Appendix B: Criterion for the thermal instability

Setting $\tau= \kappa_{\rm R} \Sigma + 8 / 3 \, \kappa_{\rm P} \Sigma + 4/\sqrt{3}$, and differentiating Eq. (A.6) while keeping $\Sigma$ constant, we obtain

 

$$\left(\frac{\partial \ln Q^-}{\partial \ln T}\right)_{\Sigma,\Omega} 4 - \frac{1}{\tau} \left[ \tau_{\rm R} b_{\rm R} - \frac{8 b_{\rm P} }{3 \, \tau_{\rm P}} \right]$$ =

$$- \frac{1}{\tau} \left[ \tau_{\rm R} a_{\rm R} - \frac{8 a_{\rm P... ...\right] \left(\frac{\partial \ln \rho }{\partial \ln T}\right)_{\Sigma,\Omega},$$ (B.1)

where $\tau_{\rm R} = \kappa_{\rm R} \Sigma$ and $\tau_{\rm P} = \kappa_{\rm P} \Sigma$ are the total Rosseland and Planck optical thicknesses respectively, and

$$\begin{eqnarray*}a_{\rm R}=\left(\frac{\partial \ln \kappa_R}{\partial \ln \rho}... ...eft(\frac{\partial \ln \kappa_P}{\partial \ln T}\right)_{\rho} . \end{eqnarray*}$$

Note that this expression does not depend on the actual viscosity prescription. For Thompson scattering, we have simply $a_{\rm R} = b_{\rm R} = a_{\rm P} = b_{\rm P} =0$, and for free-free opacity, $a_{\rm R} = a_{\rm P}=1$ and $b_{\rm R} = b_{\rm P} = -\frac{7}{2}$. For a combination of both sources, we have $a_{\rm R} = a_{\rm P}=x$ and $b_{\rm R} = b_{\rm P} = -\frac{7}{2}x$ where $x \equiv \kappa_{\rm ff} / (\kappa_{\rm ff}+\sigma_{\rm T})$.

To compute the derivative $\left(\partial \ln \rho / \partial \ln T \right)_{\Sigma,\Omega}$, we use the two different expressions of the total pressure at the midplane. First, by differentiation of Eq. (A.2), keeping $\Sigma$ and $\Omega$ constant, we have

$$\left( \frac{ \partial \ln P }{\partial \ln T} \right)_{\Sigm... ... \partial \ln \rho }{\partial \ln T} \right)_{\Sigma,\Omega} .$$ (B.2)

Note that $(\partial \ln \rho / \partial \ln T)_{\Omega,\Sigma}$ is always negative. Second, from Eq. (A.3) which is written

$$P = P_{\rm g} + \frac{Q^-}{c} \left( \frac{\tau_{\rm R} }{4} + \frac{1}{\sqrt{3}}\right)$$ (B.3)

and setting $\chi_\rho^{\rm g}=\left(\partial \ln P_{\rm g} / \partial \ln {\rm T} \right)_{\rho}$, $\chi_{\rm T}^{\rm g}=\left( \partial \ln P_{\rm g} / \partial \ln \rho \right)_{\rm T}$ and $\tilde{\beta}=P_{\rm g}/P$, we have

 

$$\left( \frac{ \partial \ln P }{\partial \ln T} \right)_{\Sigma,\Omega} \left[ \tilde{\beta} \chi_{\rm T}^{\rm g} + \left(1- \tilde{\beta}\right) a_{\rm R} \frac{\tau_{\rm R}}{\tau_{\rm R} + \frac{4}{\sqrt{3}}} \right]$$ = (B.4)

$$\times \left( \frac{ \partial \ln \rho }{\partial \ln T} \right)_{\Sigma,\Omega} + \tilde{\beta}\chi_\rho^{\rm g} + \left(1-\tilde{\beta}\right)$$

$$\times \frac{\tau_{\rm R}}{\tau_{\rm R} + \frac{4}{\sqrt{3}}} b_{... ... \right) \left(\frac{\partial \ln Q^- }{\partial \ln T}\right)_{\Sigma,\Omega}.$$

Combining now Eqs. (B.2) and (B.4), we find

 

$$-\left( \frac{ \partial \ln \rho }{\partial \ln T} \right)_{\Sigm... ...ght) a_{\rm R} \frac{\tau_{\rm R}}{\tau_{\rm R} + \frac{4}{\sqrt{3}}} \right] =$$

$$\tilde{\beta}\chi_\rho^{\rm g} + \left(1-\tilde{\beta}\right) \fr... ...\right) \left(\frac{\partial \ln Q^- }{\partial \ln T}\right)_{\Sigma,\Omega} .$$ (B.5)

