问题及解答

若曲线 $c(s)$ 是正则曲线, 且以弧长为参数, 证明

\[\kappa(s)=|\ddot{c}(s)|,\quad\tau(s)=\frac{\det(\dot{c}(s),\ddot{c}(s),\dddot{c}(s))}{\kappa^2(s)}.\]

Frenet 标架为

\[\vec{v}(s)=e_1(s)=\dot{c}(s),\quad\vec{n}(s)=e_2(s)=\frac{\ddot{c}(s)}{|\ddot{c}(s)|},\quad\vec{b}(s)=e_3(s)=\frac{\dot{c}(s)\times\ddot{c}(s)}{|\ddot{c}(s)|}.\]

若曲线 $c(t)$ 以一般参数表示, 证明

\[\kappa(t)=\frac{|\dot{c}(t)\times\ddot{c}(t)|}{|\dot{c}(t)|^3},\quad\tau(t)=\frac{\det(\dot{c}(t),\ddot{c}(t),\dddot{c}(t))}{|\dot{c}(t)\times\ddot{c}(t)|^2}.\]

Frenet 标架为

\[
\begin{aligned}
\vec{v}(t)&=e_1(t)=\frac{\dot{c}(t)}{|\dot{c}(t)|},\\
\vec{n}(t)&=e_2(t)=\frac{(\dot{c}(t)\times\ddot{c}(t))\times\dot{c}(t)}{|(\dot{c}(t)\times\ddot{c}(t))\times\dot{c}(t)|},\\
\vec{b}(t)&=e_3(t)=\frac{\dot{c}(t)\times\ddot{c}(t)}{|\dot{c}(t)\times\ddot{c}(t)|}.
\end{aligned}
\]

并证明不论用什么参数表示曲线, 曲线的曲率和挠率是不依赖于参数的. 即有

\[\kappa(s)=\kappa(t),\qquad\tau(s)=\tau(t).\]

不但如此, Frenet 标架也是一致的. 即有

\[e_i(s)=e_i(t),\qquad i=1,2,3.\]

第 $i$ 个曲率为 $K_i(t):=\frac{\omega_{i,i+1}(t)}{|\dot{c}(t)|}$.

这里不妨设曲线以弧长为参数. $c:\ I\rightarrow\mathbb{R}^3$, $s\mapsto c(s)$, $s$ 为弧长参数.

则 $e_1(s)=\dot{c}(s)$,

\[\widetilde{e}_2(s)=\ddot{c}(s)-(\ddot{c}(s)\cdot e_1(s))e_1(s)=\ddot{c}(s)\]

上面用到了 $|\dot{c}(s)|=1$, 从而 $\langle\ddot{c}(s),\dot{c}(s)\rangle=0$. 因此 $e_2(s)=\frac{\ddot{c}(s)}{|\ddot{c}(s)|}$.

\[\widetilde{e}_3(s)=\dddot{c}(s)-(\dddot{c}(s)\cdot e_1(s))e_1(s)-(\dddot{c}(s)\cdot e_2(s))e_2(s)\]

这个太复杂了, 不妨从另一个观点看 $e_3(s)$, 有

\[e_3(s)=e_1(s)\times e_2(s)=\frac{\dot{c}(s)\times\ddot{c}(s)}{|\ddot{c}(s)|}.\]

因此

\[\kappa(s)=K_1(s)=\frac{\omega_{12}(s)}{|\dot{c}(s)|}=\omega_{12}(s)=\dot{e}_1(s)\cdot e_2(s)=\ddot{c}(s)\cdot\frac{\ddot{c}(s)}{|\ddot{c}(s)|}=|\ddot{c}(s)|\]

又 $e_2(s)=\frac{\ddot{c}(s)}{|\ddot{c}(s)|}$, 故 $\ddot{c}(s)=\kappa(s)e_2(s)$. 对此式两边求导, 利用 Frenet 公式 得

\[
\begin{split}
\dddot{c}(s)&=\dot{\kappa}(s)e_2(s)+\kappa(s)\dot{e}_2(s)\\
&=\dot{\kappa}(s)e_2(s)+\kappa(s)[-\kappa(s)e_1(s)+\tau(s)e_3(s)]\\
&=\dot{\kappa}(s)e_2(s)-\kappa^2(s)e_1(s)+\kappa(s)\tau(s)e_3(s)
\end{split}
\]

