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The coordinate derivatives of the smooth two-parameter map commute. With a torsion-free Levi-Civita connection,because ${\Gamma }^a{}_{bc}$ is symmetric in its lower indices. Each member of the family is a geodesic, so $DV/Ds=0$. ConsequentlyThe commutator of covariant derivatives is the action of the Riemann curvature tensor. With the source's index and sign convention this givesThis is geodesic deviation: free fall relative acceleration measures curvature, even where an individual observer can eliminate the connection at one point.

For Earth's exterior Newtonian field, $\varphi =-GM/r$ and acceleration is $a_i=-{\partial }_i\varphi$. Nearby freely falling particles have relative acceleration $\delta a_i=-{\varphi }_{,ij}{Y}^j$. At the observation point align $z$ radially upward and $x,y$ horizontally. Since $GM/R_E^2=g$,The weak-field geodesic-deviation equation therefore identifiesIn these units time is measured in seconds as in the stated Newtonian comparison; if the zeroth coordinate is $ct$ instead, the curvature components acquire the corresponding $c^{-2}$ factor. A small freely falling cloud is squeezed in the two horizontal directions and stretched radially. These are tidal effects, not the uniform acceleration removed by a freely falling frame.

Their characteristic inverse-square time is $g/R_E$, so $\boxed{t_{\mathrm{tide}}=\sqrt{R_E/g}\simeq 8.1\times 10^2\mathrm{s}\simeq 13.4\min }$. One concrete thought experiment is an ideal satellite moving in a circular orbit just above a spherical Earth, ignoring atmosphere and terrain. Its orbital angular frequency is $\sqrt{GM/R_E^3}=\sqrt{g/R_E}$ and its period is $2\pi t_{\mathrm{tide}}\simeq 84$ minutes. Equivalently, initially nearby free fall trajectories develop order-one tidal changes on this characteristic scale. This is the surface tidal time scale of a spherical gravitating body.

The characteristic roots satisfyThey are a nonreal conjugate pair precisely when ${\mu }^{1/2}/9-16{\mu }^{7/6}<0$, or ${\mu }^{2/3}>1/144$. ThereforeIn lake-coupled Röthlisberger drainage instability, above this value the real part is ${\mu }^{1/4}/6>0$, so nonzero linear perturbations are sinusoidal oscillations with an exponentially growing envelope. Their angular frequency is $[4{\mu }^{7/6}-{\mu }^{1/2}/36{]}^{1/2}$.

Physically the feedback represents repeated lake filling and pressure increase, channel enlargement by melting, rapid drainage, pressure reduction and material creep closure. Large drainage events are glacial lake outburst floods, or jökulhlaups. The unbounded amplitude is a conclusion of the linearized equations, not a proof of infinite physical discharge or an indefinitely growing nonlinear solution. Once perturbations are comparable with the equilibrium scales, nonlinear dynamics and constraints such as lake depletion must be retained. At the threshold the two real roots coincide; below it the instability is nonoscillatory.

A spherical harmonic of angular degree $\ell$ has horizontal wavenumber $k_h=L/r$, with $L=\sqrt{\ell (\ell +1)}$. The local acoustic dispersion relation therefore gives the radial wavenumberIn the high-order regime a travelling stellar acoustic mode accumulates phase $\int k_{r}dr$ across its radial cavity. Reflection at the inner turning point and at the near-surface boundary fixes additional phases. A round trip must return the same complex amplitude, so WKB quantization givesHere $n$ counts radial oscillations, up to the choice of origin for radial order, and $\alpha$ incorporates both reflection phases. The outer acoustic reflection need not coincide exactly with the geometric stellar radius; replacing it by $R$ and absorbing its correction in $\alpha$ is the leading asymptotic approximation. The model neglects buoyancy, perturbations of self-gravity and rapid background variations on a wavelength.

