= Gas-pressure viscosity closure with mixed pressure support
{title2=$\bar\nu=\frac{3I_t}{4}(\alpha^4\kappa\mathcal R^4/\sigma)^{1/3}\Sigma^{2/3}\Omega^{-2/3}$}
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 $\partial_m^2T^4+A\Omega T=0$ in the <vertical mass coordinate of a disk>, where $A=27\alpha\kappa\mathcal R/(16\sigma)$. Set $\zeta=2m/\Sigma$, $T=T_*t(\zeta)$ and $T_*^3=A\Omega\Sigma^2/4$. With a <radiative-zero disk surface>, $t$ satisfies $(t^4)''+t=0$, $t'(0)=t(1)=0$. Writing $I_t=\int_0^1t\,d\zeta$, the <density-weighted mean kinematic viscosity> is $\bar\nu=\alpha\mathcal R T_*I_t/\Omega$, giving the displayed formula. This calculation does not assume gas-pressure domination: the closure depends on gas pressure while hydrostatic support can be mixed.
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