It is already the second week of the semester, and my “Differential Geometry” class has also been in session for two weeks. The progress is relatively slow, and we have only just reached the curvature of plane curves. At a certain point on a plane curve \bm{t}(t)=(x(t),y(t)), a unit tangent vector can be found: \bm{t}=\left(\frac{dx}{ds},\frac{dy}{ds}\right) where ds^2 = dx^2 + dy^2. After rotating this vector 90 degrees counter-clockwise, the corresponding unit normal vector \bm{n} can be defined, such that \bm{t} \cdot \bm{n} = 0.
Standard Notation
Let us use the arc length s as a parameter to describe the curve equation, \bm{t}(s)=(x(s),y(s)). A dot over a function denotes the derivative with respect to s. Now let us consider \dot{\bm{t}}. Since \bm{t}^2=1, differentiating with respect to s gives: \bm{t}\cdot\dot{\bm{t}}=0 That is to say, \dot{\bm{t}} is perpendicular to \bm{t}. Since we are only in a plane, \dot{\bm{t}} is parallel to \bm{n}. That is: \dot{\bm{t}}=\kappa \bm{n}
Similarly, \dot{\bm{n}} is parallel to \bm{t}. Furthermore, differentiating \bm{t}\cdot\bm{n}=0 yields: \dot{\bm{t}}\cdot\bm{n}+\bm{t}\cdot\dot{\bm{n}}=0 Substituting \dot{\bm{t}}=\kappa \bm{n} into the above equation gives: \dot{\bm{n}}=-\kappa \bm{t} \kappa is called the curvature of the curve at that point.
Complex Representation
The above is the standard textbook notation, but in fact, the most convenient tool for studying plane curves is complex numbers. Representing \bm{r}(s) with a parameterized complex number z(s), the two equations above can be written as a single, more concise equation: \ddot{z}(s)=i\kappa (s) \dot{z}(s)
The advantage of this notation is that, given any curvature function \kappa (s), we can find the corresponding curve: z(s)=\int e^{i\int \kappa (s)ds}ds This is concise and effective.
Additionally, let dz=ds e^{i\phi}, then: \dot{z}=e^{i\phi} Naturally: \ddot{z}=e^{i\phi}\left(i\dot{\phi}\right) Therefore, the curvature can be expressed as: \kappa=\dot{\phi}
Various Coordinate Systems
Using this, it is very convenient to derive curvature expressions in various coordinate systems. For example, when the curve is given by general parametric equations (x(t),y(t)), using a prime to denote the derivative with respect to t, we have ds=\sqrt{x'(t)^2+y'(t)^2}dt and \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, we easily obtain: \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), then z=f(\theta)e^{i\theta}, so: dz=\left(\frac{d f}{d \theta}+i f\right)e^{i\theta}d\theta Thus: 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, then: \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 something similarly convenient for three-dimensional space? I am also currently thinking about this ^_^
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