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Geodesic in cylinder

by physics_fun
Tags: cylinder, geodesic
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physics_fun
#1
Sep7-06, 02:02 AM
P: 28
Hi, i'm working on marion&thornton ch6 question 6.4.
"Show that the geodesic on the surface of a straight circular cylinder is a (partial) helix"

I used the example of the geodesic on a sphere in the book, but when i calculate the angle phi i get something like phi=b*z+c, where b and c are constants; this is a straight line?!
Or does it just mean that the 'speed' of phi doesn't change in time??
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Kurdt
#2
Sep7-06, 05:59 AM
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Phi changes linearly with z. Think about the implications of this.
physics_fun
#3
Sep7-06, 06:35 AM
P: 28
That implies the equation should be linear....and it is!
Thanks!

Kurdt
#4
Sep8-06, 05:10 AM
Emeritus
Sci Advisor
PF Gold
P: 4,980
Geodesic in cylinder

I still don't think you got what I meant. The equation you came up with shows a linear change in phi with z. Now imagine a cylinder that has a line drawn on its inside surface that changes linearly by 2pi over the total length z. The line drawn on the inside would be part of a helix.

Just making sure you can visualise that.
physics_fun
#5
Sep11-06, 04:02 AM
P: 28
I think that's just what I meant to say (my English is not always very good...)
Kurdt
#6
Sep11-06, 04:36 AM
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No problem. English is my first language and I struggle to express myself


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