Frenet frames, curvature and torsion

Try

A circular helix γ(t)=(3cos t,3sin t,2t) bends around an axis while rising. Its speed is √13, curvature is 3/13 and torsion is 2/13. Curvature alone cannot distinguish this helix from a suitably chosen plane circle; torsion records the rotation of its bending plane.

Work first with a unit-speed C³ curve on an interval where κ=||T′|| is positive. Define the principal normal N=T′/κ and binormal B=T×N. These definitions produce an oriented orthonormal frame. They are not available unchanged at zero curvature.

Predict

Predict whether torsion can be read from how far the curve rises along a chosen coordinate axis. A rotated plane circle can rise and fall while remaining planar, so coordinate height is not the invariant.

Predict what zero torsion means when κ stays positive. The binormal should stop changing, fixing a plane normal. Without the positive-curvature condition the Frenet normal may be undefined, so the argument must be qualified.

Experiment

Compare a circle, helix and straight line.

Curve	Curvature	Torsion or frame status
(3cos t,3sin t,0)	1/3	Torsion zero
(3cos t,3sin t,2t)	3/13	Torsion 2/13
(t,0,0)	Zero	Frenet normal not defined

Verify the helix values with the arbitrary-parameter formulas. The speed cubed appears in curvature, while torsion uses the squared magnitude of a cross product. Keep the two denominators distinct.

Understand

The first Frenet equation is T′=κN. Differentiate N·T=0 to obtain N′·T=−κ. Differentiate N·N=1 to get N′·N=0. Thus N′=−κT+τB, defining the torsion convention τ=N′·B. Differentiating B=T×N then gives B′=−τN.

For a regular non-unit parameter, the formulas are κ=||γ′×γ″||/||γ′||³ and τ=det(γ′,γ″,γ‴)/||γ′×γ″||², provided the cross product is nonzero. They follow by substituting the speed decomposition into the Frenet equations. For γ=(a cos t,a sin t,bt), the cross-product square is a²(a²+b²), its magnitude is a√(a²+b²) for a>0, and the triple product is a²b. Hence κ=a/(a²+b²) and τ=b/(a²+b²).

If τ=0 on an interval with κ>0, B′=0. Since γ′·B=0, the scalar γ·B is constant. The curve therefore lies in an affine plane.

Apply

Repair “Every regular curve has a Frenet frame everywhere.” A straight line has a unit tangent but no preferred principal normal. Any orthonormal pair in its normal plane can complete the tangent, and a chosen frame can rotate arbitrarily there. Such a frame is useful, but it is not determined by the Frenet construction.

For optional depth, rotation-minimizing frames can pass through some zero-curvature points where Frenet frames fail. Their governing equations and applications are different. The current theorem’s domain remains the interval with positive curvature, not every possible smooth space curve.

Continue

State smoothness, parameter choice and the nonzero-curvature region beside every torsion calculation. Continue to reconstructing curves to show that curvature and torsion determine more than isolated numerical measurements.