Changes between Version 7 and Version 8 of u/johannjc/scratchpad5


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Timestamp:
10/01/15 11:26:30 (10 years ago)
Author:
Jonathan
Comment:

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  • u/johannjc/scratchpad5

    v7 v8  
    9292where
    9393
    94 $E = A \left ( \partial_i  B_{ij} \right )$
     94$E_j = A \left ( \partial_i  B_{ij} \right )$
    9595
    96 $F = A B_{ij}$
     96$F_{ij} = A B_{ij}$
    9797
    9898
    … …  
    118118where
    119119
    120 $\alpha_0 = -\frac{2E_{jj}\Delta t}{\Delta x^2}$
     120$\alpha_0 = -\frac{2F_{ij}\delta_{ij} \Delta t}{\Delta x^2}$
    121121
    122 $\alpha_{\pm j} = \pm \frac{C_j \Delta t}{2 \Delta x} + \frac{E_{jj}\Delta t}{\Delta x^2}$
     122$\alpha_{\pm j} = \pm \frac{E_j \Delta t}{2 \Delta x} + \frac{F_{jj}\Delta t}{\Delta x^2}$
    123123
    124 $\alpha_{\pm i, \pm j} = \pm \pm \frac{E_{ij}\Delta t}{4 \Delta x^2}$
     124$\alpha_{\pm i, \pm j} = \pm \pm \frac{F_{ij}\Delta t}{4 \Delta x^2}$
    125125
    126 
    127 
    128 Now in the limit where we ignore spatial derivatives in density and magnetic field orientation, we can throw away the C terms - and if we also assume backward Euler, we can set
    129 
    130 $\psi=1$
    131 
    132 $\phi' = \frac{-\lambda}{\lambda+1}$
    133 
    134 $E_{ij}=\kappa_\parallel n b_i b_j$
    135 
    136 
    137 If we are using Crank-Nicholson, then
    138 
    139 $\psi=1/2$
    140 
    141 $\phi' = \frac{1-\lambda}{\lambda+1}$
    142 
    143 $E_{ij}=\frac{1}{2}\kappa_\parallel n b_i b_j$
    144 
    145 
    146 
    147 Now for the perpendicular term, we have
    148 
    149  $\frac{\partial T}{\partial t} = \nabla \cdot \left [ n \hat{b} \frac{-n \kappa_\perp}{B^2 \left (\Lambda+1\right )} \left ( \hat{b} \cdot \nabla T_*^{\Lambda+1} \right ) +  n \frac{n \kappa_\perp}{B^2 \left (\Lambda+1\right )}\nabla T_*^{\Lambda+1} \right ]$
    150 
    151 And performing a Taylor expansion gives
    152 
    153 $\frac{\partial T}{\partial t} = \nabla \cdot \left [ -n \hat{b} \frac{n \kappa_\perp}{B^2 \left ( \Lambda+1 \right ) } \left ( \hat{b} \cdot \nabla \left ( -\Lambda T^{\Lambda+1} + \left ( \Lambda + 1 \right ) T^{\Lambda}T_{*} \right ) \right ) +  n \frac{n \kappa_\perp}{B^2 \left ( \Lambda+1 \right ) } \nabla \left ( - \Lambda T^{ \Lambda+1} + \left ( \Lambda + 1 \right ) T^{\Lambda} T_{*} \right )  \right ] $
    154 
    155 And in Einstein notation gives
    156 
    157 $\partial_t T = - \partial_i n b_i \frac{n \kappa_\perp}{B^2 \left ( \Lambda+1 \right ) }  b_j  \partial_j \left ( -\Lambda T^{\Lambda+1} + \left ( \Lambda + 1 \right ) T^{\Lambda}T_{*} \right )  +  \partial_i n \frac{n \kappa_\perp}{B^2 \left ( \Lambda+1 \right ) }   \partial_i \left ( - \Lambda T^{ \Lambda+1} + \left ( \Lambda + 1 \right ) T^{\Lambda} T_{*} \right )  $
    158 
    159 
    160 The first term is very similar to the parallel case - and the second term can be made similar by replacing $b_i b_j$ with $\delta_{ij}$
    161 
    162 $\phi' = \frac{\phi-\psi \Lambda}{\psi \left ( \Lambda + 1 \right )}$
    163 
    164 $B_j =  \phi' C_j $   
    165 
    166 $C_j =   -\kappa_\perp \psi \partial_i  n^2 B^{-2} \left ( b_i b_j - \delta_{ij} \right ) $
    167 
    168 $D_{ij}=\phi'E_{ij}$
    169 
    170 $E_{ij} =   -\kappa_\perp \psi  n^2 B^{-2} \left ( b_i b_j - \delta_{ij} \right ) $