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Finding Geodesics What I wish to understand, is how to solve

  1. Jan 10, 2012 #1
    Finding Geodesics


    What I wish to understand, is how to solve this one:
    given this metric:
    [itex] ds^2= \frac {dt^2} {t^2}- \frac{dx^2} {t^2} [/itex]
    I have to calculate the geodesics.

    [itex] S=\int{ \frac {d} {d\lambda} \sqrt{\frac {1} {t^2} \frac {dt^2} {d\lambda^2}- \frac{1} {t^2} \frac {dx^2} {d\lambda^2}}} [/itex]

    But [itex] \lambda [/itex] here, is any parameter I choose. Therefore, L does not depend on it (right? )
    So now I want to use E-L equations:

    [itex] \frac {d}{dλ}\frac {∂L}{∂\dot t}=\frac {\partial L} {∂t},
    \frac {d} {dλ}\frac {∂L} {∂\dot x} =\frac {\partial L} {∂x} [/itex]

    When [itex] \dot t [/itex] refers to [itex] \frac {dt} {d\lambda} [/itex] etc.
    But here I get confused:
    What does L depend on?
    I'll continue my solution:

    [itex]\frac {d} {dλ} 0.5 ({{\frac {{\dot t}^2} {t^2}-\frac {{\dot x}^2} {t^2}}})^{-0.5} 2 \frac {\dot t} {t^2}= \frac {\partial L} {\partial x} [/itex]

    As I stated, L does not depend on lambda, and so:

    [itex] \frac {d} {d\lambda} {[\frac {\dot t} {t^2}]}^2 = \frac {{\dot t}^2-{\dot x}^2} {t^3} [/itex]
    So how am I supposed to derive this? Do I add another dot?
    [itex] \frac {d} {d\lambda} \dot t = \ddot t [/itex] ? and do I refer to it as an operator? Meaning- [itex] \ddot t = {\dot t}^2? [/itex]

    How do I continue? These calculations on classical mechanics were so trivial to me- but for some reason I get lost in here...
    Thank you!
     
    Last edited: Jan 10, 2012
  2. jcsd
  3. Jan 12, 2012 #2
    Re: Geodesics

    i'm sorry i'm having problems making sense of this. you seem to be wanting arc length depending on an internal parameter time(lamba) and space(lamba), but then you will need to integrate with respect to d(lamba), not d/d(lamba) for it to remain arc length. i assume if either S is a total integral or S is conserved under variation,
    d/d(lamba)S=0. now can you commute lamba variation and integration?
     
  4. Jan 14, 2012 #3
    Re: Geodesics

    The easiest Lagrangian to use would be [itex]L=g_{\alpha \beta }\frac{dx^\alpha }{d\lambda }\frac{dx^\beta }{d\lambda }[/itex]. Then you don't have to deal with the annoying square root.
     
  5. Jan 14, 2012 #4
    Re: Geodesics

    Hi... I did not understand what you meant. I integrate with respect to lambda... not d/d(lambda) ...
    Thank you
     
  6. Jan 14, 2012 #5
    Re: Geodesics

    Thank you. I would use that.
     
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