Prove that if the chord length depends only on |s-t|, then it is a line or a part of a circle.

Question

Let $\alpha : (a,b) \rightarrow \mathbb{R}^2$ be a smooth map(infinitely differentiable).

Show that if the chord length $\Vert{\alpha(s)-\alpha(t)}\Vert$ depends only on $|s-t|$, then it is a line or a part of a circle.

It comes from Shifrin's differential geometry notes.

Here is my attempt:

Since the chord length only depends on $|s-t|$, for $s-t=\delta, -\delta$, the chord length should be same.

$\Vert \alpha(s+\delta) -\alpha(s) \Vert = \Vert \alpha(s) - \alpha(s-\delta) \Vert$

Squaring both sides and viewing both sides as an inner product of themselves, respectively.

And expanding Taylor series and discarding terms with higher degree than 2 of $\delta$'s.

Then I arrived at $\alpha(s)' \cdot \alpha(s)'' =0$.

And I have no idea to do. Could you give me a hint?

Answer 1

Let $g(c)=\Vert \mathbf{\alpha}(t+c) - \mathbf{\alpha}(t)\Vert$, and we see$$g'(0)=\lim_{c\to 0}\frac{\Vert \mathbf{\alpha}(t+c) - \mathbf{\alpha}(t)\Vert}{c}=\Vert \mathbf{\alpha}'(t)\Vert.$$ This shows the curve has constant velocity.
Now we write $h(|s-t|)=\langle\mathbf{\alpha}(s)-\mathbf{\alpha}(t), \mathbf{\alpha}(s)-\mathbf{\alpha}(t)\rangle$ and take the partial derivatives, $$\begin{align} 2\langle\mathbf{\alpha}'(s),\mathbf{\alpha}(s)-\mathbf{\alpha}(t)\rangle = h'(|s-t|)\\ 2\langle-\mathbf{\alpha}'(t),\mathbf{\alpha}(s)-\mathbf{\alpha}(t)\rangle = -h'(|s-t|) \end{align}$$ where $\langle,\rangle$ is inner product and we assume $s>t$ WLOG.
Summing up above, we have $$\langle \mathbf{\alpha}'(s)-\mathbf{\alpha}'(t),\mathbf{\alpha}(s)-\mathbf{\alpha}(t)\rangle=0.$$ Now since the curve has constant velocity, the interpretation of the equation is that the two tagent vectors $\mathbf{\alpha}'(s)$ and $\mathbf{\alpha}'(t)$ both make an angle $\theta$ with the vector $\mathbf{\alpha}(s)-\mathbf{\alpha}(t)$, but in opposite direction. This is illustrated in the plot below.

Now WLOG, by rotation and translation, we may assume a point on this plane curve is at the origin with its tangent vector at that point pointing toward the $x$-axis direction.
Writing in polar coordinates, it is not hard to see the curve must satisfy $$r\frac{d\theta}{dr}=\tan\theta.$$ Solving this differential equation, we see $$\begin{align} r &= c_1 \sin\theta\\ \theta &= 0\end{align}$$ which is a circle or a line.