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In an unstratified Keplerian shearing sheet with sound speed $c$, the four linear velocity-density equations yield ${\omega }^4-({\Omega }^2+c^2{k}^2){\omega }^2+{\Omega }^2{c}^2{k}_z^2=0$. The lower-frequency branch approaches the inertial wave relation ${\omega }^2={\Omega }^2{k}_z^2/k^2$ for $ck\gg \Omega$; the upper branch is predominantly acoustic.

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

In a uniform-density incompressible flow inside a smooth rigid sphere rotating at angular speed $\Omega$, a normal mode with phase $e^{i\sigma t+im\varphi }$ has intrinsic frequency $s=\sigma +m\Omega$. For $s\ne 0,\pm 2\Omega$, eliminating velocity from the linearized Euler equations gives the displayed pressure equation, with $W=p^{\prime }/\rho$. Impermeability gives $R{W}_R+(2m\Omega /s)W+(1-4{\Omega }^2/s^2)z{W}_z=0$ on the sphere. There is no condition $W=0$: the rigid wall supplies variable normal pressure. These inertial waves are restored by Coriolis acceleration, and their pressure equation is hyperbolic for $|s|<2|\Omega |$.

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.

For the basic axisymmetric flow, ${\mathbf{u}}_0=\Omega R{\mathbf{e}}_{\varphi }$. The steady radial Euler equation for fluid motion gives $-{\Omega }^{2}R=-p_{0,R}/\rho$; the other components give $p_{0,z}=p_{0,\varphi }=0$. Thus, in the absence of an additional body force,The additive constant is undetermined without an imposed reference pressure; the rigid boundary does not prescribe a uniform boundary pressure. Its shape is fixed by the container rather than by a free-surface pressure condition.

Let $v_R,v_{\varphi },v_z$ be the small velocity amplitudes and $W=p_1/\rho$ the pressure amplitude. Linearizing the cylindrical Euler equations for an inviscid fluid about uniform rotation, the material derivative is ${\partial }_t+\Omega {\partial }_{\varphi }$. The intrinsic frequency for the stated phase convention is thereforeThe radial and azimuthal curvature terms supply the Coriolis acceleration terms, givingFor $s\ne 0,\pm 2\Omega$, solving these equations givesThe incompressible flow constraint isOn substitution, the terms proportional to $m\Omega W_R$ cancel. The remaining equation iswhich provesThis is the pressure equation for inertial modes in a rigid rotating sphere. The exceptional frequencies used in the divisions must instead be treated in the original momentum system; they are not needed for the mode below.

A fixed rigid sphere is impermeable, so its normal velocity vanishes:Using the solved amplitudes and multiplying by $s(s^2-4{\Omega }^2)/i$ givesEquivalently divide by ${\bar{\sigma }}^2$ for $\bar{\sigma }\ne 0$. Smoothness means tangential slip is allowed: there is no no-slip condition on these inviscid normal modes. Nor is there a condition $W=0$, since the wall supplies the required varying normal pressure.

For an axisymmetric oscillation, $m=0$ and $s=\sigma$. Write $\beta =1-4{\Omega }^2/{\sigma }^2$. For $W=z(A{R}^2+B{z}^2+C)$ the radial and vertical operators areThe interior pressure equation therefore requiresThe boundary equation becomesSubstituting $R^2=r_0^2-z^2$, this polynomial must vanish for every boundary height. Its constant and $z^2$ coefficients consequently giveTogether with the interior relation the latter condition is $(4+\beta )A=0$. A nontrivial nonsingular oscillatory solution needs $A\ne 0$: if $A=0$ and $\beta \ne 0$, the other equations force $B=C=0$. Hence $\beta =-4$, givingThe cubic axisymmetric inertial mode of a rigid sphere is explicitlyIts cubic pressure is regular at the axis and centre, and its reconstructed velocity is divergence free and tangent to the sphere by the preceding identities. The frequency lies within the inertial wave band $|\sigma |<2|\Omega |$: Coriolis acceleration, rather than compression or a moving surface, supplies the restoring dynamics. The amplitude $A$ is arbitrary in this linear problem.

Keep the background magnetic field, thermal gradient and wavevector fixed as $|\Omega |\to \infty$. In particular, $k_z/K$ is nonzero and of order one. Put $H=g\alpha \beta$ and $X={\omega }^2$. Multiplying the nonzero-frequency dispersion relation by ${\omega }^2$ givesthat is,The sum and product of its two roots areFor large rotation, the sum is of order ${\Omega }^2$ while the product is independent of $\Omega$. The quadratic therefore has one large root and one small root. More explicitly,The fast frequencies are consequentlywhich are inertial waves at leading order. For nonzero $a$ away from the zero-frequency boundary, the rapid-rotation slow magneto-Coriolis branch hasIts frequency is of order $|\Omega {|}^{-1}$, much smaller than the fast inertial wave frequency. Coriolis force balances most of the magnetic and buoyant forces in this slow motion, with inertia supplying only a small correction.

