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For $k_y=0$ and $k_z\ne 0$, pressure elimination gives ${\dot{v}}_x=2\Omega \chi v_y-\nu k^2{v}_x$ and ${\dot{v}}_y=-2(\Omega -A)v_x-\nu k^2{v}_y$, where $\chi =k_z^2/k^2$. The inviscid frequency is the radial epicyclic frequency times $\sqrt{\chi }$. With $w=v_x/\sqrt{\chi }$, $(\Omega -A)|w{|}^2+\Omega |v_y{|}^2$ is conserved when viscosity vanishes. In a Keplerian flow it bounds kinetic-energy amplification by four, whereas with zero rotation the same system has the algebraic lift-up effect.

// Target: turbulence.bigb

Let $c_s^2=(dP/d\Sigma {)}_0$ be the barotropic closure of a razor-thin disk sound-speed squared and ${\kappa }_r$ the radial epicyclic frequency, with ${\kappa }_r^2=2(2-q_r){\Omega }^2$ for orbital shear parameter $q_r$. For a radial axisymmetric normal mode in a uniform barotropic shearing sheet, let $U_x,U_y$ be velocity amplitudes and $\sigma$ the surface-density amplitude. Linear momentum gives $-i\omega U_x=2\Omega U_y-ik{c}_s^2\sigma /{\Sigma }_0$ and $-i\omega U_y=-{\kappa }_r^2{U}_x/(2\Omega )$; mass conservation gives $\omega \sigma =k{\Sigma }_0{U}_x$. Eliminating the amplitudes for $\omega \ne 0$ proves the displayed dispersion relation. The rotational term is epicyclic restoration; the pressure term is acoustic restoration. Without a perturbed gravitational potential, no self-gravity term belongs in the formula.

In a Cartesian frame rotating with an epicyclic guiding center, a small planar oscillation obeys $x=X\cos (\kappa t+\chi )$ and $y=-(2\Omega X/\kappa )\sin (\kappa t+\chi )$. The ellipse is elongated azimuthally when the radial epicyclic frequency satisfies $\kappa <2\Omega$. Conservation of angular momentum makes the motion around the guiding centre retrograde.

For a nonnegative, radially nonincreasing spherical mass density, ${\Omega }^2=GM(r)/r^3$ and ${\kappa }^2={\Omega }^2+4\pi G\rho (r)$. Since $\rho (r)$ does not exceed the mean enclosed density, the radial epicyclic frequency lies between $\Omega$ and $2\Omega$. General axisymmetric disks need not satisfy this bound.

Epicyclic motion is the small oscillation of a displaced fluid element or particle about a circular orbit. Its natural radial frequency is the radial epicyclic frequency.

For $\Phi =-\mu /r+{\alpha }^2\mu (3z^2-r^2)/r^5$, a nearly circular equatorial orbit has $n^2=\mu /R^3$ and azimuthal frequency ${\Omega }_c^2=n^2(1+3{\alpha }^2/R^2)$. The radial epicyclic frequency is ${\kappa }^2=Rd{\Omega }_c^2/dR+4{\Omega }_c^2$, and the vertical epicyclic frequency is ${\nu }^2={\partial }_z^2\Phi {|}_{z=0}$. Radial stability requires $R^2>3{\alpha }^2$. The nodal precession rate ${\Omega }_c-\nu$ is negative and the apsidal precession rate ${\Omega }_c-\kappa$ positive. These sign statements concern small eccentricity and inclination. For an arbitrary inclined orbit, the averaged argument of periapsis instead evolves as $(3/2)n{\alpha }^2[5{\cos }^{2}i-1]/[a^2(1-e^2{)}^2]$ and can regress.

In the local incompressible weak-field limit, take an axisymmetric wavevector $(k_r,0,k_z)$ and ${\omega }_A=\mathbf{k}\cdot \mathbf{B}/\sqrt{{\mu }_0\rho }$. Eliminating the pressure projects the radial force by $k_z^2/k^2$. This yields the displayed dispersion relation, where $\kappa$ is the radial epicyclic frequency. For a hydrodynamically stable rotation profile with outward-decreasing angular velocity, instability requires $0<{\omega }_A^2<-(k_z^2/k^2)d{\Omega }^2/d\ln r$. A purely toroidal field has no axisymmetric local field-line-bending frequency and therefore no ordinary weak-field MRI in this approximation.

