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Let $L$ be a core-scale length, $B_{\ast }$ a representative strength of the prescribed poloidal magnetic field, and $v_A=B_{\ast }/\sqrt{{\mu }_0\rho }$ the Alfvén speed. A consistent weakly damped, small-amplitude ordering isThe first ratio is the Lehnert number. It makes the wave slow relative to rotation and allows almost geostrophic flow, with the azimuthal velocity approximately constant along a rotation-axis cylinder. The small wave amplitude gives $|B_{\varphi }|\sim \sqrt{{\mu }_0\rho }|U|\ll B_{\ast }$. Poloidal advection then changes neither the background field nor the wave substantially over one wave period. If there are no-slip boundary layers, also require their damping rate $\sqrt{\nu \Omega }/L\ll \sigma$, or otherwise retain their torque; a small bulk viscosity alone does not remove that boundary torque.

In the rotating magnetohydrodynamic momentum equation, axisymmetry removes the azimuthal pressure derivative. The neglected nonlinear azimuthal advection has size $|{\mathbf{u}}_p|U/L$, and the neglected viscous term has size $\nu U/L^2$. The retained azimuthal Coriolis force is $2\Omega u_r$. A typical compatible local balance has $|u_r|\sim \sigma U/\Omega$, small compared with $U$ when the Lehnert number is small. The azimuthal Lorentz force density can be written using $\nabla \cdot {\mathbf{B}}_p=0$:The physical prefactor is $1/({\mu }_0\rho )$: the extra $i$ printed inside the first reduced momentum equation is a typographical error, also contradicted by the subsequent real-valued system.

In the resistive induction equation, neglecting poloidal advection of the perturbation and magnetic diffusion leaves the winding of the prescribed poloidal magnetic field by differential rotation. ThusThis is the small-amplitude torsional Alfvén wave approximation, with a time-independent background and no external azimuthal forcing.

We now average over the full height of a rotation-axis cylinder. The sphere is impenetrable. Integrating incompressible flow over the volume inside any such cylinder gives zero flux through its cylindrical side, since there is no flux through its spherical caps. ThereforeThis is why the Coriolis force disappears from the cylinder average; it need not vanish locally. An axisymmetric field in the insulating exterior is a potential field with $B_{\varphi }=0$, so continuity gives $B_{\varphi }=0$ on the sphere. Consequently both the end flux terms and the moving-endpoint terms in the magnetic-stress integral vanish. Expanding the divergence givesSince $U$ is independent of $z$, its integrated time derivative is $2z_b{U}_t$. Defining $T={\int }_{-z_b}^{z_b}{B}_r{B}_{\varphi }dz$ givesThe stationary background and the fact that $U/r$ depends only on $r,t$ also givesoTogether these form the cylinder-averaged torsional Alfvén wave system.

Assume $H>0$ on the cylinders carrying the wave, with regular endpoint limits. If $H$ vanishes on an interior cylinder, then $B_r$ and $T$ vanish there and that cylinder has no radial magnetic coupling; the displayed $H^{-1}$ energy term must be interpreted on the coupled region rather than literally dividing by zero. Since $H$ is time independent, differentiation of the proposed conserved energy of a cylinder-averaged torsional wave givesRegularity at the rotation axis makes the lower endpoint vanish, while vanishing magnetic torque at the collapsed outer cylinder makes the upper endpoint vanish. This provesThe magnetic term has a direct interpretation. Decompose $B_{\varphi }=(T/H)B_r+B_{\varphi }^{\perp }$, with $\int B_r{B}_{\varphi }^{\perp }dz=0$. Then $\int B_{\varphi }^{2}dz=T^2/H+\int (B_{\varphi }^{\perp }{)}^{2}dz$, and the induction equation gives $(B_{\varphi }^{\perp }{)}_t=0$. Thus, apart from a common dimensional factor and a stationary orthogonal magnetic component, the conserved quantity is the physical kinetic plus magnetic energy.

