Let Vμ be a vector field defined in a Minkowski spacetime and suppose it transforms under a Lorentz transformation V′μ=ΛμνVν. We can write this like V′μ=(eiω)μνVν I think where ω denotes a rotation in some plane spanned by indices {ρσ}, say. In 2D Euclidean space time, we can write the matrix representation of Λ as (cosωsinω−sinωcosω) and in Minkowski space this changes to the 'hyperbolic' rotation. Linearising the above yields (1ω−ω1)=Id+(0ω−ω0)=Id+ω(01−10)
Now compare with the more general treatment: V′μ=ΛμνVν≈(δμν+ωμν)Vν, where ωμν≡(ωρσSρσ)μν In 2D, the spin matrix S when acting on vectors in 2D Euclidean space time is therefore the matrix multiplying ω above, which agrees with the single generator of the SO(2) group.
If we continue with the general analysis, we obtain V′μ−Vμ=ωμνVν=ημρωρνVν=ωρσδμρησνVν Now use the antisymmetry of ωρσ gives 2(V′μ−Vμ)=ωρσ(δμρησν−δμσηρν) from which we can identify S to be δμρησν−δμσηρν. I am wondering how this agrees with the matrix I obtained above.
Many thanks.
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