The usual elastic secondary flow around a rotating sphere is inward towards the equator, then poleward beside the sphere and outward near both poles. It is superposed on the primary azimuthal rotating sphere in Stokes flow. Tensile normal stresses along curved azimuthal streamlines act as hoop stresses, pulling liquid towards the rotation axis. Incompressibility supplies the axial return flow. Unlike the primary rotation, this meridional circulation does not reverse when the rotation is reversed.

The direction assumes the usual polymeric sign $c={\psi }_1+2{\psi }_2>0$ and negligible inertia. A general second-order fluid has circulation proportional to $c$: a negative value reverses the arrows and $c=0$ eliminates the leading elastic circulation. Inertial secondary circulation is a different contribution, so slow rotation alone does not determine which contribution dominates. The sketch shows the elastic contribution for $c>0$:

For a quantitative check of the sketch, take sphere radius $a$, angular velocity $\Omega$ and $K=\Omega a^3$. The primary azimuthal velocity is $v_{\varphi }=K\sin \theta /r^2$, with $\theta$ measured from the rotation axis. Its only nonzero strain components in the spherical orthonormal basis are $E_{r\varphi }=E_{\varphi r}=-3K\sin \theta /(2r^3)$. The leading elastic secondary circulation around a rotating sphere has the stream functionwith $v_r=H(3{\cos }^2\theta -1)/r^2$ and $v_{\theta }=-H^{\prime }\sin \theta \cos \theta /r$. The bracket is $(1-a/r{)}^2(1+2a/r)>0$ outside the sphere. Thus $v_r<0$ at the equator and $v_r>0$ near the poles when $c>0$, and both components vanish on the sphere. The diagram uses these velocities, not independently chosen arrows.

Two other manifestations of the same elastic normal-stress differences are rod climbing, in which inward hoop stresses raise a free surface around a rotating rod, and die swell, in which relaxation of the axial stress and stored polymer deformation allows an extrudate to expand sideways after leaving the die. These are not explained by a scalar shear viscosity alone. A Newtonian extrudate can also swell through velocity-profile rearrangement, so the elastic contribution should not be identified with every instance of swelling.

The second-order fluid is a slow, weak-flow expansion of the objective constitutive history of an isotropic simple fluid. Material frame indifference requires objective tensor rates. For an incompressible flow, the leading isotropic stress is arbitrary pressure, the linear deviatoric term is $2\mu E$, and at the next order the independent contributions can be represented by $E^2$ and an objective rate of $E$. Isotropic scalar contributions are absorbed into the pressure. Their coefficients can be expressed through the two low-rate normal-stress differences, givingHere $\stackrel{△}{E}$ is the lower-convected derivative, not a division by $E$. This local expansion is appropriate when deformation and temporal change over the relevant viscoelastic relaxation time are small: both the Weissenberg number and the appropriate Deborah number must be small. It captures leading normal stresses in slow flows, but not arbitrary finite-strain history, rapid transients or strongly nonlinear shear thinning. In particular its restrictions are stronger than small-amplitude linear oscillatory response at arbitrary frequency.

For steady simple shear flow $\mathbf{v}=(\dot{\gamma }y,0,0)$, use component-first $L_{ij}={\partial }_j{v}_i$, so the paper's $\nabla v$ is $L^T$. The relevant tensors areSubstitution givesThus the first and second normal-stress differences are $N_1={\psi }_1{\dot{\gamma }}^2$ and $N_2={\psi }_2{\dot{\gamma }}^2$, and the shear stress is $\mu \dot{\gamma }$.

For the velocity-preservation results, the definitions imply $\stackrel{△}{E}=\stackrel{\circ }{E}+2E^2$. Consequently the non-Newtonian stress can be writtenwhere $\stackrel{\circ }{E}$ is the Jaumann derivative. Let ${\mathbf{v}}_N$ solve the steady incompressible Stokes flow problem with the prescribed boundary velocities. The supplied identity says that $\nabla \cdot {\stackrel{\circ }{E}}_N$ is curl-free. If ${\psi }_1+2{\psi }_2=0$, then $\nabla \cdot S_N=\nabla \Phi$ for a scalar $\Phi$ on the ordinary simply connected flow domain. The momentum equation at ${\mathbf{v}}_N$ isHence the Newtonian velocity itself is a solution of the second-order constitutive model, with altered pressure $p=p_N+\Phi$. For nontrivial topology one also needs the curl-free force to admit a single-valued pressure; this is implicit in the usual boundary-value problem.

