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Noether current for SO(3) rotation

  1. Mar 20, 2012 #1
    This is a problem from my theoretical physics course. We were given a solution sheet, but it doesn't go into a lot of detail, so I was hoping for some clarification on how some of the answers are derived.

    1. The problem statement, all variables and given/known data

    For the Lagrangian L=1/2(∂μTμ∅-m2T∅) derive the Noether currents and charges.


    2. Relevant equations

    jμa=∂L/(∂(∂μa))*[itex]\Phi[/itex] - [itex]\Lambda[/itex][itex]\mu[/itex]α

    Here, the lambda term is zero, because the Lagrangian is invariant under SO(3).

    a → ∅a + [itex]\Phi[/itex]εα


    3. The attempt at a solution


    We were first told to show that the above Lagrangian satisfies SO(3) symmetry (this was fine). The solution sheet then states that infintessimal transformations can be written as ∅a → ∅a-itc(Tc)abb, where (Tc)ab=-iεcab

    I could not work out how to derive this though.

    Using the above info, I can see that [itex]\Phi[/itex]ac = -i(Tc)abb, taking εα = tc

    Then I just need to calculate ∂L/∂(∂μa)
    Is this just ∂μa?? I'm not sure how to calculate this when there's 2 derivatives, one with a superscript and one with a subscript. And does the transpose affect things?
     
  2. jcsd
  3. Mar 20, 2012 #2

    fzero

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    You probably want [itex]{j^\mu}_\alpha[/itex] on the left side to match indices.

    You could either take this as a definition of the Lie algebra of the orthogonal group or else show that the finite matrices [itex]\exp (i t^a T_a)[/itex] are orthogonal. This is a generalization of the way we can parametrize 2d rotations by an angle.



    You can write

    [tex]\partial_\mu \phi^a \partial^\mu \phi_a = \delta_{ab} \eta^{\mu\nu} \partial_\mu \phi^a \partial_\nu \phi^b,[/tex]

    then use

    [tex] \frac{\partial}{\partial(\partial_\nu \phi^b) } \partial_\mu\phi^a = \delta^a_b \delta^\nu_\mu.[/tex]

    Since there are 2 factors of [itex]\partial\phi[/itex], you'll need to use the product rule which brings in a factor of 2 in the answer.
     
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