Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 44 / 3 / Solution

Integrate the reduced vertical phase-space distribution function over $v_z$. The Gaussian velocity integral cancels its normalization, givingHere the constant vertical velocity dispersion is ${\sigma }_z$, and $\Phi (0)=0$ fixes the potential reference. Neglecting the radial derivatives in the Poisson equation for Newtonian gravity leavesWith $\varphi =\Phi /{\sigma }_z^2$, $\zeta =z/z_0$ and $z_0^2={\sigma }_z^2/(8\pi G{\rho }_0)$, this becomesThus the stated dimensionless equation follows with its factor of two.

For a direct solution, multiply by $2{\varphi }^{\prime }$ and integrate from the midplane. The initial conditions giveFor $\zeta >0$ take the increasing branch and put $u=e^{\varphi /2}$. Then $4(u^{\prime }{)}^2=u^2-1$, whose smoothly increasing solution consistent with the second-order equation is $u=\cosh (\zeta /2)$. Reflection about the midplane supplies the negative-$\zeta$ branch. ThereforeSubstitution verifies $2{\varphi }^{\prime \prime }={\mathrm{sech}}^2(\zeta /2)=e^{-\varphi }$ and both boundary conditions. The smooth second-order equation has a unique solution for those initial data, so no extra constant branch from the first-integral square root is admissible.

This is the self-gravitating isothermal slab. Its scale convention matters: if its profile is instead written $\rho ={\rho }_0{\mathrm{sech}}^2(z/H)$, then $H=2z_0={\sigma }_z/\sqrt{2\pi G{\rho }_0}$. Its surface density is $\Sigma =4{\rho }_0{z}_0$. The force approaches $|{\Phi }^{\prime }|={\sigma }_z^2/z_0=2\pi G\Sigma$, consistent with the field of a plane sheet. These checks fix both the scale-height and Poisson factors.