曲率是微分几何研究的中心内容c : I → R n \boldsymbol c: I \to \mathbb R^n c : I → R n 弧长参数化下的正则曲线
# 曲率众所周知，一阶微分评价曲线的变化速度，通过已知曲线在任意一点的速度即可确定出曲线的路径，这使得一阶微分具有无可比拟的意义。在物理意义上等价于速度向量 c ′ ( s ) \boldsymbol c'(s) c ′ ( s ) 单位切向量场 (Unit Tangent Vector Field)「単位接ベクトル場」
t ( s ) : = c ′ ( s ) \boldsymbol t(s) := \boldsymbol c'(s)
t ( s ) : = c ′ ( s )
但是实际上仅有一阶微分是不够的，因为微分的过程会使得常数项消失，曲线发生偏移。因此需要切向量场 t \boldsymbol t t 加速度向量 c \boldsymbol c c t ( s ) \boldsymbol t(s) t ( s )
n 1 ( s ) : = t ′ ( s ) ∥ t ′ ( s ) ∥ \boldsymbol n_1(s) := \frac{\boldsymbol t'(s)}{\|\boldsymbol t'(s)\|}
n 1 ( s ) : = ∥ t ′ ( s ) ∥ t ′ ( s )
称 n 1 ( s ) \boldsymbol n_1(s) n 1 ( s ) c \boldsymbol c c 第 1 1 1 1 1 1 t ′ \boldsymbol t' t ′
κ 1 ( s ) : = ∥ t ′ ( s ) ∥ \kappa_1(s) := \|\boldsymbol t'(s)\|
κ 1 ( s ) : = ∥ t ′ ( s ) ∥
称 κ 1 ( s ) \kappa_1(s) κ 1 ( s ) c \boldsymbol c c 第 1 1 1 1 1 1
定义
O 1 ( s ) : = S p a n { t ( s ) , n 1 ( s ) } : = { λ 1 t ( s ) + λ 2 n 1 ( s ) ∣ λ 1 , λ 2 ∈ R } \mathcal O_1(s) := \mathrm{Span}\{\boldsymbol t(s), \boldsymbol n_1(s)\} := \{\lambda_1 \boldsymbol t(s) + \lambda_2 \boldsymbol n_1(s) \mid \lambda_1, \lambda_2 \in \mathbb R\}
O 1 ( s ) : = S p a n { t ( s ) , n 1 ( s ) } : = { λ 1 t ( s ) + λ 2 n 1 ( s ) ∣ λ 1 , λ 2 ∈ R }
为曲线 c \boldsymbol c c c ( s ) \boldsymbol c(s) c ( s ) 第一接触空间 (First Osculating Space)「第 1 1 1
命题 A 1 : = { t ( s ) , n 1 ( s ) } \mathscr A_1 := \{\boldsymbol t(s), \boldsymbol n_1(s)\} A 1 : = { t ( s ) , n 1 ( s ) } O 1 ( s ) \mathcal O_1(s) O 1 ( s )
证明
只需要证明二者正交。注意 ∥ t ( s ) ∥ = 1 \|\boldsymbol t(s)\| = 1 ∥ t ( s ) ∥ = 1 s s s
0 = d d s 1 = d d s ( t ( s ) ⋅ t ( s ) ) = 2 t ′ ( s ) ⋅ t ( s ) \begin{aligned}
0 = \frac{d}{ds} 1 &= \frac{d}{ds} (\boldsymbol t(s) \cdot \boldsymbol t(s)) \\
&= 2 \boldsymbol t'(s) \cdot \boldsymbol t(s)
\end{aligned} 0 = d s d 1 = d s d ( t ( s ) ⋅ t ( s ) ) = 2 t ′ ( s ) ⋅ t ( s )
因此 t ′ ( s ) ⋅ t ( s ) = 0 \boldsymbol t'(s) \cdot \boldsymbol t(s) = 0 t ′ ( s ) ⋅ t ( s ) = 0
n 1 ( s ) ⋅ t ( s ) = κ 1 ( s ) − 1 t ′ ( s ) ⋅ t ( s ) = 0 \boldsymbol n_1(s) \cdot \boldsymbol t(s) = \kappa_1(s)^{-1} \boldsymbol t'(s) \cdot \boldsymbol t(s) = 0
n 1 ( s ) ⋅ t ( s ) = κ 1 ( s ) − 1 t ′ ( s ) ⋅ t ( s ) = 0
□ \square □
O 1 ( s ) \mathcal O_1(s) O 1 ( s ) R n \mathbb R^n R n c \boldsymbol c c O 1 ( s ) \boldsymbol O_1(s) O 1 ( s )
于是，对上述过程进行推广，这次对第一法向量场进行微分分析。O 1 ( s ) \mathcal O_1(s) O 1 ( s ) O 1 ( s ) \mathcal O_1(s) O 1 ( s )
n 1 ′ ( s ) = α ( s ) t ( s ) + β ( s ) n 1 ( s ) ⏟ ∈ O 1 ( s ) + w ( s ) ⏟ ∈ O 1 ( s ) ⊥ , α ( s ) , β ( s ) ∈ R \boldsymbol n_1'(s) = \underbrace{\alpha(s) \boldsymbol t(s) + \beta(s) \boldsymbol n_1(s)}_{\in \mathcal O_1(s)} + \underbrace{\boldsymbol w(s)}_{\in \mathcal O_1(s)^\perp},\quad \alpha(s),\beta(s) \in \mathbb R
n 1 ′ ( s ) = ∈ O 1 ( s ) α ( s ) t ( s ) + β ( s ) n 1 ( s ) + ∈ O 1 ( s ) ⊥ w ( s ) , α ( s ) , β ( s ) ∈ R
为什么是对 n 1 \boldsymbol n_1 n 1
我们需要 w ( s ) \boldsymbol w(s) w ( s ) c \boldsymbol c c 第 2 2 2 2 2 2
n 2 ( s ) : = w ( s ) ∥ w ( s ) ∥ \boldsymbol n_2(s) := \frac{\boldsymbol w(s)}{\|\boldsymbol w(s)\|}
n 2 ( s ) : = ∥ w ( s ) ∥ w ( s )
称被移除的模长
κ 2 ( s ) : = ∥ w ( s ) ∥ \kappa_2(s) := \|\boldsymbol w(s)\|
κ 2 ( s ) : = ∥ w ( s ) ∥
为曲线 c \boldsymbol c c 第 2 2 2 2 2 2
尝试通过内积解出其他分量
α ( s ) = n 1 ′ ( s ) ⋅ t ( s ) = − n 1 ( s ) ⋅ t ′ ( s ) ⏟ d d s ( n 1 ⋅ t ) = 0 = − κ 1 ( s ) n 1 ( s ) ⋅ n 1 ( s ) = − κ 1 ( s ) β ( s ) = n 1 ′ ( s ) ⋅ n 1 ( s ) = 1 2 d d s ( n 1 ( s ) ⋅ n 1 ( s ) ) = 0 \begin{aligned}
\alpha(s) &= \boldsymbol n_1'(s) \cdot \boldsymbol t(s) \\
&= \underbrace{-\boldsymbol n_1(s) \cdot \boldsymbol t'(s)}_{\frac{d}{ds}(\boldsymbol n_1 \cdot \boldsymbol t) = 0} \\
&= -\kappa_1(s) \boldsymbol n_1(s) \cdot \boldsymbol n_1(s) = -\kappa_1(s) \\
\beta(s) &= \boldsymbol n_1'(s) \cdot \boldsymbol n_1(s) = \frac{1}{2} \frac{d}{ds} (\boldsymbol n_1(s) \cdot \boldsymbol n_1(s)) = 0
\end{aligned} α ( s ) β ( s ) = n 1 ′ ( s ) ⋅ t ( s ) = d s d ( n 1 ⋅ t ) = 0 − n 1 ( s ) ⋅ t ′ ( s ) = − κ 1 ( s ) n 1 ( s ) ⋅ n 1 ( s ) = − κ 1 ( s ) = n 1 ′ ( s ) ⋅ n 1 ( s ) = 2 1 d s d ( n 1 ( s ) ⋅ n 1 ( s ) ) = 0
那么，微分可以写为
n 1 ′ ( s ) = − κ 1 ( s ) t ( s ) + κ 2 ( s ) n 2 ( s ) \boldsymbol n_1'(s) = -\kappa_1(s) \boldsymbol t(s) + \kappa_2(s) \boldsymbol n_2(s)
n 1 ′ ( s ) = − κ 1 ( s ) t ( s ) + κ 2 ( s ) n 2 ( s )
因此可以评估的范围得到了扩张，定义曲线的 第二接触空间 (Second Osculating Space)「第 2 2 2
O 2 ( s ) : = S p a n { t ( s ) , n 1 ( s ) , n 2 ( s ) } \mathcal O_2(s) := \mathrm{Span}\{\boldsymbol t(s), \boldsymbol n_1(s), \boldsymbol n_2(s)\}
O 2 ( s ) : = S p a n { t ( s ) , n 1 ( s ) , n 2 ( s ) }
