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Let $p$ denote the fluid pressure perturbation divided by the reference mass density, and let $b=-g{\rho }^{\prime }/{\rho }_0$ be the buoyancy perturbation. The buoyancy frequency satisfies $N^2=-g{\hat{\rho }}_z/{\rho }_0$. Linearization requires small boundary slope $k|{\eta }_0|\ll 1$, small displacements compared with the vertical scales of the disturbance and background, and advection small compared with the oscillatory acceleration. For a propagating internal gravity wave this includes $|m{\eta }_0|\ll 1$; boundary slope alone is insufficient when $|m|\gg k$. The Boussinesq approximation also requires small relative mass density differences over the region of interest. These conditions must hold for the resulting disturbance, including any amplification by resonance.

The inviscid Linearized Boussinesq equations and linear kinematic boundary condition areWrite the velocity field as $(u,w)=\mathrm{Re}\{(U(z),W(z))e^{i(kx-\omega t)}\}$. Incompressibility and horizontal momentum balance giveThe vertical momentum balance therefore yieldsFor a real vertical wavenumber $m$, this gives the dispersion relationWhen $0<\omega <N$, define $\ell =k\sqrt{N^2/{\omega }^2-1}$. The radiation condition selects $m=-\ell$, because energy must leave the boundary upward. The boundary-forced internal gravity wave isIts wavevector is $\mathbf{K}=(k,m)$, its phase velocity is ${\mathbf{c}}_p=\omega \mathbf{K}/(k^2+m^2)$, and its group velocity isThus the phase velocity points down and right, while the group velocity points up and right. They are perpendicular. Internal-wave polarization makes the particle motion an oscillation along the group velocity direction, with zero first-order mean transport. The constant-phase lines of an internal gravity wave are also parallel to the group velocity. Their inclination $\vartheta$ above the horizontal satisfies $\sin \vartheta =\omega /N$.

When $\omega >N$, set $\kappa =k\sqrt{1-N^2/{\omega }^2}$. Boundedness at infinity selects the evanescent waveThe horizontal and vertical velocity components are in quadrature: fluid particles describe small ellipses, and the response decays over ${\kappa }^{-1}$. There is no upward time-averaged energy flux, because $P$ and $W$ are in quadrature. A real vertical group velocity is not defined for this evanescent wave. The pattern travels horizontally with phase velocity $\omega /k$.

For $N=0$ the response is the evanescent wave with $\kappa =k$; the ratios $\omega /N$ should not be used. At the cutoff $\omega =N>0$, the bounded harmonic solution has $W=-i\omega {\eta }_0$, $U=0$: it neither decays nor has nonzero upward group velocity. It is the limiting cutoff response, rather than a localized radiating disturbance. A uniform nonzero buoyancy frequency in an infinitely deep Boussinesq approximation is itself a local idealization of the background mass density.

The arrows distinguish the phase velocity, group velocity, and oscillatory particle motion of the internal gravity wave. The right panel shows the decay envelope and particle ellipses of the evanescent wave.

Label the two atoms in primitive cell $j$ by $u_{2j}$ and $u_{2j+1}$, with $j$ understood modulo $N$ because of the periodic boundary condition. Choose the spring between them to have constant $\lambda$. Newton's second law gives

For a longitudinal normal mode, putPeriodicity requires $e^{2iNka}=1$, so $k=\pi r/(Na)$. Since the primitive-cell length is $2a$, wavevectors differing by $\pi /a$ describe the same mode, and the first Brillouin zone may be chosen asThe amplitudes obeySetting its determinant to zero gives the alternating-spring chain dispersion relationThe lower sign is the acoustic phonon branch and the upper sign is the optical phonon branch.

At the edge of the Brillouin zone,so the frequency band gap isIt vanishes at $\alpha =1$, when the apparent two-atom primitive cell can be reduced to one atom.

Near the centre of the Brillouin zone, a Taylor expansion givesandThe group velocity of the acoustic branch at long wavelength is the speed of sound,

Let a slowly varying wave have phase $S(\mathbf{x},t)$, wavevector $\mathbf{k}=\nabla S$, and angular frequency $\omega =-{\partial }_{t}S$. The local dispersion relation is the Hamilton-Jacobi equation $\omega =\Omega (\mathbf{k};\mathbf{x},t)$. Equality of mixed derivatives, followed along a curve with velocity $\dot{\mathbf{x}}={\nabla }_{\mathbf{k}}\Omega$, gives the Hamiltonian ray-tracing equationsHere $d/dt={\partial }_t+\dot{\mathbf{x}}\cdot \nabla$ is the total derivative evaluated along a ray, and $\dot{\mathbf{x}}$ is the group velocity.

