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Use the real magnetic field $\mathcal{B}$ in the coupled equations; its leading periodic approximation will be $\mathrm{Re}(\mathbf{B}{e}^{i\omega t})$. In the conducting incompressible flow, the magnetohydrodynamic momentum equation, resistive induction equation and solenoidal magnetic-field constraint areHere $p$ is pressure, $\mathbf{j}$ is electric current density, and $\eta$ and $\nu$ are magnetic diffusivity and kinematic viscosity. The moving-conductor Ohm law gives $\mathbf{E}=-\mathbf{u}\times \mathcal{B}+\eta \nabla \times \mathcal{B}$.

In the insulating exterior, neglecting displacement current, the magnetic field is curl-free away from the wire and solenoidal. The imposed electric current density suppliesThe exterior electric field also satisfies Faraday's law. At the rigid wall, impose the no-slip boundary condition $\mathbf{u}=0$. At finite conductivity, the interface conditions for electromagnetic fields give continuous normal magnetic field, continuous tangential magnetic field for equal permeability and no prescribed singular surface current, and continuous tangential electric field. The normal electric current density is zero at an insulating wall; it is automatic for the present two-dimensional fields with current only along $z$. The induced fields decay far from the wire and wall, and initial conditions, or selection of the periodic state after transients, complete the specification. In the limiting perfect conductor approximation a surface current can emerge, permitting a tangential-field jump.

The fully coupled solution need not remain monochromatic: a harmonic magnetic field produces both a mean and a twice-frequency Lorentz force density, and the resulting velocity can generate further harmonics. The following parts consistently use the leading magnetic response with fluid motion neglected.

Let $K$ be the kinetic energy, $M$ the magnetic energy and $E=K+M$:Dot the magnetohydrodynamic momentum equation with $\mathbf{v}$ and integrate. Incompressibility and the zero boundary velocity remove the advective and pressure work. The viscous term integrates to $-\rho \nu {\int }_D|\nabla \mathbf{v}{|}^2$. ThusThe ideal magnetohydrodynamic induction equation gives, with no boundary contribution,The Lorentz force density therefore transfers energy between the field and the fluid without creating total energy. Since $\nabla \cdot \mathbf{v}=0$ and $\mathbf{v}=0$ on the boundary, another integration by parts gives $\int |\nabla \mathbf{v}{|}^2=\int |{\mathbf{\omega }}_v{|}^2$, where ${\mathbf{\omega }}_v=\nabla \times \mathbf{v}$ is vorticity. The viscous magnetic-relaxation energy identity isThe conserved nonzero magnetic helicity prevents all the field energy from disappearing. Fix a bounded inverse-curl convention for the magnetic vector potential, so that $\Vert \mathbf{A}{\Vert }_2\le C_D\Vert \mathbf{B}{\Vert }_2$. The Cauchy-Schwarz inequality gives the magnetic-energy lower bound from helicity,Consequently $E$ decreases to a finite positive limit andUnder the intended regular long-time interpretation of smooth vorticity, the dissipation cannot keep producing peaks of fixed height and arbitrarily short duration. More precisely, if $d(t)=\Vert {\mathbf{\omega }}_v(t){\Vert }_2^2$ is uniformly continuous, uniformly continuous integrable dissipation tends to zero proves $d(t)\to 0$. The Poincaré inequality and the identity above then imply $\Vert \mathbf{v}(t){\Vert }_2\to 0$.

To identify the asymptotic state, suppose the regular relaxation has a limit, or work on a compact invariant limiting set on which the energy is continuous. On such a set the energy is constant, so the displayed identity forces ${\mathbf{\omega }}_v=0$, and the zero boundary velocity forces $\mathbf{v}=0$. The induction equation is then stationary, and the momentum equation becomesThis is magnetostatic equilibrium, with nonzero field allowed and required here by conserved magnetic helicity. It is not necessarily a force-free magnetic field: a pressure gradient can balance the magnetic force. If the field develops a current sheet, equilibrium is understood with the corresponding interface force balance.

