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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 Keplerian shearing sheet, an epicyclic guiding center $x_0$ carries rotating-frame energy $-3{\Omega }^2{x}_0^2/8$, plus nonnegative radial and vertical oscillation energies. Inelastic collisions dissipate the total Jacobi energy in a shearing sheet, while momentum conservation preserves $\sum x_0$. Thus $\sum x_0^2$ must grow: the ring spreads about its fixed mean radius. The effect is an angular momentum transport process, with some particles moving inward and others outward.

The local axisymmetric density/velocity system of a pressured dust layer with linear gas drag and a fixed Keplerian shearing sheet gas flow has this cubic dispersion relation. The $\Omega /2$ azimuthal velocity coefficient includes advection of the Keplerian shear. The razor-thin disk Poisson kernel supplies the $|k|$ self-gravity term. Keeping the full amplitude determinant retains the zero-frequency branch and avoids dividing out the secular mode.

For epicyclic motion in a Keplerian shearing sheet, write $x-x_0=\mathrm{Re}(A{e}^{-i\Omega t})$. The energy of the radial harmonic oscillator per unit mass is $E_{\mathrm{epi}}={\Omega }^2|A{|}^2/2$. The full horizontal Jacobi energy in a shearing sheet also includes the negative guiding-center term $-3{\Omega }^2{x}_0^2/8$.

The epicyclic guiding center is the circular-orbit location about which a particle performs epicyclic motion. In a Keplerian shearing sheet, the conserved canonical momentum is $p_y=\dot{y}+2\Omega x=\Omega x_0/2$. The guiding center therefore labels the particle's first-order specific angular momentum offset, even while its instantaneous radial coordinate oscillates.

Let $v_x,v_y$ and $u_x,u_y$ be the dust and gas velocity perturbations after subtracting the common background shear, suppressing primes. Put $t_s>0$ for the aerodynamic stopping time. Linear axisymmetric dust perturbations about a Keplerian shearing sheet obey $D_{ϵ}{v}_x-2\Omega v_y=ϵ\Omega u_x$ and $D_{ϵ}{v}_y+\Omega v_x/2=ϵ\Omega u_y$, with $ϵ\Omega =t_s^{-1}>0$. Eliminating $v_y$ givesThus a gas epicycle $u_x=u\cos (kx-\Omega t)$, $u_y=u\sin (kx-\Omega t)/2$ supplies both $2ϵ{\Omega }^{2}u\sin (kx-\Omega t)$ and ${ϵ}^2{\Omega }^{2}u\cos (kx-\Omega t)$. Keeping the latter is necessary for an exact linear equation beyond first order in drag. The homogeneous roots are $-ϵ\Omega \pm i\Omega$, separating damping from epicyclic frequency.

The nonlinear velocity term is a divergence componentwise: $v_j{\partial }_j{v}_i={\partial }_j(v_j{v}_i)$ because $\nabla \cdot \mathbf{v}=0$. Likewise $B_j{\partial }_j{B}_i={\partial }_j(B_j{B}_i)$; their volume averages vanish. Pressure gradients, spatial Laplacians and the background-advection term $x{\partial }_y$ also average to zero by the boundary identities. The mean momentum equation is thereforeIn induction, the two nonlinear terms are divergences, $v_j{\partial }_j{B}_i={\partial }_j(v_j{B}_i)$ and $B_j{\partial }_j{v}_i={\partial }_j(B_j{v}_i)$. Averaging consequently givesThe radial and vertical mean magnetic fields stay constant, and $⟨B_y⟩(t)=⟨B_y⟩(0)-2At⟨B_x⟩(0)$. In particular an initially zero mean magnetic field stays zero.

Writing $V_i=⟨v_i⟩$, the horizontal mean motion satisfiesThis is the local radial epicyclic frequency. When the background is centrifugally stable, $\Omega (\Omega -A)>0$, the whole box supports epicyclic mean-velocity oscillations. For example,while $V_z$ is constant. A Keplerian shearing sheet has $\kappa =\Omega$. For a centrifugally unstable rotation law the same mean equations instead give exponential growth, so an oscillation interpretation requires the positive squared frequency.

