
#1
Jul1012, 11:27 AM

P: 194

Since the angular momentum vector [itex]\mathbf{J}[/itex] is just a 3vector, it transforms noncovariantly 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? 



#2
Jul1012, 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 3velocity 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]. 


Register to reply 
Related Discussions  
Vector Model of Angular Momentum  Introductory Physics Homework  1  
What is the direction of angular momentum vector of a photon?  Quantum Physics  5  
What is the angularmomentum 4vector?  Special & General Relativity  11  
Vector products Finding the direction of angular momentum  Advanced Physics Homework  7 