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Frenet-Serret Formulas 📂Geometry

Frenet-Serret Formulas

Formulas 1

If α\alpha is a unit speed curve with κ(s)≠0\kappa (s) \ne 0 then T′(s)=κ(s)N(s)N′(s)=−κ(s)T(s)+τ(s)B(s)B′(s)=−τ(s)N(s) \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. [TNB]′=[0κ0−κ0τ0−τ0][TNB] \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 nn-dimensional inner product space VV, if E={e1,⋯ ,en}E = \left\{ e_{1} , \cdots , e_{n} \right\} is an orthogonal set, then EE forms a basis of VV, and for all v∈Vv \in V v=∑k=1n<v,ek>ek v = \sum_{k=1}^{n} \left< v , e_{k} \right> e_{k}

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

The Frenet-Serret Frame {T,N,B}\left\{ T, N, B \right\} forms an orthogonal basis of R3\mathbb{R}^{3}. It is directly derived using the above lemma.

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

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

Part 2. N′(s)=−κ(s)T(s)+τ(s)B(s)N^{\prime}(s) = - \kappa (s) T(s) + \tau (s) B(s)

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

  • Part 2-1. <N′,T>=−κ\left< N^{\prime} , T \right> = -\kappa
    • Since <N,T>=0\left< N, T \right> = 0, according to Part 1 0′=<N,T>′=<N′,T>+<N,T′>  ⟹  <N′,T>=−<N,T′>  ⟹  <N′,T>=−<N,κN>=−κ∣N2∣=−κ⋅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. <N′,N>=0\left< N^{\prime} , N \right> = 0
    • Since NN is a unit vector, ∣N2∣=1\left| N^{2} \right| = 1 and by differentiating both sides 0=1′=<N,N>′=2<N,N′>  ⟹  <N,N′>=0 \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. <N′,B>=τ\left< N^{\prime} , B \right> = \tau
    • Since <N,B>=0\left< N, B \right> = 0, according to the definition of torsion τ(s):=−<B′(s),N(s)>\tau (s) := - \left< B^{\prime}(s) , N (s) \right> 0′=<N,B>′=<N′,B>+<N,B′>  ⟹  <N′,B>=−<N,B′>  ⟹  <N′,T>=τ \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′(s)=−κ(s)T(s)+τ(s)B(s) N^{\prime}(s) = - \kappa (s) T(s) + \tau (s) B(s)

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

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

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

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

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Corollaries

  • Lancret’s Theorem: For a unit speed curve α\alpha being a helix is equivalent to some constant c∈Rc \in \mathbb{R} for which τ=cκ\tau = c \kappa.
  • The curvature of the unit speed curve α\alpha being a constant κ>0\kappa > 0 and the torsion being τ=0\tau = 0 is equivalent to α\alpha being an arc of a circle with radius κ−1\kappa^{-1}.
  • α\alpha being a straight line is equivalent to all tangents of α\alpha passing through some point x0∈R3x_{0} \in \mathbb{R}^{3}.
  • Let’s take a unit speed curve α\alpha with κ≠0\kappa \ne 0.

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

  • If all normal planes of the unit speed curve α\alpha point towards some fixed point x0∈R3\mathbf{x}_{0} \in \mathbb{R}^{3}, then α\alpha lies on a sphere.

  1. Millman. (1977). Elements of Differential Geometry: p30. ↩︎