Wednesday, February 27, 2013

Generalized Helices: Curves of Constant Slope

A regular curve in space is said to have constant slope when its tangents make a constant angle with a given fixed line. Such a curve is also called a generalized helix, or even just a helix amongst the cognoscenti. The line is called the axis of the helix.


An analytic way of stating the condition is as follows: Let $A$ be a unit vector parallel to the axis $\ell$. Denote the parametrized curve by $X$ and its arclength parameter by $s$. We use standard notation for the Frenet-Serret apparatus. Then the definition means that
\[ \langle T, A \rangle = \cos \alpha . \tag{1}\]
Here $\alpha$ is the angle mentioned because $T$ and $A$ are both unit vectors.

Our goal is to describe the consequences of equation (1) for the shape of the curve.



The geometry of a curve is encoded by the curvature and torsion functions? Is there a way to characterize helices by using these functions?

Lancret's Theorem: A regular curve has constant slope if and only if the ratio $\kappa/\tau$ of curvature to torsion is constant.

Proof: We begin by differentiating equation (1) with respect to $s$. This gives us
\[ \langle \kappa N, A \rangle + \langle T, A' \rangle = 0. \]
Since $A$ is a constant, we deduce
\[ \langle N, A \rangle = 0 . \tag{2} \]
Thus $A$ must be parallel to the rectifying planes of $X$. (All of them!) Though we often picture our vectors as "free" vectors, they really are all bound with their tails at the origin, so we know that $A$ lies in $\mathrm{Span}\{ T, B\}$. Since $\{ T, B\}$ is an orthonormal basis for the rectifying plane, we can express the unit vector $A$ as a linear combination this way:
\[  A = \cos(\alpha) T + \sin(\alpha) B . \tag{3} \]
We differentiate this to find
\[ 0 = \cos(\alpha) \kappa N + \sin(\alpha) (-\tau) N = \left( \cos(\alpha) \kappa - \sin(\alpha) \tau\right) N . \]
Now, $N$ is a unit vector, so we deduce that the coefficient must vanish, from which we can deduce that
\[ \dfrac{\kappa}{\tau} = \tan(\alpha).  \tag{4} \]
a constant.

Now, suppose that $X$ is a regular curve with $\kappa/\tau = C$. We choose $\alpha$ to be the unique real number so that $\alpha \in (-\pi/2, \pi/2)$ and $\tan(\alpha) = C$. Then a little algebraic manipulation gives us that $\cos(\alpha) \kappa - \sin(\alpha)\tau = 0$. Now define a vector $A$ by
\[ A = \cos(\alpha) T + \sin(\alpha) B . \]
One can check that $A$ is a unit vector, and then
\[ A' = \dfrac{d}{ds}\left(\cos(\alpha) T + \sin(\alpha) B \right) = \cos(\alpha) \kappa N - \sin(\alpha)\tau N = 0. \]
Hence A is a constant vector, and the tangents make angle $\cos(\alpha)$ with $A$. Thus $X$ is a helix.
$\Box$

Suppose we choose coordinates so that the axis is in the direction of our third coordinate, and hence the helix is expressed in the form
\[ (x(s), y(s), \cos(\alpha) s ) .\]
This means that our upward drift is well understood, and what remains to be investigated is the plane curve $( x(s), y(s) )$.


Theorem: Let $Y$ be the projection of the helix $X$ with angle $\alpha$ onto a plane perpendicular to the axis $\ell$. The principal normal to $Y$ is parallel to the principal normal to the helix, and the curvature of $Y$ is $\kappa_{Y} = \kappa_{X} \csc^2(\alpha)$.

Proof: By definition of orthogonal projection, we know that $Y = X - \langle X, A\rangle A$. We differentiate this with respect to $s$ to find
\[ Y' = T - \langle T, A \rangle A = T - \cos(\alpha) A . \tag{5} \]
It is important to remember that $s=s_{X}$ is the arclength parameter for $X$, and not necessarily for $Y$. So when we want to compute things for $Y$ we will need to use the chain rule.

\[  \left( \dfrac{ds_{Y}}{ds}\right)^2  =  \langle Y', Y' \rangle = \langle T - \cos(\alpha) A, T - \cos(\alpha) A \rangle  \]
\[ \left( \dfrac{ds_{Y}}{ds}\right)^2= \langle T, T \rangle -2\cos(\alpha)\langle A, T \rangle + \cos^2(\alpha) \langle A, A\rangle = 1 - \cos^2(\alpha) \]
Or, more simply,
\[ \dfrac{ds_{Y}}{ds} = \sin(\alpha) . \]
So we get the very useful measurement of the speed of the projection:
\[ \dfrac{ds_{Y}}{ds} = \sin(\alpha) . \tag{6} \]

Now it is straightforward to compute derivatives for the planar projection $Y$ in terms of its arclength $s_1$. The tangent to $Y$ is

\[ T_{Y} = \dfrac{dY}{ds_{Y}} = \dfrac{ds}{ds_{Y}}\dfrac{dY}{ds} = \dfrac{1}{\sin(\alpha)} \left( T - \cos(\alpha) A \right) = \csc(\alpha) T - \cot(\alpha) A.  \]
And so we differentiate to find $N_{Y}$ and $\kappa_{Y}$:
\[ \dfrac{d}{ds_{Y}} T_{Y} = \dfrac{1}{\sin(\alpha)}  \left( \csc(\alpha)  \kappa_{X} N_{X} \right) = \csc^2(\alpha) \kappa_{X} N_{X}. \]
Now, $N_{X}$ is a unit vector, so it must also be the principal normal to $Y$, and we get the desired description of the curvature of $Y$.
$\Box$

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