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Validity and the complete angular equation. The Lagrange planetary equations here describe a planar, circular-planet, small-orbital eccentricity, weakly perturbed exterior mean-motion resonance. Require $M_{\mathrm{pl}}/M_{\star }\ll 1$, $e\ll 1$, one isolated slowly varying resonant argument, perturbation times long compared with an orbital period, and no close planetary encounters. Other resonant harmonics and short-period terms must be negligible; coefficients may be frozen only across a narrow range of semi-major axis. Take $p>0$, $q=1$ or $2$ in a reduced integer ratio. Exactly $e=0$ makes the longitude of periapsis undefined, so the displayed angular variables must then be replaced by nonsingular eccentricity vector components.

Neglecting the perturbation to the accumulated mean longitude gives $\dot{\lambda }\simeq n(a)$; this is an accumulated-phase approximation, not differentiation of a fictitious expression $n(t)t$ while discarding $t\dot{n}$. With $n^2{a}^3=G{M}_{\star }$,Differentiating, using $\dot{n}=-3n\dot{a}/(2a)$ and the supplied orbital element equations, gives the full result within this constant-coefficient model:An equivalent formula replaces the last two terms in brackets by $q^2{C}_r{e}^{q-2}[(p+q)n-p{n}_{\mathrm{pl}}-2q{C}_s]-2q^3{C}_r^2{e}^{2q-4}\cos \varphi$. No derivatives of $C_r,C_s$ are included because they were specified as constants.

A full fixed point requires $\sin \varphi =0$ and $\dot{\varphi }=0$. Hence ${\varphi }_{\ast }=0$ or $\pi$, with its corresponding detuned semi-major axis fixed byAt either point $\dot{a}=\dot{e}=0$. The longitude of periapsis still precesses, so the fixed point refers to the reduced resonant dynamics.

Linear stability analysis and encounter geometry. Linearizing the full equation about a fixed point givesFor $q=2$, $\pi$ is a center equilibrium and zero is a saddle equilibrium. For $q=1$, $\pi$ is always a center equilibrium, while zero is a saddle equilibrium ifAt sufficiently small orbital eccentricity the full truncated equations instead admit a center at zero as well; equality is a degenerate case requiring higher-order analysis. Thus an unconditional instability claim at zero does not follow from the full equation. In the usual fixed-eccentricity weak-resonance regime the inequality holds and the stable libration center is $\pi$ for either order.

The physical explanation is resonance protection. At equal mean longitudes, the conjunction direction relative to periapsis obeys $q(\Lambda -\varpi )=\varphi$ modulo $2\pi$. For $q=1$, $\varphi =0$ places conjunction near periapsis, where the exterior particle comes closest to the planet; $\varphi =\pi$ places it near apoapsis. For $q=2$, zero includes both apsidal conjunctions, one of them near periapsis, whereas $\pi$ places the two conjunction branches near quadrature. The protected arrangements give restoring kicks in the positive-$C_r$ leading-harmonic model. This geometric explanation is conditional on its non-crossing, small-orbital eccentricity approximation; it does not override the low-$e$ term retained in the full stability calculation.

Small resonant-argument librations. Define the nominal resonant semi-major axis and the displaced center byThis follows by expanding $n=n_r[1-3(a-a_r)/(2a_r)]$ and taking $\cos (\pi +\delta )=-1+O({\delta }^2)$. It gives the requested initial semi-major axis to first order in the precession-induced detuning; $e_0=e(0)$ and $\delta =\Delta \varphi$.

For the pendulum approximation of a mean-motion resonance, also require fractional changes in $e$ small across a libration, and precession terms small compared with $n$. In particular $C_r/(n_r{e}_0^3)\ll 1$ for $q=1$ makes the extra full-equation curvature negligible; for $q=2$, small fractional eccentricity changes require $C_r/(n_r{e}_0^2)\ll 1$. Freeze $e=e_0$ in the leading restoring coefficient and setThen $u=\varphi -\pi$ obeys simple harmonic motion. Using $u(0)=\delta$, $\dot{u}(0)=0$ and integrating the remaining Lagrange planetary equations givesThe omitted apsidal modulation comes from higher perturbative orders and $u^2$. All expressions are leading resonant approximations, valid while the omitted fractional changes remain small. The small-libration phase portrait iswith period $2\pi /{\omega }_0$. The semi-major axis half-width is $2a_r{\omega }_0|\delta |/(3p{n}_{\mathrm{pl}})$ and the orbital eccentricity half-width is $q{C}_r{e}_0^{q-1}|\delta |/{\omega }_0$.

