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Homework Help: Why is it that in general geodesics are paths of stationary character

  1. Mar 20, 2006 #1
    Well since the denizens of the relativity forum don't like me, I thought I might ask here see if I get better replies.


    1) Why is it that in general geodesics are paths of "stationary character" rather than minimum?

    2) http://img366.imageshack.us/img366/3280/math30016nx.jpg [Broken]

    I can't quite follow equation 11.16. Specifically how they differentiate dxm/ds in the denominator.
    Shouldn't be necessary, but for reference, the following page is http://img353.imageshack.us/img353/6488/math30024mk.jpg" [Broken].
     
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Mar 20, 2006 #2

    nrqed

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    They have been mean?

    They impose that it's an extremum (functional derivative is zero) so it could be either a min or a max.

    They do not differentiate dxm/ds..they differentiate with respect to dxm/ds. You must treat the *entire* combination dxm/ds as your variable and differentiate with respect to it (So, calling the variable x, L is essentially [itex] {\sqrt{ g_{kn} x^k x^n}}[/itex]).

    Pat
     
    Last edited by a moderator: May 2, 2017
  4. Mar 20, 2006 #3
    No they weren't mean. They just gave me the silent treatment.

    Thanks. It's a lot clearer now.
     
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