= Solution
In the <proper-eccentricity annulus>, averaging over random proper phases cancels the proper vector in the first-order angular variation. The mean line density is proportional to $1-e_f\cos\theta$, while the full ring width is independent of angle. Thus at corresponding radial locations
$$
\boxed{\frac{\Sigma_{\rm apo}}{\Sigma_{\rm peri}}\simeq\frac{1+e_f}{1-e_f}=1+2e_f+O(e_f^2).}
$$
There is a forced-apocentre enhancement in number <surface density of a disk>, together with the proper-excursion edge profile.
For a narrow <aligned eccentric ring> with locally constant $e_f$, its width has the same first-order factor $1-e_f\cos\theta$ as its line density. Their ratio is constant: \b[the constant-eccentricity aligned ring has no first-order surface-density contrast]. It still contains more particles per unit longitude at <apoapsis>, spread across a wider area.
The more general <surface density contrast in an eccentric ring> depends on the eccentricity gradient. With a smooth number distribution per unit <semi-major axis> and aligned apsides,
$$
\Sigma_{\rm peri}\propto\frac{1-e_f}{1-e_f-ae_f'}\simeq1+ae_f',\qquad
\Sigma_{\rm apo}\propto\frac{1+e_f}{1+e_f+ae_f'}\simeq1-ae_f'.
$$
Hence decreasing $e_f(a)$ can also make the aligned ring apocentre-dense; for $e_f\propto a^{-1}$ it has the same first-order contrast as the proper-phase case. A positive gradient gives a pericentre enhancement. The problem leaves this profile unspecified, so there is no universal observational distinction based solely on the two apse densities.
Resolved ring width and a measured forced-eccentricity profile can distinguish the constant-width and similar-ellipse constructions. Interpreting observed brightness additionally requires the grain <temperature>, illumination, <emissivity>, and viewing geometry; brightness is not itself particle number density. Inferences from one contrast alone can therefore be degenerate.
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