For the frequency estimate, put $V=U/r$ and eliminate $T$:Locally this is a wave equation with speedIt is the cylinder-averaged radial field component that matters, not simply the dipole measured at the Earth's surface. Taking a cylinder height of order $6\times 10^6\mathrm{m}$, $\rho \sim 10^4\mathrm{kg}{\mathrm{m}}^{-3}$, and an assumed internal cylindrical radial field $B_{r,\mathrm{rms}}\sim 0.2$--$2\mathrm{mT}$ givesFor a standing wave of radial length $L_c\sim 2$--$3.5\times 10^6\mathrm{m}$, the cyclic frequency is of order $c_T/(2L_c)$ and the angular frequency is of order $\pi c_T/L_c$. HenceGeometry and the assumed internal field can shift the estimate substantially. As a concrete calibration, the 2010 torsional-wave study inferred a cylindrical radial rms field near $2\mathrm{mT}$ and a roughly four-year crossing time, with an observed six-year recurrence. A much weaker assumed interior field instead gives the older multi-decadal estimate. The paper supplies no numerical interior field strength, so a unique numerical frequency cannot be deduced without such an assumption.

Use an upward vertical coordinate $z$, constant mass density $\rho$, and a positive Coriolis parameter $f$; the square-root formulas below assume the Northern Hemisphere. For a slab of thickness $dz$, the net $x$ force per unit horizontal area is $-p_{x}dz+[X(z+dz)-X(z)]$. Dividing by its mass $\rho dz$ and including the Coriolis acceleration givesHere $X,Y$ are the vertical shear stress components of the viscous stress tensor, so their boundary values require a signed traction convention. Linearization removes advective acceleration. Decompose the horizontal velocity into an exterior pressure response and an Ekman layer correction. On an f-plane these satisfyandThe exterior pressure response is in geostrophic balance when steady. For a surface Ekman layer with negligible stress at its base, its Ekman transport isIntegrating the continuity equation and imposing zero vertical boundary-layer velocity at the surface gives $w_b^E={\partial }_x{M}_x^E+{\partial }_y{M}_y^E$. Thus the base velocity, positive upward, isThe PDF prints the opposite sign. Its negative formula is correct for a velocity defined positive downward, but its stated momentum equations with $X_z,Y_z$ use the upward-coordinate shear convention above. This sign must be accounted for when interpreting the subsequent pumping equation.

For constant kinematic viscosity, write $W_E=u^E+i{v}^E$ and $\delta =\sqrt{2\nu /f}$. The steady Ekman layer equation is $\nu W_E^{\prime \prime }=if{W}_E$. Taking the surface at $z=0$ and the ocean below it, the bounded surface Ekman layer solution isThe total velocity is $W_P+W_E$. Prescribed wind stress determines the coefficient of this Ekman layer correction; it does not determine a relation between the wind and the independent pressure-driven velocity. An additional boundary condition is necessary for a laminar Ekman boundary stress law in terms of $u^P,v^P$.

The two printed laminar stress formulas and the positive pressure-Laplacian pumping formula are instead the standard Bottom Ekman layer relations. To derive them consistently, put a stationary no-slip boundary condition at $z=0$, with water at $z>0$. ThenConsequentlyThese are shear stress values; the actual bottom traction on the fluid has the opposite sign. Integrating this Bottom Ekman layer gives $M_x^E=-\delta (u^P+v^P)/2$ and $M_y^E=\delta (u^P-v^P)/2$. Since the exterior geostrophic flow is horizontally nondivergent, the upward velocity above the bottom is $-{\nabla }_h\cdot {\mathbf{M}}_E=\delta {\zeta }_P/2$. Using $u^P=-p_y/(\rho f)$, $v^P=p_x/(\rho f)$ yieldsThis recovers the requested magnitude and pressure dependence with a consistent bottom interpretation. The surface version would require its own specified boundary velocity and corresponding signs.

