Changes between Version 5 and Version 6 of PolarizationMaps


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Timestamp:
05/02/18 15:10:26 (7 years ago)
Author:
Jonathan
Comment:

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  • PolarizationMaps

    v5 v6  
    99Each pixel in the resulting image will be given by the integral of the function along the path
    1010
    11 $s = x_i, y_i+s \sin(\theta), s \cos(\theta)$
     11$S = x_i, y_i+s \sin(\theta), s \cos(\theta)$
    1212
    1313where $s=0$ corresponds to crossing of the $xy$ plane
     
    3434
    3535
    36 $s'=\left [ \sqrt{x_i^2+\left(s \cos(\theta) \right ) ^2}, y_i+s \sin(\theta), 0 \right ]$
     36$S'(s)=\left [ \sqrt{x_i^2+\left(s \cos(\theta) \right ) ^2}, y_i+s \sin(\theta), 0 \right ]$
     37
     38And we have $\frac{dS'}{ds}=\sqrt{ \left ( \frac{s \cos(\theta)^2}{\sqrt{x_i^2+\left ( s \cos(\theta) \right)^2} }\right ) ^2 + \sin(\theta)^2} $
     39
     40So our path integral becomes
     41
     42$\int_s f(B(x_i,y_i+s \sin(\theta), s \cos(\theta))) \frac{dS}{ds} ds = \int_{s} { f(B'(\sqrt{x_i^2+\left ( s \cos(\theta) \right )^2}, y_i+s \sin(\theta),0)) \frac{dS'}{ds} ds}$
     43
     44Where $B'$ is the rotated vector $B$  through the angle $\phi$.
     45
     46Now if $\theta=0$, this simplifies to
    3747
    3848
    39 And we have $\frac{ds'}{ds} =  ...$
     49$\int_{s} { f(B'(\sqrt{x_i^2+s^2}, y_i,0)) \frac{\sqrt{x_i^2+s^2}}{s} ds}$
    4050
    41 
    42 
    43 so we can write the integral as
    44 
    45 $\int F(x'(s)
    46 
    47 and we have $ds = \frac{\sqrt{(x_i^2+(s \cos(\theta))^2}}{s \cos(\theta)} dx + \frac{1}{\sin(\theta)} dy$
    48 
    49 which under the substitution  gives
    50 
    51 $ds = \frac{x'}{\sqrt{x'^2-x_i^2}} dx'$
    52 
    53 
    54 
    55 
    56 
    57 
    58 
    59 Assume jet is oriented along $y$ and that we are integrating along the direction $(0, sin(\theta), cos(\theta))$
    60 
    61 where $\theta$ is the angle of inclination.
    62 
    63 We are interested in calculating the integral of $B_p^2$, $B_x^2$, and $B_xB_p$ where $B_p = cos(\theta)By + sin(\theta)Bz$
    64 
    65 If $\theta = 0$ this is just $\int B_y^2 dz$, $\int B_x^2 dz$, and $\int B_x B_y dz$
    66 
    67 If $\theta /= 0$ then we have to tilt the integral and calculate $B_p$ using the expression above.
    68