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The beta barium borate crystal supplies spontaneous parametric down-conversion: a weak ultraviolet pump can produce a pair of lower-frequency photons, with their frequencies and wavevectors selected by conservation of energy and phase matching. A suitably aligned and compensated source makes the two polarization alternatives coherent, giving the stated Bell state. The crystal alone does not guarantee that every collected pair has that state.

The two polarizing beam splitters route horizontal polarization and vertical polarization to different ports. Take $H$ from arm $a$ to the trigger $T$ and $V$ to arm $c$. The ordinary beam splitter in arm $b$ is a lossless, polarization-independent 50:50 splitter, with the unused input in the vacuum state. It splits probability amplitudes, rather than measuring which output a photon takes. The half-wave plate changes the relative phase of its principal polarizations by $\pi$; orient it so that a $V$ photon in $c$ becomes $(H+V)/\sqrt{2}$ in $c^{\prime }$, up to a common phase. Finally, photodetectors absorb arriving photons and supply electrical clicks. They need not resolve photon number for the ideal four-photon, fourfold-coincidence calculation: a click at every detector already requires one photon at each.

Choose the relative wave phases so that the final polarizing beam splitter sends $H_{c^{\prime }}$ to $H_2$, $V_{c^{\prime }}$ to $V_1$, $H_d$ to $H_1$ and $V_d$ to $V_2$. Such phase conventions must be used consistently: an uncompensated relative phase would give a phase-shifted GHZ state.

Use state-dependent fluorescence. With 397 nm light on $S_{1/2}\leftrightarrow P_{1/2}$ and an 866 nm repumper on $D_{3/2}\leftrightarrow P_{1/2}$, an ion initially in $D_{3/2}$ is brought into a bright cycling manifold. The repumper prevents population trapping in $D_{3/2}$. An ion in $D_{5/2}$ remains dark while the 854 nm light is off, since it has no allowed link to this cycle. A detected bright or dark record therefore measures the qubit in its logical basis by electron shelving.

To reset, apply 854 nm light to empty the shelved state, and use 397 nm plus repumping for Doppler cooling and internal-state preparation. Doppler cooling alone generally does not reach the ground state. In the resolved-sideband and Lamb-Dicke regime, drive the red motional Raman sidebandand then optically repump to $D_{3/2}$ with little change in motional occupation. Each cycle removes approximately one vibrational quantum. The $n=0$ state is dark to this red sideband because its matrix element contains $\sqrt{n}$. Repeat with appropriate wavevector projections and frequency differences for all modes to be cooled, then optically pump into the selected $D_{3/2}$ sublevel. Small recoil during repumping is necessary for a high ground-state probability. ^[ref]

Let $H=h_{00}+\zeta$ and write the horizontal shallow water equations on the constant f-plane asKeeping only terms linear in the disturbance about rest gives the linearized shallow water equationsDefine the horizontal divergence $\delta =u_x+v_y$ and relative vorticity $q=v_x-u_y$. Taking respectively the divergence and vertical curl of the momentum equation yieldsTogether with ${\zeta }_t=-h_{00}\delta$, this provesThe nonlinear shallow-water potential vorticity is $Q=(f+q)/H$. The nonlinear vertical vorticity equation gives $D(f+q)/Dt=-(f+q)\delta$, while mass conservation gives $DH/Dt=-H\delta$; their ratio is therefore materially conserved. Expanding about rest,Advection of its constant background contributes nothing at first order, explaining why the displayed linear anomaly is conserved at each fixed position.

A further time derivative eliminates $q$ and $\zeta$ from the divergence equation:For a plane wave with horizontal wavevector $(k,l)$, write $K=\sqrt{k^2+l^2}$. More directly, substituting the plane wave into the three linear equations gives a coefficient matrix with determinant proportional to $\omega [{\omega }^2-f^2-c_0^2{K}^2]$. Thus the full linear rotating shallow-water dispersion relation has one balanced branch and two wave branches:The latter are inertia-gravity waves. The positive branch starts at $f$ with zero slope and tends to the straight line $c_{0}K$; the negative branch is its reflection. The requested sketch, including the stationary branch, is

For the positive-frequency branch, the magnitudes of the phase velocity and group velocity areThe group velocity vector is $c_0^2(k,l)/\omega$; the negative branch reverses its direction. Near $\omega =f$, namely $K{R}_D\ll 1$ with the Rossby deformation radius $R_D=c_0/f$, $c_p\sim f/K$ is large while $c_g\sim c_0^{2}K/f$ is small. The $K=0$ limit is an inertial oscillation, for which a spatial phase speed is not defined. Far above $f$, both speeds approach $c_0$, the shallow water gravity-wave speed, with $c_p>c_0$ and $c_g<c_0$ at finite $K$.