The Oort constants describe local differential galactic rotation: $A=-R{\Omega }^{\prime }/2$ and $B=-\Omega -R{\Omega }^{\prime }/2$. Thus $\Omega =A-B$, the radial epicyclic frequency obeys ${\kappa }^2=-4\Omega B$, and the epicyclic ellipse has axis ratio $\sqrt{-B/(A-B)}$.

The orbital shear parameter measures the fractional radial decrease in orbital angular frequency. A Kepler orbit has $q=3/2$, while rigid rotation has $q=0$. The radial epicyclic frequency obeys ${\kappa }_r^2=2(2-q){\Omega }^2$; this follows by differentiating the squared specific angular momentum $h^2=r^4{\Omega }^2$.

Let $r_S=2GM/c^2$. In the Paczyński-Wiita potential, circular orbit balance and the radial epicyclic frequency formula giveThus the innermost stable circular orbit lies at $r=3r_S$. Its specific orbital energy is $-c^2/16$, giving a zero-torque radiative efficiency of black-hole accretion of $1/16$. The model reproduces the Schwarzschild inner orbit exactly but approximates its relativistic energy and frequency. Abramowicz's derivation explains the potential's relation to the Schwarzschild effective potential.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 a circular mid-plane orbit, radial force balance in the Newtonian gravitational potential is $r{\Omega }^2={\Phi }_{,r}$. HenceAxisymmetry conserves the specific angular momentum $j=r^2\dot{\varphi }$. At fixed $j$, radial motion obeys $\ddot{r}=-{\partial }_r{\Phi }_{\mathrm{eff}}$, where the effective potential is ${\Phi }_{\mathrm{eff}}=\Phi +j^2/(2r^2)$. Linearize around a circular radius $r_0$, writing $r=r_0+\xi$. Since $j^2=r_0^3{\Phi }_{,r}(r_0,0)$ there,This is the radial epicyclic frequency; equivalently ${\kappa }^2=r^{-3}d(r^4{\Omega }^2)/dr$.

Reflection symmetry gives ${\Phi }_{,z}(r,0)=0$ and ${\Phi }_{,rz}(r,0)=0$, so vertical and radial displacements decouple at linear order. The vertical equation is $\ddot{z}=-{\Phi }_{,zz}(r_0,0)z$, giving the vertical epicyclic frequencyPositive squared frequencies give oscillatory stability; a negative squared frequency gives exponential instability rather than a real oscillation 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.

For axisymmetric disturbances, $k_y=0$, the wavevector is fixed, so there is no winding/swing mechanism. Take $k_z\ne 0$ and set $\chi =k_z^2/k^2$. Eliminating pressure with incompressibility givesThe axisymmetric incompressible epicyclic wave has inviscid frequencywhere ${\kappa }_r$ is the radial epicyclic frequency.

In a Keplerian flow $A=3\Omega /4$ and ${\kappa }_r=\Omega$. The axisymmetric disturbances therefore execute stable inertial/epicyclic oscillations, with viscous amplitude decay $e^{-\nu k^{2}t}$. Rotation opposes sustained radial displacement through angular-momentum conservation. There can be bounded exchanges of kinetic energy between components, but no large Reynolds-number-dependent amplification of this axisymmetric family. Indeed, with $w=v_x/\sqrt{\chi }$, inviscid motion conservesSince the physical velocity norm is $|w{|}^2+|v_y{|}^2$, Keplerian energy amplification is at most $\Omega /(\Omega -A)=4$; viscosity can only reduce that bound. A pure initial azimuthal disturbance can attain the inviscid factor four after a quarter epicyclic period. By contrast, the two-dimensional nonaxisymmetric Orr mechanism remains available in a Keplerian flow and has the ${\mathrm{Re}}^{2/3}$ gain derived above.

