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For the real scalar field, the mostly-plus Minkowski metric gives a positive kinetic energy. Its canonical momentum and Hamiltonian operator areThe Euler-Lagrange field equation is $(\square -m^2)\varphi =0$, or $({\partial }_t^2-{\nabla }^2+m^2)\varphi =0$. Canonical quantization of a real scalar field promotes the fields to Hermitian operators and replaces the equal-time Poisson brackets by canonical commutation relations:The Heisenberg equation of motion then gives $\dot{\varphi }=\pi$ and $\dot{\pi }={\nabla }^2\varphi -m^2\varphi$, so the classical field equation remains an operator identity. Products at coincident points need a regulator; equivalently, the field is an operator-valued distribution.

Resolve the free field into positive- and negative-frequency plane waves. With $E_{\mathbf{k}}=\sqrt{|\mathbf{k}{|}^2+m^2}$ and $k\cdot x=-E_{\mathbf{k}}t+\mathbf{k}\cdot \mathbf{x}$,Reality pairs the two terms by Hermitian conjugation. The equal-time canonical commutation relations are equivalent toFor example, the two mixed terms in $[\varphi ,\pi ]$ each supply half the Dirac delta function. Thus the normalization $1/\sqrt{2E_{\mathbf{k}}}$ is fixed, not optional. Each independent mode is a quantum harmonic oscillator, with $a$ and $a^{†}$ its annihilation operator and creation operator.

Choose the Fock vacuum by $a(\mathbf{k})|0⟩=0$. Repeated creation operators construct the bosonic Fock space. Substitution into the Hamiltonian operator yieldswhere a finite volume $V$ can be used as a regulator. Normal ordering sets the flat-space vacuum reference to zero. The spatial momentum operator isIn particular, $[H,a^{†}(\mathbf{k})]=E_{\mathbf{k}}{a}^{†}(\mathbf{k})$ and $[\mathbf{P},a^{†}(\mathbf{k})]=\mathbf{k}{a}^{†}(\mathbf{k})$. A one-particle state therefore has positive energy, momentum $\mathbf{k}$, and invariant mass $m$. A scalar field transforms in the trivial spin representation, so its particles have spin zero. Because the field is real there is only one set of oscillators: the particle is its own antiparticle, with no distinct conserved particle-minus-antiparticle charge. The free theory does conserve its occupation-number sum, though generic scalar interactions need not.

A momentum eigenstate extends throughout space. A localized particle is described by a superposition such asIts wave packet evolves through the phase $e^{-i{E}_{\mathbf{k}}t}$; a narrow packet has group velocity $\mathbf{v}=\mathbf{k}/E_{\mathbf{k}}$, whose magnitude is at most one. Localization and dispersion concern these superpositions, rather than classical trajectories attached to individual field modes.

Relativistic spacetime behavior is encoded by microcausality. Directly from the oscillator commutators,This is a Lorentz-invariant distribution and vanishes at equal time. Any spacelike separation can be transformed to an equal-time separation, so $[\varphi (x),\varphi (y)]=0$ for spacelike $x-y$. Thus local operations at spacelike-separated points commute. The vacuum Wightman function need not vanish outside the light cone, but that correlation does not transmit a controllable signal. The Feynman propagator iswhich propagates positive energy forward and negative energy backward in the time-ordered product. For causal response one uses the retarded Green function, whose support lies in or on the future light cone.

Finally, commuting creation operators implyThe multiparticle state is symmetric under the particle exchange operator, which is Bose statistics. For a normalized single mode, the occupation states are $(a^{†}{)}^n|0⟩/\sqrt{n!}$ with $n=0,1,2,\dots$: any number of identical bosons can occupy it. There is no Pauli exclusion principle for these particles. This construction agrees with the Spin-statistics theorem relating integer spin to bosonic exchange symmetry under relativistic locality and positive-energy assumptions. It demonstrates the required scalar case without assuming that theorem as the quantization prescription.