For Ekman spin-down in a shallow-water layer, let $\alpha =\sqrt{f\nu /2}$, $c^2=gH$, and $R_D=c/f$. Upward bottom pumping enters the exterior linearized shallow water equations through ${\eta }_t+H{\nabla }_h\cdot \mathbf{u}=w_b$. Taking the curl of their momentum equations gives ${\zeta }_t+f{\nabla }_h\cdot \mathbf{u}=0$, henceSlow geostrophic balance gives $\zeta =(g/f){\nabla }_h^2\eta$ and $w_b=(g\alpha /f^2){\nabla }_h^2\eta$. Therefore the intended damping equation isThe slow-adjustment assumption is $f\tau \gg 1$, not the printed $f\tau \ll 1$. The printed negative pumping term in continuity can alternatively describe downward extraction at the upper boundary, but it cannot be combined unchanged with the upward bottom pumping just derived. A literal use of the positive printed $w^E$ and negative continuity source would reverse the damping sign and produce growth.

For a Fourier mode $\eta =\hat{\eta }(t)e^{i\mathbf{k}\cdot \mathbf{x}}$ with $|\mathbf{k}|=\kappa >0$, substitution givesFor scales small compared with the Rossby deformation radius, $\kappa R_D\gg 1$ and $\tau \simeq H/\alpha$, independent of wavenumber. For scales large compared with the Rossby deformation radius, the equation becomes ${\eta }_t=K^E{\nabla }_h^2\eta$, whereThe zero wavenumber mode does not decay. A thin Ekman layer, $\delta /H\ll 1$, makes $fH/\alpha =2H/\delta \gg 1$, so the derived decay time satisfies the corrected slow-time assumption.

The vertical component $q_z=\zeta -f\eta /H$ is $H$ times the first-order anomaly in shallow-water potential vorticity about rest:Thus $\mathbf{q}=q_z\hat{\mathbf{z}}$ records the combination of relative vorticity and column stretching that is conserved by the unforced linearized shallow water equations. Indeed their curl and continuity equation give $(q_z{)}_t=0$. It is the persistent forcing of the balanced component during geostrophic adjustment, while inertia-gravity waves carry away the unbalanced component.

Initially $q_z=-f{\eta }_0\mathrm{sgn}(x)/H$. In the final geostrophic flow, $u=0$, $v=g{\eta }_x/f$, and $\zeta =g{\eta }_{xx}/f$. Conservation of the linearized shallow-water potential vorticity therefore givesBoundedness on both sides and continuity of $\eta ,{\eta }_x$ at the origin determine the constants:The solution is the locally attained balanced state after outgoing inertia-gravity waves have departed; the full inviscid system need not lose its globally conserved wave energy. The signed volume flux in the $y$ direction is

For double Kelvin-wave adjustment at a depth step, put $c_{\pm }=\sqrt{g{H}^{\pm }}$ and use $R_D^{\pm }=c_{\pm }/f$ on the corresponding side of the step. The PDF's far-field wording must mean $\eta \to {\eta }_0\mathrm{sgn}(x)$, rather than ${\eta }_0$ for every $x$. In the exponentially trapped outer response, cross-step geostrophic balance gives, at $y=0$,To retain the propagation of this slowly varying trapped disturbance, the along-step acceleration in $u_t-fv=-g{\eta }_x$ must also be retained. It supplies the small cross-step ageostrophic flow. Thus the two cross-step volume fluxes areHere $d\mathrm{sgn}(x)/dx=2\delta (x)$ as a distributional derivative. Equating the fluxes and dividing by $c_{+}+c_{-}$ givesThe printed forcing is missing a factor of two. It cannot follow from the printed sign-function surface profile with the usual Dirac delta distribution normalization. Pure geostrophic balance in both horizontal directions would also omit $A_t$ and could not determine the wave evolution.

The method of characteristics with zero initial amplitude gives the corrected weak solutionBehind the eastward front, $0<x<\Delta ct$, the surface elevation at the step becomes $-{\eta }_0$, matching the upstream side. All the original cross-step transport is diverted into flow along the step once the wave has passed a fixed location: $v=0$ there, with opposing along-step velocities above and below the step. This is the complete barrier in the adjusted region, and in the long-time limit at every fixed $x$. At finite time a front still carries cross-step flux: the formulas above give $H^{\pm }{v}^{\pm }=(2{\eta }_0{c}_{+}{c}_{-}/f)\delta (x-\Delta ct)$. The discontinuous outer representation does not resolve the short-scale fast adjustment at either front.