命题 A 2 : = { t ( s ) , n 1 ( s ) , n 2 ( s ) } \mathscr A_2 := \{\boldsymbol t(s), \boldsymbol n_1(s), \boldsymbol n_2(s)\} A 2 : = { t ( s ) , n 1 ( s ) , n 2 ( s ) } O 2 ( s ) \mathcal O_2(s) O 2 ( s )
证明
由于 n 2 ( s ) ∈ O 1 ( s ) ⊥ \boldsymbol n_2(s) \in \mathcal O_1(s)^\perp n 2 ( s ) ∈ O 1 ( s ) ⊥ t ( s ) , n 1 ( s ) \boldsymbol t(s), \boldsymbol n_1(s) t ( s ) , n 1 ( s )
□ \square □
同样地，对第二法向量场进行微分，得到
n 2 ′ ( s ) = γ 1 ( s ) t ( s ) + γ 2 ( s ) n 1 ( s ) + γ 3 ( s ) n 2 ( s ) ⏟ ∈ O 2 ( s ) + z ( s ) ⏟ ∈ O 2 ( s ) ⊥ , γ 1 ( s ) , γ 2 ( s ) , γ 3 ( s ) ∈ R \boldsymbol n_2'(s) = \underbrace{\gamma_1(s) \boldsymbol t(s) + \gamma_2(s) \boldsymbol n_1(s) + \gamma_3(s) \boldsymbol n_2(s)}_{\in \mathcal O_2(s)} + \underbrace{\boldsymbol z(s)}_{\in \mathcal O_2(s)^\perp}, \quad \gamma_1(s),\gamma_2(s),\gamma_3(s) \in \mathbb R
n 2 ′ ( s ) = ∈ O 2 ( s ) γ 1 ( s ) t ( s ) + γ 2 ( s ) n 1 ( s ) + γ 3 ( s ) n 2 ( s ) + ∈ O 2 ( s ) ⊥ z ( s ) , γ 1 ( s ) , γ 2 ( s ) , γ 3 ( s ) ∈ R
定义曲线 c \boldsymbol c c 第 3 3 3 3 3 3
n 3 ( s ) : = z ( s ) ∥ z ( s ) ∥ \boldsymbol n_3(s) := \frac{\boldsymbol z(s)}{\|\boldsymbol z(s)\|}
n 3 ( s ) : = ∥ z ( s ) ∥ z ( s )
定义曲线 c \boldsymbol c c 第 3 3 3 3 3 3
κ 3 ( s ) : = ∥ z ( s ) ∥ \kappa_3(s) := \|\boldsymbol z(s)\|
κ 3 ( s ) : = ∥ z ( s ) ∥
求解分量：
γ 1 ( s ) = n 2 ′ ( s ) ⋅ t ( s ) = − n 2 ( s ) ⋅ t ′ ( s ) = − κ 1 ( s ) n 2 ( s ) ⋅ n 1 ( s ) = 0 γ 2 ( s ) = n 2 ′ ( s ) ⋅ n 1 ( s ) = − n 2 ( s ) ⋅ n 1 ′ ( s ) = − κ 2 ( s ) n 2 ( s ) ⋅ n 2 ( s ) = − κ 2 ( s ) γ 3 ( s ) = n 2 ′ ( s ) ⋅ n 2 ( s ) = 1 2 d d s ( n 2 ( s ) ⋅ n 2 ( s ) ) = 0 \begin{aligned}
\gamma_1(s) &= \boldsymbol n_2'(s) \cdot \boldsymbol t(s) = -\boldsymbol n_2(s) \cdot \boldsymbol t'(s) = -\kappa_1(s) \boldsymbol n_2(s) \cdot \boldsymbol n_1(s) = 0 \\
\gamma_2(s) &= \boldsymbol n_2'(s) \cdot \boldsymbol n_1(s) = -\boldsymbol n_2(s) \cdot \boldsymbol n_1'(s) = -\kappa_2(s) \boldsymbol n_2(s) \cdot \boldsymbol n_2(s) = -\kappa_2(s) \\
\gamma_3(s) &= \boldsymbol n_2'(s) \cdot \boldsymbol n_2(s) = \frac{1}{2} \frac{d}{ds} (\boldsymbol n_2(s) \cdot \boldsymbol n_2(s)) = 0
\end{aligned} γ 1 ( s ) γ 2 ( s ) γ 3 ( s ) = n 2 ′ ( s ) ⋅ t ( s ) = − n 2 ( s ) ⋅ t ′ ( s ) = − κ 1 ( s ) n 2 ( s ) ⋅ n 1 ( s ) = 0 = n 2 ′ ( s ) ⋅ n 1 ( s ) = − n 2 ( s ) ⋅ n 1 ′ ( s ) = − κ 2 ( s ) n 2 ( s ) ⋅ n 2 ( s ) = − κ 2 ( s ) = n 2 ′ ( s ) ⋅ n 2 ( s ) = 2 1 d s d ( n 2 ( s ) ⋅ n 2 ( s ) ) = 0
那么改写微分为
n 2 ′ ( s ) = − κ 2 ( s ) n 1 ( s ) + κ 3 ( s ) n 3 ( s ) \boldsymbol n_2'(s) = -\kappa_2(s) \boldsymbol n_1(s) + \kappa_3(s) \boldsymbol n_3(s)
n 2 ′ ( s ) = − κ 2 ( s ) n 1 ( s ) + κ 3 ( s ) n 3 ( s )
构造出曲线的 第三接触空间 (Third Osculating Space)「第 3 3 3
O 3 ( s ) : = S p a n { t ( s ) , n 1 ( s ) , n 2 ( s ) , n 3 ( s ) } \mathcal O_3(s) := \mathrm{Span}\{\boldsymbol t(s), \boldsymbol n_1(s), \boldsymbol n_2(s), \boldsymbol n_3(s)\}
O 3 ( s ) : = S p a n { t ( s ) , n 1 ( s ) , n 2 ( s ) , n 3 ( s ) }
以此类推，可以得到
n i ′ ( s ) = − κ i ( s ) n i − 1 ( s ) + κ i + 1 ( s ) n i + 1 ( s ) \boldsymbol n_i'(s) = -\kappa_i(s) \boldsymbol n_{i-1}(s) + \kappa_{i+1}(s) \boldsymbol n_{i+1}(s)
n i ′ ( s ) = − κ i ( s ) n i − 1 ( s ) + κ i + 1 ( s ) n i + 1 ( s )
将上述结果汇总，可以得到曲线的 Frenet-Serret 公式：
{ t ′ ( s ) = κ 1 ( s ) n 1 ( s ) n 1 ′ ( s ) = − κ 1 ( s ) t ( s ) + κ 2 ( s ) n 2 ( s ) n 2 ′ ( s ) = − κ 2 ( s ) n 1 ( s ) + κ 3 ( s ) n 3 ( s ) ⋮ n n − 2 ′ ( s ) = − κ n − 2 ( s ) n n − 3 ( s ) + κ n − 1 ( s ) n n − 1 ( s ) n n − 1 ′ ( s ) = − κ n − 1 ( s ) n n − 2 ( s ) \begin{cases}
\boldsymbol t'(s) = \kappa_1(s) \boldsymbol n_1(s) \\
\boldsymbol n_1'(s) = -\kappa_1(s) \boldsymbol t(s) + \kappa_2(s) \boldsymbol n_2(s) \\
\boldsymbol n_2'(s) = -\kappa_2(s) \boldsymbol n_1(s) + \kappa_3(s) \boldsymbol n_3(s) \\
\vdots \\
\boldsymbol n_{n-2}'(s) = -\kappa_{n-2}(s) \boldsymbol n_{n-3}(s) + \kappa_{n-1}(s) \boldsymbol n_{n-1}(s) \\
\boldsymbol n_{n-1}'(s) = -\kappa_{n-1}(s) \boldsymbol n_{n-2}(s)
\end{cases} ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ t ′ ( s ) = κ 1 ( s ) n 1 ( s ) n 1 ′ ( s ) = − κ 1 ( s ) t ( s ) + κ 2 ( s ) n 2 ( s ) n 2 ′ ( s ) = − κ 2 ( s ) n 1 ( s ) + κ 3 ( s ) n 3 ( s ) ⋮ n n − 2 ′ ( s ) = − κ n − 2 ( s ) n n − 3 ( s ) + κ n − 1 ( s ) n n − 1 ( s ) n n − 1 ′ ( s ) = − κ n − 1 ( s ) n n − 2 ( s )
至此，A : = { t ( s ) , n 1 ( s ) , … , n n − 1 ( s ) } \mathscr A := \{\boldsymbol t(s), \boldsymbol n_1(s), \ldots, \boldsymbol n_{n-1}(s)\} A : = { t ( s ) , n 1 ( s ) , … , n n − 1 ( s ) } R n \mathbb R^n R n Frenet 标架 (Frenet Frame)「フレネ枠」 c \boldsymbol c c