Dividing the original $\dot{e}$ and $\dot{a}$ equations gives the exact invariant of their retained terms,Thus the $e$–$a$ plot is a segment of this increasing curve traversed back and forth, not a closed ellipse. Locally its slope is $q/[2(p+q)e_0{a}_c]$.

Finite amplitude and the separatrix. Keep the same weak-resonance approximation but allow $\delta$ to be finite. The leading angular equation is $\ddot{\varphi }={\omega }_0^2\sin \varphi$, or $\ddot{u}=-{\omega }_0^2\sin u$. Multiplication by $\dot{u}$ givesThis resonant pendulum energy is constant in the pendulum approximation of a mean-motion resonance. It is not an exact integral of the earlier full equation when $e$ and its restoring coefficient vary: the $\dot{\varphi }\sin \varphi$ term alone prevents that conclusion in general.

For $|\delta |<\pi$ the phase portrait has closed libration curves around $\pi$, withwhere $K$ is the complete elliptic integral of the first kind. The largest speed is $2{\omega }_0\sin (|\delta |/2)$ and the leading semi-major axis half-width is $4a_r{\omega }_0\sin (|\delta |/2)/(3p{n}_{\mathrm{pl}})$. As $|\delta |\to \pi$, the energy approaches $2{\omega }_0^2$, the separatrix through the saddles at zero and $2\pi$; the particle spends increasingly long intervals near those saddles. With $ϵ=\pi -|\delta |$,At exactly $\delta =\pi$, $\dot{\varphi }=0$, the particle stays at the unstable fixed point in the ideal model. Nonstationary separatrix trajectories approach the saddle only in infinite time; above separatrix energy, the resonant argument circulates and resonance protection is lost. These distinctions matter when describing the limiting motion.

Eigenfrequencies and mode geometry. Write $d=A_{11}-A_{22}$ and $D=\sqrt{d^2+4A_{12}{A}_{21}}>0$. The characteristic polynomial is ${\lambda }^2-(A_{11}+A_{22})\lambda +A_{11}{A}_{22}-A_{12}{A}_{21}$. Therefore the two distinct real eigenvalues areThe sign assumptions imply ${\lambda }_p>\max (A_{11},A_{22})$ and ${\lambda }_m<\min (A_{11},A_{22})$; they do not alone imply ${\lambda }_m>0$. A positive lower frequency additionally requires $\det A>0$, as in the usual nondegenerate physical Laplace-Lagrange secular matrix.

Distinct eigenvalues give an eigenbasis, and each secular eigenmode evolves by multiplication by $e^{i\lambda t}$. Hencewith nonnegative amplitudes and initial phases. The second row of each eigenvalue equation gives $z_1/z_2=(\lambda -A_{22})/A_{21}$. For the upper eigenvalue this ratio is negative, whereas for the lower one it is positive. ThusThe upper secular eigenmode has anti-aligned apsides, and the lower has aligned apsides. If one amplitude vanishes, its phase is immaterial.

The initial vector sums. The planet-1 modal vectors point at $30^{\circ }$ and $120^{\circ }$. Their resultant at $60^{\circ }$ requires $e_{1p}=\sqrt{3}{e}_{1m}$. The planet-2 vectors point at $210^{\circ }$ and $120^{\circ }$; their resultant at $150^{\circ }$ requires $e_{2m}=\sqrt{3}{e}_{2p}$. These follow by resolving components perpendicular to each resultant. In particular the specified phases constrain the matrix: $r_p/r_m=3$. They cannot be imposed on an arbitrary matrix satisfying only the sign assumptions.

Put $s=r_p{r}_m=A_{12}/A_{21}$ and let $B=e_{2p}>0$. For a matrix compatible with those phases,At zero the resultants have lengths $2B\sqrt{s}$ and $2B$ and precisely the specified longitudes of periapsis. The diagram uses $B=1$, $s=1$ as a drawing normalization, not as an additional assumption on the physical planets.

