Which curves have constant torsion?

140 Views Asked by At

I know from the Frenet-Serret equations that if a curve $r(s)$ (parametrized by arc-length) has constant torsion $\tau \neq 0$, then $$ b' = \tau n \Rightarrow b \times b' = \tau b \times n = -\tau v \Rightarrow r'= -\frac 1\tau b \times b $$ Where $n$ is the unit normal vector to the velocity $v$ and $b = v \times n$. I can also find a formula for the curvature $k$ (which I assume is non-zero everywhere on the curve) by recalling that \begin{align} n' &= -\tau b - k v \\ b'' &= -\tau(\tau b - kv) \end{align} so \begin{align} b \cdot (b' \times b'') &= -\tau^2 b \cdot( n \times (\tau b + k v)) \\ &= \tau^2 (\tau b \cdot (n \times b) + k b \cdot (n \times v)) \\ & = \tau^2k b \cdot n \times v \\ & = \tau^2 k \end{align} From this I get that $r$ has constant torsion only if $r' = -\frac 1\tau b \times b'$ where $||b|| = 1$ and $\tau ^{-2} b \cdot (b' \times b'') = k(s) \neq 0$ for all $s$. Conversely, by this question, I know that these conditions also imply the curve $r = -\frac 1 \tau \int b \times b'$ has constant torsion. My question is, are these the most specific constraints? It seems to me that they are fairly loose, all I have to do is pick some smooth trajectory on the unit sphere and satisfy $b \cdot (b' \times b'') \neq 0$ everywhere, but perhaps this last requirement narrows down the possible curves and I am not seeing how. Can we say anything more about constant torsion curves?

1

There are 1 best solutions below

0
On

If we consider surfaces it may be easier:

If $(k_1,k_2)$ are (Euler) principal curvatures (on a negative Gauss curvature then directions along which geodesic torsion is constant is given by

$$ 2 \sqrt2 \tau =\frac{k_1-k_2}{\sqrt{k_1^2+k_2^2}}.$$

This follows from Mohr's circles wit a fixed point.

Examples are: Asymptotes of a pseudosphere, central asymptotes of right/left helicoids among similar others.

The generalized Frenet Serret including $k_g$ geodesic curvature may give a solution but now I cannot derive it.