There is a mathematical qualification to the smoothness premise. Smoothness at each finite time alone does not prove uniform long-time bounds, uniform continuity of dissipation or convergence to one limiting field. The energy calculation proves finite total dissipation unconditionally for a smooth solution; the settling conclusion is the regular-relaxation argument just given. A rigorous global convergence theorem requires that additional long-time control. In particular, smooth vorticity does not require smooth electric current density in the limiting state.

Work with the complex representation and take its real part to obtain the physical magnetic field. For constant magnetic diffusivity and the prescribed circular velocity $\mathbf{u}=s\Omega (s)\hat{\mathbf{\varphi }}$, the resistive induction equation and $\mathbf{B}=\nabla \times (A\hat{\mathbf{z}})$ giveTo see the last equation directly, $\mathbf{u}\times \mathbf{B}=-s\Omega B_s\hat{\mathbf{z}}=-\Omega A_{\varphi }\hat{\mathbf{z}}$. Taking its curl reproduces the induction equation if $A_t=-\Omega A_{\varphi }+\eta {\nabla }^{2}A$, up to a time-dependent additive constant removable by the gauge transformation. This is the scalar form of advection and magnetic diffusion in cylindrical coordinates.

Put $\psi =\varphi -\Omega (s)t$. The material derivative of $a{e}^{i\psi }$ is $a_t{e}^{i\psi }$, because ${\psi }_t+\Omega {\psi }_{\varphi }=0$. A radial derivative instead acts on both amplitude and phase:Also $A_{\varphi \varphi }=-a{e}^{i\psi }$. Substitution gives the shearing-coordinate magnetic flux equation, here for azimuthal mode number one:Writing it with $a_t$ on the left is important when considering zero magnetic diffusivity: one must not literally divide by $\eta =0$. In the ideal limit, $a_t=0$, so $A(s,\varphi ,t)=A(s,\varphi -\Omega (s)t,0)$. This is magnetic flux freezing: the scalar flux pattern is carried with each rotating fluid ring. It does not make the magnetic field stationary. Indeed,so differential rotation winds the field and can make its azimuthal component grow linearly in time. Rigid rotation, for which ${\Omega }^{\prime }=0$, only rotates the pattern.

For the specified interior angular velocity, ${\Omega }^{\prime }=-{\Omega }_0/d$ and ${\Omega }^{\prime \prime }=0$ away from the circulation edge. Let $T={\Omega }_{0}t$, $x=s/d$ and $\mathrm{Rm}={\Omega }_0{d}^2/\eta$. The amplitude equation becomesFor a smooth initial amplitude on the cell scale, at $x$ bounded away from zero and the edge, the $-T^{2}a$ term dominates when $T\gg 1$. Its accumulated damping is order one at $T=O({\mathrm{Rm}}^{1/3})$, whereas the integrated terms proportional to $T$ are only $O({\mathrm{Rm}}^{-1/3})$ and the terms independent of $T$ are $O({\mathrm{Rm}}^{-2/3})$. The resulting cubic-time resistive damping in a differentially rotating cell isThe magnetic field has this leading exponential decay envelope. Its components also contain the algebraic winding factor visible in $B_{\varphi }$, so the claim concerns the cubic exponent, not an exact amplitude with no prefactor. Equivalently the radial wavenumber grows as $|k_s|\simeq |{\Omega }^{\prime }|t$ and the integrated resistive damping is ${\int }_0^t\eta k_s^{2}dt=\eta ({\Omega }^{\prime }{)}^2{t}^3/3$.

