Changes between Version 4 and Version 5 of PolarizationMaps


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Timestamp:
05/02/18 09:25:37 (7 years ago)
Author:
Jonathan
Comment:

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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.
     1Goal 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 $B_\perp$
    22
    3 $\int f(B_\perp(s), B_x(s)) ds$
     3$I = \int f(B_\perp(s), B_x(s)) ds$
    44
    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
    66
     7so $B_\perp(s) = B_y(s) \cos(\theta) + B_z(s) \sin(\theta)$
    78
    8 Each pixel will be given by the integral of the function along the path
     9Each pixel in the resulting image will be given by the integral of the function along the path
    910
    1011$s = x_i, y_i+s \sin(\theta), s \cos(\theta)$
    1112
    12 where $\delta y = \cos(\theta) \delta x$
     13where $s=0$ corresponds to crossing of the $xy$ plane
     14
     15and we have $x_i-x_{i-1} = \delta x$ and $y_i-y_{i-1} = \delta y$
     16
     17and $\delta y = \cos(\theta) \delta x$
    1318
    1419
    15 For 3D we just need to create the expressions in visit, create the lineouts, and then integrate the resulting query
     20For 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
    1621
     22For 2.5D, we need to transform the integral along $s$ (normally in the yz plane) into an integral in the $xy$ plane.
    1723
    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
     24Since 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
    2125
    2226$\left [ \sqrt{x_i^2+\left(s \cos(\theta) \right ) ^2}, y_i+s \sin(\theta), 0 \right ]$
    2327
    24 or at
     28by an angle
    2529
    26 $X'$
     30$\phi = \cos^{-1} \frac{x_i}{\sqrt{x_i^2+\left ( s \cos (\theta) \right)^2}}$
     31.
    2732
    28 where
     33So our new path is in the xy plane and is given by
    2934
    30 $X' = \left [ x', y_i+s \sin(\theta), 0 \right ]$
    3135
    32 and
     36$s'=\left [ \sqrt{x_i^2+\left(s \cos(\theta) \right ) ^2}, y_i+s \sin(\theta), 0 \right ]$
    3337
    34 $x'=\sqrt{x_i^2+(s \cos (\theta))^2}$
     38
     39And we have $\frac{ds'}{ds} =  ...$
     40
     41
    3542
    3643so we can write the integral as