Evolutes

An Evolute of a Circular Helix

In a previous post, we studied the involute of a curve. Now we will reverse the direction of the problem: Given a regular curve $\alpha$, another curve $\beta$ is called an evolute of $\alpha$ when $\alpha$ is an involute of $\beta$.

This means that

1. $\beta(s) - \alpha(s)$ lies along the tangent to $\beta$ at $\beta(s)$.
2. $\beta'(s)$ is orthogonal to $\alpha'(s)$.

Our goal is to understand what we can say about evolutes.

Basic Description of Evolutes

So, suppose that $\alpha$ is a unit speed curve with Frenet-Serret apparatus $\{\kappa, \tau, T, N, B\}$ and arclength parameter $s$. Let $\beta$ be an evolute of $\alpha$. Let's find as precise a description of $\beta$ as possible.

Begin by writing $\beta(s) = \alpha(s) + a(s) T(s) + b(s) N(s) + c(s) B(s)$. By our defining conditions, we must have that $a=0$. Thus $\beta(s) = \alpha(s) + b(s) N(s) + c(s) B(s)$. Differentiating, we find
\[ \beta' = T + b' N + b N' + c' B + c B' = (1 - b \kappa  )T +  ( b' - c \tau ) N + (b \tau + c'  ) B . \]
Again, by our defining conditions, $\langle \beta', T\rangle = 0$, so we deduce that $b = 1/\kappa$, which is the radius of curvature of $\alpha$ at $\alpha(s)$. It is traditional to denote this by $\rho$, to remind us that it is a radius---in fact, it is the radius of the osculating circle. We shall do so in what follows.

We will have to work a little harder to find a reasonable form for $c$. So far, we have
\[ \beta' = (\rho' - c \tau) N + ( \rho \tau + c' ) B.\]
Note that $\beta'$ and $\beta - \alpha$ are parallel by definition of an evolute! Since these vectors lie in the normal plane $\mathrm{span}\{N, B\}$ for $\alpha$, we deduce that the ratios of the normal to the binormal coordinates must be equal. Hence,
\[ \dfrac{\rho' - c \tau}{\rho} = \dfrac{\rho \tau - c'}{c} . \]
Some algebraic manipulation eventually yields
\[ \tau = \dfrac{ c \rho' - c \rho'}{\rho^2 + c^2} = - \dfrac{  \left( c / \rho \right) }{ 1 + \left( c / \rho \right)^2}. \]
Integrating, we find that
\[ \int_{s_0}^s \tau \, ds + C = \mathrm{arccot}\left( c / \rho \right) ,\]
which means that
\[ c = \rho \cot\left( \int_{s_0}^s \tau \, ds + C \right) \]
for some constant $C$.

So, there we have it. If $\beta$ is an evolute for the unit-speed curve $\alpha$, we can find some constant $C$ so that
\[ \beta(s) = \alpha(s) + \rho(s) N(s) + \rho(s) \cot\left( \int_{s_0}^s \tau(u) \, du + C \right) B(s) .  \tag{evolute} \]
This means that the set of all possible evolutes forms a one-parameter family, and for any particular choice of $C$, we can compute the corresponding evolute with the above formula.

Let's make a plot of a piece of a helix and part of an evolute for that helix. Note that I chose $C=\pi/2$ here. That makes the picture reasonable. Other choices lead to evolutes which are very long compared to the original curve, and the plot is hard to read. If you don't have Java working on your machine, add the code viewer='tachyon' inside the last set of parentheses. You will at least get a plot, though you won't be able to move it around.

Evolutes of Plane Curves

Consider the case where $\alpha$ is a plane curve. Then $\tau = 0$, and by Equation (evolute) above, we can write
\[ \beta = \alpha + \rho ( N + \cot(C) B)  \tag{1} \]
for some choice of constant $C$. To understand this curve better, let us compute its curvature, $\kappa_{\alpha}$, and torsion, $\tau_{\beta}$. Recall that $s$ is the arclength parameter for $\alpha$, not for $\beta$, so we will have to use the formulae for general curves. In what happens, the new feature to our work will be that $B' = - \tau N = 0$.

Preparation: Derivatives of $\beta$

To get ready to use our crazy formulae, we must compute the first three derivatives of $\beta$ with respect to $s$. We start from Equation (1) and follow our noses (we are all looking in the same direction, right?), and eventually, one finds these:

\[ \beta'  =    \rho' (N + \cot(C) B) .\]

\[ \beta'' =  - \rho' \kappa T + \rho'' (N + \cot(C) B)  \]

\[ \beta''' = (-2\rho'' \kappa - \rho'\kappa') T - \rho'\kappa^2 N + \rho'''(N + \cot(C) B) . \]
Now, it may seem odd at first that we have chosen to express things this way, instead of sticking to our basis $\{T, N, B\}$. That normally serves us so well! But these expressions will make our work below a bit easier.