This is phase mixing in magnetic flux expulsion. The circulation creates ever finer gradients, allowing weak magnetic diffusion to act much sooner than its unsheared time $t_{\eta }=d^2/\eta$:Rapid winding followed by resistive smoothing expels the imposed field from the cell interior. The maintained exterior magnetic field is redistributed into a circulation-edge boundary layer. The result is an interior high-magnetic Reynolds number transient, not a uniform assertion at the axis and the nonsmooth edge, nor the exact $t\to \infty$ solution with a continuously imposed far field. The diffusion length at ${\tau }_c$ is $\sqrt{\eta {\tau }_c}/d=3^{1/6}{\mathrm{Rm}}^{-1/3}$, identifying where a local bulk approximation can fail. The linear profile also needs smoothing at a regular rotation axis and at $s=d$ if used as a fully smooth global flow.

In photospheric magnetoconvection, circulating and diverging convection transports weak magnetic flux towards converging downflow regions. Solar granulation concentrates field in intergranular lanes, while larger-scale supergranulation helps form the solar magnetic network. Flux expulsion explains why a highly conducting photosphere can have relatively weak field in cell interiors and intermittent concentrations at their edges: high electrical conductivity delays diffusion on the original scale, but does not prevent diffusion on the fine scales created by advection. A long-lived circulation must last long enough for the expulsion time; rapidly changing cells need not reach this idealized state. Once the field becomes strong, the Lorentz force modifies the convection, so this prescribed-flow calculation alone neither determines the final field strength nor describes strong-field sunspots.

The wind integrals require a steady axisymmetric magnetohydrodynamic wind in ideal magnetohydrodynamics, with no viscous azimuthal torque or imposed toroidal loop voltage. Axisymmetry by itself is insufficient. Assume smooth connected poloidal flux surfaces with nonzero poloidal magnetic field; functions of the flux label below are understood on each such branch. In cylindrical coordinates, the given poloidal magnetic flux function obeysFor a steady ideal wind, the electric field is $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$ and $\nabla \times \mathbf{E}=0$. Its toroidal component satisfies ${\partial }_z{E}_{\varphi }=0$ and ${\partial }_s(s{E}_{\varphi })=0$, so $E_{\varphi }=C/s$. Regularity at the axis, or equivalently the absence of an imposed loop voltage in an annular domain, gives $C=0$. Thus $({\mathbf{u}}_p\times {\mathbf{B}}_p{)}_{\varphi }=0$ and ${\mathbf{u}}_p=f{\mathbf{B}}_p$. The continuity equation and $\nabla \cdot {\mathbf{B}}_p=0$ now yieldConsequently $\kappa$ is the magnetohydrodynamic mass loading. Including the toroidal magnetic field, rewrite the full velocity asSince $\mathbf{u}\times \mathbf{B}=\omega \nabla \chi$, steady ideal induction gives $0=\nabla \omega \times \nabla \chi$. Hence $\omega =\omega (\chi )$, establishingThe invariant $\omega$ is the field-line angular velocity of an axisymmetric wind, not generally the fluid's angular velocity $\Omega$. When an ideal magnetic surface is anchored to a rigidly rotating star, $\omega$ equals the stellar angular velocity. With no poloidal outflow this reduces to Ferraro's law of isorotation.

To calculate the magnetic torque, use the azimuthal momentum equation. Axisymmetric pressure and Newtonian gravity have no azimuthal components; the magnetic pressure has none either. The azimuthal acceleration and magnetic tension contain the curvature terms characteristic of cylindrical coordinates:Multiplication by $s$ gives the Maxwell torque conservation in an axisymmetric wind equationUsing $u_{\varphi }=s\Omega$, $\rho {\mathbf{u}}_p=\kappa {\mathbf{B}}_p$ and ${\mathbf{B}}_p\cdot \nabla \kappa =0$ provesThis is the total angular momentum transport per unit poloidal magnetic flux. The first term is the material contribution and the second is the contribution of the Maxwell stress tensor. The specific magnetohydrodynamic angular-momentum invariant is $L=\ell /\kappa$ when $\kappa \ne 0$; its normalization differs from the paper's $\ell$.

