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Start from $\rho ({\partial }_{t}u+u\cdot \nabla u)=-\nabla p-\rho g{e}_z$, ${\partial }_t\rho +u\cdot \nabla \rho =0$, and $\nabla \cdot u=0$. The stationary background obeys $p_0^{\prime }=-g{\rho }_0$. Linearize in small perturbations and use the Boussinesq approximation: replace the density by a reference value ${\rho }_{\ast }$ in inertia but retain the density perturbation in the buoyancy force. With $\pi =p^{\prime }/{\rho }_{\ast }$ and $b=-g{\rho }^{\prime }/{\rho }_{\ast }$,This is the Brunt–Väisälä frequency. Its assumed constancy requires a locally uniform stable density gradient in this approximation; rotation and viscosity are neglected.

For a plane wave with wavevector $K=(k,\ell ,m)$, project the momentum equation onto the space perpendicular to $K$ to eliminate pressure. Its vertical component is $-i\omega w=b[1-m^2/|K{|}^2]$, while $-i\omega b+N^{2}w=0$. Eliminating $b$ yieldsFor $0<\omega <N$, a small oscillating body radiates upward and downward conical internal gravity wave beams. In a vertical section the beams make a St Andrew's cross, with angle $\alpha$ to the horizontal satisfying $\sin \alpha =\omega /N$. Indeed the group velocity is perpendicular to the wavevector and has ratio of vertical to horizontal magnitudes $\sqrt{k^2+{\ell }^2}/|m|$. Thus wave energy follows the cones, rather than the direction normal to a fixed-radius spherical wavefront. There are no such propagating beams for $\omega >N$.

For the shear-flow calculation choose the positive intrinsic-frequency branch and $k_0>0$, conditions needed for the printed critical-level formula. The slowly varying local dispersion relation isStationarity and horizontal homogeneity conserve $\Omega$ and $k=k_0$. The Hamiltonian ray equations for a local dispersion relation giveThus $m_0<0$ stays negative and the ray propagates upward. Conservation of frequency givesIf the shear reaches the required velocity difference, its first zero is the critical level of an internal gravity wave,As $z\uparrow z_c$, the intrinsic frequency tends to zero, $m\to -\infty$, the vertical wavelength tends to zero and the vertical group velocity tends to zero. The ideal ray cannot cross this level. For finite $U_c^{\prime }>0$, $\dot{z}\sim k_0(U_c^{\prime }{)}^2(z_c-z{)}^2/N$, so it approaches the level only asymptotically in time. The disturbance develops short scales; even weak viscosity can then absorb it, and nonlinear effects can invalidate the small-amplitude approximation.

The printed sign and existence hypotheses are incomplete. On the positive-frequency branch, $k_0<0$ instead gives $\dot{m}>0$ and eventually a turning point when the intrinsic frequency reaches $N$. As a concrete counterexample to the stated universal height bound, take $N=1$, $k_0=-1$, $m_0=-3$, and $U(z)=\epsilon z$ with any small $\epsilon >0$. At the printed height $z_c=1/(\epsilon \sqrt{10})$, the intrinsic frequency is $2/\sqrt{10}<1$ and $m=-\sqrt{3/2}<0$, so the vertical group velocity is still positive and the ray crosses that height. Also, $U^{\prime }>0$ alone does not ensure that the displayed velocity difference is ever reached; an increasing bounded $U$ may have no finite critical level. The qualified derivation above gives the intended positive-$k_0$ result whenever that level exists.

Take $U_0>0$, $k=2\pi /L>0$, and represent the stationary topographic internal gravity wave by phase $kx+mz$. Its laboratory angular frequency is zero, and its intrinsic frequency is $\hat{\omega }=-k{U}_0$. The internal gravity wave dispersion relation givesThe radiation condition selects $m>0$, since this makes the vertical group velocity upward. The lower kinematic boundary condition is $w=U_0{h}_x$, so a corresponding linear wave field iswhere $\zeta$ is the vertical fluid displacement. The constant-phase lines of an internal gravity wave slope down to the right, with $dz/dx=-k/m$.

Let $K^2=k^2+m^2$. In the frame fixed to the hills the phase pattern is stationary: its phase velocity is zero, while the group velocity isEnergy travels up and downwind, with ray slope $dz/dx=m/k$. In coordinates $x^{\prime }=x-U_{0}t$ moving with the wind, the phase is $k{x}^{\prime }+mz+k{U}_{0}t$. ThusThe intrinsic phase velocity is upstream and downward; the intrinsic group velocity is upstream and upward, perpendicular to the wave vector. Adding $(U_0,0)$ to the group velocity recovers the laboratory energy ray.

Vertically propagating waves require $k{U}_0<N_0$, equivalently $\boxed{L>2\pi U_0/N_0}$. At equality the vertical group velocity vanishes; above the mountain-wave cutoff the bounded response is an evanescent wave. The linearization requires small terrain slope $k{h}_0\ll 1$ and small wave-induced displacement gradients $|m|h_0\ll 1$, often summarized by $N_0{h}_0/U_0\ll 1$. The model also neglects rotation, viscosity and compressibility; its scales must permit those approximations.

Let $\pi$ be pressure divided by the reference density and $b$ the buoyancy perturbation. The linear, inviscid Boussinesq equations about $U(z)$ areFor a stationary Fourier component $e^{ikx}$, incompressibility gives $u=i{w}^{\prime }/k$. The horizontal momentum equation then gives $\pi =i(U^{\prime }w-U{w}^{\prime })/k$, while buoyancy conservation gives $b=i{N}^{2}w/(kU)$. Substitute these into the vertical momentum equation to obtain the stationary Taylor–Goldstein equation:Hence its local WKB approximation isFor $U^{\prime \prime }$ negligible, an outgoing upward wave has $mU/k>0$. In particular $m>0$ for $k,U>0$. In an upper evanescent region choose the branch with $\mathrm{Im}m>0$ in $e^{i\int mdz}$ so the solution decays upward. These choices express the radiation condition; a reflected wave has the opposite real branch and downward group velocity.

There are two different limiting levels. A critical layer in a shear flow occurs at $U=0$, where the intrinsic frequency vanishes, $|m|\to \infty$, and upward propagation slows while the vertical wavelength contracts. Dissipation or nonlinear effects can absorb the wave and transfer its momentum to the mean flow. For a stable, slowly varying critical layer with Richardson number $\mathrm{Ri}=N_0^2/(U^{\prime }{)}^2>1/4$, the usual causal inviscid continuation suppresses transmission; it is not the same as reflection at $m=0$. When the Richardson number is small, a general absorption-only conclusion needs extra stability assumptions.

Reflection instead occurs at the turning level of a stationary internal wave $k|U|=N_0$, beyond which $m^2<0$. It is sometimes called a critical reflection height, but the wind there is nonzero. Assuming $0<k{U}_0<N_0$, the specified profiles giveThe decreasing-wind profile never encounters $m=0$ before its zero-wind layer. Both profiles have $U^{\prime \prime }=0$; a well-separated WKB description requires $N_{0}H/U_0\gg 1$, and a turning-layer transition is still required for the increasing profile.

Choose the positive-frequency branch with $k>0$, let $K=|\mathbf{k}|$, and write $\mathbf{k}=K(\cos \theta ,\sin \theta )$. The internal gravity wave dispersion relation is $\omega =Nk/K=N\cos \theta$. The phase velocity and group velocity areThey are perpendicular: the phase advances normal to the crests while energy travels along them. For a general signed $k$, the positive frequency is $N|k|/K$ and the horizontal component of group velocity changes sign with $k$.

If $\omega >N$, there is no real wave vector satisfying this relation. For prescribed real horizontal wavenumber, $m^2=k^2(N^2/{\omega }^2-1)<0$, so a bounded boundary response is an evanescent wave, rather than a freely propagating plane wave.