For completeness, solving the equation literally as printed instead gives $A={\eta }_0[\Theta (x)-\Theta (x-\Delta ct)]$. Its amplitude is half that required; $\eta (x,0,t)$ retains a jump ${\eta }_0$ at the origin after passage, so that literal equation does not imply complete blocking. The corrected equation and the literal printed equation are distinguished rather than asserting an inconsistent barrier result.

For a zonal basic geostrophic flow $U(y)$, take ${\psi }_0^{\prime }(y)=-U(y)$ and write $\psi ={\psi }_0+\varphi$. The flat-bottom shallow-water quasi-geostrophic potential vorticity gradient of the basic state isLinearizing its material conservation gives the Rossby-wave equation for a sheared zonal current,For a general nonuniform jet, a global two-dimensional plane wave is not an exact normal mode: the coefficients depend on $y$. The exact zonal normal mode problem, $\varphi =\hat{\varphi }(y)e^{i(kx-\omega t)}$, iswith appropriate transverse boundary or radiation conditions. If the jet varies slowly compared with a wavelength, a local plane wave freezes these coefficients at $y_0$. Its local dispersion relation isIn the nondivergent Barotropic Rossby wave model under a rigid-lid approximation, this becomes $\omega =kU-k(\beta -U^{\prime \prime })/(k^2+l^2)$. It is exact for uniform $U$ and otherwise a local relation. Retaining the free-surface term also requires retaining the basic surface slope in $Q_0^{\prime }$; simply adding $R_D^{-2}$ to the denominator while dropping $R_D^{-2}U$ from the numerator would describe a different prescribed-background model.

Use a one-layer, hydrostatic, homogeneous fluid with reference depth $H_0$, no forcing or friction, small Rossby number $ϵ=U/(|f_0|L)\ll 1$, slow evolution, and leading geostrophic balance. For the beta plane, take $\beta L/|f_0|=O(ϵ)$; the relative free-surface displacement is also small, $\eta /H_0=O(ϵ)$, with the Rossby deformation radius retained at the chosen horizontal scale. This excludes the equatorial limit. If a bottom elevation $h_b$ is included, additionally require $h_b/H_0=O(ϵ)$ and stationary topography. Set $h_b=0$ for a flat bottom.

The quasi-geostrophic streamfunction satisfies $\psi =g\eta /f_0$, $u=-{\psi }_y$, $v={\psi }_x$, so relative vorticity is $\zeta ={\nabla }_h^2\psi$. Expanding the shallow-water potential vorticity using actual depth $H_0+\eta -h_b$ givesConsequently the depth-scaled anomaly, the shallow-water quasi-geostrophic potential vorticity, isThe leading material derivative uses the geostrophic flow. Define $J(a,b)=a_x{b}_y-a_y{b}_x$; then conservation yields $Q_t+J(\psi ,Q)=0$. The term $-\psi /R_D^2$ represents free-surface stretching, $\beta y$ the variation of planetary rotation, and $f_0{h}_b/H_0$ topographic stretching. Under a rigid lid or at scales much smaller than the Rossby deformation radius, the free-surface term is absent or negligible and the flat-bottom Barotropic Rossby wave model has $Q={\nabla }_h^2\psi +\beta y$.

Write $\mathbf{g}=-g{\mathbf{e}}_z$. The vertical component of the balance gives hydrostatic pressure,Its horizontal gradient is ${\nabla }_{h}p=\rho g{\nabla }_{h}h$. The horizontal balance is consequentlyThus $\mathbf{u}\cdot {\nabla }_{h}h=0$: the velocity is tangent to every level set of $h$, so those contours are streamlines of the geostrophic flow.

Across a vertical section joining the contours $h_0$ and $h_0+\Delta h$, an element of horizontal distance normal to the contours is $dn=dh/|{\nabla }_{h}h|$, while $|\mathbf{u}|=(g/f)|{\nabla }_{h}h|$. The volumetric flow rate is thereforeFor dimensional analysis, $[g]=L{T}^{-2}$, $[f]=T^{-1}$ and $[h_0]=[\Delta h]=L$, so the result has dimension $L^3{T}^{-1}$, exactly that of a volume flux.

