6

The problem asks:

Let $\kappa(s)$ and $\tau(s)$ be the curvature and torsion of a curve parameterized by its arclength $s$. If

$$(\frac{1}{\kappa})^2 + [\frac{1}{\tau}\frac{d}{ds}(\frac{1}{\kappa})]^2 = constant$$

, then either this curve lies on a sphere or the its curvature $\kappa$ is constant. (Thinking in $\mathbb{R}^3$, assuming all "good conditions", like smooth curve, $\kappa \neq0$ etc.)

Since the equation looks like a inner product. I tried to construct a curve with $\frac{1}{\kappa}$ and $\frac{1}{\tau}\frac{d}{ds}(\frac{1}{\kappa})$ as its components. But it turns out that the resulting curve's curvature is not the given $\kappa$. I also tried differentiating the equations w.r.t $s$, but did not find any clues. Any ideas?

Update: although this question was marked duplicate, that post did not solve the problem completely. Here I posted my solution.

Updated 2: the following text is also posted to that post (math.SE). But be noticed that the two posts' problem differs subtlety, and my problem requires an extra step to prove.

Thanks to @Ted Shifrin and another post at math.SE, I solved this problem.

The key is to construct a vector $\alpha$ as in the link above:

We assume that $R$ is in a sphere. Then think of the vector $\alpha$ as exactly the vector that points from the centre of sphere to the curve $R$. So $\alpha$ is different from the original curve $R$ by just by a translation, hence they have the same tangent, curvature and torsion.

We can always express $\alpha$ as a linear combination (as suggested by Shifrin) of:

$$\alpha(s) = \lambda(s)T(s) + \mu(s)N(s)+\nu(s)B(s)$$

for some functions $\lambda$, $\mu$, $\nu$

(Note that $\alpha$ itself not necessarily arc length parametrised.)

Then since we assume that the $R$ is on the sphere, and that $\alpha$ is from the centre to the sphere, we require that $\alpha$ is tangent to its tangent vector (notice that if $\alpha$ does not point from the centre to the sphere, it will not be tangent to its tangent vector $T$), that is,

$$<\alpha, T>=0$$

By differentiating above, we will find the expression for $\alpha$ as shown in the post mentioned:

$$\alpha=-\frac{1}{\kappa}N-\left(\frac{1}{\kappa}\right)'\frac{1}{\tau}B,$$

We see clearly $\alpha$ is of constant length. If we can prove $r(s)= R(s) - \alpha(s)$ is a fixed point, then our proof is done.

To prove $r(s)$ is a fixed point, we prove that $\frac{d}{ds}r(s) =0$. This could be done by differentiating the original equation w.r.t $s$ (do a straight forward calculation and then compare the results of the two)

Community
  • 1
taper
  • 422
  • what do you mean by torsion here? by a definition I know it is a map from 1-forms to 2-forms (or vice versa, depending on a convention), so it should vanish on a curve – Ivan Bodhidharma Sep 28 '15 at 12:55
  • @IvanBodhidharma They mean https://en.wikipedia.org/wiki/Torsion_of_a_curve – Chappers Sep 28 '15 at 13:01
  • @IvanBodhidharma Yes as Chappers said. Also you could refer to the Frenet Formulas. – taper Sep 28 '15 at 13:14
  • @MichaelHoppe Really this is duplicated. Sorry. But I'm having trouble understanding why we can write the curve that way. – taper Sep 29 '15 at 01:35
  • If you think that you have answered the question, put your answer to that question (instead of answering here). This has the advantage that people can compare your answer to the other one in that question. –  Sep 29 '15 at 12:41
  • 1
    @JohnMa thanks for ur suggestion. I have posted it there. – taper Sep 29 '15 at 13:29

1 Answers1

2

HINT: Write $\alpha(s) = \lambda(s)T(s) + \mu(s)N(s)+\nu(s)B(s)$ for some functions $\lambda$, $\mu$, $\nu$. How can you tell when $\alpha$ lies on a sphere centered at the origin? At a general point?

Ted Shifrin
  • 115,160
  • Thanks, but how can we always be sure that this form of $\alpha$ will be arc length parametrised. – taper Sep 29 '15 at 01:34
  • Take the arclength parametrization to start with. All we're using here is that $T,N,B$ gives an orthonormal basis at each point. – Ted Shifrin Sep 29 '15 at 01:37
  • BTW, is there any parameterization possible and an image of the curve with some assumed angles at Equator? – Narasimham Sep 29 '15 at 15:28