For a single incompressible plane wave, internal-wave polarization gives ${\mathbf{u}}_1\cdot \mathbf{k}=0$. Its oscillatory displacement ${\mathbf{\xi }}_1$ is parallel to ${\mathbf{u}}_1$ and therefore also perpendicular to $\mathbf{k}$. Every spatial derivative of the plane-wave field is proportional to a component of $\mathbf{k}$, soThere is no leading Stokes drift for that single wave. The fluid parcels oscillate along straight lines within phase planes and never sample a spatial change of the wave along their displacement. This conclusion does not prohibit cross terms between different wave vectors.

Assume $\kappa ,\omega >0$ and impose upward group velocity in $z>0$. The two positive-frequency components then have wave vectors ${\mathbf{k}}_1=(\kappa ,m)$ and ${\mathbf{k}}_2=(-\kappa ,m)$ with $m<0$. Equal lengths are automatic, and orthogonality gives $m^2={\kappa }^2$. HenceThe given orthogonal-wave condition is therefore possible only at this forcing frequency. The upward radiation condition excludes the $m=+\kappa$ branch.

Use superposition with vertical amplitude $W$ in each component and internal-wave polarization:The two sets of crests have slopes $+1$ and $-1$; their energy propagates upward and outward at angles of $45^{\circ }$, while their phases travel downward. Their interference makes fixed horizontal bands of vertical-velocity nodes, at $\cos \kappa x=0$, and horizontal-velocity nodes, at $\sin \kappa x=0$.

Put $A=\kappa x$, $B=\kappa z+\omega t$. The first-order displacement relative to mean parcel labels isThe horizontal displacement correction vanishes identically, whileConsequently the formal cross-wave Stokes drift isIt has alternating downward and upward columns, no depth decay in this ideal infinite plane-wave field, and zero horizontal-period-averaged vertical transport. Each wave alone has zero drift; the interference terms create this result.

This calculation is a kinematic second-order consequence of the linear field, not a complete prediction of mean transport. In particular, an impermeable oscillating lower boundary cannot have that nonzero mean normal parcel velocity. Its first-order displacement is ${\zeta }_b=(2W/\omega )\cos (\kappa x)\sin \omega t$. Expanding the exact kinematic boundary condition on $z={\zeta }_b$ givesThis Eulerian mean velocity cancels the computed normal Stokes drift at the material boundary. Away from it, the second-order mean-flow and buoyancy equations must also be satisfied. Thus the alternating drift is a valid formal cross-wave effect, but treating it alone as persistent transport with zero Eulerian mean would be physically incomplete. A prescribed permeable boundary would require its own second-order flux specification instead.

Substitute $\varphi =e^{i(kx-\omega t)}$ into the mixed-derivative equation. Its left side is $k\omega \varphi$, so for $k\ne 0$ the dispersion relation, phase velocity and group velocity areThe graph of a function for each velocity is symmetric about $k=0$, singular at zero, and approaches the horizontal axis as $|k|$ grows. Energy packets travel right while their crests travel left.

At a distant fixed observer $x>0$, stationary components obey $x/t=c_g(k)$, so $|k|=\sqrt{\alpha t/x}$. As time increases their wavelength $2\pi /|k|$ decreases and their frequency magnitude $|\omega |=\sqrt{\alpha x/t}$ decreases; the crests pass toward decreasing $x$.

For the given initial spectrum the solution isThe value of a regular spectrum at the single point $k=0$ is irrelevant to this integral; a nonzero spatially constant mode is incompatible with the PDE. For asymptotic calculation assume the usual smooth localized spectrum, with sufficient integrability for the following stationary-phase steps.

