A Quick and Dirty Introduction to the Geometry of Curves

The picture we looked at for surfaces is actually a nice way of thinking about manifolds of any dimension. For instance, we can think of a one-dimensional curve as a map $\gamma: I \rightarrow \mathbb{R}^3$ from an interval $I = [0,T] \subset \mathbb{R}$ of the real line to $\mathbb{R}^3$. Again the differential $d\gamma$ tells us how tangent vectors get stretched out by $\gamma$, and again the induced length of a tangent vector $X$ is given by

$|d\gamma(X)| = \sqrt{d\gamma(X) \cdot d\gamma(X)}.$

Working with curves is often easier if $\gamma$ preserves length, i.e., if for every tangent vector $X$ we have

$|d\gamma(X)| = |X|.$

There are various names for such a parameterization (“unit speed”, “arc-length”, “isometric”) but the idea is simply that the curve doesn’t get stretched out when we go from $\mathbb{R}$ to $\mathbb{R}^3$ — think of $\gamma$ as a completely relaxed rubber band. This unit-speed view is also often the right one for the discrete setting where we have no notion of a base domain $I$ — from the very beginning, the curve is given to us as a subset of $\mathbb{R}^3$ and all we can do is assume that it sits there in a relaxed state.

The Frenet Frame

Suppose we have a unit-speed curve $\gamma$ and a positively-oriented unit vector $X$ on the interval $I$. Then
$T = d\gamma(X)$
is a unit vector in $\mathbb{R}^3$ tangent to the curve. Carrying this idea one step further, we can look at the change in tangent direction as we move along $\gamma$. Since $T$ may change at any rate (or not at all!) we split up the change into two pieces: a unit vector $N$ called the principal normal that expresses the direction of change, and a scalar $\kappa \in \mathbb{R}$ called the curvature that expresses the magnitude of change:

$dT(X) = \kappa N.$

One thing to realize is that $T$ and $N$ are always orthogonal. Why? Because if the change in $T$ were parallel to $T$, then it would cease to have unit length! (This argument is a good one to keep in mind any time you work with unit vector fields.) Together with a third vector $B = T \times N$ called the binormal, we end up with a very natural orthonormal coordinate frame called the Frenet frame.

How does this frame change as we move along the curve? The answer is given by the Frenet-Serret formula:
$\underbrace{\left[ \begin{array}{c} \dot{T} \\ \dot{B} \\ \dot{N} \end{array} \right]}_{\dot{Q} \in \mathbb{R}^3} = \underbrace{\left[ \begin{array}{rrr} 0 & \kappa & 0 \\ -\kappa & 0 & \tau \\ 0 & -\tau & 0 \end{array} \right]}_{A \in \mathbb{R}^3} \underbrace{\left[ \begin{array}{c} T \\ N \\ B \end{array} \right]}_{Q \in \mathbb{R}^3}.$
Here $T$, $N$, and $B$ are interpreted as row vectors, and the superscript “$\ \dot{}\$” is shorthand for “change in time,” i.e., the change in a quantity as we move along the curve. For instance, $\dot{T} = dT(X)$, where $X$ is a positively-oriented unit vector on $I$. The quantity $\tau$ is called the torsion, and describes the way the normal and binormal twist around the curve.

A concise proof of this formula was given by Cartan. First, since the vectors $T$, $N$, and $B$ are mutually orthogonal, one can easily verify that $Q Q^T = I$, i.e., $Q$ is an orthogonal matrix. Differentiating this relationship in time, we find that $\dot{Q} Q^T = -(\dot{Q} Q^T)^T$, i.e., the matrix $\dot{Q} Q^T$ is skew-symmetric. But since $A = \dot{Q} Q^T$, $A$ must also be skew-symmetric. Skew symmetry implies that the diagonal of $A$ is zero (why?) and moreover, we already know what the top row (and hence the left column) looks like from our definition of $\kappa$ and $N$. The remaining value $A_{23} = -A_{32}$ is not constrained in any way, so we simply give it a name $\tau \in \mathbb{R}$.

What do you think about this proof? On the one hand it’s easy to verify; on the other hand, it provides little geometric understanding. For instance, why does $N$ change in the direction of both $T$ and $B$, but $B$ changes only in the direction of $N$? Can you come up with more geometric arguments?

Visualizing Curvature

What’s the curvature of a circle $S$? Well, if $S$ has radius $r$ then it takes time $2\pi r$ to go all the way around the circle at unit speed. During this time the tangent turns around by an angle $2\pi$. Of course, since $T$ has unit length the instantaneous change in $T$ is described exclusively by the instantaneous change in angle. So we end up with

$\kappa = |\kappa N| = |dT(X)| = 2\pi/2\pi r = 1/r.$

In other words, the curvature of a circle is simply the reciprocal of its radius. This fact should make some intuitive sense: if we watch a circle grow bigger and bigger, it eventually looks just like a straight line with zero curvature: $\lim_{r \rightarrow \infty} 1/r = 0$. Similarly, if we watch a circle grow smaller and smaller it eventually looks like a single point with infinite curvature: $\lim_{r \rightarrow 0} 1/r = \infty$.

Now consider a smooth curve $\gamma$ in the plane. At any point $p \in \gamma$ there is a circle $S$ called the osculating circle that best approximates $\gamma$, meaning that it has the same tangent direction $T$ and curvature vector $\kappa N$. In other words, the circle and the curve agree “up to second order.” (The phrase “agree up to $n$th order” is just shorthand for saying that the first $n$ derivatives are equal.) How do we know such a circle exists? Easy: we can always construct a circle with the appropriate curvature by setting $r = 1/\kappa$. And of course, any circle has some tangent pointing in the direction $T$. Alternatively, we can consider the a circle passing through $p$ and two other points: one approaching from the left, another approaching from the right. Since these three points are shared by both $\gamma$ and $S$, the first and second derivatives will agree in the limit (consider that these points can be used to obtain consistent finite difference approximations of $T$ and $\kappa N$).

The radius and center of the osculating circle are often referred to as the radius of curvature and center of curvature, respectively. We can tell this same story for any curve in $\mathbb{R}^3$ by considering the osculating plane $T \times N$, since this plane contains both the tangent and the curvature vector.

For curves it makes little difference whether we express curvature in terms of a change in the tangent vector or a change in the (principal) normal, since the two vectors are the same up to a quarter-rotation in the osculating plane. For surfaces, however, it will often make more sense to think of curvature as the change in the normal vector, since we typically don’t have a distinguished tangent vector to work with.

October 22, 2012 | Posted in: Notes | Comments Closed

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