Preparation: funny expressions we need

Now it is not too hard to compute the different pieces of our crazy formulae.

\[ \beta' \times \beta'' = \rho' (N + \cot(C) B ) \times \left[ -\rho' \kappa T + \rho'' (N+ \cot(C) B ) \right] \]

Of course, for any vector $v$, one has $v\times v = 0$, so we only need to worry about the first term on the right hand side. Using that the Frenet framing is a right handed orthonormal basis, we obtain
\[ \beta' \times \beta'' = -(\rho')^2\kappa ( - B + \cot(C) N ) . \]
So now
\[ \langle \beta' \times \beta'' , \beta' \times \beta'' \rangle = (\rho')^4 \kappa^2 \left( 1 + \cot^2(C) \right) . \]
Also,
\[ \langle \beta', \beta' \rangle = (\rho')^2 \left( 1 + \cot^2(C)\right)^2 \]
Finally, we need the triple product $(\beta' \, \beta'' \, \beta''' )$.  Since this is a determinant, the expressions above come to the rescue. A determinant is unchanged if we add a multiple of one row to another, so we get
\[ (\beta' \, \beta'' \, \beta''' ) = \det \begin{pmatrix} \beta' \\ \beta'' \\ \beta''' \end{pmatrix} = \det\begin{pmatrix} \rho' (N+\cot(C)B) \\ - \rho' \kappa T + \rho'' (N + \cot(C) B) \\ (-2\rho'' \kappa - \rho'\kappa') T - \rho'\kappa^2 N + \rho'''(N + \cot(C) B) \end{pmatrix} \]
Removing multiples of the first row from the others, and then a multiple of the second row from the third, we obtain
\[ (\beta' \, \beta'' \, \beta''' ) = \det \begin{pmatrix} \rho' (N+\cot(C)B) \\ - \rho' \kappa T  \\ (-2\rho'' \kappa - \rho'\kappa') T - \rho'\kappa^2 N  \end{pmatrix} =  \det \begin{pmatrix} \rho' (N+\cot(C)B) \\ - \rho' \kappa T  \\  - \rho'\kappa^2 N  \end{pmatrix}\]
Now, let's pull out constants and then remove a multiple of the third row from the first row to obtain
\[ (\beta' \, \beta'' \, \beta''' ) = (\rho')^3\kappa^3 \det \begin{pmatrix} (N+\cot(C)B) \\  T  \\  N  \end{pmatrix} = (\rho')^3\kappa^3  \det \begin{pmatrix} \cot(C)B \\ T  \\ N  \end{pmatrix} = (\rho')^3\kappa^3 \cot(C) . \]

Computation of Curvature and Torsion for $\beta$:

Finally, we apply our formulae for curves which are not parametrized by arc-length. We find that the curvature is

\[ \kappa_{\beta} = \left[   \dfrac{\langle \beta'\times \beta'', \beta'\times\beta'' \rangle}{\langle \beta', \beta'\rangle^3} \right]^{1/2} = \left[ \dfrac{(\rho')^4\kappa_{\alpha}^2(1+\cot^2(C))}{(\rho')^6(1+\cot^2(C))} \right]^2 = \dfrac{\kappa_{\alpha}}{\rho' (1+\cot^2(C)) } . \]

Now we can compute the torsion
\[ \tau_{\beta} = \dfrac{(\beta' \, \beta'' \, \beta''' )}{\langle \beta'\times\beta'', \beta'\times\beta''\rangle} = \dfrac{(\rho')^3 \kappa^3 \cot(C) }{(\rho')^4\kappa^4(1+\cot^2(C))} = \dfrac{\kappa_{\alpha} \cot(C)}{ \rho' (1+\cot^2(C))}. \]

Hey, Wait A Minute!

Those expressions have a lot in common. In particular, if we look at the ratio of curvature to torsion we see

\[ \kappa_{\beta} / \tau_{\beta} = \tan(C)\]

is a constant! So by Lancret's Theorem, the evolute must be a curve of constant slope, i.e. a generalized helix. And examining the proof of Lancret's theorem, we see that the angle the helix takes against its axis is $C$.

Theorem: An evolute of a plane curve is a (generalized) helix.

[[ That needs a picture: the evolute of an ellipse with a reasonably big angle would be good ]]

Special Choice

Suppose we choose $C = \pi/2$, then $\beta = \alpha + \rho N$ lies in the same plane as $\alpha$. In fact, $\beta$ is the locus of the centers of curvature of $\alpha$. We can be more clear.

Theorem: Suppose that $\beta$ is an evolute of the planar curve $\alpha$. Then $\beta$ lies in the same plane as $\alpha$ if, and only if, $\beta$ is the locus of centers of $\alpha$. Moreover, such $\beta$ is a regular curve exactly at those points where $\kappa_{\alpha}' \neq 0$.

Proof: We already have one direction. Next, suppose that $\beta$ is an evolute of $\alpha$ which happens to lie in the same plane as $\alpha$. Then for some functions $a= a(s)$ and $b=b(s)$ and some constant $C$ we have 

\[ \beta = \alpha + a T + b N = \alpha + \rho N + \cot(C) B .\]

Since the vector $\beta - \alpha$ has a unique expression as a linear combination of the orthonormal basis $\{T, N, B\}$, we deduce that $a=0, b = \rho, C = \pi/2$. Therefore $\beta$ is the locus of centers of curvature of $\alpha$.

Now, is $\beta$ a regular curve? We can easily compute the speed from our work above:

\[ ||\beta'|| = \langle \beta', \beta' \rangle^{1/2} = \rho' \cdot (1 + 0) = \left( 1/\kappa_{\alpha}\right)' = \dfrac{-\kappa_{\alpha}'}{\kappa_{\alpha}^2}, \]

which clearly vanishes exactly at those points where $\kappa_{\alpha}' = 0$.

$\Box$

[[ I need a picture here, too: an ellipse with the planar evolute showing off the four vertices clearly.]]

The condition $\kappa'=0$ is now obviously an interesting one. Such a point is an example of a vertex of a plane curve. We might have more to say about those later. For now, the vertices of the plane curve $\alpha$ are those points which correspond to singular points on its coplanar evolute, its locus of centers of curvature.

There is a paper about evolutes in the issue of the American Mathematical Monthly that just appeared, so this serves as a small introduction. Perhaps we'll have more to say about the topic from a more geometric vein later.