Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 320 / 1 / b / Solution

Put $u=\lambda v$ and let $n$ now mean number per area. During time $T$, the two strips of impact parameters between $b$ and $b+db$, on opposite sides of the trajectory, have combined area $2uTdb$. ThusThe same straight-line impulse approximation gives a kick $2Gm/(bu)$. Summing its squared size over these encounters givesassuming $b_{\max }\gg b_{\min }$. The relevant change is now the small random speed $u$, not the full circular speed $v$. Setting $D_v{t}_{\mathrm{rel}}\sim u^2$ yieldsThe razor-thin disk relaxation integral is dominated by nearby encounters, rather than equally by logarithmic intervals. Because kicks cease to be weak near the cutoff, its numerical coefficient is an estimate.

Here the weak-deflection cutoff must use the relative speed: $b_{\min }\sim Gm/u^2=Gm/({\lambda }^2{v}^2)$, which is larger than the spherical cutoff. Thus $t_{\mathrm{rel}}\sim u/(Gmn)$. Using $n\sim N/R^2$ and $v^2\sim GNm/R$ from the virial theorem givesFor example $n=N/(\pi R^2)$ gives the order-one coefficient $\pi /8$ in this last substitution; the paper's approximate equality suppresses such structural factors. The cancellation of $N$ is the central result.

An unsoftened point-particle system confined to an infinitesimally thin plane is therefore not collisionless over orbital times for $\lambda \sim 0.1$ in this encounter model. At fixed total mass, increasing $N$ decreases $m$ and the cutoff together; the enhanced importance of close planar encounters cancels the usual benefit of increasing particle number. This is a singular zero-thickness limit. A fixed nonzero gravitational softening length or physical thickness restores a different $N$ dependence, so the conclusion is not a prohibition on collisionless numerical models of disks.