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An aligned eccentric ring consists of nested Kepler orbits with nearly common longitude of periapsis. Its width and surface density of a disk depend on both the distribution of semi-major axis values and the radial gradient of forced eccentricity.

With a fixed planetary pericentre, the linear secular equation $\dot{z}=iA(z-Z_f)$ has solution $z=Z_f+(z(0)-Z_f)e^{iAt}$. An initially circular orbit follows $z=Z_f(1-e^{iAt})$, a counterclockwise circle in the complex plane centered at the forced eccentricity $Z_f$, with free eccentricity amplitude $Z_f$. Its maximum orbital eccentricity is $2Z_f$ and its phase-averaged mean orbital eccentricity is $4Z_f/\pi$.

At each fixed semimajor axis, let $B=b_{3/2}^2(\alpha )/b_{3/2}^1(\alpha )$ and $Z_f=B{e}_{\mathrm{pl}}$. The planet's longitude of pericentre is the zero of longitude. The secular perturbation equation becomes $\dot{z}=iA(z-Z_f)$. Integrating it with the initial condition $z(0)=0$ givesThe forced eccentricity vector is the constant real $Z_f$; the free eccentricity vector is $-Z_f{e}^{iAt}$. Consequently the complex eccentricity follows a circle in the Argand plane centered at $Z_f$, with radius $Z_f$, starting at the origin and moving counterclockwise. Its initial velocity is downward, $\dot{z}(0)=-iA{Z}_f$. The secular period is $2\pi /A$, andThe longitude of pericentre is undefined at the origin, but the complex eccentricity passes smoothly through it. At $At=\pi$, $z=2Z_f$ and the orbit is maximally eccentric and aligned with the planet.

Using the small-$\alpha$ Laplace coefficient expansion givesAlso $n=n_{\mathrm{pl}}{\alpha }^{-3/2}$ to leading order in the planet-to-star mass ratio. The secular forcing by a distant outer planet is thereforeA relative $O({\alpha }^2)$ correction to $A$ can accumulate a phase error at very long times; the leading-frequency formula is not a uniform-in-time expansion.

If $a_2=a_1(1+\delta )$ with $\delta =\delta a/a_1\ll 1$, then $Z_{f,2}/Z_{f,1}=1+\delta +O({\delta }^2)$ and $A_2/A_1=1+3\delta /2+O({\delta }^2)$. The outer planetesimal therefore has a slightly larger forced eccentricity circle and a shorter secular period. Its accumulated secular phase lead is $\Delta (At)\simeq 3A_{1}t\delta /2$. Initially both circles start at the origin, but their eccentricity vectors progressively lose alignment. Linearizing the exponential in $\delta$ additionally requires $|A_{1}t\delta |\ll 1$.

The orbit crossing is caused by this accumulating differential secular phase, even though individual orbital eccentricities remain small. To first order in orbital eccentricity, the radius of an orbit at fixed inertial longitude $\theta$ isFor an infinitesimally neighboring orbit,This derivative is evaluated within the first-order orbital-shape model; derivatives of the neglected quadratic terms give relative $O(Z_f)$ corrections near crossing. The least separation over longitude is consequently $\delta a[1-|z+a{\partial }_{a}z|]$. The differential secular orbit-crossing criterion is $|z+a{\partial }_{a}z|=1$. Since $Z_f\propto a$ and $A\propto a^{3/2}$,The first term stays bounded by $4Z_f\ll 1$, whereas the second grows in proportion to time. To leading order in $Z_f$, crossing begins when $(3/2)Z_{f}At=1$. ThusMore precisely, in this first-order orbital-shape approximation the exact first crossing lies within a fractional $O(Z_f)$ window of this value, since the magnitude differs from $(3/2)Z_{f}At$ by at most $4Z_f$. Higher-order Laplace coefficients give additional corrections. The local derivation takes the neighboring-orbit limit before the long-time limit; for a finite separation it requires $\delta a/a_1\ll Z_f$ near crossing. A broad continuous disk contains such neighboring orbits. The crossings first appear in directions selected by the phase of $z+a{\partial }_{a}z$, not simultaneously at every longitude.

For the collision energy estimate adopt the phase-averaged mean orbital eccentricity $\overline{e}=4Z_f/\pi \simeq 5\alpha e_{\mathrm{pl}}/\pi$. The specified velocity estimate givesUsing the projectile kinetic energy per target mass convention for specific impact energy, equal-size and equal-density bodies have equal masses and $Q=\frac{1}{2}{v}_{\mathrm{col}}^2$. A catastrophic planetesimal collision requires $Q\gtrsim Q_D^{\ast }$, soThe numerical coefficient is only illustrative because the collision speed was specified only to order of magnitude. If the disruption threshold is instead defined using the centre of mass kinetic energy per combined mass, equal masses give $Q_R=v_{\mathrm{col}}^2/8$ and the coefficient becomes $8{\pi }^2/25$ for a threshold expressed in that convention. The robust result is a lower bound of order $Q_D^{\ast }{a}_{\mathrm{pl}}/(G{M}_{\star }{e}_{\mathrm{pl}}^2)$.

