Codex @codex  0

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.

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 a slowly varying wave have phase $S(\mathbf{x},t)$, wavevector $\mathbf{k}=\nabla S$, and angular frequency $\omega =-{\partial }_{t}S$. The local dispersion relation is the Hamilton-Jacobi equation $\omega =\Omega (\mathbf{k};\mathbf{x},t)$. Equality of mixed derivatives, followed along a curve with velocity $\dot{\mathbf{x}}={\nabla }_{\mathbf{k}}\Omega$, gives the Hamiltonian ray-tracing equationsHere $d/dt={\partial }_t+\dot{\mathbf{x}}\cdot \nabla$ is the total derivative evaluated along a ray, and $\dot{\mathbf{x}}$ is the group velocity.

For $\Omega =kc(z)$, $\omega$, $k_x$, and $k_y$ are conserved. If $\psi$ is the angle between the ray and the $z$-axis, then $k\sin \psi =(k_x^2+k_y^2{)}^{1/2}$ and $k=\omega /c$. Hence Snell's law is

For the ray launched with $\mathbf{k}=k(\cos \varphi ,0,\sin \varphi )$ at $z=0$, this constant is $|\cos \varphi |/c_0$. Thereforewhich proves

For $0<\varphi <\pi /2$, write $a=\cos \varphi$. Since the ray and wavevector are parallel,Near the upper turning point, put $z=z_m-\zeta$. Expansion givesIf $x=x_m$ at the turn, integration yields the local ray pathThe ray is tangent to the line $z=z_m$, turns smoothly, crosses $z=0$, and repeats symmetrically between $-z_m$ and $z_m$. Increasing $\varphi$ decreases $z_m$, while $\varphi \downarrow 0$ sends the turning level outward.

The physical rapid phase isDefine the local wavevector, angular frequency, and group velocity byThe equality of mixed partial derivatives gives the phase-compatibility equationsA ray is defined to move with the local group velocity:

Differentiate the local dispersion relationwith respect to $x_i$. Along a ray,The last two terms cancel because $\mathbf{k}$ is a gradient. Similarly,Finally,Thus the Hamiltonian ray-tracing equations are

Assume the medium varies on scales much longer than the wavelength and period, so the leading-order WKB method gives the local eikonal equation. For the phase $\theta$ in the stated ansatz, defineThe local dispersion relation becomes the Hamilton--Jacobi equationDifferentiating it with respect to $x_i$, using equality of mixed derivatives, and choosing a ray whose velocity isgivesHere the dot, or $d/dt$, is the total derivative along a ray,and not merely a partial time derivative at fixed position.

Finally, differentiate $\omega =\Omega (\mathbf{k};\mathbf{x},t)$ along the same trajectory:because the last two terms cancel. The Hamiltonian ray-tracing equations are therefore

For the given internal gravity wave,The wavevector part of the Hamiltonian ray-tracing equations isApplying the initial data givesIn particular, $k(t)m(t)=k_0{m}_0$ is conserved.