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Geodesic in cylinder

  1. Sep 7, 2006 #1
    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??
     
  2. jcsd
  3. Sep 7, 2006 #2

    Kurdt

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    Phi changes linearly with z. Think about the implications of this.
     
  4. Sep 7, 2006 #3
    That implies the equation should be linear....and it is!
    Thanks!:smile:
     
  5. Sep 8, 2006 #4

    Kurdt

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    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.
     
  6. Sep 11, 2006 #5
    I think that's just what I meant to say (my English is not always very good...):smile:
     
  7. Sep 11, 2006 #6

    Kurdt

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    No problem. English is my first language and I struggle to express myself :wink:
     
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