Arnold Trivium Problem 39

This is Gauss' Linking Number Formula, for two space curves $\vec{A}, \vec{B}: S^1 \to \mathbb{R}^3$

$$ \textrm{link}(A,B) = \oint_A \oint_B \frac{\vec{A}-\vec{B}}{|\vec{A}-\vec{B}|^3} \cdot (d\vec{A} \times d\vec{B})$$

In our case, $\vec{A}(t) = (\cos t, \sin t, 0)$ and $\vec{B}(t) = ( 1+ \cos 2t, \frac{1}{2}\sin t, \sin 2t)$ . How to picture these two curves:

• $\vec{A}(t)$ represents a unit circle in the $xy$ plane centered at the origin $(0,0,0)$.

• $\vec{B}(t)$ is harder to visualize but we observe a few things:

  • it projects to a circle centered at $(1,0)$ on the $xz$ plane
  • the $y$-coordinate $|B_2(t)| = |\frac{1}{2} \sin t| \leq \frac{1}{2}$.

Since the Gauss linking number is a topological invariant, we can deform the circle $\vec{A}$ to a the straight line $\{(1,t,0): t \in \mathbb{R}\} $ (or alternatively deform $\vec{B}$ to the double-circle $\vec{B}(t) = ( 1+ \cos 2t, \frac{1}{2}\sin t, \sin 2t)$) and then it is easy to check the linking number is 2 and the integral is $\color{#F76760}{\mathbf{8\pi}}$

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If we were to compute the integral directly, it would seem rather strenuous to use the triple product formula $a \cdot (b \times c)$ directly and wade through all the integrals. Instead I would introduce the family of circles:

$$ \vec{A}(s,t) = (s\cos t - s + 1, s\sin t, 0)\text{ with } s\to \infty$$

or instead $\vec{B}(s,t) = ( 1+ \cos 2t, \frac{s}{2}\sin t, \sin 2t)$ with $s \to 0$.

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CODE Here is the Python script which I used to check the Gauss Linking number formula:

import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline

N = 100
ds, dt = 1.0/N, 1.0/N


tot = 0

for s in 2*np.pi*np.arange(0,1,dt):

    for t in s + 2*np.pi*np.arange(0,1,ds):

        A  = np.array([np.cos(s), np.sin(s), 0*s])
        B  = np.array([1 + np.cos(2*t), 0.5*np.sin(t),   np.sin(2*t)])

        dA = np.array([ -1*np.sin(s),       np.cos(s), 0*s])*ds*2*np.pi
        dB = np.array([ -2*np.sin(2*t), 0.5*np.cos(t), 2*np.cos(2*t)])*dt*2*np.pi

        X  = np.cross(dA,dB)
        Y  = (A -B)/np.sum((A -B)**2)**1.5

        tot += np.sum(X*Y)

tot/np.pi

Getting an answer of -8.0000000011873595 $ = \frac{1}{\pi}\int \dots$

Here I am interpreting it as the scalar triple product.

We have vector functions $\vec A, \vec B:[0,2\pi] \to \mathbb R^3$, given by $$ \vec A(\alpha) = (\cos\alpha,\sin\alpha,0) \\ \vec B(\beta) = (2\cos^2\beta,\tfrac12 \sin\beta,\sin(2\beta)) .$$ I believe you are being asked to calculate $$ \int_{\alpha=0}^{2\pi} \int_{\beta=0}^{2\pi} \left[\left(\frac {\partial \vec A}{\partial \alpha}(\alpha) \times \frac {\partial \vec B}{\partial \beta}(\beta) \right) \cdot (\vec A(\alpha) - \vec B(\beta)) \right] \frac1{|\vec A(\alpha) - \vec B(\beta)|^3} \, d\beta \, d\alpha .$$ I used the NIntegrate function in Mathematica, and it reported an answer remarkably close to $-8\pi$. There is small chance this is a coincidence!

It looks like this is the "winding number" of one curve around the other.

