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Math Help - Geodesics on a torus

  1. #1
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    Geodesics on a torus

    Hi All, i have been asked to investigate Geodesics on a Torus and their different properties.

    Unfortunately I am getting nowhere after trying for 2 hours. I must be missing something crucial...

    Given a parameterization for a torus:

    \gamma(u,v)= \left((a+b\cos u)\cos v,(a+b\cos u)\sin v,b\sin u\right)

    Show that if at some point a geodesic is tangent to the top circle, \gamma(u=\frac{\pi}{2}), then it remains on the 'outside half' (-\frac{\pi}{2}\leq u \leq \frac{\pi}{2}) of the torus. Show also that it oscillates between the top circle and bottom circle.

    There are also some other questions that ask the same thing - ie give you a starting point, and then get you to prove some properties of the set of geodesics that satisfy the initial condition.

    I'm really stuck with this.

    Information I am trying to use: Geodesics have acceleration normal to the surface at all times.... the coefficients of the first fundamental form etc.

    Can anyone give me some help with this question or perhaps give me an idea for a method to nut out these types of questions?

    Cheers
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  2. #2
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    Re: Geodesics on a torus

    \gamma(u,v)=((a+b\cos{u})\cos{v}, (a+b\cos{u})\sin{v}, b\sin{u})
    \frac{\partial}{\partial u} = (-b\sin{u}\cos{v}, -b\sin{u}\sin{v}, b\cos{u})
    \frac{\partial}{\partial v} = (-(a+b\cos{u})\sin{v}, (a+b\cos{u})\cos{v}, 0)
    Let X=\frac{1}{(a+b\cos{u})}\frac{\partial}{\partial v}=(-\sin{v},\cos{v}, 0)}
    Let Y=\frac{1}{b}\frac{\partial}{\partial u}=(-\sin{u}\cos{v}, \sin{u}\sin{v}, \cos{u})
    Then X, Y are an othonormal frame on the torus.
    Let \bar{D} be the differential operator of ambient space
    E^3, D be the Levi Civita connection of the torus, then the
    result of D is the projection of the result of \bar{D} to the
    tangent plane. Since we have
    \bar{D}_X{X}=\frac{1}{(a+b\cos{u})}\bar{D}_{\frac{  \partial}{\partial v}}{X}
    =\frac{1}{(a+b\cos{u})}\frac{\partial}{\partial v}(-\sin{v}, \cos{v}, 0)
    =\frac{1}{(a+b\cos{u})}(-\cos{v}, -\sin{v}, 0)
    It's easy to verify that \langle \bar{D}_X{X}, X \rangle = 0
    and \langle \bar{D}_X{X}, Y \rangle = \frac{\sin{u}}{a+b\cos{u}}
    So the geodesic curvature of a v-curve is k_v=\|D_{X}X\|=\frac{\sin{u}}{a+b\cos{u}}
    Similarly we have k_u=\|D_{Y}Y\|=0. That is all the u-curves are
    geodesics.
    Using the Liouville’s formula
    for geodesic curvature
    , we have the equation
    \frac{d\theta}{ds}=-\cos{\theta}\frac{\sin{u}}{a+b\cos{u}},
    where \theta is the directed angle between the geodesic and the
    v-curve.
    When s=0, u=\pi/2, \theta=0 since the geodesic is tangent to the
    top circle. We have \frac{d\theta}{ds}=-\frac{1}{a} < 0 . So
    near the starting point \theta will be negative. This shows that
    the geodesic turns right( clock-wise) according to the v-curve. So it's
    getting down.
    Also note that |\theta| will never become \pi/2
    otherwise since the u-curves are geodesics, according to the uniqueness of
    geodesics, there is a contradiction. So \cos{\theta} is always
    positive. Before we reach the middle circle defined by u=0, \frac{d\theta}{ds} < 0, \theta keeps decreasing. When we meet u=0
    \theta achieves its minimum. After we cross u=0 \theta increases but still less than 0 so the geodesic keeps going down.
    Now notice that the torus is a geodesic symmetric space when the base point
    is on the middle circle( since we can always turn 180 degree according to
    its diameter and get it back), so the situation after we cross the u=0 line
    will be symmetric before we reach that.
    Last edited by xxp9; November 4th 2011 at 03:05 PM.
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  3. #3
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    Re: Geodesics on a torus

    the above argument didn't show that the geodesic will definitely reach the middle circle. You may need to figure it out.
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