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Thread: geodesic torsion

  1. #1
    Apr 2006

    Question geodesic torsion

    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
    Last edited by dopi; Mar 9th 2009 at 06:12 PM.
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  2. #2
    Super Member Rebesques's Avatar
    Jul 2005
    My house.
    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, \ <br />
 \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.
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