Space Curves

Torsion

Let \(\gamma\) be a unit-speed curve in \(\mathbb R^3\), \(\mathbf t = \gamma'\) be its unit tangent vector.

If \(\kappa\) is non-zero, then we define the principal normal of \(\gamma\) at the point \(\gamma(s)\) to be

\[\mathbf n(s) = \frac{1}{\kappa(s)} \mathbf t'(s)\]

Note that the principal normal is a unit vector, or normalized \(\gamma''\).

Then, define the binormal vector of \(\gamma\) at the point \(\gamma(s)\) to be \(\mathbf b = \mathbf t\times \mathbf n\), we then obtain an orthonormal, right-handed basis \(\{\mathbf t, \mathbf n, \mathbf b\}\), i.e.,

\[\mathbf b = \mathbf t\times \mathbf n, \mathbf n = \mathbf b\times \mathbf t, \mathbf t = \mathbf n \times \mathbf b\]

Then, if we differentiate \(\mathbf b\), we get

\[\begin{align*} \mathbf b' &= (\mathbf t\times \mathbf n)' \\ &=\mathbf t' \times \mathbf n + \mathbf t \times \mathbf n'\\ &= \mathbf t \times \mathbf n' &\mathbf t' = \kappa \mathbf n \text{, parallel to } \mathbf n \end{align*}\]

So that \(\mathbf b'\perp \mathbf t\). In addition, note that \(\mathbf b'\perp \mathbf b\) since \(\mathbf b\) is unit-length. Hence it must follows that

\[\mathbf b' = -\tau \mathbf n\]

We define such \(\tau\) as torsion.

Note that the torsion is only defined if the curvature is non-zero.

Formulas for Torsion

Theorem Let \(\gamma: (a,b)\rightarrow\mathbb R^3\) be a regular curve with nowhere-vanishing curvature. Then its torsion is given by

\[\tau = \frac{(\gamma' \times \gamma'')\cdot \gamma'''}{\|\gamma'\times \gamma''\|^2}\]

proof. We first sketch the proof by assuming \(\gamma\) is unit-speed so that

\[\begin{align*} -\tau \mathbf n\cdot \mathbf n &= -\mathbf n\cdot \mathbf b' \\ \tau &= \mathbf n \cdot (\mathbf t \times \mathbf n')\\ \tau &= \frac{1}{\kappa}\gamma''\cdot \big(\gamma' \times (\frac{1}{\kappa}\gamma''' - \frac{\kappa'}{\kappa^2}\gamma'')\big)\\ &= \kappa^{-2}\gamma'''\cdot (\gamma'\times \gamma'')\\ &= \frac{(\gamma' \times \gamma'')\cdot \gamma'''}{\|\gamma'\times \gamma''\|^2} \end{align*}\]

In general, we can use \(s\) to be the arc-length along \(\gamma\) so that \(\frac{d\gamma}{dt} = \frac{d\gamma}{ds}\frac{ds}{dt}\) and so on, and replaces each differentials.

Torsion for plane curves

Theorem For some \(\gamma\) be a regular curve in \(\mathbb R^3\) with nowhere vanishing curvature, \(\tau(s) = 0\) for all \(s\) IFF the image of \(\gamma\) is contained in some plane.

proof. Assume \(\gamma((a,b))\) is contained in some plane \(\{v: \mathbf v\cdot \mathbf N = d\}\) so that we have

\[\begin{align*} \gamma\cdot N &= d\\ \frac{d}{ds}(\gamma\cdot N) &= 0\\ \mathbf t \cdot \mathbf N &= 0\\ \mathbf t' \cdot \mathbf N &= 0\\ \kappa \mathbf n\cdot \mathbf N &= 0 \end{align*}\]

Therefore, both \(\mathbf t \perp\mathbf N\) and \(\mathbf n \perp\mathbf N\). Hence \(\mathbf b = \mathbf t\times \mathbf n\) is parallel to \(\mathbf N\). Since \(\mathbf N, \mathbf b\) are both unit vectors and \(\mathbf b(s)\) is smooth, we have \(\mathbf b = \mathbf N\) or \(\mathbf b = -\mathbf N\). So that \(\mathbf b' = 0\implies \tau = 0\).

Assume \(\tau = 0\) everywhere, so that \(\mathbf b' = 0\) and \(\mathbf b\) is constant. Then consider

\[\frac{d}{ds}(\gamma\cdot \mathbf n) = \gamma' \cdot \mathbf b = \mathbf t\times \mathbf b = 0\]

Hence \(\gamma\cdot \mathbf n = d\) for some constant \(d\).

