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Consider a space curve $c : I \rightarrow \mathbb{R^3}$ (regular) with $\frac{\tau}{\kappa}$ constant. (Torsion divided by curvature)

Show: a) There is a plane $A$ with : All normal vector $N(x)$ are in this plane $A$ .

So I know that I have to find a vector $cos(\phi) \cdot T(x) + sin(\phi) \cdot B(x)$ with a constant $\phi$ to solve this exercise. But how can I find such a vector? And how can I use this vector to show a)?

Second task: Now the space curve is parameterized by arc length.

Show: b) There is an isometry $J : \mathbb{R^3} \rightarrow \mathbb{R^3}$ with: $ J \circ c $ $(y)$ $=$ $(f(y),g(y), y \cdot cos(\phi))$ for two functions $f,g$ :$ I \rightarrow \mathbb{R} $. Remark: the $\phi$ is the $\phi$ from a).

So since $c$ is parameterized by arc length. I can use the following equations: $T' = \kappa \cdot N $ and $ N' = - \kappa \cdot T + \tau \cdot B $ and $B' = - \tau \cdot N $, right? I think then we need them.

Remark2:

Definition of $T,N,B$

$T(x)$ := $\frac{1}{||c'(x)||} \cdot c'(x) $.

$N(x)$ := $\frac{1}{||T'(x)||} \cdot T'(x) $.

$B(x)$ := $T(x) \times N(x)$

$\kappa(x)$ := $\frac{|| c' \times c''||(x)}{||c'(x)||^3} $. (curvature)

$\tau(x)$ := $\frac{det(c',c'',c''')(x)}{|| c' \times c''||^2(x)} $ (torsion)

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a) Assume $\kappa(s)>0$ and $\tau(s)\equiv \rho\, \kappa(s)$ for some $\rho\in{\mathbb R}$, and let $$s\mapsto\bigl({\bf t}(s),{\bf n}(s),{\bf b}(s)\bigr)$$ be the concomitant Frenet frame along the curve. Put $$\alpha:=\arctan \rho\in\left]-{\pi\over2},{\pi\over2}\right[\ ,$$ and consider the unit vector $${\bf v}(s):=\sin\alpha\,{\bf t}(s)+\cos\alpha\, {\bf b}(s)\ ,$$ which is $\perp{\bf n}(s)$ for all $s$. According to the Frenet formulas we have $$\dot{\bf v}(s)=\sin\alpha\,\dot{\bf t}(s)+\cos\alpha\,\dot{\bf b}(s)=\sin\alpha\,\kappa(s){\bf n}(s)-\cos\alpha\,\tau(s){\bf n}(s)\equiv{\bf 0}\ .$$ It follows that ${\bf v}(s)\equiv{\bf v}_0$ for some constant unit vector ${\bf v}_0$. Since ${\bf n}(s)\perp{\bf v}_0$ for all $s$ we can conclude that ${\bf n}(s)$ rotates in the plane $A$ orthogonal to ${\bf v}_0$.

b) We introduce a new orthonormal frame consisting of the vectors ${\bf n}(s)$, ${\bf p}(s)$, and ${\bf v}_0$, whereby $${\bf p}(s):={\bf n}(s)\times{\bf v}_0={\bf n}\times(\sin\alpha\,{\bf t}+\cos\alpha\,{\bf b})=-\sin\alpha\,{\bf b}+\cos\alpha\,{\bf t}$$ is lying in $A$ for all $s$ as well. Then $${\bf t}(s)=({\bf t}\cdot{\bf p}){\bf p}+({\bf t}\cdot{\bf v}_0){\bf v}_0=\cos\alpha\,{\bf p}(s)+\sin\alpha\,{\bf v}_0\ .\tag{1}$$ Similarly one computes $$\dot{\bf n}(s)=-{\kappa(s)\over\cos\alpha}{\bf p}(s),\quad \dot{\bf p}(s)={\kappa(s)\over\cos\alpha}{\bf n}(s)\ .\tag{2}$$ Instead of exhibiting an isometry $J$ of the space ${\mathbb R}^3$ we now choose a new orthonormal basis in ${\mathbb R}^3$ consisting of two vectors ${\bf v}_1$, ${\bf v}_2$ spanning $A$, and ${\bf v}_0$. If we describe the given curve $c$ in terms of this basis then we learn from $(1)$ that ${\bf t}(s)$ has a constant component $\sin\alpha$ in direction ${\bf v}_0$, hence $x_0(s)=\sin\alpha\> s$. Furthermore from $(2)$ we deduce that the vectors ${\bf n}(s)$ and ${\bf p}(s)$ rotate with (nonconstant) angular velocity $\omega(s)=\kappa(s)/\cos\alpha$ in the fixed plane $A$. Integrating $(1)$ with respect to $s$ then allows to compute $\bigl(x_1(s),x_2(s)\bigr)\in A$ as well, and in the end we have a description $s\mapsto{\bf x}(s)$ of $c$ of the desired kind.

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  • $\begingroup$ Hello Christian Blatter, thank you for your answer. To avoid misunderstandings: Why we assume that $ \kappa(s) > 0$ ? Where do we need that? My second question: I see why we consider $v(s)=ct(s)+b(s)$, but according to the exercise I could find a vector $cos(\phi) t(s) + sin(\phi) b(s)$ with fixed $\phi$ to solve part a). I see that the structure of $v(s)$ is like $cos(\phi) t(s) + sin(\phi) b(s)$, but there are differenrences. how can I ''rewrite'' your $v(s)$? I mean we have to work with this $\phi$ in part b). $\endgroup$
    – Memories
    Commented May 1, 2018 at 14:19
  • $\begingroup$ The Frenet machinery only works when $\kappa(s)\ne0$. It is standard to assume it positive. My $\alpha (={\pi\over2}-\phi)$ is better than your $\phi$ since the case $\rho=0$ is then covered as well. $\endgroup$ Commented May 1, 2018 at 14:33

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