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For a self-gravitating isothermal slab with $\rho ={\rho }_0{\mathrm{sech}}^2(z/H)$ and $H^2={\sigma }^2/(2\pi G{\rho }_0)$, hydrostatic equilibrium gives the displayed Newtonian gravitational potential. A test particle has constant in-plane velocity and conserved vertical energy $E_z={\dot{z}}^2/2+\Phi (z)-\Phi (0)$. Its vertical motion is periodic for every finite $E_z>0$, with turning height $H\mathrm{arcosh}\exp [E_z/(2{\sigma }^2)]$. Near the plane it is a harmonic oscillator of squared frequency $4\pi G{\rho }_0$; at large height the force tends to the constant $2\pi G\Sigma$, where $\Sigma =2{\rho }_{0}H$. The ideal infinite slab has no finite vertical escape energy.

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.

Write ${\sigma }_0^2=⟨v_z^2⟩=v_{\mathrm{rms}}^2/3$ for the one-dimensional thermal velocity dispersion of an isotropic monatomic ideal gas. An isothermal process keeps ${\sigma }_0$ fixed, and the vertical pressure is $P=\rho {\sigma }_0^2$. Hydrostatic equilibrium in the Newtonian gravitational potential $\Phi (z)$ givesPlanar symmetry reduces the Poisson equation for Newtonian gravity to ${\Phi }^{\prime \prime }=4\pi G\rho$. Combining the two equations yieldsThe velocity factor here is a velocity second moment, not the square of a particular particle's instantaneous vertical velocity.

Set $H^2={\sigma }_0^2/(2\pi G{\rho }_0)$ and try the symmetric self-gravitating isothermal slab profile $\rho ={\rho }_0{\mathrm{sech}}^2(z/H)$. Its logarithmic derivatives areThe chosen $H$ makes the second expression satisfy the required equation. The central conditions $\rho (0)={\rho }_0$ and ${\rho }^{\prime }(0)=0$ fix the smooth even solution uniquely. In the scale-height convention used here,The undefined constant printed as $a$ in the scale-height denominator must be $G$; this follows both from substitution and dimensions.

Integrating hydrostatic equilibrium, with the additive potential constant chosen so that $\Phi (0)=0$, givesThe orbits in a self-gravitating isothermal slab separate into horizontal and vertical motions. There is no horizontal force, so $x=x_i+v_{x}t$ and $y=y_i+v_{y}t$, with constant horizontal velocities. For collisionless test particles, or between gas collisions, conservation of energy fixes the vertical energyEvery finite $E_z>0$ has two turning points $\pm A$, whereThus the general orbit drifts uniformly in the plane while oscillating periodically across it. A particle with $E_z=0$ stays in the midplane; an orbit with nonzero horizontal velocity is generally not closed.

Near the plane, $\Phi (z)={\sigma }_0^2{z}^2/H^2+O(z^4)$, so the vertical motion is a harmonic oscillator with ${\nu }^2=2{\sigma }_0^2/H^2=4\pi G{\rho }_0$. Far from the plane, $\Phi \sim 2{\sigma }_0^2|z|/H$ and the force tends to a constant magnitude. Indeed the surface density is $\Sigma =\int \rho dz=2{\rho }_{0}H$, and that magnitude is $2\pi G\Sigma$. Large-amplitude vertical motion is therefore approximately uniformly accelerated on each side, rather than sinusoidal. Because this ideal slab extends infinitely in the horizontal directions and its potential grows without bound as $|z|\to \infty$, no finite-energy particle escapes vertically.

The Gaussian integral over the full vertical velocity line is $\sqrt{2\pi }{\sigma }_z$. Thus the mass density of the self-gravitating isothermal slab isNeglecting radial derivatives makes the Poisson equation for Newtonian gravity ${\Phi }_{zz}=4\pi G\rho (z)$. Substituting $\Phi ={\sigma }_z^2\varphi$ and $z=z_0\zeta$, with $z_0^2={\sigma }_z^2/(8\pi G{\rho }_0)$, givesMultiply by ${\varphi }^{\prime }$ and integrate. The central value and symmetry under $z\mapsto -z$ fix the constant of integration:On the positive-$\zeta$ side, Newtonian gravity makes ${\varphi }^{\prime }\ge 0$. Put $y=e^{\varphi /2}$; then $4(y^{\prime }{)}^2=y^2-1$ with $y(0)=1$. Integration gives $2\mathrm{arcosh}y=\zeta$, and even reflection supplies the other side. ThereforeDirect differentiation verifies both the equation and central conditions. Uniqueness of the smooth initial value problem rules out other branches meeting those conditions. The surface density is $4{\rho }_0{z}_0$, and the conventional slab scale height is $2z_0$; the factor of two is only a definition of the length scale.

Let ${\rho }_0$ denote the midplane mass density and choose the symmetry plane at $z=0$. Isothermal hydrostatic equilibrium and the Poisson equation for Newtonian gravity giveso $c_s^2(d^2/d{z}^2)\ln \rho =-4\pi G\rho$. For $\rho ={\rho }_0{\mathrm{sech}}^2(z/H)$, direct differentiation yieldsThus the profile is a self-gravitating isothermal slab when $H^2=c_s^2/(2\pi G{\rho }_0)$. Integrating over both sides gives $\Sigma =2{\rho }_{0}H$, henceThe corresponding potential is ${\Phi }_0(z)-{\Phi }_0(0)=2c_s^2\ln \cosh (z/H)$, whose gradient vanishes at the midplane and tends to $\pm 2\pi G\Sigma$ far from the slab. The profile parameter $H$ is not its asymptotic exponential mass density scale height: that is $H/2$.

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.