Define the poloidal Alfvén Mach number bySubstituting $\Omega =\omega +\kappa B_{\varphi }/(\rho s)$ into $\ell$ givesAt a smooth crossing of the Alfvén surface, $M_A^2=1$ and the toroidal magnetic field is finite. The Alfvén-surface regularity condition for an axisymmetric wind is thereforeOne can also expose the apparent singularity explicitly:Finite passage through the Alfvén surface requires the numerator to vanish with the denominator. Equality of the numerator zeros is necessary; a globally smooth wind must also satisfy its other dynamical critical conditions. Here $s_A$ is the cylindrical distance from the rotation axis, not automatically the spherical Alfvén radius.

For a tube carrying outward mass flux $d\dot{M}=\kappa {\mathbf{B}}_p\cdot d\mathbf{S}$, the outward angular momentum flux isThus the stellar torque is ${\dot{J}}_{\star }=-\int \omega s_A^{2}d\dot{M}$, with the integral over the escaping wind. Magnetic stresses give each mass element an effective angular-momentum lever arm equal to the cylindrical Alfvén crossing radius. In a strongly sub-Alfvénic region the fluid approximately corotates with the magnetic surface; beyond that region it can carry the accumulated angular momentum away. For an outward field and positive $\omega$, a super-Alfvénic expanding tube with $s>s_A$ has $B_{\varphi }<0$, so the magnetic stress transports positive angular momentum outwards. If $s_A$ greatly exceeds the stellar surface lever arm, a modest mass-loss rate can cause substantial wind-driven magnetic braking of a solar-type star. An unmagnetized outflow would instead carry approximately the material angular momentum fixed at its launch radius.

The pressure tensor has Cartesian components $p_{ab}=\int f(v_a-u_a)(v_b-u_b)d^{3}v$. With zero mean velocity and an isotropic galactic distribution function, off-diagonal components vanish by reflection symmetry and diagonal components are equal by rotational symmetry. ThereforeHere $p$ is a tracer number-weighted random-motion stress, rather than thermodynamic pressure per volume. Its tensor divergence is $(\mathrm{div}\mathbf{p}{)}_a={\partial }_b(p{\delta }_{ab})={\partial }_{a}p$. The time-independent, nonstreaming Jeans equation consequently reduces toThe gravitational acceleration convention is $\nabla \psi$: this $\psi$ differs by a sign from the usual Newtonian gravitational potential $\Phi$.

Let $z$ measure distance along the line of sight. Spherical density projection givesBecause isotropy makes the local second stellar velocity moment along every line of sight equal to $p/n$, and because there is no streaming motion, the projected moment isThis is the forward relation for isotropic stellar pressure deprojection.

Both inversions follow from one Abel transform calculation. Write $F(u)=N(\sqrt{u})$ and $a(v)=n(\sqrt{v})$, so $F(u)={\int }_u^{\infty }a(v)(v-u{)}^{-1/2}dv$. Compose the transforms, using Tonelli's theorem for nonnegative densities, or absolute convergence for signed profiles:Assume sufficient decay to make these integrals finite and sufficient regularity for the last derivative. Then $J^{\prime }(x)=-\pi a(x)$. Setting $x=r^2$ gives the spherical Abel deprojectionReplacing $F(u)$ by $N(\sqrt{u}){\sigma }^2(\sqrt{u})$ and $a(v)$ by $p(\sqrt{v})$ provesThese are the requested expressions, with $u=R^2$. The composed-transform argument avoids an invalid separate differentiation of a divergent lower-end kernel.

Assume axisymmetry in a thin disk orbiting a fixed central mass $M$. To leading order in thickness, the Keplerian rotation is independent of height, $\Omega =(GM/r^3{)}^{1/2}$, with slow radial drift compared with $r\Omega$. Neglect radial pressure support in the orbital balance, changes in the central potential, mass loss through the disk surfaces and applied surface torques. The Cauchy stress tensor can represent viscous or effective turbulent/magnetic transport; only its vertically integrated azimuthal component enters the conservation of angular momentum equation.

