Necessary and sufficient conditions for a helix

Question

I would like to prove that the following two statements are necessary and sufficient conditions that a curve is a helix. I know that a helix is a space curve with the property that the tangent to the curve at every point makes a constant angle with a fixed direction. I am attempting to prove it but am not sure if some of the logic is right.

(i) The Principal normal is orthogonal to a fixed direction

(ii) $\frac{\kappa}{\omega} = c$ where $\kappa$ is the curvature, $\omega$ is the torsion, and $c$ is a constant.

Now for (i) we prove the necessary condition. If my space curve is a helix, it has parametric equations $x = a \cos \theta$, $y = a \sin \theta$, $z = b\theta$. Assume without loss of generality that $a = 1$ and $b=1$. Then performing calculations I find that the principal normal $\overrightarrow{n} = \langle -\cos \theta, -\sin \theta, 0 \rangle$ which is orthogonal to any vector of the form $\langle 0, 0, a \rangle$, $a$ some constant.

For the sufficient condition, suppose $\overrightarrow{n} \cdot \overrightarrow{a} = 0$, $\overrightarrow{a}$ some vector with fixed direction. Then by the frenet- serret formulas,

$\frac{d \overrightarrow{T}}{ds} = \kappa \overrightarrow{n}$, and hence taking the dot product with $\overrightarrow{a}$ on both sides gives the equation $\frac{d \overrightarrow{T}}{ds} \cdot \overrightarrow{a} = 0$.

$(*)$ Now here's the logic. If $T'(s)$ is perpendicular to $\overrightarrow{a}$, where $T$ is my unit tangent vector, then since $T$ is perpendicular to $T'(s)$ it follows that $T$ itself makes a constant angle with $\overrightarrow{a}$. Is this bit of logic right??

Now for part (ii), the bit on necessity is easy as I just use the same helix and get that $\frac{\kappa}{\omega} = 1$. It is just the bit on sufficiency that is difficult. I have no idea how to go from $\frac{\kappa}{\omega} = c$ to the fact the tangent vector makes a constant angle to a fixed direction.

Please do not give me any full answers for the last bit, but instead pose me some questions that my motivate my understanding of the problem.

Edit: Proof of the neccesary condition for (i). If $T \cdot \bar{e} = c$, $\bar{e}$ some vector with fixed direction and $c$ a constant, then differentiating both sides you get $T'(s) \cdot \bar{e} = \kappa \bar{n} \cdot \bar{e} = 0$, by the first of the frenet serret formulas which means that the unit normal vector is perpendicular to some vector with fixed direction.

Answer 1

here's the proof:

In my proof I will use the general definition of a helix. And no concrete parametrization is needed. The definition of a helix is stated as a helix is a space curve with the property that the tangent to the curve at every point makes a constant angle with a fixed direction, and will be numbered as (0).

Some notations need to be mentioned: $\vec{t},\vec{n},\vec{b}$ are tangent vector, principal normal vector, and binormal vector as usual. Curvature and torsion are denoted as $\kappa, \omega$.

(0)$\Rightarrow$(i): Suppose there is a vector $\vec{a}$ such that $\vec{t}\cdot\vec{a}=c$ for some constant $c$, then $\frac {d\vec{t}}{ds}\cdot\vec{a}=\kappa \vec{n}\cdot\vec{a}=0$, thus $\vec{n}\cdot \vec{a}=0$, which proves statement (i).

(i)$\Rightarrow$(0): If $\vec{n}\cdot\vec{a}=0$, let $\vec{t}(0)\cdot \vec{a}=c$, then $\vec{t}(s)\cdot \vec{a}=\vec{t}(0)\cdot \vec{a}+\int_0^s \kappa(\zeta)[\vec{n}(\zeta)\cdot \vec{a}]d\zeta = c$, and we are done.

(i)$\Rightarrow$(ii): Suppose $\vec{n}\cdot \vec{a} = 0$ for some vector $\vec{a}$, then $\frac {d\vec{n}}{ds}\cdot\vec{a}=(-\kappa \vec{t} + \omega\vec{b})\cdot\vec{a}=-c\kappa + \omega(\vec{b}\cdot\vec{a})$ for some constant $c$. So it remains to show $\vec{b}\cdot\vec{a}=(\vec{t}\times \vec{n})\cdot \vec{a}$ is constant. But since we can permute the order of three vectors in the mixed product, it equals $(\vec{a}\times\vec{t})\cdot \vec{n}$, where both $\vec{a}$ and $\vec{t}$ are orthogonal to $\vec{n}$, hence it is the product of the norm of three vectors times $\sin \langle \vec{a}, \vec{t}\rangle$, which are all constants, hence we are done.

(ii)$\Rightarrow$(0): Suppose $\frac \kappa\omega = c$, let $\vec{a}=\vec{b} + \frac 1c \vec{t}$. Since $\frac {d\vec{a}}{ds}=-\omega \vec{n} + \frac 1c \kappa \vec{n}=0$, $\vec{a}$ is constant vector. Whereas $\frac {d\vec{t}}{ds}\cdot \vec{a}=0$, thus $\vec{t} \cdot \vec{a}$ is constant. This step completes the proof.

Answer 2

Hint: prove that (ii) implies (i). Remember there are always two vectors guaranteed orthogonal to the normal vector $N$: the tangent and the binormal. Using the assumption (ii), can you find a constant (non-trivial) linear combination of the tangent and binormal? One way to show that a linear combination is constant is to demonstrate its derivative is the zero vector; this lets you exploit the Serret-Frenet formulae.