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

Use the inviscid, nonrotating Boussinesq approximation, with a stable background mass density $\bar{\rho }(z)$ and reference density ${\rho }_{\ast }$. Write $B=-g{\rho }^{\prime }/{\rho }_{\ast }$ for the perturbation buoyancy and $\pi =p^{\prime }/{\rho }_{\ast }$ for the kinematic pressure. The buoyancy frequency is $N_0^2=-(g/{\rho }_{\ast }){\bar{\rho }}_z>0$. Dropping products of perturbations in the Boussinesq equations gives the Linearized Boussinesq equationsThe last equation expresses incompressible flow; the second follows by advecting the background mass density gradient. Neglecting rotation and viscosity is part of this internal gravity wave model.

For a plane internal gravity wave proportional to $\exp [i({\mathbf{k}}_h\cdot {\mathbf{x}}_h+mz-\omega t)]$, put $\kappa =|{\mathbf{k}}_h|>0$ and $K^2={\kappa }^2+m^2$. Eliminating the horizontal velocity, pressure and buoyancy from the linear equations givesFor example, the horizontal momentum and continuity equations give $\pi =-\omega mw/{\kappa }^2$; substituting $B=-i{N}_0^{2}w/\omega$ into vertical momentum yields the displayed dispersion relation. Thus an IGW has $0<|\omega |\le N_0$ in this model.

On the positive-frequency branch, the phase velocity normal to a constant-phase plane and the group velocity areConsequently $({\mathbf{k}}_h,m)\cdot {\mathbf{c}}_g=0$, and phase and group velocity are perpendicular. Equivalently, the dispersion relation is homogeneous of degree zero in the wave vector, so differentiating with respect to its scale proves the same orthogonality. Energy travels with the group velocity, along the phase planes. At the degenerate limit $m=0$, $\omega =N_0$ and the group velocity vanishes; the orthogonality statement then has this limiting interpretation.