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To test radial stability of an axisymmetric circular orbit, let $\Phi (r)$ be the time-independent gravitational potential per unit stellar mass in the midplane. Axial symmetry conserves the specific angular momentum $j=r^2\dot{\phi }$. The radial equation isA circular orbit at $r_0$ obeys $j_0^2=r_0^3{\Phi }^{\prime }(r_0)$. Choose the guiding circular orbit with the same conserved angular momentum as the perturbed star. Setting $r=r_0+x$ and expanding at fixed $j=j_0$ giveswhere the radial epicyclic frequency isHere ${\Omega }^2(r)={\Phi }^{\prime }(r)/r$ and $j_c=r^2\Omega$ refer to the family of circular orbits. They are differentiated as functions of radius only after the perturbation equation has been expanded at conserved stellar $j$.

For a radially stable circular orbit, ${\kappa }^2>0$, so to linear orderThus the orbit performs small simple harmonic radial oscillations about its guiding radius.

Symmetry alone is insufficient to guarantee stability: ${\kappa }^2<0$ gives exponentially growing displacement, and ${\kappa }^2=0$ is marginal. The intended statement assumes a stable unperturbed orbit.

For a Kepler potential $\kappa =\Omega$; for a flat circular-speed curve $\kappa =\sqrt{2}\Omega$; for solid-body rotation $\kappa =2\Omega$. Conservation of $j$ also gives $\dot{\phi }\simeq {\Omega }_0(1-2x/r_0)$, explaining the accompanying azimuthal epicycle rather than a purely radial motion in the inertial plane.

The exponent of the point-mass weight is $\beta$, with $0<\beta <1$. At the origin the unreferenced relative potential has integrand $GB{b}^{\beta -1}$, whose integral diverges at its upper endpoint. In the origin-referenced potential the two terms cancel pointwise there. More generally, first form their difference at finite cutoffs and then pass to the limit. For any fixed nonzero $(R,z)$, the small-$b$ singular term is $-b^{\beta -1}$, which is integrable because $\beta >0$. At large $b$,so the weighted difference is integrable because $\beta <1$. Hence the referenced potential is well defined, and its value at the origin is zero. It is a potential difference for an infinite astrophysical disk, not a potential with zero at infinity. In fact rescaling all lengths shows ${\psi }_{II}(\lambda R,\lambda z)={\lambda }^{\beta }{\psi }_{II}(R,z)$; its finite values approach zero at the origin even though its force is singular there.

For $R>0$, differentiation under the convergent force integral in the plane givesUse $b=R\sqrt{t/(1-t)}$, for which $db=(R/2)t^{-1/2}(1-t{)}^{-3/2}dt$. The powers of $R$, $t$, and $1-t$ then givewhere the Euler beta function evaluates the constant asThe first Gamma function argument is $(1+\beta )/2$: it follows from adding one to the exponent of $t$. Both Euler beta function arguments are positive in the stated range.

For a circular orbit in the plane, $V^2/R=-{\partial }_R{\psi }_{II}$. The prescribed circular speed therefore fixes $C=K^2$ andIndependently, the Kuzmin reflection method givesHere the power of $b$ is one larger than in the force integral, so its Euler beta function arguments differ. Substituting the normalization fixed by the circular speed yieldsThis scale-free Kuzmin superposition has nonnegative surface density, finite central enclosed mass, and infinite total mass. Its force and surface density approach the Mestel disk values as $\beta \downarrow 0$; that limiting disk requires a different potential reference at the origin.

Let $\mathcal{G}$ be the outward viscous torque in an accretion disk. For a declining angular velocity, the vertically integrated viscous shear stress givesIn a steady accretion disk without radiated angular momentum, the net inward angular-momentum flux $Fh-\mathcal{G}$ is constant and equals the swallowed flux $F{h}_0$. Thus conservation of angular momentum gives $F(h-h_0)=\mathcal{G}$. Substituting $F=2\pi r\Sigma u$ cancels the surface density and provesThis uses the stipulated phenomenological viscous shear law; it does not replace it by a different fully relativistic stress prescription.

