Lecture 02 - Torsion and Frames

#math

The torsion $τ$ of the space curve measures deviation from planarity. Thus a planar curve has zero torsion. The curvature and torsion of a curve are the intrinsic variables representing a curve.

Definition

Torsion is defined to be:

τ (t) = \frac{(r^{'} (t) \times r^{″} (t)) \cdot r^{‴} (t)}{| r^{'} (t) \times r^{″} (t) |^{2}}

Torsion essentially measures how the plane spanned by the tangent and normal of a curve $r (t)$ rotates as we go along the curve.

Frenet-Serret frame

Recall that the tangent vector of a curve is defined to be:

T (s) = r^{'} (s)

The unit normal vector is defined to be:

N (s) = \frac{r^{″}}{| r^{″} (s) |}

We define the binormal to be:

B (s) = T (s) \times N (s) .

The Frenet frame $(T, N, B)$ form an ordered triad of orthonormal vectors on the curve and they form a right handed system of axis. A straightforward application of the chain rule shows that these definitions are covariant with respect to reparameterizations. Hence, the three vectors should be conceived as being attached to the point $r (s)$ of the oriented space curve, rather than being functions of the parameter $s$ .

That is, a Frenet frame is a moving reference frame of 3 orthonormal vectors in $R^{3}$ which are used to describe a curve locally at each point $r (s)$ . In particular, in $R^{2}$ , $(T, N)$ form a moving frame whose derivatives may be expressed in terms of $(T, N)$ itself:

Note

If $r (s)$ is a planar curve (that is there is no torsion) we observe the following:

\begin{aligned} \frac{d T}{d s} & = κ N \\ \frac{d N}{d s} & = - κ T \end{aligned}

We obtain the latter equation from the fact that $T \cdot N = 0$ , thus $κ N \cdot N + \frac{d N}{d s} \cdot T = 0$ finally giving us $\frac{d N}{d s} \cdot T = - κ$ .

It is the main tool in the differential geometric treatment of curves as it is far easier and more natural to describe local properties (e.g. curvature, torsion) in terms of a local reference system than using a global one like the Euclidean coordinates.

By definition $B = T \times N$ thus:

\begin{aligned} B^{'} & = T^{'} \times N + T \times N^{'} \\ = κ N \times N + T \times N^{'} \\ = T \times N^{'} \end{aligned}

Thus $B^{'} \cdot T = 0$ . In a reference frame we have that:

\begin{aligned} B^{'} & = (B^{'} \cdot T) T + (B^{'} \cdot N) N + (B^{'} \cdot B) B \\ = (B^{'} \cdot N) N \\ = - τ n \end{aligned}

That is we have $τ = - (B^{'} \cdot N)$ .

Note that this gives us $N^{'} \cdot B = τ$ as $(B^{'} \cdot N)^{'} = 0$ .
We repeat for $N^{'}$ in the reference frame:

\begin{aligned} N^{'} & = (N^{'} \cdot T) T + (N^{'} \cdot N) N + (N^{'} \cdot B) B \\ = - κ T + τ B \end{aligned}

That is, the basic geometry of space curve embedded in 3D is described by the usual Frenet-Serret:

\begin{aligned} {(\begin{array}{c} T \\ N \\ B \end{array})}^{'} & = (\begin{array}{c} 0 & κ & 0 \\ - κ & 0 & τ \\ 0 & - τ & 0 \end{array}) (\begin{array}{c} T \\ N \\ B \end{array}) \end{aligned}

If we define $ω = τ T + κ B$ then we get:

\begin{aligned} T^{'} & = ω \times T \\ N^{'} & = ω \times N \\ B^{'} & = ω \times B \end{aligned}

Example

Let $γ (t) = (c o s (t), s i n (t), t)$ , that is the helix in $R^{3}$ .
Then we can obtain the arc length to be: $s (t) = \int_{0}^{t} \sqrt{s i n^{2} (τ) + c o s^{2} (τ) + 1} d τ$
We re-parameterize to $s$ to obtain $r (s) = (c o s (\frac{s}{\sqrt{(} 2)}), s i n (\frac{s}{\sqrt{(} 2)}), \frac{1}{\sqrt{(} 2)})$ .
Thus $T = (- \frac{\sin (\frac{s}{\sqrt{2}})}{\sqrt{2}}, \frac{\cos (\frac{s}{\sqrt{2}})}{\sqrt{2}}, \frac{1}{\sqrt{2}})$ , $N = (- c o s (\frac{s}{\sqrt{2}}), - s i n (\frac{s}{s q r t 2}), 0)$ and $B = (\frac{\sin (\frac{s}{\sqrt{2}})}{\sqrt{2}}, - \frac{\cos (\frac{s}{\sqrt{2}})}{\sqrt{2}}, \frac{1}{\sqrt{2}})$
This gives us $κ = \frac{1}{2}$ and $τ = \frac{1}{2}$

Theorem

(Hopf) For any arc length parameterized simple closed curve $r (s) = [0, L] \to R^{2}$ such that $r^{'} (0) = r^{'} (L)$ we must have:

\int_{0}^{L} κ (s) d s = \pm 2 π .

Sketch of the Proof

We could define $T (s) = (c o s (θ (s)), s i n (θ (s)))$ .
We use the frame $J T (s)$ , where $J$ is:
$J = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$
We can obtain $κ (s) = T^{'} (s) \cdot J T (s) = θ^{'} (s)$
Which finally gives us $\int_{0}^{L} κ (s) = θ (L) - θ (0)$ .