Changes between Version 4 and Version 5 of PolarizationMaps
- Timestamp:
- 05/02/18 09:25:37 (7 years ago)
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PolarizationMaps
v4 v5 1 Goal is to integrate various functions along lines of sight which is constrained to be in the y-z plane .1 Goal is to integrate various functions along lines of sight which is constrained to be in the y-z plane and has components $B_x$ and $B_\perp$ 2 2 3 $ \int f(B_\perp(s), B_x(s)) ds$3 $I = \int f(B_\perp(s), B_x(s)) ds$ 4 4 5 $ B_\perp(s) = B_y(s) \cos(\theta) + B_z(s) \sin(\theta)$5 $\theta$ is the tilt angle of the jet towards the observer 6 6 7 so $B_\perp(s) = B_y(s) \cos(\theta) + B_z(s) \sin(\theta)$ 7 8 8 Each pixel will be given by the integral of the function along the path9 Each pixel in the resulting image will be given by the integral of the function along the path 9 10 10 11 $s = x_i, y_i+s \sin(\theta), s \cos(\theta)$ 11 12 12 where $\delta y = \cos(\theta) \delta x$ 13 where $s=0$ corresponds to crossing of the $xy$ plane 14 15 and we have $x_i-x_{i-1} = \delta x$ and $y_i-y_{i-1} = \delta y$ 16 17 and $\delta y = \cos(\theta) \delta x$ 13 18 14 19 15 For 3D we just need to create the expressions in visit, create the lineouts, and then integrate the resulting query 20 For 3D we just need to create the expressions in visit, create the lineouts, and then integrate the resulting query for each x_i and y_i 16 21 22 For 2.5D, we need to transform the integral along $s$ (normally in the yz plane) into an integral in the $xy$ plane. 17 23 18 For 2.5D, we need to transform the integral along $s$ into an integral in the $xy$ plane. 19 20 the value at $x_i,y_i+s \sin(\theta), s \cos(\theta)$ is the same as the value rotated into the xy plane at 24 Since everything is axi-symmetric, the value at $x_i,y_i+s \sin(\theta), s \cos(\theta)$ can be inferred by rotating the corresponding solution at 21 25 22 26 $\left [ \sqrt{x_i^2+\left(s \cos(\theta) \right ) ^2}, y_i+s \sin(\theta), 0 \right ]$ 23 27 24 or at 28 by an angle 25 29 26 $X'$ 30 $\phi = \cos^{-1} \frac{x_i}{\sqrt{x_i^2+\left ( s \cos (\theta) \right)^2}}$ 31 . 27 32 28 where33 So our new path is in the xy plane and is given by 29 34 30 $X' = \left [ x', y_i+s \sin(\theta), 0 \right ]$31 35 32 and 36 $s'=\left [ \sqrt{x_i^2+\left(s \cos(\theta) \right ) ^2}, y_i+s \sin(\theta), 0 \right ]$ 33 37 34 $x'=\sqrt{x_i^2+(s \cos (\theta))^2}$ 38 39 And we have $\frac{ds'}{ds} = ...$ 40 41 35 42 36 43 so we can write the integral as