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 | |
| 38 | 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} $ |
| 39 | |
| 40 | So 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 | |
| 44 | Where $B'$ is the rotated vector $B$ through the angle $\phi$. |
| 45 | |
| 46 | Now if $\theta=0$, this simplifies to |
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 | | |