For the orbital calculation let a dot denote a derivative with respect to proper time. Substituting the geodesic energy and specific angular momentum first integrals into the timelike normalization in Schwarzschild spacetime givesDifferentiate with $h$ and $ϵ$ constant along the geodesic. Where $\dot{r}\ne 0$, division by $2\dot{r}$ givesThis equation also holds at turning points and on circular orbits. To justify that without dividing by zero, put $A=1-2m/r$ in the equatorial geodesic Lagrangian $L=(A{c}^2{\dot{t}}^2-A^{-1}{\dot{r}}^2-r^2{\dot{\varphi }}^2)/2$. Its radial Euler-Lagrange equation is $\ddot{r}=(A^{\prime }/2A){\dot{r}}^2-A{A}^{\prime }{c}^2{\dot{t}}^2/2+Ar{\dot{\varphi }}^2$. Substitution of the timelike normalization cancels the ${\dot{r}}^2$ terms and yields $\ddot{r}=-A^{\prime }{c}^2/2+(A/r-A^{\prime }/2)h^2/r^2$, exactly the displayed radial equation, including when $\dot{r}=0$.

A circular orbit requires $\ddot{r}=\dot{r}=0$. Thereforechoosing the positive sense of rotation. Differentiating the square is simpler than differentiating $h$:It is negative for $3m<r<6m$ and positive for $r>6m$, so the global minimum over timelike circular orbits isAs a stability check, the second radial derivative of $W=(1-2m/r)(c^2+h^2/r^2)$ at a circular orbit, holding $h$ fixed for the perturbation, is $W^{\prime \prime }=2m{c}^2(r-6m)/[r^3(r-3m)]$. Thus the exterior branch is stable, and $6m$ is the innermost stable circular orbit. It is the natural matching radius for the idealized zero-torque inner boundary condition.

Write the gravitational potential with spherical symmetry as $\Phi (r)$ and let $\mathbf{n}$ be the unit vector normal to a circular orbit, oriented in the direction of its angular velocity. Put $\mathbf{\Omega }=\Omega \mathbf{n}$ and denote the specific angular momentum by $\mathbf{h}=h\mathbf{n}$. Circular force balance and the specific orbital energy giveDifferentiating along the family of circular orbits,A change in orbital orientation contributes $hd\mathbf{n}$ to $d\mathbf{h}$, but $\mathbf{n}\cdot d\mathbf{n}=0$. ConsequentlyThis circular-orbit energy gradient accounts for radial and inclination changes together; spherical symmetry makes the orbital energy independent of the plane.

For either variable-mass particle, $\mathbf{H}=M\mathbf{h}$ and $E=M\epsilon$. Thus $d\mathbf{H}=Md\mathbf{h}+\mathbf{h}dM$, soConservation of mass and conservation of angular momentum impose $d{M}_2=-d{M}_1$, $d{\mathbf{H}}_2=-d{\mathbf{H}}_1$. Adding the two particle contributions provesTo interpret the signs, label the inner orbit 1 and the outer orbit 2. For $f(r)=\epsilon -\Omega h$ the first identity gives $f^{\prime }=-h{\Omega }^{\prime }$. The assumed outward decrease of angular velocity therefore makes $f_1<f_2$, so the mass-exchange part lowers the energy when $d{M}_1>0$: mass is transferred inward. In a coplanar, co-rotating pair, ${\Omega }_1>{\Omega }_2$ and an outward angular momentum transfer has $d{\mathbf{H}}_1=-dH\mathbf{n}$ with $dH>0$. Its contribution is $-({\Omega }_1-{\Omega }_2)dH<0$.

