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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.

The unforced planar equation of motion in a rotating frame near a circular Kepler orbit has components $\ddot{x}-2n\dot{y}-3n^{2}x=0$ and $\ddot{y}+2n\dot{x}=0$. They describe epicyclic motion and Keplerian shear; mutual gravity adds close-encounter forcing.

Reflection symmetry gives ${\Phi }_z(R,0)=0$. In a vertically stable thin disk, expand the vertical force as ${\Phi }_z(R,z)={\Omega }_z^{2}z+O(z^3)$, where ${\Omega }_z^2={\Phi }_{zz}(R,0)$ is the vertical epicyclic frequency squared. For local hydrostatic equilibrium and height-independent isothermal sound speed, $p=c_s^2\rho$ andThus $\rho ={\rho }_0\exp [-z^2/(2H^2)]$ with $H=c_s/{\Omega }_z$. Integrating the pressure and mass density through the disk gives the vertically integrated pressure of an isothermal disk:The force expansion is approximate in $H/R$, while $P=c_s^2\Sigma$ is exact for a height-independent sound speed.

In a nearly Keplerian disk, the vertical oscillation and radial epicyclic motion frequencies are close to the orbital frequency. The exact leading orbital differences are $\Omega -{\Omega }_z$ and $\Omega -\kappa$, soThus ${\omega }_1$ is the nodal precession rate of a slightly tilted orbit, and ${\omega }_2$ is the apsidal precession rate of a slightly eccentric orbit. Their signs distinguish prograde from retrograde precession. This also agrees with the sign convention in the complex tilt: a freely precessing ring obeys $W_t=i{\omega }_{1}W$.

For exactly Keplerian motion, ${\Omega }_z=\kappa =\Omega$, so ${\omega }_1={\omega }_2=0$ and $P/\Sigma =c_s^2$. In the local approximation, freeze the slowly varying coefficients and take $W,G\propto e^{i(kR-\omega t)}$. The inviscid warp equations of a nearly Keplerian disk becomeEliminating $G$ gives ${\omega }^2=k^{2}P/(4\Sigma )=k^2{c}_s^2/4$. HenceThese are bending waves of a Keplerian disk, carrying changes in orbital tilt. The local approximation needs wavelength small compared with $R$; the secular thin-disk warp equations also describe the long-vertical-wavelength regime $H\ll \lambda \ll R$.

For a nonspinning hole, $a=0$ gives ${\omega }_1=0$. A steady disk warp then satisfies $G_R=0$. The zero inner torque forces $G=0$ throughout, and the second warp equation, with positive $P$, forces $W_R=0$. The outer orientation fixes $W(R)=W_{\infty }$ everywhere. An apsidal precession term by itself therefore does not require a tilt warp.

For nonzero spin, set the time derivatives to zero. The two equations giveEliminating the internal torque yieldsUse ${\omega }_2=3\Omega /r$, ${\omega }_1=2a\Omega r^{-3/2}$, ${\Omega }^2{R}^3=GM$, and $P\simeq {\Omega }^2\Sigma H^2$. The first coefficient becomes $GM\Sigma H^{2}r/12$ and the second becomes $2aGM\Sigma r^{-3/2}$. ThereforeThis is the stationary disk warp equation under the stated weak-precession approximations. The inner condition is $W_R(R_{\mathrm{in}})=0$ when its coefficient is finite and nonzero, because $G=iGM\Sigma H^{2}r{W}_R/12$.

Let $R_g=GM/c^2$, so $R=R_{g}r$, and impose $H=ϵR$, $\Sigma \propto R^{-3/4}$. Converting both radial derivatives carefully to $r$ and dividing out the common positive factor givesor $W_{rr}+9W_r/(4r)+(24a/{ϵ}^2)r^{-9/2}W=0$. For $a>0$, introduceSince $x_r$ is proportional to $r^{-9/4}$, the first-derivative term cancels exactly, leaving $W_{xx}+W=0$. Thus $W=A\cos x+B\sin x$.