For $a^2>0$, the slow wave has real nonzero frequency whenStable thermal stratification has $H<0$ and automatically satisfies this condition. With adverse stratification, $H>0$, sufficiently strong magnetic tension along the wavevector is needed. If the inequality is reversed, $X_s<0$ and the slow branch has imaginary frequencies, including an exponentially growing mode. Equality gives a stationary limit, rather than a nonzero oscillation. If $\mathbf{B}\cdot \mathbf{k}=0$, the proposed slow wave degenerates into zero-frequency disturbances: there is no magnetic restoring coupling. The remaining nonzero branch has ${\omega }^2=(4{\Omega }^2{k}_z^2-H{h}^2)/K^2$. These qualifications also explain why zero roots introduced by clearing denominators must not be interpreted as ordinary propagating waves.

A precise stratification–rotation analogy uses constant $N$, ideal fluid, and independence of one horizontal coordinate. For the stratified system take ${\partial }_y=0$, $v=0$, and allow prescribed forces $(F_x,0,F_z)$. The possibly nonlinear Boussinesq equations areFor the unstratified system choose right-handed coordinates $(X,Y,Z)=(z,y,-x)$, rotation $2\mathbf{\Omega }=N{\mathbf{e}}_Z$, and ${\partial }_Y=0$. Its rotating Euler equations, with the static gravitational and centrifugal contributions absorbed into the background pressure, arewhere $D_R={\partial }_t+U{\partial }_X+W{\partial }_Z$. The identification isThen $D_R=D$, $P_X=p_z$, and $P_Z=-p_x$, proving the component-by-component equivalence. In particular, buoyancy perturbation maps to transverse rotating velocity, and stratified horizontal velocity maps to rotation-axis velocity. The mapping permits arbitrary prescribed forces in the two-dimensional plane, but no transverse force: a transverse force in the rotating problem would create a source in the stratified buoyancy equation. A transverse stratified force would also invalidate $v=0$. Constant $N$ is essential to this constant-rotation equivalence.

A physical free surface introduces extra structure beyond the interior equations. After subtracting hydrostatic pressure, its zero total pressure condition gives $p=g\eta$ to first order, together with the kinematic condition ${\eta }_t=w$ for the stratified vertical displacement. The mapping takes this normal velocity into $U$, whereas an actual rotating free surface normal to its physical vertical $Z$ has normal velocity $W$ and a displacement along $Z$. The image of the stratified horizontal surface is a vertical surface in the rotating coordinates. Consequently the physical free-surface problems are not analogous, although deliberately mapped rigid boundaries or mathematical boundary data can be.

For the first prescribed force, the steady linear stratified equations are satisfied byReal parts are understood. The horizontal pressure gradient balances the force; the vertical pressure gradient is balanced by buoyancy; and $w=0$ satisfies buoyancy conservation and incompressible flow. Thus neither $u(z)$ nor $b(z)$ is fixed by the steady equations alone.

Initial rest and switching on in the unbounded fluid remove this ambiguity by causal pressure adjustment by internal waves. Work with the $e^{imz}$ coefficient and let $\hat{F}(q)$ be the horizontal Fourier transform of $f^{\prime }(x)$. For an abrupt switch at time zero, elimination in the linear equations givesAt $q=0$ the bracket cancels the apparent singularity at every finite time. The persistent pressure term is the symmetric antiderivative of the force: the inverse transform of $-i\hat{F}/q$, interpreted as a Cauchy principal value, is $f(x)-\epsilon /2$. The oscillatory part propagates away. To see why its zero-frequency part does not leave an arbitrary offset, note that $\Omega (q)\sim cq$ with $c=N/m$; its long-wave contribution consists of oppositely moving fronts. At fixed $x$, the two far-field values of $f-\epsilon /2$ are $-\epsilon /2$ and $+\epsilon /2$, whose average is zero. Nonzero horizontal wavenumbers give dispersive transients that decay locally. Smooth finite-duration switching changes the transient, not this selected steady limit. ThereforeA spatially constant pressure gauge is immaterial. The long-wave group velocity $c=N/m$ gives the adjustment scale: the elapsed time since switching must satisfy $Nt\gg 1$ and $(N/m)t\gg |x|+x_1$, and exceed the switching duration. These are asymptotic conditions, not an assertion of exact equilibration after a finite time; the full dispersion relation allows tails. The same long-wave limit gives $u_{\infty }=\epsilon m{e}^{imz}/(2N)$ if desired. Equal-amplitude outward fronts leave opposite pressure offsets and the same velocity on their two sides.