Take $z$ positive upward, and write $b=\tilde{\rho }g$. For a straight steady viscous buoyant conduit, the inner and outer axial equations areRegularity gives $w_i^{\prime }(0)=0$, the wall has $w_o(R)=0$, and velocity and tangential traction are continuous at $r=a_0$: $w_i=w_o$, $\lambda \mu w_i^{\prime }=\mu w_o^{\prime }$. With no interfacial tension specified, normal traction is continuous; for this straight steady flow that sets $P_i=P_o$ at the interface. Thus the modified gradients are common in this case.

When their value is zero, integration givesConsequentlyThe large center-to-interface speed ratio explains why the inner parabolic component dominates at small viscosity ratio.

For a slowly varying radius, set $B=b-P_{i,z}$. The local inner solution has $w_i=w_a+B(a^2-r^2)/(4\lambda \mu )$, soThe shear-driven outer interface speed has scale $a^{2}B\ln (R/a)/\mu$. The pressure-driven return flow needed for zero total flux adds a velocity of order $q/R^2$ and a shear-driven contribution of order $a^{2}B/\mu$. Relative to the inner parabolic velocity these terms are $O(\lambda \ln (R/a))$ and $O((a/R{)}^2)$, which are small under the stated assumptions. Axial viscous derivatives are also smaller because the plume is slender. ThereforeFor clarity, the outer pressure-gradient estimate must include both return flux and transmitted interface shear. Their flux scales are $R^4{P}_{o,z}/\mu$, $a^2{R}^{2}B/\mu$ and $q$. Zero total flux hence impliesThis justifies the negligible outer pressure gradient. We subsequently choose its negligible pressure reference as zero, as in the reduced model.

Near the moving interface the outer radial source flow is $u_r=a{a}_t/r$: the interface's axial advection is smaller because its axial velocity is small compared with the inner core speed. Its radial strain is ${\partial }_r{u}_r=-a{a}_t/r^2$. Continuity of normal traction at $r=a$, neglecting the much smaller inner normal viscous stress, givesWith $A=a^2/a_0^2$, this is $P_i=\mu A_t/A$ after dropping $P_o$. Conservation of plume volume is $\pi (a^2{)}_t+q_z=0$.

Let the axial and velocity scales beSince $\mu U_0/(b{\ell }^2)=1$, substitution gives the conduit equationThe buoyant flux and the viscous normal-stress pressure are both retained at this scaling.

Linearizing about unit area gives $a_T+2a_Z-a_{TZZ}=0$. ThusWave crests propagate upward. The group velocity is $2(1-k^2)/(1+k^2{)}^2$: it is downward for $|k|>1$, so the phase direction is not the direction of every wave packet.

For a travelling wave $f(\zeta )$, integrating once and using its uniform far field givesDivide by $f^2$, multiply by $f^{\prime }/f$ and integrate again. A convenient potential, defined up to an additive constant, isAt the positive crest $f=\alpha f_0$, the derivative is zero. Subtracting the far-field potential yieldsThis is the solitary-wave amplitude-speed relation for the conduit equation. A nontrivial elevation wave has $\alpha >1$ and $c>2f_0$; as $\alpha \downarrow 1$, the speed tends to the long-wave speed $2f_0$. Positive area is required throughout the derivation.

Rotating Rayleigh-Bénard convection combines buoyancy-driven instability with the Coriolis force. Consider a plane layer of depth $d$ rotating uniformly about the vertical axis, heated from below. Adopt the Boussinesq approximation, fixed boundary temperatures and, for explicit formulas, impermeable stress-free boundary conditions. The conductive state of Rayleigh-Bénard convection is motionless with a linear temperature profile. The dimensionless controls are the Rayleigh number, Prandtl number and Taylor number:Here $\nu$ is kinematic viscosity and $\kappa$ thermal diffusivity. In thermal-diffusion time units, linear perturbations satisfyRotation does no direct mechanical work, since $\mathbf{u}\cdot ({\mathbf{e}}_z\times \mathbf{u})=0$, but couples vertical motion to vertical vorticity and changes the damping and oscillation balance.