# 空间曲线与平面曲线实际问题中，最常见的曲线为平面曲线与空间曲线，此时不需要将曲率和法向量等概念推广至更高维空间。简单地：
称第 1 1 1 κ 1 \kappa_1 κ 1 曲率 (curvature)「曲率」 κ \kappa κ
称第 2 2 2 κ 2 \kappa_2 κ 2 挠率 (torsion)「捩率」 τ \tau τ
称第 1 1 1 n 1 \boldsymbol n_1 n 1 主法向量 (Principal Normal Vector)「主法線ベクトル」 n \boldsymbol n n
称第 2 2 2 n 2 \boldsymbol n_2 n 2 副法向量 (Binormal Vector)「副法線ベクトル」 b \boldsymbol b b
曲率本身的定义依赖于模长，这使得向量的正反方向信息被丢失，由于 t ′ ( s ) \boldsymbol t'(s) t ′ ( s ) c ′ ( s ) \boldsymbol c'(s) c ′ ( s ) t ′ ( s ) \boldsymbol t'(s) t ′ ( s ) κ ~ ( s ) \widetilde \kappa(s) κ ( s )
t ′ ( s ) = κ ~ ( s ) n ( s ) \boldsymbol t'(s) = \widetilde \kappa(s) \boldsymbol n(s)
t ′ ( s ) = κ ( s ) n ( s )
称此处唯一确定的曲率 κ ~ ( s ) \widetilde \kappa(s) κ ( s ) 带向曲率 ，该曲率可能取正可能取负也可能为零，符号表示曲线弯曲的方向
曲线为 向法向量 方向弯曲时，带向曲率为正
曲线为 背离法向量 方向弯曲时，带向曲率为负
该符号的方向直接影响了主法向量场的方向，即
κ ~ ( s ) > 0 ⟺ { t ( s ) , n ( s ) } \widetilde \kappa(s) \gt 0 \iff \{\boldsymbol t(s), \boldsymbol n(s)\} κ ( s ) > 0 ⟺ { t ( s ) , n ( s ) } κ ~ ( s ) < 0 ⟺ { t ( s ) , n ( s ) } \widetilde \kappa(s) \lt 0 \iff \{\boldsymbol t(s), \boldsymbol n(s)\} κ ( s ) < 0 ⟺ { t ( s ) , n ( s ) }
直观上，可以按照如下说明理解两个曲率的作用
曲率 κ \kappa κ
挠率 τ \tau τ
在维度 3 3 3
{ t ′ ( s ) = κ ( s ) n ( s ) n ′ ( s ) = − κ ( s ) t ( s ) + τ ( s ) b ( s ) b ′ ( s ) = − τ ( s ) n ( s ) \begin{cases}
\boldsymbol t'(s) = \kappa(s) \boldsymbol n(s) \\
\boldsymbol n'(s) = -\kappa(s) \boldsymbol t(s) + \tau(s) \boldsymbol b(s) \\
\boldsymbol b'(s) = -\tau(s) \boldsymbol n(s)
\end{cases} ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ t ′ ( s ) = κ ( s ) n ( s ) n ′ ( s ) = − κ ( s ) t ( s ) + τ ( s ) b ( s ) b ′ ( s ) = − τ ( s ) n ( s )
改写为矩阵形式得到
d d s ( t ( s ) n ( s ) b ( s ) ) = ( 0 κ ( s ) 0 − κ ( s ) 0 τ ( s ) 0 − τ ( s ) 0 ) ( t ( s ) n ( s ) b ( s ) ) \frac{d}{ds} \begin{pmatrix} \boldsymbol t(s) \\ \boldsymbol n(s) \\ \boldsymbol b(s) \end{pmatrix} = \begin{pmatrix} 0 & \kappa(s) & 0 \\ -\kappa(s) & 0 & \tau(s) \\ 0 & -\tau(s) & 0 \end{pmatrix} \begin{pmatrix} \boldsymbol t(s) \\ \boldsymbol n(s) \\ \boldsymbol b(s) \end{pmatrix}
d s d ⎝ ⎛ t ( s ) n ( s ) b ( s ) ⎠ ⎞ = ⎝ ⎛ 0 − κ ( s ) 0 κ ( s ) 0 − τ ( s ) 0 τ ( s ) 0 ⎠ ⎞ ⎝ ⎛ t ( s ) n ( s ) b ( s ) ⎠ ⎞
这实际上是一个线性微分方程组 Y ′ = A ( s ) Y Y' = A(s) Y Y ′ = A ( s ) Y Y = e ∫ A ( s ) d s Y 0 Y = e^{\int A(s) ds} Y_0 Y = e ∫ A ( s ) d s Y 0
( t ( s ) n ( s ) b ( s ) ) = exp ( ∫ s 0 s ( 0 κ ( u ) 0 − κ ( u ) 0 τ ( u ) 0 − τ ( u ) 0 ) d u ) ( t ( s 0 ) n ( s 0 ) b ( s 0 ) ) \begin{pmatrix} \boldsymbol t(s) \\ \boldsymbol n(s) \\ \boldsymbol b(s) \end{pmatrix} = \exp\left(\int_{s_0}^s \begin{pmatrix} 0 & \kappa(u) & 0 \\ -\kappa(u) & 0 & \tau(u) \\ 0 & -\tau(u) & 0 \end{pmatrix} du\right) \begin{pmatrix} \boldsymbol t(s_0) \\ \boldsymbol n(s_0) \\ \boldsymbol b(s_0) \end{pmatrix}
⎝ ⎛ t ( s ) n ( s ) b ( s ) ⎠ ⎞ = exp ⎝ ⎛ ∫ s 0 s ⎝ ⎛ 0 − κ ( u ) 0 κ ( u ) 0 − τ ( u ) 0 τ ( u ) 0 ⎠ ⎞ d u ⎠ ⎞ ⎝ ⎛ t ( s 0 ) n ( s 0 ) b ( s 0 ) ⎠ ⎞
更简化地，考虑平面曲线
在平面上通过曲率，可以刻画出曲率圆。
ρ ( s ) : = 1 κ ( s ) \rho(s) := \frac{1}{\kappa(s)} \quad
ρ ( s ) : = κ ( s ) 1
为曲线在点 c ( s ) \boldsymbol c(s) c ( s ) 曲率半径
同时，令圆心
o ( s ) : = c ( s ) + 1 κ ~ ( s ) n ( s ) \boldsymbol o(s) := \boldsymbol c(s) + \frac{1}{\widetilde \kappa(s)} \boldsymbol n(s)
o ( s ) : = c ( s ) + κ ( s ) 1 n ( s )
那么，以 o ( s ) \boldsymbol o(s) o ( s ) ρ ( s ) \rho(s) ρ ( s ) c ( s ) \boldsymbol c(s) c ( s ) 曲率圆 。
C ( θ ) = o ( s ) + ρ ( s ) ( cos θ sin θ ) \boldsymbol C(\theta) = \boldsymbol o(s) + \rho(s) \begin{pmatrix} \cos \theta \\ \sin \theta \end{pmatrix}
C ( θ ) = o ( s ) + ρ ( s ) ( cos θ sin θ )
通过曲率圆，可以直观地理解曲率的意义：在平面上曲率蕴含的信息等价于曲线蕴含的信息 ，即唯一确定曲率可以唯一确定曲线，让我们来尝试构造这一过程
设平面曲线 c ( s ) \boldsymbol c(s) c ( s ) κ ~ ( s ) \widetilde \kappa(s) κ ( s ) t ( s ) \boldsymbol t(s) t ( s ) t ( s ) \boldsymbol t(s) t ( s )
t ( s ) = ( cos θ ( s ) sin θ ( s ) ) , t ′ ( s ) = θ ′ ( s ) ( − sin θ ( s ) cos θ ( s ) ) \boldsymbol t(s) = \begin{pmatrix} \cos \theta(s) \\ \sin \theta(s) \end{pmatrix},\quad
\boldsymbol t'(s) = \theta'(s) \begin{pmatrix} -\sin \theta(s) \\ \cos \theta(s) \end{pmatrix} t ( s ) = ( cos θ ( s ) sin θ ( s ) ) , t ′ ( s ) = θ ′ ( s ) ( − sin θ ( s ) cos θ ( s ) )
曲率的定义给出其为系数，即
κ ( s ) = d d s θ ( s ) \kappa(s) = \frac{d}{ds} \theta(s)
κ ( s ) = d s d θ ( s )
两边积分，还原出 θ ( s ) \theta(s) θ ( s )
θ ( s ) = ∫ s 0 s κ ( u ) d u + θ ( s 0 ) \theta(s) = \int_{s_0}^s \kappa(u) du + \theta(s_0)
θ ( s ) = ∫ s 0 s κ ( u ) d u + θ ( s 0 )
这样一来就获得了切向量场 t ( s ) \boldsymbol t(s) t ( s ) c ( s ) \boldsymbol c(s) c ( s )
c ( s ) = ( ∫ s 0 s cos ( ∫ s 0 u κ ( v ) d v + θ ( s 0 ) ) d u + b 1 ∫ s 0 s sin ( ∫ s 0 u κ ( v ) d v + θ ( s 0 ) ) d u + b 2 ) \boldsymbol c(s) = \begin{pmatrix}