For $\Omega =kc(z)$, $\omega$, $k_x$, and $k_y$ are conserved. If $\psi$ is the angle between the ray and the $z$-axis, then $k\sin \psi =(k_x^2+k_y^2{)}^{1/2}$ and $k=\omega /c$. Hence Snell's law is

For the ray launched with $\mathbf{k}=k(\cos \varphi ,0,\sin \varphi )$ at $z=0$, this constant is $|\cos \varphi |/c_0$. Thereforewhich proves

For $0<\varphi <\pi /2$, write $a=\cos \varphi$. Since the ray and wavevector are parallel,Near the upper turning point, put $z=z_m-\zeta$. Expansion givesIf $x=x_m$ at the turn, integration yields the local ray pathThe ray is tangent to the line $z=z_m$, turns smoothly, crosses $z=0$, and repeats symmetrically between $-z_m$ and $z_m$. Increasing $\varphi$ decreases $z_m$, while $\varphi \downarrow 0$ sends the turning level outward.

Let $D_g={\partial }_t+{\mathbf{u}}_g\cdot {\nabla }_h$, $\psi =\tilde{\Phi }/f_0$, and ${\mathbf{u}}_g=(-{\psi }_y,{\psi }_x,0)$. The hydrostatic and thermal equations combine to giveTaking the vertical curl of the horizontal momentum equations to leading quasi-geostrophic order givesThe log-pressure coordinate continuity equation gives ${\nabla }_h\cdot {\mathbf{u}}_{a,h}=-e^{z/H}{\partial }_z(e^{-z/H}w)$. Substitute $w=-f_0{D}_g{\psi }_z/N^2$. In commuting the vertical derivative with $D_g$, the extra term is ${\mathbf{u}}_{g,z}\cdot {\nabla }_h{\psi }_z=(-{\psi }_{yz},{\psi }_{xz})\cdot ({\psi }_{xz},{\psi }_{yz})=0$. Therefore the log-pressure quasi-geostrophic potential vorticity equation is

The required small parameters have distinct roles. The Rossby number ${ϵ}_R=U/(|f_0|L)$ makes inertial acceleration small relative to the leading geostrophic balance. The parameter ${ϵ}_{\beta }=\beta L/|f_0|$ lets that leading balance use a single $f_0$ while retaining the beta effect in the slower vorticity evolution. Finally, $\psi \sim UL$ and the thermal equation imply $w\sim |f_0|U^2/(N^{2}D)$. Density-weighted continuity has vertical derivative scale $1/\delta$, where $\delta =\min (D,H)$, so $|{\mathbf{u}}_{a,h}|\sim Lw/\delta$. The vertical-advection consistency criterion for log-pressure quasi-geostrophy is thusIt also controls the omitted vertical-advection ratio $wL/(UD)\sim {ϵ}_a\delta /D$. Small Rossby number alone would not ensure small ageostrophic flow for every choice of stratification and vertical scales.

For constant $N$, linearization about rest givesPutting $\psi =\mathrm{Re}[{\hat{\psi }}_c{e}^{z/(2H)}{e}^{i(kx+mz-\omega t)}]$ replaces the vertical operator by $-m^2-1/(4H^2)$. Consequently the Rossby wave in a log-pressure atmosphere has

For the forced half-space, substitution of the specified ansatz givesAssume $\beta >0$ and $k>0$, and define ${\omega }_{\ast }=-\beta k/[k^2+f_0^2/(4N^2{H}^2)]<0$.

If $\omega >0$ or $\omega <{\omega }_{\ast }$, then $m^2<0$. Boundedness excludes the growing solution and uniquely selects $\boxed{\hat{\chi }={\Psi }_b{e}^{-\sqrt{-m^2}z}}$. If $\omega ={\omega }_{\ast }$, boundedness excludes the linear-in-$z$ solution, leaving $\hat{\chi }={\Psi }_b$; this is the zero-vertical-wavenumber threshold.

If ${\omega }_{\ast }<\omega <0$, both oscillatory solutions are bounded. Boundedness alone does not give a unique propagating solution. With forcing at the bottom and no incident wave from above, impose an upward radiation condition. Sincepositive $m$ has upward group velocity. The outgoing solution is $\boxed{\hat{\chi }={\Psi }_b{e}^{imz},m=\sqrt{m^2}>0}$. The physical streamfunction includes the prescribed $e^{z/(2H)}$ density-weighting factor; it is the amplitude $\hat{\chi }$ that must remain bounded.

Thus the strict range for vertically propagating waves isAt $\omega =0$, the unforced steady interior equation requires $\beta {\psi }_x=0$, so a nonzero $k$ bottom streamfunction cannot have a time-independent solution of this ideal rest-state problem without additional dynamics.