Resolved observations of the Sun measure line-of-sight surface velocity through the Doppler effect, or brightness fluctuations. Projection on spherical harmonics identifies $\ell$ and azimuthal order $m$; a temporal Fourier transform identifies the mode angular frequencies $\omega$. The sequences of peaks of successive radial order form ridges in the frequency-degree diagram. Their ordering and comparison with a reference stellar oscillation model identify $n$; the surface image alone does not count inaccessible radial nodes. Fit the residual phase of the radial quantization relation across the observed mode set to estimate $\alpha$. A simultaneous fit with the sound-speed profile, or a calibrated reference model, is needed: $\alpha$ is not determined independently by one mode, and an integer shift of the numbering of $n$ shifts $\alpha$ oppositely. In a more accurate fit the surface phase depends on frequency. These steps constitute helioseismology.

Set $a(r)=c(r)/r$ and $w=\omega /L$. Assume $a$ decreases outwards in the region to be inverted. With the surface approximation above, the Duvall law isThus a suitably scaled mode set samples one function of $w$. Smooth the measured values before differentiating; inversion amplifies observational noise. At a finite outer reference point $r_s$, the already known contribution exterior to $r_s$ must be subtracted. For the remaining cavity, let $w_s=a(r_s)$ and $h(b)=-d\ln r/db$. Changing integration variable from radius to $b=a(r)$ givesThe boundary term vanishes because the square root vanishes at $b=w$. Now multiply the derivative by $w/\sqrt{a^2-w^2}$ and interchange the integrals. The kernel identityfollows by setting $w^2=b^2+(a^2-b^2){\sin }^2\theta$ and then integrating $1/[b^2{\cos }^2\theta +a^2{\sin }^2\theta ]$. It yields the Abel inversion of stellar acoustic travel times:Knowing the monotone relation $r(a)$ then gives $c=ar$. Observed turning points delimit the region recoverable without extrapolation; an unknown surface contribution or nonmonotone $a$ prevents this simple unique inversion.

There are genuine inconsistencies in the printed numerical example. Since $w$ and $w_0$ have dimensions of angular frequency, $F$ must have dimensions of time, but the displayed empirical right-hand side is dimensionless. Moreover its derivative supplies a factor $1/w_0$, whereas the requested inversion supplies $1/w_0^2$. The dimensionally repaired relation that produces the requested inversion integral isSubstitution in the boxed Abel transform inversion gives precisely the exponent coefficient $2a^2/(\pi w_0^2)$, with the square-root denominator inside the integral. This repairs the first normalization but does not repair the later algebra.

Indeed, put $x_s=w_s/a$ and $b=\sqrt{1-x_s^2}$. The required elementary integrals evaluate toThe finite-reference-point answer following from $F_c$ is thereforeIt still depends on $w_s$. The later printed profile suppresses that dependence, which requires the additional surface limit $w_s\to 0$, $r_s\to R$. Even in that limit it is not the result of this integral. To see this exactly, write $\mathcal{L}=\ln (R/r)$ and $y=a^2/(\pi w_0^2)$. The integral givesThe minus branch is selected by $a\to 0$ at the surface. Its decreasing-radius branch extends only to $\ln (R/r)=1/6$, where $a^2=\pi w_0^2/6$; it cannot describe $r=R/e$. The stated lower bound on the largest sampled $w$ does not fix this algebraic obstruction. In contrast, squaring and rearranging the printed closed profile would require $\mathcal{L}=18y-81y^2$. This differs identically from $2y-6y^2$, so the two printed steps cannot both follow from the same data.

The surface polytrope test can nevertheless be performed exactly, and clarifies the final discrepancy. For a plane-parallel polytropic atmosphere of index $m=3$, take $g=GM/R^2$ constant and $z=R-r$. The stellar hydrostatic equation and $p=K{\rho }^{4/3}$ give $4p/\rho =gz$, hence the surface sound speed of a polytropic star isThe exponent in the equilibrium polytropic equation of state need not equal the perturbation adiabatic exponent $\gamma$. For the corrected inversion above, expansion at $z/R\ll 1$ gives $c^2\simeq \pi w_0^{2}Rz/2$, so matching the surface polytrope fixes $w_0^2=\gamma GM/(2\pi R^3)$.