On the zero-frequency branch, geostrophic balance givesFor a nonconstant plane wave, $\hat{q}=-(g{K}^2/f)\hat{\zeta }$: relative vorticity and elevation have opposite signs. They can both be nonzero and spatially structured, whereas the horizontal divergence vanishes identically. A constant elevation is also a zero-frequency solution, with zero velocity and zero relative vorticity.

For the initially resting ridge, set ${\zeta }_0(x)=ϵh_{00}{e}^{-x^2/L^2}$ and assume an unbounded domain, uniform in $y$. Initial potential vorticity fixesIn the balanced remainder, insert $q_s=(g/f){\zeta }_s^{\prime \prime }$ to obtainThe decaying Green's function of this modified Helmholtz equation gives the Geostrophic adjustment of a Gaussian height ridge explicitly:The initial outward pressure force creates a cross-ridge current; the Coriolis acceleration turns that current and builds the along-ridge geostrophic flow. The remainder is radiated inertia-gravity waves, not a dissipative disappearance of energy.

For precision, use the Fourier transform $\hat{\zeta }(k)=\int \zeta (x)e^{-ikx}dx$. Initial rest implies ${\zeta }_t(x,0)=0$, and the conserved anomaly givesConsequentlyThese formulas satisfy both initial conditions. They also show that the waves carry no linear potential vorticity anomaly: $q=(f/h_{00})(\zeta -{\zeta }_0)$ and $\delta =-{\zeta }_t/h_{00}$ reconstruct the remaining fields, with $u$ and $v$ obtained from their derivatives. On the infinite line, dispersive wave packets propagate away and decay locally, leaving the balanced ridge and its oppositely directed flanking currents. Conservation of energy still holds globally. In a reflecting finite basin, standing waves may persist, so a permanent pointwise relaxation is not guaranteed without dissipation.

If $L\gg R_D$, most of the initial elevation remains balanced: ${\zeta }_s\simeq {\zeta }_0$ with small smoothing corrections. If $L\ll R_D$, the balanced ridge spreads to width $R_D$, and away from the original narrow coreThus the final central height is much smaller than the initial one and the wave component is substantial. The integral of the balanced elevation equals the initial integral, since the Green's function has unit integral; the oscillatory component has zero net added volume.

Let $q_j$ denote the four critical wavevectors, and write $A_j=R_j{e}^{i{\theta }_j}$. A spatial translation by $(s,t)$ acts on the amplitudes asThe two independent translation directions in phase space are $(2,2,1,1)$ and $(1,-1,2,-2)$. Their common orthogonal complement is spanned by $(2,-2,-1,1)$ and $(1,1,-2,-2)$. Hence translation invariants of Fourier-mode phases can be chosen asAway from zero amplitudes, a translation-invariant function of the phases is a function of these two invariant angles. The four magnitudes are themselves invariant. Thus an equivariant dynamical system on the eight-real-dimensional centre manifold descends, after quotienting translation symmetry, to four magnitude equations and two invariant-phase equations. The phase coordinates are singular where an amplitude vanishes; the original complex-amplitude equations remain smooth there.

The wavevector selection rule for equivariant monomials proves the cubic regularity. For a monomial ${\prod }_j{A}_j^{p_j}{\overline{A}}_j^{q_j}$ in the first equation, put $d_j=p_j-q_j$. Translation covariance requiresAt total degree $N$, additionally $\sum |d_j|\le N$ and $N-\sum |d_j|$ is even. Conversely, any integer vector satisfying these conditions yields a monomial, with any remaining even degree supplied by factors $|A_j{|}^2$. This makes the degree test exhaustive. For example, eliminate the first two entries:Checking the integer pairs $|d_3|+|d_4|\le N$, retaining only integral $d_1,d_2$ with the norm and parity conditions, givesAt degree three the only possibility is therefore $A_1|A_j{|}^2$, $j=1,\dots ,4$. Square symmetry gives the same conclusion for every other equation. All cubic terms are regular.