Assume $\beta >0$, $c=\sqrt{gH}$, and use complex normal mode amplitudes. The zonal momentum and continuity equations giveWhen ${\omega }^2\ne k^2{c}^2$, solving these two linear equations gives

For $\hat{v}=0$, a nonzero solution requires $\omega =\pm kc$ and $\hat{\eta }=\pm c\hat{u}/g$. The remaining momentum equation gives ${\hat{u}}^{\prime }=\mp \beta y\hat{u}/c$. Only the positive sign decays at both infinities, so the equatorial Kelvin wave isIts phase and group velocities are both eastward with speed $c$.

For nonzero meridional velocity, substitute the eliminated amplitudes into $-i\omega \hat{v}+\beta y\hat{u}=-g{\hat{\eta }}^{\prime }$. The result isSet $Y=y\sqrt{\beta /c}$. The decaying Hermite polynomial modes have $\hat{v}=V_n{H}_n(Y)e^{-Y^2/2}$, up to normalization, and their eigenvalue gives the equatorial shallow-water dispersion relation

For $n=0$, the quadratic in $k$ factors asThe admissible Yanai wave root is $k=\omega /c-\beta /\omega$. With $\hat{v}(0)=1$, substitution and cancellation giveThese expressions satisfy the original equations even at the isolated point where the two algebraic roots coincide and the earlier elimination formula has a zero denominator. The other root is not an independent admissible branch: its generic singular reconstruction fails the trapping conditions.

In the zonal momentum equation, the acceleration term is $-i\omega \hat{u}=({\omega }^{2}y/c)\hat{v}$, the Coriolis term is $-\beta y\hat{v}$, and the pressure term is $-ikg\hat{\eta }=k\omega y\hat{v}$. When ${\omega }^2\ll \beta c$, acceleration is small and the leading balance is between Coriolis acceleration and the pressure gradient. When ${\omega }^2\gg \beta c$, acceleration balances pressure and the Coriolis term is small; the branch tends to an eastward gravity wave.

For $n\ge 1$, the fixed-frequency zonal roots areThe radicand is nonnegative exactly when $s={\omega }^2/(\beta c)$ satisfies $s^2-(2n+1)s+1/4\ge 0$. This gives the frequency gap for higher equatorial wave modes:

For the forcing frequency $\omega =\frac{1}{2}\sqrt{\beta c}$, we have $s=1/4$. Already the $n=1$ low-frequency upper bound is $3/2-\sqrt{2}<1/4$, and that bound decreases with $n$, while the high-frequency bound increases. Hence no $n\ge 1$ mode propagates. The available real-wavenumber branches areFor the equatorial Kelvin wave, $c_{g,K}=c$. For the Yanai wave, differentiating $k=\omega /c-\beta /\omega$ givesThus both propagating responses are detected east of the localized forcing, provided it has nonzero projection onto their meridional structures. The Yanai phase travels westward at this frequency, but its wave packet and energy travel eastward. Higher modes contribute only an evanescent response near the forcing.

Take $U_0,N_0,k>0$ and assume an initially propagating mode, $k<N_0/U_0$. Put $s=N_0/U_0$ and $\delta (z)=1+\gamma z$. Neglecting $U^{\prime \prime }/U$ as stipulated, the Scorer parameter and stationary internal-wave WKB solution areThe positive root $m>0$ selects upward group velocity for the negative intrinsic frequency branch $\hat{\omega }=-kU$. The WKB approximation needs a slowly varying background relative to the local vertical wavelength, in particular $|m^{\prime }|\ll m^2$ and $|m^{\prime \prime }|\ll |m{|}^3$, away from a turning level or singular mean profile.

The local dispersion relation is $\omega =kU-Nk/(k^2+m^2{)}^{1/2}=0$. Its intrinsic and laboratory group velocities are, with $K=(k^2+m^2{)}^{1/2}$,Thus intrinsic energy propagation is upstream and upward, while laboratory energy propagation is downstream and upward, parallel to $(k,m)$. Constant-phase lines of an internal gravity wave have slope $dz/dx=-k/m$ and normal spacing $2\pi /K$; the horizontal spacing remains $2\pi /k$.

If $\gamma <0$, $m$ grows with height. The wavevector and laboratory energy ray become more vertical; phase lines become flatter and closer together. For finite nonzero $N$, the heightis a critical level of an internal gravity wave: $U\to 0$ and the vertical wavenumber diverges. The laboratory group velocity tends to zero. Importantly, $|m^{\prime }|/m^2\to |\gamma |/s$, so the formal WKB condition need not deteriorate before this level if that ratio is small. Nevertheless the stationary equation is singular at the level, the phase has no finite limit, and the horizontal perturbation velocity grows as $m^{1/2}$; the inviscid linear wave cannot be continued uniformly through it without additional physics.