Alternatively, if the later printed closed profile is adopted as a separate imposed model, its surface expansion gives $c^2\simeq \pi w_0^{2}Rz/18$. It also has the correct square-root depth dependence, but requiresAt $r=R/e$ that imposed profile then givesThe claimed coefficient $\pi /3$ is inconsistent with this substitution; $1/e\ne \pi /3$. Thus the surface scaling is demonstrable, but the empirical normalization, the requested closed inversion and the final numerical coefficient cannot all be asserted as printed. These discrepancies are visible in the original PDF, not introduced by its TeX conversion.

Let $h$ be layer thickness and $g$ gravitational acceleration. The unforced shallow water equations on the equatorial beta plane areLinearize about $u=v=0$, $h=h_0$. Put $c_0=\sqrt{g{h}_0}$ and $L=\sqrt{c_0/\beta }$, and scale $(x,y)=L(X,Y)$, $t=(L/c_0)T$, $(u,v)=c_0(U,V)$ and $h=h_0(1+\eta )$. Dropping capitals, the linearized shallow water equations becomeThe choice $\beta L^2/c_0=1$ makes both the wave speed and the coefficient of the dimensionless Coriolis parameter equal to one.

Take Fourier modes $\{u(y),v(y),\eta (y)\}{e}^{i(kx-\omega t)}$ that decay as $y\to \pm \infty$. Their equations areFor $v=0$ and nonzero $k$, the first two require ${\omega }^2=k^2$. The root $\omega =k$ gives $u=\eta$, and meridional balance becomes $u^{\prime }=-yu$. Hence the equatorial Kelvin wave isIt travels eastward at speed $c_0$ in dimensional units. The other sign gives $\eta =-u$ and $u^{\prime }=yu$, so $u\propto e^{y^2/2}$ is not trapped.

For $v\ne 0$, $\omega \ne 0$ and ${\omega }^2\ne k^2$, inversion of the first two equations gives the equatorial wave velocity polarizationSubstitute in the remaining momentum equation. Multiplication by $({\omega }^2-k^2)/i$ leavessoWriting $v=e^{-y^2/2}H(y)$ yields $H^{\prime \prime }-2y{H}^{\prime }+(\lambda -1)H=0$, the Hermite differential equation. The trapped solutions have the Hermite oscillator quantization $\lambda =2n+1$ and $H=H_n$.

For completeness, quantization follows without assuming that every acceptable solution is a polynomial. On decaying smooth functions, define $a=d/dy+y$, $a^{†}=-d/dy+y$ and $\mathcal{H}=-d^2/d{y}^2+y^2=a^{†}a+1$. Integration by parts gives $\lambda \ge 1$ and $\Vert av{\Vert }^2=(\lambda -1)\Vert v{\Vert }^2$. If $av\ne 0$, it is an eigenfunction of $\mathcal{H}$ with eigenvalue $\lambda -2$. Repeated lowering must terminate: otherwise it eventually produces an eigenvalue below one. At termination $av=0$, so the terminal eigenfunction is $e^{-y^2/2}$ and has eigenvalue one. Repeated raising gives $H_n(y)e^{-y^2/2}$ and $\lambda =2n+1$; uniqueness of a decaying solution at one end of this second-order equation fixes each mode up to scale.

Thus, away from the exceptional quotients, all fields are determined byInsert this $v$ and its derivative in the polarization formulae to obtain $u$ and $\eta$. For $n\ge 1$, the cubic ${\omega }^3-(k^2+2n+1)\omega -k=0$ supplies the low-frequency equatorial Rossby wave and two equatorial inertia--gravity wave branches. Their meridional structures are Gaussian-weighted Hermite polynomials.

For $n=0$, multiplication by $\omega$ factorizes the equatorial shallow-water dispersion relation asThe physically admissible quadratic gives the mixed Rossby-gravity wave, equivalently the Yanai wave:These fields satisfy all three original equations directly and decay at both infinities, including where the quotient formulae are singular.