Define the surface density and its mass-weighted radial velocity byand set $W_{r\varphi }=\int T_{r\varphi }dz$. Integrating mass conservation vertically givesThe azimuthal equation in cylindrical coordinates, multiplied by $r$ and combined with mass conservation, is the conservative equation for specific angular momentum $l=r^2\Omega$. The cylindrical stress tensor divergence contributes $r^{-1}{\partial }_r(r^2{T}_{r\varphi })+{\partial }_z(r{T}_{\varphi z})$. With vanishing surface torque, vertical integration therefore givesBecause $l(r)$ is fixed in time, subtracting $l$ times the mass equation yieldsDefine the effective viscosity defined by disk stress byThis sign convention treats $T_{r\varphi }$ as the shear stress appearing with $+\nabla \cdot \mathbf{T}$. For the Newtonian fluid stress tensor it reduces to the density-weighted viscosity of a disk, $\bar{\nu }={\Sigma }^{-1}\int \rho \nu dz$, rather than an unweighted height average.

Using $l=\sqrt{GMr}$ and $W_{r\varphi }=-(3/2)\bar{\nu }\Sigma \Omega$ in the conservation of angular momentum equation givesSubstitution into mass conservation proves the Keplerian viscous diffusion equation:The outward viscous torque in an accretion disk is $\mathcal{G}=-2\pi r^2{W}_{r\varphi }=3\pi \sqrt{GM}\sqrt{r}\bar{\nu }\Sigma$.

For $\bar{\nu }=Ar$ with $A>0$, make the heat-equation transform for linear-radius disk viscosityWriting $y=\sqrt{r}$, the diffusion equation becomes $g_t=(3A/4)g_{yy}$; the displayed rescaling of $y$ gives exactly $\boxed{g_t=g_{xx}}$, with the original time unchanged. The torque is $3\pi \sqrt{GM}g$, so the zero-torque inner boundary condition at the origin becomes $g(0,t)=0$ on the half-line $x>0$.

A positive similarity solution obeying this boundary condition isFor example, substitution of $g=t^{-1}F(x/\sqrt{t})$ reduces the heat equation to $F^{\prime \prime }+(\eta /2)F^{\prime }+F=0$, and $F=C\eta e^{-{\eta }^2/4}$ satisfies it and vanishes at zero. A replacement of $t$ by $t+t_0$ gives a finite-mass initial profile at time zero if desired.

To determine the global integrals, transform the mass and angular momentum:The Gaussian integrals giveThus the accreting similarity disk with viscosity proportional to radius loses mass as $t^{-1/2}$ while conserving angular momentum; its characteristic radius grows proportional to $At$. The accretion rate at the origin is $-{\dot{M}}_d=2\pi C\sqrt{3/A}{t}^{-3/2}$. Its radial velocity ${\bar{u}}_r=r/t-3A/2$ shows inward flow at small radius and outward spreading at large radius. The integrable inner density singularity is part of the idealized extension to $r=0$.

For the other profile, direct differentiation givesIt also satisfies the heat equation, but has $g(0,t)=t^{-1/2}\ne 0$ and $g_x(0,t)=0$. It therefore has a nonzero inner torque and zero mass flow at the origin, not the previous zero-torque inner boundary condition. HereAn arbitrary positive amplitude multiplies these quantities. This constant-mass disk with viscosity proportional to radius spreads outward under a central torque that supplies angular momentum, with no ongoing central mass supply. It has the reflecting Neumann boundary condition for the heat equation and can be regarded as the spreading of an initially central mass distribution driven by an applied torque. It is distinct from the accreting zero-torque solution; the ideal origin does not model a finite stellar surface or a relativistic inner disk.