No divergence of turbulent stress survives the box average, and a zero initial mean radial velocity cannot acquire a secular accretion drift from internal stresses. More generally, a steady horizontal mean flow has $V_x=V_y=0$ for the usual nondegenerate rotating disk. The epicycle has zero time-averaged radial velocity. Opposite radial faces are identified rather than connected to a central sink and an outer mass reservoir: material crossing a face re-enters the opposite one after an azimuthal shift. Thus the shearing box can model angular-momentum transport and its local stresses, but not develop a global disk accretion flow. Such accretion needs radial gradients, boundaries and mass exchange omitted from the homogeneous local model.

Take every finite positive drag coefficient $\gamma >0$ and defineThe coefficients of the characteristic polynomial are $c_3=2\gamma$, $c_2=K+{\gamma }^2$, $c_1=2L\gamma$ and $c_0=P{\gamma }^2$. The nontrivial Routh-Hurwitz criterion reduces exactly toThe printed strict chain $K>L>P>0$ makes every coefficient positive and both bracket terms positive, so it proves attraction for every positive drag coefficient.

There is, however, a boundary error in the asserted necessity. To have the bracket positive for every $\gamma >0$, its constant and slope as a function of ${\gamma }^2$ must be nonnegative, and cannot both vanish. Together with coefficient positivity this gives the exact drag-polynomial stability for all positive stopping rates:Necessity follows by taking arbitrarily small and arbitrarily large positive $\gamma$; if one bracket coefficient is negative, the inequality fails in the corresponding limit. Sufficiency follows because at least one coefficient is strictly positive. Thus $K=L>P$ and $K>L=P$ still give strict decay at each finite positive $\gamma$. They need not give a uniform stability margin in the zero- or infinite-drag limit.

In a Keplerian shearing sheet, $S=3\Omega /2$ andThe differences areHence $r>3$ establishes the requested strict chain and trapping conclusion. At $r=3$, however, $K=L={\Omega }^2$ and $P=9{\Omega }^2/16$, soAll coefficients remain positive. This is an explicit counterexample within the stated vortex model: $r=3$ also attracts particles for every finite positive drag coefficient, so the literal strict “if and only if” is false. The exact Keplerian all-positive-drag condition is $r\ge 3$. If $\gamma =0$ is included, the polynomial has zero and purely imaginary roots and there is no asymptotic attraction; “all drag coefficients” must mean positive drag. The local attraction claim presumes the particle remains in the patch, as is assured for sufficiently small perturbations of its centre.

Write the perturbation velocity as $(u,v)$, the density amplitude as ${\sigma }_1$, and ${\gamma }_d=ϵ\Omega$. Axisymmetry removes advection by the background azimuthal flow, but the radial perturbation advects the Keplerian shear: $u{\partial }_x{U}_{g,y}=-3\Omega u/2$. Combining this with the Coriolis force gives the linearized shearing sheet equationsThe razor-thin disk Poisson kernel supplies the self-gravity term. The $\Omega /2$ coefficient, rather than $2\Omega$, is essential: it includes the perturbed advection of the background velocity.

Put $A=c^2{k}^2-2\pi G{\sigma }_0|k|$ and ${\omega }^2={\Omega }^2+A$. The determinant of the three amplitude equations isExpanding it gives the dust gravitational dispersion relation with gas dragUsing a determinant avoids division by $s$ and retains the neutral/secular branch. The radial epicyclic frequency of this Keplerian shearing sheet is $\Omega$, so the three terms in ${\omega }^2$ represent rotational support, self-gravity and dust pressure.

