 Solution

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.