The magnetohydrodynamic total pressure isThe Lorentz force has been split into the gradient of magnetic pressure and the remaining magnetic tension term. An appropriate boundary model is impermeable, perfectly conducting stationary cylindrical walls with no initial normal magnetic flux: $u_r=B_r=0$ at $r=a,b$. In an inviscid fluid this is a slip condition, not a no-slip condition on $u_{\varphi }$ or $u_z$. It also makes the tangential ideal electric field vanish at the walls. Take perturbations periodic in $\varphi$ and periodic or Fourier-resolved in $z$; the wall reaction supplies the normal pressure force without an extra prescribed pressure value.

Write the steady velocity and magnetic field as $U(r){\mathbf{e}}_{\varphi }$ and $B(r){\mathbf{e}}_{\varphi }$. They satisfy both divergence constraints. Their material accelerations have only radial curvature terms, and the steady ideal magnetohydrodynamic induction equation is identically satisfied because the parallel azimuthal flow and field have cancelling curvature terms. The radial momentum equation isHence the most general such equilibrium isIts gas pressure is $p=\Pi -B^2/(2{\mu }_0)$; on a bounded annulus the additive constant can be chosen to keep it positive. Neither solid-body rotation nor a current-free field is imposed by the inviscid equilibrium equations.

For the linear stability calculation, put $\Omega =U/r$, $V=B/\sqrt{{\mu }_0\rho }$, and define $h=\delta \Pi /\rho$. Use axisymmetric normal modes proportional to $e^{ikz-i\omega t}$. Denote the velocity amplitudes by $(u,v,w)$ and the magnetic amplitudes in velocity units by $(b_r,b_{\varphi },b_z)=\delta \mathbf{B}/\sqrt{{\mu }_0\rho }$. Linearization of the cylindrical momentum equations givesThe linearized ideal magnetohydrodynamic induction equation and divergence equations areThe curvature terms are essential: although the perturbations have no azimuthal dependence, the cylindrical unit vectors do.

For $\omega \ne 0$, $b_r=b_z=0$. Introduce the radial Lagrangian fluid displacement by $u=-i\omega \xi$. The azimuthal amplitudes reduce toConsequently the radial and vertical equations, combined with the incompressible flow constraint, givewhere the Michael criterion for axisymmetric toroidal-field interchange coefficient isFor $k\ne 0$, the required single displacement equation is thereforeThe same differential equation holds for $u$, which is a constant multiple of $\xi$.

Set $y=r\xi$. Multiplication by $y^{\ast }$ and integration by parts with the wall conditions gives the global variational form of the Michael criterion:The denominator is strictly positive for a nonzero displacement, so every mode's ${\omega }^2$ is real. If $\mathcal{D}\ge 0$ everywhere, no exponentially growing mode exists. Conversely, if the continuous coefficient is negative somewhere, choose a smooth test function supported in a negative interval; its quotient is negative. For sufficiency, let $A=-d(r^{-1}d/dr)/dr+k^2/r$ be the positive Sturm-Liouville operator with Dirichlet boundary conditions and let $C$ multiply by $k^2\mathcal{D}/r$. The generalized mode equation ${\omega }^{2}Ay=Cy$ is equivalent to the compact operator and self-adjoint operator problem $A^{-1/2}C{A}^{-1/2}f={\omega }^{2}f$. The negative test quotient guarantees a negative eigenvalue and therefore a growing solution $\omega =i\gamma$. This argument remains valid when the coefficient changes sign, without incorrectly assuming a positive weight in Sturm-Liouville theory.

Thus, for any fixed nonzero vertical wavenumber,Equality is marginal. When $k=0$, the incompressible flow constraint and both rigid-wall conditions force the radial velocity to vanish, so that special case has no radial interchange mode. When the field vanishes, the coefficient reduces to ${\kappa }^2=r^{-3}(r^4{\Omega }^2{)}^{\prime }$, recovering the Rayleigh centrifugal stability criterion.

