Given a regular curve $\alpha:\mathbb{R}\to\mathbb{E}^2$ with an oriented curvature $\kappa(t) \ne 0$. Prove the evolute does not contain an open subset of a straight line.

I'm a bit stumped as to how I'm supposed to tackle this question. I mean, intuitively it feels easier to work the contraposition: assume the evolute contains a line, and then somehow end up proving the curvature vanishes.

Note: I eventually managed to construct an answer which has been confirmed to be more or less accurate by an external source.


Since $a(s)$ is regular, a Frenet-apparatus can be constructed:

$$ \begin{align} T'(s) &=\quad ||\alpha'(s)|| \cdot \kappa(s) \cdot N(s) \\ N'(s) &= -||\alpha'(s)|| \cdot \kappa(s) \cdot T(s) \\ \end{align} $$

with $R(s)=\kappa^{-1}(s)$.

The velocity of the evolute is given by

$$ \begin{align} E'(s) &= \alpha'(s) + R'(s) \cdot N(s) + R(s) \cdot N'(s) \\[1ex] &= ||\alpha'(s)|| \cdot T(s) + R'(s) \cdot N(s) - R(s) \cdot ||\alpha'(s)|| \cdot \kappa(s) \cdot T(s) \\[1ex] &= R'(s) \cdot N(s) \\[3ex] ||E'(s)|| &= ||R'(s) \cdot N(s)|| \\[1ex] &= R'(s) \end{align} $$

$a)$ Assume $R'(s) = 0$ for $s \in ] s_0, s_1 [$. Then R(s) = R is constant, which means $\alpha$ is a circle segment over $] s_0, s_1 [$. Consequently, $E(s)$ is a singleton: the center of the circle. Since a singleton is a closed set, this concludes the proof.

$b)$ Assume $R'(s) \ne 0$. This means $E(s)$ is also regular, and a Frenet-apparatus can be constructed for the evolute:

$$ \begin{align} T_E(s) &= \frac{E'(s)}{||E'(s)||} = N(s) \\[1ex] T_E'(s) &= N'(s) \\[3ex] N_E(s) &= J \cdot T_E(s) = -T(s) \\[1ex] N_E'(s) &= -T'(s) \\ \end{align} $$

Consequently, the curvature of the evolute equals

$$ \begin{align} \kappa_E(s) &= \frac{T'_E(s) \cdot N_E(s)}{||E'(s)||} = -\frac{N'(s) \cdot T(s)}{R'(s)} \\[1ex] &= \frac{||\alpha'(s)|| \cdot \kappa(s)}{R'(s)} \cdot (T(s) \cdot T(s)) \\[1ex] &= \frac{||\alpha'(s)||}{R'(s) \cdot R(s)} \ne 0\\[1ex] \end{align} $$

Since the curvature of the evolute is not zero, the evolute itself cannot be a straight line, thus concluding the proof.

  • $\begingroup$ This was confirmed correct, apart from the lack of a third case, where $R'(s) = 0$ for exactly one value of $s$. $\endgroup$ – Zeno Aug 6 '17 at 20:47

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