\displaystyle\int_{s_0}^s \cos \left(\int_{s_0}^u \kappa(v) dv + \theta(s_0)\right) du + b_1\\[12pt]
\displaystyle\int_{s_0}^s \sin \left(\int_{s_0}^u \kappa(v) dv + \theta(s_0)\right) du + b_2
\end{pmatrix} c ( s ) = ⎝ ⎜ ⎜ ⎜ ⎜ ⎛ ∫ s 0 s cos ( ∫ s 0 u κ ( v ) d v + θ ( s 0 ) ) d u + b 1 ∫ s 0 s sin ( ∫ s 0 u κ ( v ) d v + θ ( s 0 ) ) d u + b 2 ⎠ ⎟ ⎟ ⎟ ⎟ ⎞
# 非弧长参数化下的曲率推导上述定义中依赖于弧长参数化，以下推导 非弧长参数化下 的曲率公式，令曲线为一般空间曲线 c = c ( t ) \boldsymbol c = \boldsymbol c(t) c = c ( t )
由弧长参数定义以及微积分基本定理，得到
s ( t ) = ∫ t 0 t ∥ c ′ ( u ) ∣ d u ⟹ d s d t = ∥ c ′ ( t ) ∥ s(t) = \int_{t_0}^t \|\boldsymbol c'(u)| du \implies \frac{ds}{dt} = \|\boldsymbol c'(t)\|
s ( t ) = ∫ t 0 t ∥ c ′ ( u ) ∣ d u ⟹ d t d s = ∥ c ′ ( t ) ∥
将 s s s t t t
c ′ ( s ) = d c d s ( s ( t ) ) = d c d t ( t ) ⋅ d t d s \boldsymbol c'(s) = \frac{d\boldsymbol c}{ds}(s(t)) = \frac{d\boldsymbol c}{dt}(t) \cdot \frac{dt}{ds}
c ′ ( s ) = d s d c ( s ( t ) ) = d t d c ( t ) ⋅ d s d t
进一步可以计算出二阶微分
c ′ ′ ( s ) = d d s ( c ′ ( t ) ⋅ d t d s ) = c ′ ′ ( t ) ( d t d s ) 2 + c ′ ( t ) ⋅ d 2 t d s 2 \boldsymbol c''(s) = \frac{d}{ds}\left(\boldsymbol c'(t) \cdot \frac{dt}{ds}\right) = \boldsymbol c''(t) \left(\frac{dt}{ds}\right)^2 + \boldsymbol c'(t) \cdot \frac{d^2 t}{ds^2}
c ′ ′ ( s ) = d s d ( c ′ ( t ) ⋅ d s d t ) = c ′ ′ ( t ) ( d s d t ) 2 + c ′ ( t ) ⋅ d s 2 d 2 t
将 d t d s = 1 ∥ c ′ ( t ) ∥ \frac{dt}{ds} = \frac{1}{\|\boldsymbol c'(t)\|} d s d t = ∥ c ′ ( t ) ∥ 1
c ′ ′ ( s ) = ∥ c ′ ( t ) ∥ 2 c ′ ′ ( t ) − ( c ′ ( t ) ⋅ c ′ ′ ( t ) ) c ′ ( t ) ∥ c ′ ( t ) ∥ 4 = ( c ′ ( t ) ⋅ c ′ ( t ) ) c ′ ′ ( t ) − ( c ′ ( t ) ⋅ c ′ ′ ( t ) ) c ′ ( t ) ∥ c ′ ( t ) ∥ 4 = c ′ ( t ) × ( c ′ ′ ( t ) × c ′ ( t ) ) ∥ c ′ ( t ) ∥ 4 \begin{aligned}
\boldsymbol c''(s)
&= \frac{\|\boldsymbol c'(t)\|^2 \boldsymbol c''(t) - (\boldsymbol c'(t) \cdot \boldsymbol c''(t)) \boldsymbol c'(t)}{\|\boldsymbol c'(t)\|^4} \\
&= \frac{(\boldsymbol c'(t) \cdot \boldsymbol c'(t)) \boldsymbol c''(t) - (\boldsymbol c'(t) \cdot \boldsymbol c''(t)) \boldsymbol c'(t)}{\|\boldsymbol c'(t)\|^4} \\
&= \frac{\boldsymbol c'(t) \times (\boldsymbol c''(t) \times \boldsymbol c'(t))}{\|\boldsymbol c'(t)\|^4}
\end{aligned} c ′ ′ ( s ) = ∥ c ′ ( t ) ∥ 4 ∥ c ′ ( t ) ∥ 2 c ′ ′ ( t ) − ( c ′ ( t ) ⋅ c ′ ′ ( t ) ) c ′ ( t ) = ∥ c ′ ( t ) ∥ 4 ( c ′ ( t ) ⋅ c ′ ( t ) ) c ′ ′ ( t ) − ( c ′ ( t ) ⋅ c ′ ′ ( t ) ) c ′ ( t ) = ∥ c ′ ( t ) ∥ 4 c ′ ( t ) × ( c ′ ′ ( t ) × c ′ ( t ) )
取模长得到曲率
κ ( t ) = ∥ c ′ ′ ( s ( t ) ) ∥ = ∥ c ′ ( t ) × ( c ′ ′ ( t ) × c ′ ( t ) ) ∥ ∥ c ′ ( t ) ∥ 4 = ∥ c ′ ( t ) × c ′ ′ ( t ) ∥ ∥ c ′ ( t ) ∥ 3 \begin{aligned}
\kappa(t) =
\|\boldsymbol c''(s(t))\| &= \frac{\|\boldsymbol c'(t) \times (\boldsymbol c''(t) \times \boldsymbol c'(t))\|}{\|\boldsymbol c'(t)\|^4} \\
&= \frac{\|\boldsymbol c'(t) \times \boldsymbol c''(t)\|}{\|\boldsymbol c'(t)\|^3}
\end{aligned} κ ( t ) = ∥ c ′ ′ ( s ( t ) ) ∥ = ∥ c ′ ( t ) ∥ 4 ∥ c ′ ( t ) × ( c ′ ′ ( t ) × c ′ ( t ) ) ∥ = ∥ c ′ ( t ) ∥ 3 ∥ c ′ ( t ) × c ′ ′ ( t ) ∥
以下推导 非弧长参数化下 的挠率公式，即 c = c ( t ) \boldsymbol c = \boldsymbol c(t) c = c ( t )
κ ( t ) = ∥ c ′ ( t ) × c ′ ′ ( t ) ∥ ∥ c ′ ( t ) ∥ 3 \kappa(t) = \frac{\|\boldsymbol c'(t) \times \boldsymbol c''(t)\|}{\|\boldsymbol c'(t)\|^3}
κ ( t ) = ∥ c ′ ( t ) ∥ 3 ∥ c ′ ( t ) × c ′ ′ ( t ) ∥
对于挠率，需先计算 n , b , n ′ \boldsymbol n, \boldsymbol b, \boldsymbol n' n , b , n ′
c ′ ( s ) = c ′ ( t ) ∥ c ′ ( t ) ∥ , c ′ ′ ( s ) = c ′ ( t ) × ( c ′ ′ ( t ) × c ′ ( t ) ) ∥ c ′ ( t ) ∥ 4 \boldsymbol c'(s) = \frac{\boldsymbol c'(t)}{\|\boldsymbol c'(t)\|},\quad \boldsymbol c''(s) = \frac{\boldsymbol c'(t) \times (\boldsymbol c''(t) \times \boldsymbol c'(t))}{\|\boldsymbol c'(t)\|^4}
c ′ ( s ) = ∥ c ′ ( t ) ∥ c ′ ( t ) , c ′ ′ ( s ) = ∥ c ′ ( t ) ∥ 4 c ′ ( t ) × ( c ′ ′ ( t ) × c ′ ( t ) )
所以
n ( s ) = c ′ ′ ( s ) ∥ c ′ ′ ( s ) ∥ = c ′ ( t ) × ( c ′ ′ ( t ) × c ′ ( t ) ) ∥ c ′ ( t ) × ( c ′ ′ ( t ) × c ′ ( t ) ) ∥ \boldsymbol n(s) = \frac{\boldsymbol c''(s)}{\|\boldsymbol c''(s)\|}
= \frac{\boldsymbol c'(t) \times (\boldsymbol c''(t) \times \boldsymbol c'(t))}{\|\boldsymbol c'(t) \times (\boldsymbol c''(t) \times \boldsymbol c'(t))\|}
n ( s ) = ∥ c ′ ′ ( s ) ∥ c ′ ′ ( s ) = ∥ c ′ ( t ) × ( c ′ ′ ( t ) × c ′ ( t ) ) ∥ c ′ ( t ) × ( c ′ ′ ( t ) × c ′ ( t ) )
b ( s ) = c ′ ( s ) × n ( s ) = c ′ ( t ) × { c ′ ( t ) × ( c ′ ′ ( t ) × c ′ ( t ) ) } ∥ c ′ ( t ) ∥ 2 ⋅ ∥ c ′ ( t ) × ( c ′ ′ ( t ) × c ′ ( t ) ) ∥ \boldsymbol b(s) = \boldsymbol c'(s) \times \boldsymbol n(s)