将

\[e_1(s)=\dot{c}(s),\quad e_2(s)=\frac{\ddot{c}(s)}{|\ddot{c}(s)|},\quad e_3(s)=\frac{\dot{c}(s)\times\ddot{c}(s)}{|\ddot{c}(s)|},\quad\kappa(s)=|\ddot{c}(s)|\]

代入上式, 注意 $|\ddot{c}(s)|^{.}=\frac{\langle\dddot{c}(s),\ddot{c}(s)\rangle}{|\ddot{c}(s)|}$, 得

\[
\begin{split}
\dddot{c}(s)&=\frac{\langle\dddot{c}(s),\ddot{c}(s)\rangle}{|\ddot{c}(s)|}\frac{\ddot{c}(s)}{|\ddot{c}(s)|}-|\ddot{c}(s)|^2\dot{c}(s)+|\ddot{c}(s)|\tau(s)\frac{\dot{c}(s)\times\ddot{c}(s)}{|\ddot{c}(s)|}\\
&=\frac{\langle\dddot{c}(s),\ddot{c}(s)\rangle}{|\ddot{c}(s)|^2}\ddot{c}(s)-|\ddot{c}(s)|^2\dot{c}(s)+(\dot{c}(s)\times\ddot{c}(s))\tau(s),
\end{split}
\]

从而

\[
(\dot{c}(s)\times\ddot{c}(s))\tau(s)=\dddot{c}(s)-\frac{\langle\dddot{c}(s),\ddot{c}(s)\rangle}{|\ddot{c}(s)|^2}\ddot{c}(s)+|\ddot{c}(s)|^2\dot{c}(s).\tag{$*$}
\]

对该式两边右叉乘 $\dot{c}(s)$ (如果两边与 $\dddot{c}(s)$ 作内积, 则见最后的注), 得

\[
\bigl((\dot{c}(s)\times\ddot{c}(s))\times\dot{c}(s)\bigr)\tau(s)=\dddot{c}(s)\times\dot{c}(s)-\frac{\langle\dddot{c}(s),\ddot{c}(s)\rangle}{|\ddot{c}(s)|^2}\ddot{c}(s)\times\dot{c}(s)+0.
\]

易见 $(\dot{c}(s)\times\ddot{c}(s))\times\dot{c}(s)=\ddot{c}(s)$, 因此上式为

\[
\ddot{c}(s)\tau(s)=\dddot{c}(s)\times\dot{c}(s)+\frac{\langle\dddot{c}(s),\ddot{c}(s)\rangle}{|\ddot{c}(s)|^2}\dot{c}(s)\times\ddot{c}(s)
\]

两边再与 $\ddot{c}(s)$ 作内积, 即得

\[
|\ddot{c}(s)|^2\tau(s)=\det(\dddot{c}(s),\dot{c}(s),\ddot{c}(s))+\frac{\langle\dddot{c}(s),\ddot{c}(s)\rangle}{|\ddot{c}(s)|^2}\langle\dot{c}(s)\times\ddot{c}(s),\ddot{c}(s)\rangle=\det(\dot{c}(s),\ddot{c}(s),\dddot{c}(s)),
\]

因此

\[
\tau(s)=\frac{\det(\dot{c}(s),\ddot{c}(s),\dddot{c}(s))}{\kappa^2(s)}.
\]

注.

若 ($*$) 式两边与 $\dddot{c}(s)$ 作内积, 则得

\[
\det(\dot{c}(s),\ddot{c}(s),\dddot{c}(s))\tau(s)=\frac{|\dddot{c}(s)\times\ddot{c}(s)|^2}{|\ddot{c}(s)|^2}-|\ddot{c}(s)|^4
\]

从而得到恒等式

\[\bigl(\det(\dot{c}(s),\ddot{c}(s),\dddot{c}(s))\bigr)^2=|\dddot{c}(s)\times\ddot{c}(s)|^2-|\ddot{c}(s)|^6.\]

在上一回答中, 我们得到

\[
(\dot{c}(s)\times\ddot{c}(s))\tau(s)=\dddot{c}(s)-\frac{\langle\dddot{c}(s),\ddot{c}(s)\rangle}{|\ddot{c}(s)|^2}\ddot{c}(s)+|\ddot{c}(s)|^2\dot{c}(s).
\]

