Let $\gamma: (a,b) \rightarrow \mathbb{R}^2$ be a smooth regular curve such that $\forall s,t \in (a,b)$ , $||\gamma(s)-\gamma(t)||$ is a non-negative real valued function which depends only on $|t-s|$.

Show that $\gamma(t)$ has speed and curvature both constant.

Here is my attempt: I somehow have to use the constraint given of the function $||\gamma(t)- \gamma(s)||$. Since this acts on $(a,b)^2$, I want to handle a nicer funcion defined as follows: $$f(h):= ||\gamma(s+h)-\gamma(s)||^2$$ for $s\in(a,b)$ fixed. Taking the derivative with respect to $h$, denoting with $\langle \cdot , \cdot \rangle$ the Euclidean inner product on $\mathbb{R}^2$

$f'(h)= \frac{d}{dh}\langle \gamma(s+h) - \gamma(s) , \gamma(s+h) - \gamma(s) \rangle = \frac{d}{dh} [\langle \gamma(s+h), \gamma(s+h) \rangle + \langle \gamma(s), \gamma(s) \rangle -2 \langle \gamma(s+h), \gamma(s) \rangle ] = 2[\langle \frac{d}{dh}\gamma(s+h), \gamma(s+h) \rangle - \langle \frac{d}{dh}\gamma(s+h), \gamma(s) \rangle] = 2[\langle \frac{d}{dh}\gamma(s+h), \gamma(s+h) - \gamma(s) \rangle ]$

So $f'(0)=0 \quad \forall s$

I think this is somehow useful, but I don't see how to continue from here.


For the speed it is sufficient to remark that $$|\gamma'(s)|=\lim_{h\to0+}{|\gamma(s+h)-\gamma(s)|\over h}\ .$$ Here the RHS is independent of $s$, by assumption on $\gamma$.

For the curvature we have to extract the value of $\kappa(s)$ from distance measurements. We thereby may assume that $|\gamma'(s)|\equiv1$.

Consider the following model situation: $s\mapsto\gamma(s)=\bigl(x(s),y(s)\bigr)$ is a smooth arclength parametrized curve with $$x(0)=y(0)=0,\quad \dot x(0)=1,\quad \dot y(0)=0,\quad \theta(0)=0\ ,$$ where $\theta(s):={\rm arg}\bigl(\dot x(s),\dot y(s)\bigr)$ is the argument of the tangent direction. Then one has $$\eqalign{\dot x&=\cos\theta,\quad\ddot x=-\sin\theta\cdot\dot\theta,\quad x^{...}=-\cos\theta\cdot\dot\theta^2-\sin\theta\cdot\ddot\theta,\cr\dot y&=\sin\theta,\quad \ddot y=\cos\theta\cdot\dot\theta\cr}$$ identically in $s$, and therefore $$\ddot x(0)=0,\quad x^{...}(0)=-\kappa^2,\qquad \ddot y(0)=\kappa\ ,$$ where $\kappa:=\dot\theta(0)$ is the curvature of $\gamma$ at $(0,0)$. Taylor's theorem then gives $$x(s)=s-{\kappa^2\over6}s^3+O(s^4),\quad y(s)={\kappa\over2}s^2+O(s^3)\qquad(s\to0)\ .$$ It follows that $$|\gamma(s)|^2=\left(s^2-{\kappa^2\over3}s^4+O(s^5)\right)+\left({\kappa^2\over4}s^4+O(s^5)\right)=s^2-{\kappa^2\over12}s^4+O(s^5)\qquad(s\to0)\ ,$$ so that $$\lim_{s\to0}{|\gamma(s)|^2-s^2\over s^4}=-{\kappa^2\over12}\ .$$ Back to our given special curve $\gamma$ this means that $$-{\kappa^2(s)\over12}=\lim_{h\to0}{\bigl|\gamma(s+h)-\gamma(s)\bigr|^2-h^2\over h^4}\ .$$ Here the RHS is independent of $s$, hence $s\mapsto\kappa^2(s)$ is constant. By continuity this implies that $s\mapsto\kappa(s)$ is constant.

  • $\begingroup$ Thank you, this helped a lot. What if $\gamma$ is not parametrized by arc-length? Does this line of reasoning still work with more tedious calculations, or do we have to try something else? $\endgroup$ – Siupa Nov 1 '18 at 17:54
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    $\begingroup$ Part (i) shows that $\gamma$ is parametrized by arc length, up to a constant factor. Assuming $|\gamma'(s)|\equiv1$ instead of $|\gamma'(s)|\equiv c$ is no essential restriction in what follows. $\endgroup$ – Christian Blatter Nov 1 '18 at 19:26

Your derivative is right but this is not what you should look at. Fro the problem you want $\parallel \gamma^\prime(s) \parallel = cst$. So you should calculate $$\frac{\langle \gamma(s+h)-\gamma(s), \gamma(s+h)-\gamma(s) \rangle}{h^2}$$ as $h \to 0$.


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