Generalized helix curve

Question

Problem: Let $ γ(α,β): →\Bbb R^3 $ be a regular curve with torsion and curvature that are never zero. Show that $γ$ is a generalized helix curve if and only if binormal $b(s)$ has a fixed angle with a constant vector $a$ in $\Bbb R^3, a≠0 $

Definition: A regular curve $γ$ in $ℝ^3$ with curvature $> 0$ is called a generalized helix if its tangent vector makes a fixed angle $\theta$ with a fixed unit vector $a$.

This is what I've tried so far:

⟹:

$a$ is constant so $a'=0$

The curvature $\kappa > 0$

$ 0 = (a\cdot t)'= a't + at' = at' = a\kappa n$

we know that $\kappa > 0$, so:

$a\cdot n = 0$ and and $n = b x t$

so $a·(b x t) = 0$

(From here I'm stuck)

⟸:

$0 =(a·b)'=a'·b+a·b'=a·b'=-a·\tau n$

since $\tau >0$ then $-a·n =0$

(Once again I'm stuck)

Answer 1

Let $a$ be our fixed vector; assuming $\gamma(s)$ is parametrized by the arc-length $s$, which is always possible for a regular curve, we have the unit tangent vector $T$:

$T(s) = \gamma^\prime(s); \tag 1$

since $T(s)$ makes a constant angle $\theta$ with $a$, we have

$T(s) \cdot a = \vert T(s) \vert \vert a \vert \cos \theta = \vert a \vert \cos \theta, \tag 2$

since $\vert T(s) \vert = 1$; if we set

$\alpha = \dfrac{a}{\vert a \vert}, \tag 3$

then $\alpha$ is easily seen to be as unit vector and (2) yields

$T(s) \cdot \alpha = T(s) \cdot \dfrac{a}{\vert a \vert} = \cos \theta; \tag 4$

we may differentiate (4) to obtain

$T^\prime(s) \cdot \alpha = (\cos \theta)^\prime = 0, \tag 5$

since $\cos \theta$ and $\alpha$ are both constant; by the Frenet-Serret equation

$T'(s) = \kappa(s) N(s), \tag 6$

where $N(s)$ is the unit normal to $\gamma(s)$, we thus find

$\kappa(s) N(s) \cdot \alpha = 0, \tag 7$

and since by hytothesis $\kappa(s) \ne 0$ we infer that

$N(s) \cdot \alpha = 0; \tag 8$

next, letting $\phi(s)$ be the angle 'twixt $\alpha$ and the binormal to $\gamma(s)$, $B(s)$, we have

$\alpha \cdot B(s) = \vert \alpha \vert \vert B(s) \vert \cos \phi(s) = \cos \phi(s), \tag 9$

since $\vert B(s) \vert = 1$; (9) yields

$(\cos \phi(s))' = (\alpha \cdot B(s))' = \alpha \cdot B'(s); \tag{10}$

again invoking Frenet-Serret, we have

$B'(s) = -\tau(s) N(s); \tag{11}$

inserting this in (10) and also using (8) we find

$(\cos \phi(s))' = (\alpha \cdot B(s))' = \alpha \cdot (-\tau(s) N(s)) = -\tau(s) \alpha \cdot N(s) = 0; \tag{12}$

since $\cos \phi(s)$ is constant, $\phi(s)$ is constant as well. We have thus shown that the binormal vector of a generalized helix makes a constant angle with the unit vector $\alpha$.

The above argument may be reversed; if $\phi(s)$ is constant and $\tau(s) \ne 0$, then (10)-(11) show that (8) binds; from there, we need merely walk the steps backward to see that $\cos \theta(s)$, and hence $\theta(s)$, is constant.

Note Added in Edit, Wednesday 28 February 2018 8:54 PM PST: Consider equation (4); by virtue of (1), it may be written

$\gamma'(s) \cdot \alpha = \cos \theta; \tag{13}$

thus, for any $s_0, s$ in the interval of definition of $\gamma(s)$ we have

$\gamma(s) \cdot \alpha - \gamma(s_0) \cdot \alpha = \displaystyle \int_{s_0}^s \gamma'(u) \cdot \alpha \; du$ $= \displaystyle \int_{s_0}^s T(u) \cdot \alpha \; du = \int_{s_0}^s \cos \theta \; du = (\cos \theta) (s - s_0); \tag{14}$

this indicates that the component of the vector $\gamma(s)$ along an axis determined by $\alpha$, in the $\alpha$ direction, is in fact linear in $s$. If we now construct the plane $P$ normal to $\alpha$, we can describe $\gamma(s)$ in entirety by specifying two more coordinates in this plane. It would be informative, I'll warrant, to write out the equations for the Frenet-Serret framing of $\gamma(s)$ in such coordinates. Just as (4) expresses the projection of $T(s) = \gamma'(s)$ onto the $\alpha$-axis, so we would obtain corresponding equations for the projection of $\gamma(s)$ onto $P$. My curiosity is piqued. End of Note.