On the regular weakly non-Newtonian branch, Uniqueness of Stokes flow fixes the velocity correction. At first order it solves homogeneous Stokes flow, with zero boundary velocity, after the gradient force has been absorbed into pressure. Multiplying its equation by the correction and integrating gives $2\mu \int E_1:E_{1}dV=0$, hence the correction vanishes. Higher perturbative corrections vanish in turn because the previous velocities remain ${\mathbf{v}}_N$. The velocity is $\mathbf{v}={\mathbf{v}}_N$ in this second-order-fluid solution; only the pressure and stresses change. This is Newtonian velocity preservation in a second-order fluid. It does not assert global uniqueness of distant nonlinear solution branches or exact equality for a real fluid whose higher-order constitutive terms have been omitted.

For a planar incompressible velocity $(u(x,y),v(x,y),0)$, the rate-of-strain tensor has the formAll quantities are independent of $z$, so $\nabla \cdot E_N^2=\nabla (a^2+b^2)$. The extra quadratic term is therefore also a pressure gradient. Together with the supplied identity for the Jaumann derivative, the entire non-Newtonian force is a gradient for arbitrary ${\psi }_1,{\psi }_2$. The same boundary-value and weak-branch argument proves $\mathbf{v}={\mathbf{v}}_N$ for every such planar second-order-fluid flow with prescribed boundary velocities, without requiring ${\psi }_1+2{\psi }_2=0$. Specifying stresses or a free surface instead of velocities would not give the same conclusion automatically, because the extra stresses affect those boundary conditions.

The constitutive convolution is the response of a linear viscoelastic fluid close to its relaxed state. Deformations and rotations during the material's memory time must be small, so the stress is linear in the deformation history and evaluating that history at a fixed position is consistent with linearization about rest. For an isotropic incompressible flow, the deviatoric stress is proportional to the history of the rate-of-strain tensor; an arbitrary isotropic pressure is separate. Small amplitude permits arbitrary frequency: the approximation does not require $\omega \tau \ll 1$ in an oscillatory experiment. Large accumulated deformation, substantial advection of memory, and nonlinear normal-stress differences are outside this formula's scope.

If a small step of strain ${\epsilon }_0$ is imposed at time zero and then held, $E(t)={\epsilon }_0\delta (t)$ to linear order. The convolution consequently gives ${\sigma }^{\prime }(t)=2G(t){\epsilon }_0$ for $t>0$. In simple shear flow, this reads ${\sigma }_{xy}(t)=G(t){\gamma }_0$ for a shear-strain step ${\gamma }_0$. Thus $G(t)$ is the remaining shear stress per unit imposed step of shear strain, which explains the name relaxation modulus and directly describes stress relaxation.

Use the convention $E(t)=\mathrm{Re}[\hat{E}{e}^{i\omega t}]$. Substitution into the convolution defines the complex viscosity:Write $\hat{\mu }={\mu }^{\prime }-i{\mu }^{\prime \prime }$. Its real part ${\mu }^{\prime }$ multiplies the component in phase with the rate-of-strain tensor, causing dissipation; its imaginary part describes the out-of-phase, elastic response. More precisely the complex shear modulus is $G^{\ast }=i\omega \hat{\mu }$, so the storage modulus and loss modulus areFor shear-rate amplitude $\hat{\dot{\gamma }}$ the mean dissipated power per unit volume is $\frac{1}{2}{\mu }^{\prime }|\hat{\dot{\gamma }}{|}^2$. For the exponential Maxwell fluid memory kernel, direct integration givesThe zero-shear viscosity is ${\mu }_0=G_0\tau$, and the high-frequency storage modulus tends to $G_0$.

