### DIFFERENTIAL GEOMETRY COURSE

9. HELICES

Definition 81: A helix is a curve (but not a straight line) whose tangents make a fixed angle with a fixed straight line (the axis).

Proposition 82: The helices are the curves with a natural equation of the form τ/κ = c (c constant) (and the circular helices are the curves whose τ and κ are both constant and not 0).

Proof: Let v be a direction vector of the fixed straight line with lenth 1.
From t.v = c we find by differentiation n.v = 0, so v = t cos(α) + b sin(α) for a fixed α.
Differentiating again, we get 0 = (κ cos(α) - τ sin(α)) n, so τ/κ = c met c = cotg(α).

On the other hand, let τ/κ = cotg(α) (α constant).
Then b . = -τ n = (-τ/κ) t . = -cotg(α) t ., si the vector t cos(α) + b sin(α) with length 1 is constant.
If we call this vector v, then we have v.t = cos(α) = constant.

Proposition 83: We can give any helix a parametrisation of the form (x1(σ),x2(σ),σ cotg(α)), where σ is arc length of (x1(σ),x2(σ), 0), and σ = s sin(α).
Furthermore, κ2 = κ12 sin4(α), where κ1 is the curvature of (x1(σ),x2(σ), 0).

Proof: Choose coordinates in such a way that x3-as is the axis of the helix.
Then we have (with e3 = (0,0,1)): t.e3 = cos(α), α constant. Hence x3. = cos(α), so x3 = s cos(α).
The relation between s and σ follows from: 1 = (dx1/dσ)2 + (dx2/dσ)2 = (x1. 2 + x2. 2)(ds/dσ)2 = (1 - x3. 2)(ds/dσ)2 = (ds/dσ)2 sin2(α), so σ = s sin(α). Then we find the alleged parametrisation.
Furthermore, κ12 = (d2x1/dσ2)2 + (d2x2/dσ2)2 = (x1..2 + x2..2 + x3..2)(ds/dσ)4 = κ2 sin-4(α). (Here we use that dx3/dσ and ds/dσ are constant, so d2x3/dσ2 and d2s/dσ2 are 0,and also that x3.. = 0.)

Example 84: We determine a parameter trepresentation of the helix with natural equations κ = τ = (2+s2)-1.

We find tg(α)=1, σ = (1/2)s√2 and κ1 = 2 κ = (1 + σ2)-1.
As in example 27, we find (x1(σ), x2(σ)) = (∫0σ cos(arctan(σ ') dσ ', ∫0σ sin(arctan(σ ') dσ ') = (∫0σ (1 + σ ' 2)-1/2 dσ ', ∫0σ σ ' (1 + σ ' 2)-1/2 dσ ' ) = (ln(σ + √(1+σ2)), √(1+σ2)).
Hence we get the parameter representation (ln(σ + √(1+σ2)), √(1+σ2), σ) = (ln((1/2)s√2 + √(1+(1/2)s2)), √(1+(1/2)s2), (1/2)s√2).

Problem 85: Prove that x(s) is a helix if and only if det(x .., x ..., x ....) = 0, and κ ≠ 0.

Problem 86: Determine natural equations κ = κ(s) and τ = τ(s) for a helix that lies on a sphere.

Problem 87: Prove that z(t) from 47 is a helix, and determine the direction of the axis.