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For $N^2>0$, write $K=\sqrt{k_x^2+k_z^2}$ and choose either signed branch $\omega =sN{k}_x/K$, with $s=\pm 1$. These two branches cover the roots of the squared dispersion relation. The group velocity of an internal gravity wave isIf instead the positive-frequency root is chosen for both signs of $k_x$, set $s=+1$ and multiply the displayed components by $\mathrm{sgn}(k_x)$, away from the cusp $k_x=0$.

Taking the dot product with the wave vector givesThus the group velocity, which describes the travel of a narrow wave packet and its energy, is perpendicular to the wave vector, and hence perpendicular to the phase velocity ${\mathbf{c}}_p=\omega \mathbf{k}/K^2$. This is the internal-wave phase and group velocity geometry. It also follows from phase-group orthogonality for degree-zero dispersion: the frequency depends only on the direction of $\mathbf{k}$, not its magnitude. For $k_z=0$, the group velocity is zero, consistent with the stationary extremum $|\omega |=N$.

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.

Let the group velocity make an angle $\theta$ above the horizontal. A convenient right-handed choice of rotated unit vectors isThe first vector is parallel to the wavevector, the second is parallel to the group velocity, and their perpendicularity is the defining geometry of an internal-wave phase and group velocity. ThusWriting ${\Delta }_{\xi \zeta }={\partial }_{\xi }^2+{\partial }_{\zeta }^2$, the equation becomesFor an inviscid plane wave proportional to $e^{i(k\xi -\omega t)}$, its dispersion relation is $\omega =N\sin \theta$.