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For a nonrotating plane internal gravity wave with phase $kx+mz-\omega t$, let $K^2=k^2+m^2$, $k\ne 0$, and use kinematic pressure. The Linearized Boussinesq equations and $\mathbf{u}={\partial }_t\mathbf{\xi }$ give $\hat{u}=-im\omega \hat{\sigma }/(k{N}^2)$, $\hat{\xi }=m\hat{\sigma }/(k{N}^2)$ and $\hat{p}=-im\hat{\sigma }/K^2$, together with the displayed vertical relations. Buoyancy and displacement field are in phase or opposite phase; velocity and pressure are in quadrature with them. Incompressibility makes both the oscillating velocity and displacement field perpendicular to the wavevector.

Two plane internal gravity waves can each have zero Stokes drift but give a nonzero drift through interference. The displacement of one wave need not be perpendicular to the other wave's wave vector. For upward waves with wave vectors $(\kappa ,-\kappa )$ and $(-\kappa ,-\kappa )$ and vertical amplitudes $W$, the cross-wave drift is $(0,-2\kappa W^2\cos (2\kappa x)/\omega )$ with $\omega =N/\sqrt{2}$. It has zero spatially averaged vertical flux. It cannot on its own satisfy an impermeable oscillating boundary: the second-order Eulerian mean velocity supplies the necessary normal correction there.

For a slowly switched-on plane internal gravity wave in the half-space $z>0$, the method of multiple scales gives transport of its amplitude at the vertical group velocity. For ${\omega }^2=N^2{k}^2/(k^2+m^2)$, $c_{gz}=-\omega m/(k^2+m^2)$. Initially undisturbed fluid and forcing at the bottom require $c_{gz}>0$. The outgoing radiation condition therefore selects $m<0$ when $\omega >0$, although the vertical phase velocity is downward.

A slowly established upward plane internal gravity wave has $\overline{u^{\prime }{w}^{\prime }}=-m|a{|}^2/(2k)$. The mean horizontal conservation of momentum equation gives ${\overline{u}}_t=-{\partial }_z\overline{u^{\prime }{w}^{\prime }}$. Using envelope transport and an initially vanishing mean gives $\overline{u}=-m|a{|}^2/(2k{c}_{gz})$. Its energy density per unit mass is $\mathcal{E}=(k^2+m^2)|a{|}^2/(2k^2)$, so $\overline{u}=(k/\omega )\mathcal{E}$. This is the mean flow left behind the advancing wave envelope, rather than a claim that an established uniform wavetrain has nonzero local stress divergence.

For a plane internal gravity wave with vertical displacement $\zeta$, linear density advection gives ${\rho }^{\prime }=-{\bar{\rho }}_z\zeta$. In a constant stable gradient, ${\rho }_z={\bar{\rho }}_z(1-{\zeta }_z)$, so static overturning begins when the maximum of ${\zeta }_z$ exceeds one. A monochromatic displacement has maximum $|m{A}_{\zeta }|$. The criterion predicts the failure of the small-amplitude field, rather than a valid continuation of linear theory beyond overturning.

Let $p$ be kinematic pressure, with the resting hydrostatic part subtracted, and let $\sigma$ be the buoyancy perturbation about background buoyancy $N^{2}z$. The nonrotating inviscid Boussinesq equations, with no buoyancy diffusion, areHere the third equation is mass conservation for incompressible flow, and the last expresses buoyancy conservation following a parcel. Linearization about rest removes the products of disturbances. For a plane internal gravity wave proportional to $e^{i(kx+mz-\omega t)}$, the resulting Linearized Boussinesq equations becomeWith $K=(k^2+m^2{)}^{1/2}$ and $k\ne 0$, the first and third equations give $\hat{p}=-m\omega \hat{w}/k^2$. Substitution in vertical momentum and buoyancy gives the dispersion relationSince the velocity is the time derivative of the displacement field, $\hat{\mathbf{u}}=-i\omega \hat{\mathbf{\xi }}$. Thus the complete buoyancy polarization of a plane internal gravity wave isBoth velocity and displacement are perpendicular to the wavevector; pressure and velocity are in quadrature with buoyancy and displacement.

