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Space-time symmetry (Langrangian Mechanics)

  1. Dec 22, 2009 #1
    When deriving the conserved quantity in the case of space-time symmetry, a line in my notes goes from:

    [tex]\int{dt.(1+\epsilon\dot{\xi}).L[q(t+\epsilon\xi)+{\delta}q(t+\epsilon\xi)]} - \int{dt.L[q(t)+{\delta}q(t)]}[/tex]

    where L is the Lagrangian and [tex]\xi[/tex] is a function of time and both integrals are over the same time interval, to:


    I can't see how these two lines equal one another.

    How does the [tex]O(\epsilon^{2})[/tex] come about?

  2. jcsd
  3. Dec 22, 2009 #2


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    Homework Helper

    That would be the result of a Taylor expansion in [itex]\epsilon[/itex]. Also I think you're missing a factor in the last expression, and that it should be
    [tex]\int \mathrm{d}t\,\epsilon\biggl(\dot{\xi}L + \xi\frac{\mathrm{d}L}{\mathrm{d}t}\biggr) + O(\epsilon^2)[/tex]
  4. Dec 22, 2009 #3
    thanks diazona, much appreciated:

    I tried Taylor expansion as you suggested (and I did get the right answer) but I am not sure whether I have done so mathematically correctly or not. Can you check whether or not I am on the right lines...

    We have [tex]\int{dt.(1+\epsilon\dot{\xi(t)}).L[q(t+\epsilon\xi(t))+{\delta}q(t+\epsilon\xi(t))]} - \int{dt.L[q(t)+{\delta}q(t)]}[/tex]

    start with [tex]L[q(t+\epsilon\xi(t))][/tex]

    if we first Taylor expand the bit in parenthesis (ie. [tex]q(t+\epsilon\xi(t))[/tex]) we get



    This is equal to:



    [tex]\frac{{\partial}L}{{\partial}q}\frac{{\partial}q(t)}{{\partial}t} = \frac{{\partial}L}{{\partial}t}[/tex] (CANCELLING the 'dq's)


    [tex]L[q(t+\epsilon\xi(t))]=L(q)+\frac{{\partial}L}{{\partial}t}\epsilon\xi(t)+smaller terms][/tex]

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