For horizontal wavenumber $k$ and vertical mode $n$, put $m=n^2{\pi }^2$ and $a=k^2+m$. Use $w=W\sin (n\pi z)$, $\theta =\Theta \sin (n\pi z)$ and vertical vorticity $\zeta =Z\cos (n\pi z)$ with growth rate $\sigma$. Curling the momentum equation and eliminating pressure givesTheir determinant, without dividing by a possibly zero factor, is the rotating-convection growth-rate polynomialThis makes the linear stability analysis question precise: onset occurs when a root reaches zero real part and all other modes still decay.

A stationary neutral root has $\sigma =0$, givingRotation raises this stationary threshold. An oscillatory neutral root has $\sigma =i\varpi$ with $\varpi \ne 0$. Real-imaginary separation gives the oscillatory neutral curve of rotating convectionThis branch is admissible only when ${\varpi }^2>0$, requiring $P<1$ and sufficiently strong rotation. The restoring Coriolis force coupling permits an inertial/thermal oscillation whose phase-lagged buoyancy can overcome dissipation. At large $P$, temperature and momentum diffusion do not permit that overstability mechanism at primary onset, so the exchange of stabilities is stationary. The actual threshold is the minimum of the stationary and admissible oscillatory curves over all allowed modes, not an arbitrary formal value of $R_o$. Rigid plates require a different vertical eigenproblem and Ekman layers, so the explicit free-slip numbers are not universal.

At large Taylor number, the first vertical mode is selected in the ideal plane layer. Let $t=k^2,m={\pi }^2$. Minimizing the stationary curve givesThus the stationary neutral curve of rotating convection hasThe physical horizontal wavelength is $2\pi d/k_c$, hence decreases as ${\mathrm{Ta}}^{-1/6}$; its prefactor depends on the boundary convention. Thin nearly vertical cells reconcile the strong Coriolis force constraint with viscous and thermal diffusion. The oscillatory minimization replaces the right side of the wavenumber equation by $P^2\mathrm{Ta}m/(1+P{)}^2$, giving the same Taylor number exponent at fixed positive $P$. Where its frequency remains admissible,Equality is $8P^4-P-1=0$, whose positive root is approximately $P_{\ast }=0.6766$. Accordingly, for sufficiently rapid rotation in this free-slip problem, $P<P_{\ast }$ selects oscillatory onset and $P>P_{\ast }$ stationary onset. The weaker condition $P<1$ is only necessary for an oscillatory neutral mode; it does not by itself identify the first instability. Finite Taylor number, finite lateral geometry, allowed discrete wave numbers and plate conditions change the selection.

For the counterpropagating Hopf amplitudes in rotating convection near a simple oscillatory onset, the Hopf bifurcation produces slow complex amplitudes for counterpropagating roll waves. After separating the fast carrier oscillation, symmetry permits the cubic equationswith generally complex coefficients; an $i{\omega }_0{Z}_{\pm }$ term restores the fast frequency if desired. The real parts govern amplitude saturation and the imaginary parts give nonlinear frequency shifts. Write $a_s=\mathrm{Re}{g}_s$, $a_c=\mathrm{Re}{g}_c$. For a travelling wave from a supercritical bifurcation with only one amplitude nonzero, $|Z_{\pm }{|}^2=r/a_s$ with $r,a_s>0$, and the competing wave's growth rate is $r(1-a_c/a_s)$. It is amplitude-stable against that competitor when $a_c>a_s$. A standing wave has equal intensities $r/(a_s+a_c)$; provided this is positive, its intensity-difference mode is stable when $a_s>a_c$. Temporal and spatial phase symmetries leave neutral phase directions, so these are orbital/amplitude stability statements, not decay of every phase displacement.

These coefficients follow from nonlinear interactions and the Fredholm solvability condition obtained by projection onto an adjoint eigenfunction; symmetry alone cannot decide their signs. A negative saturating coefficient gives subcritical bifurcation behavior requiring higher-order terms. Spatial modulation leads to coupled complex Ginzburg–Landau equations with group velocities and diffusion; phase instabilities, mean-flow coupling and differently oriented rolls can destabilize a wave stable in the restricted two-amplitude system. A weakly nonlinear expansion of oscillations therefore predicts travelling waves or standing waves, frequency shifts, modulation and possible secondary mode competition, not a unique universal periodic state.