For upward/rightward phase velocity, choose $k,m>0$ and $\omega =Nk/K$. Differentiating this dispersion relation gives the group velocity, while the vector phase velocity points along the wavevector:Their dot product is zero and ${\mathbf{c}}_p+{\mathbf{c}}_g=(N/K,0)$. Draw ${\mathbf{c}}_g$ from the tip of ${\mathbf{c}}_p$: the right-triangle geometry of internal-wave velocities has horizontal hypotenuse $N/K$. This hypotenuse is distinct from the horizontal phase velocity $\omega /k$.

At an instant with real complex amplitude $\hat{\sigma }=A$, write $\varphi =kx+mz-\omega t$. The spatial fields areThe constant-phase lines of an internal gravity wave slope down to the right, with $dz/dx=-k/m$. Buoyancy maxima lie a quarter wavelength from pressure maxima; the oscillating displacement and velocity run along the phase lines. The group velocity points down and right along them, despite upward/rightward motion of the wave phase.

Let $P=\overline{{\xi }_{x}u+{\zeta }_{x}w}$, using horizontal averaging over a wavelength. This is the displacement pseudomomentum of an internal gravity wave, whose sign convention is opposite to $k\overline{E}/\omega$. For complex coefficients, $\overline{ab}=\frac{1}{2}\mathrm{Re}(\hat{a}{\hat{b}}^{\ast })$. The above wave polarization consequently givesSince $c_{gz}=-\omega m/K^2$, these obey $\overline{{\zeta }_{x}p}=c_{gz}P$. With weak kinematic viscosity, each leading monochromatic velocity component satisfies ${\nabla }^{2}u=-K^{2}u$, ${\nabla }^{2}w=-K^{2}w$. The dissipative term is therefore $-\nu K^{2}P$, so the supplied averaged balance reduces toFor constant $k,m,N$ and a steady wave envelope, $c_{gz}{P}_z=-\nu K^{2}P$. As $P$ is proportional to squared wave amplitude, viscous attenuation of an internal gravity wave givesHere $c_{gz}<0$, so amplitude decreases below a source at $z_s$: the source is above the receiving level. The same conclusion follows directly because downward group velocity carries energy from the source downward. Upward phase velocity does not locate the source.

For a slowly varying mean shear flow $U(z)$, the local intrinsic frequency is $\hat{\omega }=\omega -kU(z)$. A local WKB approximation uses ${\hat{\omega }}^2=N^2{k}^2/(k^2+m(z{)}^2)$, with varying vertical wavenumber and group velocity. The appropriate transported quantity is wave action, or horizontal wave pseudomomentum $kE/\hat{\omega }$; its vertical flux has viscous loss, and the changing polarization factors must be retained. In particular, $P/|\hat{\sigma }{|}^2$ is no longer constant.

Near a critical level of an internal gravity wave, $\hat{\omega }\to 0$, so $|m|\sim N|k|/|\hat{\omega }|$ and $|c_{gz}|\sim {\hat{\omega }}^2/(N|k|)$. The damping per vertical distance, of order $\nu K^2/|c_{gz}|$, grows rapidly. The waves are strongly attenuated over a narrow altitude range and transfer momentum to the mean flow there or before reaching the critical level. The inviscid WKB approximation eventually fails; viscosity and possibly internal-wave breaking regulate the small scales.

Use $p$ for disturbance pressure divided by the constant inertial density. The ideal, nonrotating Linearized Boussinesq equations areHere $N^2>0$ is a prescribed time-independent buoyancy frequency squared. Taking the scalar product of momentum with velocity gives ${\partial }_t(|\mathbf{u}{|}^2/2)=-\nabla \cdot (p\mathbf{u})+\sigma w$, using incompressibility. Multiplying the buoyancy equation by $\sigma /N^2(z)$ gives ${\partial }_t[{\sigma }^2/(2N^2)]=-\sigma w$. No spatial derivative of $N$ is needed in this cancellation. ThusThis is positive disturbance energy per unit inertial density; its flux is pressure work.