The outer boundary $r\to \infty$ is $x\to 0$, giving $A=W_{\infty }$. The specified inner-radius relation is exactly $x_{\mathrm{in}}=\pi$. The inner zero-torque condition $W_x(\pi )=0$ then gives $B=0$. Hence the stationary bending warp of a Kerr disk isThe tilt magnitude is $\beta (r)=|W_{\infty }||\cos x|$. Far out it tends to the imposed tilt. Moving inward, it decreases to zero atthen grows again with opposite signed tilt, reaching $W(r_{\mathrm{in}})=-W_{\infty }$. All tilt vectors lie on the same line in the complex $W$ plane. Their azimuth is $\arg W_{\infty }$ outside the node and differs by $\pi$ inside; azimuth is undefined at the aligned node itself. The tilt vector remains smooth there. Thus this is an oscillatory standing bending warp, with equal outer and inner tilt magnitudes and reversed inner orientation, rather than monotone inner alignment.

The prescribed real inner radius assumes $a>0$: for $a<0$, its square root is not real, so that particular boundary prescription does not apply. With a separately chosen positive inner radius and negative spin, put $x=4\sqrt{24|a|}{r}^{-5/4}/(5ϵ)$. The equation becomes $W_{xx}-W=0$, and the same outer tilt and zero inner torque giveThis negative-spin solution decreases smoothly toward the inner disk without a sign reversal. The specified positive-spin solution is the cosine profile above; the $a=0$ case is the flat solution already derived.

Use cylindrical coordinates and tracer density $\nu (R,z)$. A steady axisymmetric Jeans equation, with no mean radial or vertical flow, isWrite $\overline{v_R^2}={\sigma }_R^2$, $\overline{v_{\varphi }^2}={\bar{v}}_{\varphi }^2+{\sigma }_{\varphi }^2$, and $V_c^2=R{\partial }_R\Phi$. Rearrangement gives the stellar asymmetric drift equationFor an isotropic velocity ellipsoid, all three dispersions equal $\sigma$ and the cross stress vanishes. Therefore the required general result isThe positive root is for a prograde galactic disk. A declining radial stress gives $A>0$: random motions supply some support, so mean stellar rotation is slower than cold-gas circular rotation. For $A{\sigma }^2\ll V_c^2$, $V_c-{\bar{v}}_{\varphi }\simeq A{\sigma }^2/(2V_c)$. The condition $\sigma <V_c$ alone is insufficient; one also needs $A{\sigma }^2\le V_c^2$.

There is no universal numerical coefficient expressible from $V_c$ and $\sigma$ alone. For constant velocity dispersion and $\nu \propto R^{-p}$, $A=p$. Thus $p=2$ gives the often-used estimate ${\bar{v}}_{\varphi }=\sqrt{V_c^2-2{\sigma }^2}$, while $p=1$ gives a different answer with the same $V_c$ and $\sigma$. For an exponential galactic disk tracer with $\nu \propto e^{-R/R_d}$ and ${\sigma }^2\propto e^{-R/R_{\sigma }}$, the exponential-disk asymmetric drift has $A=R(R_d^{-1}+R_{\sigma }^{-1})$.

Real galactic disks generally have anisotropic velocity dispersions, often ${\sigma }_R>{\sigma }_{\varphi }>{\sigma }_z$. In the small-random-motion epicyclic motion limit for a flat galaxy rotation curve, ${\sigma }_{\varphi }^2/{\sigma }_R^2\simeq 1/2$, so even the non-gradient term matters. The velocity ellipsoid can tilt, and bars or spiral structure can violate steady axisymmetry. Age-dependent galactic disk heating makes the lag population dependent. Gas has finite turbulent and thermal stresses too, though a cold gas disk often approximates $V_c$ much better than a hot stellar population.

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.