The Küppers–Lortz instability is a different route to time dependence: it destabilizes steady saturated rolls against oblique roll perturbations. For stationary-roll amplitudes of orientations ${\varphi }_j$, a leading competition system hasA pure roll has $|A_i{|}^2=r/g_0$ with $g_0>0$. An infinitesimal new roll at relative angle $\vartheta$ grows atFor sufficiently strong rotation in appropriate boundary and Prandtl number regimes, some finite oblique angle has $g(\vartheta )<g_0$, so a steady roll is unstable arbitrarily close above its stationary onset. Rotation is handed and allows $g(\vartheta )\ne g(-\vartheta )$, so replacement of one roll by another can favor a definite cyclic sense. Three or more competing orientations can form a heteroclinic cycle; whether it attracts depends on contraction/expansion rates and other modes. Noise, spatially varying domains and modulation can turn this competition into repeated orientation switching and irregular patterns.

The invading rolls are three-dimensional disturbances even when the original straight roll is described by a two-dimensional section. The finite-angle Küppers–Lortz instability mechanism should also be distinguished from the small-angle instability of rotating convection rolls mediated by large-scale mean flow at finite Prandtl number. Numerical thresholds and favored angles depend on mechanical boundaries and material parameters; the essential criterion is the cross-coupling relative to self-saturation. Rotation both changes primary onset and wavelength, and can prevent the resulting steady roll pattern from remaining a stable nonlinear state.

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.

For a two-dimensional internal gravity wave with fixed horizontal wave number $k\ne 0$, the vertical velocity amplitude satisfiesThis follows from the Linearized Boussinesq equations just as in part (a), allowing the background buoyancy frequency to vary. The WKB approximation gives locallyIt requires slowly varying stratification compared with the local vertical wavelength, in particular $|m_z|\ll m^2$, and a small-amplitude wave in the inviscid, nonrotating regime. Propagation requires $N>|\omega |$; where $N<|\omega |$, the local vertical wave number is imaginary and the solution is evanescent. Turning levels and abrupt changes need a separate connection calculation. Finite viscosity, background flow, rotation or nonlinear effects require modifications of this model.

The ray tracing equations use the local dispersion relation as a Hamiltonian:The stationary, horizontally uniform background conserves the frequency and horizontal wave number. In two dimensions the group velocity gives $dx/dz=-m/k$. For the specified buoyancy frequency and $\omega =1$, $m^2=256k^2{z}^2$. Hence for $z>0$ the two geometric internal-wave rays through the specified point areEach curve has both directed propagation senses, giving four local directed rays. Their formal intersections with $z=0$ are $x=-8$ and $x=8$.

At $z=0$, $N=\omega$ and $m=0$: this is a double-zero internal-wave turning level, since $m^2$ touches zero without changing sign. The ray becomes vertical geometrically, but its group velocity tends to zero. Taking $k>0$ without loss of generality, the downward branch hasThe formal travel time diverges logarithmically as $z\to 0$. Meanwhile $|m_z|/m^2=1/(16|k|z^2)$ grows, so the WKB approximation fails before that limit; its predicted $|m{|}^{-1/2}$ amplitude cannot be extrapolated to infinity. The spatial ray reaches a finite limiting position, but finite-wavelength behaviour near the zero must be found beyond WKB. There is no evanescent half-space supplied by this particular $N(z)$, so ordinary simple-turning-point reflection cannot be assumed solely from the local ray construction.

On a sphere of radius $R$ rotating at angular speed $\Omega$, the Coriolis parameter is $f=2\Omega \sin \varphi$. Near latitude ${\varphi }_0$, write $y=R(\varphi -{\varphi }_0)$, use locally eastward and northward Cartesian coordinates, and expandThe beta plane retains the first northward variation of planetary vorticity while neglecting higher latitude dependence and metric curvature. A midlatitude local calculation assumes $L/R\ll 1$ and $|\beta |L/|f_0|\ll 1$; near the equator the distinct equatorial beta plane has $f_0=0$, so the midlatitude low-frequency reduction used below does not apply.