In non-rotating shear, $\Omega =0$ while $A$ remains nonzero. The axisymmetric solution isCross-stream motion transports the basic shear and generates an azimuthal streak: this is the lift-up effect. Inviscid azimuthal velocity grows linearly and its energy quadratically, rather than oscillating. With viscosity, the gain for a suitable initial cross-stream disturbance scales as $[A/(\nu k^2){]}^2$ at large Reynolds number, appreciably stronger than the two-dimensional ${\mathrm{Re}}^{2/3}$ scaling. For instance, $C_y=0$ and fixed $\chi >0$ give the leading optimized gain $4\chi A^2/(e^2{\nu }^2{k}^4)$ at $\tau \simeq 1/(\nu k^2)$. Neither mechanism is an exponentially growing hydrodynamic normal mode. Rotation bounds axisymmetric lift-up in Keplerian shear; without rotation, lift-up gives strong algebraic transient growth. If $k_z=0$ as well as $k_y=0$, incompressibility forces $v_x=0$ and removes this coupling altogether.

For a bound Kepler orbit, the semimajor axis is half the long diameter of the ellipse; if $r_p,r_{mathrma}$ are its apsidal distances, $a=(r_p+r_{mathrma})/2$. The orbital eccentricity is $e=(r_{mathrma}-r_p)/(r_{mathrma}+r_p)$, with $0\le e<1$ and semiminor axis $b=a\sqrt{1-e^2}$. The gravitating centre is a focus, displaced by $ae$ from the ellipse's geometric centre.

Choose a reference plane, a positive normal and a fixed reference direction in that plane. The ascending node is where the orbit crosses toward positive normal displacement. The longitude of ascending node $\Omega$ is the reference-plane angle from the fixed direction to that node. The argument of periapsis $\omega$ is the angle from the ascending node to periapsis, measured in the orbital plane in the direction of motion. The orbital inclination $i$ is the angle between the reference normal and the orbital angular-momentum normal, with $0\le i\le \pi$. The diagram separates the ellipse's shape from these orientation angles.

Let $\mu =GM$ and use actions per unit test-particle mass. In the prograde spherical-action chart underlying the stated Hamiltonian, $J_{\varphi }=L_z$ and $J_{\theta }=L-L_z$, so $J_{\theta }+J_{\varphi }=L$. The radial action then satisfies $J_r+L=\sqrt{\mu a}$, consistently with $H_K=-\mu /(2a)$. DefineIf $(w_r,w_{\theta },w_{\varphi })$ are the old angles, a type-two generating function isDifferentiation with respect to the old angles recovers $J_r,J_{\theta },J_{\varphi }$, and differentiation with respect to the new actions gives the canonical transformationIts triangular integer action matrix has determinant one, so it preserves the angle periodicities. In the new coordinates,These are the Delaunay variables:With the usual choices of angle origins, $w_1=\ell$ is mean anomaly, $w_2=\omega$ is argument of periapsis, and $w_3=\Omega$ is longitude of ascending node. The generator $\mathcal{G}=L$ rotates the orbit within its plane, whereas $\mathcal{H}=L_z$ rotates that plane about the reference normal; this explains the angular identification. The fast angle obeys Kepler's equation $\ell =\mathcal{E}-e\sin \mathcal{E}$, where $\mathcal{E}$ is eccentric anomaly.

For retrograde orbits, signed $J_{\varphi }=L_z$ gives $J_{\theta }=L-|L_z|$ and the spherical Hamiltonian uses $J_r+J_{\theta }+|J_{\varphi }|$. The stated plus-sign form is therefore a prograde chart, not a global signed-action formula. The Delaunay variables still use $\mathcal{H}=L_z$; on the retrograde branch the generating function is $(\mathcal{L}-\mathcal{G})w_r+(\mathcal{G}+\mathcal{H})w_{\theta }+\mathcal{H}{w}_{\varphi }$. At $e=0$, periapsis is undefined, and at $i=0$ or $\pi$ the node is undefined: these are angle-chart degeneracies, not physical singularities.