Write the horizontal velocity amplitudes as $(u,v)$ and magnetic amplitudes as $(b_x,b_y)$. The perturbation is horizontally uniform, divergence-free, and has no vertical velocity, so the unperturbed density and pressure are consistent at linear order. In the Keplerian shearing sheet, the horizontal components of the linearized ideal magnetohydrodynamic equations areUse the undivided equations at a zero of $F$. If $F^{\prime \prime }=-k^{2}hF$, the magnetic-force coefficient becomes $ik{v}_A^2$; the stratification cancels. The ideal magnetohydrodynamic induction equation similarly reduces toThe azimuthal induction term is field stretching by the background differential rotation. The four amplitudes obey the homogeneous systemIts determinant must vanish for a nonzero normal mode. Put $A=v_A^2{k}^2$. For nonzero $s$, elimination first gives $(s^2+A)u-2\Omega sv=0$ and $s(s^2+A)v+\Omega (s^2-3A)u/2=0$, whose solvability condition is $(s^2+A{)}^2+{\Omega }^2(s^2-3A)=0$. The original determinant extends the same polynomial to marginal $s=0$. ThusThis is the ideal magnetorotational dispersion relation with the midplane Alfvén speed $v_A=B_0/\sqrt{4\pi {\rho }_0}$. The vertical structure enters through the admissible eigenvalues $k$, rather than through a different horizontal dispersion polynomial. No division by a vanishing growth rate is required in the determinant derivation.

Use perturbations $\mathbf{v},\mathbf{b},p,\vartheta$ for velocity, magnetic field, magnetohydrodynamic total pressure and buoyancy displacement variable. SetThis is advection by the background Keplerian shearing sheet. Subtracting the equilibrium before retaining first-order terms gives the nine linearized equations:The azimuthal coefficient $\Omega /2$ combines the Coriolis acceleration with perturbation advection of the background shear. The $-3\Omega b_x/2$ term is the winding of radial magnetic field by that shear. The tidal force cancels when the equations at a fixed position are subtracted; it does not add a separate Eulerian perturbation force. Magnetic pressure is already included in $p$, leaving only linear magnetic tension. In this normalization $\vartheta$ has dimensions of length because $D_0\vartheta =v_x$.

Choose the positive-frequency branch and real radial amplitude $u$. The azimuthal amplitude of the inertial-acoustic wave is $U_y=-i\Omega u/(2\omega )$, so taking real parts givesFor $|k|c_s/\Omega \ll 1$, $\omega =\Omega [1+O((k{c}_s/\Omega {)}^2)]$. Thus, to leading long-wavelength order,The quarter-cycle phase shift and the factor $1/2$ are the epicyclic motion relation in a Keplerian shearing sheet. This approximation drops the small frequency detuning; over times $|\omega -\Omega |t\gtrsim 1$ that phase approximation is not uniform. The associated fractional surface density of a disk perturbation is $\sigma /{\Sigma }_0=(ku/\omega )\cos (kx-\omega t)$, so linearization also requires $|ku/\omega |\ll 1$. The source's arbitrary real $u$ sets the wave phase/amplitude within that small-perturbation regime.

The razor-thin disk approximation uses vertically integrated pressure and surface density of a disk. To close the printed equations, interpret the barotropic equation of state as $P=P(\Sigma )$, or supply an equivalent fixed vertical conversion from the printed $\rho$ to $\Sigma$. Without that conversion, the literal $P=P(\rho )$ does not close this two-dimensional system. Set $c_s^2=(dP/d\Sigma {)}_0>0$, and write $\Sigma ={\Sigma }_0+\sigma$, $\mathbf{u}={\mathbf{u}}_0+{\mathbf{u}}^{\prime }$, $h^{\prime }=P^{\prime }/{\Sigma }_0=c_s^2\sigma /{\Sigma }_0$. Axisymmetric flow here means ${\partial }_y=0$, so background advection vanishes. The background shear enters through ${\mathbf{u}}^{\prime }\cdot \nabla {\mathbf{u}}_0=-3\Omega u_x^{\prime }{\mathbf{e}}_y/2$. The complete linearized equations areThese are the compressible Keplerian shearing sheet equations for an inertial-acoustic wave. No perturbed self-gravity is present in the given force law, so no gravitational potential perturbation is to be added. This closure is the barotropic closure of a razor-thin disk.

For the Keplerian shearing sheet background $\mathbf{U}=-3\Omega x{\mathbf{e}}_y/2$, the only spatial derivative is ${\partial }_x{U}_y=-3\Omega /2$. Since $U_x=0$ and all fields are independent of $y$, $\mathbf{U}\cdot \nabla \mathbf{U}=0$ and $\nabla \cdot \mathbf{U}=0$.