For a Keplerian disk, ${\kappa }^2={\Omega }^2$. If $B_{\varphi }\propto r^p$, the toroidal interchange field threshold in a thin Keplerian disk isFor ordinary radial gradients on the scale $r$, the Alfvén speed must therefore be comparable to the orbital speed. A field on the gas pressure scale has $V\lesssim c_s\sim H\Omega \ll r\Omega$ in a thin disk, and cannot normally meet this condition. Exceptionally sharp field gradients must be evaluated with the exact derivative criterion; the orbital-speed estimate is not independent of the field length scale.

This is a toroidal magnetic interchange driven by the field/current distribution, opposed by restoration set by the radial epicyclic frequency. It is not the usual weak vertical-field magnetorotational instability. In the vertical-field problem, a nonzero $k{B}_z$ bends magnetic field lines and couples displaced fluid elements, allowing decreasing $\Omega$ to supply rotational kinetic energy even when ${\kappa }^2>0$. Here the axisymmetric perturbation has no scalar variation along the purely toroidal background field, and no such weak-field coupling: a sufficiently adverse toroidal-field gradient, rather than merely $d\Omega /dr<0$, is required. Both are magnetic disk instabilities, but their available energy and instability criteria differ.

Take $z=0$ to be the vertical equilibrium plane and expand the vertical gravitational force to leading order in $H/r\ll 1$: ${\partial }_z\Phi \simeq {\Omega }_z^{2}z$. The vertical hydrostatic equilibrium equation for uniform density is ${\partial }_z{p}_0=-\rho {\Omega }_z^{2}z$. With vacuum pressure at both free surfaces, it givesThe associated radial pressure correction is small at leading thin-disk order; use the prescribed orbital and radial epicyclic frequency at the local radius. Here ${\Omega }_z$ is the vertical epicyclic frequency.

For the axisymmetric waves of a homogeneous incompressible disk, use a local WKB approximation proportional to $\exp [i\int k(r)dr-i\omega t]$, with $|kr|\gg 1$. Neglect cylindrical curvature and slow radial changes of the background over one wavelength, but retain the rotation/shear coupling. Let $(u,v,w)$ be the velocity perturbations and $h=p^{\prime }/\rho$. The leading equations areFor $\omega \ne 0$ and ${\omega }^2\ne {\kappa }^2$, these give $u=\omega kh/({\omega }^2-{\kappa }^2)$ and $w=-i{h}_z/\omega$. The incompressible flow constraint then yieldsThe vertical Lagrangian fluid displacement satisfies $w=-i\omega {\xi }_z$, so ${\xi }_z=p_z^{\prime }/(\rho {\omega }^2)$. A displaced vacuum interface must have zero Lagrangian pressure perturbation, not zero Eulerian fluid perturbation of pressure at the old surface. Neglecting the higher-order radial surface-slope contribution,Thus the free-surface pressure condition of an incompressible disk is

Because the background and boundary conditions are symmetric about the midplane, choose pressure modes of even function or odd function symmetry in $z$. For ${\omega }^2>{\kappa }^2$, put $q=k\omega /\sqrt{{\omega }^2-{\kappa }^2}$ and $Q=qH$. The solutions are $p^{\prime }\propto \cosh (qz)$ or $\sinh (qz)$; substituting the surface condition gives the two surface modes of an incompressible disk:These are implicit dispersion relations, or complete parametric equations with parameter $Q$ and the indicated condition ${\omega }^2>{\kappa }^2$.

