|r{|}^2$ is the quantum reflection probability, and $T=J_{\mathrm{tr}}/J_{\mathrm{inc}}$ is the transmission probability. The net exterior probability current equals the interior probability current; it is not lost as the wavefunction decreases. A physical incident wave packet supplies finite total probability, while the stationary state describes the energy component used to compute its transmitted flux.

For $u_t=u_x$, the recurrence $U_m^{n+1}-U_m^{n-1}={\sum }_{j=\pm 1,\pm 2}{a}_j{U}_{m+j}^n$ has fourth-order normalized consistency of a numerical method at fixed $\mu =k/h$ ifTo derive this, match moments $M_r={\sum }_j{a}_j{j}^r$ to the temporal Taylor series: $M_0=M_2=M_4=0$, $M_1=2\mu$, $M_3=2{\mu }^3$. The even conditions force antisymmetry; the two odd conditions then determine the displayed coefficients. The normalized residual begins with $h^4({\mu }^2-1)({\mu }^2-4)u_{xxxxx}/120$. Its amplification polynomial of a multilevel finite difference scheme is $G^2-2iA(\theta )G-1$, where $A=a_1\sin \theta +a_2\sin 2\theta$. At $\mu =1/2$, $|A|\le 3/4$, giving a uniform power bound. At $\mu =3/2$, $A(\pi /3)=19\sqrt{3}/32>1$, and wave packets on the growing polynomial root prove Cauchy linear instability.

For a free particle of mass $m>0$ in nonrelativistic quantum mechanics, choose the potential energy zero and writeApplying the momentum operator $-i\hslash \nabla$ returns $\mathbf{p}\psi$; this fixes the sign convention. The angular frequency is measured relative to this chosen energy zero. Adding a constant energy changes only a time-dependent global phase, leaving the spatial de Broglie wavelength unchanged. No rest-energy term is included in this nonrelativistic quantum mechanics convention.

The plane wave has $|\psi {|}^2=|A{|}^2$ and probability current $\mathbf{j}=(\mathbf{p}/m)|A{|}^2$. Its real part can be drawn as a sine or cosine, but that graph is a component of a probability amplitude, not a transverse displacement. On infinite space this is a generalized momentum eigenstate, not a state satisfying wavefunction normalization. A narrow wave packet instead moves with group velocity ${\nabla }_{\mathbf{k}}\omega =\mathbf{p}/m$; the phase velocity $\omega /|\mathbf{k}|=|\mathbf{p}|/(2m)$ is not its transport speed.

The equality of canonical momentum and kinetic momentum here assumes a field-free region with the magnetic vector potential chosen zero. For a charged particle in a magnetic field, the kinetic momentum is $\mathbf{\Pi }=\mathbf{p}-q\mathbf{A}$. An arbitrary canonical phase gradient must not then be substituted into $m\mathbf{v}$; the free particle formula needs its stated assumptions.

A gravity-capillary wave has restoring contributions from both gravity and surface tension. Consider an initially resting, horizontally unbounded liquid layer of constant depth $h$, with uniform mass density $\rho$, incompressible flow, inviscid flow and irrotational flow. The upper gas exerts constant pressure and has negligible inertia. Assume a clean interface with constant surface tension $\gamma$: no surfactant stresses, imposed current or wind. The wave amplitude $a$ must satisfy $ka\ll 1$ and $a\ll h$, so linearized boundary conditions apply. Deep water means $kh\gg 1$, rather than a particular depth in metres. ^[ref]

Let $z=0$ be the undisturbed free surface, with liquid below it. Potential flow gives a velocity potential $\varphi$ with $\mathbf{u}=\nabla \varphi$, and incompressible flow gives the Laplace equation. The linearized kinematic boundary condition for a free-surface graph and the Unsteady Bernoulli equation, including gravity and the Young–Laplace equation, giveHere $\eta (x,t)$ is the interface displacement; the signed small-slope pressure jump is $p_{\mathrm{liquid}}-p_{\mathrm{gas}}=-\gamma {\eta }_{xx}$. A wave crest has ${\eta }_{xx}<0$, so its capillary pressure is positive. Choose a right-travelling Fourier mode of positive wavenumber $k$ and angular frequency $\omega$:The factor $e^{kz}$ decays into the deep liquid, satisfies the Laplace equation, and makes ${\varphi }_z={\eta }_t$ at $z=0$. Substituting in the dynamic boundary condition leavesThis derivation displays the two restoring contributions with the same sign. Dimensional analysis also checks the result: $\gamma /\rho$ has units ${\mathrm{m}}^3{\mathrm{s}}^{-2}$, so both $gk$ and $(\gamma /\rho )k^3$ have units ${\mathrm{s}}^{-2}$. ^[ref]

