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.