Changes between Version 5 and Version 6 of PolarizationMaps

Timestamp:

05/02/18 15:10:26 (8 years ago)

Author:

Jonathan

Comment:

—

PolarizationMaps

@@ -9 +9 @@
 Each pixel in the resulting image will be given by the integral of the function along the path
 
-$s = x_i, y_i+s \sin(\theta), s \cos(\theta)$
+$S = x_i, y_i+s \sin(\theta), s \cos(\theta)$
 
 where $s=0$ corresponds to crossing of the $xy$ plane
@@ -34 +34 @@
 
 
-$s'=\left [ \sqrt{x_i^2+\left(s \cos(\theta) \right ) ^2}, y_i+s \sin(\theta), 0 \right ]$
+$S'(s)=\left [ \sqrt{x_i^2+\left(s \cos(\theta) \right ) ^2}, y_i+s \sin(\theta), 0 \right ]$
+
+And 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} $
+
+So our path integral becomes
+
+$\int_s f(B(x_i,y_i+s \sin(\theta), s \cos(\theta))) \frac{dS}{ds} ds = \int_{s} { f(B'(\sqrt{x_i^2+\left ( s \cos(\theta) \right )^2}, y_i+s \sin(\theta),0)) \frac{dS'}{ds} ds}$
+
+Where $B'$ is the rotated vector $B$  through the angle $\phi$.
+
+Now if $\theta=0$, this simplifies to
 
 
-And we have $\frac{ds'}{ds} =  ...$
+$\int_{s} { f(B'(\sqrt{x_i^2+s^2}, y_i,0)) \frac{\sqrt{x_i^2+s^2}}{s} ds}$
 
-
-
-so we can write the integral as
-
-$\int F(x'(s)
-
-and we have $ds = \frac{\sqrt{(x_i^2+(s \cos(\theta))^2}}{s \cos(\theta)} dx + \frac{1}{\sin(\theta)} dy$
-
-which under the substitution  gives
-
-$ds = \frac{x'}{\sqrt{x'^2-x_i^2}} dx'$
-
-
-
-
-
-
-
-Assume jet is oriented along $y$ and that we are integrating along the direction $(0, sin(\theta), cos(\theta))$
-
-where $\theta$ is the angle of inclination.
-
-We are interested in calculating the integral of $B_p^2$, $B_x^2$, and $B_xB_p$ where $B_p = cos(\theta)By + sin(\theta)Bz$
-
-If $\theta = 0$ this is just $\int B_y^2 dz$, $\int B_x^2 dz$, and $\int B_x B_y dz$
-
-If $\theta /= 0$ then we have to tilt the integral and calculate $B_p$ using the expression above.
-