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An inviscid uniform-density self-gravitating cylinder in vacuum, with no rotation or surface tension, has the displayed axisymmetric dispersion relation, where $\kappa =|k_z|R$. To derive it, let $W=\psi +\pi /\rho =ϵ{\Pi }_0{I}_0(kr)e^{i(kz-\omega t)}$. The linear kinematic boundary condition gives ${\Pi }_0={\omega }^2/[k{I}_0^{\prime }(\kappa )]$. The Lagrangian pressure perturbation vanishes on the moving free surface, giving $\pi (R)=2\pi G{\rho }^{2}R\eta$. Substitute the axisymmetric gravitational potential of a displaced cylindrical interface and use $I_0^{\prime }=I_1$ to obtain the relation. Its nonzero neutral root is ${\kappa }_c\simeq 1.0667$: wavelengths longer than $2\pi R/{\kappa }_c\simeq 5.89R$ have ${\omega }^2<0$ and grow. The fastest-growing mode has $\kappa \simeq 0.58$, so its wavelength is about $11R$. Negative ${\omega }^2$ must be interpreted through an imaginary frequency, not through a real oscillatory ansatz.

In a thin constant-density disk with free surfaces $z=\pm H$, neglect slow radial gradients and let perturbations vary as $e^{ikr-i\omega t}$. With $h=p^{\prime }/\rho$, the local inviscid equations are $-i\omega u-2\Omega v=-ikh$, $-i\omega v+{\kappa }^{2}u/(2\Omega )=0$, $-i\omega w=-h_z$ and $iku+w_z=0$. Eliminating velocity gives $h_{zz}=k^2{\omega }^{2}h/({\omega }^2-{\kappa }^2)$. This local WKB approximation assumes $H/r\ll 1$ and $|kr|\gg 1$ and has both surface and inertial wave branches.

// Target: astrophysics.bigb

For a clean inviscid bubble a distance $d\gg a$ below a gravity-restored approximately flat free surface, the surface displacement scales as $F/(\rho g{d}^2)=O(a^3/d^2)$. Zero normal velocity and zero tangential traction are imposed to leading order by an opposite normal image Stokeslet at distance $2d$ from the bubble. Its incident velocity is $-F/(8\pi \mu d)$, which gives the displayed correction at fixed buoyancy. The image can be represented by a same-radius inviscid drop of density $2\rho$ in liquid of density $\rho$. Its deformation-producing strain is $O(F/(\mu d^2))$, so the leading fractional capillary deformation is $O(\mathrm{Bo}(a/d{)}^2)$; small Bond number is the conventional stronger sufficient criterion.

A capillary wave is a wave along a fluid interface whose restoring stress is predominantly supplied by surface tension. A curved free surface requires a pressure jump by the Young–Laplace equation; increasing its corrugation also costs surface energy. In the ideal limit, inertia carries the interface through its flat position and the resulting oscillation propagates. When gravity also contributes appreciably, use the combined gravity-capillary wave model. Within the capillary wave regime, shortening a ripple makes it travel faster. ^[ref]

The laser-diffraction measurement of capillary waves below turns this mechanism into a test: set the frequency, measure the wavelength independently with light, and ask whether the dispersion relation follows. The theory is useful both for predicting how ripple wave packets separate and for inferring surface tension from their motion.

A uniform-mass density self-gravitating cylinder has ${\Phi }_R=2\pi G{\rho }_{0}R$. Radial balance with the displayed rotation law gives $p_R={\rho }_0(k^2{R}^3-2\pi G{\rho }_{0}R)$. At central pressure ${\pi }^2{G}^2{\rho }_0^3/k^2$, pressure becomes ${\rho }_0{k}^2(R^2-R_0^2{)}^2/4$ and the free surface has zero effective gravity. The positive radius uses $|k|$ if either rotation direction is allowed.

Take a reservoir with constant horizontal cross-sectional area $A$, one small orifice of area $a$, no inflow, and constant stipulated discharge coefficient $C_d>0$. Let the initial head be $H>0$. For constant mass density $\rho$, the stored mass above the outlet level is $\rho Ah$ and outgoing mass flow rate is $\rho Q$. Mass conservation cancels $\rho$ and gives the initial value problemThe downward speed of the free surface is $Q/A$, smaller than the ideal jet speed by the factor $C_{d}a/A$. Thus a small opening relative to the reservoir supports the negligible-surface-velocity assumption.