At finite depth, a flat impermeable bottom at $z=-h$ requires ${\varphi }_z=0$. Replacing $e^{kz}$ by the normalized vertical factor $\cosh (k(z+h))/\cosh (kh)$ makes the surface ratio ${\varphi }_z/\varphi =k\tanh (kh)$. Eliminating the mode amplitudes then givesThus the same model supplies its own depth check: use the deep-water formula only when $\tanh (kh)$ is sufficiently close to one for the required accuracy. For example, $kh=3$ reduces $\omega$ by about $0.248\%$ relative to deep water, while $kh=1$ reduces it by about $12.7\%$. These percentages are calculations from this factor. ^[ref]

In deep water the phase velocity magnitude follows a wave crest, while the group velocity magnitude follows the wave envelope of a narrow wave packet. Differentiating the positive-frequency dispersion relation givesFor $k{\ell }_c\ll 1$, the surface gravity wave limit has $c_g\simeq c_p/2$. For $k{\ell }_c\gg 1$, the deep-water capillary-wave dispersion relation has $c_g\simeq 3c_p/2$: a wave crest slips backward relative to the faster wave envelope, even while both advance. Neither speed is the oscillating fluid element's own velocity. ^[ref]

There is a particularly neat coincidence at the crossover. Differentiate $c_p^2=g/k+\gamma k/\rho$: its unique minimum is at $k_c=\sqrt{\rho g/\gamma }=1/{\ell }_c$, exactly where the restoring terms are equal. ThereforeFor the stipulated parameters in capillary length, the minimum is $0.231\mathrm{m}{\mathrm{s}}^{-1}$. At $\lambda =3\mathrm{mm}$, direct substitution gives $f=\omega /(2\pi )\simeq 131\mathrm{Hz}$, $c_p\simeq 0.394\mathrm{m}{\mathrm{s}}^{-1}$ and $c_g\simeq 0.580\mathrm{m}{\mathrm{s}}^{-1}$. This short wavelength is capillary dominated, but its group velocity is still slightly below the pure-capillary ratio $3c_p/2$ because gravity has not vanished.

The group refractive index is $n_g=c/v_g$ for a transparent isotropic propagation branch with forward group velocity $v_g=d\omega /dk>0$. Differentiating the local wavenumber $k=n(\omega )\omega /c$ givesThe last formula gives the propagation group delay along a path whose geometry is fixed as frequency varies; additional apparatus wave phases require their own frequency derivatives. For a narrow wave packet with little loss and negligible reshaping, it approximates the transit of its wave envelope. In the nondispersive case ${\partial }_{\omega }n=0$, it reduces to $\mathrm{OPL}/c$. An optical path length at one frequency contains no information about ${\partial }_{\omega }n$, so in a medium with optical dispersion it cannot by itself determine the group delay. ^[ref]

For an original local model near ${\omega }_{\ast }$, choose two equal geometric paths of length $L$, with $n_A(\omega )=3/2$ and $n_B(\omega )=3/2+(\omega -{\omega }_{\ast })/(10{\omega }_{\ast })$. At ${\omega }_{\ast }$ both optical path lengths equal $3L/2$, and matched launch and interface references give equal carrier wave phases. But $n_{g,A}=3/2$ and $n_{g,B}=8/5$, so their group delays differ by $L/(10c)$. This local stipulated model is a counterexample to identifying equal wave phase with equal pulse arrival; it asserts no material response outside the narrow band considered. Strong loss or pulse reshaping requires a fuller propagation model, and group delay need not time the earliest causal response.