The vector difference also identifies the inclination-reducing exchange. For a fixed small magnitude of admissible angular momentum exchange, its energy contribution is most negative whenby the Cauchy-Schwarz inequality. This gives $d{E}_{\mathrm{angular}}=-s|{\mathbf{\Omega }}_1-{\mathbf{\Omega }}_2{|}^2$. Let $q={\mathbf{n}}_1\cdot {\mathbf{n}}_2=\cos \iota$ and $H_i=|{\mathbf{H}}_i|$, taking fixed particle masses for this exchange. Projecting $d{\mathbf{H}}_i$ perpendicular to its own direction givesThereforeThe downhill exchange reduces the relative orbital inclination. Its inner angular momentum component is ${\mathbf{n}}_1\cdot d{\mathbf{H}}_1=-s({\Omega }_1-{\Omega }_{2}q)<0$; in the aligned limit the outer orbit receives that angular momentum directly. For strongly inclined orbits, part of the exchange instead cancels differently directed angular momenta as the planes align. Together these gradients explain the tendencies toward inward mass transfer, outward angular momentum transport and coplanarity.

Choose a reference circular orbit at radius $r_0$, rotate the frame with $\mathbf{\Omega }=\Omega (r_0){\mathbf{e}}_z$, and introduce local Cartesian coordinates $x=r-r_0$, $y=r_0(\varphi -\Omega t)$ and vertical coordinate $z$. On scales small compared with $r_0$, neglect curvature of the coordinate directions but retain the leading differential rotation. The background velocity relative to the rotating frame isThus $A=3\Omega /4$ for a Keplerian disk. The shearing sheet retains Coriolis acceleration and the linear tidal acceleration. The radial shearing-sheet tidal potential is $-2\Omega A{x}^2$, whose force $4\Omega Ax{\mathbf{e}}_x$ balances the background Coriolis acceleration. If vertical tidal gravity is included, its equilibrium contribution is balanced by a background pressure $p_0(z)$ and disappears when that equilibrium is subtracted.

Write the total velocity as $\mathbf{u}={\mathbf{u}}_0+\mathbf{v}$. In the rotating-frame magnetohydrodynamic momentum equation, background advection is $({\mathbf{u}}_0\cdot \nabla )\mathbf{v}=-2Ax{\partial }_y\mathbf{v}$ and advection of the background by the perturbation is $(\mathbf{v}\cdot \nabla ){\mathbf{u}}_0=-2A{v}_x{\mathbf{e}}_y$. The uniform-viscosity diffusion of the linear background shear is zero. For a solenoidal magnetic field the Lorentz force per unit mass isAt constant density, absorb the magnetic-pressure gradient into $\psi =(p-p_0)/\rho +B^2/(2{\mu }_0\rho )$. Subtracting the background force balance then gives the stated momentum equation, with the remaining nonlinear advection $\mathbf{v}\cdot \nabla \mathbf{v}$, Coriolis acceleration $2\mathbf{\Omega }\times \mathbf{v}$, magnetic tension and viscous diffusion.

For constant finite electrical conductivity ${\sigma }_e$, put $\eta =({\mu }_0{\sigma }_e{)}^{-1}$. The resistive induction equation with uniform magnetic diffusivity is ${\partial }_t\mathbf{B}+\mathbf{u}\cdot \nabla \mathbf{B}=\mathbf{B}\cdot \nabla \mathbf{u}+\eta {\nabla }^2\mathbf{B}$. Substituting the background shear supplies the term $-2A{B}_x{\mathbf{e}}_y$ and the same background-advection operator as in the momentum equation. The two further conditions areThey express incompressibility and the absence of magnetic monopoles, respectively.

Now take real plane waves with common phase $\vartheta =\mathbf{k}(t)\cdot \mathbf{x}$ and real wavevector. The solenoidal conditions becomeEvery field depends on position only through $\vartheta$. Hence for either transverse vector field $\mathbf{P}$ and either wave field $\mathbf{Q}$, $(\mathbf{P}\cdot \nabla )\mathbf{Q}=(\mathbf{P}\cdot \mathbf{k}){\partial }_{\vartheta }\mathbf{Q}=0$. This proves that all the displayed quadratic advection and tension terms vanish pointwise, including products involving conjugate amplitudes. These waves are therefore exact solutions at finite amplitude, not merely a linearization.

