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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.

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.

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.