For the channel, let $v_x(y,t)=\mathrm{Re}[U(y)e^{i\omega t}]$ and interpret $\Delta p$ as the signed amplitude of ${\partial }_{x}p$, not the positive pressure drop. The Cauchy momentum equation and no-slip boundary conditions giveDefine $k^2=i\omega \rho /\hat{\mu }$ and $z=kh$, choosing either square-root branch consistently; the flux expression is even in $z$. The symmetric solution isIts integral across the channel gives the oscillatory channel flux of a linear Maxwell fluid:This includes inertia while using the linear constitutive response. The pressure-gradient amplitude must be small enough that the resulting deformation stays in that response regime.

As $\omega \to 0$, $\tanh z/z=1-z^2/3+2z^4/15+\cdots$. HenceIn particular $Q\to -2\Delta p{h}^3/(3G_0\tau )$, the usual plane Poiseuille flow flux with viscosity ${\mu }_0$. The fluid has enough time to relax and behaves as a Newtonian fluid at leading order.

For fixed positive $h,\tau ,\rho ,G_0$, let $c=\sqrt{G_0/\rho }$. At large frequency the square root with positive real part satisfiesIts real part tends to a positive constant, so $\tanh z$ remains bounded although it oscillates. Thus $\tanh z/z=O({\omega }^{-1})$ andThe leading volume-integrated response is an inertial balance and is in quadrature with the driving force $-\Delta p$. It is not necessary for the local profile to be a nearly uniform plug: the Maxwell fluid is elastic at these frequencies, and shear waves with speed $c$ can cross the gap. The oscillatory part of $U$ largely cancels on integration. Small attenuation allows resonance-like enhancements at finite frequencies; the leading flux limit still holds for fixed nonzero relaxation and gap parameters. A Newtonian thin viscous-layer argument would not justify the Maxwell profile here.

Finally take the different ordering $\omega \tau \gg 1$ but $\rho {\omega }^2{h}^2/G_0\ll 1$. Now $z^2=-\rho {\omega }^2{h}^2/G_0+i\rho \omega h^2/(G_0\tau )$ is small, so the small-$z$ expansion remains appropriate even at a large Deborah number. The result isHere inertia is negligible throughout the narrow gap. Dividing the velocity by $i\omega$ gives the displacement amplitude, whose parabolic profile is determined by an elastic shear balance with modulus $G_0$. The velocity therefore grows like frequency for a fixed displacement response. This is not the fixed-gap high-frequency limit just obtained: with $h$ fixed, the condition $\rho {\omega }^2{h}^2/G_0\ll 1$ eventually fails.

The quasi-geostrophic approximation isolates slowly evolving, nearly balanced motions in a rotating, stably stratified fluid. Ocean eddies and mid-latitude atmospheric disturbances often have this character: a leading Coriolis force–pressure balance suppresses rapid horizontal acceleration, while hydrostatic balance suppresses rapid vertical acceleration. Small ageostrophic flow nevertheless supplies the vertical motion that changes the leading vorticity. The approximation filters fast inertia-gravity waves while retaining nonlinear transport, vertical shear and Rossby waves.

Use a reference density ${\rho }_0$ and remove the background hydrostatic pressure. Let $\pi =p^{\prime }/{\rho }_0$ and let the upward buoyancy perturbation be $b=-g{\rho }^{\prime }/{\rho }_0$. A vertically varying reference density stratification has $N^2(z)>0$. Under the Boussinesq approximation, the inviscid and adiabatic equations areHere $D/Dt={\partial }_t+{\mathbf{u}}_H\cdot {\nabla }_H+w{\partial }_z$. Density variations are small compared with ${\rho }_0$ but retained in the vertical buoyancy force. This is a useful local, incompressible description; a compressible atmosphere needs the corresponding weighted or pressure-coordinate version.

