Equation solved in Betti's code (SI units and Temperature in Joules)
\rho C^{\prime}_{v}\frac{\partial K_{B}T}{\partial t}=\frac{\partial}{\partial x}\left[\kappa_{0}(K_{B}T)^{n}\frac{\partial (K_{B}T)}{\partial x}\right]
where K_{B} is Boltzmann constant and C^{\prime}_{v}=C_{v}/K_{B} as in Betti's document and C_{v} is the normal specific heat capacity.
And the flux is
q_{0}=\kappa_{0}(K_{B}T)^{n}\frac{\partial (K_{B}T)}{\partial x_{SI}}
To convert this to cgs units we write
10^{3}\rho_{cgs}\frac{\partial T}{\partial t}=\frac{\kappa_{0}K_{B}^{n}}{C^{\prime}_{v}}\frac{\partial}{10^{-2}\partial x_{cgs}}\left[T^{n}\frac{\partial T}{10^{-2}\partial x_{cgs}}\right]
That is
\frac{1}{\gamma-1}\rho_{cgs}\frac{\partial T}{\partial t}=\frac{10\kappa_{0}K_{B}^{n}}{(\gamma-1)C^{\prime}_{v}}\frac{\partial}{\partial x_{cgs}}\left[T^{n}\frac{\partial T}{\partial x_{cgs}}\right]
In AstroBEAR we define
\kappa_{1}=\frac{10\kappa_{0}K_{B}^{n}}{(\gamma-1)C^{\prime}_{v}}
and
q^{*}_{cgs}=\kappa_{1}T^{n}\frac{\partial T}{\partial x_{cgs}}
Comparing the definition of q_{0} and q^{*}_{cgs} we have q_{0}=10(\gamma-1)q^{*}_{cgs}C^{\prime}_{v}K_{B}
In Betti's data, C^{\prime}_{v}=7.816e26 and \gamma=5/3 so
q_{0}=\frac{2}{3}*7.816*1.38*10^{4}*q^{*}_{cgs}
