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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 05:12 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 $\displaystyle \tau\neq 0$. By continuity, $\displaystyle \tau\neq 0$ over an interval $\displaystyle I$.

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

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