# Necessary and sufficient conditions for a helix

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.

• $(a\cos\;t\quad a\sin\;t\quad bt)^T$ is a helix, but it's not the only helix.... use the definition "the tangent to the curve at every point makes a constant angle with a fixed direction", not some parametrization you know. Commented Apr 25, 2011 at 3:33
• +1 for showing what you've tried and for asking for hints. Commented Apr 25, 2011 at 6:22
• @J.M. I don't understand what you mean by $\langle a \cos t, a \sin t, bt \rangle ^{T}$. So I need to use the general definition of a helix? But then is it not that the most general form of the equation of a helix is $\langle a \cos t, a \sin t, bt \rangle$?
– user38268
Commented Apr 25, 2011 at 8:04
• @J.M. @Jesse Madnick Hi I've corrected the proof of the necessary condition for (i). I've added it into the above. However, could some one point out if there are any flaws in the logic for $(*)$? Thanks.
– user38268
Commented Apr 25, 2011 at 8:11
• @J.M. @Jesse Madnick Hi Here's my take on a proof that if my curve is a helix, then $\kappa / \omega$ = constant. If it is a helix, then $\kappa \bar{n} \cdot e = 0$ from above. So differentiating both sides and using the third of the frenet-serret formulas, I deduce that $-\kappa c + \omega (T \times \bar{n} \cdot \bar{e}) = 0.$ So here's thing: if $(T \times \bar{n} \cdot \bar{e})$ is a constant, then I'm done. However, how can I deduce this from the above? I am guessing that as $n \cdot e = 0$, then $n \times e = e$ in some fixed direction and hence $T$ dot this is a constant. Please help
– user38268
Commented Apr 25, 2011 at 8:32

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.

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.