How can I calculate the square of the Pauli-Lubanski pseudovector?

  • #1
tannhaus
2
0
TL;DR Summary
I need to calculate the square of the Pauli-Lubanski pseudovector in a rest frame such that the results is proportional to the square of the spin operator.
Hello there, recently I've been trying to demonstrate that, $$\textbf{W}^2 = -m^2\textbf{S}^2$$ in a rest frame, with ##W_{\mu}## defined as $$W_{\mu} = \dfrac{1}{2}\varepsilon_{\mu\alpha\beta\gamma}M^{\alpha\beta}p^{\gamma}$$ such that ##M^{\mu\nu}## is an operator of the form $$ M^{\mu\nu}=x^{\mu}p^{\nu} - x^{\nu}p^{\mu} + \frac{i}{2}\Sigma^{\mu\nu}$$ and ##S^i## defined as $$S_i = \varepsilon^{ijk}\frac{i}{2}\Sigma^{jk}$$ Where ##\Sigma^{\mu\nu} = [\beta^{\mu}, \beta^{\nu}]##. I've managed to show that ##\textbf{S}^2 = \dfrac{1}{2}\Sigma^{ij}\Sigma_{ij}## but I can't for my life work out the necessary result. Any sort of light towards this is very welcome!
 
Physics news on Phys.org
  • #2
Your expressions are manifestly covariant. "Spin" for a massive particle is, however, most easily to interpret in the rest frame of the particle. So to have some intuitive picture, it's best to calculate it within this frame, and this is simply defined by ##(p^{\mu})=(m c,0,0,0)##. In this frame you have a pretty intuitive interpretation of "spin" and the Pauli Lubanski vector (the latter one being the only viable definition of spin in relativistic physics, where in general a unique split of total angular momentum into "spin" and "orbital" is not possible). For more on "classical spin" in relativity, see Sect. 1.8 in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

That becomes much clearer in the context of relativistic QFT and a detailed analysis of the representation theory of the Poincare group, where the Pauli Lubanski vector is the generator for little-group transformations, and the little group for massive-particle representations is the rotation group (or its covering group SU(2)) as defined in the rest frame of the particle. The quantities in other frames is then given by the (rotation free) Lorentz boosts from the rest frame of the particle to an arbitrary frame, where it's moving. For details see

https://itp.uni-frankfurt.de/~hees/publ/lect.pdf

(particularly Appendix B).
 

Similar threads

  • Special and General Relativity
Replies
5
Views
1K
  • Special and General Relativity
Replies
3
Views
2K
  • Special and General Relativity
Replies
1
Views
2K
  • Advanced Physics Homework Help
Replies
2
Views
689
  • Science and Math Textbooks
Replies
7
Views
776
  • Special and General Relativity
Replies
6
Views
2K
  • Beyond the Standard Models
Replies
1
Views
1K
Replies
1
Views
990
Replies
5
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
1K
Back
Top