# 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 a subset $$I = [0,T] \subset \mathbb{R}$$ 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 point 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 rubber band that is completely relaxed. This unit-speed view is also the right one for the discrete setting where we often have no notion of a base domain $$I$$ — the curve is “born” in $$\mathbb{R}^3$$ 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 unit vector $$\hat{X} \in \mathbb{R}$$ oriented along the positive direction. Then

$T = d\gamma(\hat{X})$

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:

$\kappa N = dT(\hat{X}).$

The scalar $$\kappa \in \mathbb{R}$$ is called the curvature and expresses the magnitude of change, while the vector $$N \in \mathbb{R}^3$$ is called the principal normal and expresses the direction of change.

One thing to notice is that $$T$$ and $$N$$ are always orthogonal. Why? Simply because if $$T$$ changed along the direction of $$T$$ itself then it would no longer have unit length! Therefore one often defines a third vector $$B = T \times N$$ called the binormal, which completes the orthonormal Frenet frame $$(T,N,B)$$ of $$\gamma$$.

Finally, how does $$B$$ change along $$\gamma$$? Here our story ends since the change in the binormal can be expressed in terms of the normal: $$dB(\hat{X}) = -\tau N$$. The quantity $$\tau \in \mathbb{R}$$ is called the torsion, and describes how much the Frenet frame “twists” around $$T$$ as we travel along $$\gamma$$.

Note that for a curve with torsion the normal $$N$$ also twists around the tangent — in particular $$dN = -\kappa T + \tau B$$. Altogether we get the Frenet-Serret formulas, which describe the motion of the Frenet frame:

$\begin{array}{rcc} dT &=& \kappa N \\ dN &=& -\kappa T + \tau B \\ dB &=& -\tau N \end{array}$

Since $$T$$, $$N$$, and $$B$$ have unit length at each point, we can visualize them as curves on the unit sphere $$S^2 \subset \mathbb{R}^3$$, just as we did with the Gauss map. In this context the sphere is often given a special name: the (tangent-, principal normal-, or binormal-) indicatrix. Later on we’ll see that this “indicatrix” picture has some beautiful geometric consequences.

Visualizing Curvature

Imagine that $$\gamma$$ sits in the plane. At any given point 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 $$\kappa N$$. How do we know such a circle exists? Well, we can certainly find a point with tangent $$T$$ on a circle of any radius. We can then adjust the size of $$S$$ until it has the desired curvature. Alternatively, one can construct $$S$$ by considering a circle passing through three consecutive points on $$\gamma$$, which approaches the osculating circle $$S$$ as the points move closer together. Since these points yield the same approximations of first- and second- derivatives along both $$\gamma$$ and $$S$$ (using finite differences), the tangents and curvatures of the two curves will agree as the points converge.

In any case, we can use the geometry of $$S$$ to express the curvature of $$\gamma$$. If $$S$$ has radius $$r$$ then it takes time $$2\pi r$$ to go all the way around the circle at unit speed, and 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 change in angle. So we end up with $$\kappa = |\kappa N| = |dT(\hat{X})| = 2\pi/2\pi r = 1/r$$: 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 from a fixed viewpoint, it eventually looks just like a straight line.

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 consider the tilt of the tangent vector to find the curvature or the tilt of the (principal) normal vector since the two are related via a rotation by $$\pi/2$$ in the osculating plane. For surfaces, however, it will make more sense to consider the change in normal vector, since we typically don’t have a distinguished tangent vector at any given point.

October 11, 2011 | Comments Closed