# Implicit energy update

FargoCPT code (old):

$\alpha = 1 +  2 H \frac{\sigma_R}{c} 4 (\frac{m_\mu (\gamma - 1)}{R \Sigma})^4 e^3$

$\frac{\partial e}{\partial t} = \frac{-p\nabla u + Q^+ - Q^-}{\alpha}$



## What is $\alpha$

### Equation to solve: $\frac{\partial (e + E_{rad})}{\partial t} = -p\nabla u + Q^+ - Q^- - \nabla F$
  
$\frac{\partial e}{\partial t} = - p \nabla u + Q^+$

$\frac{\partial E_{rad}}{\partial t} = - Q^- - \nabla F$


$E_{rad} = 2 H \frac{\sigma_R}{c} T^4$

$T = \frac{m_\mu(\gamma - 1)}{R\Sigma} e$

$E_{rad} = 2 H \frac{\sigma_R}{c} (\frac{m_\mu (\gamma - 1)}{R \Sigma})^4 e^4$


$\frac{\partial E_{rad}}{\partial t} = 2 H \frac{\sigma_R}{c} 4 (\frac{m_\mu (\gamma - 1)}{R \Sigma})^4 e^3 \frac{\partial e}{\partial t}$

$\frac{\partial (e + E_{rad})}{\partial t} = -p\nabla u + Q^+ - Q^- - \nabla F$

$\frac{\partial e}{\partial t} +  2 H \frac{\sigma_R}{c} 4 (\frac{m_\mu (\gamma - 1)}{R \Sigma})^4 e^3 \frac{\partial e}{\partial t}= -p\nabla u + Q^+ - Q^- - \nabla F$

$\frac{\partial e}{\partial t}(1 +  2 H \frac{\sigma_R}{c} 4 (\frac{m_\mu (\gamma - 1)}{R \Sigma})^4 e^3)= -p\nabla u + Q^+ - Q^- - \nabla F$


$\alpha = 1 +  2 H \frac{\sigma_R}{c} 4 (\frac{m_\mu (\gamma - 1)}{R \Sigma})^4 e^3$

$\frac{\partial e}{\partial t} = \frac{-p\nabla u + Q^+ - Q^- - \nabla F} {\alpha}$


## Questions

### Is this operation splitting still correct?
$\frac{\partial e}{\partial t} = \frac{-p\nabla u + Q^+ - Q^-}{\alpha}$

$\frac{\partial e}{\partial t} = -p\nabla u$

$\frac{\partial e}{\partial t} = \frac{Q^+ - Q^-}{\alpha}$

###  What is the actually emitted energy?
$Q^-$ or $\frac{Q^-}{\alpha}$