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A time-independent ray Hamiltonian conserves the stationary-observer absolute frequency. A horizontally uniform background also conserves horizontal wavenumber. The intrinsic frequency $\hat{\omega }=\omega -kU(z)$ nevertheless changes when the ray moves through shear. These follow from the Hamiltonian ray-tracing equations and are essential to identifying a critical level of an internal gravity wave.

A critical level is a height at which the intrinsic frequency of an internal gravity wave vanishes. For finite positive buoyancy frequency, its local vertical wavenumber diverges as $\hat{\omega }\to 0$. The inviscid Taylor–Goldstein equation is singular there; viscosity, diffusion or nonlinear wave breaking can regularize the resulting small scales.

For a uniform vertical shear flow $U=sz$ with $s>0$, constant buoyancy frequency $N>0$ and horizontal wavenumber $k>0$, choose the initially rising branch $m_0<0$. The ground-frame angular frequency is the conserved ray HamiltonianThe internal-wave ray tracing equations give $\dot{k}=0$ and $\dot{m}=-ks$, hence $m=m_0-kst$ and $z=(\omega -\hat{\omega })/(ks)$. As $t\to \infty$, the vertical wavenumber tends to negative infinity, the intrinsic frequency tends to zero and the height approaches the critical level $z_c=\omega /(ks)$. For $\theta =\arctan (|m|/k)$, the laboratory ray slope isThe second term is mean advection divided by vertical group velocity. The zero-shear case has no finite $z_c$ and must instead be treated as propagation in a uniform current.

For a stationary topographic internal gravity wave in a uniform horizontal velocity $U>0$ and constant buoyancy frequency $N>0$, a nonzero horizontal wavenumber $k$ gives intrinsic frequency $\hat{\omega }=-Uk$. The nonrotating Boussinesq approximation givesVertical propagation therefore requires $0<|k|<N/U$. At $|k|=N/U$, the vertical wavenumber vanishes and so does the vertical group velocity; this grazing limit carries no finite upward energy flux for a finite-amplitude plane mode. For $|k|>N/U$, the vertical wavenumber is imaginary and the bounded response is an evanescent wave. A hill pattern with very short horizontal wavelength can consequently produce a local disturbance without radiating an upward internal gravity wave.

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.

Assume $U>0$, $k=\alpha >0$, constant $N$, negligible rotation, dissipation and reflection from infinity, small $k{\eta }_0$ and $m{\eta }_0$, and no wave overturning. The terrain is stationary, so laboratory frequency is zero and intrinsic frequency is $\hat{\omega }=-kU$. Propagation requires $kU<N$. PutThe kinematic boundary condition at the linearized bed is $w(x,0)=U{\eta }_x=-kU{\eta }_0\sin (kx)$. The upward radiation condition selects the positive $m$ branch because $c_{gz}=-\hat{\omega }m/(k^2+m^2)>0$. ThusThe internal-wave polarization and every other perturbation follow from this vertical velocity and the linear equations.

In the frame moving with the fluid, $x^{\prime }=x-Ut$, the phase is $k{x}^{\prime }+mz+kUt$, andThe intrinsic energy ray slopes upward and upstream, $dz/d{x}^{\prime }=-k/m$, along a constant-phase line of an internal gravity wave. At fixed intrinsic frequency the spatial wave equation is hyperbolic, with the two characteristic curve line slopes $dz/d{x}^{\prime }=\pm k/m$; the radiation condition selects the upward ray of the appropriate horizontal propagation branch.

In the terrain frame, phase lines remain fixed and the pattern phase velocity is zero. A transported packet hasIts energy path slopes upward and downstream, $dz/dx=m/k$, and need not coincide with a stationary phase line. One should not add $(U,0)$ to the chosen normal intrinsic phase velocity and call that the normal ground-frame phase velocity: tangential motion along a phase line is arbitrary; zero laboratory frequency makes the entire crest pattern stationary. The sketch distinguishes stationary crests, intrinsic characteristics and transported energy.

If $kU>N$, $m$ is imaginary and the bounded field is an evanescent wave; at $kU=N$ the vertical group velocity vanishes and an ordinary upward radiation interpretation degenerates.

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 $b=-g{\rho }^{\prime }/{\rho }_0$ 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.

Write $\theta =|\Theta |=\arctan (|m|/k)$, so $\sigma =N\cos \theta$ with $0<\theta <\pi /2$. The intrinsic internal-wave phase and group velocity calculation givesThe observer-frame horizontal ray velocity is $U+c_{gx}$. Dividing it by $c_{gz}$ proves the internal-wave ray in uniform vertical shear:The angle increases toward $\pi /2$ and the vertical group speed tends to zero near the critical level.