For constant $N$, write a plane internal gravity wave with wave vector $\mathbf{K}=({\mathbf{k}}_h,m)$ and phase $\theta ={\mathbf{k}}_h\cdot {\mathbf{x}}_h+mz-\omega t$. Let $k_h=|{\mathbf{k}}_h|>0$ and $K^2=k_h^2+m^2$. In complex amplitudes, incompressibility and horizontal momentum giveConsequently $\hat{p}=-\omega m\hat{w}/k_h^2$. Substitution into vertical momentum, $-i\omega \hat{w}=-im\hat{p}+\hat{\sigma }$, yieldsThe same dispersion relation follows directly from the displacement equation: its two terms become $K^2{\omega }^2\hat{\zeta }$ and $-N^2{k}_h^2\hat{\zeta }$. The positive-frequency branch is $\omega =N{k}_h/K$; the other branch is its sign reversal.

Choose real vertical-displacement amplitude $A$. The internal-wave polarization can be writtenVelocity and displacement lie along the phase planes, perpendicular to $\mathbf{K}$, and are a quarter cycle out of phase. Buoyancy is opposite in sign to vertical displacement. In the requested two-dimensional sketch take $k<0$, $m<0$ and $\omega >0$: the phase velocity $\omega \mathbf{K}/K^2$ points down and left. Then $\mathbf{e}=(-m/k,1)$ points up and left, and positive $p$ accompanies positive $w$ and negative $u$. Negative-pressure bands have the opposite velocity. Multiplication by pressure therefore transports energy up and left in both signs of band.

Differentiating the positive-frequency dispersion relation gives the group velocityIn particular $\mathbf{K}\cdot {\mathbf{c}}_g=0$. Since $|\mathbf{e}{|}^2=K^2/k_h^2$, the wavelength averages areUsing $N^2={\omega }^2{K}^2/k_h^2$ in the last expression provesFor $k,m<0$ the group velocity points up and left, perpendicular to the down-left phase velocity. The pressure sign pattern, rather than the direction of moving crests, determines the energy-flux direction shown below.

For the moving-boundary problem, first take the fluid to occupy $z>0$ with the boundary near $z=0$. An outgoing boundary-forced internal gravity wave must have $c_{gz}>0$. With the specified $k>0$ and $\omega >0$, the radiation condition therefore selectsIf instead the fluid occupies $z<0$, radiation requires $c_{gz}<0$ and the positive root for $m$. The fluid half-space is needed to select a unique sign; the following formulas use the upper half-space.

For slowly modulated internal-wave boundary forcing, put $\zeta =A(T,Z)e^{i\theta }$, $T=\mu t$, $Z=\mu z$. Acting on $A$ by ${\partial }_t\mapsto -i\omega +\mu {\partial }_T$ and ${\partial }_z\mapsto im+\mu {\partial }_Z$, the displacement equation has leading term $(K^2{\omega }^2-N^2{k}^2)A$. Its order-$\mu$ term isThus characteristics carry the boundary amplitude into the fluid:Real parts are understood. To the same leading order,Slow-envelope corrections to these polarization formulas are order $\mu$. At $T=2$, the amplitude is $ϵ$ for $0\le Z\le c_{gz}$, falls smoothly across $c_{gz}<Z<2c_{gz}$, and is zero beyond $2c_{gz}$. This translating ramp is also shown in the figure.

The mean boundary work per unit horizontal area is $\overline{W}={\rho }_{00}\overline{p(0)w(0)}$ at quadratic disturbance order. The small boundary displacement permits evaluation at the unperturbed plane. From the flux calculation,The finite causal wave train has energy integralDifferentiation, or integration of the local energy law with zero flux beyond the front, proves

Finally, distinguish the oscillatory velocity from the second-order Eulerian mean $\overline{u}$. The displacement polarizations giveThe supplied mean-flow relation therefore yields ${\overline{u}}_t=(k/\omega ){\overline{E}}_t+O({\mu }^2)$ at quadratic wave order. Equivalently, the horizontal momentum flux is $\overline{uw}=(k/\omega )c_{gz}\overline{E}$, and envelope-energy transport gives the same acceleration through ${\overline{u}}_t=-{\partial }_z\overline{uw}$.

