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Angular Momentum Conservation (Lorentz transf. + Noether Th.)

  1. Oct 19, 2014 #1
    1. The problem statement, all variables and given/known data

    Prove $$ j^{\mu} = j_ {EXTERIOR}^{\mu} + j_ {INTERIOR}^{\mu}$$. Writing $$j_ {EXTERIOR}^{\mu}$$ in terms of the energy-momentum tensor. Prove $$j_ {EXTERIOR}^{\mu}$$ is related to the Orbital Momentum and $$j_ {INTERIOR}^{\mu}$$ to the spin.

    Sorry, for the lane shifts.

    2. Relevant equations

    $$ x' \longrightarrow x $$
    $$ x \longrightarrow \Lambda^{-1}x $$

    $$ \Lambda = \mathbb{1} - i\epsilon I_D \implies D[\Lambda] = \mathbb{1} - i\epsilon I_R $$

    This yields:

    $$ \phi' (x) = D[\Lambda]\phi(\Lambda^{-1}x) = \phi(x) - i\epsilon I_R\phi(x) + i\epsilon(I_Dx)^{\mu}\partial_{\mu}\phi(x)$$

    So in Taylor expansion:

    $$\frac {\delta \phi'}{\delta \epsilon} = -iI_R\phi + i(I_Dx)^{\mu}\partial_{\mu}\phi$$

    $$\frac {\delta x'}{\delta x} = -i\epsilon I_Dx$$

    The conserved current formula:

    $$j^{\mu}= \frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi)} \frac {\delta \phi'}{\delta \epsilon} + \mathcal{L}\frac {\delta x'^{\mu}}{\delta \epsilon} $$

    3. The attempt at a solution

    Ok, I tried a few things. Trying to use all was given in the course notes but I am always crashing with the next:

    The $$\frac {\delta x'}{\delta x} = -i\epsilon I_Dx$$ seems to be wrong for me. If I try this:

    $$\frac {\delta x'}{\delta \epsilon} = -i I_Dx$$ Because this must come from the $$\Lambda^{-1}x $$ right?

    And setting two indices to the I_R and I_D matrices (there are 6 for each one corresponding to the Lorentz transf. in each angle or rapidity), then I reach the expresion for the internal part, which corresponds to the spin:

    $$-iI_{\alpha \beta}( \frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi)}\phi) $$

    plus, $$ iI_{\alpha \beta}x^{\beta}T_{\beta}^{\alpha} $$ Which must be wrong! because the indices of x and T must be different to set the antisymmetric part and then prove the orbital momentum conservation.


  2. jcsd
  3. Oct 20, 2014 #2


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    Staff Emeritus
    Science Advisor

    This problem is worked out in this paper:

    The notation is different, and they are assuming particles with spin (so that the angular momentum has a spin part and and "orbital" part).
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