Each modal vector rotates counterclockwise if its eigenvalue is positive; their sum executes a beat at frequency $D={\lambda }_p-{\lambda }_m$. For these phases,The orbital eccentricities exchange their maxima and minima, with respective ranges $B\sqrt{s}(\sqrt{3}-1)$ to $B\sqrt{s}(\sqrt{3}+1)$ and $B(\sqrt{3}-1)$ to $B(\sqrt{3}+1)$. The longitude of periapsis is the argument of each displayed vector sum; it does not in general advance uniformly at either individual eigenfrequency. The complex eccentricity trajectories are generally quasiperiodic; they close only for commensurate modal frequencies. This example has a dominating upper mode in planet 1 and a dominating lower mode in planet 2, so their apsidal phases have different winding rates.

Amplitude product and the angular-momentum convention. Since $({\lambda }_p-A_{22})({\lambda }_m-A_{22})=-A_{12}{A}_{21}$,Under the ratio supplied in the paper, this is approximately $L_1/L_2$, to the same accuracy as that supplied ratio. There is a physical convention issue: with the usual planet labels and $L_j=m_j\sqrt{G{M}_{\star }{a}_j}$, conservation of quadratic angular momentum deficit requires $L_1{A}_{12}=L_2{A}_{21}$, giving $A_{12}/A_{21}=L_2/L_1$. This follows directly by differentiating $\frac{1}{2}(L_1|z_1{|}^2+L_2|z_2{|}^2)$. Thus the paper's stated angular-momentum ratio is reversed for the conventional column-vector equation $\dot{z}=iAz$. The exact algebraic product above is independent of that naming discrepancy.

Tidal dissipation and damping. Assume the tidal dissipation contributes only the specified linear eccentricity damping, without changing the conservative matrix or adding apsidal precession. ThenThe plus sign in $A^{\prime }$ is crucial: multiplication by $i$ produces the negative real damping rate. Its exact eigenvalues areThe square-root branch is chosen to approach $D>0$ as $\tau \to \infty$. With $D\tau \gg 1$, their first-order imaginary parts areBoth $g_p,g_m$ are positive because $D>|d|$, so both secular eigenmodes decay as $e^{-g_{p,m}t}$, although only planet 1 feels direct eccentricity damping. The coupling transmits that damping to planet 2. Their precession frequencies change only at second order; their eigenvectors acquire small complex corrections, so exact alignment or anti-alignment becomes a small apsidal phase lag. The more slowly damped mode eventually dominates, unless its initial coefficient vanishes: it is the lower aligned mode for $d>0$ and the upper anti-aligned mode for $d<0$. At $d=0$ both leading damping rates equal $1/(2\tau )$.

The stated slow-secular condition presumes a nonzero positive ${\lambda }_m$. It need not imply $D\tau \gg 1$ near a nearly degenerate pair, so the exact square-root formula is the appropriate answer there. Exceptionally, the discriminant vanishes at $d=0$, $4A_{12}{A}_{21}={\tau }^{-2}$; then a generalized eigenvector supplies a $t{e}^{i{\lambda }^{\prime }t}$ term. There is still decay. Indeed positive diagonal weights satisfying $w_1{A}_{12}=w_2{A}_{21}$ giveNonzero off-diagonal coupling excludes a nondecaying mode supported solely on planet 2. This checks the damping signs without relying on the weak-damping expansion.

Write $R_p$ for the physical planet radius and $R_c=\eta R_{H1}$ for the debris-cloud radius; the source uses $R_1$ for both even though their stated size orderings are opposite. A central transit duration of a small planet on a nearly circular Kepler orbit is $D_p\simeq 2R_{\star }/(n_1{a}_1)=P_1{R}_{\star }/(\pi a_1)$. Kepler's third law, with $M_{\star }=4\pi {\rho }_{\star }{R}_{\star }^3/3$, givesA noncentral planetary exoplanet transit is shorter, so within the small-radius and nearly circular approximations, a duration requiringis too long for the planet alone and favors extended debris. For a central cloud transit with $R_c\gg R_{\star }$, the path length is approximately $2R_c$. Using the Hill radius $R_{H1}=a_1(M_1/(3M_{\star }){)}^{1/3}$ gives $D_1/P_1\simeq R_c/(\pi a_1)$, hence the Hill-cloud transit mass estimateThis equality assumes a central cloud chord. An orbit described merely as nearly edge-on need not have central cloud crossings. If its projected cloud impact distance is $s$, the measured half-chord is $\sqrt{R_c^2-s^2}$, so the boxed mass is multiplied by $[1-(s/R_c{)}^2{]}^{-3/2}$. It is a lower estimate when that geometry is unknown.

