Visualising Torsion of a Curve

Question

For a space curve $\dot b = -\tau n$ where $\tau$ is the torsion. This says that $\dot b$ is parallel to $n$. Let’s assume a circle lies in $xy-plane$ and a circular helix is spiralling up along $z-axis$. Since $\dot b$ is parallel to $n$ it will lie in the $xy-plane$, now what makes the torsion zero for a circle and nonzero for a circular helix? As a point moves along a circle as well along a helix why is torsion zero in one case and not zero in the other? Or to put it this way, we say torsion tells that how much a curve deviates from being a plane curve, if so how does one see this from $\dot b = -\tau n$? If $\dot b$ is parallel to $n$ why can’t we write $\dot b = -\tau n$ for a plane curve, say circle?

Answer 1

Given a regular curve $\boldsymbol{\alpha}(t)$ (therefore of class $C^1$ and with $\boldsymbol{\alpha}'(t) \ne \boldsymbol{0}$), tangent, normal and binormal versors are defined as:

$$ \mathbf{T}(t) = \frac{\boldsymbol{\alpha}'(t)}{||\boldsymbol{\alpha}'(t)||}\,, \; \; \; \; \; \; \mathbf{N}(t) = \frac{\mathbf{T}'(t)}{||\mathbf{T}'(t)||} \, (\text{when} \; \mathbf{T}'(t) \ne \boldsymbol{0})\,, \; \; \; \; \; \; \mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t)\,. $$

So, starting from these definitions, in turn we define the curvature and the torsion:

$$ \kappa(t) = \frac{||\mathbf{T}'(t)||}{||\boldsymbol{\alpha}'(t)||}\,, \; \; \; \; \; \; \tau(t) = -\frac{\mathbf{B}'(t) \cdot \mathbf{N}(t)}{||\boldsymbol{\alpha}'(t)||}\,, $$

which respectively measure how far a curve moves away from being a straight line and from being contained in a plane.

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In particular, given a regular curve parameterized as:

$$ \boldsymbol{\alpha}(t) := \left(r\,\cos t, \; r\,\sin t, \; c\,t\right) \; \; \; \text{with} \; t \in [0,\,2\pi) $$

we get:

$$ \begin{aligned} & \boldsymbol{\alpha}'(t) = \left(-r\,\sin t, \; r\,\cos t, \; c\right); \\ & ||\boldsymbol{\alpha}'(t)|| = \sqrt{r^2+c^2}\;; \\ & \mathbf{T}(t) = \left(-\frac{r}{\sqrt{r^2+c^2}}\,\sin t, \; \frac{r}{\sqrt{r^2+c^2}}\,\cos t, \; \frac{c}{\sqrt{r^2+c^2}}\right); \\ & \mathbf{T}'(t) = \left(-\frac{r}{\sqrt{r^2+c^2}}\,\cos t, \; -\frac{r}{\sqrt{r^2+c^2}}\,\sin t, \; 0\right); \\ & ||\mathbf{T}'(t)|| = \frac{r}{\sqrt{r^2+c^2}}\;; \\ & \mathbf{N}(t) = \left(-\cos t, \; -\sin t, \; 0\right); \\ & \mathbf{B}(t) = \left(\frac{c}{\sqrt{r^2+c^2}}\,\sin t, \; -\frac{c}{\sqrt{r^2+c^2}}\,\cos t, \; \frac{r}{\sqrt{r^2+c^2}}\right); \\ & \mathbf{B}'(t) = \left(\frac{c}{\sqrt{r^2+c^2}}\,\cos t, \; \frac{c}{\sqrt{r^2+c^2}}\,\sin t, \; 0\right); \end{aligned} $$

from which:

$$ \kappa(t) = \frac{r}{r^2+c^2}\,, \; \; \; \; \; \; \tau(t) = \frac{c}{r^2+c^2}\,, $$

i.e.

• if $c = 0$ (circle) we have $\tau(t) = 0$, then we deduce that it is a plane curve;
• if $c \ne 0$ (helix) we have $\tau(t) \ne 0$, then we deduce that it isn't a plane curve.

Answer 2

We need to look at how Frenet-Serret relations are derived. Imagine a circular band or belt in a horizontal plane . All vectors $n$ normal to the band remain in a plane. and derivative $\dot b $ of bi-normal that is vertical up rotating in the normal plane of $(n,b)$ is always directed perpendicular to $ n$. Dot product of unit vectors gives a zero torsion.

If the band twists, angle between$(\dot b, n) $ becomes less or more than $90^{\circ}$ in the normal plane of $(n,b)$ by an out-of-plane rotation of $b$ vector so that their dot product is non-zer0...by vector definition torsion is non-zero.

Answer 3

Of course it's true for a plane curve that $\dot b = -\tau n$. Since $b$ is constant, we deduce that $\tau = 0$. There's no contradiction here.