$$\label{e:psfin}
  {\ensuremath{|p,s\rangle}}' = U(\Lambda){\ensuremath{|p,s\rangle}} = \sqrt{\frac{(\Lambda p)^0}{p^0}}D_{s' s}
  (W(\Lambda, p)){\ensuremath{|\Lambda p,s'\rangle}},$$ and $$\label{e:wiggen}
  W(\Lambda,p)=L^{-1}(\Lambda p)\Lambda L(p)$$ is a Wigner transformation which leaves $k^\mu$ invariant[@wigner], and $D(W)$ represents its action on the state (summation convention over the repeated index ${s'}$ is assumed). The explicit form of $L(p)$ is dependent on the class of the four-momenta. For massive particles, $p^\mu p_\mu = -m^2 <
0$, and a convenient choice for the standard vector is $k^\mu = (m,{\bm 0})$. It is then obvious that the set of Wigner transformations leaving $k^\mu$ unchanged is just the rotation group $SO(3)$. Furthermore, $L(p)$ can then be taken as the pure Lorentz boost $$\begin{aligned}
\label{e:L}
  L^0_{~0}(p)&=&\cosh\chi \\
  L^0_{~i}(p)&=&L^i_{~0}(p)={\ensuremath{\hat p}}_i\sinh\chi \\
  L^i_{~j}(p)&=&\delta^i_{~j}+(\cosh\chi-1){\ensuremath{\hat p}}^i{\ensuremath{\hat p}}_j.\end{aligned}$$ with $\tanh\chi = \frac{{\ensuremath{{\ensuremath{\vert {\ensuremath{\textbf p}}\vert}}}}}{\sqrt{{\ensuremath{{\ensuremath{\vert {\ensuremath{\textbf p}}\vert}}}}^2 + m^2}}$. In this parametrization, $L(p) =\exp(-i\chi{\hat{\textbf p}}\cdot{\bf K})$ where $K^{i} = M^{0i}$ is the boost generator[@weibqft].

We may rely upon the analytic nature of Lie groups for the computation of the Wigner angle, and the corresponding infinitesimal Wigner rotation can be evaluated by Taylor expansion. Details of the rest of the computations are delegated to Appendix A. The end result is that the infinitesimal Wigner rotation of a massive particle is $$\begin{aligned}
\label{e:wrg}
  W(\Lambda,p) &=& 