= \frac{\boldsymbol c'(t) \times \{\boldsymbol c'(t) \times (\boldsymbol c''(t) \times \boldsymbol c'(t))\}}{\|\boldsymbol c'(t)\|^2 \cdot \|\boldsymbol c'(t) \times (\boldsymbol c''(t) \times \boldsymbol c'(t))\|}
b ( s ) = c ′ ( s ) × n ( s ) = ∥ c ′ ( t ) ∥ 2 ⋅ ∥ c ′ ( t ) × ( c ′ ′ ( t ) × c ′ ( t ) ) ∥ c ′ ( t ) × { c ′ ( t ) × ( c ′ ′ ( t ) × c ′ ( t ) ) }
n ′ ( s ) = d n d t ( t ) ⋅ d t d s = d d t N ∥ N ∥ ⋅ 1 ∥ c ′ ( t ) ∥ ( N : = c ′ ( t ) × ( c ′ ′ ( t ) × c ′ ( t ) ) ) = ∥ N ∥ 2 ⋅ N ′ − ( N ⋅ N ′ ) ⋅ N ∥ N ∥ 3 ⋅ ∥ c ′ ( t ) ∥ = N × ( N ′ × N ) ∥ N ∥ 3 ⋅ ∥ c ′ ( t ) ∥ \begin{aligned}
\boldsymbol n'(s)
&= \frac{d\boldsymbol n}{dt}(t) \cdot \frac{dt}{ds} \\
&= \frac{d}{dt} \frac{N}{\|N\|} \cdot \frac{1}{\|\boldsymbol c'(t)\|} \quad (N := \boldsymbol c'(t) \times (\boldsymbol c''(t) \times \boldsymbol c'(t))) \\
&= \frac{\|N\|^2 \cdot N' - (N \cdot N') \cdot N}{\|N\|^3 \cdot \|\boldsymbol c'(t)\|} \\
&= \frac{N \times (N' \times N)}{\|N\|^3 \cdot \|\boldsymbol c'(t)\|}
\end{aligned} n ′ ( s ) = d t d n ( t ) ⋅ d s d t = d t d ∥ N ∥ N ⋅ ∥ c ′ ( t ) ∥ 1 ( N : = c ′ ( t ) × ( c ′ ′ ( t ) × c ′ ( t ) ) ) = ∥ N ∥ 3 ⋅ ∥ c ′ ( t ) ∥ ∥ N ∥ 2 ⋅ N ′ − ( N ⋅ N ′ ) ⋅ N = ∥ N ∥ 3 ⋅ ∥ c ′ ( t ) ∥ N × ( N ′ × N )
因此
τ ( t ) = n ′ ( s ( t ) ) ⋅ b ( s ( t ) ) = [ N × ( N ′ × N ) ] ⋅ [ c ′ ( t ) × { c ′ ( t ) × ( c ′ ′ ( t ) × c ′ ( t ) ) } ] ∥ N ∥ 4 ⋅ ∥ c ′ ( t ) ∥ 3 = 1 ∥ N ∥ 4 ⋅ ∥ c ′ ( t ) ∥ 2 ⋅ { N × ( N ′ × N ) } ⋅ ( c ′ ( t ) × N ) = ( N ⋅ c ′ ( t ) ) { N × ( N ′ × N ) } − { c ′ ( t ) ⋅ ( N ′ × N ) } ⋅ N 2 ∥ N ∥ 4 ⋅ ∥ c ′ ( t ) ∥ 2 = c ′ ( t ) ⋅ ( N ′ × N ) ∥ N ∥ 2 ⋅ ∥ c ′ ( t ) ∥ 2 (Since N ⋅ c ′ ( t ) = 0 ) = N ′ ⋅ ( c ′ ( t ) × N ) ∥ c ′ ( t ) ∥ 4 ⋅ ∥ c ′ ′ ( t ) × c ′ ( t ) ∥ 2 \begin{aligned}
\tau(t) &= \boldsymbol n'(s(t)) \cdot \boldsymbol b(s(t)) \\
&= \frac{[N \times (N' \times N)] \cdot [\boldsymbol c'(t) \times \{\boldsymbol c'(t) \times (\boldsymbol c''(t) \times \boldsymbol c'(t))\}]}{\|N\|^4 \cdot \|\boldsymbol c'(t)\|^3} \\
&= \frac{1}{\|N\|^4 \cdot \|\boldsymbol c'(t)\|^2} \cdot \{N \times (N' \times N)\} \cdot (\boldsymbol c'(t) \times N) \\
&= \frac{(N \cdot \boldsymbol c'(t)) \{N \times (N' \times N)\} - \{\boldsymbol c'(t) \cdot(N' \times N)\} \cdot N^2}{\|N\|^4 \cdot \|\boldsymbol c'(t)\|^2} \\
&= \frac{\boldsymbol c'(t) \cdot (N' \times N)}{\|N\|^2 \cdot \|\boldsymbol c'(t)\|^2} \quad \text{(Since $N \cdot \boldsymbol c'(t) = 0$)} \\
&= \frac{N' \cdot (\boldsymbol c'(t) \times N)}{\|\boldsymbol c'(t)\|^4 \cdot \|\boldsymbol c''(t) \times \boldsymbol c'(t)\|^2} \\
\end{aligned} τ ( t ) = n ′ ( s ( t ) ) ⋅ b ( s ( t ) ) = ∥ N ∥ 4 ⋅ ∥ c ′ ( t ) ∥ 3 [ N × ( N ′ × N ) ] ⋅ [ c ′ ( t ) × { c ′ ( t ) × ( c ′ ′ ( t ) × c ′ ( t ) ) } ] = ∥ N ∥ 4 ⋅ ∥ c ′ ( t ) ∥ 2 1 ⋅ { N × ( N ′ × N ) } ⋅ ( c ′ ( t ) × N ) = ∥ N ∥ 4 ⋅ ∥ c ′ ( t ) ∥ 2 ( N ⋅ c ′ ( t ) ) { N × ( N ′ × N ) } − { c ′ ( t ) ⋅ ( N ′ × N ) } ⋅ N 2 = ∥ N ∥ 2 ⋅ ∥ c ′ ( t ) ∥ 2 c ′ ( t ) ⋅ ( N ′ × N ) (Since N ⋅ c ′ ( t ) = 0) = ∥ c ′ ( t ) ∥ 4 ⋅ ∥ c ′ ′ ( t ) × c ′ ( t ) ∥ 2 N ′ ⋅ ( c ′ ( t ) × N )
整理分子部分
N ′ = c ′ ′ × ( c ′ ′ × c ′ ) + c ′ × ( c ′ ′ ′ × c ′ + c ′ ′ × c ′ ′ ⏟ 0 ) = − ( c ′ ′ ( t ) ⋅ c ′ ′ ( t ) + c ′ ( t ) ⋅ c ′ ′ ′ ( t ) ) ⏟ coefficient on c ′ ( t ) c ′ ( t ) + ( c ′ ′ ( t ) ⋅ c ′ ( t ) ) ⏟ coefficient on c ′ ′ ( t ) c ′ ′ ( t ) + ( c ′ ( t ) ⋅ c ′ ( t ) ) ⏟ coefficient on c ′ ′ ′ ( t ) c ′ ′ ′ ( t ) c ′ ( t ) × N = c ′ ( t ) × [ c ′ ( t ) × ( c ′ ′ ( t ) × c ′ ( t ) ) ] = c ′ ( t ) × { ( c ′ ( t ) ⋅ c ′ ( t ) ) c ′ ′ ( t ) − ( c ′ ( t ) ⋅ c ′ ′ ( t ) ) c ′ ( t ) } = ( c ′ ( t ) ⋅ c ′ ( t ) ) ⏟ coefficient on production ( c ′ ( t ) × c ′ ′ ( t ) ) \begin{aligned}
N' &= \boldsymbol c'' \times (\boldsymbol c'' \times \boldsymbol c') + \boldsymbol c' \times (\boldsymbol c''' \times \boldsymbol c' + \underbrace{\boldsymbol c'' \times \boldsymbol c''}_{0}) \\
&= \underbrace{-(\boldsymbol c''(t) \cdot \boldsymbol c''(t) + \boldsymbol c'(t) \cdot \boldsymbol c'''(t))}_{\text{coefficient on } \boldsymbol c'(t)} \boldsymbol c'(t) + \underbrace{(\boldsymbol c''(t) \cdot \boldsymbol c'(t))}_{\text{coefficient on } \boldsymbol c''(t)} \boldsymbol c''(t) + \underbrace{(\boldsymbol c'(t) \cdot \boldsymbol c'(t))}_{\text{coefficient on } \boldsymbol c'''(t)} \boldsymbol c'''(t) \\
\boldsymbol c'(t) \times N &= \boldsymbol c'(t) \times [\boldsymbol c'(t) \times (\boldsymbol c''(t) \times \boldsymbol c'(t))] \\
&= \boldsymbol c'(t) \times \{(\boldsymbol c'(t) \cdot \boldsymbol c'(t)) \boldsymbol c''(t) - (\boldsymbol c'(t) \cdot \boldsymbol c''(t)) \boldsymbol c'(t)\} \\
&= \underbrace{(\boldsymbol c'(t) \cdot \boldsymbol c'(t))}_{\text{coefficient on production}} (\boldsymbol c'(t) \times \boldsymbol c''(t))