两边与 $\dddot{c}(s)$ 作内积, 得

\[
\bigl((\dot{c}(s)\times\ddot{c}(s))\cdot\dddot{c}(s)\bigr)\tau(s)=|\dddot{c}(s)|^2-\frac{\langle\dddot{c}(s),\ddot{c}(s)\rangle^2}{|\ddot{c}(s)|^2}+|\ddot{c}(s)|^2\langle\dot{c}(s),\dddot{c}(s)\rangle.
\]

注意 $\dot{c}(s)\perp\ddot{c}(s)$, 因此

\[
0=\frac{d}{ds}\langle\dot{c}(s),\ddot{c}(s)\rangle=\langle\ddot{c}(s),\ddot{c}(s)\rangle+\langle\dot{c}(s),\dddot{c}(s)\rangle
\]

这推出 $\langle\dot{c}(s),\dddot{c}(s)\rangle=-|\ddot{c}(s)|^2$, 因此

\[
\begin{split}
\det(\dot{c}(s),\ddot{c}(s),\dddot{c}(s))\tau(s)&=\frac{|\dddot{c}(s)|^2|\ddot{c}(s)|^2-\langle\dddot{c}(s),\ddot{c}(s)\rangle^2}{|\ddot{c}(s)|^2}-|\ddot{c}(s)|^4\\
&=\frac{|\dddot{c}(s)\times\ddot{c}(s)|^2}{|\ddot{c}(s)|^2}-|\ddot{c}(s)|^4
\end{split}
\]

结合之前关于挠率的公式:

\[\tau(s)=\frac{\det(\dot{c}(s),\ddot{c}(s),\dddot{c}(s))}{|\ddot{c}(s)|^2},\]

得

\[\bigl(\det(\dot{c}(s),\ddot{c}(s),\dddot{c}(s))\bigr)^2=|\dddot{c}(s)\times\ddot{c}(s)|^2-|\ddot{c}(s)|^6.\]

对于一般参数 $t$, $e_1(t)=\frac{\dot{c}(t)}{|\dot{c}(t)|}$,

\[
\widetilde{e}_2(t)=\ddot{c}(t)-(\ddot{c}(t)\cdot e_1(t))e_1(t)=\ddot{c}(t)-\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle}{|\dot{c}(t)|^2}\dot{c}(t)
\]

\[
\begin{split}
|\widetilde{e}_2(t)|^2&=\biggl|\ddot{c}(t)-\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle}{|\dot{c}(t)|^2}\dot{c}(t)\biggr|^2\\
&=|\ddot{c}(t)|^2-2\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle}{|\dot{c}(t)|^2}\langle\ddot{c}(t),\dot{c}(t)\rangle+\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle^2}{|\dot{c}(t)|^4}|\dot{c}(t)|^2\\
&=|\ddot{c}(t)|^2-\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle^2}{|\dot{c}(t)|^2}\\
&=\frac{|\ddot{c}(t)|^2|\dot{c}(t)|^2-\langle\ddot{c}(t),\dot{c}(t)\rangle^2}{|\dot{c}(t)|^2}\\
&=\frac{|\ddot{c}(t)\times\dot{c}(t)|^2}{|\dot{c}(t)|^2}
\end{split}
\]

因此,

\[
|\widetilde{e}_2(t)|=\frac{|\ddot{c}(t)\times\dot{c}(t)|}{|\dot{c}(t)|.}
\]

又

\[
\begin{split}
\dot{e}_1(t)&=\frac{\ddot{c}(t)|\dot{c}(t)|-\dot{c}(t)(|\dot{c}(t)|)\'_t}{|\dot{c}(t)|^2}\\
&=\frac{\ddot{c}(t)|\dot{c}(t)|-\dot{c}(t)\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle}{|\dot{c}(t)|}}{|\dot{c}(t)|^2}\\
&=\frac{\ddot{c}(t)|\dot{c}(t)|^2-\dot{c}(t)\langle\ddot{c}(t),\dot{c}(t)\rangle}{|\dot{c}(t)|^3},
\end{split}
\]