For the dynamical calculation use ${\mu }_2=M_2/M_{\star }$, $A=G{M}_2/a_2$, and $\mathcal{R}=Af(\alpha )e_1\cos {\varphi }_1$. The derivatives in the supplied Lagrange planetary equations hold the other osculating orbital elements fixed:Since $A/(n_1{a}_1^2)=n_1{\mu }_2\alpha$, the physical mean-longitude equation for a first-order resonant term isThe source's epoch notation needs care when $n_1$ varies. Literally differentiating ${\lambda }_1=n_1(t)t+{ϵ}_1(t)$ adds $t{\dot{n}}_1$; one cannot retain that definition and also identify the printed ${\dot{ϵ}}_1$ with ${\dot{\lambda }}_1-n_1$. A consistent version writes ${\lambda }_1={\int }_0^t{n}_1(s)ds+{ϵ}_{\mathrm{int}}$ and assigns the displayed correction to ${\dot{ϵ}}_{\mathrm{int}}$. Equivalently, the canonical mean-longitude equation for a first-order resonant term follows from the disturbing function directly. The result for $g$ is the intended physical equation under that convention.

Away from resonant-argument libration, neglect changes in $e_1,\alpha ,{\varpi }_1$ on one slow cycle to this order. With $n_0=2\pi /P_1$ the reference mean motion, the signed slow frequency isThis is the near-resonant transit-timing superperiod. Choose ${\varphi }_1(0)=0$. Differentiating $n_1\propto a_1^{-3/2}$ with the semi-major axis equation givesLet $C=3j{\mu }_2{n}_0^2\alpha f{e}_1$ and $n_s=n_1(0)$. Direct integration giveswith zero initial mean longitude. The constant and linear terms are absorbed into the fitted transit ephemeris: the measured mean frequency is $\bar{n}=n_s-C/\omega$. This distinction avoids mistaking an arbitrary initial frequency for the long-term fitted period.

For $|\Delta |\ll 1$, the double-integrated frequency term dominates the direct $g$ term provided $g/f$ stays finite. Relative to the fitted linear ephemeris, ${\lambda }_1(t_k)=2\pi k$ givesHere $P_1$ denotes the fitted mean period at the retained order. The signed-frequency convention fixes the phase sign. Replacing $\omega$ by $2\pi /P_t>0$ changes the sign of the signed $A_t$ when needed; a nonnegative amplitude hasUsing a signed $P_t=2\pi /\omega$ instead gives the source's signed proportionality $A_t/P_t\simeq 3{\mu }_2\alpha f{e}_1/(2\pi \Delta )$. The leading implicit transit equation can equally use $P_{1}k$ inside the sine, since the difference is higher order in the perturbation. This is the eccentricity-enhanced near-resonant transit-timing variation; its derivation assumes small longitude oscillations, ${\mu }_2{e}_1\ll {\Delta }^2$ at fixed resonance coefficients, and the effectively fixed eccentricity used in this truncated model. Dominance over forced-eccentricity timing terms typically also requires $e_1\gg |\Delta |$ at fixed $j$. A full transit calculation also includes forced orbital eccentricity and the conversion from mean longitude to the actual sky-crossing angle. Those are lower order in this eccentricity-enhanced regime, rather than identically absent; broader near-resonant formulas are derived by Lithwick, Xie and Wu.

Inside a mean-motion resonance, ${\varphi }_1$ undergoes resonant-argument libration and its evolution must be solved together with $a_1,e_1,{\varpi }_1$. The constant circulation frequency $\omega$ and the associated ${\Delta }^{-2}$ expansion then fail. At approximately fixed eccentricity, the resonant resonant pendulum approximation has libration frequency of order $j{n}_0\sqrt{{\mu }_2\alpha |f|e_1}$, up to numerical factors. The transit-timing variation follows that libration period and depends on its libration amplitude of a resonant argument. Exactly at a resonant equilibrium there need be no libration signal; near a separatrix the period grows, and eccentricity dynamics can modify the simple estimate. Resonant amplitudes remain controlled by the bounded resonant motion rather than diverging as the circulating detuning tends to zero.

