我们来证明
\[\dot{e}_1(t)\cdot e_2(t)=0\Rightarrow\dot{e}_1(t)=0\]
回忆, 标架场 $\{e_i(t)\}$ 是采用 Gram-Schmidt 正交化得来的. 因此 $e_1(t)=\frac{\dot{c}(t)}{|\dot{c}(t)|}$,
\[\widetilde{e}_2(t):=\ddot{c}(t)-(\ddot{c}(t)\cdot e_1(t))e_1(t),\]
而 $e_2(t)=\frac{\widetilde{e}_2(t)}{|\widetilde{e}_2(t)|}$. 因此由 $\dot{e}_1(t)\cdot e_2(t)=0$ 得
\[
\frac{\ddot{c}(t)|\dot{c}(t)|-\dot{c}(t)(|\dot{c}(t)|)^{.}}{|\dot{c}(t)|^2}\cdot\frac{\widetilde{e}_2(t)}{|\widetilde{e}_2(t)|}=0,
\]
这推出
\[
\begin{split}
0&=\biggl(\ddot{c}(t)|\dot{c}(t)|-\dot{c}(t)\frac{\langle\dot{c}(t),\ddot{c}(t)\rangle}{|\dot{c}(t)|}\biggr)\cdot\biggl(\ddot{c}(t)-\frac{\langle\dot{c}(t),\ddot{c}(t)\rangle}{|\dot{c}(t)|^2}\dot{c}(t)\biggr)\\
&=|\ddot{c}(t)|^2|\dot{c}(t)|-\frac{\langle\dot{c}(t),\ddot{c}(t)\rangle^2}{|\dot{c}(t)|}-\frac{\langle\dot{c}(t),\ddot{c}(t)\rangle^2}{|\dot{c}(t)|}+\frac{\langle\dot{c}(t),\ddot{c}(t)\rangle^2|\dot{c}(t)|^2}{|\dot{c}(t)|^3}\\
&=|\ddot{c}(t)|^2|\dot{c}(t)|-\frac{\langle\dot{c}(t),\ddot{c}(t)\rangle^2}{|\dot{c}(t)|},
\end{split}
\]
从而
\[|\ddot{c}(t)||\dot{c}(t)|=|\langle\dot{c}(t),\ddot{c}(t)\rangle|\]
注意到 $\dot{c}(t)\neq 0$, 故推出 $|\ddot{c}(t)|=|\ddot{c}(t)||\cos\alpha(t)|$, 其中 $\alpha(t)$ 指 $\ddot{c}(t)$ 与 $\dot{c}(t)$ 的夹角. 若 $\ddot{c}(t)\neq 0$, 则 $\ddot{c}(t)=k(t)\dot{c}(t)$. 即 $\dot{c}(t)$ 与 $\ddot{c}(t)$ 线性相关, 因此根据 $\widetilde{e}_2(t)$ 的定义, $e_2(t)\perp e_1(t)$. 而此时
\[
\begin{split}
\dot{e}_1(t)&=\frac{\ddot{c}(t)|\dot{c}(t)|-\dot{c}(t)\frac{\langle\dot{c}(t),\ddot{c}(t)\rangle}{|\dot{c}(t)|}}{|\dot{c}(t)|^2}\\
&=\frac{k(t)\dot{c}(t)|\dot{c}(t)|-\dot{c}(t)\frac{\langle\dot{c}(t),k(t)\dot{c}(t)\rangle}{|\dot{c}(t)|}}{|\dot{c}(t)|^2}\\
&\equiv 0
\end{split}
\]
注意, 此时不是通过下式来定义 $\widetilde{e}_2(t)$ 的.
\[
\begin{split}
\widetilde{e}_2(t)&=\ddot{c}(t)-(\ddot{c}(t)\cdot e_1(t))e_1(t)\\
&=k(t)\dot{c}(t)-(k(t)\dot{c}(t)\cdot\frac{\dot{c}(t)}{|\dot{c}(t)|})\frac{\dot{c}(t)}{|\dot{c}(t)|}\\
&=k(t)\dot{c}(t)-k(t)\dot{c}(t)=0,
\end{split}
\]
于是 $e_2(t)=0$.
变换参数, 使得 $\dot{c}(s)=e_1(t)$, 从而推出 $\ddot{c}(s)=\dot{e}_1(t)t\'(s)\equiv 0$, 故而推出 $c$ 是直线. 若仍从参数 $t$ 的角度, 则可认为 $c$ 在作变速直线运动, 故不一定有 $\ddot{c}(t)\equiv 0$.