Changes between Version 8 and Version 9 of PolarizationMaps

Timestamp:

05/04/18 14:25:38 (8 years ago)

Author:

Jonathan

Comment:

—

PolarizationMaps

@@ -30 +30 @@
 $\phi = \cos^{-1} \frac{x_i}{\sqrt{x_i^2+\left ( s \cos (\theta) \right)^2}}$
 .
+
+So our integral is now
+
+$\int f( \sqrt{x_i^2+\left(s \cos(\theta) \right ) ^2}, y_i+s \sin(\theta), 0 ) ds$
+
+
+Now, if $\theta = 0$, this simplifies to
+
+$\int f( \sqrt{x_i^2+s ^2}, y_i, 0 ) ds$
+
+and we can undergo a chance of variables
+
+$ x' = \sqrt{x_i^2+s^2}$
+
+and
+
+$dx' = \frac{s}{\sqrt{x_i^2+s^2}} = \frac{\sqrt{x'^2-x_i^2}}{x'} ds$
+
+so our integral becomes
+
+$2 \int_{x_p}^{\infty} f( x', y_i, 0 ) \frac{x'}{\sqrt{x'^2-x_i^2} } dx'$
+
+
+and $B_\perp$ is unchanged, while $B_x = B_\phi \cos(\theta) = B_\phi \frac{x_i}{x'}$
+
+
+
+
+[[CollapsibleStart]]
 
 So our new path is in the xy plane and is given by 
@@ -71 +100 @@
 and $B_x'(x,y_i,0)=B_\phi(x,y_i,0) \frac{x_i}{x}$
 
+[[CollapsibleEnd]]