I'm trying to introduce myself to QFT following these lectures by David Tong. I've started with lecture 1 (Classical Field Theory) and I'm trying to prove that under an infinitesimal Lorentz transformation of the form
Λμν=δμν+ωμν,
where ω is antisymmetric, the variation of the Lagrangian density L is
δL=−∂μ(ωμνxνL).
Using L=L(ϕ,∂μϕ), I've tried computing δL directly using
δϕ=−ωμνxν∂μϕ
[which I obtained earlier computing explicitly ϕ(x)→ϕ(Λ−1x)], however, I get δL=−∂μ(ωμνxνL)−∂L∂(∂μϕ)ωσμ∂σϕ The extra term arises when I compute ∂μ(δϕ)=−ωσν[δνμ∂σϕ+xν∂μσϕ]=−ωσμ∂σϕ−ωσνxν∂μσϕ
[because I'm assuming ∂μ(δϕ)=δ(∂μϕ)]; I thought I'd get rid of it just replacing ϕ with ∂σϕ in (1.52), however ∂μ(δϕ)=δ(∂μϕ) should still hold, ain't it? I also tried using (the previous expression to) 1.27 in the lectures, namely that the derivatives of the field transform as
∂μϕ(x)→(Λ−1)νμ∂νϕ(Λ−1x),
but I still get (to the first order in ω),
(Λ−1)νμ∂νϕ(Λ−1x)=(δνμ−ωνμ)∂νϕ(xσ−ωσρxρ)=(δνμ−ωνμ)[∂νϕ(x)−ωσρxρ∂σνϕ(x)]=∂μϕ−ωσρxρ∂σμϕ−ωνμ∂νϕ
I'm resisting the idea that ωνμ∂νϕ=0, but I don't understand what I'm doing wrong.
Answer
Provided that L is a Lorentz scalar, the quantity ∂L/∂(∂μϕ) has to carry an upper index. Since L is a function of ϕ and ∂μϕ, the only object that can give such an index is ∂μϕ. Hence ∂L∂(∂μϕ)∝∂μϕ. Then, ∂L∂(∂μϕ)ωσμ∂σϕ∝ωσμ∂σϕ∂μϕ=ωσμ∂σϕ∂μϕ=0. The last expression vanishes because ∂σϕ∂μϕ is symmetric under the interchange of indices while ωσμ is antisymmetric.
I actually don't understand why Tong didn't simply write δL=−ωμνxν∂μL. After all, L should have the same transformation rule as ϕ because they are both Lorentz scalars. One can verify the above equation by noting that δL=−∂μ(ωμνxνL)=−ωμνxν∂μL−ωμμL, and that ωμμ=ημρωμρ=0 because ημρ is symmetric and ωμρ is antisymmetric.
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