Complex-Number Representation of the Curvature of Plane Curves

平面曲线的曲率的复数表示 苏剑林 · 2014-03-04 · Mathematics curvature complex number analysis differential geometry

The second week of the semester has already begun, and my Differential Geometry course has been running for two weeks as well. The pace is fairly slow — we've only just gotten to the curvature of plane curves. At any point on a plane curve $\boldsymbol{t}(t)=(x(t),y(t))$ we can pick out a unit tangent vector.

$$\boldsymbol{t}=\left(\frac{dx}{ds},\frac{dy}{ds}\right)$$

where $ds^2 =dx^2+dy^2$. Rotating this vector counterclockwise by 90 degrees, we can then define the corresponding unit normal vector $\boldsymbol{n}$, namely $\boldsymbol{t}\cdot\boldsymbol{n}=0$.

The conventional approach

Let us use the arc length $s$ as the parameter describing the curve equation, $\boldsymbol{t}(s)=(x(s),y(s))$, where a dot over a function denotes differentiation with respect to $s$. Consider $\dot{\boldsymbol{t}}$; since $\boldsymbol{t}^2=1$, differentiating with respect to s gives

$$\boldsymbol{t}\cdot\dot{\boldsymbol{t}}=0$$

That is to say, $\dot{\boldsymbol{t}}$ is perpendicular to $\boldsymbol{t}$; and since we're working purely in the plane, $\dot{\boldsymbol{t}}$ must be parallel to $\boldsymbol{n}$. That is,

$$\dot{\boldsymbol{t}}=\kappa \boldsymbol{n}$$

Similarly, $\dot{\boldsymbol{n}}$ is parallel to $\boldsymbol{t}$. Differentiating $\boldsymbol{t}\cdot\boldsymbol{n}=0$ then gives

$$\dot{\boldsymbol{t}}\cdot\boldsymbol{n}+\boldsymbol{t}\cdot\dot{\boldsymbol{n}}=0$$

Substituting $\dot{\boldsymbol{t}}=\kappa \boldsymbol{n}$ into the above yields

$$\dot{\boldsymbol{n}}=-\kappa \boldsymbol{t}$$

$\kappa$ is called the curvature of the curve at that point.

Complex-number representation

The above is the standard textbook treatment, but in fact the most convenient tool for studying plane curves is still complex numbers. If we represent $\boldsymbol{r}(s)$ by a parametrized complex number $z(s)$, then the two equations above can be combined into one, much more compact, expression:

$$\ddot{z}(s)=i\kappa (s) \dot{z}(s) $$

The advantage of writing things this way is that, given an arbitrary curvature function $\kappa (s) $, we can immediately recover the corresponding curve:

$$z(s)=\int e^{i\int \kappa (s)ds}ds $$

This is both concise and effective.

Furthermore, let us set $dz=ds e^{i\phi}$; then

$$\dot{z}=e^{i\phi}$$

and naturally

$$\ddot{z}=e^{i\phi}\left(i\dot{\phi}\right)$$

so the curvature can be expressed as

$$\kappa=\dot{\phi}$$

Various coordinate systems

This representation makes it very easy to derive the curvature expressions in various coordinate systems. For instance, if the curve is given by a general parametric equation $(x(t),y(t))$, and we denote differentiation with respect to t by a prime, so that $ds=\sqrt{x'(t）^2+y'(t)^2}dt,\phi=\arctan\left(\frac{y'(t)}{x'(t)}\right)$, then

$$\frac{d\phi}{ds}=\frac{\frac{y''(t)}{x'(t)}-\frac{y'(t)x''(t)}{[x'(t)]^2}}{1+\left(\frac{y'(t)}{x'(t)}\right)^2}\div \left(\frac{ds}{dt}\right)$$

Substituting and simplifying readily gives

$$\kappa=\frac{y''(t) x'(t)-x''(t) y'(t)}{[x'(t)^2+y'(t)^2]^{3/2}}$$

In polar coordinates, let $r=f(\theta)$, so that $z=f(\theta)e^{i\theta}$; then

$$dz=\left(\frac{d f}{d \theta}+i f\right)e^{i\theta}d\theta$$

hence

$$ds=\sqrt{f^2+\left(\frac{d f}{d \theta}\right)^2}d\theta$$

and since $\phi=\arctan\frac{f}{\left(\frac{d f}{d \theta}\right)}+\theta$, we get

$$\frac{d\phi}{ds}=\left[\frac{1-\left(\frac{d^2 f}{d \theta^2}\right) f/\left(\frac{d f}{d \theta}\right)^2}{1+f^2/\left(\frac{d f}{d \theta}\right)^2}+1\right]\div \left(\frac{d s}{d \theta}\right)$$

Substituting and simplifying gives

$$\kappa=\frac{2\left(\frac{d f}{d \theta}\right)^2+f^2-\left(\frac{d^2 f}{d \theta^2}\right)f}{\left[\left(\frac{d f}{d \theta}\right)^2+f^2\right]^{3/2}}$$

Is there an equally convenient tool for three-dimensional space? I'm still pondering that one myself ^_^