I would like to compute the linking number of a link in $\mathbb{R}^3$ which consists of two disjoint curves $c_a$ and $c_b$.
Define $z=x_1+\sqrt{-1}x_2$ and $v=x_3+\sqrt{-1}x_4$ where $(x_1,x_2,x_3, x_4)\in\mathbb{R}^4$ so that $\mathbb{R}^4\cong\mathbb{C}^2$. Let $a,b\in\mathbb{C}$ and let $L_a=\{(z,v)\in\mathbb{C}^2|v=az\}\subset\mathbb{C}^2$ and $L_b=\{(z,v)\in\mathbb{C}^2|v=bz\}\subset\mathbb{C}^2$ be two complex lines in the complex plane. Also let $C_a=S^3\cap L_a$ and $C_b=S^3\cap L_b$ where $S^3$ is the 3-sphere.
How do I calculate the integral (the linking number) $$L(c_a, c_b) = \frac{1}{4\pi}\int_{0}^{2\pi}\int_{0}^{2\pi}\frac{(c_a(t)-c_b(s),\dot{c_a}(t), \dot{c_b}(s))}{|c_a(t)-c_b(s)|^3}dtds$$ where $c_a=p(C_a)$ and $c_b=p(C_b)$ and $p$ is the stereographic projection defined by $$p(x_1, x_2, x_3, x_4)=\frac{1}{1-x_4}(x_1, x_2, x_3, 0)$$ if $L(c_a, c_b)\neq 0$ (the orientation of curves is not important)?
The linking number is 1. The curves $C_{a},C_{b}$ are both fibres of the Hopf fibration $F : S^{3} \rightarrow S^{2}$. It is known (see https://en.wikipedia.org/wiki/Hopf_fibration) that any two fibres of the Hopf fibration map to Hopf link in $\mathbb{R}^{3}$ (after stereographic projection) and hence have linking number $1$ .