Two generators of the square dihedral group act as follows. Reflection in the $x$ axis gives $(A_1,A_2,A_3,A_4)\mapsto (A_2,A_1,A_4,A_3)$, while interchange of $x,y$ gives $(A_3,{\overline{A}}_4,A_1,{\overline{A}}_2)$. Rotation by $\pi$ conjugates every amplitude, forcing the coefficients in a steady normal form of a dynamical system to be real. With real unfolding parameter $\sigma$ and real coefficients $c_1,c_2,c_3,c_4$, defineThe most general cubic normal form is ${\dot{A}}_j=A_j{G}_j+O(|A{|}^5)$. There are no further coefficient identifications: the square-group stabilizer of one of these oblique wavevectors is trivial, so its three couplings to the other magnitudes are independent. At cubic order ${\dot{R}}_j=R_j{G}_j$ and all phases, including ${\chi }_1,{\chi }_2$, are constant.

A periodic pattern breaks continuous translation symmetry: translating it changes its phase without changing its shape or its energy. Consequently a uniform phase change is neutral. A slowly varying phase change alters the local spacing or orientation and can grow even while ordinary uniform amplitude perturbations decay. These are long-wavelength sideband instabilities. The relevant wavelengths are long compared with the carrier pattern, so their leading dynamics is often a phase modulation equation. Positive phase diffusion smooths spacing errors; negative phase diffusion amplifies them. Near onset, an amplitude equation computes this diffusion explicitly. Away from onset, one must calculate it from the full nonlinear periodic pattern.

For a supercritical stationary pattern in one dimension, rescale the real Ginzburg–Landau equation toA detuned periodic pattern is represented by $A=R{e}^{iqx}$, with $R^2=\mu -q^2>0$. Perturb its magnitude and phase by $A=(R+u)e^{i(qx+\varphi )}$. Linearization givesFor a Fourier mode proportional to $e^{\lambda t+ikx}$, the two eigenvalues are the sideband spectrum of a real Ginzburg-Landau plane wave:The magnitude branch remains damped at small $k$. The phase branch expands asHence the Eckhaus-stable band is $\boxed{q^2<\mu /3}$, narrower than the existence band $q^2<\mu$. This Eckhaus instability is longitudinal: it compresses and dilates the pattern along its wavevector. The coupling to the relaxing magnitude is crucial; examining phase diffusion while holding the magnitude fixed would miss the destabilizing $2q^2/R^2$ correction. At the stability boundary the $k^2$ coefficient vanishes, and the next term is negative, $-2q^4{k}^4/R^6$. Beyond the boundary, a phase distortion can create amplitude zeros and phase slips, changing the number of periods. These nonlinear events are possible outcomes, not conclusions of the linear calculation.

A fixed periodic domain allows only discrete sideband wavenumbers. Indeed, the displayed spectrum is unstable only if $0<k^2<6q^2-2\mu$. A domain too short to support such a Fourier mode can suppress the nominal long-wave instability. Thus an infinite-domain stability band and a finite-domain threshold must be distinguished.

The zigzag instability concerns transverse bending of rolls in an isotropic two-dimensional system. Let $k_c$ be the preferred carrier wavenumber, use an $x$-directed carrier, and denote its slow longitudinal detuning by $q$. Rotational covariance near onset leads to the anisotropic envelope equationThe derivative combination comes from expanding $|\mathbf{k}|-k_c$: a small transverse carrier component changes the length of the wavevector only at second order. A straight-roll envelope $A=R{e}^{iqX}$ has $g{R}^2=\mu -\xi q^2$. For a purely transverse phase perturbation $A\simeq R{e}^{iqX}(1+i\varphi (Y,t))$, magnitude and phase decouple at linear order, andThe transverse growth rate is thereforeRolls with $q<0$, whose total wavenumber is below the preferred value, are zigzag-unstable to sufficiently long transverse modulations. Bending increases their local wavevector length and can relieve this mismatch. Rolls with $q>0$ are transversely stable in this near-onset model; at $q=0$ the restoring force first appears at fourth spatial order. Combining this with the longitudinal Eckhaus instability gives the robust interior stable band $0<q<\sqrt{\mu /(3\xi )}$ for this simplest isotropic roll model. Anisotropy, a pinned orientation, additional mean fields, or other pattern symmetries can modify this conclusion.

