Boost Angular Momentum Vector: Interpreting the Result

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SUMMARY

The discussion focuses on the transformation of the angular momentum vector \(\mathbf{J}\) under Lorentz boosts, specifically highlighting its non-covariant nature. The commutation relation \([J_i, K_j] = i\epsilon_{ijk}J_k\) illustrates the interaction between angular momentum and the boost vector \(\mathbf{K}\). The transformation of \(\mathbf{J}\) under a finite boost is given by \(\mathbf{J} \rightarrow \gamma\left[\mathbf{J} - \left(\frac{\gamma}{\gamma+1}(\mathbf{\beta} \cdot \mathbf{J})\mathbf{\beta} - \mathbf{\beta} \times \mathbf{K}\right)\right]\). The discussion concludes that interpreting angular momentum as a bivector simplifies the transformation, yielding the relation \({J'}^{cd} = L_a^c L_b^d J^{ab}\).

PREREQUISITES
  • Understanding of Lorentz transformations in special relativity
  • Familiarity with angular momentum as a bivector
  • Knowledge of commutation relations in quantum mechanics
  • Basic concepts of 3-vectors and boost vectors
NEXT STEPS
  • Study the properties of bivectors in the context of special relativity
  • Learn about the implications of Lorentz transformations on physical quantities
  • Explore the mathematical framework of commutation relations in quantum mechanics
  • Investigate the role of boost vectors in relativistic physics
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Physicists, particularly those specializing in theoretical physics, students of special relativity, and anyone interested in the mathematical foundations of angular momentum in relativistic contexts.

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Since the angular momentum vector [itex]\mathbf{J}[/itex] is just a 3-vector, it transforms non-covariantly under Lorentz transformations -- more specifically, boosts generated by [itex]\mathbf{K}[/itex]. Indeed, the commutator reads [itex][J_i,\,K_j]=i\epsilon_{ijk}J_k[/itex].

Under a finite boost, I find the angular momentum vector gets mixed up with the 'boost vector'
[tex]\mathbf{J}\rightarrow\gamma\left[\mathbf{J}-\left(\frac{\gamma}{\gamma+1}(\mathbf{\beta}\cdot \mathbf{J})\mathbf{\beta}-\mathbf{\beta}\times\mathbf{K}\right)\right][/tex]

(c.f. the Lorentz transformation of the electric field). How do I interpret this result? In which direction does the new angular momentum vector point? It depends on the boost vector?
 
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The problem is that angular momentum is not a vector. It's a bivector.

What precisely is this boost vector you speak of? Edit: you mean the vector along the 3-velocity of the frame we're boosting into?

At any rate, it's much more elegant to consider angular momentum as a bivector. Then, you just get the result,

[tex]{J'}^{cd} = L_a^c L_b^d J^{ab}[/tex]

where [itex]J^{ab} = x^a p^b - p^a x^b[/itex].
 
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