For the perturbed planet orbit, the radial and vertical frequencies mean small oscillations about a circular equatorial orbit of cylindrical radius $R$. Write $r=\sqrt{R^2+z^2}$ and retain the stated quadrupole potential. On the equator,At fixed axial specific angular momentum $L_z$, the radial effective potential is $\varphi (R,0)+L_z^2/(2R^2)$. Its second derivative at circular balance $L_z=R^2{\Omega }_c$ gives the radial epicyclic frequencyReflection symmetry decouples vertical motion to first order. Expanding the potential in $z$ givesso the vertical epicyclic frequency isThe circular orbit is radially stable only for $R^2>3{\alpha }^2$. For $\alpha \ne 0$ in that domain, $\nu >{\Omega }_c>\kappa$. Vertical node crossings repeat at frequency $\nu$ while azimuth advances at ${\Omega }_c$, so the line of nodes has nodal precession rate ${\Omega }_c-\nu <0$. Radial periapsis passages repeat at $\kappa$, giving apsidal precession rate ${\Omega }_c-\kappa >0$. To leading quadrupole order, with $n=\sqrt{\mu /R^3}$,This is oblate-quadrupole epicyclic precession. Here $\varpi$ denotes the nearly equatorial longitude of periapsis, not $\omega$ alone.

The precession signs above have a small-eccentricity, small-inclination domain. They are not a universal claim for every inclined Kepler ellipse. To see the distinction with the same Delaunay variables, average the perturbation over an unperturbed orbit. Since $z/r=\sin i\sin (\omega +f)$ and $dt=r^{2}df/\mathcal{G}$, the term linear in $e\cos f$ integrates to zero and the ${\sin }^2(\omega +f)$ average in the $r^{-3}$ integral is $1/2$. ThusDifferentiating at fixed other actions yields the first-order secular ratesTheir nearly equatorial sum reproduces the prograde $\dot{\varpi }$. For a polar orbit, however, the node is fixed and $\dot{\omega }<0$. This supplies a concrete counterexample to extending the requested forward periapsis precession beyond the intended epicyclic regime.

For an equatorial circular orbit in the Paczyński-Wiita potential, radial force balance givesIts specific angular momentum is $j=R^2\Omega =\sqrt{GM}{R}^{3/2}/(R-R_S)$. Perturbing the circular radius at fixed $j$ gives the radial epicyclic frequency squared ${\kappa }^2=R^{-3}d(j^2)/dR$. ThereforeCircular motion is radially stable for $R>3R_S$, marginal at $3R_S$, and unstable between $R_S$ and $3R_S$. Thus the inner edge of the nearly circular thin accretion disk is $R_{\mathrm{in}}=3R_S$, joining a plunging region of a black-hole accretion disk inside the innermost stable circular orbit. This conclusion concerns the circular thin-disk model, not an assertion that all gas disappears inside this radius.

Take the inward mass accretion rate $\dot{M}$ as positive, and write the positive outward viscous torque in an accretion disk asSteady conservation of angular momentum makes the net inward flux $C=\dot{M}j-\mathcal{G}$ independent of radius. Impose a zero-torque inner boundary condition at $3R_S$, appropriate to an approximately stress-free transition to plunging motion. ThenThis is the steady accretion disk with arbitrary rotation law angular-momentum balance.

The local viscous dissipation integrated through the full disk thickness is $D=\nu \Sigma (R{\Omega }^{\prime }{)}^2=-\mathcal{G}{\Omega }^{\prime }/(2\pi R)$. It is the power per unit planar area summed over both radiating faces. Sincesubstitution gives the two-face dissipation of a Paczyński-Wiita disk:In particular $D$ vanishes at the stress-free inner edge and is nonnegative outside it because $j$ increases there. A one-face radiative flux is $D/2$; the total accretion luminosity is therefore $L={\int }_{3R_S}^{\infty }2\pi RDdR$, with no additional factor of two.

For a direct evaluation put $x=R/R_S$. ThenThe first integral is $7/4$, using $(3x-1)/(x-1{)}^3=3/(x-1{)}^2+2/(x-1{)}^3$. For the second,so it equals $2/[\sqrt{3}(3-1)]=1/\sqrt{3}$. HenceThis is the radiative efficiency of black-hole accretion within this pseudo-Newtonian model.