因此,

\[
\begin{split}
\kappa(t)&=\frac{\dot{e}_1(t)\cdot e_2(t)}{|\dot{c}(t)|}\\
&=\frac{1}{|\dot{c}(t)|}\biggl[\frac{\ddot{c}(t)|\dot{c}(t)|^2-\dot{c}(t)\langle\ddot{c}(t),\dot{c}(t)\rangle}{|\dot{c}(t)|^3}\cdot\frac{\ddot{c}(t)-\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle}{|\dot{c}(t)|^2}\dot{c}(t)}{\frac{|\ddot{c}(t)\times\dot{c}(t)|}{|\dot{c}(t)|}}\biggr]\\
&=\frac{1}{|\dot{c}(t)|^3|\ddot{c}(t)\times\dot{c}(t)|}\biggl[|\ddot{c}(t)|^2|\dot{c}(t)|^2-\langle\ddot{c}(t),\dot{c}(t)\rangle^2\biggr]\\
&=\frac{|\ddot{c}(t)\times\dot{c}(t)|^2}{|\dot{c}(t)|^3|\ddot{c}(t)\times\dot{c}(t)|}\\
&=\frac{|\ddot{c}(t)\times\dot{c}(t)|}{|\dot{c}(t)|^3}.
\end{split}
\]

对于一般参数 $t$, $\tau(t)=K_2(t)=\frac{\omega_{23}(t)}{|\dot{c}(t)|}$, $\omega_{23}(t)=\dot{e}_2(t)\cdot e_3(t)$. 下面详细计算这些量.

\[
e_2(t)=\frac{\ddot{c}(t)-\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle}{|\dot{c}(t)|^2}\dot{c}(t)}{\frac{|\ddot{c}(t)\times\dot{c}(t)|}{|\dot{c}(t)|}}=\frac{|\dot{c}(t)|}{|\ddot{c}(t)\times\dot{c}(t)|}\ddot{c}(t)-\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle}{|\dot{c}(t)||\ddot{c}(t)\times\dot{c}(t)|}\dot{c}(t),
\]

因此

\[
\begin{split}
\dot{e}_2(t)&=\biggl(\frac{|\dot{c}(t)|}{|\ddot{c}(t)\times\dot{c}(t)|}\biggr)\'_t \ddot{c}(t)+\frac{|\dot{c}(t)|}{|\ddot{c}(t)\times\dot{c}(t)|}\dddot{c}(t)\\
&\quad -\biggl[\biggl(\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle}{|\dot{c}(t)||\ddot{c}(t)\times\dot{c}(t)|}\biggr)\'_t\dot{c}(t)+\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle}{|\dot{c}(t)||\ddot{c}(t)\times\dot{c}(t)|}\ddot{c}(t)\biggr]
\end{split}
\]

其中

\[
\begin{split}
\biggl(\frac{|\dot{c}(t)|}{|\ddot{c}(t)\times\dot{c}(t)|}\biggr)\'_t &=\frac{|\ddot{c}(t)|\'_t |\ddot{c}(t)\times\dot{c}(t)|-|\dot{c}(t)|(|\ddot{c}(t)\times\dot{c}(t)|)\'_t}{|\ddot{c}(t)\times\dot{c}(t)|^2}\\
&=\frac{1}{|\ddot{c}(t)\times\dot{c}(t)|^2}\biggl[\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle}{|\dot{c}(t)|}|\ddot{c}(t)\times\dot{c}(t)|-|\dot{c}(t)|\frac{\langle(\ddot{c}(t)\times\dot{c}(t))\'_t,\ddot{c}(t)\times\dot{c}(t)\rangle}{|\ddot{c}(t)\times\dot{c}(t)|}\biggr]\\
&=\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle}{|\dot{c}(t)||\ddot{c}(t)\times\dot{c}(t)|}-\frac{|\dot{c}(t)|}{|\ddot{c}(t)\times\dot{c}(t)|^3}\bigl\langle\dddot{c}(t)\times\dot{c}(t),\ddot{c}(t)\times\dot{c}(t)\bigr\rangle
\end{split}
\]