If $\gamma >0$, $m$ decreases to zero atThe wavevector and laboratory ray become horizontal, the phase lines become vertical, and the normal spacing increases. Above $z_t$, $m^2<0$ and the disturbance is an evanescent wave. Here $|m^{\prime }|/m^2$ diverges, so the WKB approximation fails in a turning region and must be replaced by a local connection solution.

Keep $\mathbf{A}=(0,Bx,0)$ and choose the scalar potential $\varphi =-Ex$. ThenFor $p_z=0$, completing the square givesHence the spectrum on ${\mathbb{R}}^3$ is

For a translationally invariant dispersion, the group velocity is $v_y={\hslash }^{-1}\partial E_n/\partial k$. Thereforein agreement with the electric-cross-magnetic-field drift $\mathbf{E}\times \mathbf{B}/B^2$.

In the rectangle, $k=2\pi r/L_y$ as before. The energy now changes with $r$, so the guiding-centre degeneracy is lifted, while the number of states in each tilted band remains approximately $|qB|A/(2\pi \hslash )$. Adjacent states within a formerly degenerate level have the electric-field splitting of a Landau level in a rectangleThus this is the ground-to-first-excited gap when the lowest Landau band contains at least two allowed guiding centres and $\Delta E_{\mathrm{electric}}<\hslash {\omega }_c$. Without that implicit weak-field or large-sample condition, the exact full spectral gap is

The cutoff occurs at $k=0$:For $k>0$, the phase velocity and group velocity areTheir product is $c_{p,n}{c}_{g,n}=c_S^2$. On a plot against positive $k$, $c_{p,n}$ decreases from $+\infty$ to $c_S$, while $c_{g,n}$ increases from zero to $c_S$; larger mode number shifts both transitions to larger wavenumber.

The physical rapid phase isDefine the local wavevector, angular frequency, and group velocity byThe equality of mixed partial derivatives gives the phase-compatibility equationsA ray is defined to move with the local group velocity:

Differentiate the local dispersion relationwith respect to $x_i$. Along a ray,The last two terms cancel because $\mathbf{k}$ is a gradient. Similarly,Finally,Thus the Hamiltonian ray-tracing equations are

In a homogeneous, time-independent medium, $\partial \Omega /\partial x_i=0$ and $\partial \Omega /\partial t=0$. Therefore $\mathbf{k}$ and $\omega$ are constant along each ray. The group velocity ${\mathbf{c}}_g={\nabla }_{\mathbf{k}}\Omega$ is consequently constant, andso every ray is a straight line.

For $\omega =0$, let $s=|\mathbf{x}|$ denote distance along a ray from the origin and letThe phase equation givesBoth vectors on the right are constant along the ray, soThe final formula in the question denotes the physical phase $\Theta =\theta /\epsilon$ simply by $\theta$, giving its stated expression.

The stationary contribution travels with group velocityso $k_0=4V^2/(9S^2)$. The conditionis therefore equivalent, up to a constant factor, to $k_0ϵ\ll 1$: the dominant wavelength is much longer than the impacted interval. Its Fourier transform can then be replaced bywhich is the total initial vertical velocity integrated over space. The far field cannot resolve $W$ and $ϵ$ separately and depends only on their product $Wϵ$.

Without self-gravity and for $k>0$,The phase velocity and group velocity areso

With self-gravity restored,The product changes sign atFor $k>k_{\mathrm{crit}}$, crests and a localized wave packet travel in the same radial direction. For $k<k_{\mathrm{crit}}$, the group velocity and phase velocity have opposite signs: the envelope and its wave energy propagate opposite to the individual crests. At $k_{\mathrm{crit}}$ the group velocity vanishes and neighboring wave components cause the packet to spread.

Let every perturbation be proportional to $\exp [i(kx+mz-\omega t)]$ and define the intrinsic frequencyEliminating pressure with incompressibility, and then eliminating buoyancy, gives the internal gravity wave dispersion relationThe vertical group velocity isThus energy propagates upward when $\hat{\omega }m<0$ and downward when $\hat{\omega }m>0$. Equivalently, on the positive-intrinsic-frequency branch upward propagation requires $m<0$, while on the negative branch it requires $m>0$.

At $O(ϵ\gamma )$, the slowly varying wave envelope obeyswhere the group velocity obtained from $\omega =Nk/(k^2+m^2{)}^{1/2}$ isIt satisfies $\mathbf{k}\cdot {\mathbf{c}}_g=0$. If initially $\mathbf{k}\cdot \hat{\nabla }\tilde{\psi }=0$, the envelope structure varies across the wavevector and translates unchanged in the direction ${\mathbf{c}}_g$ at speedThus internal-wave crests propagate with phase velocity parallel to $\mathbf{k}$, while the packet and its energy propagate perpendicular to $\mathbf{k}$ with the group velocity.