Frenet-Serret Equations

Theorem Let \(\gamma\) be a unit speed curve in \(\mathbb R^3\) with nowhere vanishing curvature. Then

\[\begin{bmatrix}\mathbf t' \\\mathbf n' \\\mathbf b' \end{bmatrix} = \begin{bmatrix}0&\kappa&0\\-\kappa&0&\tau\\0&-\tau&0\end{bmatrix} \begin{bmatrix}\mathbf t \\\mathbf n \\\mathbf b \end{bmatrix} = \begin{bmatrix}\kappa\mathbf n \\-\kappa \mathbf t + \tau \mathbf b \\-\tau\mathbf n \end{bmatrix}\]

proof. We have already obtained \(\mathbf t', \mathbf n'\), for \(\mathbf n'\) we have

\[\mathbf n' = \mathbf b'\times \mathbf t + \mathbf b\times \mathbf t' = -\tau\mathbf n\times \mathbf t + \kappa\mathbf b \times \mathbf n = \tau \mathbf b - \kappa\mathbf t\]

Example: k = c, t = 0

Claim For \(\gamma\) be a unit-speed curve in 3D, if \(\kappa\) is a constant and \(\tau = 0\) for all \(s\), then, \(\gamma\) parameterizes a circle or an arc of a circle.

proof. Since \(\tau = 0\), let \(\Pi\) be the plane that \(\gamma\) lies, and \(\Pi\) has its plane normal be \(\mathbf b\).
By FS equation, \(\mathbf n' = -\kappa\mathbf t + \tau\mathbf b = -\kappa \mathbf t\). Then,

\[\frac{d}{ds}(\gamma + \frac{1}{\kappa}\mathbf n) = \mathbf t + \frac{1}{\kappa}\mathbf n' = \mathbf t- \mathbf t = 0\]

So that \(\gamma + \kappa^{-1}\mathbf n\) is a constant, and

\[\|\gamma - (\gamma + \frac{1}{k}\mathbf n)\| = \|-\frac{1}{k}\mathbf n\| = \kappa^{-1}\]

So that \(\gamma\) lies on the sphere \(S\) with center \(\gamma + \frac{1}{k}\mathbf n\) and radius \(\kappa^{-1}\) and the intersection of \(\Pi\).

Space Curve via k and t

Lemma Let \(A\) be a skew-symmetric \(3\times 3\) matrix (\(a_{ij} = -a_{ji}\)). Let \(v_1,v_2,v_3\) be smooth functions of a parameter \(s\) satisfying the differential equations \(v_i' = \sum_{j=1}^3 a_{ij}v_j\) and suppose that for some \(s_0\), \(v_1(s_0), v_2(s_0), v_3(s_0)\) are orthonormal. Show \(\forall s. v_1(s), v_2(s), v_3(s)\) are orthonormal.

proof. We aim to show that \(v_i\cdot v_j = 1\) for all \(i\neq j\). Note that

\[\begin{align*} \frac{d}{ds}(v_i\cdot v_j) &= v_i'\cdot v_j + v_i\cdot v_j' \\ &= \sum_{k=1}^3 a_{ik}v_k\cdot v_j + \sum_{k=1}^3 a_{jk}v_k \cdot v_i \\ &= \sum_{k=1}^3 a_{ik}v_k\cdot v_j - \sum_{k=1}^3 a_{kj}v_k \cdot v_i \end{align*}\]

Have a unique solution given that the condition satisfies for \(s=s_0\).

Theorem For \(k:\mathbb R\rightarrow \mathbb R^{>0}\) and \(t:\mathbb R\rightarrow \mathbb R\) be two smooth function, there is a unit-speed curve in \(\mathbb R^3\) whose curvature is \(k\) and torsion is \(t\).

proof. Following from FS equations, we want to solve for functions \(\mathbf T, \mathbf N, \mathbf B\) for some point \(s_0\) s.t. \(\mathbf T(s_0) = e_1, \mathbf N(s_0) = e_2, \mathbf B(s_0) = e_3\). Note that the FS matrix is skew-symmetric so that by lemma we have a unique solution for \(\mathbf T,\mathbf N, \mathbf B\) s.t. they are orthonormal for all \(s\). Then, let

\[\gamma(s) = \int_{s_0}^s \mathbf T(u)du\]

and we can verify all of \(\mathbf T, \mathbf N, \mathbf B, \gamma\) are unit-length.

Theorem And all curves with the same curvature and torsion are isomorphic to each other.