Let horizontal and vertical lengths be $L,H$, speed be $U$, and take the slow time $L/U$. Take $f_0>0$ for definiteness; use $|f_0|$ in the scale estimates in the opposite hemisphere. The principal small parameter is the Rossby number $\mathrm{Ro}=U/(f_{0}L)\ll 1$ at a latitude where $f_0\ne 0$. For the hydrostatic, strongly stratified regime, $H/L\ll 1$ and $f_0/N\ll 1$. Retaining both horizontal vorticity and vortex stretching gives the Burger numberThe geostrophic balance pressure scale is $f_{0}UL$ and the buoyancy perturbation scale is $B=f_{0}UL/H$. Hence $B/(N^{2}H)=\mathrm{Ro}/\mathrm{Bu}\ll 1$: buoyancy perturbations produce small displacements relative to the background stratification scale. The ageostrophic flow is $O(\mathrm{Ro}U)$ and incompressibility then gives $w=O(\mathrm{Ro}UH/L)$. Vertical acceleration is smaller than hydrostatic terms by order ${\mathrm{Ro}}^2(f_0/N{)}^2$ in this scaling. Weak forcing or diffusion can be added at the slow order, but neither is present here.

Define the quasi-geostrophic streamfunction by $\psi =\pi /f_0$. Leading geostrophic balance and hydrostatic balance implyDifferentiating horizontally and vertically gives the thermal-wind balanceThus vertical shear is determined by horizontal buoyancy gradients. The homogeneous Taylor–Proudman theorem is recovered only when those gradients vanish. Stable stratification allows sloping density surfaces and different geostrophic flow at different levels, without abandoning leading balance.

The Prandtl ratio of scales makes that modification quantitative. Horizontal and vertical terms in balanced potential-vorticity inversion have sizes $\psi /L^2$ and $(f_0^2/N^2)\psi /H^2$. Equality givesFor constant $N$, $Z=(N/|f_0|)z$ makes the inversion operator an ordinary three-dimensional Laplacian. Equivalently, an interior balanced disturbance of horizontal size $L$ naturally penetrates a depth of order $|f_0|L/N$. With $N\gg |f_0|$ this is a thin, vertically sheared structure, rather than the arbitrarily tall homogeneous columns suggested by strict Taylor–Proudman theorem behaviour. This is an aspect-ratio statement, not the diffusivity Prandtl number.

To allow a weak meridional variation of the Coriolis parameter, use the beta plane $f=f_0+\beta y$, with $\beta L/f_0=O(\mathrm{Ro})$, or equivalently $\beta L^2/U=O(1)$. The small correction to the leading balance is dynamically important on the slow time. Let $\xi ={\nabla }_H^2\psi$ be vertical relative vorticity and let $D_g={\partial }_t+{\mathbf{u}}_g\cdot {\nabla }_H$. At the first nontrivial order, the vertical vorticity equation and buoyancy equation reduce toRelative-vorticity stretching and tilting are higher order here; planetary-vorticity stretching $f_0{w}_z$ must remain. Eliminate $w$ to expose a materially conserved balanced scalar. Because ${\mathbf{u}}_{g,z}=(-b_y/f_0,b_x/f_0)$, the commutator term ${\mathbf{u}}_{g,z}\cdot {\nabla }_{H}b$ is exactly zero. For $N=N(z)$ this givesAdding the two equations proves the quasi-geostrophic potential-vorticity equationAn additive constant such as $f_0$ changes no dynamics. The horizontal streamfunction advection bracket can express this as $Q_t+{\psi }_x{Q}_y-{\psi }_y{Q}_x=0$. Although small departures from balance generated this evolution equation, the leading geostrophic flow itself advects $Q$ nonlinearly. The stretching term couples horizontal motion to vertical displacement of density surfaces.

This gives the practical advection-and-inversion description. Start with interior three-dimensional quasi-geostrophic potential vorticity, advect it with ${\mathbf{u}}_g$, and recover $\psi$ fromThe stratified quasi-geostrophic inversion is elliptic for positive $N^2$. It must be supplied with lateral and vertical boundary conditions and any circulation or mean-flow constraints; interior potential vorticity alone is not sufficient. At rigid horizontal lids, $w=0$ makes boundary buoyancy satisfy $D_{g}b=0$, and $b=f_0{\psi }_z$ supplies the vertical derivative data for inversion. Boundary buoyancy perturbations can therefore support flow even when the interior anomaly is zero. Once $\psi$ is found, its horizontal derivatives give velocity, its vertical derivative gives buoyancy, and the thermodynamic equation diagnoses $w$. This balanced evolution excludes the independent fast-wave initial data present in the full equations.

