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

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SUMMARY

The discussion focuses on proving the relationship between the exterior and interior components of the conserved current in the context of angular momentum conservation, specifically through Lorentz transformations and Noether's theorem. The exterior current, \( j_{EXTERIOR}^{\mu} \), is linked to orbital momentum, while the interior current, \( j_{INTERIOR}^{\mu} \), corresponds to spin. The participants analyze the conserved current formula and its derivation using the energy-momentum tensor and variations in the fields, highlighting specific equations and transformations involved in the proof.

PREREQUISITES
  • Understanding of Noether's theorem in theoretical physics
  • Familiarity with Lorentz transformations and their mathematical representation
  • Knowledge of conserved currents and energy-momentum tensors
  • Proficiency in Taylor expansions and functional derivatives
NEXT STEPS
  • Study the derivation of conserved currents in Noether's theorem
  • Explore the implications of Lorentz transformations on physical systems
  • Learn about the energy-momentum tensor and its role in field theory
  • Investigate the relationship between spin and angular momentum in quantum mechanics
USEFUL FOR

The discussion is beneficial for theoretical physicists, graduate students in physics, and researchers focusing on quantum field theory and conservation laws in particle physics.

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Homework Statement



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.

Homework 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:

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

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.Help?
 
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