Acting with $D_0={\partial }_t-2Ax{\partial }_y$ on the phase yieldsCanceling its position-dependent part requiresThe remaining amplitude equations areDifferentiating the velocity constraint determines the pressure amplitude:The resistive induction equation preserves its constraint because $\dot{\mathbf{k}}\cdot \tilde{\mathbf{B}}=2A{k}_y{\tilde{B}}_x$ cancels the corresponding shear term in $\mathbf{k}\cdot \dot{\tilde{\mathbf{B}}}$. The specified magnetic field has no spatially uniform part. Thus there is no background magnetic field producing an additional linear magnetic-tension coupling. The scalar $\psi$ is the modified pressure; the physical pressure can include a constant and a second spatial harmonic from $-B^2/(2{\mu }_0)$ while $\psi$ retains the stipulated plane-wave form. This is an exact zero-mean magnetic shearing wave.

For decay, take the real part of the scalar product of the velocity-amplitude equation with its complex conjugate. Pressure does no work because $\mathbf{k}\cdot \tilde{\mathbf{v}}=0$, and the real part of Coriolis work is zero. ThereforeThe resistive induction equation similarly givesusing $2|u_x{u}_y|\le |\mathbf{u}{|}^2$. Integrating these differential inequalities, defineThenA non-axisymmetric wave has $k_y\ne 0$. For nonzero differential shear and positive $\nu ,\eta$, the negative cubic terms dominate the linear upper bound on shear work. ConsequentlyThis cubic-exponent decay of a nonaxisymmetric shearing wave allows transient energy growth before the winding to short radial wavelengths makes diffusion dominant. The magnetic amplitudes can also be written explicitly aswhich confirms the damping independently. Positivity of the dissipative coefficients is necessary: with $k_z=0$, an inviscid vertical velocity amplitude can remain constant, and an ideal vertical magnetic amplitude can remain constant when $\eta =0$. The asserted decay applies to the viscous, finitely conducting model; taking an ideal limit before the long-time limit changes the conclusion.

Inside the uniform sphere, $V_c^2=3{\sigma }^2{r}^2/r_0^2$. Thus, with $X=V_c/(\sqrt{2}\sigma )$, part (b) givesThe Gaussian integral needed for the lowered Maxwellian velocity distribution isSubstitution $v=\sqrt{2}\sigma y$ therefore yieldsSince $V_c<v_e$ throughout the core, the integrand is positive and so is $I(X)$ for $X>0$. Using $m{n}_0={\rho }_0$, part (a)'s massive-test-body drag magnitude isThe slow drift of a circular orbit is determined by torque, not by equating its changing circular speed to the tangential drag. Here $V_c=\omega r$, with constant $\omega =\sqrt{4\pi G{\rho }_0/3}$. The specific angular momentum $j=r{V}_c=\omega r^2$ obeysHence the positive local decay timescale isThere is a genuine factor-of-two discrepancy with the requested denominator $32{\pi }^2$. That denominator would give $V_c/(2D)$, which is the timescale for a flat galaxy rotation curve, not this uniform-density core. It also follows from the incorrect replacement ${\dot{V}}_c=-D$, which drops the $V_c\dot{r}$ term in $d(r{V}_c)/dt$. The literal signed definition $r/(2\dot{r})$ is negative for an inward drift; it equals $-{\tau }_{+}$. A positive decay time requires the absolute value or a minus sign.

This is the slow circular inspiral under tangential drag calculation. Its assumptions require $D$ small enough for nearly circular adjustment and a meaningful large Coulomb logarithm in stellar dynamics; the leading-log approximation is not uniform at $V_c\to 0$. Also, integrating the truncated distribution all the way to $v_e$ does not give exactly $n_0$. The model uses $n_0$ as the stated core normalization; exact self-consistency of a sharply truncated uniform core with that distribution is not implied.

Set $\mu =GM>0$ and interpret $L$ in the displayed equation as specific angular momentum, $L=r^2\dot{\psi }$, rather than total angular momentum. The gravitational potential is per unit mass and $a$ has dimensions of length. Spherical symmetry fixes the orbital plane and conserves the vector angular momentum, while the autonomous specific orbital energy is also conserved.