For oscillatory patterns the coefficients of the amplitude equation are generally complex. Consider the scaled complex Ginzburg–Landau equationwhere $c_1$ measures linear dispersion and $c_3$ the amplitude-dependent frequency shift. The uniform oscillation is $A=e^{-i{c}_{3}t}$. Put $A=e^{-i{c}_{3}t}(1+u+iv)$, with real infinitesimal $u,v$. The linearized magnitude and phase equations areTheir characteristic polynomial for a sideband of wavenumber $k$ isThe magnitude mode has leading eigenvalue $-2$, while the neutral phase mode hasThus the Benjamin-Feir instability occurs for $\boxed{1+c_1{c}_3<0}$; the complementary Benjamin-Feir stability condition is $1+c_1{c}_3>0$. Both dispersive couplings matter: the phase perturbs the magnitude through $c_1$, and the magnitude feeds back into phase through $c_3$. In the unstable regime, higher spatial derivatives regularize sufficiently short disturbances. On a finite domain, an unstable mode must satisfy $0<k^2<-2(1+c_1{c}_3)/(1+c_1^2)$. At equality of the phase-diffusion coefficient to zero, the leading phase damping is fourth order, not exponential growth at order $k^2$.

Oscillatory plane waves with a nonzero carrier detuning also have a generalized Eckhaus boundary. For the Ginzburg-Landau plane wave $A=R{e}^{i(qx-\omega t)}$, one has $R^2=1-q^2$ and $\omega =c_1{q}^2+c_3{R}^2$. Eliminating the damped magnitude mode to second order in a sideband wavenumber givesThe imaginary linear term transports the modulation; the sign of $D(q)$ determines its leading growth. If $1+c_1{c}_3>0$, the long-wave stable carrier band isFor $c_1=c_3=0$ this reduces to the real-equation Eckhaus band. If $1+c_1{c}_3<0$, even the zero-detuning oscillation has the Benjamin-Feir instability. Nonlinear outcomes can include smooth phase modulation, defect production or phase turbulence, depending on the equation and parameters.

The general nonlinear-pattern problem starts from a periodic solution $u_0(x)$ of the full field equations, not from a small-amplitude envelope. Linearize about it to obtain an operator $L$ with periodic coefficients. Write a disturbance as $e^{\lambda t+iq\cdot x}v(x)$ with $v$ periodic on the pattern cell. The Bloch stability operatoracts on this fixed cell and determines the entire sideband spectrum. Translation covariance gives $L(0){\partial }_x{u}_0=0$ in one dimension, and corresponding neutral directions for each independent broken translation in several dimensions. Oscillatory base patterns require the analogous Floquet theory calculation over their temporal period, or a rotating frame where applicable.

For a simple translational zero eigenvalue, the phase-diffusion calculation can be made explicit. Expandwhere $v_0=u_0^{\prime }$ and $L_0{v}_0=0$. Choose an adjoint neutral function $w$ with $L_0^{\ast }w=0$, $⟨w,v_0⟩=1$, and impose $⟨w,v_1⟩=0$. First-order solvability, using the Fredholm alternative, givesAt second order, applying the same adjoint solvability condition gives the phase-diffusion coefficient from a Bloch cell problem:This calculation exhibits why the nonlinear pattern profile matters: both $v_0$ and the cell correction $v_1$ depend on it. A negative $D$ signals a long-wave instability provided the rest of the spectrum is stable. With several phase directions, or an extra neutral conserved mean, one obtains a coupled modulation system and diffusion matrices rather than this scalar formula. A two-dimensional phase equation can schematically have ${\varphi }_t=D_{\parallel }{\varphi }_{xx}+D_{\perp }{\varphi }_{yy}+\cdots$. Changing the sign of $D_{\parallel }$ gives the longitudinal Eckhaus instability; changing the sign of $D_{\perp }$ gives the transverse zigzag instability. For oscillatory waves, nonvariational magnitude-phase coupling gives the Benjamin-Feir instability.

