 Solution

Assume a steady, locally plane-symmetric galactic disk, with negligible radial tilt term in the vertical Jeans equation. Write $\rho (z)$ for the stellar mass density, and $K_z={\partial }_z\Phi$ for the upward gradient of the Newtonian gravitational potential; the actual vertical gravitational acceleration is $-K_z$. Constant vertical velocity dispersion givesIntegrating the plane-parallel Poisson equation for Newtonian gravity from the symmetry plane gives $K_z(z)=4\pi G{\int }_0^z\rho (z^{\prime })d{z}^{\prime }$. Outside most of the mass, this tends to $2\pi G\Sigma$ for $z>0$, where the surface density includes both sides of the galactic disk. Treating this field as constant and integrating the vertical Jeans equation yieldsThis exponential is the thin-sheet approximation or the outer tail, not the exact profile at every height. To see the distinction without an additional assumption, set $\rho ={\rho }_0{e}^{-u}$ with $u=(\Phi -\Phi (0))/{\sigma }^2$. The Poisson equation for Newtonian gravity becomes $u^{\prime \prime }=(4\pi G{\rho }_0/{\sigma }^2)e^{-u}$, with $u(0)=u^{\prime }(0)=0$, and is solved bySince $\Sigma =2{\rho }_0{z}_0$, the exact self-gravitating isothermal slab has $z_0={\sigma }^2/(\pi G\Sigma )$. At large $|z|/z_0$, ${\mathrm{sech}}^2(z/z_0)\sim 4e^{-2|z|/z_0}$, recovering precisely the boxed scale height; near the plane its slope vanishes, whereas an exponential has a cusp.