At $x=Vt$, $V>0$, the phase $\Phi (k)=Vk+\alpha /k$ has stationary points $k=\pm k_0$, $k_0=\sqrt{\alpha /V}$. Their values are $\pm 2\sqrt{\alpha V}$ and their second derivatives are $\pm 2\alpha /k_0^3$. Expanding quadratically and using the oscillatory Gaussian integral gives the stationary phase leading termFor real data this is twice the real part of the positive-frequency term. If a stationary coefficient vanishes, the formula instead asserts that its contribution has smaller order. At $x=-Vt$, the phase derivative is $-V-\alpha /k^2<0$, so there is no real stationary point and no corresponding $t^{-1/2}$ wave packet. For smooth compactly supported spectra away from zero, repeated integration by parts gives decay faster than any power of $t$. More general spectra require endpoint/tail control; absence of stationary points alone does not justify that stronger rate for arbitrary $A$.

Use an acoustic velocity potential $\phi$ with $\mathbf{u}=\nabla \phi$ and $p^{\prime }=-{\rho }_0{\phi }_t$. It obeys the wave equation ${\phi }_{tt}=c_0^2{\nabla }^2\phi$. The rigid walls require zero normal velocity, hence Neumann boundary conditions on the transverse variables. Separation givesDefine ${\kappa }_{mn}^2=m^2/{\alpha }^2+n^2/{\beta }^2$. The dispersion relation, phase velocity and group velocity for positive $k,\omega$ areFor a dispersive mode, $\omega (0)=c_0{\kappa }_{mn}>0$, $c$ decreases from infinity to $c_0$, and $c_g$ increases from zero to $c_0$. The $(0,0)$ mode has $\omega =c_{0}k$ and $c=c_g=c_0$: it propagates without dispersion.

Let $\mathcal{A}={\pi }^2\alpha \beta$ be the cross-sectional area. The appropriate time and cross-sectional average at fixed $x$ isWrite $F_{mn}(y,z)$ for the product of cosines and $Q={\mathcal{A}}^{-1}\int F_{mn}^{2}dydz$. Its transverse eigenvalue and wall conditions give $\int |{\nabla }_{\perp }{F}_{mn}{|}^2={\kappa }_{mn}^2\int F_{mn}^2$ by integration by parts. The kinetic energy density and compressive acoustic energy density are ${\rho }_0|\mathbf{u}{|}^2/2$ and $(p^{\prime }{)}^2/(2{\rho }_0{c}_0^2)$. Time averaging each squared real oscillation contributes a factor $1/2$, soThis holds for every dispersive mode, and also for the zero mode.

For localized smooth initial data, a dispersive mode contains integrals of the form $\int a(k)e^{it[kV-\omega (k)]}dk$ along $x=Vt$. When $0<V<c_0$, a stationary point exists at $k_{\ast }={\kappa }_{mn}V/\sqrt{c_0^2-V^2}$. It is nondegenerate because ${\omega }^{\prime \prime }(k)=c_0{\kappa }_{mn}^2/(k^2+{\kappa }_{mn}^2{)}^{3/2}>0$. The stationary phase method therefore gives an oscillatory amplitude $O(t^{-1/2})$ for each excited dispersive mode, and for their sum under sufficient mode summability.

At $V=c_0$, no dispersive mode has a finite stationary wavenumber; with smooth localized data these contributions decay, while the nondispersive right-moving pulse retains its original amplitude. Thus a general disturbance with nonzero zero-mode component has an undiminished $O(1)$ pulse on the sound-speed ray. If that component is absent, this surviving contribution is absent too. For $V>c_0$, finite propagation speed puts a compactly supported initial disturbance outside this ray for all sufficiently large $t$, so the field is eventually zero. Rapidly localized, rather than strictly compact, data give a correspondingly negligible tail instead.

A plane wave in the rest frame of the medium has phase $\mathbf{k}\cdot \mathbf{x}-{\Omega }_0(\mathbf{k})t$. Under the Galilean transformation ${\mathbf{x}}^{\prime }=\mathbf{x}-\mathbf{U}t$, this becomes $\mathbf{k}\cdot {\mathbf{x}}^{\prime }-[{\Omega }_0(\mathbf{k})-\mathbf{U}\cdot \mathbf{k}]t$. ThereforeHere ${\mathbf{c}}_g$ is the group velocity in the moving frame.