于是,

\[
\begin{split}
\dot{e}_2(t)&=-\biggl(\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle}{|\dot{c}(t)||\ddot{c}(t)\times\dot{c}(t)|}\biggr)\'_t\dot{c}(t)\\
&\quad -\frac{|\dot{c}(t)|}{|\ddot{c}(t)\times\dot{c}(t)|^3}\bigl\langle\dddot{c}(t)\times\dot{c}(t),\ddot{c}(t)\times\dot{c}(t)\bigr\rangle\ddot{c}(t)\\
&\quad +\frac{|\dot{c}(t)|}{|\ddot{c}(t)\times\dot{c}(t)|}\dddot{c}(t)
\end{split}
\]

其中

\[
\begin{split}
&\biggl(\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle}{|\dot{c}(t)||\ddot{c}(t)\times\dot{c}(t)|}\biggr)\'_t\\
=&\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle \'_t|\dot{c}(t)||\ddot{c}(t)\times\dot{c}(t)|-\langle\ddot{c}(t),\dot{c}(t)\rangle(|\dot{c}(t)||\ddot{c}(t)\times\dot{c}(t)|)\'_t}{|\dot{c}(t)|^2|\ddot{c}(t)\times\dot{c}(t)|^2}\\
=&\frac{1}{|\dot{c}(t)|^2|\ddot{c}(t)\times\dot{c}(t)|^2}\biggl\{\bigl(\langle\dddot{c}(t),\dot{c}(t)\rangle+\langle\ddot{c}(t),\ddot{c}(t)\rangle\bigr)|\dot{c}(t)||\ddot{c}(t)\times\dot{c}(t)|\\
&-\langle\ddot{c}(t),\dot{c}(t)\rangle\biggl[\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle}{|\dot{c}(t)|}|\ddot{c}(t)\times\dot{c}(t)|+|\dot{c}(t)|\frac{\langle\dddot{c}(t)\times\dot{c}(t),\ddot{c}(t)\times\dot{c}(t)\rangle}{|\ddot{c}(t)\times\dot{c}(t)|}\biggr]\biggr\}\\
=&\frac{1}{|\dot{c}(t)||\ddot{c}(t)\times\dot{c}(t)|}\biggl[\langle\dddot{c}(t),\dot{c}(t)\rangle+|\ddot{c}(t)|^2-\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle^2}{|\dot{c}(t)|^2}-\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle}{|\ddot{c}(t)\times\dot{c}(t)|^2}\langle\dddot{c}(t)\times\dot{c}(t),\ddot{c}(t)\times\dot{c}(t)\rangle\biggr]\\
=&\frac{1}{|\dot{c}(t)||\ddot{c}(t)\times\dot{c}(t)|}\biggl[\langle\dddot{c}(t),\dot{c}(t)\rangle+\frac{|\ddot{c}(t)\times\dot{c}(t)|^2}{|\dot{c}(t)|^2}-\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle}{|\ddot{c}(t)\times\dot{c}(t)|^2}\Bigl[\langle\dddot{c}(t),\ddot{c}(t)\rangle|\dot{c}(t)|^2-\langle\dddot{c}(t),\dot{c}(t)\rangle\langle\ddot{c}(t),\dot{c}(t)\rangle\Bigr]\biggr]\\
=&\frac{1}{|\dot{c}(t)||\ddot{c}(t)\times\dot{c}(t)|}\biggl[\frac{|\ddot{c}(t)\times\dot{c}(t)|^2}{|\dot{c}(t)|^2}+\langle\dddot{c}(t),\dot{c}(t)\rangle+\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle^2}{|\ddot{c}(t)\times\dot{c}(t)|^2}\langle\dddot{c}(t),\dot{c}(t)\rangle-\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle}{|\ddot{c}(t)\times\dot{c}(t)|^2}\langle\dddot{c}(t),\ddot{c}(t)\rangle|\dot{c}(t)|^2\biggr]\\
=&\frac{1}{|\dot{c}(t)||\ddot{c}(t)\times\dot{c}(t)|}\biggl[\frac{|\ddot{c}(t)\times\dot{c}(t)|^2}{|\dot{c}(t)|^2}+\langle\dddot{c}(t),\dot{c}(t)\rangle\frac{|\ddot{c}(t)|^2|\dot{c}(t)|^2}{|\ddot{c}(t)\times\dot{c}(t)|^2}-\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle\langle\dddot{c}(t),\ddot{c}(t)\rangle}{|\ddot{c}(t)\times\dot{c}(t)|^2}|\dot{c}(t)|^2\biggr]\\
=&\frac{1}{|\dot{c}(t)||\ddot{c}(t)\times\dot{c}(t)|}\biggl[\frac{|\ddot{c}(t)\times\dot{c}(t)|^2}{|\dot{c}(t)|^2}+\frac{|\dot{c}(t)|^2}{|\ddot{c}(t)\times\dot{c}(t)|^2}\Bigl(\langle\dddot{c}(t),\dot{c}(t)\rangle\langle\ddot{c}(t),\ddot{c}(t)\rangle-\langle\dddot{c}(t),\ddot{c}(t)\rangle\langle\ddot{c}(t),\dot{c}(t)\rangle\Bigr)\biggr]\\
=&\frac{1}{|\dot{c}(t)||\ddot{c}(t)\times\dot{c}(t)|}\biggl[\frac{|\ddot{c}(t)\times\dot{c}(t)|^2}{|\dot{c}(t)|^2}+\frac{|\dot{c}(t)|^2}{|\ddot{c}(t)\times\dot{c}(t)|^2}\bigl\langle\dddot{c}(t)\times\ddot{c}(t),\dot{c}(t)\times\ddot{c}(t)\bigr\rangle\biggr]
\end{split}
\]

