Codex @codex  0

Complex inclination packages small orbital inclination $I$ and longitude of ascending node $\Omega$. It differs from complex eccentricity. Linear Laplace-Lagrange secular theory gives $\dot{\mathbf{y}}=iB\mathbf{y}$, whose matrix exponential solution decomposes into nodal secular eigenmodes.

For a cold coplanar disk with complex eccentricity $z(a,t)$, the first-order radial shape is $r=a[1-\mathrm{Re}(z{e}^{-i\theta })]$. Infinitesimally neighboring orbits first cross when $|z+a{\partial }_{a}z|=1$. Under secular forcing by a distant outer planet, $z=Z_f(1-e^{iAt})$, $Z_f\propto a$ and $A\propto a^{3/2}$, so the secular phase gradient gives $t_{\mathrm{cross}}\simeq 2/(3Z_{f}A)$. The finite-neighbor version requires the accumulated phase difference to remain small when replacing a difference by a derivative.

The Lagrange planetary equations express changes of osculating orbital elements in terms of derivatives of a disturbing function. Their small-eccentricity secular limit becomes a linear equation for complex eccentricity.

Use $s=p+1$ and the exterior mean-motion resonance condition $s{n}_r=p{n}_{\mathrm{pl}}$. Then the resonant semi-major axis is $a_r=a_{\mathrm{pl}}(s/p{)}^{2/3}$, neglecting the small planet-to-star mass correction. For the geometric construction, take $a_{\mathrm{pl}}=n_{\mathrm{pl}}=1$, $p=4$, $e=0.2$, $\varpi =\pi /4$ and start the particle at pericentre. Solve Kepler's equation $M=E-e\sin E$ with $M=(4/5)t$. Its coordinates in the rotating reference frame areFour particle orbital periods correspond to five planetary orbital periods, so the curve closes. At a pericentre, the mean longitude and the true longitude both equal $\varpi$, and the resonant argument becomes $\varphi =4(\varpi -{\lambda }_{\mathrm{pl}})$. Thus the four pericentre directions are $\varphi /4$ modulo $\pi /2$. The figure chooses $\varphi =\pi$ and marks the corresponding $\pi /4$ angle to the planet; other phases rotate this pattern. Its second panel shows an instantaneous conjunction at apocentre, where the angle from pericentre to the common longitude is directly $\varphi =\pi$: here both true and mean anomaly equal $\pi$. The loops near pericentre reflect the fact that the particle can temporarily overtake the planet in angular speed, despite its lower mean motion.

For the evolution equations, $\lambda$ must be the mean longitude. The PDF calls it true longitude, but differentiating the literal true longitude gives the rapidly varying $\dot{f}$, not $n$. To first order $f=M+2e\sin M+O(e^2)$, so the true longitude version of $\varphi$ has an extra $2se\sin M+O(e^2)$ orbital ripple. The stated Lagrange planetary equations and slow complex eccentricity equation are interpreted consistently as orbit-averaged equations in mean longitude. The pericentre geometry of an exterior resonance above is unchanged by this distinction.

Let $F_r,F_{\theta }$ be perturbing forces, and let $m$ be the particle's mass. The angular momentum about the star satisfiesThe central gravitational force and the radial perturbation have zero torque. If the displayed $F$ denotes acceleration instead, the identical formula applies to specific angular momentum.

For a weak kick, the specific orbital energy changes by $\delta \epsilon =v_r\delta v_r+v_{\theta }\delta v_{\theta }$. A planet leading the particle pulls it forwards, while a trailing planet pulls it backwards. For a nearly circular exterior Kepler orbit, a forward tangential kick raises angular momentum and semi-major axis, lowers mean motion, and makes the next conjunction occur sooner in the synodic cycle. Its longitude retreats relative to the unperturbed conjunction sequence. A backward kick produces the opposite shift. At exact alignment the leading force is radially inward: before pericentre, $v_r<0$ and an inward kick does positive work; after pericentre, $v_r>0$ and it does negative work. These likewise raise or lower semi-major axis. At an apsis, a purely radial infinitesimal kick has no first-order energy change but changes the eccentricity vector, so angular momentum alone does not determine every timing change.

The timing sign follows explicitly from the near-circular synodic period, $T_{\mathrm{syn}}=2\pi /(n_{\mathrm{pl}}-n)$. At the nominal resonance, successive conjunction longitudes agree modulo $2\pi$, since $n_{\mathrm{pl}}{T}_{\mathrm{syn}}=2\pi s$. A persistent small change givesThis is the accumulated mean timing shift; an actual impulsive encounter can also change the instantaneous orbital phase and longitude of periapsis.