For acoustic waves, ${\Omega }_0=c_0|\mathbf{k}|$ and take $\mathbf{U}=M{c}_0{\mathbf{e}}_x$. A steady wave has $\omega =0$, requiring $\cos \beta =1/M$, where $\beta$ is the angle of $\mathbf{k}$ to the flight direction. Its ray velocity is $c_0\hat{\mathbf{k}}-M{c}_0{\mathbf{e}}_x$, with backward component $-c_0(M^2-1)/M$ and transverse magnitude $c_0\sqrt{M^2-1}/M$. Thus the angle $\theta$ of the ray to the backward flight direction obeysThe axial symmetry gives the Mach cone behind the aircraft.

For unsteady emitted disturbances, all ray velocities lie on the sphere ${\mathbf{c}}_g=-\mathbf{U}+c_0\hat{\mathbf{k}}$. Since $M>1$, the origin lies outside this velocity sphere and its tangent cone has exactly the same semi-angle. All other rays are strictly inside the cone. Equivalently a signal emitted $\tau$ time units ago has a wavefront sphere centered at $-\mathbf{U}\tau$ with radius $c_0\tau$; these spheres are enclosed by their Mach-cone envelope. Consequently the unsteady wavefield is confined to the cone's interior, with the steady envelope on its boundary.

For capillary waves put $a=\sqrt{T/\rho }$, so ${\Omega }_0=a{k}^{3/2}$. The positive-frequency steady condition is $a\sqrt{k}=U\cos \beta$, requiring $\cos \beta >0$. ThenandIts endpoint traces the circlesince $c_{gx}/U=-1/4+(3/4)\cos 2\beta$ and $c_{gy}/U=(3/4)\sin 2\beta$. As $\beta$ runs through $(-\pi /2,\pi /2)$ this traces the circle except its leftmost limiting point, corresponding to the zero-wavenumber limit. The circle surrounds the origin, so every direction except the precisely backward axis is attained by a nonzero-wavenumber steady ray; the backward direction is approached as $k\downarrow 0$. Thus the closure of the steady ray field extends all around the insect, with no excluded angular sector such as the acoustic Mach-cone exterior.

Write $\mathbf{u}={\partial }_t\mathbf{\xi }$. In the uniform equilibrium, the linearized ideal magnetohydrodynamic equations reduce toIntegrating from the undisplaced reference state, and using the uniform background magnetic field, givesThese are the perturbations induced by the fluid displacement; independent time-independent changes to the reference state are excluded. The linear Lorentz force density is $(\nabla \times \delta \mathbf{B})\times \mathbf{B}/{\mu }_0$. Its magnetic pressure and magnetic tension parts giveFor a Fourier mode, put $s=\mathbf{k}\cdot \mathbf{\xi }$, $q=\mathbf{B}\cdot \mathbf{\xi }$ and $K=\mathbf{k}\cdot \mathbf{B}$. Then $\delta p=-i\gamma ps$ and $\delta \mathbf{B}=i(K\mathbf{\xi }-\mathbf{B}s)$. Substitution givesIf $s=q=0$, the fluid displacement is perpendicular to both the wave vector and the magnetic field. Only magnetic tension restores it, andThis is the Alfvén wave. The Alfvén velocity is the field-directed propagation vector: $\omega =\pm \mathbf{k}\cdot {\mathbf{v}}_a$. The signed phase velocity normal to the wavefront is $\pm v_a\cos \theta \hat{\mathbf{k}}$; the group velocity is $\pm {\mathbf{v}}_a$. The distinction matters for oblique propagation. For generic directions the Alfvén wave has one transverse polarization, with $\delta \rho =\delta p=0$.

