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Correct way for symmetry transform

  1. Dec 13, 2013 #1
    1. The problem statement, all variables and given/known data

    Hi
    I'm trying to understand how symmetry transform works.
    Suppose a lagrangian [itex]L = q^{-2}[/itex]
    (actually it had another kinetic member, but I dont need it for my question here)

    The matching action [itex]S = \int dt q^{-2}[/itex]

    We were told that it has the next symmetry
    [itex]t \rightarrow at[/itex]
    [itex]q \rightarrow a^{-1/2}q[/itex]
    I tried to figure how, but I cant get it.

    3. The attempt at a solution

    I thought the correct way is this
    [itex]T = at[/itex]
    [itex]Q(T) = a^{-1/2}q(T)[/itex]

    and now I put it in the action
    [itex]S = \int dt q^{-2} = \int (dT/a) (q(T))^{-2} = \int (dT/a) (a^{+1/2}*Q(T))^{-2} = \int (dT/a) (a^{-1})*(Q(T))^{-2} = a^{-2}*S ≠ S[/itex]

    Is that the correct way to do it? Obviously I could change the power of a in t/q transform and it would fix it, is that it?

    Thanks a lot for reading
    Tamir
     
  2. jcsd
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