 Polytropic vertical structure of a self-gravitating disk

For a locally plane-parallel thin disk, assume a polytropic equation of state $P=K{\rho }^{1+1/n}$ with $K>0$ and polytropic index $n>0$. Let ${\rho }_0$ be the midplane mass density and $c_0^2=(1+1/n)K{\rho }_0^{1/n}$ the midplane squared sound speed. Take the external vertical gravitational acceleration to be $-{\Omega }^{2}z$, as for a local Keplerian disk.

The specific enthalpy is $\mathcal{H}=(n+1)K{\rho }^{1/n}$. Define $w=\mathcal{H}/c_0^2=n(\rho /{\rho }_0{)}^{1/n}$ and $\zeta =\Omega z/c_0$. Hydrostatic equilibrium gives $d\mathcal{H}/dz=-{\Omega }^{2}z-d{\Phi }_d/dz$, while the plane-parallel Poisson equation gives $d^2{\Phi }_d/d{z}^2=4\pi G\rho$. Combining them yieldsThis $Q_0$ is a midplane-density parameter; it is not the surface-density Toomre parameter. Midplane symmetry imposes $w(0)=n$ and $w^{\prime }(0)=0$. The first positive zero of $w$ defines the model's surface, where the mass density vanishes.

In the negligible-self-gravity limit $Q_0\to \infty$, $w=n-{\zeta }^2/2$ and the half-thickness is $\sqrt{2n}{c}_0/\Omega$. This finite polytropic surface differs from an isothermal Gaussian atmosphere with no sharp edge.^[ref]