# Geodesics on a torus

• Oct 23rd 2011, 06:45 PM
mathswannabe
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:

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

Show that if at some point a geodesic is tangent to the top circle, $\displaystyle \gamma(u=\frac{\pi}{2})$, then it remains on the 'outside half' $\displaystyle (-\frac{\pi}{2}\leq u \leq \frac{\pi}{2})$ 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
• Nov 4th 2011, 02:48 PM
xxp9
Re: Geodesics on a torus
$\displaystyle \gamma(u,v)=((a+b\cos{u})\cos{v}, (a+b\cos{u})\sin{v}, b\sin{u})$
$\displaystyle \frac{\partial}{\partial u} = (-b\sin{u}\cos{v}, -b\sin{u}\sin{v}, b\cos{u})$
$\displaystyle \frac{\partial}{\partial v} = (-(a+b\cos{u})\sin{v}, (a+b\cos{u})\cos{v}, 0)$
Let $\displaystyle X=\frac{1}{(a+b\cos{u})}\frac{\partial}{\partial v}=(-\sin{v},\cos{v}, 0)}$
Let $\displaystyle Y=\frac{1}{b}\frac{\partial}{\partial u}=(-\sin{u}\cos{v}, \sin{u}\sin{v}, \cos{u})$
Then $\displaystyle X, Y$ are an othonormal frame on the torus.
Let $\displaystyle \bar{D}$ be the differential operator of ambient space
$\displaystyle E^3$, D be the Levi Civita connection of the torus, then the
result of D is the projection of the result of $\displaystyle \bar{D}$ to the
tangent plane. Since we have
$\displaystyle \bar{D}_X{X}=\frac{1}{(a+b\cos{u})}\bar{D}_{\frac{ \partial}{\partial v}}{X}$
$\displaystyle =\frac{1}{(a+b\cos{u})}\frac{\partial}{\partial v}(-\sin{v}, \cos{v}, 0)$
$\displaystyle =\frac{1}{(a+b\cos{u})}(-\cos{v}, -\sin{v}, 0)$
It's easy to verify that $\displaystyle \langle \bar{D}_X{X}, X \rangle = 0$
and $\displaystyle \langle \bar{D}_X{X}, Y \rangle = \frac{\sin{u}}{a+b\cos{u}}$
So the geodesic curvature of a v-curve is $\displaystyle k_v=\|D_{X}X\|=\frac{\sin{u}}{a+b\cos{u}}$
Similarly we have $\displaystyle k_u=\|D_{Y}Y\|=0$. That is all the u-curves are
geodesics.
Using the Liouville’s formula
for geodesic curvature
, we have the equation
$\displaystyle \frac{d\theta}{ds}=-\cos{\theta}\frac{\sin{u}}{a+b\cos{u}}$,
where $\displaystyle \theta$ is the directed angle between the geodesic and the
v-curve.
When s=0, $\displaystyle u=\pi/2, \theta=0$ since the geodesic is tangent to the
top circle. We have $\displaystyle \frac{d\theta}{ds}=-\frac{1}{a} < 0$. So
near the starting point $\displaystyle \theta$ 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 $\displaystyle |\theta|$ will never become $\displaystyle \pi/2$
otherwise since the u-curves are geodesics, according to the uniqueness of
geodesics, there is a contradiction. So $\displaystyle \cos{\theta}$ is always
positive. Before we reach the middle circle defined by u=0, $\displaystyle \frac{d\theta}{ds} < 0, \theta$ keeps decreasing. When we meet u=0
$\displaystyle \theta$ achieves its minimum. After we cross u=0 $\displaystyle \theta$ 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.
• Nov 4th 2011, 04:22 PM
xxp9
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.