Set $\xi =\tanh (z/H)$, so $h=1-{\xi }^2$ and $d/dz=(1-{\xi }^2)H^{-1}d/d\xi$. The vertical equation becomesFor $|\xi |<1$, division by the common factor gives the Legendre differential equation with eigenvalue $k^2{H}^2$. Requiring boundedness at both surfaces $\xi =\pm 1$ selectsThe second independent solution is unbounded at an endpoint. These are the bounded vertical modes of a sech-squared magnetized disk. The division by $H$ is outside the square root, as in the original PDF; the converted TeX places it incorrectly. The magnetic perturbation involves $F_n^{\prime }$, which tends to zero at either surface.

The constant $n=0$ function has $k=0$. The specified magnetic ansatz contains $1/(ik)$ and must not be applied to it. A uniform horizontal velocity with no magnetic perturbation is an epicyclic motion, not a growing magnetic mode; the zero-wavenumber formal degeneracy does not provide growing MRI at arbitrary field strength. Nontrivial magnetically coupled vertical modes have $n\ge 1$.

For $A>0$, the magnetorotational instability grows when $A<3{\Omega }^2$. Indeed, the quadratic in $s^2$ then has a negative constant term, and one positive root. For $A\ge 3{\Omega }^2$, the two roots are nonpositive, since their sum is negative, their product nonnegative, and their discriminant is ${\Omega }^4+16{\Omega }^{2}A>0$. The smallest nonzero vertical eigenvalue is $k_1^2=2/H^2$, so all vertical modes are stable ifThe strict inequality requested gives stability, and equality is marginal for $n=1$. A sufficiently strong field increases magnetic tension. The unstable MRI needs a sufficiently long vertical wavelength to exchange angular momentum without excessive restoring tension; bounded vertical structure imposes a smallest nonzero effective wavenumber. Beyond the finite-thickness magnetorotational instability criterion, no allowed magnetic mode is long enough to remain unstable.

In a stationary axisymmetric gravitational potential, conservation of angular momentum gives $L_z=R^2\dot{\varphi }$. Eliminating $\dot{\varphi }$ introduces the effective potentialFor an equatorial circular orbit, radial balance is ${\partial }_R{\Phi }_{\mathrm{eff}}(R_c,0)=0$, givingVertical balance also requires ${\Phi }_{,z}(R_c,0)=0$.

For epicyclic motion, assume a twice differentiable stationary gravitational potential symmetric under $z\mapsto -z$, small displacements $|x|,|z|\ll R_c$, and a stable circular orbit. Choose the epicyclic guiding center using the conserved $L_z$. Reflection symmetry makes ${\Phi }_{,Rz}(R_c,0)=0$, eliminating radial-vertical coupling at first order. Expanding the equations about $(R_c,0)$ givesThe derivatives are evaluated at the guiding centre, and stability requires ${\kappa }^2,{\nu }^2>0$. Without midplane symmetry, a mixed Hessian term can couple the two oscillations. The radial epicyclic frequency and vertical epicyclic frequency are the frequencies of these independent linear oscillations.

Set $\Omega =L_z/R_c^2$. Along the family of equatorial circular orbits, ${\Phi }_{,R}=R{\Omega }^2$. Differentiating this relation and adding the centrifugal contribution givesTo interpret the common frequency range, let $q=d\log v_c/d\log R$. Since $v_c=R\Omega$, one has ${\kappa }^2=2(1+q){\Omega }^2$. A Keplerian disk has $q=-1/2$ and $\kappa =\Omega$, a flat galaxy rotation curve has $q=0$ and $\kappa =\sqrt{2}\Omega$, and solid-body rotation has $q=1$ and $\kappa =2\Omega$. Typical galactic rotation curves lie between these slopes. Thus $\Omega \lesssim \kappa \lesssim 2\Omega$ is a useful galactic range, not a theorem for every possible axisymmetric potential. As a precise sufficient example, epicyclic frequency bounds for monotone spherical density follow from ${\kappa }^2={\Omega }^2+4\pi G\rho$ and $0\le \rho \le 3M/(4\pi R^3)$.