Write $H_{\pm }=H_0\pm d$, with the upper sign referring to $y>0$, and use an f-plane with Coriolis parameter $f_0>0$ for the stated directions. The formulas retain the signed $f_0$; a positive Rossby deformation radius means $R_{\pm }=\sqrt{g{H}_{\pm }}/|f_0|$. Let $\rho$ be the constant fluid density.

Far from the depth step the problem is independent of $y$. The linearized shallow water equations imply conservation of $v_x-f_0\eta /H_{\pm }$. Initially this is $-f_0{\eta }_0\mathrm{sgn}(x)/H_{\pm }$. In geostrophic balance, $u_s=0$ and $v_s=g{\eta }_{s,x}/f_0$, soBoundedness and matching of height and its derivative at $x=0$ give geostrophic adjustment of a surface-height jump:The potential energy release per unit length in $y$ is the finite difference between two individually infinite energies:For comparison, the final kinetic energy is $\rho g{\eta }_0^2{R}_{\pm }/2$, so the energy radiated in inertia-gravity waves is $\rho g{\eta }_0^2{R}_{\pm }$. Integrating the depth-weighted velocity over $x$ gives the volume transportsFor $f_0,{\eta }_0>0$, $Q_{-}$ is towards the step and $Q_{+}$ is away from it. Their difference is $4g{\eta }_{0}d/f_0$; a global steady state cannot simply join the two far-field currents without additional transport along the step.

To obtain the rotating shallow-water height equation with variable depth, set $\zeta =v_x-u_y$ and $J_H={\partial }_x(Hv)-{\partial }_y(Hu)$. Direct differentiation of the original momentum and continuity equations givesConsequentlyAt the step, $H_y=2d\delta (y)$ is a Dirac delta distribution; the original matching conditions are used rather than assuming a smooth depth there. Velocity elimination raises the time order: initially ${\eta }_t=0$ and ${\eta }_{tt}=g{\nabla }_h\cdot (H{\nabla }_h\eta )$, as well as the prescribed initial height.

For a decaying harmonic disturbance, the height equation on each constant-depth side givesSolving the two momentum equations yieldsContinuity of $H\hat{v}$ at the step gives the implicit step-trapped topographic Rossby wave dispersion relationThis derivation assumes ${\omega }^2\ne f_0^2$ and positive decay rates; it describes the low-frequency trapped branch, rather than the radiating inertia-gravity waves.

In the quasi-geostrophic approximation, replace $f_0^2-{\omega }^2$ by $f_0^2$. The resulting explicit frequency, and then its leading small-$d/H_0$ form, areFor $d>0$, the phase velocity has the sign of $f_0$. Thus propagation is eastward in the Northern Hemisphere, with the shallower side on the right. Differentiating the leading dispersion relation givesIn the long-wave limit the group velocity and phase velocity coincide:

To find the slow amplitude, integrate the low-frequency height equation across the step. This gives ${\partial }_t[H{\eta }_y]=f_0[H]{\eta }_x$, where brackets mean the upper-side minus lower-side value. For the given exponentially trapped outer profile, $[H{\eta }_y]=2H_{0}A/R_D$ and ${\eta }_x(x,0)=2{\eta }_0\delta (x)-A_x$. Hence long-wave transport along a depth step obeysThe method of characteristics, interpreted for the Dirac delta distribution source, givesFor $c_T>0$ this is $2{\eta }_0$ in $0<x<c_{T}t$ and zero outside: a surface depression is left in the wake of the wave. The sharp $x$-jump is a long-wave outer approximation. The fast adjustment region of width $O(R_D)$ near the origin, and the detailed wavefront, are unresolved. Keeping a smoothed background ${\eta }_s(x)$ instead would give $A_t+c_T{A}_x=c_T{\eta }_s^{\prime }(x)$ and $A={\eta }_s(x)-{\eta }_s(x-c_{T}t)$ at this order.

