Gas-pressure viscosity closure with mixed pressure support ($\bar{\nu }=\frac{3I_t}{4}({\alpha }^4\kappa {\mathcal{R}}^4/\sigma {)}^{1/3}{\Sigma }^{2/3}{\Omega }^{-2/3}$)

Let $\mathcal{R}=k/({\mu }_m{m}_H)$, take constant opacity $\kappa$, and prescribe the dynamic viscosity $\mu =\alpha p_g/\Omega =\alpha \mathcal{R}\rho T/\Omega$ in a Keplerian disk. The pressure supporting the disk may contain both gas pressure and radiation pressure. Combining viscous dissipation and radiative diffusion gives ${\partial }_m^2{T}^4+A\Omega T=0$ in the vertical mass coordinate of a disk, where $A=27\alpha \kappa \mathcal{R}/(16\sigma )$. Set $\zeta =2m/\Sigma$, $T=T_{\ast }t(\zeta )$ and $T_{\ast }^3=A\Omega {\Sigma }^2/4$. With a radiative-zero disk surface, $t$ satisfies $(t^4{)}^{\prime \prime }+t=0$, $t^{\prime }(0)=t(1)=0$. Writing $I_t={\int }_0^{1}td\zeta$, the density-weighted mean kinematic viscosity is $\bar{\nu }=\alpha \mathcal{R}{T}_{\ast }{I}_t/\Omega$, giving the displayed formula. This calculation does not assume gas-pressure domination: the closure depends on gas pressure while hydrostatic support can be mixed.