我们证明曲线的曲率和挠率与参数表示无关. 即有 $\kappa(s)=\kappa(t)$, $\tau(s)=\tau(t)$. 这里 $s$ 是弧长参数, $t$ 是一般参数.

考虑曲线 $c:[0,\ell]\rightarrow\mathbb{R}^3$, $s\mapsto c(s)$, $\ell=\text{length}(c)$. 设 $s:[a,b]\rightarrow[0,\ell]$ 是一参数变换, $t\mapsto s(t)$. 因此有复合映射 $t\mapsto c(s(t))$. 简记 $\dot{c}(t)=\frac{d}{dt}c(s(t))$, $\dot{c}(s)=\frac{d}{ds}c(s)$. 因此,

\[\dot{c}(t)=\dot{c}(s)\cdot\frac{ds}{dt}=\dot{c}(s)|\dot{c}(t)|.\]

即 $\dot{c}(s)=\frac{\dot{c}(t)}{|\dot{c}(t)|}$.

下面证明:

\[\ddot{c}(s)=\frac{\ddot{c}(t)}{|\dot{c}(t)|^2}-\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle}{|\dot{c}(t)|^4}\dot{c}(t),\]

\[
\dddot{c}(s)=\frac{\dddot{c}(t)}{|\dot{c}(t)|^3}-\frac{3\langle\ddot{c}(t),\dot{c}(t)\rangle}{|\dot{c}(t)|^5}\ddot{c}(t)-\biggl[\frac{\langle\dddot{c}(t),\dot{c}(t)\rangle+|\ddot{c}(t)|^2}{|\dot{c}(t)|^5}-\frac{4\langle\ddot{c}(t),\dot{c}(t)\rangle^2}{|\dot{c}(t)|^7}\biggr]\dot{c}(t)
\]

计算过程:

\[
\begin{split}
\ddot{c}(s)&=\biggl(\frac{\dot{c}(t)}{|\dot{c}(t)|}\biggr)\'_t\cdot\frac{dt}{ds}\\
&=\frac{\ddot{c}(t)|\dot{c}(t)|-\dot{c}(t)(|\dot{c}(t)|)\'_t}{|\dot{c}(t)|^2}\cdot\frac{1}{\frac{ds}{dt}}\\
&=\frac{\ddot{c}(t)|\dot{c}(t)|-\dot{c}(t)\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle}{|\dot{c}(t)|}}{|\dot{c}(t)|^3}\\
&=\frac{\ddot{c}(t)}{|\dot{c}(t)|^2}-\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle}{|\dot{c}(t)|^4}\dot{c}(t).
\end{split}
\]