Finally, a gradient in $f$ supplies a restoring mechanism even in an otherwise uniform fluid. Linearizing about rest with constant $N$, the quasi-geostrophic potential-vorticity equation becomesSubstitution of $\psi \propto e^{i(kx+ly+mz-\omega t)}$ gives the Rossby wave dispersionFor $\beta >0$, its zonal phase speed is westward relative to a resting background. A meridionally displaced parcel retains potential vorticity, so it acquires a relative-vorticity anomaly that induces the velocity tending to return the disturbance. Background shear and stratification modify the full potential-vorticity gradient; they can alter propagation or permit instability. On a constant f-plane the planetary contribution vanishes, but gradients of relative vorticity, boundary buoyancy or topography can still support balanced wave motion. Close enough to the equator that $f_0$ vanishes, this mid-latitude quasi-geostrophic approximation and its scaling must be replaced.

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.

Use pressure per unit constant density, $\pi$, with the gravitational potential and centrifugal potential absorbed into it. Strict geostrophic balance is the exact balancewith material acceleration omitted, for a homogeneous incompressible flow. Writing $f=2\Omega \ne 0$, its components are $fv={\pi }_x$, $fu=-{\pi }_y$ and ${\pi }_z=0$. Differentiating the first two equations with respect to $z$ gives $u_z=v_z=0$. Alternatively, the vector calculus identityshows, on taking the curl of the balance, that $(\mathbf{\Omega }\cdot \nabla )\mathbf{u}=0$. Thus $w_z=0$ too, proving the Taylor–Proudman theorem:The physical pressure still includes its background hydrostatic pressure; it is the reduced pressure $\pi$ that is independent of $z$. Between horizontal impermeable boundaries, the vertically constant $w$ is zero.

For boundaries $z=z_b(x,y)$ and $z=z_t(x,y)$, impermeability condition gives $w_b={\mathbf{u}}_H\cdot {\nabla }_H{z}_b$ and $w_t={\mathbf{u}}_H\cdot {\nabla }_H{z}_t$. A nearly columnar flow crossing contours of $H=z_t-z_b$ must therefore have $w_t-w_b\simeq {\mathbf{u}}_H\cdot {\nabla }_{H}H\ne 0$. Exact columnarity must give way to small corrections from ageostrophic flow. The leading horizontal geostrophic flow can remain nearly independent of $z$, and the leading reduced pressure has ${\pi }_z=0$, but $w_z$ is now nonzero and exact three-dimensional strict geostrophic balance no longer holds. For small slopes, the dominant vertical vorticity equation iswhere $q=v_x-u_y$ is relative vorticity. Equivalently, $D[(q+f)/H]/Dt\simeq 0$. Crossing geostrophic contours stretches or squashes columns: increasing $H$ produces cyclonic relative vorticity, decreasing $H$ produces anticyclonic relative vorticity. The full vorticity equation also contains tilting terms, neglected at this columnar order.

For the circular shallowing, fluid arriving from upstream has potential vorticity $f/h$. Its height is $h(1-ϵ)$ inside, so conservation givesThis uses the stated column-stretching description and initially open streamlines. To obtain the leading disturbance velocity field, use the conventional streamfunction definition $u=-{\Psi }_y$, $v={\Psi }_x$, and write $\Psi =-Uy+\psi (r)$. Then $q={\nabla }_H^2\psi =r^{-1}(r{\psi }_r{)}_r$. Regularity at $r=0$ gives ${\psi }_r=-fϵr/2$ inside. Outside, ${\psi }_r=B/r$; matching tangential velocity at $r=a$ fixes $B=-fϵa^2/2$. Choose the irrelevant additive constant so thatBoth $\psi$ and ${\psi }_r$ match at $a$, so there is no extra vorticity sheet on the rim. This is the circular topographic anticyclone. The radial expression specifies the disturbance; the complete streamfunction is $\Psi =-Uy+\psi$, since a radial function alone cannot represent the required uniform far-field current. Reversing the streamfunction sign convention reverses the displayed proportionality constant, but leaves the physical velocity unchanged.

Differentiation gives the full relative velocity field in the rotating frame:away from the narrow transition across the change of depth. The disturbance turns clockwise for $fϵ>0$ and decays as $1/r$. The horizontal field is the leading small-$ϵ$ columnar inversion: the exact variable-depth transport across a sharp rim needs higher-order corrections, rather than an exactly two-dimensional velocity continuous across a literal vertical step.