For a homogeneous shallow layer, the hydrostatic approximation gives $p=p_{\mathrm{atm}}+\rho g(\eta -z)$ and hence $p_x/\rho =g{\eta }_x$, $p_y/\rho =g{\eta }_y$, independently of depth. The approximation follows from the small aspect ratio $H/L$ and neglect of vertical acceleration. Differentiate the linear horizontal momentum equations in $z$. Their depth derivatives obeyStarting from rest gives $u_z=v_z=0$ initially, and the unique solution remains zero. This establishes depth independence of hydrostatic shallow-water flow; an arbitrary pre-existing shear would not be removed merely by taking the hydrostatic approximation. Integrating the continuity equation between the rigid bottom and the moving surface gives ${\eta }_t+H(u_x+v_y)=0$ at linear order. With the depth-integrated shallow-water transports $U=Hu$, $V=Hv$, one obtains

It is useful to make the coefficient approximation in the height reduction explicit. Put ${\mathcal{F}}_y={\partial }_t^2+f(y{)}^2$. Differentiating the two momentum equations in time and eliminating the other transport givesSince ${\partial }_y({\mathcal{F}}_{y}V)={\mathcal{F}}_y{V}_y+2f\beta V$, taking their horizontal divergence and using the continuity equation givesApplying ${\mathcal{F}}_y$ proves the exact variable-Coriolis shallow-water height equationAt the reference latitude, or after the usual local freezing of undifferentiated $f$ factors to $f_0$, this gives the height relation written with ${\mathcal{F}}_0={\partial }_t^2+f_0^2$. With $f=f_0+\beta y$ over a finite region, that constant-coefficient version is a local approximation, not an exact identity. Keeping ${\mathcal{F}}_y$ as above avoids silently commuting a variable Coriolis parameter through a spatial derivative.

For the slow Rossby wave branch, take $|\omega |/|f_0|\ll 1$, approximate ${\mathcal{F}}_0$ by $f_0^2$, and discard the two time-derivative terms on the right compared with $-f_0^2{\eta }_x$. This is the regular midlatitude long-time ordering, with nondegenerate zonal variation. If $a=f_0/c$, the result isThe printed low-frequency equation has the opposite right-hand sign. The minus sign follows directly from the preceding eliminated equation and is also the sign needed for the printed isofrequency-circle centre. An independent check uses geostrophic balance: $\psi =g\eta /f_0$, $(u,v)=(-{\psi }_y,{\psi }_x)$, and the shallow-water quasi-geostrophic potential vorticity is ${\nabla }_h^2\psi -a^2\psi +\beta y$. Linear potential-vorticity conservation gives $({\nabla }_h^2\psi -a^2\psi {)}_t+\beta {\psi }_x=0$.

For $\eta =A{e}^{i(kx+ly-\omega t)}$, the shallow-water Rossby-wave dispersion relation isFor fixed $l$ and $\beta >0$, it is odd in $k$, zero at $k=0$, negative for $k>0$, and tends to zero from below as $k\to +\infty$. Its minimum occurs at $k=\sqrt{l^2+a^2}$ with ${\omega }_{\min }=-\beta /[2\sqrt{l^2+a^2}]$. Thus both long and short waves have small frequency. The zonal phase velocity is westward, $\omega /k<0$, whereas the zonal group velocity isIt changes sign at the frequency minimum.

At fixed nonzero frequency, completing the square yields the Rossby-wave isofrequency circleThe radius is real only if $|\omega |\le |\beta |/(2|a|)$. For $\beta >0$, the branch with $k>0$ has $\omega <0$ and a circle centred on the positive $k$-axis. If instead the printed plus-sign wave equation is taken literally with this same Fourier convention, its dispersion is $\omega =+\beta k/(k^2+l^2+a^2)$ and its circle is centred at $(+\beta /(2\omega ),0)$. These two conventions cannot be mixed.

