Two curves with Same curvature but are not isometric.

Question

question: Give an example of two space curves with the same curvature but are not isometric to each other(there is no isometry between them).

I am using the Elementary differential geometry book by Presley. I really only know how to do curvature of curves with parametrization having only one variable, i.e. a circle parametrized by $\gamma(t)$. Are there any examples you can give that are one variable parameterizations???

Thanks!!!

Answer 1

Let $\alpha:(-\infty, \infty)\to \mathbb{R}^3$ be the circle $\alpha(u)=(4\cos u/4, 4\sin u/4, 0)$ and $\beta:(-\infty, \infty)\to \mathbb{R}^3$, the helix $\beta(u)=(\cos u/\sqrt{2}, \sin u/\sqrt{2}, u/\sqrt{2})$.

Observe that $\alpha'(u)= (-\sin u/4, \cos u/4,0)$ and $\beta(u)=(-(1/\sqrt{2})\sin u/\sqrt{2},(1/\sqrt{2})\cos u/\sqrt{2},1/\sqrt{2})$ which are unitary and therefore the curves are parametrized by arclength.

Now, $\alpha''(u)=(-1/4\cos u/4, -1/4\sin u/4, 0)$ and therefore its curvature is $1/2$. On the other hand $\beta''(u)=(-1/2\cos (u/\sqrt{2}), -1/2\sin u/\sqrt{2}, 0)$ and therefore the curvature of $\beta$ is $1/2$.

However, $\alpha$ and $\beta$ cannot be isometric since the $\alpha$ is contained in a plane, while, $\beta$ has non-zero torsion.