The along-step geostrophic flow is $u=-g{\eta }_y/f_0$. Integrating separately over the two sides givesIn the wake, $T_{-}=Q_{-}$ and $T_{+}=-Q_{+}$. The shallow-side transport is eastward and the deep-side transport westward, with magnitudes equal to the respective far-field currents. The net along-step transport is $T_{-}+T_{+}=-4g{\eta }_{0}d/f_0$, balancing their mismatch. These comparisons use the stated leading long-wave and small-depth-contrast approximations.

Represent a horizontal vector $(u,v)$ by the complex number $W=u+iv$, let $s=\mathrm{sgn}f$, and define the two Ekman layer depths and drag coefficients byThe atmospheric and oceanic departures from their respective geostrophic flow satisfyThe solutions that decay away from the ice areandHere $W_a^{\infty }$ and $W_o^{\infty }$ denote the atmospheric and oceanic geostrophic velocities.

The viscous stress exerted on the ice by the atmosphere and ocean is, respectively,Because the ice is an infinitesimally thin, freely moving sheet, its horizontal force balance is ${\tau }_a+{\tau }_o=0$. The common Ekman turning factor cancels, leavingThus the ice moves along the weighted mean of the two geostrophic currents. In particular, as ${\nu }_a\to 0$ one has $r_a\to 0$ and $W_i\to W_o^{\infty }$: an atmosphere with vanishing viscosity transmits no finite stress to the ice.

The atmospheric Ekman transport relative to its geostrophic current isThe atmospheric stress isIf the two geostrophic currents are parallel but unequal, both directions are obtained by rotating their velocity difference: in the Northern Hemisphere the stress lies $45^{\circ }$ anticlockwise from $W_a^{\infty }-W_o^{\infty }$, while the atmospheric transport lies $45^{\circ }$ clockwise from its negative. Both rotations reverse in the Southern Hemisphere. If the two currents are identical, the shear, stress, and relative Ekman transport all vanish.

Potential-vorticity dynamics. The quasi-geostrophic potential-vorticity equation expresses material conservation of a balanced potential vorticity by a horizontally nondivergent geostrophic flow. Its terms represent relative vorticity, planetary vorticity and the layer-thickness/free-surface contribution. The Rossby deformation radius $L_d$ measures the scale at which that thickness contribution competes with the horizontal-vorticity term. The quasi-geostrophic approximation filters fast acoustic waves and gravity waves, making the slower balanced evolution accessible to early digital forecasts. The 1950 barotropic-vorticity forecasting study demonstrated practical twenty-four-hour forecasts from observed mid-tropospheric fields, while identifying discretization and omitted baroclinicity as error sources.^[ref]

Near the north pole put $\delta =\pi /2-\theta$ and $r\simeq a\delta$. ThenUsing projected rather than arc distance gives the same leading polar expansion. It is a north-polar expression; the sign structure is changed consistently at the opposite pole.

In cylindrical coordinates, horizontal advection is $u_r{\partial }_{r}Q+(u_{\varphi }/r){\partial }_{\varphi }Q$. Substituting the specified streamfunction givesThe factors of $1/r$ are required by cylindrical geometry.

An exact traveling eigenfunction. This is a polar gamma-plane Rossby wave. The radial equation implies ${\nabla }_h^2\psi -\psi /L_d^2=-{\alpha }^2\psi$, so $Q=f-{\alpha }^2\psi$. Since $J(\psi ,\psi )=0$, its nonlinear self-advection vanishes exactly, regardless of amplitude. Also $f_r=-\gamma r$, giving $J(\psi ,f)=\gamma {\psi }_{\varphi }$. HenceFor $\psi =F(r)\cos (\omega t-n\varphi )$, this yieldsThe wave is retrograde relative to the balanced mean flow, since $\omega /n<0$ in this convention.

Multiplying the radial equation by $r^2$ givesFor $K^2={\alpha }^2-L_d^{-2}>0$, setting $y=Kr$ gives the standard Bessel equation with derivatives with respect to $y$. ThusRegularity at the pole excludes $Y_n$ for the usual nonnegative integers. If $K^2<0$, the corresponding general solutions are modified Bessel functions; at $K=0$ they are $r^{\pm n}$, with a constant and $\ln r$ for $n=0$. The printed Bessel hint mixes its independent-variable labels; using $y=Kr$ and derivatives in $y$ resolves that notation.