📂Geometry

Reparameterization and the Tools of Frenet-Serret

Definition

Let’s call $\beta : [a,b] \to \mathbb{R}^{3}$ a regular curve. The arc length reparametrization $t = t(s)$ satisfies $s(t) = \int_{a}^{t} \left| \beta^{\prime}(t) \right| dt$, and the Frenet-Serre apparatus of the unit speed curve $\alpha (s) := \beta \left( t (s) \right)$ $$\left\{ \kappa_{\alpha} \left( s(t) \right), \tau_{\alpha} \left( s(t) \right) , T_{\alpha} \left( s(t) \right) , N_{\alpha} \left( s(t) \right), B_{\alpha} \left( s(t) \right) \right\}$$ is defined as the Frenet-Serre apparatus of $\beta$.

Explanation

The definition of the Frenet-Serre apparatus was generalized for regular curves. This is a common method used throughout mathematics, where the unit speed is enforced if not already present. For a bijective arc length reparametrization, one can consider $\alpha$, which is not $\beta$ itself but incorporates its geometry.

Notation

$${{ df } \over { ds }} = f^{\prime} \quad \text{and} \quad {{ df } \over { dt }} = \dot{f}$$

Dot $\dot{}$ and prime $'$ both indicate differentiation, but in the context of differential geometry, symbols are distinguished as above. Usually, $s$ represents the parameter of a unit speed curve, and $t = t(s)$ is the parameter of the curve after arc length reparametrization. Additionally, for representing the scalar triple product, the following ternary symbols are used for cross product $\times$ and dot product $\left< \cdot, \cdot \right>$. $$\left[ \mathbf{u} , \mathbf{v} , \mathbf{w} \right] := \left< \mathbf{u} \times \mathbf{v} , \mathbf{w} \right>$$

Theorem ¹

If $\beta (t)$ is a regular curve

• (a): $\displaystyle T = {{ \dot{\beta} } \over { \left| \dot{\beta} \right| }}$
• (b): $\displaystyle B = {{ \dot{\beta} \times \ddot{\beta} } \over { \left| \dot{\beta} \times \ddot{\beta} \right| }}$
• (c): $N = B \times T$
• (d): $\displaystyle \kappa = {{ \left| \dot{\beta} \times \ddot{\beta} \right| } \over { \left| \dot{\beta} \right|^{3} }}$
• (e): $\displaystyle \tau = {{ \left[ \dot{\beta} , \ddot{\beta}, \overset{\cdot\cdot\cdot}{\beta} \right] } \over { \left| \dot{\beta} \times \ddot{\beta} \right|^{2} }}$

Proof

Frenet-Serre Formulas: If $\alpha$ is a unit speed curve such that $$\begin{align*} T^{\prime}(s) =& \kappa (s) N(s) \\ N^{\prime}(s) =& - \kappa (s) T(s) + \tau (s) B(s) \\ B^{\prime}(s) =& - \tau (s) N(s) \end{align*}$$

Following the Notation introduced, we can easily deduce: $${{ ds } \over { dt }} = \dot{s} = \left| \dot{\beta} \right| = {{ d } \over { dt }} \int_{a}^{t} \left| \beta^{\prime}(t) \right| dt$$ Especially, $\dot{T}$ is according to the Frenet-Serre formula $$\dot{T} = {{ dT } \over { dt }} = {{ dT } \over { ds }} {{ ds } \over { st }} = \kappa N \dot{s}$$

(a)

$$\begin{align*} \dot{\beta} =& \alpha^{\prime} \dot{s} \\ =& T \dot{s} \\ =& T \left| \dot{\beta} \right| \end{align*}$$ Going beyond $\left| \dot{\beta} \right|$ $$T = {{ \dot{\beta} } \over { \left| \dot{\beta} \right| }}$$

(d)

For convenience, we first prove $(d)$. $$\dot{\beta} = \dot{s} T$$ Differentiating both sides by $t$ gives $$\begin{align*} \ddot{\beta} =& \ddot{s} T + \dot{s} \dot{T} \\ =& \ddot{s} T + \dot{s}^{2} \dot{T}^{\prime} \\ =& \ddot{s} T + \kappa \dot{s}^{2} N \end{align*}$$ Thus $$\begin{align*} \left| \dot{\beta} \times \ddot{\beta} \right| =& \left| \dot{s} T \times \left( \ddot{s} T + \kappa \dot{s}^{2} N \right) \right| \\ =& \left| \kappa \dot{s}^{3} B \right| \\ =& \kappa \dot{s}^{3} \end{align*}$$ As $\dot{s} = \left| \dot{\beta} \right|$, then $$\kappa = {{ \left| \dot{\beta} \times \ddot{\beta} \right| } \over { \left| \dot{\beta} \right|^{3} }}$$

(b)

From the proof of (d), if $\kappa \ne 0$ then $$\begin{align*} B =& {{ \dot{\beta} \times \ddot{\beta} } \over { \kappa \dot{s}^{3} }} \\ =& {{ \dot{\beta} \times \ddot{\beta} } \over { \left| \dot{\beta} \times \ddot{\beta} \right| }} \end{align*}$$

(c)

It’s obvious.

(e)

Differentiating $\ddot{\beta} = \ddot{s} T + \dot{s} \dot{T}$ once more with $t$ gives $$\begin{align*} \overset{\cdot\cdot\cdot}{\beta} =& \overset{\cdot\cdot\cdot}{s} T + \ddot{s} \dot{T}+\left(\kappa \dot{s}^{2}\right)^{\cdot} N+\kappa \dot{s}^{2} \dot{N} \\ =& \overset{\cdot\cdot\cdot}{s} T+\dot{s} \ddot{s} T^{\prime} + \left(\kappa \dot{s}^{2}\right)^{\cdot} N + \kappa \dot{s}^{3} N^{\prime} \\ =& \overset{\cdot\cdot\cdot}{s} T+\kappa \dot{s} \ddot{s} N+\left(\kappa \dot{s}^{2}\right)^{\cdot} N-\kappa^{2} \dot{s}^{3} T+\kappa \tau \dot{s}^{3} \mathbf{B} \\ =& \left(\overset{\cdot\cdot\cdot}{s}-\kappa^{2} \dot{s}^{3}\right) T+\left(\kappa \dot{s} \ddot{s}+\left(\kappa \dot{s}^{2}\right)^{\cdot}\right) N+\kappa \tau \dot{s}^{3} \mathbf{B} \end{align*}$$ As the scalar triple product is $\left[ \dot{\beta} , \ddot{\beta}, \overset{\cdot\cdot\cdot}{\beta} \right] = \left< \dot{\beta} \times \ddot{\beta} , \overset{\cdot\cdot\cdot}{\beta} \right>$ and with $T \perp B$, $N \perp B$, one finds $$\begin{align*} \left[ \dot{\beta} , \ddot{\beta}, \overset{\cdot\cdot\cdot}{\beta} \right] =& \left< \dot{\beta} \times \ddot{\beta} , \overset{\cdot\cdot\cdot}{\beta} \right> \\ =& \left< \kappa \dot{s}^{3} B , \overset{\cdot\cdot\cdot}{\beta} \right> \\ =& \tau \left( \kappa \dot{s}^{3} \right)^{2} \\ =& \tau \left| \dot{\beta} \times \ddot{\beta} \right|^{2} \end{align*}$$ In conclusion $$\tau = {{ \left[ \dot{\beta} , \ddot{\beta}, \overset{\cdot\cdot\cdot}{\beta} \right] } \over { \left| \dot{\beta} \times \ddot{\beta} \right|^{2} }}$$

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