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QFT - conserved charge.

  1. Aug 28, 2012 #1
    1. The problem statement, all variables and given/known data
    From the Lagrangian density

    [tex] L = \frac{1}2 \partial_\mu \phi_a \partial^\mu \phi_a - \frac{1}2 \phi_a \phi_a,[/tex]

    where a = 1,2,3 and the transformation

    [tex] \phi_a \to \phi _a + \theta \epsilon_{abc} n_b \phi_c [/tex]

    show that one gets the conserved charges

    [tex]Q_a = \int d^3x \epsilon_{abc}\dot{\phi}_b \phi_c.[/tex]


    2. Relevant equations
    The transformation is a symmetry of the Lagrangian so by Noethers theorem
    we got a conserved current which is given by

    [tex]j^\mu = \frac{\partial L}{\partial(\partial_\mu \phi_a)} \delta \phi_a = \partial^\mu \phi_a \epsilon_{abc} n_b \phi_c[/tex]


    3. The attempt at a solution
    The obvious conserved charge is just

    [tex] Q = \int d^3x j^0 = \int d^3 x \dot \phi_a \epsilon_{abc} nb \phi_c [/tex]
    but this is not the 3 different charges in the expression for Q_a. There is no normal vector n in that expresion and the time differentiated field has got b-index instead of an a index.
    How can one get from the conserved current to the expression for these charges?
    [
     
    Last edited: Aug 28, 2012
  2. jcsd
  3. Aug 28, 2012 #2

    vela

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    Your expression for the conserved current doesn't look correct. In the middle term, you sum over a, but in the expression on the right, a is a free index.
     
  4. Aug 28, 2012 #3
    I agree and I have corrected it now. That was just a typo. Do you have any suggestion on how to go from the corrected expression to the conserved charge?
     
  5. Aug 28, 2012 #4

    TSny

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    Is there a typo here for the subscript on the left side?
    Is the Lagrangian invariant under the transformation for any choice of the vector nb? If so, you should be able to get three independent conserved charges by choosing three independent vectors for n. For example, what would the conserved charge be if you choose n1 = 1 and n2 = n3 = 0?
     
  6. Aug 28, 2012 #5
    Yes there was yet another typo. I have corrected that one too now. That's true.
    Hmm.. The conserved current satisfy

    [tex] \partial_\mu (\partial^\mu \phi_a \epsilon_{abc} n_b \phi_c) = 0[/tex]

    I agree with you that one should be able to get 3 independent conserved charges (and also currents), so what if i chose all components to be zero except the a'th component?
    I.e. choose

    [tex] n_b = \delta _{ba}.[/tex]

    Would it then be correct of me to write

    [tex] \partial_\mu (\partial^\mu \phi_a \epsilon_{abc} \delta_{ab} \phi_c) =\partial_\mu ( \epsilon_{abc} \partial^\mu \phi_b \phi_c)= 0[/tex]

    so that i get

    [tex](j^\mu)_a = \epsilon_{abc} \partial^\mu \phi_b \phi_c?[/tex]
     
  7. Aug 28, 2012 #6

    TSny

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    You don't want to use a subscript "a" here, since you are already using that symbol as a dummy summation index in jμ.

    See, the subscript "a" is occurring three times here on the left. That's not good.

    You might try again by letting [tex] n_b = \delta _{bd}.[/tex] so that you're letting nd be the nonzero component of n.

    You can actually go to your result for Q in your original post: [tex] Q = \int d^3x j^0 = \int d^3 x \dot \phi_a \epsilon_{abc} nb \phi_c [/tex] and make this substitution rather than starting way back at the expression for jμ.
     
    Last edited: Aug 28, 2012
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