Changes between Version 7 and Version 8 of u/bliu/AblativeRT/Equations


Ignore:
Timestamp:
06/05/15 11:20:46 (10 years ago)
Author:
Baowei Liu
Comment:

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  • u/bliu/AblativeRT/Equations

    v7 v8  
    66$$C_{v}\rho_{cgs}\frac{\partial T}{\partial t}=\frac{\partial}{\partial x_{cgs}}\left[\kappa_{1}T^{n}\frac{\partial T}{\partial x_{cgs}}\right]$$
    77
    8 where T is in cgs unit or Kelvin and $C_{v}$ is the specific heat capacity.
     8where everything is in cgs unit or Kelvin and $C_{v}$ is the specific heat capacity.
    99
    10 Here
    11  $$\kappa_{1}=\frac{10\kappa_{0}K_{B}^{n}}{(\gamma-1)}$$
    12 In current version of AstroBEAR, we have
    13 $$\frac{\partial E}{\partial T}\frac{\partial T}{\partial t}=\frac{10\kappa_{0}K_{B}^{n}}{(\gamma-1)C_{v}}\frac{\partial}{\partial x_{cgs}}\left[T^{n}\frac{\partial T}{\partial x_{cgs}}\right]$$
     10In order to find the relationship between $\kappa_{1}$ and $\kappa_{0}$ in [http://www.pas.rochester.edu/~bliu/AblativeRT/Docs/RBetti.pdf Betti's document], we have to rewrite the first equation in wiki:TempEquations to make temperature as in Kelvin instead of $K_{B}T$ which is in Joule:
     11$$C_{v}\rho_{SI}\frac{\partial T}{\partial t}=\frac{\partial}{\partial x_{SI}}\left[\kappa_{0}K_{B}^{n+1}T^{n}\frac{\partial T}{\partial x_{SI}}\right]$$
     12
     13where $C_{v}$ is in unit of '''Joule per kg per kelvin'''.
     14and
     15$$ q_{0}=\kappa_{0}(K_{B}T)^{n}\frac{\partial K_{B}T}{\partial x_{SI}} $$
     16
     17or
     18$$ q_{0}=10^{2}\kappa_{0}(K_{B}T)^{n+1}\frac{\partial T}{\partial x_{cgs}} $$
     19
     20By converting everything but the Boltzmann constant in the above equation to cgs units we have
     21
     22$$10^{3}10^{-4}C_{v}\rho_{cgs}\frac{\partial T}{\partial t}=\frac{\partial}{\partial 10^{-2}x_{cgs}}\left[\kappa_{0}K_{B}^{n+1}T^{n}\frac{\partial T}{\partial 10^{-2}x_{cgs}}\right]$$
     23Here
     24or
     25$$C_{v}\rho_{cgs}\frac{\partial T}{\partial t}=\frac{\partial}{\partial  x_{cgs}}\left[\kappa_{0}10^{5}K_{B}^{n+1}T^{n}\frac{\partial T}{\partial x_{cgs}}\right]$$
     26
     27where $C_{v}$ is in unit of '''erg per g per kelvin''' and 1 '''erg per g per kelvin''' = $10^{-4}$ '''Joule per kg per kelvin'''
     28
     29So the $\kappa_{1}$ in the new version of the code should be.
     30 $$\kappa_{1}=10^{5}\kappa_{0}K_{B}^{n+1}$$
     31and $K_{B}$ is in SI unit?
     32
     33And since we use the subroutine getdEdT to get the energy derivative so
     34$$\frac{\partial E}{\partial T}\frac{\partial T}{\partial t}=\frac{\partial}{\partial x_{cgs}}\left[\kappa_{1}T^{n}\frac{\partial T}{\partial x_{cgs}}\right]$$
    1435
    1536And same as blog:bliu01092014, we define
    1637
    17 $$ \kappa_{1}=\frac{10\kappa_{0}K_{B}^{n}}{(\gamma-1)C_{v} } $$
    18 $$ q^{*}_{cgs}=\kappa_{1}T^{*}\frac{\partial T}{\partial x_{cgs}}$$
     38$$ \kappa_{1}=10^{5}\kappa_{0}K_{B}^{n+1} $$
     39$$ q^{*}_{cgs}=10^{5}\kappa_{0}K_{B}^{n+1}T^{n}\frac{\partial T}{\partial x_{cgs}}$$
    1940
    2041And so same as blog:bliu01092014, we have
    2142
    22 In Betti's data, $C_{v}=7.816e26$ and $\gamma=5/3$ so
    23 $$q_{0}=\frac{2}{3}*7.816*1.38*10^{4}*q^{*}_{cgs}$$
     43In Betti's data, $C^{\prime}_{v}=7.816e26$, $\kappa_{0}=3.734E69$ $q_{0}=5.876E18$ so
     44$$\kappa_{1}=10^{5}*3.734E69*(1.38EK_{B}^{n+1} $$ 
     45$$q^{*}_{cgs}=10^{3}*q_{0}=5.876E21$$ 
    2446
    25472. Scales for converting from cgs to cu