The Frenet Frame
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Recall from the previous lecture that we defined the curvature of a smooth regular curve as follows. First we reparametrised by arclength (which is always possible when the curve is regular) and then defined the curvature vector \(k(s)\) and the curvature \(\kappa(s)\) by
\[
\begin{aligned}
k(s) & = T'(s) \in \mathbb{R}^3 \\
\kappa(s) & = \left| k(s) \right| = \left| T'(s) \right| \in \mathbb{R} .
\end{aligned}
\]
We also showed that \(k(s) \cdot T(s) = 0\), and so the curvature vector is always orthogonal to the tangent vector.
In today's lecture we will define the Frenet frame, an orthonormal coordinate frame that is intrinsic to a path \(p : I \rightarrow \mathbb{R}^3\).
Definition. (Inflection point)
A point \(p(s_0) \in \mathbb{R}^3\) is a point of inflection if and only if \(\kappa(s_0) = 0\).
Example 1. (Straight line)
Consider a straight line \(p : \mathbb{R} \rightarrow \mathbb{R}^3\) parametrised by arclength
\[
p(s) = \frac{1}{\sqrt{a^2 + b^2 + c^2}} \left( as, bs, cs \right) .
\]
Then
\[
\begin{aligned}
T(s) = p'(s) & = \frac{1}{\sqrt{a^2 + b^2 + c^2}} \left( a, b, c, \right) \\
\Rightarrow \quad k(s) = T'(s) & = (0, 0, 0) \\
\Rightarrow \kappa(s) & = 0 \quad \text{for all \(s \in \mathbb{R}\)} .
\end{aligned}
\]
Therefore we see that a straight line has zero curvature (as we expect) and therefore every point is an inflection point.
Example 2. (Cubic curve)
Now consider the cubic curve \(p : \mathbb{R} \rightarrow \mathbb{R}^3\) given by the following parametrisation
\[
p(t) = \left( t, t^3, 0 \right) .
\]
It turns out that reparametrising this by arclength and then computing the curvature is a long calculation. We will see in a future lecture that the curvature is
\[
\kappa(t) = \frac{6|t|}{\left( 1 + 9t^4 \right)^{3/2}}
\]
Therefore the point \(p(0) = (0, 0, 0)\) is a point of inflection.
We can also see this in the following diagram. The curvature vector \(k(t)\) is the green vector which is orthogonal to the tangent vector and which points in the direction that the path is curving. We see that at \(t=0\), the curvature vector changes from one side of the curve to the other, and since it is continuous then the magnitude must be zero when \(t = 0\).
You can click and drag the point \(P\) to move it along the cubic curve, and see how the curvature vector changes. Note that the magnitude \(\kappa = |k|\) becomes large when the path becomes "more curved" and the magnitude is small when the path "straightens out" at the ends, and at the point \(t = 0\).
The principal normal vector and the binormal vector
If \(p(s)\) is not a point of inflection, then we can define two more vectors which, in addition to the unit tangent vector \(T(s)\), make up an orthonormal basis at the point \(p(s)\).
Definition. (Principal normal vector and binormal vector)
Let \(p : I \rightarrow \mathbb{R}^3\) be a smooth regular curve parametrised by arclength. If \(p(s)\) is not a point of inflection, then the principal normal vector is defined to be
\[
N(s) = \frac{k(s)}{|k(s)|} = \frac{T'(s)}{|T'(s)|} .
\]
The binormal vector is defined to be
\[
B(s) = T(s) \times N(s) .
\]
Note.
- By definition, \(|N(s)| = 1\), and since \(T'(s) \cdot T(s)\), then \(N(s) \cdot T(s) = 0\), so the principal normal vector \(N(s)\) is a unit vector which is orthogonal to \(T(s)\).
- Since \(T(s)\) and \(N(s)\) are orthogonal unit vectors, then their cross product is also a unit vector. Therefore, by properties of the cross product, the binormal vector \(B(s)\) is a unit vector which is orthogonal to both \(T(s)\) and \(N(s)\).
The Frenet frame
Definition. (Frenet frame)
Let \(p : I \rightarrow \mathbb{R}^3\) be a smooth regular path parametrised by arclength. If \(p(s)\) is not a point of inflection, then the Frenet frame is the triple of vectors
\[
\left( T(s), N(s), B(s) \right)
\]
which form an orthonormal frame at the point \(p(s)\).
We can see this in the following example of the helix.
Example. (The Frenet frame for the helix)
Recall that \(p : \mathbb{R} \rightarrow \mathbb{R}^3\) given by
\[p(s) = \left( r \cos \left( \frac{s}{\sqrt{r^2+c^2}} \right), r \sin \left( \frac{s}{\sqrt{r^2+c^2}} \right), \frac{cs}{\sqrt{r^2+c^2}} \right)\]
represents a helix parametrised by arclength.
The unit tangent vector is
\[ T(s) = p'(s) = \left( - \frac{r}{\sqrt{r^2+c^2}} \sin \left( \frac{s}{\sqrt{r^2+c^2}} \right), \frac{r}{\sqrt{r^2+c^2}} \cos \left( \frac{s}{\sqrt{r^2+c^2}} \right), \frac{c}{\sqrt{r^2+c^2}} \right) \]
and we can differentiate again to obtain the curvature vector
\[ k(s) = T'(s) = \left( - \frac{r}{r^2+c^2} \cos \left( \frac{s}{\sqrt{r^2+c^2}} \right), -\frac{r}{r^2+c^2} \sin \left( \frac{s}{\sqrt{r^2+c^2}} \right), 0 \right) \]
Therefore the curvature is \[ \kappa(s) = \frac{r}{\sqrt{r^2+c^2}} \] and the unit normal vector is \[N(s) = \left( -\cos \left( \frac{s}{\sqrt{r^2+c^2}} \right), -\sin \left( \frac{s}{\sqrt{r^2+c^2}} \right) , 0 \right) .\]
The binormal vector is then
\[ B(s) = T(s) \times N(s) = \left( \frac{c}{\sqrt{r^2 + c^2}} \sin \left( \frac{s}{\sqrt{r^2+c^2}} \right), -\frac{c}{\sqrt{r^2 + c^2}} \cos \left( \frac{s}{\sqrt{r^2+c^2}} \right), \frac{r}{\sqrt{r^2+c^2}} \right). \]
The following picture shows the Frenet frame \(T, N, B\) for the helix, and how it changes as the point \(P\) moves along the curve.
Next time
We will define the osculating circle and the osculating plane, and then use these to define the torsion of a curve.