For $0<{\omega }^2<{\kappa }^2$, put $s=k\omega /\sqrt{{\kappa }^2-{\omega }^2}$ and $S=sH$. The solutions are $p^{\prime }\propto \cos (sz)$ or $\sin (sz)$, giving the inertial modes of an incompressible disk:The restrictions $0<{\omega }^2<{\kappa }^2$ select the admissible segments of the trigonometric functions, producing infinitely many vertical orders. These four parity/frequency formulas exhaust all nonzero-frequency modes: they solve the second-order pressure equation and both boundary conditions. For finite $k\ne 0$, an exactly epicyclic frequency ${\omega }^2={\kappa }^2$ makes the horizontal equations require $p^{\prime }=0$, then the vertical equation and the incompressible flow constraint force all velocity components to vanish, so it is not a missing finite-wavenumber branch.

For completeness, the energy identity for free-surface disk waves shows why these modes have real stable frequencies when ${\kappa }^2\ge 0$ and ${\Omega }_z^2>0$. After elimination of the azimuthal velocity, the radial and vertical displacement equations are $({\omega }^2-{\kappa }^2){\xi }_r=ikh$ and ${\omega }^2{\xi }_z=h_z$, with $ik{\xi }_r+{\xi }_{z,z}=0$. Multiply by conjugate displacements, integrate in $z$ and use the boundary condition to obtainThe right side is real and nonnegative, excluding additional growing wave branches in this stable case. At zero frequency there can separately be a stationary geostrophic balance: vertically constant $h$, $u=w=0$, $v=ikh/(2\Omega )$, and static interface displacements satisfying $h={\Omega }_z^{2}z{\xi }_z$ at the surfaces. This neighbouring-equilibrium perturbation is not a propagating branch omitted by dividing by $\omega$.

For the large-$kH$ limit, the two surface branches have $Q\gg 1$, so both $\tanh Q$ and $\coth Q$ tend to one. Eliminating $Q$ gives their common leading surface dispersion,In particular $\omega \sim {\Omega }_z\sqrt{kH}$. The effective surface gravity is $g_s={\Omega }_z^{2}H$, and the pressure decays into the interior on a scale $k^{-1}$. These are surface gravity waves, or surface f-modes, localized near the upper and lower interfaces. Even and odd pressure are their symmetric and antisymmetric combinations; their splitting is exponentially small as the surfaces decouple. Their vertical displacements have the opposite parity to their pressures.

For each fixed inertial vertical order, large $kH$ means $\omega \ll \kappa$. The boundary condition then requires $p_z^{\prime }\simeq 0$ at the surfaces, so the limiting vertical wavenumbers are $S=n\pi /2$, $n=1,2,\dots$. ThusEven $n$ have even pressure, odd $n$ odd pressure. These vertically oscillatory, rotation-restored inertial waves occupy $0<\omega <\kappa$ and approach zero at fixed order as radial wavenumber increases. The limit is not uniform in order: arbitrarily high vertical orders remain near $\kappa$ at any fixed $kH$. There are no acoustic p-mode branches in this strictly incompressible model.

For ${\Omega }_z>\kappa >0$, the two surface curves are unique. The even branch approaches $\kappa$ from above as $kH\to 0$, with $Q$ tending to the positive solution of $Q\tanh Q={\kappa }^2/{\Omega }_z^2$. The odd branch approaches ${\Omega }_z$ because $Q\coth Q\to 1$ as $Q\to 0$. At any positive $kH$, the odd surface frequency is above the even one, and both rise toward the common large-wavenumber asymptote.

The inertial branches all approach $\kappa$ from below as $kH\to 0$. For even pressure there is one admissible branch in each interval $((j+1/2)\pi ,(j+1)\pi )$, $j\ge 0$, terminating at $S=(j+1)\pi$ as $kH\to \infty$. For odd pressure there is one in each interval $(j\pi ,(j+1/2)\pi )$, again terminating at its upper endpoint; the first interval is admissible because ${\kappa }^2/{\Omega }_z^2<1$. Their curves decrease toward zero as shown in the sketch. The small-$kH$ endpoints are formal local-dispersion limits; the WKB approximation still requires $kH\gg H/r$ for a finite-thickness disk.