In the frame moving horizontally at $c=\omega /k$, the boundary pattern is stationary and the initial mean current is $-c$. Subtract that background's kinetic energy before integrating over the infinite half-space. The internal-wave work in a phase-speed frame is thenIts rate of change isThe mean-current cross term supplies the quadratic energy loss in this moving frame; the lab-frame mean kinetic energy alone is fourth order in wave amplitude. The equality is at the leading slow and quadratic orders used throughout the forcing calculation.

For a nonrotating inviscid Boussinesq fluid, start from $\nabla \cdot \mathbf{u}=0$, $D\rho /Dt=0$ andThe resting background has ${\bar{p}}_z=-g\bar{\rho }$ and $N^2=-g{\bar{\rho }}_z/{\rho }_0>0$. Write $\rho =\bar{\rho }(z)+{\rho }^{\prime }$, $p=\bar{p}+p^{\prime }$ and $\sigma =-g{\rho }^{\prime }/{\rho }_0$. Keeping first-order terms givesThe PDF has a genuine sign error: its momentum equation uses a minus sign before $\sigma \hat{\mathbf{z}}$ despite defining $\sigma =-g{\rho }^{\prime }/{\rho }_0$. A lighter parcel has $\sigma >0$ and accelerates upward, so the plus sign is necessary. Retaining both printed signs would yield negative squared wave frequency and exponential growth in a supposedly stable fluid; no oscillatory stable-wave dispersion relation follows from that inconsistent system. Alternatively one can define a downward density variable $\tilde{\sigma }=g{\rho }^{\prime }/{\rho }_0$, but then its density equation must have ${\tilde{\sigma }}_t=+N^2{w}^{\prime }$.

For the consistent system, take a plane internal gravity wave proportional to $e^{i(kx+mz-\omega t)}$, with $k\ne 0$. Incompressibility gives $ku+mw=0$. Project the momentum equation perpendicular to the wavevector, eliminating pressure: $-i\omega w=k^2\sigma /(k^2+m^2)$. Combining with $-i\omega \sigma =-N^{2}w$ givesOn a smooth signed-frequency branch, the phase velocity and group velocity areTheir scalar product is zero. Equivalently, the dispersion relation is homogeneous of degree zero in $(k,m)$, so Euler homogeneous function theorem gives $(k,m)\cdot {\nabla }_{(k,m)}\omega =0$. The phase velocity moves crests normal to their phase lines; the group velocity moves wave packets and energy along those lines. Their perpendicularity explains why following individual crests does not reveal the direction of energy propagation.

For $0<\omega <N$, put $\alpha =\omega /\sqrt{N^2-{\omega }^2}$. The dispersion relation permits $k=\pm \alpha m$. On the positive-frequency branch, for $k>0$,The radiation condition therefore selects $k=\alpha m$, since energy must travel into the fluid. Substitution into the internal-wave polarization relations gives the complex fieldsThe physical fields are their real parts; $v=0$. The vertical component of group velocity is negative, so this positive-$m$ plane internal gravity wave carries energy down and right, although its phase propagates in a different direction.

In the Boussinesq approximation, density departures from a constant reference density ${\rho }_0$ are small enough to neglect in inertia and mass continuity, but their gravitational effect is retained. Write the total buoyancy acceleration as $b=-g(\rho -{\rho }_0)/{\rho }_0$, choose a hydrostatic background $\bar{b}(z)$, and set $\sigma =b-\bar{b}$. Thus $\sigma$ is the buoyancy perturbation, positive for a parcel lighter than the background at its present height. The squared buoyancy frequency isStable stratification has $N^2>0$. A small vertical displacement $\xi$ of a parcel conserving its buoyancy gives $\sigma =-N^2\xi$, explaining the restoring acceleration and the name of $N$.