So suppose you have two closed curves $C_1$ and $C_2$, each of which are boundaries of surfaces $S_1$ and $S_2$. Let's suppose that the curve $C_2$ never intersects $S_1$. Then I assert that $$ \oint_{\vec A \in C_1} \oint_{\vec B \in C_2} \frac{((\vec A-\vec B) \times d\vec B) \cdot d\vec A}{|\vec A-\vec B|^3} = 0 .$$ To see this, interchange the two integrals, and apply Stoke's Theorem on the integral with respect to $\vec A$, to get $$ \oint_{\vec B \in C_2} \int_{\vec X\in S_1} \text{curl}_{\vec X} \left(\frac{(\vec X-\vec B)\times d\vec B}{|\vec X-\vec B|^3} \right) \cdot \vec\nu(\vec X) \, dA(X) ,$$ where $\vec\nu(\vec X)$ is unit normal to the surface $S_1$ at $\vec X$, and $dA$ is the surface measure on $S_1$. Using standard vector calculus identities (and I might have some sign errors here), one obtains that $$ \text{curl}_{\vec X} \left(\frac{(\vec X-\vec B)\times d\vec B}{|\vec X-\vec B|^3} \right) = - \text{div}_{\vec X} \left(\frac{\vec X-\vec B}{|\vec X-\vec B|^3} \right) d\vec B + d\vec B \cdot \nabla_{\vec X}\left(\frac{\vec X-\vec B}{|\vec X-\vec B|^3} \right) .$$ The first term on the right hand computes to $0$. So after exchanging the integrals again, we are left with $$ \int_{\vec X \in S_1} \left[ \oint_{\vec B \in C_2} d\vec B \cdot \nabla_{\vec X}\left(\frac{\vec X-\vec B}{|\vec X-\vec B|^3} \right)\right] \cdot \vec\nu(\vec X) \, dA(\vec X).$$ The inner integral becomes $$ - \oint_{\vec B \in C_2} d\vec B \cdot \nabla_{\vec B}\left(\frac{\vec X-\vec B}{|\vec X-\vec B|^3} \right) ,$$ and we see that each component is a path integral of the gradient of a scalar, and hence also $0$.

I assert that in general the answer is invariant under smooth homotopies of the curves. Suppose you have three closed curves $C_1$, $\tilde C_1$ and $C_2$ such that $C_1$ can be deformed to $\tilde C_1$ without crossing any point of $C_2$. Let $S_1$ be the surface traced by the deformation of $C_1$ to $\tilde C_1$. Then using similar computations, you should get $$ \oint_{\vec A \in C_1} \oint_{\vec B \in C_2} \frac{((\vec A-\vec B) \times d\vec B) \cdot d\vec A}{|\vec A-\vec B|^3} = \oint_{\vec A \in \tilde C_1} \oint_{\vec B \in C_2} \frac{((\vec A-\vec B) \times d\vec B) \cdot d\vec A}{|\vec A-\vec B|^3} .$$

Now reduce your integral to some more simple integral by, for example, making one of the curves a straight line, and the other go twice around in a perfect circle perpendicular to the straight line, and centered along a point in the line.

Here is 'physical' derivation for Gauss's linking number formula. In contrast with David H's remark above, I will talk in terms of magnetic circulation rather than the magnetic force between two current-carrying wires.

Suppose I have two closed curves, with paths $C_1,C_2$ respectively. We can generate a magnetic field by running a uniform current $I$ through $C_1$, which by the Biot-Savart law is $$\mathbf{B}_{1}(\mathbf{r})=\frac{\mu_0 I}{4\pi}\oint_{C_1}\frac{d\mathbf{l}_1\times (\mathbf{r}-\mathbf{r}_1)}{|\mathbf{r}-\mathbf{r}_1|^3}.$$ The resulting circulation of this magnetic field relative to $C_2$ is $$\oint_{C_2}\mathbf{B}_1(\mathbf{r})\cdot d\mathbf{l}=\frac{\mu_0 I}{4\pi}\oint_{C_2}\oint_{C_1}\frac{\mathbf{r}_2-\mathbf{r}_1}{|\mathbf{r}_2-\mathbf{r}_1|^3}\cdot (d\mathbf{l}_2\times d\mathbf{l}_1)$$ where the cyclic invariance of the scalar triple product has been used to rearrange the integrand.

But Ampere's law informs us that such a circulation integral equal $\mu_0 I_{enc}$ where $I_{enc}$ is the current enclosed by $C_2$; since the current is uniform , this amounts to $I_{enc}=N_{12} I$ where $N_{12}$ is the linking number of $C_1$ relative to $C_2$. Solving for this linking number then finally gives Gauss's formula: $$\boxed{N_{12}=\frac{1}{4\pi}\oint_{C_2}\oint_{C_1}\frac{\mathbf{r}_2-\mathbf{r}_1}{|\mathbf{r}_2-\mathbf{r}_1|^3}\cdot (d\mathbf{l}_2\times d\mathbf{l}_1)}$$ Consequently your integral indeed amounts to finding the linking number, a task which the other answers have elaborated in sufficient detail.