Uniform advecting radiation in diffusive limit

In this test, we simulation an advecting uniform gas where radiation and matter are in thermal equilibrium in the co-moving frame. Following the Lorentz tranform, the initial radiation energy and flux in the lab frame to first order in \(v/c\) are \(E_r = a_r T^4\) and \(F_r = \frac{4}{3} v E_r\).

Parameters

\[\begin{aligned} \begin{align} T_0 = 10^7~{\rm K} \\ \rho_0 = 1.2 ~{\rm g~cm^{-3}}, \mu = 2.33 ~m_{\rm H} \\ \kappa_P=\kappa_R=100 \mathrm{~cm}^2 \mathrm{~g}^{-1} \\ v_{x,0} = 10 ~{\rm km~s^{-1}} \\ E_{r,0} = a_r T_0^4 \\ F_{x,0} = \frac{4}{3} v_{x,0} E_{r,0} \\ t_{\rm end} = 4.8 \times 10^{-5} ~{\rm s} \end{align} \end{aligned}\]

Results

With \(O(\beta \tau)\) terms:

The radiation temperature and matter temperatures, along with the exact solution.

The matter velocity, along with the exact solution.

Without \(O(\beta \tau)\) terms:

The radiation temperature and matter temperatures, along with the exact solution.

The matter velocity, along with the exact solution.

Physics

In the transport equation, both the radiation energy and flux are unchanged because the radiation flux and pressure are uniform. In the matter-radiation exchange step, the source term is zero since the radiation and matter are in equilibrium. Finally, the flux is updated following

\[\mathbf{F}_{r}^{(t+1)} = \frac{\mathbf{F}_{r}^{(t)} + \Delta t \left[ \rho \kappa_P \left(\frac{4 \pi B}{c}\right) \mathbf{v}c + \rho \kappa_F (\mathbf{v} :\mathsf{P}_r) c \right] }{1+\rho \kappa_{F} {c} \Delta t}.\]

With \(F_{r}^{(t)} = 4 v E_{r}^{(t)} / 3\), and \(\kappa_P=\kappa_R=\kappa\), we have

\[\mathbf{F}_{r}^{(t+1)} = \frac{\frac{4}{3} v E_r^{(t)} + \Delta t \left[ \rho \kappa E_r^{(t)} \mathbf{v}c + \rho \kappa \mathbf{v} (\frac{1}{3}E_r^{(t)}) c \right] }{1+\rho \kappa {c} \Delta t} = \frac{4}{3} v E_r^{(t)} = F_{r}^{(t)}\]

Therefore, \(F_r\) remains constant. This demonstrates that the code is invariant under Lorentz transformation.

We can also show that, with the \(O(\beta \tau)\) terms in the matter-radiation exchange step, the space-like component of the radiation four-force vanishes:

\[\begin{aligned} \begin{align} -G &= -\rho \kappa_F \frac{\mathbf{F}_r}{c} + \rho \kappa_P\left(\frac{4 \pi B}{c}\right) \frac{\mathbf{v}}{c}+\rho \kappa_F \frac{\mathbf{v} :\mathsf{P}_r}{c} \\ &= -\rho \kappa \frac{4}{3} E_r v / c + \rho \kappa E_r v / c+ \rho \kappa \frac{1}{3} E_r v / c \\ &= 0 \end{align} \end{aligned}\]