To obtain the other magnetohydrodynamic waves, take the scalar products with $\mathbf{k}$ and $\mathbf{B}$:The determinant condition for a nonzero pair $(s,q)$ isThus the fast magnetosonic wave and slow magnetosonic wave haveTheir fluid displacements lie in the plane of $\mathbf{k}$ and $\mathbf{B}$ and are generally compressive. In the fast magnetosonic wave, gas and magnetic pressure provide the stronger restoring combination; in the slow magnetosonic wave, their perturbations oppose one another. The phase speeds depend on direction. They satisfy $v_{\mathrm{sl}}\le \min (c_s,v_a)$ and $v_{\mathrm{f}}\ge \max (c_s,v_a)$. For parallel propagation the two speeds are $c_s$ and $v_a$, with a degeneracy between a transverse branch and the Alfvén wave. For perpendicular propagation $v_{\mathrm{f}}^2=c_s^2+v_a^2$ and $v_{\mathrm{sl}}=0$; the latter is a nonpropagating limiting disturbance. These special directions require interpreting the polarizations by continuity rather than assuming three distinct nonzero frequencies.

For the positive-frequency internal gravity wave branch, $\omega =N{K}_h/K$. Differentiating with respect to the components of the wave vector gives the group velocity,The wavefront-normal phase velocity is ${\mathbf{c}}_p=\omega \mathbf{k}/K^2$. Directly,The negative-frequency branch changes both propagation signs but not their orthogonality. Another proof is that $\omega$ is homogeneous of degree zero in the wave vector, so Euler's homogeneous-function identity gives $\mathbf{k}\cdot {\nabla }_{\mathbf{k}}\omega =0$. Energy propagates along constant-phase surfaces, rather than normal to them. When $m=0$, $\omega =N$ and ${\mathbf{c}}_g=0$; the scalar-product result remains true, but there is no nonzero group-velocity direction to describe geometrically.

Write $w(x,t)$ for the vertical displacement of the sea ice, and take the undisturbed water surface as $z=0$, with water occupying $z<0$. The elastic plate has areal mass $m={\rho }_{i}h$ and bending stiffnessHere $E$ is Young's modulus and $\eta$ is Poisson's ratio. We neglect in-plane prestress, viscosity and plate shear deformation, and linearize about hydrostatic equilibrium. These are important assumptions: perfect elasticity alone does not specify every term in a floating-plate model.

For a plane wave $w=\hat{w}{e}^{i(kx-\omega t)}$, $k>0$, potential flow in deep water has velocity potential $\varphi =\hat{\varphi }{e}^{kz}{e}^{i(kx-\omega t)}$. This solves Laplace's equation and decays downwards. The kinematic boundary condition gives $k\hat{\varphi }=-i\omega \hat{w}$. Linearizing the water pressure at the displaced interface gives an upward excess loadThe elastic plate equation is $m{w}_{tt}+D{w}_{xxxx}=p$. Substitution and multiplication by $k$ therefore give the flexural-gravity wave dispersion relationEquivalently, with $\alpha ={\rho }_{i}h/{\rho }_w$ and $\beta =D/{\rho }_w$,This determines the positive wavenumber implicitly for prescribed positive angular frequency. It is unambiguous: the derivative of ${\omega }^2$ is $(g+5\beta k^4+4\alpha \beta k^5)/(1+\alpha k{)}^2>0$, while ${\omega }^2$ runs from zero to infinity.

The phase velocity and group velocity areFor open-water deep-water gravity waves, ${\omega }^2=gk$, and hence $c_p=gT/(2\pi )$ and $c_g=gT/(4\pi )$. At large period the ice-covered curves approach these straight lines. At shorter period, plate bending raises the speeds, so both ice-covered curves turn upward as period decreases. In the bending regime with negligible plate inertia, $\omega \propto k^{5/2}$ and $c_g\simeq (5/2)c_p$; in the formal plate-inertia-dominated limit, $\omega \propto k^2$ and $c_g\simeq 2c_p$. The latter extrapolation eventually leaves thin-plate validity and should not be read as a prediction at arbitrarily small wavelength.

