Homologous vertical structure of a constant-opacity alpha disk

Let $\mathcal{R}=k/({\mu }_m{m}_p)$ and choose $h^6=3\alpha \kappa {\Sigma }^2{\mathcal{R}}^4/(16\sigma {\Omega }^5)$. Set $z=h\zeta$, $\rho =(\Sigma /h)D$, $p=\Sigma {\Omega }^{2}hP$, $T=({\Omega }^2{h}^2/\mathcal{R})\Theta$ and $F=\alpha \Sigma {\Omega }^3{h}^{2}Q$. Vertical hydrostatic equilibrium, viscous dissipation, radiative diffusion and the ideal gas law reduce to $P^{\prime }=-D\zeta$, $Q^{\prime }=9P/4$, ${\Theta }^{\prime }=-DQ/{\Theta }^3$, $P=D\Theta$. Midplane symmetry gives $Q(0)=0$; a regular radiative-zero surface has $P=\Theta =0$ at a universal ${\zeta }_s$, with $2{\int }_0^{{\zeta }_s}Dd\zeta =1$. All physical parameters have disappeared. Uniqueness of this normalized boundary-value problem makes every physical profile a rescaling of one dimensionless solution.

// Target: astrophysics.bigb

For a constant-opacity, gas-pressure-supported radiative disk surface at $z=H$, imposing $p=T=0$ does not impose $F=0$. If the outgoing surface flux is finite and positive, dividing hydrostatic balance by the temperature gradient gives $dp/dT\sim 16\sigma {\Omega }^{2}H{T}^3/(3\kappa F_s)$. Consequently $p\propto T^4$, $\rho \propto T^3$, and the diffusion equation gives $T\propto H-z$. Thus the mass density also vanishes regularly while the surface radiates a finite flux. This ideal boundary replaces the finite-temperature photospheric transition; its accuracy is not controlled solely by the disk aspect ratio.

// Target: astrophysics.bigb

The density-weighted mean kinematic viscosity of the universal vertical profile is $\bar{\nu }=\alpha \Omega h^2{\int }_{-{\zeta }_s}^{{\zeta }_s}Pd\zeta$. Substitution of the height scale gives $\bar{\nu }=C{\alpha }^{4/3}(\kappa /\sigma {)}^{1/3}{\mathcal{R}}^{4/3}{\Sigma }^{2/3}{\Omega }^{-2/3}$. For Keplerian rotation, ${\Omega }^{-2/3}=(GM{)}^{-1/3}r$. The positive numerical constant depends only on the dimensionless vertical profile and the convention for $\alpha$.

// Target: astrophysics.bigb