\[
\begin{split}
\dddot{c}(s)&=\biggl(\frac{\ddot{c}(t)}{|\dot{c}(t)|^2}\biggr)\'_t\cdot\frac{1}{|\dot{c}(t)|}-\biggl(\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle}{|\dot{c}(t)|^4}\dot{c}(t)\biggr)\'_t\cdot\frac{1}{|\dot{c}(t)|}\\
&=\frac{\dddot{c}(t)|\dot{c}(t)|^2-\ddot{c}(t)(|\dot{c}(t)|^2)\'_t}{|\dot{c}(t)|^4}\cdot\frac{1}{|\dot{c}(t)|}-\biggl[\biggl(\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle}{|\dot{c}(t)|^4}\biggr)\'_t\cdot\dot{c}(t)+\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle}{|\dot{c}(t)|^4}\cdot\ddot{c}(t)\biggr]\cdot\frac{1}{|\dot{c}(t)|}\\
&=\frac{\dddot{c}(t)|\dot{c}(t)|^2-\ddot{c}(t)2\langle\ddot{c}(t),\dot{c}(t)\rangle}{|\dot{c}(t)|^5}-\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle}{|\dot{c}(t)|^5}\ddot{c}(t)\\
&\quad-\frac{1}{|\dot{c}(t)|}\frac{\bigl(\langle\dddot{c}(t),\dot{c}(t)\rangle+|\ddot{c}(t)|^2\bigr)|\dot{c}(t)|^4-\langle\ddot{c}(t),\dot{c}(t)\rangle 4|\dot{c}(t)|^3\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle}{|\dot{c}(t)|}}{|\dot{c}(t)|^8}\cdot\dot{c}(t)\\
&=\frac{\dddot{c}(t)}{|\dot{c}(t)|^3}-\frac{3\langle\ddot{c}(t),\dot{c}(t)\rangle}{|\dot{c}(t)|^5}\ddot{c}(t)-\biggl[\frac{\langle\dddot{c}(t),\dot{c}(t)\rangle+|\ddot{c}(t)|^2}{|\dot{c}(t)|^5}-\frac{4\langle\ddot{c}(t),\dot{c}(t)\rangle^2}{|\dot{c}(t)|^7}\biggr]\dot{c}(t)
\end{split}
\]

因此,

\[
\dot{c}(s)\times\ddot{c}(s)=\frac{\dot{c}(t)}{|\dot{c}(t)|}\times\biggl(\frac{\ddot{c}(t)}{|\dot{c}(t)|^2}-\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle}{|\dot{c}(t)|^4}\dot{c}(t)\biggr)=\frac{\dot{c}(t)\times\ddot{c}(t)}{|\dot{c}(t)|^3}
\]

由于 $\dot{c}(s)\perp\ddot{c}(s)$, 故

\[\kappa(t)=\frac{|\dot{c}(t)\times\ddot{c}(t)|}{|\dot{c}(t)|^3}=|\dot{c}(s)\times\ddot{c}(s)|=|\ddot{c}(s)|=\kappa(s).\]

另外, 若记 $\dddot{c}(s)=\frac{\dddot{c}(t)}{|\dot{c}(t)|^3}+a\ddot{c}(t)+b\dot{c}(t)$, 则

\[
\begin{split}
\det(\dot{c}(s),\ddot{c}(s),\dddot{c}(s))&=\langle\dot{c}(s)\times\ddot{c}(s),\dddot{c}(s)\rangle\\
&=\langle\frac{\dot{c}(t)\times\ddot{c}(t)}{|\dot{c}(t)|^3},\frac{\dddot{c}(t)}{|\dot{c}(t)|^3}+a\ddot{c}(t)+b\dot{c}(t)\rangle\\
&=\frac{\langle\dot{c}(t)\times\ddot{c}(t),\dddot{c}(t)\rangle}{|\dot{c}(t)|^6},
\end{split}
\]

又

\[
\begin{split}
|\ddot{c}(s)|^2&=\biggl\langle\frac{\ddot{c}(t)}{|\dot{c}(t)|^2}-\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle}{|\dot{c}(t)|^4}\dot{c}(t),\frac{\ddot{c}(t)}{|\dot{c}(t)|^2}-\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle}{|\dot{c}(t)|^4}\dot{c}(t)\biggr\rangle\\
&=\frac{|\ddot{c}(t)|^2}{|\dot{c}(t)|^4}-2\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle^2}{|\dot{c}(t)|^6}+\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle^2}{|\dot{c}(t)|^8}|\dot{c}(t)|^2\\
&=\frac{|\ddot{c}(t)|^2}{|\dot{c}(t)|^4}-\frac{\langle\ddot{c}(t),\dot{c}(t)\rangle^2}{|\dot{c}(t)|^6}\\
&=\frac{|\ddot{c}(t)|^2|\dot{c}(t)|^2-\langle\ddot{c}(t),\dot{c}(t)\rangle^2}{|\dot{c}(t)|^6}\\
&=\frac{|\ddot{c}(t)\times\dot{c}(t)|^2}{|\dot{c}(t)|^6},
\end{split}
\]

