Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question on MTW (not HW).

  1. Jan 26, 2009 #1
    The question is:"Show that the Guassian curvature R of the surface of a cylinder is zero by showing that geodesics on that surface suffer no geodesic deviation.
    Give an independent argument for the same conclusion by employing the formula
    [tex]R=\frac{1}{\rho_1 \rho_2}[/tex] where [tex]\rho_1[/tex] and [tex]\rho_2[/tex] are the principal radii of curvature at the point in question wrt the enveloping euclidean 3-dimensional space."

    Now if I write down the deviation geodesic equation I get:
    [tex]\frac{d^2\chi}{ds^2}+R\chi=0[/tex] where chi is the distance between geodesics, now because the cylinder has a quasi rectangular shape, the geodesics which start parallel stay parallel thus there is no geodesic deviation, and R=0 cause [tex]\frac{d^2\chi}{ds^2}=0[/tex] and xsi is linear wrt s chi=as+b so R=0.
    Is this just plain mambo jambo from my behalf or there's something genuine here?

    I am not sure about the second argument, which point is in question here?

    Any hints are appreciated.
  2. jcsd
  3. Jan 28, 2009 #2
  4. Jan 29, 2009 #3
    Any hints?
  5. Feb 2, 2009 #4
    Can someone move my post to advanced physics HW?, perhaps there my post will get my attention it deserves, or not.
  6. Feb 2, 2009 #5
    It's a bit mumbo-jumboish, because you've essentially just re-stated the premise of the question. MTW ask you to show that geodesics on the surface of a cylinder suffer no geodesic deviation, and your answer is "because the cylinder has a quasi rectangular shape, the geodesics which start parallel stay parallel thus there is no geodesic deviation". It isn't clear what you mean when you say "the cylinder has a quasi rectangular shape", nor have you explained why geodesics on a surface with such a shape suffer no geodesic deviation.

    Maybe what you mean is that we can "unroll" a cylinder and lay it flat on a table, without changing any of the internal surface distances, so it is a metrically flat surface, but this is really a third argument for flatness. It depends on what kind of answer you want. Most likely MTW expect you to determine the geodesic equations for a cylindrical surface and show explicitly that there is no geodesic deviation. That would be "something genuine".

    Well, the minimum radius of a cylindrical surface is the usual raduis, whereas the radius perpendicular to that is infinite, and 1/infinity = 0, so the Gaussian curvature of the surface is zero.
  7. Feb 3, 2009 #6
    Thank you for your post, I thought this thread of mine will get lost.

    Anyway, back to topic, so I gather from your reply that I need to find parametric equation of the cylinder and from there show that the deviation is null, correct?

    Well finding this parametric coordinates of the cylinder wouldn't be a problem.

    If I have more questions I will ask them later one, thanks again for your reply.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Question on MTW (not HW).
  1. MTW wrong ? (Replies: 2)

  2. MTW questions (Replies: 12)

  3. MTW Exercise 21.16 (Replies: 0)

  4. MTW Excercise 7.1 (Replies: 2)