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

Timestamp:

06/05/15 11:20:46 (10 years ago)

Author:

Baowei Liu

Comment:

—

u/bliu/AblativeRT/Equations

@@ -6 +6 @@
 $$C_{v}\rho_{cgs}\frac{\partial T}{\partial t}=\frac{\partial}{\partial x_{cgs}}\left[\kappa_{1}T^{n}\frac{\partial T}{\partial x_{cgs}}\right]$$
 
-where T is in cgs unit or Kelvin and $C_{v}$ is the specific heat capacity.
+where everything is in cgs unit or Kelvin and $C_{v}$ is the specific heat capacity.
 
-Here 
- $$\kappa_{1}=\frac{10\kappa_{0}K_{B}^{n}}{(\gamma-1)}$$
-In current version of AstroBEAR, we have
-$$\frac{\partial E}{\partial T}\frac{\partial T}{\partial t}=\frac{10\kappa_{0}K_{B}^{n}}{(\gamma-1)C_{v}}\frac{\partial}{\partial x_{cgs}}\left[T^{n}\frac{\partial T}{\partial x_{cgs}}\right]$$
+In 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:
+$$C_{v}\rho_{SI}\frac{\partial T}{\partial t}=\frac{\partial}{\partial x_{SI}}\left[\kappa_{0}K_{B}^{n+1}T^{n}\frac{\partial T}{\partial x_{SI}}\right]$$
+
+where $C_{v}$ is in unit of '''Joule per kg per kelvin'''.
+and 
+$$ q_{0}=\kappa_{0}(K_{B}T)^{n}\frac{\partial K_{B}T}{\partial x_{SI}} $$
+
+or
+$$ q_{0}=10^{2}\kappa_{0}(K_{B}T)^{n+1}\frac{\partial T}{\partial x_{cgs}} $$
+
+By converting everything but the Boltzmann constant in the above equation to cgs units we have
+
+$$10^{3}10^{-4}C_{v}\rho_{cgs}\frac{\partial T}{\partial t}=\frac{\partial}{\partial 10^{-2}x_{cgs}}\left[\kappa_{0}K_{B}^{n+1}T^{n}\frac{\partial T}{\partial 10^{-2}x_{cgs}}\right]$$
+Here
+or 
+$$C_{v}\rho_{cgs}\frac{\partial T}{\partial t}=\frac{\partial}{\partial  x_{cgs}}\left[\kappa_{0}10^{5}K_{B}^{n+1}T^{n}\frac{\partial T}{\partial x_{cgs}}\right]$$
+
+where $C_{v}$ is in unit of '''erg per g per kelvin''' and 1 '''erg per g per kelvin''' = $10^{-4}$ '''Joule per kg per kelvin'''
+
+So the $\kappa_{1}$ in the new version of the code should be.
+ $$\kappa_{1}=10^{5}\kappa_{0}K_{B}^{n+1}$$
+and $K_{B}$ is in SI unit?
+
+And since we use the subroutine getdEdT to get the energy derivative so
+$$\frac{\partial E}{\partial T}\frac{\partial T}{\partial t}=\frac{\partial}{\partial x_{cgs}}\left[\kappa_{1}T^{n}\frac{\partial T}{\partial x_{cgs}}\right]$$
 
 And same as blog:bliu01092014, we define
 
-$$ \kappa_{1}=\frac{10\kappa_{0}K_{B}^{n}}{(\gamma-1)C_{v} } $$
-$$ q^{*}_{cgs}=\kappa_{1}T^{*}\frac{\partial T}{\partial x_{cgs}}$$
+$$ \kappa_{1}=10^{5}\kappa_{0}K_{B}^{n+1} $$
+$$ q^{*}_{cgs}=10^{5}\kappa_{0}K_{B}^{n+1}T^{n}\frac{\partial T}{\partial x_{cgs}}$$
 
 And so same as blog:bliu01092014, we have
 
-In Betti's data, $C_{v}=7.816e26$ and $\gamma=5/3$ so
-$$q_{0}=\frac{2}{3}*7.816*1.38*10^{4}*q^{*}_{cgs}$$
+In Betti's data, $C^{\prime}_{v}=7.816e26$, $\kappa_{0}=3.734E69$ $q_{0}=5.876E18$ so
+$$\kappa_{1}=10^{5}*3.734E69*(1.38EK_{B}^{n+1} $$  
+$$q^{*}_{cgs}=10^{3}*q_{0}=5.876E21$$  
 
 2. Scales for converting from cgs to cu