Finally, from Eqs. (B.1) and (B.5) and after rearrangement

 

$$\left(\frac{\partial \ln \rho }{\partial \ln T}\right)_{\Sigma,\Omega} \left[4(1-\tilde{\beta}) -(1- \tilde{\beta}) \frac{\tau_{\rm R} b_{\rm R} - \frac{8 \, b_{\rm P} }{3 \, \tau_{\rm P}}}{\tau} \right.$$ =

$$\left. + \tilde{\beta} \chi_\rho^{\rm g} + (1- \tilde{\beta}) \frac{\tau_{\rm R}}{\tau_{\rm R} + \frac{4}{\sqrt{3}}} b_{\rm R} \right]$$

$$\times \left[ (1-\tilde{\beta}) \frac{\tau_{\rm R} a_{\rm R} - \f... ..._{\rm P} }{3 \, \tau_{\rm P}}}{\tau} - \left( \frac{1}{1+\zeta} \right. \right.$$

$$\left. \left. + \tilde{\beta} \chi_{\rm T}^{\rm g} + \left(1- \ti... ... R} \frac{\tau_{\rm R}}{\tau_{\rm R} + \frac{4}{\sqrt{3}} } \right)\right]^{-1}$$ (B.6)

and so Eq. (B.1) becomes

 

$$\left(\frac{\partial \ln Q^-}{\partial \ln T}\right)_{\Sigma,\Omega} \left[ \tilde{\beta} \chi_\rho^{\rm g} + \left(1- \tilde{\beta}\right) \frac{\tau_{\rm R}}{\tau_{\rm R} + \frac{4}{\sqrt{3}}} b_{\rm R} \right.$$ =

$$\left. + \left( \frac{1}{1+\zeta} + \tilde{\beta} \chi_{\rm T}^{\rm g} + \left(1-\tilde{\beta}\right) \right. \right.$$

$$\left. \left. \times a_{\rm R} \frac{\tau_{\rm R}}{\tau_{\rm R} +... ...u_{\rm R} a_{\rm R} - \frac{8 \, a_{\rm P} }{3 \, \tau_{\rm P}}}\right) \right]$$

$$\times \left[\frac{\tau }{ \tau_{\rm R} a_{\rm R} - \frac{8 \, a_... ...\left( \frac{1}{1+\zeta} + \tilde{\beta} \chi_{\rm T}^{\rm g} + \right. \right.$$

$$\left. \left. \times \left(1-\tilde{\beta}\right) a_{\rm R} \frac... ...R}}{\tau_{\rm R} + \frac{4}{\sqrt{3}} } \right) + \tilde{\beta} -1\right]^{-1}.$$ (B.7)

It is then easy to derive from Eq. (B.7) the following asymptotic expressions:

• in the optically thick limit ( $\tau \sim \tau_{\rm R}$ and )
  

  $$-\left(\frac{\partial \ln Q^- }{\partial \ln T}\right)_{\Sigm... ...})}{ \tilde{\beta} \chi^{\rm g}_{\rm T} + \frac{1}{1+\zeta}} ;$$ (B.8)

  
  

• in the radiative pressure dominated limit ( $\tilde{\beta} \sim 0$)
  

  $$-\left(\frac{\partial \ln Q^- }{\partial \ln T}\right)_{\Sigma,\Omega} = b_{\rm R} - 4 - 4 a_{\rm R} (1+\zeta) ;$$ (B.9)

  
  

• in the gas pressure dominated limit ( $\tilde{\beta} \sim 1$)
  

  $$-\left(\frac{\partial \ln Q^- }{\partial \ln T}\right)_{\Sigm... ...mega} = b_{\rm R} - 4 - a_{\rm R} \frac{1 + \zeta}{2+\zeta} ;$$ (B.10)

  
  

• in the optically thin limit ( $\tau \sim \frac{8}{3 \, \tau_{\rm P}}$ and $\tau_{\rm P} \ll 1$)
  
   

  $$\left(\frac{\partial \ln Q^- }{\partial \ln T}\right)_{\Sigma,\Omega} =$$ -

  $$b_{\rm P} - 4 + a_{\rm P} \frac{\tilde{\beta} \chi^{\rm g}_{\rho}... ...\beta})a_{\rm P} +\tilde{\beta} \chi^{\rm g}_{\rm T} + \frac{1}{1+\zeta}} \cdot$$ - (B.11)