# Standard boost, particles with mass M > 0

Weinberg considers (p.68 QFT Vol. 1) particles with mass M > 0. The Little Group is SO(3). He wants to calculate the rotation
W(Λ,p) ≡ L-1(Λp) Λ L(p). He says that for this we need to choose a standard boost L(p) which carries the four momentum from
kμ = (0,0,0,M) to pμ. He then shows the expressions for the components of L(p). What I dont understand is that because the spatial components of kμ all vanish, i.e. ki = 0, why does it matter to choose a specific set of expressions for (L(p))i j?

vanhees71
Gold Member
You can choose any Lorentz transformation that maps ##k^{\mu}## into ##p^{\mu}##. The most convenient (standard) choice is to use the uniquely defined rotation free Lorentz boost, i.e., a boost with velocity ##\vec{p}/p^0=\vec{p}/\sqrt{M^2+\vec{p}^2}##.

For more details have a look at appendix B of my QFT manuscript:

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

kent davidge and dextercioby