# Value of $p^{2}$ for little groups

I am looking at Weinberg, The Quantum Theory of Fields, Volume 1 page 66.

In this table, the author mentions various little groups of the Lorentz group. Orthochronous Lorentz transformations must leave $$p^2$$ and the sign of $$p^{0}$$ for $$p^{2} \leq 0$$ invariant. Shouldn't (c) and (d) then be $$(\kappa,0,0,\kappa)$$ and $$(-\kappa,0,0,\kappa)$$ respectively? And shouldn't $$p^2 = -N^2$$ in part (e)? $$p$$ and $$k$$ are related by $$p^{\mu} = L^{\mu}_\phantom{\mu}{\nu} k^{\nu}$$.

• Weinberg puts the zeroth component in the last place. The $p^2 < 0$ cases is already dealt with in (a),(b). Part (e) has $p^2 > 0$. Feb 15 at 10:20
• Please do not post images. Use MathJax instead! Feb 15 at 10:37

Weinberg writes four-vector in the order $$1,2,3,0$$ with $$p^2:= \vec{p}^2-(p^0)^2$$ (c.f. p. xxv "Notations").
(c) $$p=(0,0,\kappa,\kappa)$$ with $$\kappa \gt 0$$ $$\; \Rightarrow \;$$ $$p^0=\kappa \gt 0$$ and $$p^2=0$$, but $$p^\prime=(\kappa,0,0,\kappa)$$ would also be possible.
(d) $$p=(0,0,\kappa,-\kappa)$$ with $$\kappa \gt 0$$ $$\; \Rightarrow \;$$ $$p^0=-\kappa \lt 0$$ and $$p^2=0$$, but $$p^\prime =(-\kappa,0,0,\kappa)$$ has $$p^0=\kappa \gt 0$$.
(e) $$p=(0,0,N,0)$$ with $$N\ne 0\; \Rightarrow \;$$ $$p^2=N^2 \gt 0$$