Use a Cartesian frame rotating with the epicyclic guiding center, with $x$ pointing radially outwards and $y=R_c(\varphi -\Omega t)$ in the direction of rotation. To first order, conservation of angular momentum givesThe radial harmonic oscillator solution and its azimuthal integral areA constant in $y$ merely changes the azimuthal origin of the guiding centre. The epicyclic ellipse obeys $x^2/X^2+y^2/Y^2=1$. In the usual frequency range it is elongated azimuthally, and the star travels clockwise when $x$ points right and $y$ up: its small motion relative to the prograde guiding centre is retrograde.

For the Oort constants, subtract the defining expressions to obtain $\Omega =A-B$ and insert $R{\Omega }^{\prime }=-2A$ into the radial epicyclic frequency formula:The solar-neighbourhood values give $\Omega =26.5\mathrm{km}{\mathrm{s}}^{-1}\mathrm{kp}{\mathrm{c}}^{-1}$, $\kappa =\sqrt{1272}\mathrm{km}{\mathrm{s}}^{-1}\mathrm{kp}{\mathrm{c}}^{-1}$, andThe solar epicyclic ellipse is therefore about $1.49$ times longer azimuthally than radially.

A complete radial oscillation takes $T_r=2\pi /\kappa$. The oscillatory part of $\dot{\varphi }$ has zero average over this interval, so the epicyclic azimuthal advance isThis describes the advance between consecutive radial turning points of the same type; it need not be a full $2\pi$ revolution. For the Sun,The azimuthal advance estimate uses the same linear epicyclic motion approximation as the axis ratio.

Put $A=(Gm/h^3)F(hk)$. The two amplitude equations have coefficient matrixIts determinant gives the dispersion relationWriting $q=A/{\Omega }^2\ge 0$, its squared growth rates areand the four growth rates are both square roots of each displayed value. DefineFor $0<q<q_{-}$, the discriminant is positive, the sum of the squared rates is $q-1<0$, and their product is $2q(3-q)>0$; both squared rates are negative. These modes have purely imaginary $s$. For $q_{-}<q<q_{+}$, the squared rates form a nonreal complex conjugate pair; their square roots include roots with positive real part. This is overstability, meaning growing oscillations. For $q\ge q_{+}$, at least the plus squared rate is positive because $q>1$, so a real growing root exists. Consequently every $q>q_{-}$ is exponentially unstable.

Take the maximizing mode $hk=\pi$, so $q=(Gm/(h^3{\Omega }^2))F(\pi )$. It is unstable wheneverThis is also the threshold for the onset of exponential growth among all lattice modes, since $0\le F\le F(\pi )$. At equality the maximizing mode has repeated imaginary roots and can have algebraic growth; it is a marginal collision of oscillatory modes, not a strict-decay assertion. The $k=0$ mode has $F=0$, an epicyclic motion sector and a zero-frequency epicyclic guiding center sector with possible secular shear drift. These neutral or algebraic behaviors are distinct from the strict exponential instability established by the requested inequality.

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.

For a normal mode proportional to $e^{ikx-i\omega t}$, the linear equations giveFor $\omega \ne 0$, eliminate $U_y=-i\Omega U_x/(2\omega )$ and $\hat{\sigma }=k{\Sigma }_0{U}_x/\omega$. The density-wave dispersion relation in a non-self-gravitating disk isThe ${\Omega }^2$ term is the radial epicyclic frequency squared for Keplerian rotation: Coriolis force and background shear restore an orbital displacement. The $c_s^2{k}^2$ term is acoustic restoration by a pressure gradient. Long wavelengths are mainly epicyclic motion; short wavelengths are mainly acoustic waves. The given model contains no disk self-gravity term.

For completeness, the full three-variable determinant is proportional to $\omega ({\omega }^2-{\Omega }^2-c_s^2{k}^2)$. Its additional $\omega =0$ branch has $U_x=0$ and $2\Omega U_y=ik{c}_s^2\hat{\sigma }/{\Sigma }_0$. This is a stationary pressure/Coriolis balance, the zero-frequency balanced mode of an axisymmetric disk, rather than a propagating density wave. At $k=0$ it reduces to a uniform surface-density change. Dividing by $\omega$ is legitimate only for the wave branches just derived.

