Lorentz Currents
A global lorentz transformation xμ→Λ νμxν should produce a conserved current K. Now we try to derive such a current.
Consider a infinitesimal lorentz transformation
Λ νμ=I νμ+21ωαβJ ναβμ+⋯
where J is antisymmetric in the first two indices :
J ναβμ=−J νβαμ
then δxμ is the difference between such a small spin and the original xμ, that is
δxμ=21ωαβJ ναβμxν.
Now we turn to look at the arbitrary current Jμ. Such a current must satisfy ∂μJμ=0, which lead to
∂μ[Lδxμ+∂(∂μϕ)∂Lδϕ]=0
it is then obvious that
δϕ=ϕ(xμ)−ϕ(Λ νμxν)=−21ωαβJ ναβμxν∂μϕ
then the whole equation becomes
∂μ[∂(∂μϕ)∂L∂λϕωαβJ ναβλxν−LωαβJ ναβμxν]=0
consider an antisymmetric tensor gμν. We will have the following:
gμνxμxν=[21(gμν+gνμ)+21(gμν−gνμ)]xμxν=21(gμνxμxν−gνμxμxν)=21gμν(xμxν−xνxμ).
Then the whole current-conservation-equation will become
∂μωνλ[∂(∂μϕ)∂L∂λϕxν−∂(∂μϕ)∂L∂νϕxλ−Lgλμxν+Lgνμxλ]=0∂μωνλ[(∂(∂μϕ)∂L∂λϕ−Lgλμ)xν−(∂(∂μϕ)∂L∂νϕ−Lgνμ)xλ]=0∂μωνλ[Tμλxν−Tμνxλ]=0ωνλ(Tμλxν−Tμνxλ)=Jμ
in which we used the fact that
J ναβμ=Jλναβgμλ=(δλαδνβ−δναδλβ)gμλ.
Having ω as an arbitrary parameter, the part inside the brackets is a conserved tensor as well, namely
∂μ(Tμλxν−Tμνxλ)=0⟹∂μKμνλ=0
where the Kμνλ is our Lorentz Current.