For the orbit equation for combined inverse-square and inverse-cube attraction, defineA regular nonradial bounded oscillation requires $L\ne 0$ and $L^2-2\mu a>0$. Solving the linear Binet equation then givesFor a noncircular bound orbit $0<e<1$; its specific orbital energy is $E={\mu }^2(e^2-1)/[2(L^2-2\mu a)]$. The effective potential $[L^2-2\mu a]/(2r^2)-\mu /r$ has a centrifugal barrier and a minimum in this case. If $L^2-2\mu a\le 0$, the oscillator frequency is zero or imaginary and there is no regular finite-radius bound oscillation: the attractive centre lacks a centrifugal barrier. Unbound branches with $e\ge 1$ are not closed ellipselike trajectories.

Successive pericentres differ in angle by $2\pi /\nu$. The full phase space state returns after $p$ radial oscillations and $q$ revolutions precisely whenwhere $p,q$ are coprime positive integers; the second formula applies when $a\ne 0$. Thus $a>0$ requires $0<p/q<1$ and $L^2>2\mu a$, whereas $a<0$ requires $p/q>1$. An irrational $\nu$ advances the apsidal direction through a dense set of angles and gives a nonclosed rosette, with only the four regular spherical invariants needed to label its two-dimensional closure.

To see the additional resonant label explicitly, let $u_c=\mu /(L^2-2\mu a)$ and $w=u-u_c-i{u}^{\prime }/\nu$. Then $w=C{e}^{i\nu (\psi -{\psi }_0)}$, so on a family with fixed resonant $L$,Its argument fixes the remaining apsidal phase modulo the symmetry of the closed curve; its magnitude is already determined by specific orbital energy and $L$. Choosing reference axes in the fixed orbital plane makes this single-valued under $\psi \mapsto \psi +2\pi$. This is the fifth independent orbit label on the resonant family described by closed orbits with an inverse-cube perturbation.

For $a\ne 0$, $\nu$ varies with $L$. The orbital resonance condition holds only on selected angular momentum surfaces, so this construction is not a fifth global smooth integral on an open set of arbitrary nearby $L$. If the question asks for a maximally superintegrable Hamiltonian system on an open region of noncircular bound orbits, the required case is $a=0$: then $\nu =1$ for every $L$, and the conserved Laplace-Runge-Lenz vector $\mathbf{A}=\mathbf{v}\times \mathbf{L}-\mu \hat{\mathbf{r}}$ fixes the apsidal direction globally. The relations $\mathbf{A}\cdot \mathbf{L}=0$ and $A^2={\mu }^2+2E{L}^2$ leave five independent constants. For the usual orbit-wise interpretation, noncircular bound orbits with rational $\nu$ have five isolating labels; irrational $\nu$ has four. Circular orbits are closed for any allowed $\nu$ but have no independent apsidal phase, so the generic noncircular count cannot simply be applied to them.

For completeness, regular unbound branches are different. When $e\ge 1$ and $\nu >0$, positive radius restricts the radial phase to $|\nu (\psi -{\psi }_0)|<\arccos (-1/e)\le \pi$. The apsidal invariant on an unbound central-force branch isThe principal argument is continuous on this scattering branch because it never reaches an apocentre across the negative-real-axis cut. Thus the periapsis direction can be a fifth single-valued scattering label for any positive $\nu$, including irrational $\nu$; the rationality condition applies to the noncircular bound case. For plunging or collision branches with $L^2-2\mu a\le 0$, local trajectory labels can still be defined away from the singular centre, but the preceding periodic-orbit criterion is inapplicable and no smooth continuation through $r=0$ is asserted.

Write $c=\cosh \varphi$. The proposed function is $u=(c-2)/(3M(c+1))$. Differentiating using $c^{\prime }=\sinh \varphi$ and $(c^{\prime }{)}^2=c^2-1$ verifies $u^{\prime \prime }+u-3M{u}^2=0$. Its physical branches have $u>0$, namely $|\varphi |>{\varphi }_0=\mathrm{arcosh}2$, andOn the positive branch it comes from infinity at ${\varphi }_0$ and spirals toward $r=3M$ from outside as $\varphi \to \infty$. The negative branch gives the reversed orbit after reflection. The interval where $u<0$ does not describe positive physical radius.

