Exploring the Frenet Frame of a Curve in R3

[MxwllsPersuasns]

Homework Statement

The Frenet frame of a curve in R 3 . For a regular plane curve (and more generally for a regular curve on a 2-dimensional surface - e.g. the 2-sphere above) we could construct a unique adapted frame F. This is not the case for curves in higher dimensional spaces. Besides the curve being regular we need more conditions to ensure the existence of a unique adapted frame, which then will give invariants of the curve, which in turn reconstruct the curve up to Euclidean motions. Let γ : I → R 3 be an arclength parametrized curve. Then T = γ' has unit length.

(i) Show that < γ'', T >= 0. Thus, provided that γ'' is nowhere vanishing, we can define N := γ''/||γ''|| and obtain a moving basis {T, N, T × N}. A regular space curve for which γ'' is nowhere vanishing is called a Frenet curve.

(ii) Let γ be an arclength parametrized Frenet curve. Define the curvature function to be κ := ||γ 00|| > 0 and the torsion function τ :=< T ×N, N' >. Show that the adapted frame F = (T, N, T × N): I → SO(3, R) and calculate A = F^-1F' in terms of κ and τ .

(iii) If γ is am arclength parametrized plane curve, we can regard it as a space curve. Show that this space curve has τ ≡ 0. Also prove the converse: if a space curve has τ ≡ 0 then it lies in a plane in R³ .

(iv) Show that given κ: I → R, τ (t) > 0 for all t ∈ I, and τ : I → R smooth, there exists a unique (up to Euclidean motion) Frenet curve in R³ whose curvature and torsion are κ and τ respectively.

(v) Classify the Frenet space curves which have curvature and torsion constant

Homework Equations

The Attempt at a Solution

Please feel free to answer any part you want, I shall start with i)...

Part i) So I defined γ' as (x', y', z')^T and γ'' as (x'', y'', z'')^T and when I take the dot product i get {x''x' + y''y' + z''z'}. Nothing jumps out at me as to how I can argue this equals 0, though I have a few inclinations...

So I'm thinking that the dot product of two orthogonal vectors is always 0 and thus γ' and γ'' must be orthogonal though I'm not sure how exactly i can show that as it's not a general fact that the derivative of a curve is orthogonal to that curve (its supposed to be tangent at a particular point). The only other thing I can think of is using the fact that gamma is arc length parameterized but I can't immediately see how that leads to 0. Any help is greatly appreciated.

Part ii) So basically this part asks us to construct the 3x3 matrix with column vectors {T, N, T x N} and then show that this matrix is part of the Rotation group (Special Orthogonal in 3D) by i) showing that the matrix times its transpose equal 1 and then that the determinant of the matrix is 1. Correct?

Part iii) For this part I imagine we need to look at the def of torsion < T x N, N'> and notice that either Tx N or N' must be identically zero given that we've flattened down into the plane. Thought I can't quite imagine which one it is.

That's probably enough for now.. Really hoping for some interesting insights and help on this tough problem. Thanks guys/gals!

 

[kurt.physics]

Edit:For part i) I think I've got it. The vector γ' is the unit vector in the direction of the curve, and γ'' is the unit vector in the direction of the curvature. Since the magnitude of the curvature is given by ||γ 00||, then the dot product of γ' and γ'' is equal to the magnitude of the curvature, which is always non-negative.