The Frenet-Serret Apparatus for Space Curves

The most important tool in classical differential geometry of curves is the orthonormal framing on the curve provided by the tangent, normal, and binormal vectors to the curve, and the Frenet-Serret equations which control the evolution of that framing.

Let's introduce what these things are, and how they help us define and compute the curvature and torsion of the curve. This will be very computational. We will fill in interpretations of the quantities we describe in other posts.



Let $\gamma (s)$ be a unit-speed curve in space. This means that our curve has been reparametrized so that time is equal to arc-length. We shall use primes to denote derivatives with respect to the arclength parameter $s$.

The tangent vector $T$ to $\gamma$ is defined in the usual way:
$$T(s)={\gamma }^{\prime }(s).$$
Since we are assuming that $\gamma$ is unit speed, we have that
$$||T(s)||=1,$$
for all values of $s$. This has the standard set of interpretations as in multivariable calculus: the tangent vector is the velocity vector of the motion, and its norm is the speed of motion.

Now we would like to have a sense of how the curve is bending, or curving, in space. This should somehow be about a second derivative. Indeed, the curvature of $\gamma$ is
$$\kappa (s)=||T^{\prime }(s)||.$$
It is also helpful to have a sense of which way the curve is turning, so we also want to track the normal vector $N$, defined by
$$N(s)={\displaystyle \frac{T^{\prime }(s)}{||T^{\prime }(s)||}}.$$
By the way, we can rewrite this to look like
$$\begin{array}{}\text{(1)}& T^{\prime }(s)=\kappa (s)N(s).\end{array}$$
What is important here? Well, the definition makes it clear that $N$ is always a unit vector. OK, you caught me, if $T^{\prime }=0$, then the normal vector is not defined. But what happens then? Go ahead. I'll wait. (Hint. Hint. What is the curvature?)

That's right. Aren't you clever. So, we will make a blanket assumption that $T^{\prime }\ne 0$, so the normal vector is defined.

Another important thing is to note that the normal vector is always orthogonal to the tangent vector. How do you see this? You have to use a simple, but important trick. In fact, this trick is used all over classical differential geometry, so it would be better to call it a method. Note that $1=||T(s)|{|}^2=⟨T(s),T(s)⟩$.

Differential geometers often use angle brackets like that for a dot product. It is not so crucial here in ${\mathbb{R}}^3$, but in later study it is really common. Now I can't help it. So, angle brackets mean dot product of the two vectors contained.

So, $1=⟨T,T⟩$. (Oh, yeah. We often suppress the parameter because we get tired of typing it.) Now, take a derivative. You obtain
$$0=⟨T^{\prime },T⟩+⟨T,T^{\prime }⟩=2⟨T^{\prime },T⟩=2\kappa ⟨N,T⟩.$$
And that is it. You deduce that $N$ and $T$ are orthogonal. Wonderful News!

Having a pair of orthogonal vectors in ${\mathbb{R}}^3$ is a nice thing, but it is much nicer to have a whole orthonormal basis of ${\mathbb{R}}^3$. How can we get one? Using the cross product. Basic properties of the cross product tell us that $T\times N$ will be another unit vector, and it will be orthogonal to both $T$ and $N$. So we define another vector called the binormal
$$B=T\times N.$$

This is a glorious situation: for each value of the parameter $s$, we have an orthonormal basis $\{T,N,B\}$ of ${\mathbb{R}}^3$ which somehow is sensitive to the geometry of the curve! As $\gamma$ goes running around, this framing turns and spins. Keeping track of this turning and spinning gives us lots of information about the curve itself, and that is the point of the Frenet-Serret apparatus.

How does this framing move about as we travel along the curve? We want some expressions for $T^{\prime }$,  $N^{\prime }$ and $B^{\prime }$. The first we already have: equation (1).

Because I know how this works out, I will now skip and consider $B^{\prime }$. This will be a little easier to handle than $N^{\prime }$. We start by differentiating the defining equation for $B$. This yields
$$B^{\prime }=T^{\prime }\times N+T\times N^{\prime }=\kappa N\times N+T\times N^{\prime }=T\times N^{\prime }.$$
Describing this directly doesn't seem the most convenient. But we can use the fact that $\{T,N,B\}$ forms an orthonormal basis of ${\mathbb{R}}^3$. Using the fundamental nature of orthonormal bases, we can write
$$B^{\prime }=⟨B^{\prime },T⟩T+⟨B^{\prime },N⟩N+⟨B^{\prime },B⟩B.$$
What remains is to sort out what those coefficients are. Let's take the ``easy'' one: $⟨B^{\prime },B⟩=0$ for exactly the same reasons that applied to $T$ above.

Also, since a cross product is orthogonal to both of the vector inputs, we deduce that $⟨B^{\prime },T⟩=⟨T\times N^{\prime },T⟩=0$.

This leaves us the convenient fact that $B^{\prime }$ must be a scalar multiple of $N$. We now define a quantity called torsion and denoted by $\tau (s)$ by the relationship
$$\begin{array}{}\text{(2)}& B^{\prime }=-\tau N.\end{array}$$
Given where we started, one can see that $\tau$ is described as
$$\tau =-⟨B^{\prime },N⟩=⟨B,N^{\prime }⟩.$$
That last equality follows from differentiating $⟨B,N⟩=0$. (This little argument is fundamental to classical differential geometry!)

Now, let's go back and tackle $N^{\prime }$. Again, we will try to describe $N^{\prime }$ by understanding its dot products with the orthonormal basis vectors $T$, $N$ and $B$. A third repeat of ``the little argument" shows that $⟨N^{\prime },N⟩=0$. If we start with $⟨T,N⟩=0$ and try the same line, we quickly find that
$$0=⟨T^{\prime },N⟩+⟨T,N^{\prime }⟩=⟨\kappa N,N⟩+⟨T,N^{\prime }⟩.$$
From that we rearrange to deduce that $⟨N^{\prime },T⟩=-\kappa$. And of course, just a minute ago we found that $⟨N^{\prime },B⟩=\tau$. If we put all of that together we find
$$\begin{array}{}\text{(3)}& N^{\prime }=-\kappa T+\tau B\end{array}$$

Now we have it. The Frenet-Serret equations are
$$\begin{array}{}\text{(FSE)}& T^{\prime }=\kappa N,N^{\prime }=-\kappa T+\tau B,B^{\prime }=-\tau N.\end{array}$$

Often, one sees these equations organized into the form of 3-by-3 matrices. Write $T,N,B$ as the columns of a matrix, and you should get
$$\left(\begin{array}{ccc}|& |& |\\ T^{\prime }& N^{\prime }& B^{\prime }\\ |& |& |\end{array}\right)=\left(\begin{array}{ccc}|& |& |\\ T& N& B\\ |& |& |\end{array}\right)\left(\begin{array}{ccc}0& -\kappa & 0\\ \kappa & 0& -\tau \\ 0& \tau & 0\end{array}\right).$$

As an aside, let me note that it is very important that the matrix on the right-hand side of this is skew symmetric. This is the tip of an iceberg where we reinterpret differential geometry as a subject living on principal bundles, and that matrix helps us see that the structure of the relevant bundle here involves the Lie group $\mathrm{SO}(3)$ and its Lie algebra of skew-symmetric 3-by-3 matrices.