Near ${\varphi }_0$, $\cosh \varphi -2=\sqrt{3}(\varphi -{\varphi }_0)+O((\varphi -{\varphi }_0{)}^2)$. Thus $r\sim 3\sqrt{3}M/(\varphi -{\varphi }_0)$, while $\sin (\varphi -{\varphi }_0)\sim \varphi -{\varphi }_0$. HenceThis is the perpendicular offset of the incoming asymptote and agrees with $|L/E|$ from the circular orbit. The orbit is the critical separatrix between scattering and capture.

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.

Independently, the specific orbital energy of a circular particle isThus $E(\infty )=0$ and $E(3R_S)=-GM/(8R_S)=-c^2/16$. The energy released per unit accreted mass is exactly $c^2/16$, confirming the integrated accretion luminosity. The Paczyński-Wiita circular orbit reproduces the Schwarzschild inner radius but gives an approximate binding energy; the efficiency derived here is for the stated potential.

Put $M=M_1+M_2$. In a circular orbit, the distances from the centre of mass are $r_1=a{M}_2/M$ and $r_2=a{M}_1/M$. Summing the components' angular momentum givesUse Kepler's third law ${\Omega }^2{a}^3=GM$ and $\Omega =2\pi /P_{\mathrm{orb}}$ to obtain the circular-binary orbital angular momentumThe numerator contains orbital period, not the erroneous $M_{\mathrm{orb}}$ token in the converted TeX.

The total mass changes as $\dot{M}=(1-f){\dot{M}}_1$. With the prescribed wind carrying specific angular momentum $\lambda J/M$, the angular-momentum loss law is $\dot{J}/J=\lambda \dot{M}/M$. For constant $\lambda$, integration gives $J\propto M^{\lambda }$. Cubing the orbital formula then proves the binary period invariant for a constant wind angular-momentum factor:This integration does not require $f$ to be constant, although its instantaneous value enters the following response coefficients. In the conservative case $f=1$, both $M$ and $J$ are constant and the usual period-product invariant for conservative mass transfer is recovered.

Kepler's law and the given Roche lobe approximation imply $R_L\propto M_1^{1/3}{P}_{\mathrm{orb}}^{2/3}$, since the total-mass factor cancels. Using the period law therefore givesBecause ${\dot{M}}_2=-f{\dot{M}}_1$, the instantaneous Roche-lobe radius response exponent isFor $x=\log (R/R_L)$, subtracting this response from the prescribed stellar radius evolution givesThe overflow-rate model requires $f>0$ and is used for small positive overfill. It gives ${\dot{M}}_1/M_1=-x/(f{t}_{\mathrm{dyn}})$. Hence the nuclear-driven Roche-overfill relaxation equation is

Let the bracket be $A(q)$. On the short response time, freeze the slowly changing masses and coefficients. If $A>0$, the local solution isThe equilibrium donor mass-loss rate is ${\dot{M}}_1/M_1=-1/(A{t}_{\mathrm{nuc}})$. Thus the required stabilizing mass-ratio condition isWhen $A$ is of order one and $t_{\mathrm{dyn}}\ll t_{\mathrm{nuc}}$, the overfill is small and relaxation is fast, while mass transfer proceeds on a nuclear timescale. A positive $A$ arbitrarily close to zero does not by itself ensure this timescale separation or small overfill.

For an explicit threshold put $B=n+5/3$ and $C_w=2\delta (1-f)/3$. The inequality isIf $B>0$ and $f>0$, its positive root givesFor $f=1$ this reduces to $q<(n+5/3)/2$, the conservative mass-transfer critical mass ratio. For other parameter signs the unsimplified inequality is the general condition.

If $A<0$, a larger overfill causes further growth rather than restoration, on the short scale $f{t}_{\mathrm{dyn}}/|A|$. Transfer runs away towards a dynamical episode, potentially a common envelope or merger; the small-overfill model then fails. At $A=0$ the overfill grows linearly under nuclear expansion and there is no steady regulated contact. These conclusions use the stipulated donor-radius response; an actual dynamical episode requires its adiabatic response and hydrodynamics.