Christopher edited Temperature [1].tex  almost 10 years ago

Commit id: 2f868fb9b19c8d0ac990dbf9be7030824282e315

deletions | additions      

       

\begin{equation} %\label{eqn:F(T_ex)}  F(T_{ex}) = 2.31 * 10^{14} \frac{T_{ex} + 0.92}{1 - e^{-5.53/t_{ex}}} \frac{1}{J(T_{ex}) - J(T_{bg})}  \end{equation}  and assuming that the gas is a black body  \begin{equation} %\label{eqn:rayleigh jean}  T_B = \frac{B_v c^2}{2 v^2 k T}  \end{equation} 

\begin{equation} %\label{eqn:J(T)}  J(T) = \frac{hv}{k(e ^ {\frac{hv}{kT}} - 1)}  \end{equation}  $M$ is the mass, $[{\rm H}_2/^{12}{\rm CO}]$ the mass ratio between $H_2$ and $^{12}CO$ (assumed to be $1/(8.6 * 10 ^{-5})$, $\mu_m$ the mean molecular mass of $H_2$, $A$ the area of emission (1 pixel), $T_{ex}$ the assumed temperature of excitation, $T_{bg}$ the background temperature (assumed to be 0K), $T_B$ the brightness temperature(???), temperature,  $\Delta v$ the velocity resolution. resolution, k the Boltzman constant, $v$ the frequency of observation (115.3Ghz for $^{12}CO$, 110.2Ghz for $^{13}CO$).  (Bourke et al. 1997)  {\bf Every symbol here must be defined. What is v, what is alpha, A, etc? Explain it to yourself a year ago. If that you wouldn't understand, the reader probably won't either. State what this equation assumes. Also how do you get $T_B$ from the datacube I gave you? Give the equation you use. }