To first order, the initial-phase correction in $M_0$ can be omitted inside a term already multiplied by $e$. ThusFor $|a-a_g|/a\ll 1$ and $|\delta \lambda |\ll 1$, expand the exact rotating reference frame coordinates. Products of the small radial and angular displacements are second order. This givesAt $t=0$ the two sine terms cancel, as required. Equal mean motion gives a closed local epicyclic motion with radial-to-tangential amplitude ratio $1:2$; unequal mean motion adds along-track drift. The local expansion also requires $|(n-n_g)t|\ll 1$, so small frequency mismatch alone does not justify it for arbitrarily large times.

Treat $n^2=N^2/{\Omega }^2$ as a signed small parameter, with $|n^2|\ll 1$, as printed in the original PDF. At $n^2=0$, the dispersion relation factors asThus the two epicyclic motion modes have $s_0=\pm i\Omega$, while the thermal energy mode of a shearing sheet hasIt decays on the thermal diffusion time ${\beta }^{-1}$. For completeness, its next correction is $s_E=-\beta +\beta N^2/({\beta }^2+{\Omega }^2)+O(\Omega n^4)$ at fixed $\beta /\Omega$.

For either epicyclic motion root, the implicit function theorem applied to the dispersion relation givesHenceSince $\beta >0$, the oscillations undergo overstability precisely whenThe convective overstability growth rate tends to zero both for very slow diffusion and for very fast diffusion. Differentiating $\beta /({\beta }^2+{\Omega }^2)$ shows that its maximum occurs at $\beta =\Omega$, corresponding to $k^2=\Omega /\xi$. ThereforeThis is the growth rate of the perturbation amplitude; a quadratic perturbation energy grows at twice that rate. The convective overstability arises when an adverse radial specific entropy gradient couples to epicyclic motion with a finite thermal conduction lag.

Put $r_S=2R_g$. On the midplane, circular orbit balance gives $r{\Omega }^2={\partial }_r\Phi =GM/(r-r_S{)}^2$. The specific angular momentum, specific orbital energy and orbital shear parameter of a Paczyński-Wiita circular orbit are thereforeFor a small radial displacement at fixed specific angular momentum, the effective potential stability criterion gives the radial epicyclic frequencyHere ${\kappa }_r$ is the question's epicyclic $\kappa$. Its square is negative for $2R_g<r<6R_g$, so a radial displacement grows instead of undergoing epicyclic motion. The orbit at $r_{\mathrm{in}}=6R_g$ is marginal, and the outer orbits are stable to small radial displacements. At this inner edge,This efficiency belongs to the Paczyński-Wiita potential; it is an approximation to the relativistic black-hole result.

The horizontally invariant magnetized shearing-sheet equations are linear in the horizontal fields, so perturbations about the equilibrium satisfy the same equations. The fixed surface boundary conditions require $\delta B_x=\delta B_y=0$ at $z=\pm H$. Choose a normal mode withwhere the vertical wavenumber is $k=m\pi /(2H)$, $m=1,2,\dots$. With the Alfvén frequencythe four amplitude equations becomeEliminating the velocities leavesA nonzero amplitude requires the determinant of this system to vanish, yielding the ideal magnetorotational dispersion relationEquivalently, with ${\kappa }_r^2=2(2-q){\Omega }^2$,As a quadratic equation for $s^2$, its discriminant is ${\kappa }_r^4+16{\Omega }^2{\omega }_a^2>0$. If $0<{\omega }_a^2<2q{\Omega }^2$, its constant term is negative, so one root $s^2$ is positive and there is an exponentially growing mode. If ${\omega }_a^2>2q{\Omega }^2$, both the constant term and the coefficient of $s^2$ are positive, giving two negative roots and only oscillatory modes. Equality is marginal.