These mechanisms extend beyond onset, but the near-onset numerical boundaries do not: their coefficients must be replaced by those obtained from the actual nonlinear pattern and its cell problems. Long-wave linear stability is also only one part of stability; finite-wavenumber eigenvalues, additional modes, defects, and boundary constraints can intervene. The unifying calculation is the splitting of translation-induced neutral modes at small modulation wavenumber, together with the coupling to the relaxing magnitude and any other slow fields.

Choose a reference circular orbit at radius $r_0$, rotate the frame with $\mathbf{\Omega }=\Omega (r_0){\mathbf{e}}_z$, and introduce local Cartesian coordinates $x=r-r_0$, $y=r_0(\varphi -\Omega t)$ and vertical coordinate $z$. On scales small compared with $r_0$, neglect curvature of the coordinate directions but retain the leading differential rotation. The background velocity relative to the rotating frame isThus $A=3\Omega /4$ for a Keplerian disk. The shearing sheet retains Coriolis acceleration and the linear tidal acceleration. The radial shearing-sheet tidal potential is $-2\Omega A{x}^2$, whose force $4\Omega Ax{\mathbf{e}}_x$ balances the background Coriolis acceleration. If vertical tidal gravity is included, its equilibrium contribution is balanced by a background pressure $p_0(z)$ and disappears when that equilibrium is subtracted.

Write the total velocity as $\mathbf{u}={\mathbf{u}}_0+\mathbf{v}$. In the rotating-frame magnetohydrodynamic momentum equation, background advection is $({\mathbf{u}}_0\cdot \nabla )\mathbf{v}=-2Ax{\partial }_y\mathbf{v}$ and advection of the background by the perturbation is $(\mathbf{v}\cdot \nabla ){\mathbf{u}}_0=-2A{v}_x{\mathbf{e}}_y$. The uniform-viscosity diffusion of the linear background shear is zero. For a solenoidal magnetic field the Lorentz force per unit mass isAt constant density, absorb the magnetic-pressure gradient into $\psi =(p-p_0)/\rho +B^2/(2{\mu }_0\rho )$. Subtracting the background force balance then gives the stated momentum equation, with the remaining nonlinear advection $\mathbf{v}\cdot \nabla \mathbf{v}$, Coriolis acceleration $2\mathbf{\Omega }\times \mathbf{v}$, magnetic tension and viscous diffusion.

For constant finite electrical conductivity ${\sigma }_e$, put $\eta =({\mu }_0{\sigma }_e{)}^{-1}$. The resistive induction equation with uniform magnetic diffusivity is ${\partial }_t\mathbf{B}+\mathbf{u}\cdot \nabla \mathbf{B}=\mathbf{B}\cdot \nabla \mathbf{u}+\eta {\nabla }^2\mathbf{B}$. Substituting the background shear supplies the term $-2A{B}_x{\mathbf{e}}_y$ and the same background-advection operator as in the momentum equation. The two further conditions areThey express incompressibility and the absence of magnetic monopoles, respectively.

Now take real plane waves with common phase $\vartheta =\mathbf{k}(t)\cdot \mathbf{x}$ and real wavevector. The solenoidal conditions becomeEvery field depends on position only through $\vartheta$. Hence for either transverse vector field $\mathbf{P}$ and either wave field $\mathbf{Q}$, $(\mathbf{P}\cdot \nabla )\mathbf{Q}=(\mathbf{P}\cdot \mathbf{k}){\partial }_{\vartheta }\mathbf{Q}=0$. This proves that all the displayed quadratic advection and tension terms vanish pointwise, including products involving conjugate amplitudes. These waves are therefore exact solutions at finite amplitude, not merely a linearization.

Acting with $D_0={\partial }_t-2Ax{\partial }_y$ on the phase yieldsCanceling its position-dependent part requiresThe remaining amplitude equations areDifferentiating the velocity constraint determines the pressure amplitude:The resistive induction equation preserves its constraint because $\dot{\mathbf{k}}\cdot \tilde{\mathbf{B}}=2A{k}_y{\tilde{B}}_x$ cancels the corresponding shear term in $\mathbf{k}\cdot \dot{\tilde{\mathbf{B}}}$. The specified magnetic field has no spatially uniform part. Thus there is no background magnetic field producing an additional linear magnetic-tension coupling. The scalar $\psi$ is the modified pressure; the physical pressure can include a constant and a second spatial harmonic from $-B^2/(2{\mu }_0)$ while $\psi$ retains the stipulated plane-wave form. This is an exact zero-mean magnetic shearing wave.

