问题及解答

曲线 $\Gamma:\ \gamma=\gamma(t)$ 的球面标线是指由曲线的切向量、主法线向量和副法线向量的末端点形成的这三种曲线, 由于它们均是单位向量, 故都位于单位球面上. 若 $s$ 是曲线 $\gamma$ 的弧长参数, 则它们是

切线球面标线 $\Gamma_1:\ \gamma_1=\vec{v}(s)=e_1(s)=\dot{\gamma}(s)$.

主法线球面标线 $\Gamma_2:\ \gamma_2=\vec{n}(s)=e_2(s)=\frac{\ddot{\gamma}(s)}{|\ddot{\gamma}(s)|}$.

副法线球面标线 $\Gamma_3:\ \gamma_3=\vec{b}(s)=e_3(s)=\frac{\dot{\gamma}(s)\times\ddot{\gamma}(s)}{|\ddot{\gamma}(s)|}$.

若 $s_1,s_2,s_3$ 分别记指 $\gamma_1$, $\gamma_2$, $\gamma_1$ 的弧长参数, 则有

\[ds_1=\kappa(s)ds,\quad ds_2=\sqrt{\kappa^2(s)+\tau^2(s)}ds,\quad ds_3=|\tau(s)|ds.\]

这三条曲线的曲率分别是

\[\kappa_1=\sqrt{1+(\frac{\tau}{\kappa})^2},\quad\kappa_2=\sqrt{1+\frac{(\kappa\dot{\tau}-\dot{\kappa}\tau)^2}{(\kappa^2+\tau^2)^3}},\quad\kappa_3=\sqrt{1+(\frac{\kappa}{\tau})^2}.\]

挠率分别是

\[\tau_1=\frac{\kappa\dot{\tau}-\dot{\kappa}\tau}{\kappa(\kappa^2+\tau^2)},\]

\[\tau_2=\frac{(\kappa^2+\tau^2)\frac{d}{ds}(\kappa\dot{\tau}-\dot{\kappa}\tau)-\frac{3}{2}(\kappa\dot{\tau}-\dot{\kappa}\tau)\frac{d}{ds}(\kappa^2+\tau^2)}{(\kappa^2+\tau^2)^3+(\kappa\dot{\tau}-\dot{\kappa}\tau)^2},\]

\[\tau_3=\frac{\kappa\dot{\tau}-\dot{\kappa}\tau}{\tau(\kappa^2+\tau^2)}.\]

我们仍用点号记对参数求导, 即 $\dot{e}_1(s_1)=\frac{d}{ds_1}e_1(s_1)$. 根据复合函数求导, 有

\[\dot{e}_1(s_1)=\dot{e}_1(s)\cdot\frac{ds}{ds_1}=\frac{\ddot{\gamma}(s)}{\frac{ds_1}{ds}}.\]

由于 $s_1$ 是曲线 $e_1(s_1)$ 的弧长参数, $|\dot{e}_1(s_1)|=1$, 因此 $\frac{ds_1}{ds}=|\ddot{\gamma}(s)|=\kappa(s)$, 即有 $ds_1=\kappa(s)ds$.

类似的,

\[\dot{e}_2(s_2)=\dot{e}_2(s)\cdot\frac{ds}{ds_2}=\frac{-\kappa(s)e_1(s)+\tau(s)e_3(s)}{\frac{ds_2}{ds}}.\]

由于 $|\dot{e}_2(s_2)|=1$, 故

\[\frac{ds_2}{ds}=|\dot{e}_2(s)|=|-\kappa(s)e_1(s)+\tau(s)e_3(s)|=\sqrt{\kappa^2+\tau^2},\]

即有

\[ds_2=\sqrt{\kappa^2+\tau^2}ds.\]

\[\dot{e}_3(s_3)=\dot{e}_3(s)\cdot\frac{ds}{ds_3}=\frac{-\tau(s)e_2(s)}{\frac{ds_3}{ds}}.\]

由于 $|\dot{e}_3(s_3)|=1$, 故

\[\frac{ds_3}{ds}=|\dot{e}_3(s)|=|-\tau(s)e_2(s)|=|\tau(s)|,\]

即有

\[ds_3=|\tau|ds.\]

计算曲线 $e_1(s_1)$ 的曲率, 由于 $s_1$ 是其弧长参数, 因此, 计算公式同以前一样. 即 $\kappa_1(s_1)=|\ddot{e}_1(s_1)|$.

根据 $\dot{e}_1(s_1)=\dot{e}_1(s)\frac{ds}{ds_1}$, 两边对 $s_1$ 求导, 得

\[
\begin{split}
\ddot{e}_1(s_1)&=\frac{d}{ds_1}\dot{e}_1(s)\cdot\frac{ds}{ds_1}+\dot{e}_1(s)\cdot\frac{d^2 s}{ds_1^2}\\
&=\ddot{e}_1(s)(\frac{ds}{ds_1})^2+\dot{e}_1(s)\frac{d^2 s}{ds_1^2},
\end{split}
\]

由 $ds_1=\kappa(s)ds$, 得 $\frac{ds}{ds_1}=\frac{1}{\kappa}$, 且

\[
\begin{split}
\frac{d^2 s}{ds_1^2}&=\frac{d}{ds_1}(\frac{1}{\kappa})\\
&=\frac{d}{ds}(\frac{1}{\kappa})\frac{ds}{ds_1}\\
&=-\frac{\dot{\kappa}(s)}{\kappa^2}\cdot\frac{1}{\kappa}\\
&=-\frac{\dot{\kappa}(s)}{\kappa^3}
\end{split}
\]

故

\[\ddot{e}_1(s_1)=\frac{1}{\kappa^2}\ddot{e}_1(s)-\frac{\dot{\kappa}(s)}{\kappa^3}\dot{e}_1(s).\]

又

\[
\begin{split}
|\ddot{e}_1(s_1)|^2&=\Bigl\langle\frac{1}{\kappa^2}\ddot{e}_1(s)-\frac{\dot{\kappa}(s)}{\kappa^3}\dot{e}_1(s),\frac{1}{\kappa^2}\ddot{e}_1(s)-\frac{\dot{\kappa}(s)}{\kappa^3}\dot{e}_1(s)\Bigr\rangle\\
&=\frac{1}{\kappa^4}|\ddot{e}_1(s)|^2-2\frac{\dot{\kappa}(s)}{\kappa^5}\langle\ddot{e}_1(s),\dot{e}_1(s)\rangle+\frac{{\dot{\kappa}(s)}^2}{\kappa^6}|\dot{e}_1(s)|^2
\end{split}
\]

下面计算其中这些项.