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Let $F:{\Bbb R}^2\to {\Bbb R}^2$ be such that $$ F(x,y)=\left(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2}\right). $$ Suppose we have to one-to-one $C^1$ curves: $\gamma_j:[0,1]\to{\Bbb R}^2$, such that $$ \gamma_j(0)=p, \ \ \gamma_j(1)=q, \quad j=1,2, $$ for some $p,q\in{\Bbb R}^2\setminus\{(0,0)\}$. Assume furthermore that $\gamma_j(t)\not=(0,0)$ and $\gamma_j'(t)\not= 0$ for all $t\in[0,1]$, and $\gamma_1((0,1))\cap\gamma_2((0,1))=\emptyset$. Show that $$ \int_{\Gamma_1}F\cdot dr-\int_{\Gamma_2}F\cdot dr=2k\pi $$ where $k=0,1,-1$, and $\Gamma_j=\gamma_j([0,1])$.

When $p=(1,0)$, $q=(-1,0)$, let $\Gamma_1$ be the upper semi-circle and $\Gamma_2$ the lower one. Then one can show explicitly that $k=\pm 1$.How can one generalize this case?

Some thoughts:

  • I think the result depends on whether $(0,0)$ is inside the closed curve $\Gamma_1\cup\Gamma_2$ or not, which would give $k$ is zero or nonzero.
  • A quick calculation renders that $$ F=\nabla f $$ where $f(x,y)=\arctan(y/x)$ when $x\not=0$. But this is not always true for ${\Bbb R^2}\setminus\{(0,0)\}$.
  • Geometrically, $F$ is the unit outer normal on the unit circle. But I don't see how this might be used.
Yiorgos S. Smyrlis
  • 83,933

1 Answers1

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Clearly, $$ \int_{\Gamma_1}F\cdot dr-\int_{\Gamma_2}F\cdot dr=\int_{\gamma}\frac{-y\,dx+x\,dy}{x^2+y^2}, $$ where $\gamma=\Gamma_1-\Gamma_2$, meaning $\gamma$ is the union of the two curves, with the second in the opposite direction, and hence $\gamma$ is a simple closed curve. But $$ \int_{\gamma}\frac{-y\,dx+x\,dy}{x^2+y^2}=\mathrm{Im}\int_{\gamma}\frac{dz}{z}, $$ where $z=x+\mathrm{i}y$. Now $$ \frac{1}{2\pi\mathrm{i}}\int_{\gamma}\frac{dz}{z}, $$ is the the index of $\gamma$ around $z=0$, or $\mathrm{Ind}_\gamma(0)$. If $0$ is not in the interior of $\gamma$, then $\mathrm{Ind}_\gamma(0)=0$. Otherwise it is equal to 1 or -1 depending on the orientation of $\gamma$.

Yiorgos S. Smyrlis
  • 83,933