因此,

\[
\begin{split}
\tau(s)&=\frac{\det(\dot{c}(s),\ddot{c}(s),\dddot{c}(s))}{\kappa^2(s)}\\
&=\frac{\langle\dot{c}(s)\times\ddot{c}(s),\dddot{c}(s)\rangle}{|\ddot{c}(s)|^2}\\
&=\frac{\frac{\langle\dot{c}(t)\times\ddot{c}(t),\dddot{c}(t)\rangle}{|\dot{c}(t)|^6}}{\frac{|\ddot{c}(t)\times\dot{c}(t)|^2}{|\dot{c}(t)|^6}}\\
&=\frac{\langle\dot{c}(t)\times\ddot{c}(t),\dddot{c}(t)\rangle}{|\ddot{c}(t)\times\dot{c}(t)|^2}\\
&=\tau(t).
\end{split}
\]

根据上面的计算,

\[e_1(s)=\dot{c}(s)=\frac{\dot{c}(t)}{|\dot{c}(t)|}=e_1(t),\]

\[e_3(s)=\frac{\dot{c}(s)\times\ddot{c}(s)}{|\ddot{c}(s)|}=\frac{\frac{\dot{c}(t)\times\ddot{c}(t)}{|\dot{c}(t)|^3}}{\frac{|\dot{c}(t)\times\ddot{c}(t)|}{|\dot{c}(t)|^3}}=\frac{\dot{c}(t)\times\ddot{c}(t)}{|\dot{c}(t)\times\ddot{c}(t)|}=e_3(t),\]

既然 $e_1(s)=e_1(t)$, $e_3(s)=e_2(t)$, 从而必有 $e_2(s)=e_2(t)$. 当然也可以直接验证. 由上面已算的结果, 有

\[\dot{c}(t)\times\ddot{c}(t)=|\dot{c}(t)|^3\dot{c}(s)\times\ddot{c}(s)\]

从而

\[
\begin{split}
(\dot{c}(t)\times\ddot{c}(t))\times\dot{c}(t)&=|\dot{c}(t)|^3(\dot{c}(s)\times\ddot{c}(s))\times|\dot{c}(t)|\dot{c}(s)\\
&=|\dot{c}(t)|^4(\dot{c}(s)\times\ddot{c}(s))\times\dot{c}(s)
\end{split}
\]

注意到 $\dot{c}(s)\perp\ddot{c}(s)$, 因此可设 $(\dot{c}(s)\times\ddot{c}(s))\times\dot{c}(s)=\alpha\ddot{c}(s)$, $\alpha > 0$. 又 $|(\dot{c}(s)\times\ddot{c}(s))\times\dot{c}(s)|=|\ddot{c}(s)|$, 故 $\alpha=1$, 即 $(\dot{c}(s)\times\ddot{c}(s))\times\dot{c}(s)=\ddot{c}(s)$. 因此

\[(\dot{c}(t)\times\ddot{c}(t))\times\dot{c}(t)=|\dot{c}(t)|^4\ddot{c}(s).\]

这推出

\[e_2(s)=\frac{\ddot{c}(s)}{|\ddot{c}(s)|}=\frac{(\dot{c}(t)\times\ddot{c}(t))\times\dot{c}(t)}{|(\dot{c}(t)\times\ddot{c}(t))\times\dot{c}(t)|}=e_3(t),\]

并且不妨验证一下正确性:

\[|(\dot{c}(t)\times\ddot{c}(t))\times\dot{c}(t)|=|\dot{c}(t)|^4\cdot|\ddot{c}(s)|=|\dot{c}(t)|^4\frac{|\ddot{c}(t)\times\dot{c}(t)|}{|\dot{c}(t)|^3}=|\dot{c}(t)|\cdot|\ddot{c}(t)\times\dot{c}(t)|.\]