22 | | and I am assuming $\nu_0 = \frac{c}{\lambda_0}$ |
| 18 | Now to get the line profile in terms of velocities, we can start with the line profile in terms of frequencies: |
| 19 | |
| 20 | $\phi ( \nu - \nu' ) = \frac{\Gamma}{4 \pi^2 (\nu - \nu')^2 + \left ( \Gamma / 2 \right )^2}$ |
| 21 | |
| 22 | and realizing that this is a distribution, we can use |
| 23 | |
| 24 | $\phi ( v - v') dv' = \phi ( \nu - \nu') d \nu$ |
| 25 | |
| 26 | where |
| 27 | |
| 28 | $\nu = \nu_0 + \nu_0 \frac{v}{c}$ |
| 29 | |
| 30 | so |
| 31 | |
| 32 | $ \nu - \nu' = \frac{v - v'}{\lambda_0}$ |
| 33 | |
| 34 | and |
| 35 | |
| 36 | $d \nu = \frac{dv}{\lambda_0}$ |
| 37 | |
| 38 | to get |
| 39 | |
| 40 | $ \phi ( v-v' ) =\frac{\Gamma \lambda_0}{4 \pi^2 (v - v')^2 + \left ( \Gamma \lambda_0 / 2 \right )^2}$ |
| 41 | |
| 42 | |
| 43 | Putting this all together we have... |
| 44 | |
| 45 | $\tau_i = \frac{\lambda_0}{\Delta v}\displaystyle \int_{v_i-\frac{1}{2}\Delta v}^{v_i + \frac{1}{2}\Delta v} \sigma_{v0} \int N(v') \frac{\Gamma \lambda_0}{4 \pi^2 (v - v')^2 + \left ( \Gamma \lambda_0 / 2 \right )^2} dv' dv$ |
| 46 | |
| 47 | Now if we bin the column densities into velocity bins, we have |
| 48 | |
| 49 | $N_j = \displaystyle \int_{v_j-\frac{1}{2}\Delta v}^{v_j + \frac{1}{2}\Delta v} N (v') dv'$ |
| 50 | |
| 51 | and |
| 52 | |
| 53 | $ \int N(v') \frac{\Gamma \lambda_0}{4 \pi^2 (v - v')^2 + \left ( \Gamma \lambda_0 / 2 \right )^2} dv' = \sum N_j \frac{\Gamma \lambda_0}{4 \pi^2 (v - v_j)^2 + \left ( \Gamma \lambda_0 / 2 \right )^2}$ |
| 54 | |
| 55 | and we have (wave arms in air for a bit) |
| 56 | |
| 57 | $\tau_i = \frac{\lambda_0}{\Delta v} \sigma_{v0} \sum N_j \displaystyle \int_{v_i-\frac{1}{2}\Delta v}^{v_i + \frac{1}{2}\Delta v} \sigma_{v0} \frac{\Gamma \lambda_0}{4 \pi^2 (v - v_j)^2 + \left ( \Gamma \lambda_0 / 2 \right )^2} dv$ |
| 58 | |
| 59 | or |
| 60 | |
| 61 | $\tau_i = \frac{\lambda_0}{\Delta v} \sigma_{v0} \sum N_j C_{ij}$ |
| 62 | |
| 63 | where |
| 64 | |
| 65 | $C_{ij}=\displaystyle \int_{v_i-\frac{1}{2}\Delta v}^{v_i + \frac{1}{2}\Delta v} \sigma_{v0} \frac{\Gamma \lambda_0}{4 \pi^2 (v - v_j)^2 + \left ( \Gamma \lambda_0 / 2 \right )^2} dv$ |
| 66 | |
| 67 | |
| 68 | Now $\Delta v$ is of order 20 km/s and $\Gamma \lambda_0$ is 76 m/s... so if $i == j$, then the integral may as well go to infinity and we have |
| 69 | |
| 70 | $C_{ii} = 1$ |
| 71 | |
| 72 | and for $i \ne j$ |
| 73 | |
| 74 | $C_{ij} = \frac{\Gamma \lambda_0}{4 \pi^2 (v_i-v_j)^2} \Delta v$ |
| 75 | |
| 76 | so we have |
| 77 | |
| 78 | $\tau_i = \frac{\sigma_{v0} N_i \lambda_0}{\Delta v} + \displaystyle \sum_{i \ne j} N_j \sigma_{v0} \frac{\Gamma \lambda_0^2}{4 \pi^2 (v_i-v_j)^2} $ |
| 79 | |
| 80 | |
| 81 | Or at least that's how I get agreement with their equation 11... |
| 82 | |
| 83 | |
| 84 | So to implement this approach, I plan on creating projections (integrations along LOS) of $\kappa_i$ where |
| 85 | |
| 86 | $\kappa_i(n,v) = \delta_{ij}\frac{\sigma_{v0} n \lambda_0}{\Delta v} + \left ( 1 - \delta_{ij} \right ) n \sigma_{v0} \frac{\Gamma \lambda_0^2}{4 \pi^2 (v_i-v_j)^2}$ |
| 87 | |
| 88 | where $j$ corresponds to the bin that the LOS cell velocity $v$ lies within |