The wave-action conservation law fixes the prescribed upward flux. For a nonzero packet, $B>0$, and the given flux relation impliesApply the monochromatic internal-wave overturning criterion, using $|m|=k\tan \theta$. After multiplying by the positive trigonometric factors, the exact instability condition isAt marginal overturning near a critical level, $\sin \theta \simeq 1$, so the wave-action criterion for critical-level overturning givesWith fixed $B,k,N$, this is the requested quarter-power order estimate; the prefactor supplies the dimensions suppressed in that notation. It is an onset balance, not a replacement for $\cos \theta =\sigma /N$. Combining the two relations instead gives ${\cos }^3\theta =(B{k}^3/N^2)\sin \theta$ at onset. Since $m^2{A}_{\zeta }^2$ diverges as ${\sigma }^{-3}$ toward $z_c$, any nonzero packet flux eventually violates the linear overturning criterion before reaching that level, within this nondissipative ray model.

Take a common constant buoyancy frequency $N$ and put $k=k_T>0$. Define $\eta$ to be vertical parcel displacement; the kinematic boundary condition is $w=ik{U}_j{\eta }_j$. In either uniform region the exact Taylor–Goldstein equation givesNo WKB approximation is needed. If the earlier $N(z)$ is genuinely nonconstant, the exact regional equation is instead ${\eta }_j^{\prime \prime }+[N(z{)}^2/U_j^2-k^2]{\eta }_j=0$; its fundamental solutions, not plane exponentials, must be used. For example, let $C_1(0)=1$, $C_1^{\prime }(0)=0$, $S_1(0)=0$, $S_1^{\prime }(0)=1$, and let $f_2$ be an outgoing upper solution. Then ${\eta }_1=h_0{C}_1+D{S}_1$, ${\eta }_2=A{f}_2$, with $D,A$ determined by $h_0{C}_1(H)+D{S}_1(H)=A{f}_2(H)$ and $U_1^2[h_0{C}_1^{\prime }(H)+D{S}_1^{\prime }(H)]=U_2^{2}A{f}_2^{\prime }(H)$. This gives the exact arbitrary-profile field implicitly, while a simple angle-only coefficient requires further stratification assumptions.

Under the common constant-$N$ assumption, propagating incident internal gravity waves requireFor co-directed upper flow, the outgoing propagating branch exists for $0<U_2<N/k$ and has $m_2>0$. The radiation condition chooses $m_j>0$ because the laboratory vertical group velocity is positive on the negative intrinsic frequency branch. Define $\tan {\theta }_j=m_j/k$, so $\cos {\theta }_j=k{U}_j/N$.

At the velocity jump, the material-interface jump conditions for stratified inviscid shear flow require continuity of displacement and of fluid pressure. Vertical velocity itself is generally discontinuous: $w_j=ik{U}_j\eta$. In terms of displacement the pressure is $p_j=U_j^2{\eta }_j^{\prime }$. The displacement impedance for an internal wave at a velocity jump is $Z_j=U_j^2{m}_j$. An upward incident component of complex displacement amplitude $I$ at $H$, a reflected component $RI$ and a transmitted component $TI$ therefore obeyConsequentlyFor a one-pass incident displacement $h_0$, a real wave field is the imaginary part ofThe incident vertical-displacement amplitude is $h_0$, the reflected amplitude is $|R|h_0$ and the transmitted amplitude is $|T|h_0$; the sign of $R$ encodes a phase reversal. The other fields follow from $w=ikU\eta$, $u=-w^{\prime }/(ik)$, $b=-N^2\eta$ and $p=U^2{\eta }^{\prime }$. If a perfectly reflecting terrain boundary is enforced after the return wave arrives, replace the incident coefficient $h_0$ by $h_0/[1+R{e}^{2i{m}_{1}H}]$. The one-pass amplitudes use the permitted neglect of repeated terrain reflections.

For common $N$, $Z_j=(N^2/k)\cos {\theta }_j\sin {\theta }_j$. The physical vertical-displacement transmission coefficient is thereforeThe printed coefficient has a different denominator and cannot be obtained for this standard displacement convention. For example, when $U_2=U_1$, there is no interface scattering and the vertical amplitude stays $h_0$, whereas the printed expression gives $h_0/\cos {\theta }_1$. A total displacement magnitude would instead be the vertical amplitude divided by $\cos {\theta }_2$; with unequal velocities it still does not give the printed expression. The transmission formula needs an explicit correction or a different, presently unspecified amplitude/interface model.

