= 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 $R^2z_0$, 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,
$$
\boxed{\frac{t_{\rm disk}}{t_{\rm sphere}}\sim\lambda^3\frac{z_0}{R}\frac{\log(R/b_{\min})}{\log(z_0/b_{\min})}.}
$$
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 $q=K/\varepsilon$, with simulation duration $K t_{\rm dyn}$ and allowed fractional diffusion $\varepsilon$. Combining the paper's spherical normalization and its comparison formula gives the leading criterion
$$
\boxed{\frac{\lambda^3hN}{8\log(z_0/b_{\min})}\gtrsim q.}
$$
The 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 $v^2\sim GNm/R$, $z_0/b_{\min}\sim h\lambda^2N$. Hence the implicit requirement is $N/\log(h\lambda^2N)\gtrsim8q/(\lambda^3h)$, in the regime $h\lambda^2N\gg1$. For $\lambda=h=0.1$, the <stellar relaxation time> is of order $10^{-4}$ of the spherical value apart from logarithms; $N\sim10^8$ gives a <time> of order 100 crossing times in this estimate. Negligible relaxation over that duration requires an additional margin. \b[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>.
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