# Boosting the angular momentum vector

1. Jul 10, 2012

### TriTertButoxy

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

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

(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?

2. Jul 10, 2012

### Muphrid

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,

$${J'}^{cd} = L_a^c L_b^d J^{ab}$$

where $J^{ab} = x^a p^b - p^a x^b$.

Last edited: Jul 10, 2012