\end{aligned} N ′ c ′ ( t ) × N = c ′ ′ × ( c ′ ′ × c ′ ) + c ′ × ( c ′ ′ ′ × c ′ + 0 c ′ ′ × c ′ ′ ) = coefficient on c ′ ( t ) − ( c ′ ′ ( t ) ⋅ c ′ ′ ( t ) + c ′ ( t ) ⋅ c ′ ′ ′ ( t ) ) c ′ ( t ) + coefficient on c ′ ′ ( t ) ( c ′ ′ ( t ) ⋅ c ′ ( t ) ) c ′ ′ ( t ) + coefficient on c ′ ′ ′ ( t ) ( c ′ ( t ) ⋅ c ′ ( t ) ) c ′ ′ ′ ( t ) = c ′ ( t ) × [ c ′ ( t ) × ( c ′ ′ ( t ) × c ′ ( t ) ) ] = c ′ ( t ) × { ( c ′ ( t ) ⋅ c ′ ( t ) ) c ′ ′ ( t ) − ( c ′ ( t ) ⋅ c ′ ′ ( t ) ) c ′ ( t ) } = coefficient on production ( c ′ ( t ) ⋅ c ′ ( t ) ) ( c ′ ( t ) × c ′ ′ ( t ) )
所以二者做内积时，仅 c ′ ′ ′ ( t ) \boldsymbol c'''(t) c ′ ′ ′ ( t )
N ′ ⋅ ( c ′ ( t ) × N ) = ∥ c ′ ( t ) ∥ 4 c ′ ′ ′ ( t ) ⋅ ( c ′ ( t ) × c ′ ′ ( t ) ) = ∥ c ′ ( t ) ∥ 4 det ( c ′ ( t ) , c ′ ′ ( t ) , c ′ ′ ′ ( t ) ) N' \cdot (\boldsymbol c'(t) \times N) = \|\boldsymbol c'(t)\|^4 \boldsymbol c'''(t) \cdot (\boldsymbol c'(t) \times \boldsymbol c''(t))
= \|\boldsymbol c'(t)\|^4 \det(\boldsymbol c'(t), \boldsymbol c''(t), \boldsymbol c'''(t))
N ′ ⋅ ( c ′ ( t ) × N ) = ∥ c ′ ( t ) ∥ 4 c ′ ′ ′ ( t ) ⋅ ( c ′ ( t ) × c ′ ′ ( t ) ) = ∥ c ′ ( t ) ∥ 4 det ( c ′ ( t ) , c ′ ′ ( t ) , c ′ ′ ′ ( t ) )
带回得到
τ ( t ) = det ( c ′ ( t ) , c ′ ′ ( t ) , c ′ ′ ′ ( t ) ) ∥ c ′ ( t ) × c ′ ′ ( t ) ∥ 2 \tau(t) = \frac{\det(\boldsymbol c'(t), \boldsymbol c''(t), \boldsymbol c'''(t))}{\|\boldsymbol c'(t) \times \boldsymbol c''(t)\|^2}
τ ( t ) = ∥ c ′ ( t ) × c ′ ′ ( t ) ∥ 2 det ( c ′ ( t ) , c ′ ′ ( t ) , c ′ ′ ′ ( t ) )
结论：对于一般曲线 c ( t ) : I → R 3 \boldsymbol c(t): I \to \mathbb R^3 c ( t ) : I → R 3
κ ( t ) = ∥ c ′ ( t ) × c ′ ′ ( t ) ∥ ∥ c ′ ( t ) ∥ 3 , τ ( t ) = det ( c ′ ( t ) , c ′ ′ ( t ) , c ′ ′ ′ ( t ) ) ∥ c ′ ( t ) × c ′ ′ ( t ) ∥ 2 \kappa(t) = \frac{\|\boldsymbol c'(t) \times \boldsymbol c''(t)\|}{\|\boldsymbol c'(t)\|^3}, \quad
\tau(t) = \frac{\det(\boldsymbol c'(t), \boldsymbol c''(t), \boldsymbol c'''(t))}{\|\boldsymbol c'(t) \times \boldsymbol c''(t)\|^2} κ ( t ) = ∥ c ′ ( t ) ∥ 3 ∥ c ′ ( t ) × c ′ ′ ( t ) ∥ , τ ( t ) = ∥ c ′ ( t ) × c ′ ′ ( t ) ∥ 2 det ( c ′ ( t ) , c ′ ′ ( t ) , c ′ ′ ′ ( t ) )
# 刚体变换下的曲率不变性给定正交矩阵 A ∈ O ( 3 ) A \in O(3) A ∈ O ( 3 ) b ∈ R 3 \boldsymbol b \in \mathbb R^3 b ∈ R 3 T : R 3 → R 3 T: \mathbb R^3 \to \mathbb R^3 T : R 3 → R 3
T ( x ) = A x + b T(\boldsymbol x) = A\boldsymbol x + \boldsymbol b
T ( x ) = A x + b
称 T T T R 3 \mathbb R^3 R 3 刚体变换 (Rigid Transformation)「合同変換」
当 ∣ A ∣ = 1 |A| = 1 ∣ A ∣ = 1 T T T 正刚体变换
当 ∣ A ∣ = − 1 |A| = -1 ∣ A ∣ = − 1 T T T 反刚体变换
对于刚体变换 T T T
∥ T ( x ) − T ( y ) ∥ = ∥ A x + b − ( A y + b ) ∥ = ∥ A ( x − y ) ∥ = ( x − y ) T A T A ( x − y ) = ∥ x − y ∥ \begin{aligned}
\|T(\boldsymbol x) - T(\boldsymbol y)\| &= \|A\boldsymbol x + \boldsymbol b - (A\boldsymbol y + \boldsymbol b)\| \\
&= \|A(\boldsymbol x - \boldsymbol y)\| \\
&= \sqrt{(\boldsymbol x - \boldsymbol y)^T A^T A (\boldsymbol x - \boldsymbol y)} \\
&= \|\boldsymbol x - \boldsymbol y\|
\end{aligned} ∥ T ( x ) − T ( y ) ∥ = ∥ A x + b − ( A y + b ) ∥ = ∥ A ( x − y ) ∥ = ( x − y ) T A T A ( x − y ) = ∥ x − y ∥
这意味着刚体变换保持了 R 3 \mathbb R^3 R 3
设 c : I → R 3 \boldsymbol c: I \to \mathbb R^3 c : I → R 3 κ ( s ) , τ ( s ) \kappa(s), \tau(s) κ ( s ) , τ ( s )
设 c ~ : I → R 3 \widetilde{\boldsymbol c}: I \to \mathbb R^3 c : I → R 3 T ∘ c T \circ \boldsymbol c T ∘ c c ~ ( s ) = A c ( s ) + b \widetilde{\boldsymbol c}(s) = A\boldsymbol c(s) + \boldsymbol b c ( s ) = A c ( s ) + b κ ~ ( s ) , τ ~ ( s ) \widetilde \kappa(s), \widetilde \tau(s) κ ( s ) , τ ( s )
对 c ~ \widetilde{\boldsymbol c} c
c ~ ′ ( s ) = A c ′ ( s ) c ~ ′ ′ ( s ) = A c ′ ′ ( s ) c ~ ′ ′ ′ ( s ) = A c ′ ′ ′ ( s ) \begin{aligned}
\widetilde{\boldsymbol c}'(s) &= A\boldsymbol c'(s) \\
\widetilde{\boldsymbol c}''(s) &= A\boldsymbol c''(s) \\
\widetilde{\boldsymbol c}'''(s) &= A\boldsymbol c'''(s)
\end{aligned} c ′ ( s ) c ′ ′ ( s ) c ′ ′ ′ ( s ) = A c ′ ( s ) = A c ′ ′ ( s ) = A c ′ ′ ′ ( s )
代入曲率和挠率的公式
κ ~ ( s ) = ∥ c ~ ′ ( s ) × c ~ ′ ′ ( s ) ∥ ∥ c ~ ′ ( s ) ∥ 3 = ∥ A c ′ ( s ) × A c ′ ′ ( s ) ∥ ∥ A c ′ ( s ) ∥ 3 = ∥ A ( c ′ ( s ) × c ′ ′ ( s ) ) ∥ ∥ A c ′ ( s ) ∥ 3 = κ ( s ) τ ~ ( s ) = det ( c ~ ′ ( s ) , c ~ ′ ′ ( s ) , c ~ ′ ′ ′ ( s ) ) ∥ c ~ ′ ( s ) × c ~ ′ ′ ( s ) ∥ 2 = det ( A c ′ ( s ) , A c ′ ′ ( s ) , A c ′ ′ ′ ( s ) ) ∥ A ( c ′ ( s ) × c ′ ′ ( s ) ) ∥ 2 = ∣ A ∣ ⋅ det ( c ′ ( s ) , c ′ ′ ( s ) , c ′ ′ ′ ( s ) ) ∥ c ′ ( s ) × c ′ ′ ( s ) ∥ 2 = ∣ A ∣ ⋅ τ ( s ) \begin{aligned}
\widetilde \kappa(s) &= \frac{\|\widetilde{\boldsymbol c}'(s) \times \widetilde{\boldsymbol c}''(s)\|}{\|\widetilde{\boldsymbol c}'(s)\|^3} \\
