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Boosting the angular momentum vector

by TriTertButoxy
Tags: angular, boosting, momentum, vector
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TriTertButoxy
#1
Jul10-12, 11:27 AM
P: 194
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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Muphrid
#2
Jul10-12, 12:08 PM
P: 834
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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