To test the other factor, set $\omega =-k\ne 0$. The original equations give $u+\eta =-iyv/k$ and $v^{\prime }=-yv$. Put $\delta =u-\eta$; meridional momentum then givesDecay of $u,\eta$ at both infinities requires the integral of the right side to vanish. Hence $C=0$ unless $k^2=1/2$. Ordinarily the extra factor supplies no trapped mode. At the exceptional root of the mixed Rossby-gravity mode, $k^2=1/2$, it coincides with the quadratic root and gives the same physical fields, not an additional branch. This distinction avoids discarding a regular physical mode at the crossing.

These arguments concern nonzero wave angular frequency. At $k=0$, the oscillatory nonzero-$v$ modes have $\omega =\pm \sqrt{2n+1}$, while zero-frequency $v=0$ geostrophic states must be considered in the original equations rather than in expressions divided by $\omega$.

Use the displacement-potential convention $u_x={\partial }_x\varphi -{\partial }_z\psi$, $u_z={\partial }_z\varphi +{\partial }_x\psi$ and harmonic dependence $e^{-i\omega t}$. Choosing the opposite sign for the shear potential changes its column sign but not the P-SV directional impedance matrix. Write $p=k_{\alpha }>0$, $s=k_{\beta }>0$, $D=k^2+ps$, and $H=\rho {\omega }^2-2\mu k^2$. At a common point the forward potential wave amplitudes giveThe traction formulas follow from ${\sigma }_{xz}=\mu (u_{x,z}+u_{z,x})$ and ${\sigma }_{zz}=\lambda \nabla \cdot \mathbf{u}+2\mu u_{z,z}$, using $(\lambda +2\mu ){\omega }^2/{\alpha }^2=\rho {\omega }^2$ and $\mu (s^2-k^2)=H$. The velocity matrix has determinant ${\omega }^{2}D>0$, so elimination gives the P-SV directional impedance matrixIts off-diagonal entries are real and antisymmetric. ThereforeWith real-part phasors, the outward $z$ flux is $-\frac{1}{2}\mathrm{Re}({\mathbf{v}}^{†}\mathbf{t})$, sofor every nonzero forward displacement field. Mass density and angular frequency are essential to this flux coefficient; both appear correctly in the PDF.

Place the four fixed charges at $(sa,ta)$ with $s,t\in \{-1,1\}$, and write the moving particle's position as $(x,y)$. Introduce its mass $m>0$, which is needed for an oscillation frequency but is not named in the question. Let $C=Qq/(4\pi {ϵ}_0)>0$. By Coulomb's law and superposition, the moving particle's potential energy isThe electrostatic potential itself is $V/q$. Put $r^2=x^2+y^2$ and ${\delta }_{st}=-(sx+ty)/a+r^2/(2a^2)$. Near the centre, a Taylor expansion givesOn summing over the four corners, linear terms cancel. To quadratic order, $\sum {\delta }_{st}=2r^2/a^2$ and $\sum {\delta }_{st}^2=4r^2/a^2+O(r^4/a^4)$, since $\sum st=0$. Reflection symmetry removes all odd-degree terms. HencewhereThus the centre has zero planar force and a positive-definite planar Hessian matrix $\kappa I_2$. It is a strict local minimum of the constrained potential energy, so it is a stable equilibrium. More explicitly, conservation of the total energy confines sufficiently small-energy motions within a small neighbourhood of this isolated local minimum.

Keeping only the quadratic potential, the two coordinates obey independent small oscillations:Both planar modes have the same angular frequency, and the ordinary frequency isThe planar constraint is essential to this constrained electrostatic equilibrium. Along the perpendicular axis, $V(0,0,z)=4C/(2a^2+z^2{)}^{1/2}=V(0,0)-C{z}^2/(\sqrt{2}{a}^3)+O(z^4)$, so the same stationary point is unstable to unconstrained normal displacements. That also checks the cancellation of the three-dimensional Hessian trace required by Laplace's equation away from charges; it does not contradict the positive-definite restriction to the given plane.

