📂Geometry

Frenet-Serret Formulas

Formulas ¹

If $\alpha$ is a unit speed curve with $\kappa (s) \ne 0$ then $$ \begin{align*} T^{\prime}(s) =& \kappa (s) N(s) \\ N^{\prime}(s) =& - \kappa (s) T(s) + \tau (s) B(s) \\ B^{\prime}(s) =& - \tau (s) N(s) \end{align*} $$

Description

In matrix form, it can be expressed as follows. $$ \begin{bmatrix} T \\ N \\ B \end{bmatrix} ^{\prime} = \begin{bmatrix} 0 & \kappa & 0 \\ - \kappa & 0 & \tau \\ 0 & - \tau & 0 \end{bmatrix} \begin{bmatrix} T \\ N \\ B \end{bmatrix} $$

Derivation

Lemma: In an $n$-dimensional inner product space $V$, if $E = \left\{ e_{1} , \cdots , e_{n} \right\}$ is an orthogonal set, then $E$ forms a basis of $V$, and for all $v \in V$ $$ v = \sum_{k=1}^{n} \left< v , e_{k} \right> e_{k} $$

Differentiation of Inner Products: $$\left< f, g \right>^{\prime} = \left< f^{\prime}, g \right> + \left< f, g^{\prime} \right>$$

The Frenet-Serret Frame $\left\{ T, N, B \right\}$ forms an orthogonal basis of $\mathbb{R}^{3}$. It is directly derived using the above lemma.

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Part 1. $T^{\prime}(s) = \kappa (s) N(s)$

From the definition of the normal vector, since $N(s) = {{ T^{\prime}(s) } \over { \kappa (s) }}$ then $$ T^{\prime}(s) = \kappa (s) N(s) $$

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Part 2. $N^{\prime}(s) = - \kappa (s) T(s) + \tau (s) B(s)$

According to the lemma $$ N^{\prime}(s) = \left< N^{\prime} , T \right> T + \left< N^{\prime} , N \right> N + \left< N^{\prime} , B \right> B $$

• Part 2-1. $\left< N^{\prime} , T \right> = -\kappa$
  • Since $\left< N, T \right> = 0$, according to Part 1 $$ \begin{align*} & 0^{\prime} = \left< N , T \right>^{\prime} = \left< N^{\prime} , T \right> + \left< N , T^{\prime} \right> \\ \implies& \left< N^{\prime} , T \right> = - \left< N , T^{\prime} \right> \\ \implies& \left< N^{\prime} , T \right> = - \left< N , \kappa N \right> = - \kappa \left| N^{2} \right| = - \kappa \cdot 1 \end{align*} $$
• Part 2-2. $\left< N^{\prime} , N \right> = 0$
  • Since $N$ is a unit vector, $\left| N^{2} \right| = 1$ and by differentiating both sides $$ \begin{align*} & 0 = 1^{\prime} = \left< N , N \right>^{\prime} = 2 \left< N , N^{\prime} \right> \\ \implies& \left< N , N^{\prime} \right> = 0 \end{align*} $$
• Part 2-3. $\left< N^{\prime} , B \right> = \tau$
  • Since $\left< N, B \right> = 0$, according to the definition of torsion $\tau (s) := - \left< B^{\prime}(s) , N (s) \right>$ $$ \begin{align*} & 0^{\prime} = \left< N , B \right>^{\prime} = \left< N^{\prime} , B \right> + \left< N , B^{\prime} \right> \\ \implies& \left< N^{\prime} , B \right> = - \left< N , B^{\prime} \right> \\ \implies& \left< N^{\prime} , T \right> = \tau \end{align*} $$

This leads to the following. $$ N^{\prime}(s) = - \kappa (s) T(s) + \tau (s) B(s) $$

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Part 3. $B^{\prime}(s) = - \tau (s) N(s)$

According to the lemma $$ B^{\prime}(s) = \left< B^{\prime} , T \right> T + \left< B^{\prime} , N \right> N + \left< B^{\prime} , B \right> B $$

• Part 3-1. $\left< B^{\prime} , T \right> = 0$
  • Since $\left< T, B \right> = 0 = \left< N, B \right>$, according to Part 1 $$ 0 = \left< T^{\prime}, B \right> + \left< T, B^{\prime} \right> = \kappa \left< N, B \right> + \left< T, B^{\prime} \right> = \left< T, B^{\prime} \right> $$
• Part 3-2. $\left< B^{\prime} , N \right> = -\tau$
  • According to the definition of torsion and the symmetry of the inner product $$ \left< B^{\prime} , N \right> = \left< N , B^{\prime} \right> = - \tau $$
• Part 3-3. $\left< B^{\prime} , B \right> = 0$
  • Since $\alpha$ is assumed to be a unit speed curve, $B = T \times N$ is also a unit vector. Similarly to Part 2-2 $$ 0 = \left< B^{\prime} , B \right> $$

This leads to the following. $$ B^{\prime}(s) = - \tau (s) N(s) $$

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Corollaries

• Lancret’s Theorem: For a unit speed curve $\alpha$ being a helix is equivalent to some constant $c \in \mathbb{R}$ for which $\tau = c \kappa$.
• The curvature of the unit speed curve $\alpha$ being a constant $\kappa > 0$ and the torsion being $\tau = 0$ is equivalent to $\alpha$ being an arc of a circle with radius $\kappa^{-1}$.
• $\alpha$ being a straight line is equivalent to all tangents of $\alpha$ passing through some point $x_{0} \in \mathbb{R}^{3}$.
• Let’s take a unit speed curve $\alpha$ with $\kappa \ne 0$.

$\alpha$ lying on a plane is equivalent to all tangent planes being parallel.

• If all normal planes of the unit speed curve $\alpha$ point towards some fixed point $\mathbf{x}_{0} \in \mathbb{R}^{3}$, then $\alpha$ lies on a sphere.