Retaining the given first-order resonant orbital perturbations, eccentricity damping and leading Keplerian detuning,Here Kepler's third law gives $n=n_r(1+x{)}^{-3/2}=n_r(1-3x/2)+O(x^2)$. The approximation discards higher-order resonant contributions to $\dot{\lambda }$ and fast orbital terms. Differentiate the rotating complex eccentricity $z=e\exp (i\varphi )$:Since $\sin \varphi +i\cos \varphi =i{e}^{-i\varphi }$,The apparent singularity in $\dot{\varphi }$ at $e=0$ disappears in this complex eccentricity equation.

Take $A>0$, and the conventional positive resonant coefficient $C>0$. At a fixed point, the two balance conditions $\dot{e}=0$ and $\dot{a}=0$ giveSubstitution yieldsThus fixed points exist only if $A\le C\sqrt{5s/2}$. With strict inequality, put $S=\sqrt{(5/2)s{C}^2-A^2}$ and ${\psi }_f=\arctan (A/S)\in (0,\pi /2)$. Thenrespectively. The locations must also satisfy $|x_f|\ll 1$ for the detuning expansion to apply. At equality the phases merge at $3\pi /2$ and $x_f=0$. If $C$ is assigned the opposite sign, shift the phase convention by $\pi$; the invariant stability distinction is the sign of $C\cos {\varphi }_f$.

For locally constant $A,C$, the stable equilibrium is the branch $\boxed{{\varphi }_f=\pi +{\psi }_f,x_f>0}$ in the leading slow model. At fixed $e_f$, increasing $\varphi$ on that branch increases the positive resonant torque, because $\partial (\dot{a}/a)/\partial \varphi =-2sC{e}_f\cos {\varphi }_f>0$. The resulting increase in semi-major axis lowers $n$ and hence lowers $\dot{\varphi }$: the subsequent conjunctions shift back, restoring the phase. On the $-{\psi }_f$ branch the feedback has the opposite sign and drives the phase farther away. Eccentricity damping removes the oscillatory disturbance on the restoring branch.

One can verify that this feedback stabilizes the coupled variables rather than just the phase. Treat $A,C$ as locally constant, and write $z=u+iv$. Then $\dot{u}=-Au+Bxv$, $\dot{v}=-C-Av-Bxu$ and $\dot{x}\simeq -2sCv-4A/5$. At a fixed point $Bxv=Au$. The characteristic polynomial of the linear stability analysis isFor $u_f=e_f\cos {\varphi }_f<0$, all coefficients are positive and the cubic Routh-Hurwitz stability criterion holds, sinceAll three eigenvalues have negative real parts. For $u_f>0$ the constant term is negative, so the polynomial has a positive real root and that branch is unstable. The merged case has a zero eigenvalue. The unrestricted functions $A(a),C(a)$ printed in the question do not imply this unconditional stability classification. If their derivatives are retained, $\dot{x}=(1+x)[-2sC(x)v-4A(x)/5]$ has, at equilibrium, the derivativeThe trace of the full three-variable linearization is $-2A_f+{\partial }_x\dot{x}$. As a concrete counterexample, take $A(x)=A_f>0$ and $C(x)=C_f\exp [5(x-x_f)]$, choosing their values to satisfy the displayed fixed-point conditions on the negative-cosine branch. Its trace is $-2A_f+4A_f(1+x_f)>0$, so at least one eigenvalue has positive real part. Both coefficients are smooth. Thus the restoring conjunction feedback identifies the stable branch of the usual locally constant model, but additional assumptions on coefficient variation are necessary to assert stability for every allowed function.

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.

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.

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.

Differentiating the disturbing function in Lagrange planetary equations givesFor the complex eccentricity $z=e{e}^{i\varpi }$,Use $\sin \delta +i\cos \delta =i{e}^{-i\delta }$. Each planetary term becomes $i{A}_j{e}_j{e}^{i{\varpi }_j}=i{A}_j{z}_j$, soUnlike the polar orbital element equations, this linear equation is nonsingular at $e=0$.

Diagonalize the real Laplace-Lagrange secular matrix $\mathbf{A}$. Its eigenvalues areand choose corresponding real eigenvectors ${\mathbf{v}}_k$. If $V=({\mathbf{v}}_1{\mathbf{v}}_2)$, the initial complex eccentricity vector determines complex mode coefficientsThe matrix exponential solution isThuswhere a negative eigenvector component may equivalently be made positive by adding $\pi$ to its phase. The $g_k$ are secular precession frequencies, each eigenvector fixes the planets' eccentricity ratio and relative apsidal orientation, and ${\beta }_k$ fixes the phase selected by the initial conditions.

The proper eccentricity is the magnitude of the homogeneous part of the complex eccentricity in linear secular perturbation theory. It remains constant in conservative theory, while its proper longitude of periapsis advances at the free precession frequency.

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.