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

For finite-thickness disk relaxation, assume a well-separated range $b_{\min }\ll z_0\ll R$ and approximately virialized self-gravity. The three factors have distinct origins. Replacing the characteristic encounter speed by $u=\lambda v$ shortens the diffusion time by ${\lambda }^3$: the variance-production rate has one inverse power of speed, and the target random velocity squared has two powers. Concentrating the same number of stars into volume of order $R^2{z}_0$, rather than $R^3$, raises the number volume density by $R/z_0$ and shortens the time by $z_0/R$. Finally, only separations below the thickness sample a three-dimensional encounter geometry. The logarithmic upper cutoff becomes $z_0$, and therefore the inverse-logarithm time acquires $\log (R/b_{\min })/\log (z_0/b_{\min })$. More distant encounters have the planar geometry and provide a convergent, non-logarithmic correction. A smaller Coulomb logarithm in stellar dynamics actually partly offsets the first two shortenings.

Thus, using the same cutoff for this comparison,For actual point particles the disk cutoff is $Gm/(\lambda v{)}^2$, whereas the spherical one is $Gm/v^2$; one must use the appropriate cutoff in each logarithm rather than silently identifying them. For a softened calculation it is instead of order the larger of the weak-deflection scale and the softening length. The formula is not valid when $z_0$ approaches that cutoff.

Write $h=z_0/R$ and $q=K/\epsilon$, with simulation duration $K{t}_{\mathrm{dyn}}$ and allowed fractional diffusion $\epsilon$. Combining the paper's spherical normalization and its comparison formula gives the leading criterionThe spherical comparison logarithm cancels when its numerical convention and a common comparison cutoff are used; order-one shape factors remain approximate. In an unsoftened disk with $v^2\sim GNm/R$, $z_0/b_{\min }\sim h{\lambda }^{2}N$. Hence the implicit requirement is $N/\log (h{\lambda }^{2}N)\gtrsim 8q/({\lambda }^{3}h)$, in the regime $h{\lambda }^{2}N\gg 1$. For $\lambda =h=0.1$, the stellar relaxation time is of order $10^{-4}$ of the spherical value apart from logarithms; $N\sim 10^8$ gives a time of order 100 crossing times in this estimate. Negligible relaxation over that duration requires an additional margin. A disk generally needs far more particles than a round galaxy for the same relaxation tolerance; no universal minimum follows without duration, thickness and softening. Collective spiral or bending responses are a separate source of evolution even in a collisionless stellar system.