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

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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 $p^{\mu} = L^{\mu}_\phantom{\mu}{\nu} k^{\nu}$.

  • 3
    $\begingroup$ Weinberg puts the zeroth component in the last place. The $p^2 < 0$ cases is already dealt with in (a),(b). Part (e) has $p^2 > 0$. $\endgroup$
    – Prahar
    Feb 15 at 10:20
  • 1
    $\begingroup$ Please do not post images. Use MathJax instead! $\endgroup$
    – Hyperon
    Feb 15 at 10:37

1 Answer 1


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 \;$ $p^2=N^2 \gt 0$

  • $\begingroup$ Right, that makes sense. Thank you. $\endgroup$
    – saad
    Feb 15 at 10:39

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