Take $U>0$ and $ϵ>0$. The greatest opposing zonal disturbance speed is $Ca$, attained at $(0,-a)$. If $Ca<U$, then $u>0$ everywhere and every particle advances in $x$, precluding closed streamlines. If $Ca>U$, the interior stagnation point $(0,-U/C)$ is a center, since the interior contours are circles about that point. The exterior stagnation point $(0,-C{a}^2/U)$ is a saddle: linearization there has real eigenvalues of opposite sign. Closed contours around the center are bounded by a separatrix through the saddle. The two points meet at the rim at onset, giving the closed-streamline threshold for a circular topographic anticyclone:At this threshold $U/(fa)=ϵ/2\ll 1$, consistent with small Rossby number. Equality is the onset, not a finite-area closed cell. Beyond onset, upstream data alone no longer determine the potential vorticity of trapped fluid; the threshold follows from the continuation of the supplied open-flow model.

Let the surface radius be $a$, let $\Delta$ denote epicentral angle in radians, and put $\eta (r)=r/v(r)$. The spherical elastic ray invariant gives $\sin i=p/\eta$. Along either radial leg, if $l$ is arclength and $\theta$ polar angle in the ray plane,For a turning ray with minimum radius $r_p$ defined by $\eta (r_p)=p$, the outgoing and incoming legs therefore giveThe square-root turning singularity is integrable for a regular turning point.

Introduce the reduced travel timeIts integrand vanishes at the moving lower endpoint, so differentiation gives ${\tau }^{\prime }(p)=-\Delta (p)$. Consequentlyon every regular travel-time branch. This is also the endpoint variation of travel time: the tangential endpoint slowness multiplied by the surface radius is $p$. Angles expressed in degrees would introduce the corresponding conversion factor; the displayed equality uses radians.

For Herglotz–Wiechert inversion, assume $\eta$ is strictly increasing so that $r=r(\eta )$ is single-valued, and set ${\eta }_a=a/v(a)$ and $h(\eta )=d\log r/d\eta$. The angular integral becomes an Abel integral equation:For any $0<\mu <{\eta }_a$, multiply by $(p^2-{\mu }^2{)}^{-1/2}$ and integrate. Interchanging the triangular region $\mu <p<\eta <{\eta }_a$ givesThe inner integral equals $\pi /2$: the substitution $p^2={\mu }^2+({\eta }^2-{\mu }^2){\sin }^2\theta$ reduces it to ${\int }_0^{\pi /2}d\theta$. Thus the full constructive inversion isThe surface radius and surface slowness fix the normalization. Varying $\mu$ reconstructs radius and speed parametrically; the centre is recovered by a limit when the data extend that far. The calculation assumes smooth regular turning rays and complete branch data, not merely the first arrival at each epicentral angle.

For a check, a constant speed $v_0$ gives ${\eta }_a=a/v_0$, $\Delta (p)=2\arccos (p/{\eta }_a)$ and $T(p)=2\sqrt{a^2-p^2{v}_0^2}/v_0$. These satisfy $dT/d\Delta =p$, and the inversion returns $r(\mu )=v_0\mu$.

If $\eta$ decreases outwards on an interval, then it increases as a surface ray travels inward through that interval. Such a ray cannot turn there, since turning requires the decreasing inward value of $\eta$ to reach $p$. Rays that penetrate this interval can traverse it and turn deeper, but the map $\eta \mapsto r$ now has several branches. The angular data constrain combinations of those branch contributions, rather than the single function $h(\eta )$ used above. This is a hidden turning-ray zone from nonmonotone spherical slowness. The stated single-valued inversion fails and does not uniquely locate the speed profile in that interval without additional information. Missing turning rays and possible multiple travel-time branches are the obstruction, not a failure of the algebraic Abel-kernel identity.

