Convenient Formulae for Curvature and Torsion

The Frenet-Serret apparatus is a useful and important way to deal with the geometry of differentiable curves, with the curvature and torsion helping us to measure the way a curve bends and twists around in space. Understanding $\kappa$ and $\tau$ is important and will be our key to the shape of the curve, so it will be important that we can compute them efficiently. The direct way is always available: simply compute the whole framing $\{T, N, B\}$ and find $\kappa$ and $\tau$ when computing some pair of $T'$, $N'$ and $B'$. But this is very time consuming. Can we find a faster way?

Problem: Find formulations for computing the curvature and torsion of a unit-speed curve $x(s)$ that involves only the derivatives $x'$, $x''$ and $x'''$ and easily computed linear algebra operations.

As usual, a prime denotes differentiation with respect the arclength parameter. This parameter will be suppressed throughout this post.

Curvature:

This one pretty direct. By definition, we have that $\kappa^2 = \langle T', T' \rangle$. But $T = x'$ for a unit-speed curve, so

\[ \kappa^2 = \langle x'', x'' \rangle  \tag{curvature} \]

Torsion:

This will take just a little bit of work, and a fancy object called the triple product. As usual, we will make free use of the Frenet-Serret equations (FSE) when necessary. We start with the definition.

\[ \tau = - \langle N, B' \rangle =  - \langle N, (T\times N)' \rangle .\]

By the Leibniz rule for the cross product and the fact that a cross product is orthogonal to its inputs

\[ \tau = - \langle N, T'\times N \rangle - \langle N, T \times N' \rangle  = - \langle N, T \times N'\rangle .\]

Now one is tempted to use the Frenet-Serret equations here. But we want to get rid of references to $T$, $N$ and $B$! Instead, apply the basic definitions to see that 

\[ \tau = - \langle N, T\times N \rangle  = - \langle (\kappa^{-1} x'') ,  x' \times (\kappa^{-1}x'')' \rangle . \]

Now we will simplify the display of this formula, if not its actual computation, by making use of the triple product of three vectors in $\mathbb{R}^3$. Given three vectors $a, b, c \in \mathbb{R}^3$, their triple product is the real number

\[ (a\, b \, c) = \langle a , b\times c\rangle  = \det \begin{pmatrix} |& | & | \\ a & b & c \\ | & | & | \end{pmatrix},\]
where the vectors are interpreted as columns of a 3-by-3 matrix to compute the determinant.

The interpretation as a determinant is very useful, and direct from the standard formal method of computing a cross product. Note that the triple product can be interpreted as the oriented volume of the parallelepiped spanned by the three vectors.

Applying this, we find that 

\[ \tau = \dfrac{ (x' \, x'' \, x''')}{\kappa^2} = \dfrac{(x' \, x'' \, x''')}{\langle x'', x'' \rangle}   \tag{torsion}\]

This fits the bill. We have an expression for each of the curvature and the torsion that we can compute by taking derivatives of the original curve and applying simple linear algebra operations: the dot product and the cross product.