Geodesics of the cylinder

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April 21st 2011, 06:37 AM
hmmmm

Geodesics of the cylinder

How do I find the geodesics of the cylinder.
I know that they will be a helix between the two points on the cylinder. As the cylinder is locally isomorphic to the plane and the geodesic on a plane is a straight line. So if we wrapped the plane into a cylinder the straight line would become a helix.

Thanks for any help
April 22nd 2011, 01:07 AM
xxp9

Consider a parametrization r(u,v) = (cosu, sinu, v), it is easy to verify that the first fundamental form I of the surface is I=du^2+dv^2, so r is an isometry. So a geodesic on the cylinder corresponds to a straight line on the (u, v) plane.
April 22nd 2011, 02:18 AM
hmmmm

yeah cool thanks thats what I was thinking along the lines of thanks very much.
May 2nd 2011, 01:53 PM
hmmmm

Ok so it took me a bit to get back to this but is this good? (im a little unclear as to how geodesics map over an isomorphism)

If we start with the cylinder parameterized as follows

$\left ( \begin{array}{c} cos(u) \\ sin(u) \\ v \end{array} \right )$

then by finding the partial with respect to u and v we get the first fundamental form the same as the surface (the 2x2 identity matrix) so the two surface are isomorphic.

We can then say that geodesics of the cylinder are the geodesics of the plane (under the mapping that maps the plane to the cylinder) (Im a little uncertain of this, I know that the minimum distance between two points would be the same (after being mapped) but i donjt know if the geodesics are?)

The geodesics of the plane are straight lines so if we split these into three categories: horizontal, vertical, and diagonal lines as follows

A horizontal line $\left ( \begin{array}{c} x \\ 0 \\ 0 \end{array} \right ) \rightarrow\left ( \begin{array}{c} cos(x) \\ sin(x) \\ 0 \end{array} \right )$ (a horizontal circle round the cylinder)

A vertical line $\left ( \begin{array}{c} 0 \\ y \\ 0 \end{array} \right ) \rightarrow \left ( \begin{array}{c} 0 \\ 0 \\ y \end{array} \right )$ (A verticle line up the cylinder)

A diagonal line $\left ( \begin{array}{c} x \\ y \\ 0 \end{array} \right ) \rightarrow \left ( \begin{array}{c} cos(x) \\ sin(x) \\ y \end{array} \right )$ (A helix round the cylinder)

Is this ok or am I way off with what I am dong here?

Thanks for any help
May 2nd 2011, 05:52 PM
xxp9

You should parametrize a line(curve) using only one variable. A line in the (u,v) plane is ( u0, v0) + t w, where w=(w1, w2) is the vector of the direction of that line. That is, a line in the plane is u = u0 + t w1, v = v0 + t w2.
So a geodesics of the cylinder is ( cos(u0+t w1), sin(u0+t w1), v0+t w2). Though this geodesics isn't necessarilly a segment( i.e. shortest path).