For the force depending instead on $y$, any steady linear horizontal momentum balance requires $p_x=f^{\prime }(x)e^{i\ell y}$ and $p_y=0$. Their mixed derivatives disagree wherever $f^{\prime }\ne 0$, since ${\partial }_y{p}_x=i\ell f^{\prime }(x)e^{i\ell y}$ but ${\partial }_x{p}_y=0$. Hence no steady linear velocity field can satisfy the equations, irrespective of whether it is $x$-independent. This statement concerns the linearized problem; nonlinear inertial balances were excluded by the assumptions.

For an unsteady solution put $\mathbf{u}=(-t{\varphi }_y,t{\varphi }_x,0)$ and take $\sigma =0$. Its horizontal divergence is zero. The pressure derivatives needed in momentum balance areTheir compatibility is exactlyThus the Poisson equation supplies an actual pressure and the proposed velocity solves both momentum equations; the vertical momentum and buoyancy equations hold because $w=\sigma =0$. This flow grows linearly in time, so it remains a valid small-disturbance prediction only while its velocities are small.

As $x_1\to 0$ with total force $\epsilon$ fixed, $f^{\prime }\to \epsilon \delta (x)$. Set $\varphi =\Phi (x)e^{i\ell y}$. Away from the source ${\Phi }^{\prime \prime }-{\ell }^2\Phi =0$, and decay at both ends requires $\Phi =A{e}^{-\ell |x|}$. Integrating across the source gives the jump ${\Phi }^{\prime }(0^{+})-{\Phi }^{\prime }(0^{-})=-i\ell \epsilon$. Since this jump is $-2\ell A$, $A=i\epsilon /2$, soThis verifies the distributional source as well as the decay condition.

The corresponding rotating problem is not expected to reproduce this evanescent accelerating flow. Its forcing varies in $Y$, precisely the coordinate whose independence was required in the stratification–rotation analogy. Moreover, $x$ maps to the rotation-axis coordinate $-Z$. At small Rossby number, the Taylor–Proudman theorem gives ${\partial }_Z{\mathbf{u}}_R=0$ in the unforced exterior regions. Such nonzero columns cannot decay with $|Z|$, hence cannot reproduce an $e^{-\ell |x|}$ tail. There is also a direct unsteady obstruction to copying the proposed flow: in a homogeneous fluid rotating about the mapped $x$ axis, a mode with nonzero axial wavenumber has inertial wave frequency of magnitude $N|q|/\sqrt{q^2+{\ell }^2}$, rather than the zero-frequency horizontal vortical response of the stratified problem. The steady or slowly varying exterior rotating response must be columnar or accompanied by outgoing inertial waves; the decaying stratified solution is outside the partial analogy.

In a thin disk dominated by a central stellar mass, the vertical gravitational potential is harmonic to leading order, so its vertical acceleration is $-{\Omega }_0^{2}z$. For a locally isothermal background, $p=c_s^2\rho$ with $c_s$ independent of height. Vertical hydrostatic equilibrium therefore givesThis is the isothermal vertical structure in stellar gravity. On the upper half of the disk the downward gravity magnitude is $g={\Omega }_0^{2}z>0$; the density profile extends evenly through the midplane.

For an adiabatically displaced parcel kept in pressure equilibrium, the parcel's logarithmic density gradient is ${\gamma }^{-1}d\log p/dz$, whereas the environmental density gradient is $d\log \rho /dz$. A small upward displacement $\eta$ consequently produces a density contrast $\delta \rho /\rho =[{\gamma }^{-1}d\log p/dz-d\log \rho /dz]\eta$. Its buoyancy acceleration is $-g\delta \rho /\rho =-N^2\eta$. Thus the Brunt–Väisälä frequency isBoth background gradients equal $-z/H^2$ in the isothermal atmosphere, soAn isothermal background is therefore stably stratified for adiabatic perturbations with $\gamma >1$, but not buoyant for isothermal perturbations. In applying the adiabatic equation of state to a stratified atmosphere, its proportionality relates Lagrangian parcel changes: $\Delta p=(\gamma p/\rho )\Delta \rho$. Eulerian perturbations generally obey $p^{\prime }+\eta p_z=(\gamma p/\rho )({\rho }^{\prime }+\eta {\rho }_z)$ instead. Thus the printed simple proportionality must be interpreted as a parcel relation for the buoyancy calculation. For $\gamma =1$ in this isothermal background, the gradient terms cancel and the Eulerian proportionality holds too.