For $h>0$, separation of variables or the chain rule gives $d\sqrt{h}/dt=-k/2$. Integrating yieldsThe model reaches zero head in finite time even though volumetric flow rate tends to zero. Extending the squared expression beyond $T$ would make head grow again and violate the original ordinary differential equation. The zero-head continuation enforces no further outflow. For an elevated side opening, zero head means the surface has reached that opening, leaving fluid below it; $T$ is not the time to remove every drop from such a tank.

For an original synthetic example, stipulateThe initial volume above outlet level is $AH=0.0250{\mathrm{m}}^3=25.0\mathrm{L}$. The initial ideal jet speed is $2.215\mathrm{m}{\mathrm{s}}^{-1}$, the initial volumetric flow rate is $Q_0=0.1373\mathrm{L}{\mathrm{s}}^{-1}$, and the predicted zero-head time is $T=364.1\mathrm{s}$. These are consequences of stipulated parameters, not experimental measurements. At $t=T/2=182.1\mathrm{s}$, head is $H/4=0.0625\mathrm{m}$, stored volume is $6.25\mathrm{L}$, and volumetric flow rate is $Q_0/2=0.06866\mathrm{L}{\mathrm{s}}^{-1}$. Half the initial volume has already left at $t=T(1-1/\sqrt{2})=106.7\mathrm{s}$.

An independent mass conservation check integrates the outgoing volumetric flow rate:Multiply both sides by mass density to obtain discharged mass. A dimensional analysis check gives $[k]={\mathrm{m}}^{1/2}{\mathrm{s}}^{-1}$, $[Q]={\mathrm{m}}^3{\mathrm{s}}^{-1}$ and $[T]=\mathrm{s}$. The triangular area under $Q(t)$ explains why the initial-rate estimate $AH/Q_0$ is only half the actual model time. Doubling $A$ doubles $T$; doubling $a$ or $C_d$ halves $T$, within the same assumptions.

This finite-time result is an idealized prediction. The constant $C_d$ and quasi-steady approximation need not survive very small head or the initial startup. Appreciable viscosity in a narrow outlet tube requires a different pressure balance, as in Poiseuille flow; appreciable surface tension can prevent zero head altogether.

An external gravity wave displaces a fluid's free surface and moves nearly uniformly through the depth in the long-wave limit. In a layer of depth $H$, its leading shallow-water approximation speed is $\sqrt{gH}$.

For a deep inviscid liquid above a gas of negligible inertia, an inverted free surface has dispersion relation ${\omega }^2=q(\sigma q^2/\rho -g)$, $q=|k|$. The Rayleigh-Taylor instability affects $0<q<k_c=\sqrt{\rho g/\sigma }$. A bounded aperture can suppress it if its smallest allowed wavenumber is at least $k_c$. The capillary length is ${\ell }_c=k_c^{-1}$ and the marginal wavelength is $2\pi {\ell }_c$. A pinned one-dimensional span of width $L$ with a half-sine lowest mode instead has $k_{min}=\pi /L$, giving $L\le \pi {\ell }_c$. A numerical maximum container size therefore also requires aperture geometry and contact-line boundary conditions; the marginal wavelength alone is not a universal diameter.

The front velocity is the propagation speed of the earliest boundary of a newly introduced disturbance's support. It asks when an initially unaffected observer can first respond, whereas group velocity asks how a narrow wave envelope translates. A detector threshold defines a practical arrival time, but can miss arbitrarily small earlier responses; a packet peak is another distinct timing convention. In a causal dielectric model, a high-angular frequency contour argument gives exactly zero field for $t<x/c$, even when phase velocity exceeds $c$. This is a statement about that model's relativistic causality, not a universal identification of group velocity with front velocity.^[ref]

For the stipulated Klein-Gordon equation with compactly supported real Cauchy data, front velocity is bounded by $c$, and $c$ is the model's maximal speed of influence. A short energy method outline usesIntegration outside an expanding interval shows that zero initial energy there remains zero. Thus initial support in $[-a,a]$ cannot influence $x>a$ before $t=(x-a)/c$. The positive energy density includes the restoring term ${\omega }_0^2{u}^2/2$: this adds oscillation behind the front without extending its domain of dependence. A particular disturbance may fail to excite the leading boundary, so the maximal speed is not a promise of a detectable response for every source.

