Torsion at a Point on a Curve

Recall that geometrically, the curvature $\kappa$ of a curve $C$ represented the rate of change of the direction of the unit tangent vector as a point traverses the curve. We will now look at another property of space curves known as their torsion which is the rate of change of the direction of the unit binormal vector.

Definition: Let $\vec{r}(s) = (x(s), y(s), z(s))$ be a vector-valued function with an arc length parameterization that traces the smooth curve $C$. Then for all $s$ such that $\kappa (s) \neq 0$, the Torsion of $C$ at $\vec{r}(s)$ is the value $\tau (s)$ such that $\frac{d \hat{B}}{ds} = - \tau (s) \hat{N} (s)$.

Often times it is not easy to calculate the torsion using the formula above as it may be difficult to obtain an arc-length parameterization of $\vec{r}(t) = (x(t), y(t), z(t))$. The following theorem will let us compute the torsion at $t$ for any general parameterization of $C$.

Theorem 1: Let $\vec{r}(t) = (x(t), y(t), z(t))$ be a vector-valued function that traces the smooth curve $C$. Then the torsion of $C$ is also given by the following formula $\tau (t) = \frac{( \vec{r'}(t) \times \vec{r''}(t)) \cdot \vec{r'''}(t)}{\| \vec{r'}(t) \times \vec{r''}(t) \|^2}$.

The formula in theorem 1 can also be written in terms of velocity and acceleration, and jerk functions, that is $\vec{r'}(t) = \vec{v}(t)$ and $\vec{r''}(t) = \vec{a}(t)$ and $\vec{r'''}(t) = \vec{j}(t)$, and so the torsion can be calculated with the formula $\tau (t) = \frac{\vec{v}(t) \times \vec{a}(t)) \cdot \vec{j}(t)}{\| \vec{v}(t) \times \vec{a}(t) \|^2}$. Do NOT to confuse $\vec{j}(t)$ with the standard basis vector $(0, 1, 0)$!

Before we look at the proof to theorem 1, we will be using two Frenet-Serret formulas which we haven't yet discussed yet. That is $\frac{d\hat{T}}{ds} = \kappa \hat{N}$ and $\frac{d\hat{N}}{ds} = - \kappa \hat{T} + \tau \hat{B}$.

• Proof: Let $\vec{v} = \vec{v}(t) = \vec{r'}(t)$, $\vec{a} = \vec{a}(t) = \vec{r''}(t)$, $\vec{j} = \vec{j}(t) = \vec{r'''}(t)$ and $v = v(t) = \| \vec{v}(t) \| = \| \vec{r'}(t) \|$.

• First notice that $\vec{v} = \frac{d \vec{r}}{dt} = \frac{d \vec{r}}{ds} \frac{ds}{dt} = v \hat{T}$.

• If we differentiate $\vec{v}$, we get $\vec{a}$. Applying the product rule, we have that:

(1)

\begin{align} \quad \frac{d\vec{v}}{dt} = \vec{a} = \frac{dv}{dt} \hat{T} + v \frac{d\hat{T}}{dt} \\ \quad \frac{d\vec{v}}{dt} = \vec{a} = \frac{dv}{dt} \hat{T} + v \frac{d\hat{T}}{ds} \frac{ds}{dt} \\ \quad \frac{d\vec{v}}{dt} = \vec{a} = \frac{dv}{dt} \hat{T} + v \kappa \hat{N} v \\ \quad \frac{d\vec{v}}{dt} = \vec{a} = \frac{dv}{dt} \hat{T} + v^2 \kappa \hat{N} \end{align}

• Now compute the cross product $\vec{v} \times \vec{a}$. Note that$\vec{T} \times \vec{T} = 0$ and by the Frenet-Serret formula $\frac{d\hat{T}}{ds} = \kappa \hat{N}$, so it follows that:

(2)

\begin{align} \quad \vec{v} \times \vec{a} = \left [ v \hat {T} \right ] \times \left [ \frac{dv}{dt} \hat{T} + v^2 \kappa \hat{N} \right ] \\ \quad \vec{v} \times \vec{a} = \left [ v \hat {T} \right ] \times \left [ \frac{dv}{dt} \hat{T} \right ] + \left [ v \hat {T} \right ] \times \left [ v^2 \kappa \hat{N} \right ] \\ \quad \vec{v} \times \vec{a} = \left [ v \hat {T} \right ] \times \left [ v^2 \kappa \hat{N} \right ] \\ \quad \vec{v} \times \vec{a} = v^3 \kappa [ \hat{T} \times \hat{N}] \\ \quad \vec{v} \times \vec{a} = v^3 \kappa \hat{B} \end{align}

