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Let $\alpha (t)$ be a curve such that $|\alpha'(t)|=1$ for all $t\in\mathbb R$. Assume $k(t)\neq 0$, $k'(t)\neq 0$ (whereas $k=|\alpha''(t)|$ is the curvature) and $\tau(s)\neq 0$, whereas $\tau$ is the torsion.

Prove: The trace of $\alpha$ lies on a sphere $\Leftrightarrow$ $\frac{1}{k^2} +\frac{1}{(k'\cdot\tau)^2}=$const.$>0$.

I know this somehow works by using the Frenet-Serret-equations, but I don't really know how to do this proof. Can anyone help me out? Thanks!

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1 Answer 1

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Your equation is incorrect, the correct condition is

$$\frac{1}{\kappa^2} + \left(\frac{\dot{\kappa}}{\tau\kappa^2}\right)^2 = \text{constant}$$

I will only show the $\Leftarrow$ part here.

Let $s$ be the arc length parametrization and $\vec{t}(s), \vec{n}(s), \vec{b}(s)$ be the vectors appear in Frenet Serret equations. Define $$\vec{\beta}(s) = \vec{\alpha}(s) + \frac{1}{\kappa(s)}\vec{n}(s) - \frac{\dot{\kappa}(s)}{\tau(s)\kappa(s)^2} \vec{b}(s)\tag{*1}$$ Differentiate it with respect to $s$, we get:

$$\begin{align} \frac{d}{ds}\vec{\beta}(s) = & \vec{t} - \frac{\dot{\kappa}}{\kappa^2}\vec{n} + \frac{1}{\kappa}(-\kappa \vec{t} + \tau \vec{b} ) - \frac{d}{ds}\left(\frac{\dot{\kappa}}{\tau\kappa^2}\right)\vec{b} -\frac{\dot{\kappa}}{\tau\kappa^2}(-\tau\vec{n})\\ = & \left(\frac{\tau}{\kappa} -\frac{d}{ds}\left(\frac{\dot{\kappa}}{\tau\kappa^2}\right)\right) \vec{b}\\ = & \frac{\tau\kappa^2}{\dot{\kappa}}\left(\frac{\dot{\kappa}}{\kappa^3} - \frac{\dot{\kappa}}{\tau\kappa^2}\frac{d}{ds}\left(\frac{\dot{\kappa}}{\tau\kappa^2}\right) \right) \vec{b}\\ = & -\frac{\tau\kappa^2}{2\dot{\kappa}}\frac{d}{ds}\left(\frac{1}{\kappa^2} + \left(\frac{\dot{\kappa}}{\tau\kappa^2}\right)^2\right) \vec{b}\\ = & \vec{0} \end{align}$$ This implies $\vec{\beta}(s) = \vec{\beta}(0)$ is a constant. From this, we get

$$ \vec{\alpha} - \vec{\beta}(0) = -\frac{1}{\kappa}\vec{n} + \frac{\dot{\kappa}}{\tau\kappa^2} \vec{b} \quad\implies\quad \left|\vec{\alpha} - \vec{\beta}(0)\right|^2 = \frac{1}{\kappa^2} + \left(\frac{\dot{\kappa}}{\tau\kappa^2}\right)^2 = \text{constant}. $$ i.e $\vec{\alpha}(s)$ lies on a sphere with $\beta(0)$ as center.

Motivation of above proof

You may wonder how can anyone figure out the magic formula in $(*1)$. If you work out the $\Rightarrow$ part of the proof where $\alpha(s)$ lies on a sphere centered at $\vec{c}$, you should obtain a bunch of dot products between $\vec{\alpha}(s) - \vec{c}$ and $\vec{t}(s)$, $\vec{n}(s)$ and $\vec{b}(s)$. In particular, you should get: $$\begin{cases} \vec{t} \cdot (\vec{\alpha} - \vec{c}) & = 0\\ \vec{n} \cdot (\vec{\alpha} - \vec{c}) & = -\frac{1}{\kappa}\\ \vec{b} \cdot (\vec{\alpha} - \vec{c}) &= \frac{\dot{\kappa}}{\tau\kappa^2} \end{cases}$$

Using these, you can express the center $\vec{c}$ in terms of $\kappa, \tau$ like what we have in $(*1)$. If the curve does lie on a sphere, then the "center" should not move as $s$ changes. The proof of the $\Leftarrow$ part above is really using the given condition to verify the "center" so defined doesn't move.

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  • $\begingroup$ Thank you for your answer! I tried to do the $\Rightarrow$ part, but I can't even calculate the dot products you pointed out. E.g. $\vec t\cdot (\vec\alpha-\vec c) = \vec t\cdot\vec\alpha - \vec t\cdot\vec c$, but I don't know any formula to go on from here. $\endgroup$
    – dinosaur
    Oct 24, 2013 at 13:37
  • $\begingroup$ What I also don't understand is: Why does $\vec\beta'(s)=0$ imply $\vec b(s)=\vec b(0)$? Or does this actually have to be $\vec\beta(s)=\vec\beta(0)$? $\endgroup$
    – dinosaur
    Oct 24, 2013 at 13:49
  • $\begingroup$ @dinosuar. It is a typo, should be $\vec{\beta}(s) = \vec{\beta}(0)$. About the $\implies$ part, look at this answer to a related question. It contains the derivation of the 3 dot products in the middle of the steps. $\endgroup$ Oct 24, 2013 at 14:01
  • $\begingroup$ Ah, I didn't think about using the equation of the sphere. Thank you very much! $\endgroup$
    – dinosaur
    Oct 24, 2013 at 14:23

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