0
$\begingroup$

Let $\gamma$ be a unit-speed $C^{\infty}$ parameterized curve in $\Bbb R^3$. Define $C(s) = \frac1{\sqrt2} (\gamma(s) + \frac1{\sigma} B(s))$. Where $(T,N,B)$ is the Frenet-Serret frame of $\gamma$ at a point $\gamma(s)$. Suppose that $\sigma$, the torsion of $\gamma$, is constant and that $\sigma = k \sqrt2$ where $k$ is the curvature of $\gamma$.

I found that, if $\sigma_C$ and $k_C$ denote the torsion and curvature of $\gamma$ (resp.), then $\sigma_C = k$ and $k_C = \sigma$.

In other words, I found that the curvature of $\gamma$ is the torsion of $C$ and that the torsion of $\gamma$ is the curvature of $C$.

Can someone verify whether or not this is true?

$\endgroup$
1
$\begingroup$

Somewhat strange question, but, yes, $C$ turns out to be parametrized by arclength. The curvature of $C$ is $k\sqrt2=\sigma$ and the torsion of $C$ is $k$. (Basically, one needs to write the Frenet frame of $C$ in terms of that of $\gamma$, which I'm sure you must have done correctly.)

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you. Is there any geometric interpretation to this whole thing? Or is this just some algebraic manipulation to get a funny statement "the torsion of $\gamma$ is the curvature of $C$, and the torsion of $C$ is the curvature of $\gamma$"? $\endgroup$ – user258700 Apr 2 '16 at 17:00
  • 1
    $\begingroup$ Well, you're starting with a right circular helix (constant curvature and torsion) and this creates an associated right circular helix with the curvature and torsion swapped. For a right circular helix $\gamma(s)=\big(a\cos(s/c),a\sin(s/c),bs/c\big)$, with $c=\sqrt{a^2+b^2}$, we have $k=a/c^2$ and $\sigma = b/c^2$. So you can work out what's going on in your case if you want. I've never seen this exercise before. It seems rather an oddity to me. $\endgroup$ – Ted Shifrin Apr 2 '16 at 17:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy