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A collisionless stellar system is one for which discrete gravitational encounters change its distribution negligibly over the time under study, usually $T\ll t_{\mathrm{rel}}$. Stars can orbit and respond collectively to a varying smooth gravitational potential even in this regime; collisionless does not mean force-free or stable against collective gravitational instabilities. Its smooth distribution follows the Collisionless Boltzmann equation.

Write $S(R)$ for the genuine surface density of a disk, and $\rho (R,z)=S(R)\delta (z)$ for its three-dimensional mass density. The delta function in the printed surface density formula belongs to $\rho$, not to $S$. Introduce a reference length $R_{\ast }$ to make logarithms dimensionless. The intended model is the infinite, self-gravitating, scale-free Mestel disk, with no extra source or imposed gravitational field.

The circular speed satisfies $v_c^2=R{\partial }_R\Phi (R,0)$. Thus the flat galaxy rotation curve givesA reflection-symmetric harmonic function with this midplane boundary value isVerify this continuation using the Poisson equation for Newtonian gravity. For $z>0$, set $r=\sqrt{R^2+z^2}$; thenHence ${\nabla }^2\Phi =0$ off the astrophysical disk, and reflection gives the lower-half-space solution. Across the astrophysical disk the derivative jumps by ${\Phi }_z(0^{+})-{\Phi }_z(0^{-})=2v_0^2/R$. The distributional Poisson equation for Newtonian gravity therefore givesAt the origin the enclosed disk mass tends to zero linearly with radius, so there is no additional central point mass. This verifies the Mestel disk potential-density pair. The astrophysical disk has infinite total mass and a logarithmic Newtonian gravitational potential, so it is not an isolated finite-mass model with Newtonian gravitational potential zero at infinity. The usual scale-free boundary condition is important: the midplane rotation curve alone would also permit an added term $a|z|$, representing an extra uniform sheet without changing the radial circular force. That contribution is excluded in the intended Mestel disk model.

Use a mass-weighted planar galactic distribution function, so $F(\mathbf{x},\mathbf{v})d^{2}x{d}^{2}v$ is the stellar mass in a small planar phase space element. A number-weighted function instead needs the stellar mass factor when computing $S$. Assume a steady collisionless stellar system and isotropy in the two in-plane velocity components. Put $u=|\mathbf{v}{|}^2/2$ and write $F=f(\mathbf{x},u)$. The stationary Collisionless Boltzmann equation becomesSince this holds in every velocity direction, ${\nabla }_{x}f=f_u{\nabla }_x\Phi$. In coordinates $(\mathbf{x},E)$ with $E=u+\Phi$, this is exactly ${\nabla }_{x}f{|}_E=0$. Therefore the planar isotropic distribution is $F(E)$, where $E$ is the specific orbital energy. Stationarity is essential; instantaneous isotropy alone would not imply this result.

Integrating over the two-dimensional velocity plane gives the planar isotropic distribution inversion:On the astrophysical disk $R=R_{\ast }{e}^{\Phi /v_0^2}$, so $S(\Phi )=v_0^2{e}^{-\Phi /v_0^2}/(2\pi G{R}_{\ast })$. Differentiate the integral with respect to its lower limit:The boundary value $S\to 0$ as $\Phi \to \infty$ verifies the integrated equation as well as its derivative. Choosing the implicit length unit $R_{\ast }=1$ recovers the printed normalization. Changing the additive energy zero changes this prefactor accordingly.

At a fixed radius, the normalized velocity density isIt is a product of centered Gaussian distributions. Differentiating the supplied Gaussian integral with respect to its coefficient gives the second moments, and odd moments vanish. Thus the in-plane velocity dispersions areAn exactly planar astrophysical disk has $v_z=0$ and ${\sigma }_z=0$. The nonzero circular speed is a property of the force field, not a statement that this hot stellar distribution has net rotation.

