geodesic torsion

dopi · #1 $Old$ March 8th, 2009, 10:39 PM

let a(t) be a regular curve which lies on the unit sphere in R^3

a(t) . a(t) = 1 for all t

i want to show that the geodesic torsion Tg(t) vanishes for such a curve.

i can use the fact that a(t) can be interpreted as a unit normal N(t) to the surface along he curve.

can anyone help me with this question? thanks

Rebesques · #2 $Old$ June 7th, 2009, 08:01 PM

Let's see... Reparametrize for arclength, and let the torsion $\tau\neq 0$ . By continuity, $\tau\neq 0$ over an interval $I$ .

If the Frenet-Serret frame is $\{t,\eta,b\}$ , then $b$ is also the position vector. We therefore have $b''=k\eta, \ b'=-\tau\eta, \ \eta'=-kt+\tau b$ where $k$ is the curvature of $a$ . These last equations give $k\eta=(-\tau\eta)'$ or $\eta'+\psi\eta=0$ , where $\psi=1+\tau'/k$ .

This DE is linear and thus solvable throughout I. Solve to get $\eta=ce^{\int\psi}$ , for some constant vector $c$ . This means $b'=c\int ke^{\int\psi}$ , and the parametrization gives $|b'|=\left|c\int ke^{\int\psi}\right|=1$ , differentiating which gives $ke^{\int\psi}=0$ . This implies $k=0$ over I, a contradiction as the curve lies on the sphere.