For decay, take the real part of the scalar product of the velocity-amplitude equation with its complex conjugate. Pressure does no work because $\mathbf{k}\cdot \tilde{\mathbf{v}}=0$, and the real part of Coriolis work is zero. ThereforeThe resistive induction equation similarly givesusing $2|u_x{u}_y|\le |\mathbf{u}{|}^2$. Integrating these differential inequalities, defineThenA non-axisymmetric wave has $k_y\ne 0$. For nonzero differential shear and positive $\nu ,\eta$, the negative cubic terms dominate the linear upper bound on shear work. ConsequentlyThis cubic-exponent decay of a nonaxisymmetric shearing wave allows transient energy growth before the winding to short radial wavelengths makes diffusion dominant. The magnetic amplitudes can also be written explicitly aswhich confirms the damping independently. Positivity of the dissipative coefficients is necessary: with $k_z=0$, an inviscid vertical velocity amplitude can remain constant, and an ideal vertical magnetic amplitude can remain constant when $\eta =0$. The asserted decay applies to the viscous, finitely conducting model; taking an ideal limit before the long-time limit changes the conclusion.

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.

Because $s$ is arc length, $|d\mathbf{x}/ds|=1$. Thus $\mathbf{y}=\mathbf{n}/\alpha$, where $\mathbf{n}$ is the unit tangent to the elastic ray. In an isotropic medium this is also the wavefront normal. Consequently $\mathbf{y}$ is the vector slowness, of magnitude $1/\alpha$; locally $\mathbf{y}=\nabla T=\mathbf{k}/\omega$, with $T$ the travel time and $\mathbf{k}$ the wavevector.

For depth-dependent wave speed $\alpha (z)$, the ray equation gives $d{y}_x/ds=d{y}_y/ds=0$. The horizontal component of $d\mathbf{x}/ds$ therefore has a fixed direction, and the elastic ray lies in the vertical plane determined by that direction and its initial point. A vertical elastic ray is the limiting case. If $i$ is the angle to the vertical, its conserved horizontal slowness givesAt a horizontal interface, matching the tangential wavevector at fixed frequency gives the same invariant on both sides: this is Snell's law.

For radial wave speed $\alpha (r)$, take the cross product of position with slowness. The ray equations givesince the second vector is radial. A nonzero constant $\mathbf{L}=\mathbf{x}\times \mathbf{y}$ is perpendicular to every position on the elastic ray, so the ray lies in a fixed plane through the centre. Radial rays obey the same conclusion with a nonunique plane. The spherical elastic ray invariant is thereforewhere $i$ is measured from the radial direction. Across a spherical interface the radius is common and tangential slowness is continuous, so this form of Snell's law survives an interface discontinuity.

In a homogeneous sphere a P wave follows the straight chord between the two surface points. If their principal angular separation is $0\le \Delta \le \pi$, its length is $2R\sin (\Delta /2)$. For constant P wave speed $c_P$,Now put $a=R^{\prime }$ and denote the mantle and core P wave speeds by $c_m>c_c$. Equal incidence and reflection angles put the core-boundary reflection point on the angular bisector. Each mantle leg has length $L$ given by $L^2=R^2+a^2-2Ra\cos (\Delta /2)$. Hence the core-reflected travel time isThe stated domain is essential. At the reflection point the incoming line approaches from the mantle only if $R\cos (\Delta /2)\ge a$. Otherwise it has already crossed the core boundary. The mantle-only direct chord also exists only for $\Delta \le {\Delta }_g$, with $T_D=2R\sin (\Delta /2)/c_m$. In this interval,so the reflected arrival is later. Its intercept is $2(R-a)/c_m$, whereas the direct intercept is zero. At grazing incidence the branches meet at $T=2\sqrt{R^2-a^2}/c_m$, and both have slope $dT/d\Delta =a/c_m$. A continuation of the homogeneous-sphere direct formula beyond this endpoint is not a mantle-only arrival in the modified model.