Let $P=p^{\prime }/{\rho }_0$ be pressure with the background hydrostatic pressure removed, $\mathbf{u}=(u,v,w)$, and $D/Dt={\partial }_t+\mathbf{u}\cdot \nabla$ the material derivative. The ideal, nonrotating Boussinesq equations areThe last equation is equivalent to $Db/Dt=0$. There is neither viscosity nor buoyancy diffusion. The reference density occurs in the pressure normalization, while the small density anomaly occurs only through buoyancy.

For the two-dimensional disturbance, linearization about rest givesA plane internal gravity wave proportional to $e^{i(kx+mz-\omega t)}$ therefore satisfiesThese internal-wave polarization relations and the dispersion relation determine the individual responses in parts i and ii. A radiating response has no incoming internal gravity wave from $x=+\infty$; an evanescent response is bounded there. The harmonic solutions describe the response after excluding arbitrary initial transients.

For the localized oscillatory boundary, define $\alpha =\omega /\sqrt{N^2-{\omega }^2}$ and $I(z)=ϵ{\int }_0^{\infty }{\hat{u}}_0(m)e^{imz}dm=f(z)+ig(z)$. Both signs of vertical wavenumber must radiate into $x>0$. Their horizontal wavenumber is $k=\alpha |m|$, so the positive-$m$ contribution involves $Z_1=z+\alpha x$ and the negative-$m$ contribution involves $Z_2=z-\alpha x$. Consequently the complex horizontal velocity isTaking its real part provesThe positive-$m$ internal-wave polarization has $w=-\alpha u$ and the negative-$m$ one has $w=+\alpha u$. HenceIn particular, $2f(z)=ϵu_0(z)$, so the boundary condition is recovered exactly. This is two-beam radiation from a localized oscillating boundary.

Take a smooth localized profile, for example $u_0(z)=e^{-z^2/H^2}$. Its effective support is a neighbourhood of width a few $H$; increasing that neighbourhood makes the tails arbitrarily small. The boundary displacement is $ϵu_0(z)\sin \omega t/\omega$ to first order, so at $t=0$ it is flat and moving at maximum speed. At that instant,There are two narrow oblique beams centered on $z=-\alpha x$ and $z=+\alpha x$. Away from their overlap, the velocity is parallel to the relevant beam, in the directions $(1,-\alpha )$ and $(1,+\alpha )$. The figure uses $\alpha =3/4$ and measures velocities in units of $ϵ$.

At other phases the $g$ terms contribute. For the positive-frequency convention above, $g$ is the Hilbert transform of $f$; it generally has longer tails than $f$. Thus even when the boundary is nearly stationary, the interior need not be stationary or confined to the bright cores shown here. Half a period later all velocities reverse. The beam directions remain the same, but the instantaneous velocity profile and its quadrature component change.

A short pulse has a broad frequency spectrum. Frequencies below $N$ produce outgoing internal gravity wave packets with frequency-dependent ray slopes $\pm \omega /\sqrt{N^2-{\omega }^2}$; frequencies above $N$ contribute an evanescent near-boundary response. Moreover the group velocity depends on wavelength as well as direction. The radiated packet therefore spreads into a dispersive fan or wake, with different frequencies traveling at different angles and different scales separating along those directions. After the forcing ends the propagating packet travels away, rather than maintaining the two permanently driven monochromatic beams.

In the Boussinesq approximation, replace density by a constant reference density ${\rho }_0$ in inertia and incompressible flow, retaining the small density departure only in the gravitational force. Write $\rho =\bar{\rho }(z)+{\rho }^{\prime }$ and subtract the resting hydrostatic pressure. The upward buoyancy perturbation acceleration and squared buoyancy frequency areThus stable density stratification has $N^2>0$. Let $p$ henceforth denote the pressure anomaly divided by ${\rho }_0$. The ideal, nonrotating Boussinesq equations areThe buoyancy equation follows by materially conserving the total buoyancy $b_0(z)+\sigma$, where $b_0^{\prime }=N^2$; it does not assume constant $N$.

