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Geodesic math help

  1. Apr 4, 2007 #1
    1. The problem statement, all variables and given/known data

    Show that a regular curve on a smooth surface is a geodesic and an asymptotic curve if and only if it is a segment of a straight line.

    3. The attempt at a solution

    I did the <= implication, which is quite easy. I can't get the other one.
  2. jcsd
  3. Apr 4, 2007 #2


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    A unit speed curve [itex]\gamma[/itex] is a segment of a straight line iff [itex]\ddot{\gamma} = 0[/itex], and all regular curves can be reparametrised to be unit speed. What are the definitions of "geodesic" and "asymptotic curve"?
    Last edited: Apr 4, 2007
  4. Apr 4, 2007 #3
    Suppose [tex]\alpha[/tex] a unit speed curve, then it is a geodesic if the covariant derivative [tex]D\alpha/dt=0[/tex] and it is an asymptotic curve if [tex]II(\alpha '(s))=0[/tex] for all s, where II is the second fundamental form and the normal curvature k_n at a point on the curve. Since [tex]k_n=k<n,N>=0[/tex] where k is the curvature of [tex]\alpha[/tex] at a point, n the curve's normal and N the surface's normal, we have to conclude that k=0 or <n,N>=0. This is where I am stuck. What if <n,N>=0?
  5. Apr 4, 2007 #4


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    My book says that a curve is a geodesic if it's second derivative is zero or if it's second derivative is perpendicular to the surface. Since the second derivative is kn, this is equivalent to saying that kn = aN for some real a.

    k<n,N> = <kn,N> = <aN,N> = a<N,N>

    if the curve is asymptotic, then k<n,N> = 0, which would imply a = 0, implying that the second derivative is 0, which I claimed to be equivalent to saying that the curve is part of a line segment. So somehow, you need to show that your book's definition of "geodesic" is equivalent to my book's.
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