Within the inner-disk approximation, catastrophic collisions require sufficiently large $a_1/a_{\mathrm{pl}}$, not arbitrarily small radii. Although the Keplerian speed rises inward, the secularly induced orbital eccentricity falls faster: $v_{\mathrm{col}}^2\propto \alpha$. The lower bound must be much smaller than one to leave a domain compatible with $\alpha \ll 1$. A small planet orbital eccentricity, a large disruption threshold, or a distant planet can eliminate that domain. The planet mass controls the time to crossing but cancels from the eventual forced eccentricity amplitude and this collision-energy criterion. The threshold alone does not ensure that crossing has occurred within the disk's available lifetime.

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.

Put $B_i={\sum }_j{A}_j{e}_{ji}$. For $A\ne g_i$, the solution of the linear secular forcing of a test particle equation isThe initial condition gives $e_p=|z(0)-z_f(0)|$ and ${\beta }_p=\arg [z(0)-z_f(0)]$. The proper eccentricity $e_p$ is the constant amplitude of the homogeneous response; its proper longitude of periapsis is $At+{\beta }_p$. The forced eccentricity is the driven vector $z_f(t)$, not necessarily a constant magnitude.

On an Argand diagram, $z-z_f$ describes a circle of radius $e_p$. Its center $z_f$ can itself move under the planetary modes, so $z$ need not trace one fixed circle in the inertial complex plane. At a secular resonance $A=g_i$, that mode instead produces $i{B}_{i}t{e}^{i(At+{\beta }_i)}$, and the undamped linear response grows until the approximation fails.

For one isolated perturbing planet, its Kepler orbit has fixed longitude of periapsis, so $z_1$ is constant and its secular forcing frequency is zero. A constant forced eccentricity satisfies $0=iA{z}_f+i{A}_1{z}_1$. Since $A=-A_{1}f({\alpha }_1)$,This assumes $A\ne 0$. If an external process made the planet precess at frequency $g$, the particular solution would instead be $z_f=A_1{z}_1/(g-A)$; the stated formula relies on the isolated single-planet case.

If the proper eccentricity is negligible and the semi-major axis ranges over $a\pm \Delta a$, particles lie on nearly aligned forced Kepler orbits. For a locally constant forced vector, these are geometrically similar ellipses, represented to first order by circles of radius $a^{\prime }$ centered at $-a^{\prime }{\mathbf{e}}_f$.

The boundaries of the aligned eccentric ring are thereforeIt is narrower at forced periapsis and wider at forced apoapsis, unlike the constant-width proper-eccentricity annulus. Its centers shift slightly between the two edges, as shown in the right panel of the figure. The condition $e_p/e_f\ll \Delta a/a$ makes the proper radial excursion negligible compared with the variation in forced center location.

The forced vector need not actually be constant across the interval. More generallyFor aligned apsides this becomes $2\Delta a[1-(e_f+a{e}_f^{\prime })\cos \theta ]$, provided the orbits remain nested. The radial gradient of forced eccentricity matters for the density comparison even in a narrow ring.

Insert the planets' two secular eigenmodes and define their forcing strengthsSolving the first-order linear ordinary differential equation givesThe first term is the particle's freely precessing eccentricity. The remaining terms are its forced eccentricity, phase-locked to the planets' modes. In the complex plane, their vector sum makes the eccentricity and longitude of periapsis oscillate. A denominator becomes small at a secular resonance $A=g_k$; the nonresonant formula then ceases to be uniform.

For a common semi-major axis $a$ and proper eccentricity $e_p$, randomly distributed proper longitudes of periapsis give an annulus of radii $a(1-e_p)$ and $a(1+e_p)$ about the point displaced from the star by minus $a$ times the forced eccentricity vector, to first order in orbital eccentricity.

For a test particle interior to a distant eccentric planet, write $\alpha =a/a_p\ll 1$. The leading Laplace coefficients are $b_{3/2}^1=3\alpha$ and $b_{3/2}^2=15{\alpha }^2/4$. Thus its forced eccentricity is $Z_f=5\alpha e_p/4$ and its secular frequency is $A=(3/4)n_p(m_p/M){\alpha }^{3/2}$. More distant inner particles have larger forced orbital eccentricities and faster secular precession, producing a growing gradient of complex eccentricity.