&= \frac{\|A\boldsymbol c'(s) \times A\boldsymbol c''(s)\|}{\|A\boldsymbol c'(s)\|^3} \\
&= \frac{\|A(\boldsymbol c'(s) \times \boldsymbol c''(s))\|}{\|A\boldsymbol c'(s)\|^3} = \kappa(s) \\
\widetilde \tau(s) &= \frac{\det(\widetilde{\boldsymbol c}'(s), \widetilde{\boldsymbol c}''(s), \widetilde{\boldsymbol c}'''(s))}{\|\widetilde{\boldsymbol c}'(s) \times \widetilde{\boldsymbol c}''(s)\|^2} \\
&= \frac{\det(A\boldsymbol c'(s), A\boldsymbol c''(s), A\boldsymbol c'''(s))}{\|A(\boldsymbol c'(s) \times \boldsymbol c''(s))\|^2} \\
&= \frac{|A| \cdot \det(\boldsymbol c'(s), \boldsymbol c''(s), \boldsymbol c'''(s))}{\|\boldsymbol c'(s) \times \boldsymbol c''(s)\|^2} = |A| \cdot \tau(s)
\end{aligned} κ ( s ) τ ( s ) = ∥ c ′ ( s ) ∥ 3 ∥ c ′ ( s ) × c ′ ′ ( s ) ∥ = ∥ A c ′ ( s ) ∥ 3 ∥ A c ′ ( s ) × A c ′ ′ ( s ) ∥ = ∥ A c ′ ( s ) ∥ 3 ∥ A ( c ′ ( s ) × c ′ ′ ( s ) ) ∥ = κ ( s ) = ∥ c ′ ( s ) × c ′ ′ ( s ) ∥ 2 det ( c ′ ( s ) , c ′ ′ ( s ) , c ′ ′ ′ ( s ) ) = ∥ A ( c ′ ( s ) × c ′ ′ ( s ) ) ∥ 2 det ( A c ′ ( s ) , A c ′ ′ ( s ) , A c ′ ′ ′ ( s ) ) = ∥ c ′ ( s ) × c ′ ′ ( s ) ∥ 2 ∣ A ∣ ⋅ det ( c ′ ( s ) , c ′ ′ ( s ) , c ′ ′ ′ ( s ) ) = ∣ A ∣ ⋅ τ ( s )
结论：
曲率 κ \kappa κ
挠率 τ \tau τ
同时也等价于：给定曲率和挠率，可以在刚体变换下唯一确定曲线，该结论称为 曲线论基本定理 (Fundamental Theorem of Curves)「曲線論の基本定理」
实际上，对于 Frenet 标架，由于 ∥ A x ∥ = ∥ x ∥ \|A\boldsymbol x\| = \|\boldsymbol x\| ∥ A x ∥ = ∥ x ∥
t ~ ( s ) = c ~ ′ ( s ) ∥ c ~ ′ ( s ) ∥ = A c ′ ( s ) ∥ A c ′ ( s ) ∥ = A t ( s ) n ~ ( s ) = c ~ ′ ′ ( s ) ∥ c ~ ′ ′ ( s ) ∥ = A c ′ ′ ( s ) ∥ A c ′ ′ ( s ) ∥ = A n ( s ) b ~ ( s ) = t ~ ( s ) × n ~ ( s ) = A t ( s ) × A n ( s ) = A ∣ A ∣ ( t ( s ) × n ( s ) ) = A ∣ A ∣ b ( s ) \begin{aligned}
\widetilde{\boldsymbol t}(s) &= \frac{\widetilde{\boldsymbol c}'(s)}{\|\widetilde{\boldsymbol c}'(s)\|} = \frac{A\boldsymbol c'(s)}{\|A\boldsymbol c'(s)\|} = A\boldsymbol t(s) \\
\widetilde{\boldsymbol n}(s) &= \frac{\widetilde{\boldsymbol c}''(s)}{\|\widetilde{\boldsymbol c}''(s)\|} = \frac{A\boldsymbol c''(s)}{\|A\boldsymbol c''(s)\|} = A\boldsymbol n(s) \\
\widetilde{\boldsymbol b}(s) &= \widetilde{\boldsymbol t}(s) \times \widetilde{\boldsymbol n}(s) = A\boldsymbol t(s) \times A\boldsymbol n(s) = A |A| (\boldsymbol t(s) \times \boldsymbol n(s)) = A |A| \boldsymbol b(s)
\end{aligned} t ( s ) n ( s ) b ( s ) = ∥ c ′ ( s ) ∥ c ′ ( s ) = ∥ A c ′ ( s ) ∥ A c ′ ( s ) = A t ( s ) = ∥ c ′ ′ ( s ) ∥ c ′ ′ ( s ) = ∥ A c ′ ′ ( s ) ∥ A c ′ ′ ( s ) = A n ( s ) = t ( s ) × n ( s ) = A t ( s ) × A n ( s ) = A ∣ A ∣ ( t ( s ) × n ( s ) ) = A ∣ A ∣ b ( s )
即有
( t ~ ( s ) n ~ ( s ) b ~ ( s ) ) = A ( t ( s ) n ( s ) b ( s ) ) \begin{pmatrix} \widetilde{\boldsymbol t}(s) \\ \widetilde{\boldsymbol n}(s) \\ \widetilde{\boldsymbol b}(s) \end{pmatrix} = A \begin{pmatrix} \boldsymbol t(s) \\ \boldsymbol n(s) \\ \boldsymbol b(s) \end{pmatrix}
⎝ ⎛ t ( s ) n ( s ) b ( s ) ⎠ ⎞ = A ⎝ ⎛ t ( s ) n ( s ) b ( s ) ⎠ ⎞
示例 R 3 \mathbb R^3 R 3
C : c ( s ) = ( 1 2 ∫ 0 s cos u 2 d u 1 2 ∫ 0 s sin u 2 d u s 2 ) , s ∈ ( 0 , ∞ ) C: \boldsymbol c(s) = \begin{pmatrix}
\displaystyle\frac{1}{\sqrt{2}} \int_0^s \cos u^2 du \\[12pt]
\displaystyle\frac{1}{\sqrt{2}} \int_0^s \sin u^2 du \\[12pt]
\displaystyle\frac{s}{\sqrt{2}}
\end{pmatrix},\quad s \in (0, \infty) C : c ( s ) = ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎛ 2 1 ∫ 0 s cos u 2 d u 2 1 ∫ 0 s sin u 2 d u 2 s ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎞ , s ∈ ( 0 , ∞ )
设 c \boldsymbol c c κ ( s ) \kappa(s) κ ( s ) τ ( s ) \tau(s) τ ( s ) c ~ \widetilde{\boldsymbol c} c
κ ~ ( s ) = κ ( s ) \widetilde{\kappa}(s) = \kappa(s) κ ( s ) = κ ( s ) τ ~ ( s ) = − τ ( s ) \widetilde{\tau}(s) = -\tau(s) τ ( s ) = − τ ( s ) \begin{pmatrix} \widetilde{\boldsymbol c}(1) & \widetilde{\boldsymbol c}'(1) & \widetilde{\boldsymbol c}''(1) \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 2 & -1 & 0 \\ 3 & 0 & \sqrt{2} \end
解
按照曲线论基本定理，我们需要寻找一个正交变换
c ~ ( s ) = A c ( s ) + b , ∣ A ∣ = − 1 , b ∈ R 3 \widetilde{\boldsymbol c}(s) = A \boldsymbol c(s) + \boldsymbol b,\quad |A| = -1,\ \boldsymbol b \in \mathbb R^3
c ( s ) = A c ( s ) + b , ∣ A ∣ = − 1 , b ∈ R 3
将曲线进行两次微分
c ′ ( s ) = ( 1 2 cos s 2 1 2 sin s 2 1 2 ) , c ′ ′ ( s ) = ( − 2 s sin s 2 2 s cos s 2 0 ) \boldsymbol c'(s) = \begin{pmatrix} \frac{1}{\sqrt{2}} \cos s^2 \\ \frac{1}{\sqrt{2}} \sin s^2 \\ \frac{1}{\sqrt{2}} \end{pmatrix},\quad
\boldsymbol c''(s) = \begin{pmatrix} -\sqrt{2} s \sin s^2 \\ \sqrt{2} s \cos s^2 \\ 0 \end{pmatrix} c ′ ( s ) = ⎝ ⎜ ⎜ ⎛ 2 1 cos s 2 2 1 sin s 2 2 1 ⎠ ⎟ ⎟ ⎞ , c ′ ′ ( s ) = ⎝ ⎛ − 2 s sin s 2 2 s cos s 2 0 ⎠ ⎞