Take $U_0>0$, $k=2\pi /L>0$, and represent the stationary topographic internal gravity wave by phase $kx+mz$. Its laboratory angular frequency is zero, and its intrinsic frequency is $\hat{\omega }=-k{U}_0$. The internal gravity wave dispersion relation givesThe radiation condition selects $m>0$, since this makes the vertical group velocity upward. The lower kinematic boundary condition is $w=U_0{h}_x$, so a corresponding linear wave field iswhere $\zeta$ is the vertical fluid displacement. The constant-phase lines of an internal gravity wave slope down to the right, with $dz/dx=-k/m$.

Let $K^2=k^2+m^2$. In the frame fixed to the hills the phase pattern is stationary: its phase velocity is zero, while the group velocity isEnergy travels up and downwind, with ray slope $dz/dx=m/k$. In coordinates $x^{\prime }=x-U_{0}t$ moving with the wind, the phase is $k{x}^{\prime }+mz+k{U}_{0}t$. ThusThe intrinsic phase velocity is upstream and downward; the intrinsic group velocity is upstream and upward, perpendicular to the wave vector. Adding $(U_0,0)$ to the group velocity recovers the laboratory energy ray.

Vertically propagating waves require $k{U}_0<N_0$, equivalently $\boxed{L>2\pi U_0/N_0}$. At equality the vertical group velocity vanishes; above the mountain-wave cutoff the bounded response is an evanescent wave. The linearization requires small terrain slope $k{h}_0\ll 1$ and small wave-induced displacement gradients $|m|h_0\ll 1$, often summarized by $N_0{h}_0/U_0\ll 1$. The model also neglects rotation, viscosity and compressibility; its scales must permit those approximations.

The rotation introduces angular frequency $\Omega$ and, after combining the two blades, harmonics such as $2\Omega$. The acoustic compact-source approximation requires the propagation time $a/c_0$ to be small compared with ${\Omega }^{-1}$. ThusIt also bounds every blade element's Mach number by $\mu$. Consequently $|1-M_r|=1+O(\mu )$, and its leading value is one. The separate acoustic far field condition is $\Omega R/c_0\gg 1$.

Choose the positive rotation sense so that a first blade at phase $\alpha$ has radial and tangential unit vectorsIntegrating the given line force from $r=0$ to $a$ yields ${\mathcal{F}}_1=F{e}_{\varphi }(\Omega \tau )+D{e}_x$. Its axial component is constant, while ${\dot{\mathcal{F}}}_1=-F\Omega e_r$. Since $n=\cos \theta e_x+\sin \theta e_y$, the compact acoustic dipole sound from this blade isThe other blade has phase $\alpha +\pi$ and contributes the opposite rotating force. At a common compact retarded time, their total force is $2D{e}_x$, soThis cancellation calls for the next source-delay correction; it does not mean that the complete moving-source field vanishes.

Write $w(x,t)$ for the vertical displacement of the sea ice, and take the undisturbed water surface as $z=0$, with water occupying $z<0$. The elastic plate has areal mass $m={\rho }_{i}h$ and bending stiffnessHere $E$ is Young's modulus and $\eta$ is Poisson's ratio. We neglect in-plane prestress, viscosity and plate shear deformation, and linearize about hydrostatic equilibrium. These are important assumptions: perfect elasticity alone does not specify every term in a floating-plate model.

For a plane wave $w=\hat{w}{e}^{i(kx-\omega t)}$, $k>0$, potential flow in deep water has velocity potential $\varphi =\hat{\varphi }{e}^{kz}{e}^{i(kx-\omega t)}$. This solves Laplace's equation and decays downwards. The kinematic boundary condition gives $k\hat{\varphi }=-i\omega \hat{w}$. Linearizing the water pressure at the displaced interface gives an upward excess loadThe elastic plate equation is $m{w}_{tt}+D{w}_{xxxx}=p$. Substitution and multiplication by $k$ therefore give the flexural-gravity wave dispersion relationEquivalently, with $\alpha ={\rho }_{i}h/{\rho }_w$ and $\beta =D/{\rho }_w$,This determines the positive wavenumber implicitly for prescribed positive angular frequency. It is unambiguous: the derivative of ${\omega }^2$ is $(g+5\beta k^4+4\alpha \beta k^5)/(1+\alpha k{)}^2>0$, while ${\omega }^2$ runs from zero to infinity.