For an $x$-periodic flow of period $P$, define $\bar{a}=P^{-1}{\int }_0^{P}adx$. An infinite-domain average with vanishing end contributions works as well. The necessary pressure condition is no mean zonal pressure gradient, $\overline{p_x}=0$: periodic pressure satisfies it, whereas a pressure containing a term proportional to $x$ does not. Use incompressible flow to write the zonal momentum equation in conservative form,Averaging its zonal derivative terms yields the exact Reynolds stress relation

For two-dimensional small disturbances about rest, the Linearized Boussinesq equations are $u_t=-p_x$, $w_t=-p_z+\sigma$, ${\sigma }_t=-N^{2}w$, and $u_x+w_z=0$. Differentiate the two momentum equations to eliminate pressure: $(u_z-w_x{)}_t=-{\sigma }_x$. Differentiating in $x$ and using $u_{xz}=-w_{zz}$ gives ${\nabla }^2{w}_t={\sigma }_{xx}$. One further time derivative givesThe equation remains valid for $N=N(z)$. No vertical derivative of $N$ was taken: ${\partial }_x^2(N^{2}w)=N^2{w}_{xx}$.

For constant $N$ and $k\ne 0$, a plane internal gravity wave hasBoth signs give the prescribed lower-boundary velocity. The radiation condition excludes energy arriving from infinity. At positive frequency, the vertical group velocity isFor a source below the fluid, energy must propagate upward, so the outgoing solution has $m<0$, even though its vertical phase velocity $\omega /m$ is downward.

To obtain this selection causally, let $T=\mu t$ and $Z=\mu z$, and replace the constant amplitude by $a(Z,T)$. The method of multiple scales treats $(z,t,Z,T)$ as independent during differentiation: the carrier derivatives hold $Z,T$ fixed, while $a_T$ holds $Z$ fixed and $a_Z$ holds $T$ fixed. Only after differentiating do we restrict to $T=\mu t$, $Z=\mu z$. Acting on the complex wave, the operators becomeWith $K^2=k^2+m^2$, the wave equation has residualThe zeroth-order term is the dispersion relation; the first-order term vanishes precisely whenIf $a_0(T)$ is the switched-on boundary amplitude, with $a_0(T)=0$ for $T<0$, the outgoing envelope is $a(Z,T)=a_0(T-Z/b)$ for $b>0$. Its characteristics carry the boundary signal into initially undisturbed fluid. If $b<0$, characteristics instead arrive from the upper half-space; zero initial disturbances and no incoming signal cannot establish the proposed boundary-forced wavetrain. This is the internal-wave envelope radiation condition.

The leading incompressible flow polarization gives $u^{\prime }=-(m/k)w^{\prime }$. Hence, to second order in amplitude and leading order in the slow modulation,There is no meridional transport here. Envelope transport gives ${\partial }_t|a{|}^2=-b{\partial }_z|a{|}^2$, so an initially vanishing mean develops asThese are the leading expressions for internal-wave momentum deposition; higher slow-modulation corrections to the stress are beyond the displayed approximation. In particular, a spatially uniform fully established wavetrain has zero stress divergence, but the front has already produced its mean flow.

The mean wave energy density per unit mass follows from ${\sigma }^{\prime }=-i{N}^2{w}^{\prime }/\omega$:Thus $\bar{u}=(k/\omega )\mathcal{E}$. In the original frame the mean kinetic energy is ${\bar{u}}^2/2=O(a^4)$. In a frame translating at $c=\omega /k$, the undisturbed velocity is $-c$, and the change in mean kinetic energy is $-c\bar{u}+O(a^4)=O(a^2)$. Indeed $\mathcal{E}-c\bar{u}=0$ at this order behind the front. Mean-flow energy is therefore essential at the same order as wave energy in the translating frame. This follows also from the frame transformation of energy, $E\mapsto E-cP$ up to the fixed background constant, and is why conservation of momentum cannot be omitted from the moving-frame conservation of energy budget.