首先求解原曲线位于 s = 1 s = 1 s = 1
t ( 1 ) = c ′ ( 1 ) = ( 1 2 cos 1 1 2 sin 1 1 2 ) n ( 1 ) = c ′ ′ ( 1 ) ∥ c ′ ′ ( 1 ) ∥ = ( − sin 1 cos 1 0 ) b ( 1 ) = t ( 1 ) × n ( 1 ) = ( − 1 2 cos 1 − 1 2 sin 1 1 2 ) \begin{aligned}
\boldsymbol t(1) &= \boldsymbol c'(1) = \begin{pmatrix} \frac{1}{\sqrt{2}} \cos 1 \\ \frac{1}{\sqrt{2}} \sin 1 \\ \frac{1}{\sqrt{2}} \end{pmatrix} \\
\boldsymbol n(1) &= \frac{\boldsymbol c''(1)}{\|\boldsymbol c''(1)\|} = \begin{pmatrix} -\sin 1 \\ \cos 1 \\ 0 \end{pmatrix} \\
\boldsymbol b(1) &= \boldsymbol t(1) \times \boldsymbol n(1) = \begin{pmatrix} -\frac{1}{\sqrt{2}} \cos 1 \\ -\frac{1}{\sqrt{2}} \sin 1 \\ \frac{1}{\sqrt{2}} \end{pmatrix}
\end{aligned} t ( 1 ) n ( 1 ) b ( 1 ) = c ′ ( 1 ) = ⎝ ⎜ ⎜ ⎛ 2 1 cos 1 2 1 sin 1 2 1 ⎠ ⎟ ⎟ ⎞ = ∥ c ′ ′ ( 1 ) ∥ c ′ ′ ( 1 ) = ⎝ ⎛ − sin 1 cos 1 0 ⎠ ⎞ = t ( 1 ) × n ( 1 ) = ⎝ ⎜ ⎜ ⎛ − 2 1 cos 1 − 2 1 sin 1 2 1 ⎠ ⎟ ⎟ ⎞
同样地，求解目标曲线位于 s = 1 s = 1 s = 1
t ~ ( 1 ) = c ~ ′ ( 1 ) = ( 0 − 1 0 ) n ~ ( 1 ) = c ~ ′ ′ ( 1 ) ∥ c ~ ′ ′ ( 1 ) ∥ = ( 0 0 1 ) b ~ ( 1 ) = t ~ ( 1 ) × n ~ ( 1 ) = ( − 1 0 0 ) \begin{aligned}
\widetilde{\boldsymbol t}(1) &= \widetilde{\boldsymbol c}'(1) = \begin{pmatrix} 0 \\ -1 \\ 0 \end{pmatrix} \\
\widetilde{\boldsymbol n}(1) &= \frac{\widetilde{\boldsymbol c}''(1)}{\|\widetilde{\boldsymbol c}''(1)\|} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \\
\widetilde{\boldsymbol b}(1) &= \widetilde{\boldsymbol t}(1) \times \widetilde{\boldsymbol n}(1) = \begin{pmatrix} -1 \\ 0 \\ 0 \end{pmatrix}
\end{aligned} t ( 1 ) n ( 1 ) b ( 1 ) = c ′ ( 1 ) = ⎝ ⎛ 0 − 1 0 ⎠ ⎞ = ∥ c ′ ′ ( 1 ) ∥ c ′ ′ ( 1 ) = ⎝ ⎛ 0 0 1 ⎠ ⎞ = t ( 1 ) × n ( 1 ) = ⎝ ⎛ − 1 0 0 ⎠ ⎞
因此，建立方程 A ( t ( 1 ) n ( 1 ) b ( 1 ) ) = ( t ~ ( 1 ) n ~ ( 1 ) − b ~ ( 1 ) ) A \begin{pmatrix} \boldsymbol t(1) & \boldsymbol n(1) & \boldsymbol b(1) \end{pmatrix} = \begin{pmatrix} \widetilde{\boldsymbol t}(1) & \widetilde{\boldsymbol n}(1) & -\widetilde{\boldsymbol b}(1) \end{pmatrix} A ( t ( 1 ) n ( 1 ) b ( 1 ) ) = ( t ( 1 ) n ( 1 ) − b ( 1 ) ) ( t ( 1 ) n ( 1 ) b ( 1 ) ) \begin{pmatrix} \boldsymbol t(1) & \boldsymbol n(1) & \boldsymbol b(1) \end{pmatrix} ( t ( 1 ) n ( 1 ) b ( 1 ) )
A = ( 0 0 1 − 1 0 0 0 1 0 ) ( 1 2 cos 1 1 2 sin 1 1 2 − sin 1 cos 1 0 − 1 2 cos 1 − 1 2 sin 1 1 2 ) = ( − 1 2 cos 1 − 1 2 sin 1 1 2 − 1 2 cos 1 − 1 2 sin 1 − 1 2 − sin 1 cos 1 0 ) A = \begin{pmatrix} 0 & 0 & 1 \\ -1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}
\begin{pmatrix}
\frac{1}{\sqrt{2}} \cos 1 & \frac{1}{\sqrt{2}} \sin 1 & \frac{1}{\sqrt{2}} \\
-\sin 1 & \cos 1 & 0 \\
-\frac{1}{\sqrt{2}} \cos 1 & -\frac{1}{\sqrt{2}} \sin 1 & \frac{1}{\sqrt{2}}
\end{pmatrix} = \begin{pmatrix}
-\frac{1}{\sqrt{2}} \cos 1 & -\frac{1}{\sqrt{2}} \sin 1 & \frac{1}{\sqrt{2}} \\
-\frac{1}{\sqrt{2}} \cos 1 & -\frac{1}{\sqrt{2}} \sin 1 & -\frac{1}{\sqrt{2}} \\
-\sin 1 & \cos 1 & 0
\end{pmatrix} A = ⎝ ⎛ 0 − 1 0 0 0 1 1 0 0 ⎠ ⎞ ⎝ ⎛ 2 1 cos 1 − sin 1 − 2 1 cos 1 2 1 sin 1 cos 1 − 2 1 sin 1 2 1 0 2 1 ⎠ ⎞ = ⎝ ⎛ − 2 1 cos 1 − 2 1 cos 1 − sin 1 − 2 1 sin 1 − 2 1 sin 1 cos 1 2 1 − 2 1 0 ⎠ ⎞
最后反解出 b \boldsymbol b b
b = c ~ ( 1 ) − A c ( 1 ) = ( 1 2 3 ) − ( − 1 2 cos 1 − 1 2 sin 1 1 2 − 1 2 cos 1 − 1 2 sin 1 − 1 2 − sin 1 cos 1 0 ) ( 1 2 ∫ 0 1 cos u 2 d u 1 2 ∫ 0 1 sin u 2 d u 1 2 ) \boldsymbol b = \widetilde{\boldsymbol c}(1) - A \boldsymbol c(1)
=\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}
-\begin{pmatrix}
-\frac{1}{\sqrt{2}} \cos 1 & -\frac{1}{\sqrt{2}} \sin 1 & \frac{1}{\sqrt{2}} \\
-\frac{1}{\sqrt{2}} \cos 1 & -\frac{1}{\sqrt{2}} \sin 1 & -\frac{1}{\sqrt{2}} \\
-\sin 1 & \cos 1 & 0
\end{pmatrix}
\begin{pmatrix}
\frac{1}{\sqrt{2}} \int_0^1 \cos u^2 du \\
\frac{1}{\sqrt{2}} \int_0^1 \sin u^2 du \\
\frac{1}{\sqrt{2}}
\end{pmatrix} b = c ( 1 ) − A c ( 1 ) = ⎝ ⎛ 1 2 3 ⎠ ⎞ − ⎝ ⎛ − 2 1 cos 1 − 2 1 cos 1 − sin 1 − 2 1 sin 1 − 2 1 sin 1 cos 1 2 1 − 2 1 0 ⎠ ⎞ ⎝ ⎜ ⎜ ⎛ 2 1 ∫ 0 1 cos u 2 d u 2 1 ∫ 0 1 sin u 2 d u 2 1 ⎠ ⎟ ⎟ ⎞
注：就该问题而言，至此可以算是解答完毕，无需求出 b \boldsymbol b b
∫ 0 1 cos u 2 d u = ∫ 0 1 ( ∑ n = 0 ∞ ( − 1 ) n u 4 n ( 2 n ) ! ) d u = ∑ n = 0 ∞ ( − 1 ) n ( 2 n ) ! ∫ 0 1 u 4 n d u = ∑ n = 0 ∞ ( − 1 ) n ( 2 n ) ! ( 4 n + 1 ) ≈ 0.9 \int_0^1 \cos u^2 du = \int_0^1 \left( \sum_{n=0}^{\infty} \frac{(-1)^n u^{4n}}{(2n)!} \right) du = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} \int_0^1 u^{4n} du = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!(4n+1)} \approx 0.9
∫ 0 1 cos u 2 d u = ∫ 0 1 ( n = 0 ∑ ∞ ( 2 n ) ! ( − 1 ) n u 4 n ) d u = n = 0 ∑ ∞ ( 2 n ) ! ( − 1 ) n ∫ 0 1 u 4 n d u = n = 0 ∑ ∞ ( 2 n ) ! ( 4 n + 1 ) ( − 1 ) n ≈ 0 . 9
sin \sin sin □ \square □