The Coriolis acceleration exactly balances the tidal term:With constant pressure and $\theta =0$, the buoyancy displacement variable equation also has zero advection, zero source $U_x$ and zero diffusion. Thereforeis a steady solution. Its absolute vorticity is ${\mathbf{\omega }}_0=(\Omega /2){\mathbf{e}}_z$, since the background relative vorticity is $-3\Omega {\mathbf{e}}_z/2$.

For the Keplerian shearing sheet velocity $\mathbf{U}=-(3/2)\Omega x{\mathbf{e}}_y$, its material acceleration vanishes because $\mathbf{U}\cdot \nabla \mathbf{U}=0$. The $x$ component of the Coriolis acceleration is $-3{\Omega }^{2}x$, which is cancelled by $-\nabla {\Phi }_t=+3{\Omega }^{2}x{\mathbf{e}}_x$ with the signs in the stated equation; constant $P_0$ supplies no force. The flow is also incompressible.

Let $\mathbf{u}=\mathbf{U}+{\mathbf{u}}^{\prime }$ and $h^{\prime }=P^{\prime }/\rho$. Axisymmetry removes $\mathbf{U}\cdot \nabla {\mathbf{u}}^{\prime }$, while ${\mathbf{u}}^{\prime }\cdot \nabla \mathbf{U}=-(3/2)\Omega u_x^{\prime }{\mathbf{e}}_y$. Keeping first-order terms givesDifferentiate the momentum equations in time and use incompressibility to eliminate $h^{\prime }$. The radial velocity obeysFor a plane wave proportional to $e^{i(k_{x}x+k_{z}z-\omega t)}$, this gives the inertial wave dispersion relation

When $k_x=0$ and $k_z\ne 0$, incompressibility forces $u_z^{\prime }=0$, and the vertical momentum equation then forces $h^{\prime }=0$. The remaining motion has $\omega =\pm \Omega$ and satisfies $u_y^{\prime }=\mp i{u}_x^{\prime }/2$: each horizontal layer executes an epicyclic motion, with the phase varying vertically but no pressure or vertical-velocity perturbation.

For a Keplerian shearing sheet, the shear rate is $S=3\Omega /2$ and the shearing-sheet tidal potential isA uniform steady solution isThe Coriolis acceleration of this linear shear flow balances the radial tidal acceleration. The only nonzero background component of the viscous stress tensor that matters is $T_{xy}=-\nu ({\Sigma }_0){\Sigma }_{0}S$, which is spatially constant. Hence $\nabla \cdot \mathbf{T}=0$: a local uniform patch has no stress gradient or torque divergence to drive an accretion flow.

For ideal axisymmetric radial displacements in a Keplerian shearing sheet, the hydrodynamic squared oscillation frequency is $N_r^2+{\kappa }^2$, with radial epicyclic frequency $\kappa =\Omega$. Exponential instability occurs when $N_r^2+{\kappa }^2<0$. This is the radial special case of rotating-fluid convective stability, rather than the complete set of three-dimensional Solberg–Høiland criteria.

For a razor-thin fluid disk with sound speed or effective random speed $c$, surface density $\Sigma$ and radial epicyclic frequency $\kappa$, $Q=c\kappa /(\pi G\Sigma )$ compares pressure and rotational support with self-gravity. The local axisymmetric fluid Toomre stability criterion is $Q\ge 1$. For a Keplerian shearing sheet, $\kappa =\Omega$. Collisionless stellar disks use a different numerical normalization; gas drag can permit secular growth even when this fluid parameter exceeds unity.

Besides the two propagating inertial-acoustic wave branches, a uniform barotropic Keplerian shearing sheet has a stationary branch balancing pressure against Coriolis acceleration. For radial wavenumber $k\ne 0$, setting $\omega =0$ in continuity gives $U_x=0$, and radial momentum gives the displayed azimuthal balance. The full determinant is proportional to $\omega ({\omega }^2-{\Omega }^2-c_s^2{k}^2)$. At $k=0$ the stationary branch is a uniform column-density change; it is not a travelling density wave.