Use the usual minor and major symmetries of the elastic stiffness tensor, positive density, and strong ellipticity in elasticity. For the plane elastic wave $u=a{e}^{ik(n\cdot x-vt)}$, substitution into momentum balance gives the acoustic tensor problemThe major symmetry makes $\Gamma$ a real symmetric matrix. Moreover $a^{\mathsf{T}}\Gamma a=c_{ijpq}{a}_i{n}_j{a}_p{n}_q>0$ for nonzero $a$ under strong ellipticity in elasticity. Hence its three eigenvalues are positive and the three positive wave speeds are real. Symmetry alone proves only reality of $v^2$: an unstable isotropic tensor with negative shear modulus has negative transverse $v^2$. The requested reality therefore uses stability as part of the physical elastic-solid assumption.

The elastic slowness surface consists of three sheets, with $s=n/v_{\alpha }(n)$, satisfyingIt is centrosymmetric. Sheets can meet at degenerate polarizations; they need not be spherical or convex in an anisotropic solid. The group velocity is normal to the corresponding smooth sheet: for a sheet tangent $ds$, differentiation of $a^{\mathsf{T}}[c_{ijpq}{s}_j{s}_q-\rho I]a=0$ gives $a_i{c}_{ijpq}{a}_p{s}_{q}d{s}_j=0$. The coefficient vector is proportional to the group velocity, so its scalar product with every tangent vanishes.

Fix horizontal slowness $p=(n_1/c,n_2/c)$ and put $q=s_3$. DefineThe symbol $p$ on a tensor index is an index, whereas $p_{\alpha }$ is a horizontal slowness component. The vertical slowness equation is the vertical slowness sextic of an anisotropic solidThe leading coefficient is $\det T>0$, so it is a degree-six polynomial with real coefficients. Nonreal roots occur in conjugate pairs, leaving between zero and six real roots counted with multiplicity. Generically the counts are $0,2,4,6$; a grazing repeated root can give an odd count of distinct real roots. Geometrically these are the intersections of a vertical line of fixed horizontal slowness with the three sheets.

All four generic counts are actually possible in a stable material. For example, take $\rho =1$ and orthotropic Voigt stiffnessesThe stiffness matrix is positive definite. For horizontal slowness $(p,0)$, the three diagonal acoustic equations areThe choices $p=0.1,0.3,0.8,1.2$ give respectively six, four, two and zero real vertical roots.

For the general depth-dependent amplitude, substitution of the horizontal harmonic factor yieldsFor six simple roots $q_{\ell }$, choose nonzero null vectors $a_{\ell }$ of the corresponding matrix. The complete solution isComplex vertical roots give evanescent waves. At defective repeated roots the usual polynomial-in-depth factors multiplying exponentials must be included, or the solution can be defined by a limiting procedure.

To classify the modes, take $x_3$ positive downwards and $\omega >0$. The reduced traction for a mode isMechanical energy crosses a plane at rate $-{\sigma }_{i3}{u}_{i,t}$. Averaging the real harmonic fields uses $⟨\mathrm{Re}(f{e}^{-i\omega t})\mathrm{Re}(g{e}^{-i\omega t})⟩=\mathrm{Re}(f{g}^{\ast })/2$. With velocity amplitude $-i\omega a_{\ell }$ and traction amplitude $i\omega b_{\ell }$, the time-averaged vertical acoustic energy flux isFor a real propagating mode, write its physical wavevector as $k=\omega (p_1,p_2,q)$. Differentiate $a^{\mathsf{T}}\Gamma (k)a=\rho \omega (k{)}^2{a}^{\mathsf{T}}a$ in $k_3$. The derivatives of $a$ cancel by the eigenvalue equation, while $a^{\mathsf{T}}({\partial }_{k_3}\Gamma )a=2\omega a^{\mathsf{T}}(Tq+R)a$. Thus its vertical group velocity and flux areThus the elastic mode classification by vertical energy flux is upgoing if $J_3<0$, downgoing if $J_3>0$. In anisotropy, the sign of vertical phase slowness alone need not classify energy propagation. For evanescent modes, select $\mathrm{Im}q>0$ for decay into a downward half-space and $\mathrm{Im}q<0$ for decay into an upward half-space. Away from grazing and degeneracy, continuation from zero horizontal slowness, with a small radiation damping if needed, separates three outgoing modes on each side. Superpose only the appropriate three to construct the most general upgoing or downgoing field; limiting modes handle exceptional cases.