Now take isothermal perturbations, so $\gamma =1$ and $N^2=0$. The two supplied wave equations reduce toFor the nonsingular oscillatory case $\omega \ne 0$ and ${\omega }^2\ne {\Omega }_0^2$, substitute $u_z={\xi }_z/{\omega }^2$. With $x=z/H$, the resulting Hermite differential equation isThe exceptional frequencies must be treated through the original velocity-pressure equations rather than division by these factors.

The physical vertical condition is finite perturbation energy when integrated with the Gaussian density. In particular the pressure contribution is proportional to $\int e^{-x^2/2}|\xi {|}^{2}dx$ for nonzero frequency. Generic nonterminating solutions have an exponentially growing component at at least one vertical end and fail this condition; acceptable modes are the terminating Hermite polynomials. This is a finite-energy condition, not pointwise boundedness of the velocity in the vanishing-density atmosphere.

To prove the allowed values of $\alpha$, let the nonzero polynomial have degree $n$ and leading coefficient $a_n$. The highest power in ${\xi }_{xx}-x{\xi }_x+\alpha \xi$ has coefficient $(\alpha -n)a_n$. Hence $\alpha =n\in \{0,1,2,\dots \}$. Conversely, writing $\xi ={\sum }_j{a}_j{x}^j$ gives the recurrenceFor $\alpha =n$, choose the parity of $n$ and iterate until the series terminates at degree $n$. Thus all allowed Hermite vertical modes of an isothermal disk havewhere ${\mathrm{He}}_n$ is the Probabilists' Hermite polynomial. In particular, ${\mathrm{He}}_0=1$ and ${\mathrm{He}}_1=x$.

Let $K=kH$ and $s={\omega }^2/{\Omega }_0^2$. Eliminating the denominator for regular modes givesFor $n=0$, the nonzero-frequency solution isIts pressure variable is independent of height and $u_z=0$. These are inertial-acoustic waves, restored by epicyclic motion and pressure; at short radial wavelength they become ordinary acoustic waves. The multiplied dispersion polynomial also has $s=0$. This is not an extra oscillatory solution represented by nonzero $\xi$, since $\xi =i\omega p^{\prime }/\rho$ vanishes at zero frequency for finite pressure perturbation. The full horizontal equations have a zero-frequency balanced mode of an axisymmetric disk, with no radial velocity and azimuthal Coriolis acceleration balancing a radial pressure gradient. It must be recovered without dividing by $\omega$, rather than counted as a propagating wave. At $k=0$, the propagating branch limits to an epicyclic oscillation at $|\omega |={\Omega }_0$, another value at which the eliminated form must be interpreted by a limit.

For $n=1$, solve $(s-1{)}^2=K^{2}s$. The two branches areBoth signs of frequency occur on each squared-frequency branch. The upper branch is acoustic in the large-$|K|$ limit, $|{\omega }_{+}|\sim |k|c_s$; the lower branch is an inertial wave with $|{\omega }_{-}|\sim {\Omega }_0/|K|$. At $|K|\ll 1$, both frequencies approach ${\Omega }_0$, reflecting the equality of the radial and vertical epicyclic frequencies in a Keplerian disk. Here $\xi \propto z$, so $u_z={\xi }_z/{\omega }^2$ is independent of height: these modes have a coherent vertical displacement of the disk column, rather than a vertical breathing motion.

For a local non-axisymmetric disturbance proportional to $e^{i(kR+m\varphi -\omega t)}$, the material frequency is $\hat{\omega }=\omega -m{\Omega }_0$. In the same local radial-wave approximation, replace $\omega$ by $\hat{\omega }$. For $n=1,m=1$, this givesThe four frequencies can equivalently be written $\omega /{\Omega }_0=1\pm (\sqrt{K^2+4}\pm |K|)/2$, with independent choices of the two signs. This is a local dispersion relation, not a claim that Doppler substitution supplies an exact global non-axisymmetric solution in a differentially rotating disk. The radial-wave ordering neglects azimuthal gradients compared with radial gradients; $|k|R\gg 1$ can coexist with $|k|H\ll 1$ in a thin disk.