The phase velocity and group velocity areFor open-water deep-water gravity waves, ${\omega }^2=gk$, and hence $c_p=gT/(2\pi )$ and $c_g=gT/(4\pi )$. At large period the ice-covered curves approach these straight lines. At shorter period, plate bending raises the speeds, so both ice-covered curves turn upward as period decreases. In the bending regime with negligible plate inertia, $\omega \propto k^{5/2}$ and $c_g\simeq (5/2)c_p$; in the formal plate-inertia-dominated limit, $\omega \propto k^2$ and $c_g\simeq 2c_p$. The latter extrapolation eventually leaves thin-plate validity and should not be read as a prediction at arbitrarily small wavelength.

The plotted parameters are illustrative rather than measured at the observation site. They give a group-velocity minimum of a flexural-gravity wave near $15.1\mathrm{m}{\mathrm{s}}^{-1}$ at period $17.6\mathrm{s}$. There is no arbitrarily slow wave-energy branch under a continuous elastic sheet. Energy put into a localized disturbance travels away at at least this minimum group velocity; a slowly moving wind system cannot retain a wave packet indefinitely beneath itself. This reduces the opportunity for sustained local growth compared with slow, short open-water waves, and the continuous cover also prevents direct wind forcing of an exposed water surface. Incoming long swell can still propagate.

A group-velocity minimum of a flexural-gravity wave is not by itself a universal minimum wind speed for wave generation. A steadily translating forcing pattern requires a phase velocity matching its translation speed, so the minimum of $c_p$, not $c_g$, supplies the corresponding resonance threshold. Random wind forcing, dissipation and aerodynamic coupling must be specified before making an absolute generation claim.

The sea ice acts as a frequency-selective filter. Over a long path, wave attenuation in sea ice is generally much stronger for shorter surface gravity waves, through repeated scattering, internal ice losses and water-side dissipation. Their amplitude can fall below the tiltmeter's detection level even if the source initially generated them. Thus the long-period swell survives while the shorter-period tail does not.

The disappearance near 14 s is an attenuation and detectability limit, not a forbidden-frequency interval of the ideal elastic plate. The flexural-gravity wave dispersion relation admits a real positive wavenumber for every positive angular frequency. In an impulsive-source picture, shorter-period deep-water gravity waves also arrive later because their group velocity is smaller, but delayed arrival alone does not explain a persistent observed cutoff.

In classical charged particle in a uniform magnetic field motion, the Lorentz force and Newton's second law giveAssume $\mathbf{B}\ne 0$ and choose an orthonormal coordinate system with $\mathbf{B}=B{\mathbf{e}}_z$, $B=|\mathbf{B}|$. Write $\Omega =qB/m$, retaining the sign of the electric charge. The velocity equations areThus the parallel velocity is constant, and the perpendicular velocity rotates at constant angular frequency. Choosing the time origin and perpendicular axes conveniently givesIntegrating the velocity, one obtainsThe first two coordinates describe a circle of radius $m{v}_{\perp }/(|q|B)$, while the third advances uniformly. This is helical motion in a uniform magnetic field: the helix has axis parallel to $\mathbf{B}$ through the guiding centre. Its cyclotron frequency isand its signed angular velocity vector about that axis is $-q\mathbf{B}/m$. The minus sign follows from the ordering $\mathbf{v}\times \mathbf{B}$ in the Lorentz force. The helix degenerates to a circle when $v_{\parallel }=0$ and to a straight line when $v_{\perp }=0$. If $q=0$ or $B=0$, there is only straight-line motion. In a relativistic extension with fixed Lorentz factor $\gamma$, the frequency becomes $|q|B/(\gamma m)$; the classical result above uses $m\mathbf{v}$ as momentum.

