= Solution
A <small-angle gravitational encounter> between a test <star> and a field <star> of mass $m$ gives a transverse kick $\Delta v_\perp\simeq2Gm/(bu)$, where $b$ is the impact parameter and $u$ the relative speed. For number density $n$, encounters in $b,b+db$ occur at rate $2\pi b\,db\,nu$. Independent kicks add in mean square, giving the <random-walk derivation of stellar relaxation>
$$
\frac{d\langle\Delta v^2\rangle}{dt}\simeq\frac{8\pi G^2m^2n}{u}\int_{b_{\min}}^{b_{\max}}\frac{db}{b}=\frac{8\pi G^2m^2n}{u}\ln\Lambda.
$$
The <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 $u^2$:
$$
t_{\mathrm{rel}}\simeq\frac{u^3}{8\pi G^2m^2n\ln\Lambda}.
$$
Using $n\sim3N/(4\pi r_s^3)$ gives $t_{\mathrm{rel}}\sim[N/(6\ln N)]r_s/u$. With a diameter-crossing convention $T_{\mathrm{cross}}\sim2r_s/u$, this is $[N/(12\ln N)]T_{\mathrm{cross}}$, conventionally rounded to
$$
\boxed{T_{\mathrm{relax}}\simeq0.1\frac{N}{\ln N}T_{\mathrm{cross}}.}
$$
The 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 $N\sim10^{10}$ and $T_{\mathrm{cross}}\sim10^8$ years, the estimate is about $4\times10^{15}$ 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 $dM=F(\mathbf x,\mathbf v,t)d^3x\,d^3v$, so $\rho=\int F\,d^3v$. 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
$$
\partial_tF+\mathbf v\cdot\nabla_{\mathbf x}F+\nabla\psi\cdot\nabla_{\mathbf v}F=0.
$$
For a steadily rotating <galactic bar>, write the inertial polar angle as $\phi_{\mathrm I}=\phi+\Omega_b t$. The inertial radial and angular equations are $\ddot R-R\dot\phi_{\mathrm I}^2=\psi_R$ and $R\ddot\phi_{\mathrm I}+2\dot R\dot\phi_{\mathrm I}=\psi_\phi/R$. Substitution gives the <equations of motion in a rotating frame>:
$$
\boxed{\ddot R-R\dot\phi^2=\psi_R+2R\dot\phi\Omega_b+\Omega_b^2R,\qquad R\ddot\phi+2\dot R\dot\phi=\frac{\psi_\phi}{R}-2\dot R\Omega_b.}
$$
Here the <velocities> are measured in the bar frame. Set $v_R=\dot R$, $v_\phi=R\dot\phi$ and define the positive-force <rotating-frame relative effective potential>
$$
\boxed{\psi_{\mathrm{eff}}=\psi+\frac12\Omega_b^2R^2.}
$$
The centrifugal sign is positive in this relative-potential convention. The characteristics are
$$
\dot R=v_R,\quad\dot\phi=v_\phi/R,\quad\dot v_R=2\Omega_bv_\phi+v_\phi^2/R+\psi_{\mathrm{eff},R},\quad\dot v_\phi=-2\Omega_bv_R-v_Rv_\phi/R+\psi_{\mathrm{eff},\phi}/R.
$$
Consequently the <rotating-frame collisionless Boltzmann equation> is
$$
\boxed{\begin{aligned}
0={}&F_t+v_RF_R+\frac{v_\phi}{R}F_\phi+\left(2\Omega_bv_\phi+\frac{v_\phi^2}{R}+\psi_{\mathrm{eff},R}\right)F_{v_R}\\
&+\left(-2\Omega_bv_R-\frac{v_Rv_\phi}{R}+\frac{\psi_{\mathrm{eff},\phi}}R\right)F_{v_\phi}.
\end{aligned}}
$$
For these planar equations, integrate over $dv_R\,dv_\phi$ and interpret $\rho$ as the planar or vertically integrated <mass density>. Define $\rho\langle A\rangle=\int A F\,dv_R\,dv_\phi$. Boundary terms in <velocity> vanish for a sufficiently decaying distribution.
Let the <velocity> accelerations above be $A_R,A_\phi$. Their <velocity> divergence is $\partial_{v_R}A_R+\partial_{v_\phi}A_\phi=-v_R/R$. Thus <integration by parts> in the zeroth moment contributes $\rho\langle v_R\rangle/R$, producing the <cylindrical continuity equation for a rotating stellar system>
$$
\boxed{\partial_t\rho+\frac1R\partial_R[R\rho\langle v_R\rangle]+\frac1R\partial_\phi[\rho\langle v_\phi\rangle]=0.}
$$
This 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 $R\,dR\,d\phi\,dv_R\,dv_\phi$.
For the radial first moment, multiply the <rotating-frame collisionless Boltzmann equation> by $v_R$. Its <velocity> integrations are $\int v_RA_RF_{v_R}\,d^2v=-\rho\langle A_R\rangle$ and $\int v_RA_\phi F_{v_\phi}\,d^2v=\rho\langle v_R^2\rangle/R$. Hence the radial <cylindrical Jeans equations in a rotating frame> are
$$
\boxed{\begin{aligned}
\partial_t(\rho\langle v_R\rangle)&+\partial_R(\rho\langle v_R^2\rangle)+\frac1R\partial_\phi(\rho\langle v_Rv_\phi\rangle)\\
&-2\Omega_b\rho\langle v_\phi\rangle+\frac\rho R(\langle v_R^2\rangle-\langle v_\phi^2\rangle)=\rho\psi_{\mathrm{eff},R}.
\end{aligned}}
$$
For the azimuthal first moment, the $A_R$ <velocity> integral vanishes, while $\partial_{v_\phi}(v_\phi A_\phi)=\psi_{\mathrm{eff},\phi}/R-2\Omega_bv_R-2v_Rv_\phi/R$. This gives
$$
\boxed{\begin{aligned}
\partial_t(\rho\langle v_\phi\rangle)&+\partial_R(\rho\langle v_Rv_\phi\rangle)+\frac1R\partial_\phi(\rho\langle v_\phi^2\rangle)\\
&+2\Omega_b\rho\langle v_R\rangle+\frac{2\rho\langle v_Rv_\phi\rangle}{R}=\frac\rho R\psi_{\mathrm{eff},\phi}.
\end{aligned}}
$$
The <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.
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