For $U_2>N/k$, choose $m_2=i\kappa$ with $\kappa =(k^2-N^2/U_2^2{)}^{1/2}$: the transmitted response decays as $e^{-\kappa (z-H)}$ and the same impedance equations give a complex $R,T$, with $|R|=1$. At $U_2=0$ the steady equation is singular. Oppositely directed upper flow formally supports propagating waves when $0<|U_2|<N/k$, but upward radiation then requires $m_2<0$ and signed $Z_2$; the positive-angle co-directed formula is not valid there. A real velocity jump can also have Kelvin-Helmholtz instability, which is excluded from this formal steady scattering calculation.

For the rapidly varying wave phase, define $k_i={\partial }_i\theta$ and $\omega =-{\partial }_t\theta$, with the common scaling by $\epsilon$ absorbed into these local variables. A packet moves with group velocity $c_{g,i}=\partial \Omega /\partial k_i$. Along that trajectory, $d/dt={\partial }_t+c_g\cdot {\nabla }_x$. Compatibility gives ${\partial }_t{k}_i=-{\partial }_i\omega$ and ${\partial }_i{k}_j={\partial }_j{k}_i$. Differentiating the local dispersion relation then cancels its implicit $k$ derivatives, yieldingThe last equality follows also by differentiating $\omega =\Omega (k,x,t)$ along the ray and cancelling the two Hamiltonian cross terms. The partial derivatives of $\Omega$ hold its other arguments fixed.

A mean flow adds the Doppler shift: the intrinsic frequency is $\hat{\omega }=\omega -kU(z)$, soon the positive-intrinsic-frequency branch launched here. The medium is stationary and independent of $x$, so $\omega$ and $k$ are constant. Initially $m=0,U=0$, hence $\omega =N$. Moreover $\dot{m}=-k{U}^{\prime }<0$, and then $\dot{z}=-Nkm/(k^2+m^2{)}^{3/2}>0$. The initial vertical group velocity is zero, but $m$ immediately becomes negative and the ray accelerates upwards.

If $U$ reaches $N/k$ at a finite height $z_c$, this is the critical level of an internal gravity wave, with $U(z_c)=\omega /k$. Below it,Assuming $0<U^{\prime }(z_c)<\infty$, put $\delta =z_c-z$. Then $N-kU\sim k{U}^{\prime }(z_c)\delta$, givingThus the ray cannot cross this level in finite time within the geometrical-optics approximation. The hypothesis $U^{\prime }>0,U(0)=0$ alone does not guarantee a finite $z_c$: for example $U(z)=\frac{N}{2k}(1-e^{-z})$ on $z\ge 0$ never reaches $N/k$. Existence of a finite critical level is an additional assumption. Very close to one, diverging wavenumber can invalidate the inviscid, linear, slowly varying approximation; the result is the formal ray prediction.

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.

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 $m>0$.

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

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

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

For a stationary internal gravity wave, the Taylor–Goldstein equation gives $m^2=N^2/U^2-U^{\prime \prime }/U-k^2$. In a propagating interval with slowly varying $m$, the WKB approximation has amplitude proportional to $m^{-1/2}$ and phase derivative $m$. The conditions $|m^{\prime }|\ll m^2$ and $|m^{\prime \prime }|\ll |m{|}^3$ fail at a regular turning level. With $U>0$, the upward branch has $m>0$ and negative intrinsic frequency; laboratory group velocity is $U(k^2,km)/(k^2+m^2)$ when the curvature term is negligible.

A fixed terrain obstacle forces a zero laboratory-frequency internal gravity wave through the kinematic boundary condition $w=U{h}_x$. Its intrinsic frequency is $-kU$. Propagation, radiation and the transient arrival of its permanent component are distinct questions.

With weak kinematic viscosity $\nu$, zero buoyancy diffusion and locally unchanged plane internal gravity wave polarization, the displacement pseudomomentum of an internal gravity wave obeys $P_t+{\partial }_z(c_{gz}P)=-\nu K^{2}P$. For constant wavenumbers and a stationary wave envelope, $P\propto |\hat{\sigma }{|}^2$ proves the displayed amplitude decay. Decay is along the group velocity, so $c_{gz}<0$ describes a source above the receiving level. In a slowly varying shear flow, use wave action and the changing intrinsic frequency rather than treating $P/|\hat{\sigma }{|}^2$ as constant.

For a small-amplitude conservative wave packet in a slowly varying fluid flow, its intrinsic energy divided by intrinsic frequency is transported along its group velocity, with the basic advection added in the laboratory frame. A stationary background conserves laboratory angular frequency, but a changing mean velocity changes the intrinsic one through the Doppler effect; intrinsic energy alone need not be conserved. Geometrical spreading and dissipation enter the action-flux balance.