The plotted parameters are illustrative rather than measured at the observation site. They give a group-velocity minimum of a flexural-gravity wave near $15.1\mathrm{m}{\mathrm{s}}^{-1}$ at period $17.6\mathrm{s}$. There is no arbitrarily slow wave-energy branch under a continuous elastic sheet. Energy put into a localized disturbance travels away at at least this minimum group velocity; a slowly moving wind system cannot retain a wave packet indefinitely beneath itself. This reduces the opportunity for sustained local growth compared with slow, short open-water waves, and the continuous cover also prevents direct wind forcing of an exposed water surface. Incoming long swell can still propagate.

A group-velocity minimum of a flexural-gravity wave is not by itself a universal minimum wind speed for wave generation. A steadily translating forcing pattern requires a phase velocity matching its translation speed, so the minimum of $c_p$, not $c_g$, supplies the corresponding resonance threshold. Random wind forcing, dissipation and aerodynamic coupling must be specified before making an absolute generation claim.

Use a common emission time $t_0$ and a path length $L$, and treat each observed period as a narrow wave packet. For deep-water gravity waves, its group velocity is $gT/(4\pi )$, not the phase velocity $gT/(2\pi )$. Therefore the dispersive swell source inversion givesThus period decreases hyperbolically, while frequency $1/T$ increases linearly. The elapsed time between the two observations is $40\mathrm{h}40\min =146400\mathrm{s}$, soThe frequency slope is $2.6021\times 10^{-7}\mathrm{Hz}{\mathrm{s}}^{-1}$. Initially $dT/dt=-0.843\mathrm{s}{\mathrm{h}}^{-1}$; at 14 s it has slowed to $-0.184\mathrm{s}{\mathrm{h}}^{-1}$. The inferred emission was $C/30=35\mathrm{h}35\min$ before the first arrival, namely 19 March at 1710 UT. Both observed endpoints give this same time.

A northward-propagating swell with this distance scale points to a remote energetic storm south of the high Arctic rather than local wind acting on continuous sea ice. Along a meridional route, 3,000 km is about $27^{\circ }$ of latitude, putting the source on the scale of the northern North Atlantic and adjacent open seas. A route through Fram Strait is plausible, but longitude and refraction are not supplied, so no particular storm or unique source position follows. Long swell arriving first and steadily increasing frequency are the expected signatures of remote wave dispersion. The distance and time are conditional on an approximately impulsive source and open-water propagation speeds; passage through ice, currents and finite storm duration produce corrections.

The sea ice acts as a frequency-selective filter. Over a long path, wave attenuation in sea ice is generally much stronger for shorter surface gravity waves, through repeated scattering, internal ice losses and water-side dissipation. Their amplitude can fall below the tiltmeter's detection level even if the source initially generated them. Thus the long-period swell survives while the shorter-period tail does not.

The disappearance near 14 s is an attenuation and detectability limit, not a forbidden-frequency interval of the ideal elastic plate. The flexural-gravity wave dispersion relation admits a real positive wavenumber for every positive angular frequency. In an impulsive-source picture, shorter-period deep-water gravity waves also arrive later because their group velocity is smaller, but delayed arrival alone does not explain a persistent observed cutoff.

For the positive-frequency branch and $K^2=k^2+m^2$, the wavefront-normal phase velocity and the group velocity areTheir scalar product vanishes since $k(k{m}^2)+m(-m{k}^2)=0$. Equivalently, the frequency is homogeneous of degree zero in the wave vector, and hence $\mathbf{k}\cdot {\nabla }_{\mathbf{k}}\omega =0$.

The vertical sign relation isIt is negative for the usual geophysical ordering $N>|f|$ and nonzero $k,m$. Thus vertical phase and energy propagation are opposite under that ordering; negative-frequency waves obey the same product relation. The PDF does not state this ordering, so its unconditional direction assertion needs qualification. For example, $N=1$, $f=2$, $k=m=1$ gives positive vertical components on the positive-frequency branch. If $N=|f|$ the group velocity vanishes, and in axial limiting directions one vertical component can be zero. Orthogonality is valid throughout, while strictly opposite nonzero vertical components need the stated nondegeneracy and $N>|f|$.