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General question about E-L in revativity

  1. Jan 10, 2012 #1
    Hello! I ask this question in this forum because I wish to be given a general explanation.
    Hope it is OK

    given this metric:
    [itex] ds^2= \frac {dt^2} {t^2}- \frac{dx^2} {t^2} [/itex]
    I wish 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!
  2. jcsd
  3. Jan 10, 2012 #2


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    Science Advisor

    noamriemer, There are several different equivalent Lagrangians that may be used to calculate the geodesics. The easiest, I believe, is to use L = gμν dxμ/ds dxν/ds. This eliminates dealing with the square root. So for your case, L = t-2 (dt/ds)2 - t-2 (dx/ds)2. You let t' = dt/ds and x' = dx/ds for short, and write the equations as, for example, d/ds(∂L/∂t') - ∂L/∂t = 0.
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