Reversing every retrograde star folds the azimuthal Gaussian to a half-normal distribution. Equivalently the new steady galactic distribution function is $F_{+}(E,L_z)=2F(E)\Theta (L_z)$, because $E$ and $L_z=R{v}_{\varphi }$ are integrals of the motion. Its density and even velocity moments are unchanged. Its streaming velocity isThis maximally prograde stellar distribution still has radial motion and a spread of azimuthal speeds; it does not place every star on a circular orbit. In particular $⟨v_{\varphi }^2{⟩}_{+}=v_0^2$ while ${\sigma }_{\varphi ,+}^2=v_0^2(1-2/\pi )$. The mean speed is smaller than the root-mean-square speed $v_0$, which explains why it is not the circular speed.

A small-angle gravitational encounter between a test star and a field star of mass $m$ gives a transverse kick $\Delta v_{\perp }\simeq 2Gm/(bu)$, where $b$ is the impact parameter and $u$ the relative speed. For number density $n$, encounters in $b,b+db$ occur at rate $2\pi bdbnu$. Independent kicks add in mean square, giving the random-walk derivation of stellar relaxationThe Coulomb logarithm in stellar dynamics has $b_{\max }$ of order system size $r_s$, and $b_{\min }$ of order the strong-deflection scale $Gm/u^2$. For a virialized $N$-star system, $u^2\sim GNm/r_s$, so $\Lambda$ is of order $N$. The stellar relaxation time is the time for accumulated velocity variance to become comparable with $u^2$:Using $n\sim 3N/(4\pi r_s^3)$ gives $t_{\mathrm{rel}}\sim [N/(6\ln N)]r_s/u$. With a diameter-crossing convention $T_{\mathrm{cross}}\sim 2r_s/u$, this is $[N/(12\ln N)]T_{\mathrm{cross}}$, conventionally rounded toThe coefficient is approximate: spatial profile, velocity averages and crossing-time convention change it by factors of order unity. The robust result is $N/\ln N$ crossing times.

For $N\sim 10^{10}$ and $T_{\mathrm{cross}}\sim 10^8$ years, the estimate is about $4\times 10^{15}$ years, enormously longer than a Hubble time. Most galaxies therefore behave as collisionless stellar systems. Dense nuclei and star clusters can relax more rapidly. Negligible stellar encounters do not mean negligible collective gravitational instabilities.

Define the mass-weighted galactic distribution function by $dM=F(\mathbf{x},\mathbf{v},t)d^{3}x{d}^{3}v$, so $\rho =\int F{d}^{3}v$. Hamiltonian gravitational motion preserves phase space volume by the Liouville theorem in Hamiltonian mechanics. In the collisionless regime no encounter term redistributes stars between neighbouring phase space trajectories, so $dF/dt=0$. With acceleration $\nabla \psi$, the Collisionless Boltzmann equation is

For a steadily rotating galactic bar, write the inertial polar angle as ${\varphi }_{\mathrm{I}}=\varphi +{\Omega }_{b}t$. The inertial radial and angular equations are $\ddot{R}-R{\dot{\varphi }}_{\mathrm{I}}^2={\psi }_R$ and $R{\ddot{\varphi }}_{\mathrm{I}}+2\dot{R}{\dot{\varphi }}_{\mathrm{I}}={\psi }_{\varphi }/R$. Substitution gives the equations of motion in a rotating frame:Here the velocities are measured in the bar frame. Set $v_R=\dot{R}$, $v_{\varphi }=R\dot{\varphi }$ and define the positive-force rotating-frame relative effective potentialThe centrifugal sign is positive in this relative-potential convention. The characteristics areConsequently the rotating-frame collisionless Boltzmann equation isFor these planar equations, integrate over $d{v}_{R}d{v}_{\varphi }$ and interpret $\rho$ as the planar or vertically integrated mass density. Define $\rho ⟨A⟩=\int AFd{v}_{R}d{v}_{\varphi }$. Boundary terms in velocity vanish for a sufficiently decaying distribution.

Let the velocity accelerations above be $A_R,A_{\varphi }$. Their velocity divergence is ${\partial }_{v_R}{A}_R+{\partial }_{v_{\varphi }}{A}_{\varphi }=-v_R/R$. Thus integration by parts in the zeroth moment contributes $\rho ⟨v_R⟩/R$, producing the cylindrical continuity equation for a rotating stellar systemThis is conservation of mass: change of density balances flux through the radial and azimuthal sides of a small cylindrical element. The factor $R$ is geometric; the physical phase space measure is $RdRd\varphi d{v}_{R}d{v}_{\varphi }$.