• Therefore the unit binormal vector $\hat{B}$ goes in the same direction as $\vec{v} \times \vec{a}$ since $\tau > 0$ and $v > 0$ and so $\hat{B} = \frac{\vec{v} \times \vec{a}}{\| \vec{v} \times \vec{a} \|}$. Also note that $\| \vec{v} \times \vec{a} \| = v^3 \kappa$ since $\hat{B}$ has unit length.

• If we were to differentiate $\vec{a}$ we would have many terms as we would have to calculate:

(3)

\begin{align} \quad \frac{d\vec{a}}{dt} = \vec{j} = \frac{d}{dt} \left ( \frac{dv}{dt} \hat{T} \right) + \left ( v^2 \kappa \hat{N} \right ) \end{align}

• Notice that when we differentiate this, there is a single term containing $\hat{B}$ which occurs in evaluating through the product rule $v^2 \kappa \frac{d\hat{N}}{dt} = v^3 \kappa \frac{d \hat{N}}{ds} = v^3 \kappa (\tau \hat{B} - \kappa \hat{T})$ by the Frenet-Serret formula $\frac{d\hat{N}}{ds} = \tau \hat{B} - \kappa \hat{T}$, and so for some scalars $\lambda, \mu$ we have that $\frac{da}{dt} = \lambda \hat{T} + \mu \hat{N} + v^3 \kappa \tau \hat{B}$.

• And so finally we have (nothing that $\hat{B} \cdot \hat{T} = 0$ and $\hat{B} \cdot \hat{N} = 0$ since these vectors are orthogonal, and $\vec{B} \cdot \vec{B} = \| \hat{B} \| ^2 = 1$ since $\hat{B}$ is a unit vector):

(4)

\begin{align} \quad (\vec{v} \times \vec{a}) \cdot \vec{j} = \left ( v^3 \kappa \hat{B} \right ) \cdot \left ( \lambda \hat{T} + \mu \hat{N} + v^3 \kappa \tau \hat{B} \right ) \\ \quad (\vec{v} \times \vec{a}) \cdot \vec{j} = v^6 \kappa^2 \tau \\ \quad (\vec{v} \times \vec{a}) \cdot \vec{j} = (v^3 \kappa )^2 \tau \\ \quad (\vec{v} \times \vec{a}) \cdot \vec{j} = \| \vec{v} \times \vec{a} \|^2 \tau \\ \end{align}

• Isolating for $\tau$ we get $\tau = \frac{(\vec{v} \times \vec{a}) \cdot \vec{j}}{\| \vec{v} \times \vec{a} \|^2}$ which of course can be written as $\tau (t) = \frac{(\vec{r'}(t) \times \vec{r''}(t)) \cdot \vec{r'''}(t)}{\| \vec{r'}(t) \times \vec{r''}(t) \| ^2}$. $\blacksquare$

Remark: You should notice that in the formula for torsion, $\tau(t) = \frac{( \vec{r'}(t) \times \vec{r''}(t)) \cdot \vec{r'''}(t)}{\| \vec{r'}(t) \times \vec{r''}(t) \|^2}$, that $(\vec{r'}(t) \times \vec{r''}(t)) \cdot \vec{r'''}(t)$ is the scalar triple product between $\vec{r'}(t)$, $\vec{r''}(t)$, and $\vec{r'''}(t)$. Therefore, we know that $(\vec{r'}(t) \times \vec{r''}(t)) \cdot \vec{r'''}(t) = \begin{vmatrix} x'(t) & y'(t) & z'(t)\\ x''(t) & y''(t) & z''(t)\\ x'''(t) & y'''(t) & z'''(t) \end{vmatrix}$. Often times, evaluating this determinant can be useful for calculating the numerator for the torsion formula, however, it is still necessary to calculate the cross product $\vec{r'}(t) \times \vec{r''}(t)$ and then its magnitude.