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Curves with zero torsion are planar

Recall from the previous lecture that we defined the torsion of a curve in terms of the Frenet frame as follows.

Definition. (Torsion)
Let

$p : I \rightarrow \mathbb{R}^3$ be a smooth regular path parametrised by arclength, and suppose that

$p(s)$ is not a point of inflection. The torsion is the function

$\tau(s)$ such that

$B'(s) = \tau(s) N(s) .$ Equivalently, since

$N(s)$ is a unit vector, we have

$\tau(s) = B'(s) \cdot N(s) .$
We then described the intuition behind the torsion: that it measures the change in the osculating plane of the curve. If the osculating plane is moving off its axis (more precisely: the binormal vector

$B(s)$ is moving) then we expect the torsion to be non-zero.

In this lecture we will make this idea precise by proving that the torsion is zero if and only if the curve remains in a fixed plane.

Recall that a plane is determine by a point

$p_0 \in \mathbb{R}^3$ and a unit normal vector

$\hat{n}$ . The equation of the plane is then

$(p - p_0) \cdot \hat{n} = 0 .$ Equivalently, the vector

$p$ is on the plane if and only if

$p-p_0$ is parallel to the plane, and hence perpendicular to the normal vector.

Definition. (Planar curve)
A curve

$p : I \rightarrow \mathbb{R}^3$ is planar if and only if it is contained in a fixed plane. Equivalently, if

$\hat{n}$ is a unit normal vector to the plane, then

$p : I \rightarrow \mathbb{R^3}$ is contained in this plane if and only if for any

$s_0 \in I$ we have

$\left( p(s) - p(s_0) \right) \cdot \hat{n} = 0$ for all

$s \in I$ .

Theorem.
If

$p : I \rightarrow \mathbb{R}^3$ has no inflection points, then

$p$ is planar if and only if

$\tau(s) = 0$ .

Proof. First we show that

$p$ is planar implies that the torsion is zero.

If

$p : I \rightarrow \mathbb{R}^3$ is planar for a plane with normal vector

$\hat{n}$ , then

$p'(s) \cdot \hat{n} = 0$ and

$p''(s) \cdot \hat{n} = 0$ . Therefore, since

$p(s)$ has no inflection points, then

$T(s) \cdot \hat{n} = 0 \quad \text{and} \quad N(s) \cdot \hat{n} = 0 .$ Therefore the osculating plane spanned by

$T(s)$ and

$N(s)$ is the plane orthogonal to

$\hat{n}$ , and so

$\begin{aligned} B(s) & = T(s) \times N(s) = \pm \hat{n} = constant \\ \Rightarrow \quad B'(s) & = 0 \\ \Leftrightarrow \quad \tau(s) & = 0 . \end{aligned}$ Therefore we have shown that

$p$ is planar implies that the torsion is zero. Conversely, now suppose that the torsion is zero. Then

$B'(s) = \tau(s) N(s) = 0 \quad \Leftrightarrow \quad B(s) = B(s_0) = constant.$ Since the Frenet frame is orthonormal, then

$0 = T(s) \cdot B(s) = T(s) \cdot B(s_0)$ . Define

$\hat{n} = B(s_0) = B(s)$ for all

$s \in I$ (this makes sense since

$B(s)$ is constant), and note that

$\frac{d}{ds} \left( p(s) \cdot \hat{n} \right) = p'(s) \cdot B(s) = T(s) \cdot B(s) = 0$ Therefore

$p(s) \cdot \hat{n}$ is constant. If we fix some

$s_0 \in I$ , then this implies that

$\begin{aligned} p(s) \cdot \hat{n} & = p(s_0) \cdot \hat{n} \quad \text{for all $s \in I$} \\ \Leftrightarrow \quad \left( p(s) - p(s_0) \right) \cdot \hat{n} & = 0 \quad \text{for all $s \in I$} , \end{aligned}$ and so

$p : I \rightarrow \mathbb{R}^3$ is planar.

Therefore we have shown that the torsion is zero implies that the curve is planar, which completes the proof. q.e.d.

Next time

We will define the notions of orientation and signed curvature.