Appendix A: Vertically averaged equations for a steady state Keplerian $\beta$-disc

In a steady state Keplerian accretion disc where the viscosity is given by Eq. (2), the mass accretion rate $\dot{M}$ is linked to the surface density $\Sigma$ by the relation

$$\dot{M} = 3 \, \pi \beta \Omega R^2 \Sigma,$$ (A.1)

where $\Omega$ is the Keplerian angular velocity and R the radius. The surface density is $\Sigma = 2 \rho H$, with $\rho$ being the mass density and H the total pressure scale height, which is determined from the equation of hydrostatic equilibrium

$$\frac{P}{\rho H} = \Omega^2 H (1 + \zeta) \quad \hbox{with} \quad \zeta = {4 \pi G \rho \over \Omega^2} ,$$ (A.2)

where $\zeta$ represents the effect of self-gravity. The total pressure P is the sum of the gas pressure $P_{\rm g}$ and of the radiation pressure $P_{\rm r}$:

$$P = P_{\rm g} + P_{\rm r} = \frac{P_{\rm g}}{\tilde{\beta}} \cdot$$ (A.3)

We draw $P_{\rm g}$ from the perfect gas equation

$$P_{\rm g} = \frac{\rho k T}{\mu {\rm m}_{\rm H}},$$ (A.4)

where T is the midplane temperature and $\mu$ the mean mass per particle in units of the proton mass. The disc temperature in the equatorial plane is fixed by the balance Q⁺=Q^- between viscous heating (Frank et al. 1992)

$$Q^+ = \frac{3\, \Omega^2 \dot{M}}{8\, \pi}$$ (A.5)

and radiative losses (Hubeny 1990)

$$Q^- = \frac{8\, \sigma T^4}{3\, \kappa_{\rm R} \rho H + \frac{2}{\kappa_{\rm P} \rho H} +2 \sqrt{3}},$$ (A.6)

where $\kappa_{\rm R}$ and $\kappa_{\rm P}$ are respectively the Rosseland and Planck mean opacities (in cm²g^-1). The general expression for the radiative pressure is then

$$P_{\rm r}=\frac{Q^-}{c}\left( \frac{1}{2} \kappa_{\rm R} \rho H+\frac{1}{\sqrt{3}}\right) .$$ (A.7)

Note that in the optically thick limit the radiation pressure tends to its LTE value (labeled with an asterisk)

$$P_{\rm r}=\frac{4 \sigma T^4}{3 c} \equiv P_{\rm r}^* .$$ (A.8)

The above equation set can in fact be reduced to a system of two non linear algebraic equations with $\Sigma$ as the unknown. From Eqs. (A.1)-(A.7), the first equation is

$$P^* - \pi G \Sigma^2 - \frac{\pi^6 \beta^6 G^4 M^4}{256 \, \d... ...\, c \kappa_{\rm R} \kappa_{\rm P} {\mathcal S}(\Sigma) } = 0,$$ (A.9)

where the function ${\cal S}(\Sigma)$ is

$${\mathcal S}(\Sigma) = \frac{\Sigma^2}{2} + \frac{2 \Sigma}{\... ...appa_{\rm R}} +\frac{4}{3 \kappa_{\rm R} \kappa_{\rm P}} \cdot$$ (A.10)

The second equation is obtained from Eqs. (A.1), (A.5) and (A.6)

$$\frac{\sigma \dot{M}^5 T^4}{3^{14} \, \pi^{11} \, \kappa_{\rm R}} = G^4 M^4 \beta^6 {\mathcal S}(\Sigma) \Sigma^5 .$$ (A.11)

We can follow the method described in Huré (1998) to solve the system of Eqs. (A.9) and (A.11). Note that the solution is valid in all cases: optically thick or thin, pressure dominated by gas or radiation, self-gravitating or not.

A.1 The optically thick limit

If we set ${\mathcal S}(\Sigma) = \frac{1}{2} \Sigma^2$ in Eqs. (A.9) and (A.11), and omit the last term of Eq. (A.9) we recover the optically thick limit ( $\tau \gg 1$), and there is a single equation to solve:

$$\dot{M}^{12/7}+\lambda_1 \dot{M}^{2/7}+\lambda_0 = 0 .$$ (A.12)

The coefficients $\lambda_0$ and $\lambda_1$ are respectively

  

$$\lambda_0 16 \left( \frac{\pi^{5/6} \beta \sigma M^{2/3}}{3 \sqrt{G} \kappa_{\rm R}}\right)^{6/7} \frac{T^{24/7}}{\rho}\quad \hbox{and}$$ =

$$\lambda_1 - \frac{3}{4} \left( \frac{3 \kappa_{\rm R} \pi \beta^6 \sqrt{\pi G} M^4}{\sigma}\right)^{2/7} \frac{P^*}{T^{8/7}}\cdot$$ = (A.13)

Note that the solution of Eq. (A.12) must satisfy $\tau \gg 1$.

A.2 The optically thin limit

In the optically thin limit ( $\tau \ll 1$), ${\mathcal S}(\Sigma) = 4 / 3 \kappa_{\rm R} \kappa_{\rm P}$ and the last term of Eq. (A.9) is identical to $P_{\rm r}^*$. There is also a single equation to solve, namely

 

$$\left[4 \, \pi G \rho + \left( \frac{16 \sigma \kappa_{\rm P}}{3 \beta G^{2/3} M^{2/3}}T^4 \right)^{6/5} \right] \dot{M}^2$$

$$\qquad\quad - \frac{3 \, k \rho^2}{\mu {\rm m}_{\rm H} T^{3/5}} \left(\frac{6 \pi^5 \beta^6 G^4 M^4}{\sigma \kappa_{\rm P}}\right)^{2/5} = 0$$ (A.14)

and its solution must be such that $\tau \ll 1$.