In the center of mass frame, the velocities are ${\mathbf{v}}_1=(M_2/M)\mathbf{v}$ and ${\mathbf{v}}_2=-(M_1/M)\mathbf{v}$. Hence the kinetic energy is $T=\frac{1}{2}\mu v^2$, while the Newtonian gravitational potential energy is $U=-G{M}_1{M}_2/r=-GM\mu /r$. The two-body orbital energy is thereforeThe relative equation of Newtonian gravity is $\ddot{\mathbf{r}}=-GM\mathbf{r}/r^3$. It is a central force, so conservation of angular momentum confines the motion to a plane and preserves the specific angular momentum $h=|\mathbf{r}\times \mathbf{v}|=r^2\dot{\theta }$.

For completeness, derive the polar equation of a Kepler orbit. Let $w=1/r$ and use primes for $\theta$ derivatives. Then $\dot{r}=-h{w}^{\prime }$ and $\ddot{r}-r{\dot{\theta }}^2=-h^2{w}^2(w^{\prime \prime }+w)$. The radial equation gives the Binet equationChoose the angular origin at closest approach. Its solution is $w=(GM/h^2)(1+e\cos \theta )$, soFor the ellipse, the orbital eccentricity satisfies $0\le e<1$. Its extreme separations are $r_p=l/(1+e)$ and $r_a=l/(1-e)$. The semi-major axis is half their sum, giving $a=l/(1-e^2)$ and thus $l=a(1-e^2)$. The true anomaly $\theta$ runs through a full $2\pi$; the endpoints are the same position, and for a circular Kepler orbit the angular origin is arbitrary.

The radial and transverse relative speeds are $\dot{r}=(GM/h)e\sin \theta$ and $r\dot{\theta }=h/r$. Substituting these into the specific orbital energy givesTherefore the conserved orbital energy isThe negative specific orbital energy and fixed angular momentum characterize the bound Kepler orbit; neither should be assumed unchanged through a mass-ejecting explosion.

Initially the circular Kepler orbit has $r=a$ and $v^2=GM/a$. Treat the supernova kick in a binary star as impulsive: the relative position does not change, the companion's velocity is unchanged during the impulse, and the new neutron star receives $\mathbf{u}$ with $|\mathbf{u}|=\alpha v$. Thus ${\mathbf{v}}^{\prime }=\mathbf{v}+\mathbf{u}$, and with $\psi$ the angle between these vectors before the kick,Write $M^{\prime }=M_1^{\prime }+M_2$, $\beta =M^{\prime }/M$ and ${\mu }^{\prime }=M_1^{\prime }{M}_2/M^{\prime }$. The post-explosion specific orbital energy isUsing ${\epsilon }^{\prime }=-G{M}^{\prime }/(2a^{\prime })$ givesHere $a^{\prime }>0$ for a bound Kepler orbit. For a hyperbolic Kepler orbit, this formula uses the signed energy parameter $a^{\prime }<0$, rather than the positive geometric magnitude of the hyperbola's semi-major axis. At the parabolic Kepler orbit boundary, $1/a^{\prime }=0$.

The kick-direction binary survival criterion is bound if $S<2\beta$ and unbound with positive asymptotic speed if $S>2\beta$. To prove the printed sufficient disruption condition, choose a perpendicular kick, $\cos \psi =0$: then $S=1+{\alpha }^2$, so $M^{\prime }<\frac{1}{2}(1+{\alpha }^2)M$ produces positive specific orbital energy. This is sufficient, not the sharp existence threshold. Sincethe complete kick-direction binary survival criterion isThe second inequality follows by taking a kick directly opposite to $\mathbf{v}$. All directions remain bound if $2\beta >(1+\alpha {)}^2$, while every direction has positive escape energy if $2\beta <(1-\alpha {)}^2$. For $\alpha >0$, the marginal direction obeys $\cos \psi =(2\beta -1-{\alpha }^2)/(2\alpha )$ when the right-hand side lies in $[-1,1]$. Equalities give marginal parabolic Kepler orbits for the relevant extreme direction. With zero kick, losing more than half the original total mass unbinds the circular Kepler orbit.

For an escaping pair, the Newtonian gravitational potential energy approaches zero at infinite separation. Conservation of the post-explosion two-body orbital energy then gives $\frac{1}{2}{\mu }^{\prime }{V}^2={\mu }^{\prime }{\epsilon }^{\prime }$, or the asymptotic relative speed of a disrupted binaryThis is the relative recession speed, not the velocity of the new center of mass or either star's individual velocity in the original inertial frame. A marginal parabolic Kepler orbit separates with $V=0$ at infinity.