For reflection of a Rossby wave at a meridional wall, the stationary wall preserves frequency, and its translation invariance in $y$ preserves the tangential wavenumber. Thus ${\omega }_r={\omega }_i$, $l_r=l_i$. Both $k$ values solveTheir sum and product giveAn incident wave in $x>0$ travels toward the wall in group velocity, not necessarily in phase velocity. For $\beta >0$, $0<k_i<\sqrt{l_i^2+a^2}$ has $c_{gx,i}<0$, while its partner $k_r>\sqrt{l_i^2+a^2}$ has $c_{gx,r}>0$. The double-root case has zero normal group velocity and does not describe a wave packet incident on the wall.

At leading quasi-geostrophic approximation, the impermeability condition is $u=-g{\eta }_y/f_0=0$. For $l_i\ne 0$, the boundary condition givesThe reflected height therefore has equal amplitude and a phase change of $\pi$. Equality here concerns the height or quasi-geostrophic streamfunction amplitudes in the reduced model. At the degenerate $l_i=0$, geostrophic no-normal-flow is automatically satisfied and alone does not determine their ratio; the usual homogeneous wall-streamfunction condition supplies $A_r=-A_i$ if imposed. Retaining the small ageostrophic transport changes the boundary condition towhich tends to $-1$ in the regular low-frequency ordering but is not generally of unit modulus. In particular, $l_i=0$ gives $A_r/A_i=-k_i/k_r$ in that more complete boundary relation. Thus the printed equal-amplitude assertion requires the nondegenerate leading quasi-geostrophic approximation, or an explicit homogeneous wall condition; it is not a general exact shallow-water reflection law.

The linear elastic wave equation is $\rho {\mathbf{u}}_{tt}=(\lambda +\mu )\nabla (\nabla \cdot \mathbf{u})+\mu \Delta \mathbf{u}$. For a shear-horizontal wave, take $\mathbf{u}=(0,v(x,z,t),0)$ with no $y$ dependence. Its divergence vanishes, andRigid fixed boundaries require $v=0$ at $z=0,h$. The modes are $v=A\sin (q_{n}z)\cos (kx-{\omega }_{n}t)$ with $q_n=n\pi /h$, $n=1,2,\dots$, andThere is no $n=0$ displacement mode with these clamped boundary conditions. For $k>0$, the phase velocity and group velocity areThe phase velocity decreases from infinity to $c_s$, while the group velocity increases from zero to $c_s$; their product is $c_s^2$.

For a mode of frequency $\omega$, the appropriate time and cross-sectional average is $⟨A⟩=\omega /(2\pi h){\int }_0^{2\pi /\omega }{\int }_0^{h}Adzdt$, per unit transverse width. Its average kinetic energy density is $\rho A^2{\omega }^2/8$. The only nonzero strains are $e_{xy}=e_{yx}=v_x/2$ and $e_{yz}=e_{zy}=v_z/2$, while $e_{kk}=0$. Thus the elastic energy density is $\mu (v_x^2+v_z^2)/2$, whose average is $\mu A^2(k^2+q_n^2)/8$. The dispersion relation proves equality of average kinetic and elastic energies.

For the ray asymptotics of a clamped elastic waveguide mode, expand the localized initial displacement in transverse sine modes and Fourier transform in $x$. For a displacement released from rest, each mode evolves with $\cos ({\omega }_n(k)t)$. At $x=Vt$ the oscillatory phases are $t(kV\pm {\omega }_n(k))$. If $0<|V|<c_s$, one phase has a stationary point withThe stationary phase method gives a generic oscillatory amplitude proportional to $t^{-1/2}$ for each contributing mode. If the initial Fourier coefficient vanishes at the stationary point, that mode can decay faster; smooth localized data make the modal sum well behaved. If $|V|>c_s$, there is no stationary point. Repeated integration by parts gives rapid decay for smooth rapidly decaying data; for initially compactly supported displacement, finite propagation speed makes the displacement on that ray exactly zero after a sufficiently long time. These are the requested subsonic and super-shear-speed regimes.

