Let $\alpha$ be a regular curve of curvature and torsion not zero. Show that $\alpha$ is helix if and only if $\frac{k}{\tau}$ is constant.

Question

Definition. A regular curve $\alpha: I \to \mathbb R^3$ is a helix if there is a unit vector $v$ that forms a constant angle with $\alpha'(t), \forall t \in I$.

We can assume $\alpha$ parameterized by arc length. If $\alpha$ is a helix, then there is a unit vector $v$ such that $\langle \alpha'(s), v \rangle$ is constant. So $\langle \alpha''(s),v \rangle = 0$, that is $k(s)\langle n(s), v \rangle = 0$. As $k(s) \ne 0$, it follows that $v$ belongs to the plan determined by $t(s)$ and $b(s)$, for each $s \in I $. So be

$$ v= \cos\theta(s)\,t(s) + \sin\theta(s)\,b(s)$$

Differentiating and using Frenet's formulas, we get

$$0 = -\sin\theta(s)\theta'(s)t(s) + (k(s)\cos\theta(s)+\tau(s)\sin\theta(s))n(s) + \cos\theta(s)\theta'(s)b(s)$$

Therefore,$\forall s \in I$,

$$\sin\theta(s)\theta'(s) = 0,$$ $$\cos\theta(s)\theta'(s) = 0,$$ $$k(s)\cos\theta(s) + \tau(s) \sin\theta(s) = 0$$

The first two equations determine $\theta'(s) = 0, \forall \in I$. Therefore, $\theta(s)$ is constant. Also, the constant $\cos\theta$ is non-zero, otherwise we would have $\tau(s)=0$, which contradicts the hypothesis. It follows from the third equality that $\frac{k}{\tau}$ is constant. Conversely, if $\frac{k}{\tau}$ is constant, we set $\theta$ such that $tg\theta = -\frac{k}{\tau}$. So $$v=cos\theta t(s) + sin \theta b(s)$$ is a constant unit vector and $\forall s \in, \langle t(s), v \rangle = cos\theta $ is constant. So $\alpha$ is helix.

Apparently the answer is correct but incomplete, I'm not able to complete it.

Thanks for any help.

Answer 1

I see you're probably reading Keti Tenenblat's book - she has the (good) habit of leaving this kind of calculation for the reader to work out. The only thing that's missing (as Ted Shifrin pointed out) from the proof is showing that $v(s)$ is actually a constant vector (because that's what's asked in the definition of a helix). This is actually by design (because of the choice of $\theta$). Notice that from the relation $$\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = -\frac{\kappa(s)}{\tau(s)}$$ we obtain $$-\tau(s) \frac{\sin(\theta)}{\cos(\theta)} = \kappa(s) \implies \cos(\theta) \kappa(s) = - \sin(\theta)\tau(s) $$

Now, recalling the Frenet frame relations: $$t'(s) = \kappa(s) n(s), \text{ and } b'(s) = \tau(s) n(s)$$ we get (recall that $\theta$ is by definition constant) $$\begin{aligned} v'(s) &= \cos(\theta) t'(s) + \sin(\theta) b'(s) \\ &= (\cos(\theta) \kappa(s) + \sin(\theta) \tau(s)) n(s) \\ &= (- \sin(\theta)\tau(s) + \sin(\theta)\tau(s)) n(s) \\ &= 0 \end{aligned}$$ so $v(s) \equiv v$ is indeed a constant vector and the proof is complete.

Answer 2

The curve $\alpha: \vec{r}(s)$ is parametrized by its arclenght. The associated vector functions $\vec{T}$, $\vec{N}$, $\vec{B}$ and Frenet-Serret formulas are explained in here: https://en.wikipedia.org/wiki/Frenet%E2%80%93Serret_formulas. In the solution, we will show these functions without the parameter $s$.

Assume first that ${\kappa}{\tau}=c$ is a constant. Then the first and third Frenet-Serret formulas $\frac{\vec{T}}{ds}=\kappa\vec{N}$ and $\frac{\vec{B}}{ds}=-\tau\vec{N}$ give $\frac{\vec{T}+c\vec{B}}{ds}=\vec{0}$ and hence $\vec{T}+c\vec{B}=\vec{v_0}$ is a constant vector. But then, since $\vec{T}\bot\vec{B}$ and $\vec{T}\cdot\vec{T}=1$ ve have $\vec{T}\cdot\vec{v_0}=1$. That is, the tangent vector makes a constant angle with $\vec{v_0}$. We can normalize $\vec{v_0}$ to obtain a unit vector with the same property.

Conversely, assume that the tangent vector makes a constant angle with a constant vector $\vec{v_0}$. Then $\vec{T}\cdot\vec{v_0}=k_0$ for some constant $k_0$. Taking derivative of both sides, we get $\frac{d\vec{T}}{ds}\cdot\vec{v_0}=0$. Then, if $\kappa$ is not zero, from the first Frenet-Serret formula, we deduce that $\vec{N}\cdot\vec{v_0}=0$ and hence $a\vec{T}+b\vec{B}=\vec{v_0}$ for some constants $a,b$. If $\kappa$ is zero, alternatively, we can use the third formula since $\vec{B}$ makes a constant angle with $\vec{v_0}$ too. Taking the derivative of the last equation we obtained, we get $(a\kappa -b\tau)\vec{N}=\vec{0}$. Hence, $a\kappa -b\tau=0$ and ${\kappa}{\tau}=\frac{a}{b}$ is a constant. By the way, $b$ must be non-zero, by the assumption of the problem that $\tau$ is non-zero.