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A radiation-supported vertical disk has vertical hydrostatic equilibrium supplied mainly by radiation pressure. In a constant-opacity, optically thick layer in gravity $-{\Omega }^{2}z$, radiative diffusion gives $F=c{\Omega }^{2}z/\kappa$. This statement applies at a fixed radius and does not resolve the optically thin surface.

A laminar column can contract through nearly hydrostatic equilibrium when its cooling time greatly exceeds its vertical dynamical timescale of a disk. For cooling $A{T}^{1+\beta }$ and fixed column mass, a similarity solution has $\rho \propto {\eta }^{-1}$, $T\propto {\eta }^2$, $P\propto \eta$, and vertical velocity $u_z=\dot{\eta }z/\eta$. The coefficient $C$ also depends on the initial mass density normalization.

For $P_0>0$, $\alpha \ne 0$, $T\ge T_0$ and physical $T>0$, the atmospheric pressure-temperature profile $P=P_0\exp (\alpha \sqrt{T-T_0})$ is equivalent toThe branch restriction is essential: squaring does not create a second physical branch. Its temperature gradient is $dT/d\log P=2\sqrt{T-T_0}/\alpha$. Thus, in hydrostatic equilibrium, $\alpha <0$ gives an atmospheric thermal inversion, while $\alpha >0$ gives outward cooling. At $T=T_0$ the inverse branch has zero derivative; $\alpha =0$ gives constant pressure and cannot describe a nontrivial hydrostatic vertical interval.

For a gas-supported alpha disk with $\kappa \propto {\rho }^{1/3}{T}^{10}$, compare rates at fixed surface density of a disk, radius, $\alpha$ and composition on the thermal timescale of an accretion disk. Assume that the vertical dynamical timescale of a disk is shorter than that thermal time, and that the viscous timescale is longer; the separation is controlled for $\alpha \ll 1$ and small disk aspect ratio. Vertical hydrostatic equilibrium gives $H\propto T^{1/2}$, $\rho \propto T^{-1/2}$, so $\kappa \propto T^{59/6}$. Viscous heating satisfies $Q^{+}\propto T$ and radiative cooling satisfies $Q^{-}\propto T^4/(\kappa \Sigma )\propto T^{-35/6}$. At equilibrium a positive temperature perturbation changes the net heating by $(41/6)(Q_0/T)\delta T>0$. With positive effective heat capacity, it grows rather than relaxes. This is a local thermal statement and uses the fixed-column, rather than fixed-volume-density, comparison.

A uniform-density spherical hydrostatic equilibrium model with zero surface pressure has $P(r)=2\pi G{\rho }^2(R^2-r^2)/3$, $M=4\pi \rho R^3/3$ and gravitational energy $-3G{M}^2/(5R)$. A perfect gas can have this mechanical structure with varying temperature and entropy; the imposed equilibrium density does not require incompressible perturbations.

A constant-density gas sphere in hydrostatic equilibrium has $g={\omega }_d^{2}r$ and $p=\rho {\omega }_d^2(R^2-r^2)/2$. Its stellar oscillations can be expanded using regular solid harmonics and radial displacement amplitudes. Uniform density does not imply uniform specific entropy, so this equilibrium can be radially stable and still exhibit convective instability.

For positive finite opacity and nonvacuum ideal gas satisfying radiative diffusion, zero flux implies $dT/dz=0$. Write $P=\mathcal{R}\rho T$, where $\mathcal{R}=k_B/({\mu }_m{m}_p)$ for constant mean molecular weight ${\mu }_m$ in proton-mass units. If a polytropic equation of state $P=K{\rho }^{\gamma }$ with constant $K>0$ is also imposed, then $T=(K/\mathcal{R}){\rho }^{\gamma -1}$. For $\gamma \ne 1$, constant temperature would force constant density, contradicting vertical hydrostatic equilibrium $P^{\prime }=-\rho {\Omega }^{2}z$ on an open interval away from the midplane. The familiar finite polytropic slab for $\gamma >1$ is therefore a solution of the hydrostatic/adiabatic closure, not of this finite-opacity zero-radiative-flux system. For $\gamma =1$, the consistent isothermal profile is $\rho ={\rho }_m\exp [-{\Omega }^2{z}^2/(2\mathcal{R}{T}_m)]$, with no finite surface.