For brevity write $b_1=b_{3/2}^{(1)}(\alpha )$, $b_2=b_{3/2}^{(2)}(\alpha )$, and $K=G{M}_p\alpha /(4a_p)$. The quadratic disturbing function is $\mathcal{R}=K[\frac{1}{2}{b}_1{e}^2-b_2{e}_{p}e\cos ({\varpi }_p-\varpi )]$. Differentiation in Lagrange planetary equations givesThe separate longitude of pericentre equation is singular at $e=0$, because a circular orbit has no distinguished periapsis. The complex eccentricity, $z=e{e}^{i\varpi }$, is regular there. Using $z_p=e_p{e}^{i{\varpi }_p}$ and $\dot{z}=e^{i\varpi }(\dot{e}+ie\dot{\varpi })$ givesSince $n{a}^2=\sqrt{G{M}_{\star }a}$ and $n_p=\sqrt{G{M}_{\star }/a_p^3}$,The approximations are planar, small-orbital eccentricity, first order in the planet-to-star mass ratio, and secular averaging away from important mean-motion resonances. The semimajor axis values remain constant at this order.

For a planet with fixed orbit, set $g=A{b}_1$ and $z_f=(b_2/b_1)z_p$. Then $\dot{z}=ig(z-z_f)$, so $z-z_f=(z(0)-z_f)e^{igt}$. With $z(0)=0$,On the Argand diagram, the complex eccentricity traces a circle centred on the forced eccentricity $z_f$, of radius $|z_f|$. For an interior test particle, $0<\alpha <1$ and $g>0$, so the motion is counterclockwise, starting at the origin. The rotating radius $z-z_f$ is the free eccentricity. The orbital eccentricity ranges from zero to $2|z_f|$, attaining its maximum at half a secular period. If $z_p$ is placed on the positive real axis, the initial tangent points downward, which is consistent with counterclockwise motion from the leftmost point of this circle.

Now let the planetary mass grow as $M_p(t)=M_{p}t/t_p$ for $0\le t\le t_p$, while the planet's orbital elements and the test particle's semimajor axis remain fixed. Both the precession and forcing coefficients acquire the same factor $t/t_p$. Their ratio, and hence $z_f$, remain constant. ThusandThis is the secular phase integral for proportional mass growth. The Argand diagram circle and the free eccentricity amplitude are unchanged; only the rate of traversal changes. The angular speed about $z_f$ starts at zero and rises linearly to $g$. The early displacement is quadratic in time rather than linear. Since $t_p=2\pi /g$, $\Phi (t_p)=\pi$, soA planet already at its final mass would instead return the particle to $z=0$ at $t_p$. After growth stops, the circle continues at rate $g$, with $z(t)=z_f[1+e^{ig(t-t_p)}]$ for $t\ge t_p$. Slow mass growth does not damp the free eccentricity here: the fixed centre of the circle and the conservative equation preserve its amplitude.

For several planets with fixed masses, Laplace-Lagrange secular theory gives $\dot{\mathbf{z}}=i\mathbf{Az}$ for the planets andfor a massless particle, with real secular coupling coefficients ${\nu }_j$. Resolve planetary complex eccentricities into secular eigenmodes, $z_j(t)={\sum }_k{E}_{jk}{e}^{i{g}_{k}t}$. Away from secular resonances, the test-particle solution has the formIts Argand diagram trajectory is a superposition of rotations: generally a quasiperiodic curve with beats, rather than a circle with one fixed centre. Even if the planetary orbits are artificially held fixed, their combined forcing shifts the centre to the sum of the individual forced eccentricities. At $g_k=g_0$, the resonant contribution is proportional to $t{e}^{i{g}_{0}t}$ in linear theory; its growing amplitude eventually invalidates the small-orbital eccentricity approximation.

If one planetary mass changes, the Laplace-Lagrange secular matrix, its eigenvalues, its eigenvectors, and the particle's forced response generally change with time. Planetary modes need not remain independent. Frequencies drift and a secular resonance can sweep across the particle, changing or exciting its free eccentricity. Very slow evolution away from resonances permits an approximate adiabatic mode description; rapid evolution or a resonance passage can mix modes. The simple substitution of a phase integral works only in the special proportional-growth case above, where the entire particle equation is multiplied by one scalar and its forced centre stays fixed. It does not apply to a general many-planet system.

