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Deep in an optically thick grey atmosphere, write $I_{\nu }=B_{\nu }+\delta I_{\nu }$ and retain the first spatial-gradient correction in the transfer equation:Angular and frequency integration then gives the radiative diffusion fluxFor a thin plane-parallel atmosphere, constant Rosseland mean opacity $\kappa$, negligible external irradiation, and radius nearly equal to $R$, radiative equilibrium gives $F=L/(4\pi R^2)$. ThereforeWith hydrostatic balance $dP/dz=-\rho g$, the equivalent pressure form iswhere $g=GM/R^2$.

For a perfect gas, $c_s^2\sim kT/({\mu }_m{m}_p)$. At the margin of the Toomre stability criterion,Integrating viscous heating through the disk and applying radiative diffusion givesEquating these fluxes yieldsThe kinematic viscosity is $\bar{\nu }\sim \alpha c_s^2/\Omega$, soBecause a Keplerian accretion disk has $\Omega \propto r^{-3/2}$,

The first equation is vertical hydrostatic equilibrium: the pressure gradient balances the vertical gravity of the central mass, ${\Omega }^{2}z$, and the disk's own potential ${\Phi }_d$. The second is the plane-parallel Poisson equation for disk self-gravity. The third balances the vertical increase of radiative flux against local viscous heating in a Keplerian alpha disk, and the fourth is the optically thick radiative diffusion law.

With $P/\rho \sim c_s^2$, $\Sigma \sim \rho H$, and all terms in hydrostatic balance comparable,The first comparison gives $H\sim c_s/\Omega$; inserting it into the second gives $\Omega c_s\sim G\Sigma$. Henceup to the order-one constants deliberately omitted by the scaling argument.