A Gaussian function is nonzero arbitrarily far away, so the Gaussian envelope in a Klein-Gordon medium has no sharp initial support boundary. Its small distant tails are already in the initial data. Likewise, keeping only a narrow interval of the frequency spectrum cannot reproduce an exact abrupt onset. Cutting off a Gaussian tail changes its Fourier transform and the accompanying initial velocity; it is a different initial-value problem. The exponential initial-velocity tail for the Klein-Gordon equation is another concrete example of a pre-existing tail outside the main disturbance.

The water-tank measurement in phase velocity records free surface elevation with calibrated resistance probes and uses a ruler and stopwatch for an independent check. The inspected protocol calls the front or back of a group a timing feature. That visible edge establishes an envelope transit estimate, not the first nonzero disturbance in the sense of front velocity.^[ref] A proposed additional control would repeat the transit fit at several detection thresholds and train durations, using matched probe calibration and the same sampling rate. Dependence on the threshold diagnoses wave envelope reshaping or limited sensitivity; records should be compared before reflections return. Such a test could expose limitations of a group velocity estimate, but would not establish an exact causal front or validate the stipulated Klein-Gordon equation for water waves.

A constant-depth layer initially at rest with a step in its free surface adjusts by radiating inertia-gravity waves. Conservation of linearized shallow-water potential vorticity leaves the bounded balanced profile ${\eta }_s={\eta }_0\mathrm{sgn}(x)(1-e^{-|x|/R_D})$, where $R_D=\sqrt{gH}/|f|$ is the Rossby deformation radius. The released potential energy per unit transverse length is $3\rho g{\eta }_0^2{R}_D/2$; the final kinetic energy is $\rho g{\eta }_0^2{R}_D/2$, leaving $\rho g{\eta }_0^2{R}_D$ in radiated waves.

A gravity-capillary wave has restoring contributions from both gravity and surface tension. Consider an initially resting, horizontally unbounded liquid layer of constant depth $h$, with uniform mass density $\rho$, incompressible flow, inviscid flow and irrotational flow. The upper gas exerts constant pressure and has negligible inertia. Assume a clean interface with constant surface tension $\gamma$: no surfactant stresses, imposed current or wind. The wave amplitude $a$ must satisfy $ka\ll 1$ and $a\ll h$, so linearized boundary conditions apply. Deep water means $kh\gg 1$, rather than a particular depth in metres. ^[ref]

Let $z=0$ be the undisturbed free surface, with liquid below it. Potential flow gives a velocity potential $\varphi$ with $\mathbf{u}=\nabla \varphi$, and incompressible flow gives the Laplace equation. The linearized kinematic boundary condition for a free-surface graph and the Unsteady Bernoulli equation, including gravity and the Young–Laplace equation, giveHere $\eta (x,t)$ is the interface displacement; the signed small-slope pressure jump is $p_{\mathrm{liquid}}-p_{\mathrm{gas}}=-\gamma {\eta }_{xx}$. A wave crest has ${\eta }_{xx}<0$, so its capillary pressure is positive. Choose a right-travelling Fourier mode of positive wavenumber $k$ and angular frequency $\omega$:The factor $e^{kz}$ decays into the deep liquid, satisfies the Laplace equation, and makes ${\varphi }_z={\eta }_t$ at $z=0$. Substituting in the dynamic boundary condition leavesThis derivation displays the two restoring contributions with the same sign. Dimensional analysis also checks the result: $\gamma /\rho$ has units ${\mathrm{m}}^3{\mathrm{s}}^{-2}$, so both $gk$ and $(\gamma /\rho )k^3$ have units ${\mathrm{s}}^{-2}$. ^[ref]

At finite depth, a flat impermeable bottom at $z=-h$ requires ${\varphi }_z=0$. Replacing $e^{kz}$ by the normalized vertical factor $\cosh (k(z+h))/\cosh (kh)$ makes the surface ratio ${\varphi }_z/\varphi =k\tanh (kh)$. Eliminating the mode amplitudes then givesThus the same model supplies its own depth check: use the deep-water formula only when $\tanh (kh)$ is sufficiently close to one for the required accuracy. For example, $kh=3$ reduces $\omega$ by about $0.248\%$ relative to deep water, while $kh=1$ reduces it by about $12.7\%$. These percentages are calculations from this factor. ^[ref]

