next up previous contents index
Next: The LVG Approximation Up: The radiative transfer equation Previous: The radiative transfer equation

   
The Source Function

Any program that attempts to solve the radiative transfer equation needs some way of calculating the source function. Using the definition in equation 3.2 and the expressions for $\alpha_{\nu}$ and $j_{\nu}$given in equations 2.25 & 2.26 gives

\begin{displaymath}S_\nu=\frac{\frac{h\nu}{4\pi}\frac{dn_u}{d\nu}A_{ul}}{\frac{h...
...\left(\frac{dn_l}{d\nu}B_{lu}
-\frac{dn_u}{d\nu}B_{ul}\right)}
\end{displaymath}

Now using the relationships between the Einstein coefficients given in equations 2.23 & 2.24 this can be reduced to

 \begin{displaymath}
S_\nu=\frac{2h\nu^3}{c^2}\left[\frac{1}{\frac{g_u}{g_l}\frac{dn_l}{d\nu}\frac{d\nu}{dn_u}-1}\right]
\end{displaymath} (3.7)

The only remaining problem here is the $\frac{dn_l}{d\nu}\frac{d\nu}{dn_u}$ term which can be considered as $\frac{dn_l}{dn_u}$. Differentiating equation 2.27 (the definition of excitation temperature) shows that

\begin{displaymath}\frac{dn_u}{dn_l}=\frac{g_u}{g_l}e^{-\frac{h\nu}{kT_{\rm EX}}}=\frac{n_u}{n_l}
\end{displaymath}

So substituting this into equation 3.7 finally yields

 \begin{displaymath}
S_\nu=\frac{2h\nu^3}{c^2}\left[\frac{1}{\frac{g_u}{g_l}\frac{n_l}{n_u}-1}\right]
\end{displaymath} (3.8)




1999-04-12