For the local phase $S$, define $k_i={\partial }_{i}S$ and $\omega =-{\partial }_{t}S$. The dispersion relation is the Hamilton-Jacobi equation $S_t+\Omega (\nabla S,\mathbf{x},t)=0$. Differentiate it in space and follow a ray with group velocity ${\dot{x}}_i={\Omega }_{k_i}$; symmetry of the phase Hessian cancels the terms involving ${\partial }_i{k}_j$. Thus the ray-tracing equations areHere $d/dt={\partial }_t+\dot{\mathbf{x}}\cdot \nabla$ means differentiation along the moving ray; subscripts on $\Omega$ mean partial derivatives holding its other independent arguments fixed.

The depth is stationary and independent of $y$, so $\omega ={\omega }_{\infty }$ and $l$ are constant. Choose the incident direction with positive $l$; the other sign reflects the picture. In the far deep water,Since $\Omega$ depends isotropically on the horizontal wavevector, the group-velocity ratio is $l/k$. Therefore the ray equation isNear shore, $\kappa h\to 0$, so the shallow-water dispersion relation gives $\kappa \sim {\omega }_{\infty }/(\sqrt{g\alpha }{x}^{p/2})$ and $k\sim -\kappa$. Integrating the ray slope yieldsFor $p<2$, the near-shore phase is $S\sim ly-{\omega }_{\infty }t-{\omega }_{\infty }{x}^{1-p/2}/[\sqrt{g\alpha }(1-p/2)]$. A constant-phase wave crest thus hasIts tangent becomes parallel to the shoreline, while rays approach perpendicular to it. The exponent is between zero and one, giving the curved crests sketched below for $p=1$.

These are the formal geometrical optics asymptotics. For fixed $\alpha$ and $p<2$, $\kappa x\to 0$ arbitrarily close to the shore, so the slowly varying wave approximation eventually fails there; the sketch describes the ray-theory regime before that final inner region.

Take $ck\gg \Omega$, retaining a finite low-frequency branch and a nonzero wavevector. In the dispersion relation, neglect ${\omega }^4$ against $c^2{k}^2{\omega }^2$ and neglect ${\Omega }^2$ against $c^2{k}^2$ in their coefficient. The inertial wave limit isChoose the signed branches $\omega =ϵ\Omega k_z/k$, $ϵ=\pm 1$. The phase velocity normal to the phase planes and the group velocity areTherefore $\mathbf{k}\cdot {\mathbf{c}}_g=ϵ\Omega (k_z/k-k_z{k}^2/k^3)=0$, soThe geometric reason is that this limiting dispersion relation is homogeneous of degree zero in $\mathbf{k}$: its frequency depends on direction, not magnitude. Euler homogeneous function theorem gives $\mathbf{k}\cdot {\nabla }_{\mathbf{k}}\omega =0$. Exact orthogonality holds for this inertial wave limit; finite compressibility introduces corrections. At a zero-frequency degeneracy the usual propagating-wave interpretation needs a limiting treatment.

The WKB approximation treats the vertical wavenumber as locally constant over one vertical wavelength. Write $w=\Re \{A(z)e^{ikx+i{\int }^{z}m(s)ds}\}$, with $|m^{\prime }|\ll m^2$ and $|A^{\prime }|\ll |mA|$. For the stationary internal-wave WKB solution, $A\propto m^{-1/2}$ away from singular or turning levels. A small terrain slope and small perturbations are also needed for Linearized Boussinesq equations.

Let $p$ be the perturbation fluid pressure divided by reference mass density, and $b$ the buoyancy perturbation. The steady Linearized Boussinesq equations and incompressibility condition areFor an $e^{ikx}$ component, elimination gives $u=-w^{\prime }/(ik)$, $b=-N^{2}w/(ikU)$ and $p=(U{w}^{\prime }-U^{\prime }w)/(ik)$. Substitution into vertical momentum gives the exact stationary Taylor–Goldstein equationThe leading WKB approximation therefore identifiesThe kinematic boundary condition fixes the horizontal wavenumber to $k=k_T$ for the sinusoidal forcing.