For the radial first moment, multiply the rotating-frame collisionless Boltzmann equation by $v_R$. Its velocity integrations are $\int v_R{A}_R{F}_{v_R}{d}^{2}v=-\rho ⟨A_R⟩$ and $\int v_R{A}_{\varphi }{F}_{v_{\varphi }}{d}^{2}v=\rho ⟨v_R^2⟩/R$. Hence the radial cylindrical Jeans equations in a rotating frame areFor the azimuthal first moment, the $A_R$ velocity integral vanishes, while ${\partial }_{v_{\varphi }}(v_{\varphi }{A}_{\varphi })={\psi }_{\mathrm{eff},\varphi }/R-2{\Omega }_b{v}_R-2v_R{v}_{\varphi }/R$. This givesThe Jeans equations are local momentum-balance equations. The second moments include streaming momentum flux and random-velocity stress, while gravity, centrifugal acceleration and Coriolis acceleration supply the frame-dependent forces. They do not by themselves close the full distribution: a stress prescription or a galactic distribution function is also needed.

The stellar relaxation time is the time on which accumulated discrete gravitational encounters change a typical star's velocity by an amount comparable to its original velocity. It concerns two-body relaxation, rather than the much faster orbital evolution in a smooth gravitational potential. A collisionless stellar system requires this time to greatly exceed the duration being studied.

In the straight-line impulse approximation, an encounter at relative speed $u$ has transverse acceleration $Gmb/(b^2+u^2{t}^2{)}^{3/2}$. ThusA small-angle gravitational encounter needs $\Delta v_{\perp }\ll u$. The transition to order-one deflections is therefore $b\sim Gm/u^2$, ignoring factors such as the equal-mass relative deflection. For the spherical estimate $u\sim v$,This is the lower cutoff of the weak-scattering estimate; closer encounters actually occur and require strong-scattering treatment.

The number of encounters in time $T$ with impact parameters in $[b,b+db]$ is $2\pi bdbnuT$. Independent random transverse directions make mean kicks cancel while their variances add. ConsequentlyEach logarithmic interval contributes equally, giving the Coulomb logarithm in stellar dynamics. Defining a deflection time by variance $v^2$ and setting $u=v$ gives $v^3/(8\pi G^2{m}^{2}n\log \Lambda )$. The paper instead uses an order-one normalization three quarters of this estimate:Both have the same physical scaling. The specified characteristic speed, the word “comparable”, the velocity-distribution average and strong-encounter cutoff do not fix that numerical coefficient uniquely. The crude equal-speed impulse calculation must not be claimed to determine $3/(32\pi )$ exactly.

For a self-gravitating, approximately virialized system, the virial theorem gives $v^2\sim GNm/R$. With $n=3N/(4\pi R^3)$, $\Lambda \sim N$ and the stellar crossing time $t_{\mathrm{dyn}}=R/v$, substitution into the paper's convention givesAn externally dominated gravitational potential or a different structural constant changes this substitution. For an N-body simulation lasting $T=K{t}_{\mathrm{dyn}}$, a tolerable fractional velocity-squared diffusion $\epsilon \ll 1$ requires $T/t_{\mathrm{rel}}\lesssim \epsilon$, henceThere is no unique smallest $N$ without $K$, the error tolerance and the force prescription. For scale, equality at $t_{\mathrm{rel}}/t_{\mathrm{dyn}}=100$ occurs near $N=7.1\times 10^3$; $N=10^5$ gives about $1.1\times 10^3$ crossing times. A calculation lasting 100 crossing times therefore needs substantially more than the first threshold to have negligible relaxation. Gravitational softening can raise the effective cutoff and reduce artificial scattering, but it also sets the spatial force resolution.

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.

For a stationary collisionless stellar system confined to a plane, velocity isotropy makes the galactic distribution function depend on speed through $u=v^2/2$. The Collisionless Boltzmann equation implies ${\nabla }_{x}f=f_u{\nabla }_x\Phi$, hence $f=F(E)$ with specific orbital energy $E=u+\Phi$. This conclusion requires stationarity as well as isotropy.