In deep water the phase velocity magnitude follows a wave crest, while the group velocity magnitude follows the wave envelope of a narrow wave packet. Differentiating the positive-frequency dispersion relation givesFor $k{\ell }_c\ll 1$, the surface gravity wave limit has $c_g\simeq c_p/2$. For $k{\ell }_c\gg 1$, the deep-water capillary-wave dispersion relation has $c_g\simeq 3c_p/2$: a wave crest slips backward relative to the faster wave envelope, even while both advance. Neither speed is the oscillating fluid element's own velocity. ^[ref]

There is a particularly neat coincidence at the crossover. Differentiate $c_p^2=g/k+\gamma k/\rho$: its unique minimum is at $k_c=\sqrt{\rho g/\gamma }=1/{\ell }_c$, exactly where the restoring terms are equal. ThereforeFor the stipulated parameters in capillary length, the minimum is $0.231\mathrm{m}{\mathrm{s}}^{-1}$. At $\lambda =3\mathrm{mm}$, direct substitution gives $f=\omega /(2\pi )\simeq 131\mathrm{Hz}$, $c_p\simeq 0.394\mathrm{m}{\mathrm{s}}^{-1}$ and $c_g\simeq 0.580\mathrm{m}{\mathrm{s}}^{-1}$. This short wavelength is capillary dominated, but its group velocity is still slightly below the pure-capillary ratio $3c_p/2$ because gravity has not vanished.

A gravity wave is a fluid disturbance restored by gravity or buoyancy. A surface gravity wave displaces a free surface, while an internal gravity wave displaces density surfaces within a stratified fluid. Rotation can additionally contribute to restoration, giving an inertia-gravity wave.

A steady incompressible jet at atmospheric pressure conserves Bernoulli's equation and volume flux. Its radius contracts as $R=a[u/u_0{]}^{-1/2}$. After impact, a thin nearly horizontal axisymmetric film has $2\pi rhv=\pi a^2{u}_0$. Evaluating Bernoulli's equation on its free surface gives $v^2=u_0^2+2g(H-h)$, hence $a^4/(4r^2{h}^2)=1+2g(H-h)/u_0^2$. This inviscid thin-film description applies away from the impact region and does not describe a dissipative hydraulic jump.

The Lagrangian fluid displacement $\mathbf{\xi }$ compares neighbouring fluid trajectories. Linear velocity perturbations about a steady flow $\mathbf{U}$ obey $\delta \mathbf{u}=({\partial }_t+\mathbf{U}\cdot \nabla )\mathbf{\xi }-(\mathbf{\xi }\cdot \nabla )\mathbf{U}$. For a background scalar $f_0$, its material perturbation is $\Delta f=\delta f+\mathbf{\xi }\cdot \nabla f_0$. In an axisymmetric normal mode about cylindrical rotation, the radial and vertical velocities are $-i\omega {\xi }_r$ and $-i\omega {\xi }_z$. The displacement of a free surface is needed to impose the Lagrangian pressure perturbation condition.

// Target: astrophysics.bigb

A liquid in rigid rotation about a vertical axis is at rest in its rotating reference frame. Constant surface pressure makes $gz-{\Omega }^2{r}^2/2$ constant along the free surface, giving the displayed paraboloid. Comparing with $z-z_0=r^2/(4f)$ gives the focal length $f$. The result assumes uniform gravitational acceleration, a vertical rotation axis and negligible surface-tension corrections. It explains how a rotating liquid can form a reflecting surface or a mould for a mirror.

For an inverted free surface, gravity promotes a displacement instead of restoring it. The surface tension term still resists curvature. Hence the gravity-capillary wave relation becomes, with $q=|k|$,For $0<q<k_c$, where $k_c=\sqrt{\rho g/\sigma }$, ${\omega }^2<0$. Writing the displacement as $\eta \propto e^{st+ikx}$ gives the Rayleigh-Taylor instability growth rateShorter wavelengths, $q>k_c$, remain oscillatory because surface tension dominates. The marginal scale is determined by the capillary length,A sufficiently small aperture excludes the long unstable modes: this is finite-aperture stabilization of an inverted liquid surface. The estimate requested is therefore a container opening of order one centimetre, with $1.7\mathrm{cm}$ the marginal full wavelength rather than a geometry-independent maximum diameter.

The exact largest opening cannot be fixed from the stated infinite-surface dispersion relation alone. It requires the aperture shape and contact-line boundary conditions. For example, a one-dimensional span of width $L$ whose contact line is pinned at both ends has a half-sine lowest mode with $q_{\min }=\pi /L$. In that simple model,Different geometries change the numerical factor. This is a small-disturbance stability estimate with negligible air inertia and no leakage or contact-line failure; a guarantee that a real inverted container retains its water needs those additional boundary details.

The first argument mixes a material fluid element with a fixed spatial volume. For a material volume element following the fluid, its volume is generally time dependent:Therefore the material derivative cannot be taken outside the product with $\delta V$ as in the proposed cancellation. Instead, mass conservation statesThe momentum balance for a material fluid element, using Newton's second law, is consequentlyFor a fixed control volume, fluid crosses its boundary and carries momentum; a momentum flux term must be included. Its local conservative form is ${\partial }_t(\rho \mathbf{u})+\nabla \cdot (\rho \mathbf{u}\otimes \mathbf{u})=-\nabla p$, equivalent to the displayed Euler equations after using the continuity equation. The erroneous equation omits the term $\rho \mathbf{u}\nabla \cdot \mathbf{u}$ needed to convert a material derivative of momentum density into this conservative form.

For the second task, take a homogeneous fluid with constant mass density $\rho$, prescribed gravitational acceleration $\mathbf{g}$, incompressible flow and inviscid flow. Its equations areUnder the replacement $\rho \mapsto \tilde{\rho }=C\rho$, the same velocity solves these equations if $p\mapsto \tilde{p}=Cp$, up to a spatially constant pressure offset. This is density scaling invariance of incompressible Euler flow:The comparison assumes the same geometry and compatible velocity or gravitational free surface conditions; an independently specified nonuniform boundary pressure would supply additional information. The gravitational field is taken as prescribed. If the fluid's own self-gravity must be included, its field changes with mass density, so the invariance need not hold. A single freely moving particle's density-independent gravitational acceleration does not justify neglecting that distinction.

Treat the cylinder as infinitely long, with a vacuum exterior and no surface tension. Let ${\Phi }_0$ be its gravitational potential. Integrating the gravitational Poisson equation over a coaxial cylinder of radius $r<R$ and length $L$, then applying the divergence theorem, givesThus the inward gravitational acceleration is $-2\pi G\rho r\hat{\mathbf{r}}$. Hydrostatic equilibrium requiresImpose zero pressure on the free surface, $p_0(R)=0$, and integrate inward:This is the hydrostatic self-gravitating incompressible cylinder. Its central pressure is $\pi G{\rho }^2{R}^2$, and the units of the expression are $\mathrm{Pa}$. A nonzero external pressure would add that constant to $p_0(r)$.

The vanishing total pressure must be imposed at the displaced free surface, not at fixed $r=R$. The first-order Taylor expansion isSince $p_0(R)=0$ and $p_0^{\prime }(R)=-2\pi G{\rho }^{2}R$, the Lagrangian pressure perturbation condition isThe supplied gravitational perturbation and the combined potential, evaluated at $R$, then giveHere $K_0$ is the Modified Bessel function of the second kind. Equating mode amplitudes and using part (c) yieldsRearrange to obtain the dispersion relation for the axisymmetric gravitational instability of an incompressible cylinder:Every quantity on the right is dimensionless, consistent with $G\rho$ having units ${\mathrm{s}}^{-2}$. The stabilizing term $1/2$ comes from displacing the background hydrostatic pressure profile; omitting the Lagrangian pressure perturbation would lose it.

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.

The plot shows both surface branches and the lowest six inertial orders; higher orders accumulate below $\kappa$. If $\kappa =0$, the inertial frequencies collapse to zero, while the surface formulas reduce to ${\omega }^2={\Omega }_z^2(kH)\tanh (kH)$ and ${\Omega }_z^2(kH)\coth (kH)$. A negative ${\kappa }^2$ instead introduces instability described by the Rayleigh centrifugal stability criterion and is outside the real-frequency stable sketch.