Radiative drag is only one of several loss mechanisms. Around a star, radiation-pressure blowout can eject small fragments; its $\beta =1/2$ threshold applies specifically to zero-kick release from a circular parent orbit. Stellar-wind drag and gas drag can drive planetary migration, while sublimation destroys grains approaching high-temperature regions. Collisional cascades destroy or fragment grains and can feed the unbound size range. Planetary scattering can cause ejection, collision with a planet, or a stellar impact; resonant trapping of dust can instead delay planetary migration.

For circumplanetary orbits, collisions with the planet or its satellites, disruption in collisions, and escape under stellar tidal forces are additional losses. Orbits near or outside the Hill sphere need not remain planet-bound. Radiation pressure on circumplanetary dust can excite planetocentric orbital eccentricity or unbind very small grains; it need not act only through slow Poynting–Robertson drag. For charged grains, the Lorentz force in stellar or planetary magnetic fields can alter or destabilize an orbit. Shadowing of circumplanetary dust changes the radiation-force average and can reduce the quoted decay rate. Which mechanism dominates depends on grain size and composition, environment, orbit orientation and the available collision or gas density.

In the weak, non-crossing, leading-eccentricity model the exterior resonant coefficient has positive sign: write its disturbing function as $R_{\mathrm{res}}=B_q{e}_c^q\cos {\varphi }_{bc}$, with $B_q>0$ for the orders considered. The physical feedback can be seen for first order. Before an outer conjunction near apoapsis, the faster inner planet pulls the outer one backwards; after passing it pulls forwards. On the outward side, the nearer, stronger earlier pull wins, reducing the outer semi-major axis and increasing its mean motion. Thus ${\varphi }_{bc}<\pi$ is driven upward. On the inward side, the later forward pull wins and drives ${\varphi }_{bc}>\pi$ downward. The paired impulses cancel at the symmetric centre.

For higher order, sum the feedback over all $q$ conjunction phases. The first surviving angular Fourier harmonic is $\sin {\varphi }_{bc}$; the lower Fourier harmonics cancel over the equally spaced directions. A local Hamiltonian reduction with canonical momentum $J$ hasThe negative curvature of the Kepler orbit Hamiltonian gives the displayed sign. Linear stability analysis at $\pi$ gives a restoring acceleration. The symmetric libration centres of an eccentricity resonance are thereforeThe corresponding conjunction phases relative to periapsis are $\{\pi \}$, $\{\pi /2,3\pi /2\}$ and $\{\pi /3,\pi ,5\pi /3\}$. They suppress repeatedly close passages and provide resonance protection.

These are expected centres of the specified leading-harmonic model, not centres fixed solely by the integer order for arbitrary orbital eccentricity and masses. Asymmetric resonant-argument libration can occur when higher Fourier harmonics matter. For example, if $R=A\cos \varphi +B\cos 2\varphi$, $A>0$, $B>A/4$, the symmetric $\pi$ equilibrium loses stability and stable centres satisfy $\cos \varphi =-A/(4B)$. Thus further dynamical information is needed beyond the PDF's unrestricted eccentric orbit to assert unique centres. Exterior asymmetric branches are documented by Winter and Murray, “Resonance and chaos. II. Exterior resonances and asymmetric libration”.

Use conventional mean longitudes $\lambda$ and take the positive integers $p,q$ in lowest terms. Rotational invariance requires the angular coefficients of a resonant argument to sum to zero. Since the inner planet has zero orbital eccentricity, the eccentricity-type term involving its undefined apsidal angle has zero amplitude. The surviving outer-eccentricity argument isIts leading disturbing function term is proportional to $e_c^q\cos {\varphi }_{bc}$. Near the period commensurability, $(p+q)n_c-p{n}_b$ is small, so the resonant argument is slow and repeated perturbations can add coherently. Trapping produces resonant-argument libration, rather than circulation, with $(p+q)n_c-p{n}_b-q{\dot{\varpi }}_c\simeq 0$.

A near-integer period ratio alone does not prove mean-motion resonance: the diagnostic is bounded resonant-argument libration. At $e_c=0$ this eccentric resonant term vanishes and ${\varpi }_c$ is undefined. If $p,q$ have a common factor, reduce the ratio first; the unreduced $q$ is not the actual lowest resonance order.