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Christer Watson edited Analysis.tex
over 9 years ago
Commit id: 81ba97eb9a6eec4d5da3d35af887213c8f9cd7cf
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\begin{equation}
B_{mod} =B_0 \frac{h^2\nu^{3+\beta}}{c^3}\frac{1}{e^{\frac{h\nu}{kT}}-1}
\end{equation}
where $\beta$ is assumed to be 2. B$_0$ and T were taken as free parameters and a Levenberg–Marquardt algorithm was used to find the best-fit. The
fitting was done to the flux density in Janskies, so B$_0$ carries these units. The total column density is calculated using:
\begin{equation}
N_{tot} = \frac{I}{B_\nu(860 GHz) \mu m_H \kappa R_d}\\
B_\nu = \frac{2 h \nu^3}{c^2 (e^\frac{h \nu}{kT}-1)}\\
I =
1.085\times 10^4 3.73\times 10^{-16} B_{mod}(860 GHz)\left(\frac{18.6"}{\theta}\right)^2\\
\kappa = \kappa_{1.3mm} \left(\frac{\lambda}{1.3mm}\right)^{-\beta}\\
\end{equation}
where
we have made 3.73 x 10$^{-16}$ converts the surface brightness from Jy/sq arcsec to SI units. We make the following assumptions: $\kappa_{1.3mm}$ = 0.11 $\frac{m^2}{kg}$, appropriate for ice-covered dust grains from OH94, $\theta$=15.0", the beamsize of the GBT at 49 GHz, the mean molecular weight $\mu$ = 2.3 and mass to dust ratio $R_d$ = 100.