Take the oscillation to be $z(t)=L\cos \omega t$; a different initial phase does not affect a period average. In the mass-density normalization of the quadrupole formula, the second mass moment tensor of the point mass has only $I_{33}=m{z}^2$ nonzero. Its trace-free mass quadrupole moment is thereforeThe constant part does not radiate. Since $d^3({\cos }^2\omega t)/d{t}^3=4{\omega }^3\sin 2\omega t$, the tensor contraction isUsing $⟨{\sin }^2(2\omega t)⟩=1/2$ in the quadrupole formula gives the averaged positive radiated powerThe mass quadrupole moment oscillates at twice the source's angular frequency, and the source energy loss has the opposite sign. The expression uses the leading slow-motion weak-field approximation, $\omega L\ll c$. With SI stress-energy tensor components, the mass-density integrand is $T_{00}/c^2$; the displayed $c^{-5}$ quadrupole formula uses that mass normalization. This is the point-mass contribution requested by the model: an apparatus maintaining an accelerating mass would also contribute its own mass quadrupole moment.

False. Adiabatic invariance of an orbital action preserves actions under sufficiently slow regular evolution, not the ratios of their frequencies. With action-angle variables, ${\Omega }_i=\partial H/\partial J_i$ can change even when $J_i$ does not. A closed bound orbit requires commensurate radial and angular frequencies; rational commensurability is not an adiabatic invariant.

For an explicit counterexample, slowly change the physical potential from Kepler form towith a small, slowly varying positive $\alpha$. The radial epicyclic frequency and circular angular frequency at a nearly circular orbit areInitially their ratio is one. For generic nonzero $\alpha$, $\Omega /\kappa$ is irrational, producing the Kepler-harmonic epicyclic precession. An initially slightly eccentric ellipse with nonzero radial action therefore becomes a precessing nonclosed orbit, while its radial action remains approximately fixed and its angular momentum remains exactly fixed by spherical symmetry.

Adiabatic actions do not preserve orbital closure. Exceptional circular or resonantly commensurate final orbits do not rescue a claim about every initially closed orbit.

For a normal mode, ${\partial }_t\mapsto -i\omega$, ${\partial }_x\mapsto ik$, and ${\partial }_x^2\mapsto -k^2$. Substitution into the linear complex Ginzburg-Landau equation givesTherefore the dispersion relation isFor real $k$, its temporal growth rate is ${\omega }_i=\mu -k^2$, with unique maximum $\mu$ at $k=0$. Its real angular frequency is $Uk+c_d{k}^2$, displaying the separate advection and wave dispersion terms.

Use the geometrical optics conventionBecause ${\partial }_t=\epsilon {\partial }_T$ and ${\partial }_x=\epsilon {\partial }_X$, differentiating $e^{i\theta /\epsilon }$ gives an order-one physical angular frequency and order-one physical wavenumbers. Absorb a constant leading amplitude phase into $\theta$ and take $A_0$ real; for a complex amplitude the corresponding quadratic density is $|A_0{|}^2$.

Substitution into the wave equation gives the eikonal equation for a variable-speed wave at leading order:Choose the positive-frequency branch $\omega =\Omega =cK$, $K=\sqrt{k^2+{\ell }^2}$; the negative branch can be treated in exactly the same way. The imaginary first-order terms giveMultiplying by $A_0$ puts this wave-action transport for a variable-speed wave in conservation form:Its density is $\omega A_0^2$ and its flux is that density times the group velocity. Independently, equality of mixed wave phase derivatives givesThis phase compatibility in geometrical optics is essential when converting the differentiated dispersion relation into equations along rays.

For completeness let $\omega =\Omega (\mathbf{k};\mathbf{X},T)$ be a general local dispersion relation. Define ray tracing by $d\mathbf{X}/dT={\mathbf{c}}_g={\nabla }_{\mathbf{k}}\Omega$. Using the chain rule in ${\nabla }_{\mathbf{X}}\omega$ and the symmetry ${\partial }_{X_i}{k}_j={\partial }_{X_j}{k}_i$ givesLikewise ${\omega }_T={\Omega }_T+{\mathbf{c}}_g\cdot {\mathbf{k}}_T$ cancels its advective term because ${\mathbf{k}}_T=-\nabla \omega$. Hence the Hamiltonian ray equations for a local dispersion relation areThe derivatives on the right hold the wavevector fixed. For this wave equation they reduce to