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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.

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    One way to go about the "if $\mathbf T\cdot\mathbf v=c$ then $\kappa/\tau$ is constant" direction of Lancret is to verify that $\mathbf v$ is coplanar to the plane spanned by the tangent and the binormal, and then differentiate the associated relation accordingly. To disabuse you of the notion that only curves of the form $(a\cos\;t\quad a\sin\;t\quad bt)^T$ are helices, note that the torsion-curvature ratio of any planar curve is constant. However, $(a\cos\;t\quad a\sin\;t\quad bt)^T$ is *the* curve where the curvature and torsion are individually constant.2011-04-25

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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.