Show that the knowledge of the vector function $n=n(s)$ of a curve $\alpha:I\rightarrow \mathbb{R^3}$ with nonzero torsion everywhere, determines the curvature $k(s)$ and the torsion $\tau (s)$ of $\alpha$.

Notes: $n$ is the normal versor to $\alpha$.

Attempt: I tried using Frenet-Serret formulas, and then using the vector product between $n$ and $n'$, but it seems like I can't get to any result.


2 Answers 2


The Question as posed here is not solvable (as is indicated by the comments in the answer above). One needs the knowledge of the function $\frac{\kappa}{\tau}$ at one point $t_0$.

Lets give a counterexample:

Consider the Helix

$$ c(s) := (a \cdot \cos(s) , a \cdot \sin(s) , b \cdot s) \text{ for } s \in \mathbb{R}$$

with $a^2 +b^2 = 1$, $a,b>0$. Then $c(s)$ ist parametrized by arclength.

The normal vector is given by

$$ n(s) = (- \cos(s) , -\sin(s), 0) \text{ for all } s\in \mathbb{R}. $$

One has in general $\kappa=a, \tau= -b$ and $\frac{\kappa}{\tau}=-a/b$.

The choices $a_1=b_1= 1/\sqrt{2}$ and $a_2= 1/2 ,b_2= \sqrt{3}/2$ give two different curves parametrized by arclength. Both curves have the same normal vector for all times and non-vanishing torsion. And both curves have different curvature and torsion.

  • $\begingroup$ Nice counterexample. $\endgroup$ Sep 26, 2021 at 5:35

Hint: This is rather a long calculation. If you need any more steps feel free to ask.

The idea though is to show that $$\frac{(n \wedge n') \cdot n''}{|n'|^2} = \frac{(\frac{\kappa}{\tau})'}{(\frac{\kappa}{\tau})^2 + 1} : = a (s)$$


$$\int a(s) ds = \arctan \left(\frac{\kappa}{\tau}\right)$$

Thus $\kappa/\tau$ can be determined, plus we have that $\kappa > 0$ then we may get the sign of $\tau$. Finally, use

$$|n'|^2 = |- \kappa t - \tau b|^2 = \kappa ^2 + \tau^2$$

to determine $\kappa^2$ and $\tau^2$.

  • $\begingroup$ ok everything is clear, but aren't you ignoring the costant when you integrate $a(s)$ ? $\endgroup$
    – shevaar
    Feb 12, 2016 at 19:09
  • $\begingroup$ This is a skecth of the solution. What do you think? $\endgroup$ Feb 12, 2016 at 19:12
  • $\begingroup$ It seems to work for me, the only problem it's I can't understand why you can just ignore the integration constant. $\endgroup$
    – shevaar
    Feb 12, 2016 at 19:17
  • $\begingroup$ It's a constant, you will still be able to find $\frac{\kappa}{\tau}$. Plus, you may choose a "chunk" of the curve to integrate (definite integral) over. The main idea though is to find $\kappa$ and $\tau$ in terms of $n(s)$. $\endgroup$ Feb 12, 2016 at 19:22
  • $\begingroup$ @AaronMaroja I won't be able to find $\frac{\kappa}{\tau}$ with the constant, i.e; estimate $\frac{\kappa}{\tau}$ in the equation $\int_{a}^{t} a(s)ds= \arctan\left( \frac{\kappa(t)}{\tau(t)}\right) −\arctan \left( \frac{\kappa (a)}{\tau (a)} \right) $. Can you explain how to do this manipulation? $\endgroup$ Aug 13, 2017 at 23:00

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