Write $w=\omega /{\Omega }_0$. For the two low inertial-frame frequency branches, $|w|\ll 1$, while the material frequency remains near $-{\Omega }_0$. Expanding the dispersion relation givesThereforeFor signed $k$, the two leading branches are conventionally labelled by $\pm kH/2$. These are bending waves of a Keplerian disk: an $m=1$ vertical displacement varies once around an annulus and corresponds to a small tilt of its orbital plane. A radial variation of the tilt is a warp. The waves transport the warp radially at group speeds $c_s/2$ in opposite directions. At zero radial wavenumber their low-frequency limit is a stationary rigid tilt; the apparent slow pattern does not mean that the orbiting gas has slow vertical motion in its own frame.

For constant density $\rho$ and constant angular velocity $\mathbf{\Omega }$, the Euler equations for an inviscid fluid in the rotating frame isThe last two forces are gravity and centrifugal acceleration; $2\mathbf{\Omega }\times \mathbf{u}$ is the Coriolis force term on the left. Incorporate both conservative forces into the modified pressureso the right side is $-\nabla \Pi$.

At latitude $\lambda$ on a spherical Earth of radius $R$, the outward local vertical component of centrifugal acceleration is ${\Omega }_E^{2}R{\cos }^2\lambda$. Its ratio to gravity is at most ${\Omega }_E^{2}R/g\simeq 3.5\times 10^{-3}$, using ${\Omega }_E\simeq 7.3\times 10^{-5}{\mathrm{s}}^{-1}$, $R\simeq 6.4\times 10^6\mathrm{m}$, and $g\simeq 9.8{\mathrm{m}\mathrm{s}}^{-2}$. Thus this component is small compared with gravity everywhere on Earth.

For a velocity scale $U$ and length scale $\ell$, nonlinear advection scales as $U^2/\ell$, whereas the Coriolis term scales as $2\Omega U$. Their ratio is the Rossby number $U/(2\Omega \ell )$. If it is small, for example by increasing rotation at fixed $U,\ell$, neglect nonlinear advection in perturbations about a fluid at rest. The conservative equilibrium forces have already been absorbed into pressure; with a rotation-timescale unsteady motion the leading equations areThese are the linear equations governing inertial waves.

For a nonzero wavevector $\mathbf{k}$, substitute the plane wave into part (c). Each time derivative contributes $-i\omega$, each Laplacian contributes $-|\mathbf{k}{|}^2$, and each second $z$ derivative contributes $-k_z^2$. A nonzero amplitude therefore requiresIf $\theta$ is the angle between the wavevector and the rotation axis, $\cos \theta =k_z/|\mathbf{k}|$, andThis is the inertial wave dispersion relation. Incompressibility also requires $\mathbf{k}\cdot {\mathbf{U}}_0=0$; the original first-order equations select the corresponding transverse polarizations. A horizontal wavevector gives the zero-frequency geostrophic case.

Substitute the axisymmetric Fourier mode into the linearized equations. With hats denoting amplitudes,For a nondegenerate nonzero frequency, eliminate the azimuthal velocity and useContinuity then givesEquivalently, taking the determinant of the original four amplitude equations avoids division by zero and gives the full dispersion relationThe two squared frequencies areThe discriminant is $({\Omega }^2-c^2{k}^2{)}^2+4{\Omega }^2{c}^2{k}_x^2\ge 0$. For $c^2>0$, both squared frequencies are nonnegative. The high-frequency branch is predominantly acoustic and the low-frequency branch is an inertial wave. Zero-frequency limiting modes, including $k_z=0$, are retained by the determinant calculation.

Take $ck\gg \Omega$, retaining a finite low-frequency branch and a nonzero wavevector. In the dispersion relation, neglect ${\omega }^4$ against $c^2{k}^2{\omega }^2$ and neglect ${\Omega }^2$ against $c^2{k}^2$ in their coefficient. The inertial wave limit isChoose the signed branches $\omega =ϵ\Omega k_z/k$, $ϵ=\pm 1$. The phase velocity normal to the phase planes and the group velocity areTherefore $\mathbf{k}\cdot {\mathbf{c}}_g=ϵ\Omega (k_z/k-k_z{k}^2/k^3)=0$, soThe geometric reason is that this limiting dispersion relation is homogeneous of degree zero in $\mathbf{k}$: its frequency depends on direction, not magnitude. Euler homogeneous function theorem gives $\mathbf{k}\cdot {\nabla }_{\mathbf{k}}\omega =0$. Exact orthogonality holds for this inertial wave limit; finite compressibility introduces corrections. At a zero-frequency degeneracy the usual propagating-wave interpretation needs a limiting treatment.