Assume small-amplitude, nonrotating, inviscid Boussinesq approximation motion about a fluid at rest, with stable background stratification. In either uniform layer let $b=-g{\rho }^{\prime }/{\rho }_{\ast }$ be the buoyancy perturbation and $N^2=-(g/{\rho }_{\ast }){\rho }_0^{\prime }(z)>0$. The linear equations areFor a plane internal gravity wave proportional to $e^{i(kx+mz-\omega t)}$, continuity gives $u=-mw/k$, and horizontal momentum gives $p={\rho }_{\ast }\omega u/k$. Substituting these into vertical momentum and buoyancy evolution givesChoose $k>0$, $\omega >0$. Its group velocity isThus downward energy propagation has $m>0$. If $\theta$ is the acute angle between the ray and the vertical in its direction of propagation, then $\tan \theta =|m|/k$, soThe phase velocity is parallel to $(k,m)$, while ${\mathbf{c}}_g\cdot (k,m)=0$: phase and group velocities are perpendicular, and their vertical components have opposite signs. Propagating incident waves require $0<\omega <N_1$; both layers support propagating waves only when $0<\omega <\min (N_1,N_2)$. At or above $N_2$, the lower layer cannot carry a propagating downward internal wave.

For the amplitude convention used below, $\eta$ multiplies the unit particle-displacement vector along the ray, with positive horizontal component. That vector is $(\sin \theta ,-\cos \theta )$ for a downward wave and $(\sin \theta ,+\cos \theta )$ for an upward wave. Stating this polarization convention is important when assigning a sign to a reflection coefficient.

Let $b=-g{\rho }^{\prime }/{\rho }_0$ be the buoyancy perturbation and $p$ the pressure perturbation divided by ${\rho }_0$. For stable stratification, $N>0$. The nonrotating Linearized Boussinesq equations areEliminating $u,p,b$ gives $({\partial }_x^2+{\partial }_z^2)w_{tt}+N^2{w}_{xx}=0$. Since $w={\zeta }_t$, a nonzero-frequency plane wave obeys the same equation for its displacement. Substituting its phase yields the dispersion relation for a plane internal gravity wave:Thus the frequency depends on the wavevector direction rather than its magnitude.

Advection of the background density gives ${\rho }^{\prime }=-\zeta {\bar{\rho }}_z$ to first order. With constant ${\bar{\rho }}_z<0$, the instantaneous density gradient is ${\rho }_z={\bar{\rho }}_z(1-{\zeta }_z)$. A region has unstable density stratification when this becomes positive, namely when ${\zeta }_z>1$. The maximum of ${\zeta }_z$ is $|m{A}_{\zeta }|$, so the monochromatic internal-wave overturning criterion isEquality gives a locally vanishing gradient. This is the prediction of the displacement field extrapolated to overturning; the small-amplitude approximation itself ceases to be reliable there.

For the rising packet, distinguish its conserved absolute frequency $\omega$ from its actual intrinsic frequency $\sigma =\omega -kU(z)$. The printed terminology calls $\omega$ intrinsic while also assigning it to a stationary observer; the stationary-observer interpretation is the one consistent with the displayed Doppler shift. On the positive-frequency branch, the ray Hamiltonian isThe Hamiltonian ray-tracing equations giveThe last identity follows also by differentiating the Hamiltonian along its canonical trajectory: the spatial and wavevector terms cancel in pairs. Thus absolute-frequency conservation in steady shear gives constant $k$, constant $\omega$, and constant stationary-observer horizontal phase speed $c_x=\omega /k$. In contrast, $\sigma =\omega -ksz$ decreases as the packet rises. At its initial height,This is the critical level of an internal gravity wave. In fact $m(t)=m_0-kst$ and $z(t)=[\omega -\sigma (k,m(t))]/(ks)$, so the inviscid ray approaches $z_c$ as $t\to \infty$, rather than reaching it at a finite time.