Tag Archives: vortex line

Area of the symmetric difference of two disks

This post goes back to high school: the area $\delta_d$ of the symmetric difference of two $d$-dimensional disks when one center is shifted a little bit. Let’s start with $d=1$. So we have two intervals $[-r,r]$ and $[x-r,x+r]$. It is easy to see the symmetric difference is of length $\delta_1(x)=2x$.

Then we move to $d=2$: two disks $L$ and $R$ of radius $r$ and center distance $x=2a. So the angle $\theta(x)$ satisfies $\cos\theta=\frac{a}{r}$.

The symmetric difference is the union of $R\backslash L$ and $L\backslash R$, which have the same area: $\displaystyle (\pi-\theta)r^2+2x\sqrt{r^2-x^2}-\theta r^2=2(\frac{\pi}{2}-\arccos\frac{x}{r})r^2+2x\sqrt{r^2-x^2}$. Note that the limit
$\displaystyle \lim_{x\to0}\frac{\text{area}(\triangle)}{2a}=\lim_{a\to0}2\left(\frac{r^2}{\sqrt{1-\frac{a^2}{r^2}}}\cdot\frac{1}{r}+\sqrt{r^2-a^2}\right)=4r$.
So $\delta_2(x)\sim 4rx$.

I didn’t try for $d\ge3$. Looks like it will start with a linear term $2d r^{d-1}x$.

—————–

Now let ${\bf r}(t)=(a\cos t,\sin t)$ be an ellipse with $a>1$, and ${\bf r}'(t)=(-a\sin t,\cos t)$ be the tangent vector at ${\bf r}(t)$. Let $\omega$ be the angle from ${\bf j}=(0,1)$ to ${\bf r}'(t)$.
Let $s(t)=\int_0^t |{\bf r}'(u)|du$ be the arc-length parameter and $K(s)=|{\bf l}''(s)|$ be the curvature at ${\bf l}(s)={\bf r}(t(s))$. Alternatively we have $\displaystyle K(t)=\frac{a}{|{\bf r}'(t)|^{3}}$.

The following explains the geometric meaning of curvature:

$\displaystyle K(s)=\frac{d\omega}{ds}$, or equivalently, $K(s)\cdot ds=d\omega$. $(\star)$.

Proof. Viewed as functions of $t$, it is easy to see that $(\star)$ is equivalent to $K(t)\cdot \frac{ds}{dt}=\frac{d\omega}{dt}$.

Note that $\displaystyle \cos\omega=\frac{{\bf r}'(t)\cdot {\bf j}}{|{\bf r}'(t)|}=\frac{\cos t}{|{\bf r}'(t)|}$. Taking derivatives with respect to $t$, we get
$\displaystyle -\sin\omega\cdot\frac{d\omega}{dt}=-\frac{a^2\sin t}{|{\bf r}'(t)|^3}$. Then $(\star)$ is equivalent to

$\displaystyle \frac{a}{|{\bf r}'(t)|^{3}}\cdot |{\bf r}'(t)|=\frac{a^2\sin t}{\sin\omega\cdot |{\bf r}'(t)|^3}$, or
$\displaystyle \sin\omega\cdot |{\bf r}'(t)|=a\sin t$. Note that $\displaystyle \sin^2\omega=1-\cos^2\omega=1-\frac{\cos^2 t}{|{\bf r}'(t)|^2}$. Therefore $\displaystyle \sin^2\omega\cdot |{\bf r}'(t)|^2= |{\bf r}'(t)|^2-\cos^2 t=a^2\sin^2 t$, which completes the verification.