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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.

Take the sky frame and disk frame to be right-handed orthonormal bases, with positive angular displacement measured from $\hat{X}$ toward $\hat{Y}$. In particular, positive $Z$ is toward the observer, so material crossing the sky plane with positive $\dot{Z}$ is at the ascending node. Write $c_d=\cos {\Omega }_d$, $s_d=\sin {\Omega }_d$, $C_d=\cos I_d$, and $S_d=\sin I_d$. The disk frame's first unit vector lies along its ascending node, and its second has positive $Z$ component. The orbital-frame rotation from inclination and node therefore givesEach column is a disk-frame unit vector in sky coordinates, so $\mathbf{X}=T\mathbf{x}$ and the inverse is $\mathbf{x}=T^{\mathsf{T}}\mathbf{X}$. At the disk's ascending node, the $Z$ component of its tangential velocity is positive, confirming the sign of the tilt.

Let $T_p=R_Z({\Omega }_p)R_X(I_p)$ define a provisional planet frame whose first axis is the planet's sky-plane ascending node. Its normal is ${\hat{z}}_p=(\sin {\Omega }_p\sin I_p,-\cos {\Omega }_p\sin I_p,\cos I_p)$. Expressing that normal in the disk orthonormal basis, with $\Delta \Omega ={\Omega }_p-{\Omega }_d$, givesThe last component is the dot product of the two orbital normals, hence the cosine of the mutual inclination. The mutual ascending node points along ${\hat{z}}_d\times {\hat{z}}_p$: motion there has positive disk-normal component. SetThen $\hat{x}\prime =(A\hat{x}+B\hat{y})/\sqrt{A^2+B^2}$, and the quadrant-correct longitude of ascending node is ${\Omega }_m=atan2(B,A)$. ThusThe tangent alone cannot fix the quadrant, and ${\Omega }_m$ is undefined for exactly coincident orbital planes.

For the two rotation routes, define $\psi$ as the oriented angle, within the planet's orbital plane, from its sky-plane ascending node to its disk-plane ascending node. If ${\hat{x}}_p,{\hat{y}}_p$ are the first two columns of $T_p$, take $\psi =atan2(\hat{x}\prime \cdot {\hat{y}}_p,\hat{x}\prime \cdot {\hat{x}}_p)$. Both routes describe the same planet orthonormal basis:The left route goes through the disk frame; the right goes through the sky-node planet frame. Comparing their third columns gives the two boxed relations, while comparison of first columns fixes the extra in-plane angle.

Put $I_d=\pi /2+{\delta }_d$ and $I_p=\pi /2+{\delta }_p$. A second-order Taylor expansion gives $\cos I_m=1-[({\delta }_p-{\delta }_d{)}^2+\Delta {\Omega }^2]/2+O({\delta }^4)$, and therefore the mutual inclination of nearly edge-on orbits satisfiesAll expansion angles are in radians. For plotting in degrees this is $I_m\simeq \sqrt{(I_p-87^{\circ }{)}^2+(4^{\circ }{)}^2}$. The given planet has exact $I_m\simeq 4.119^{\circ }$ and ${\Omega }_m\simeq 76.044^{\circ }$. The exact minimum occurs at $I_p=atan2(\sin I_d\cos \Delta \Omega ,\cos I_d)\simeq 86.993^{\circ }$, with $I_{m,\min }\simeq 3.995^{\circ }$. At $I_p=85^{\circ }$ and $95^{\circ }$, the second-order values are ${\sqrt{20}}^{\circ }\simeq 4.472^{\circ }$ and ${\sqrt{80}}^{\circ }\simeq 8.944^{\circ }$.

For the disk evolution, neglect disk self-gravity, collisions that align the planes, and back-reaction on the planet. In the initial disk plane, use the complex inclination $w=I{e}^{i\Omega }$ for a ring and $w_p=I_m{e}^{i{\Omega }_m}$ for the planet. Linear Laplace-Lagrange secular theory has forced inclination $w_p$ and solution $w(a,t)=w_p[1-e^{-i\nu (a)t}]$ for initially flat rings. For an inner circular perturber, the Laplace coefficient formula is $\nu =n(M_p/M_{\star })\alpha b_{3/2}^{(1)}(\alpha )/4$, with $\alpha =a_p/a$. Its quadrupole approximation uses $b_{3/2}^{(1)}\simeq 3\alpha$, yieldingIn the $(I\cos \Omega ,I\sin \Omega )$ plot referenced to the initial disk, every ring runs clockwise on the same circle centered at $w_p$, starting at the origin. Inner rings advance more quickly. If the axes are instead referenced to the planet's plane, the relative tilt is $w-w_p=-w_p{e}^{-i\nu t}$: these circles are centered at the origin, and the mutual inclination is constant. Both conventions describe the same nodal precession; they should not be mixed.

At an intermediate epoch, rings with $\nu t\gtrsim 1$ have tilted substantially while outer rings with $\nu t\ll 1$ remain near the original disk plane. This radial variation is a planet-induced debris-disk warp. The characteristic affected radius grows like $t^{2/7}$. Conservative differential nodal precession preserves each ring's tilt relative to the planet, so it does not by itself align every orbit; an unresolved inner region can acquire a mean plane near the planetary plane through phase mixing. The secular interpretation is also described in Wyatt's planetary dynamics lectures.

For small $\alpha$, expand the defining integrand of the Laplace coefficient:HenceThe constant term integrates to zero and ${\int }_0^{2\pi }{\cos }^{2}xdx=\pi$, givingThere is no quadratic term: its factors are combinations of $\cos x$ and ${\cos }^{3}x$, which both integrate to zero over a period. The linear approximation is used for the distant planet, not for the closely spaced inner pair.

Keep the inner-pair Laplace coefficient unexpanded because $\alpha =a_1/a_2$ is close to one. Put $\ell =L_1/L_2$ andUsing the small-argument Laplace coefficient for the distant planet and $n_2/n_1={\alpha }^{3/2}$ givesThe reverse-coupling expressions follow from the angular momentum identity $L_j{B}_{jk}=L_k{B}_{kj}$. If desired, $L_1/L_2=(M_1/M_2)\sqrt{\alpha }$, $L_1/L_3=(M_1/M_3)\sqrt{\beta }$, and $L_2/L_3=(M_2/M_3)\sqrt{\beta /\alpha }$ express all ratios in the supplied quantities. The diagonal entries are $B_{11}=-(B_{12}+B_{13})$, $B_{22}=-(B_{21}+B_{23})$, and $B_{33}=-(B_{31}+B_{32})$.

For comparable masses and closely spaced inner orbits, $\alpha \simeq 1$, $\ell \sim 1$, and $L_1/L_3,L_2/L_3=O(\sqrt{\beta })$. ThusThe distant couplings $B_{13},B_{23}$ scale as ${\beta }^3$ relative to a fixed inner architecture, and the reverse couplings have an additional $\sqrt{\beta }$ suppression.

Expanding the disturbing function in orbital elements gives a Fourier series of the formwhereThe radial coefficients are combinations of Laplace coefficients. Rotational and reflection symmetry impose the D'Alembert characteristic: the integer coefficients sum to zero, nodal coefficients obey a parity rule, and each harmonic begins at the corresponding order in eccentricities and inclinations.

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.