I have a super basic and stupid question about the Lorentz invariance of the Polyakov action (cannot skip the disclaimer..) Sp[X,γ]=−14πα′∫∞−∞dτ∫l0dσ(−γ)1/2γab∂aXμ∂bXμ
I may write the action as (using metric tensor γab) Sp[X,γ]=−14πα′∫∞−∞dτ∫l0dσ(−γ)1/2∂aXμ∂aXμ
The Lagrangian is obviously invariant under proper Lorentz transformation. ∂aXμ∂aXμ is a Lorentz scalar. dτdσ transforms with a determinant equals 1 for proper Lorentz transformation. (−γ)1/2 also transforms with a determinant equals 1.
But for the range of integration [0,l], there is length contraction under a boost. Why the action is still invariant under Lorentz transformation? (or I completely mistaken something....)
Answer
You are confusing the spacetime Lorentz symmetry and the world sheet Lorentz symmetry.
The spacetime Lorentz symmetry only acts on the fields locally on the world sheet, Xμ(σ,τ)→ΛμνXν(σ,τ).
On the other hand, the world sheet theory also locally has the SO(1,1) Lorentz symmetry on the world sheet which only acts on the two σ,τ coordinates i.e. on the Latin indices a,b. This symmetry is broken if the world sheet is compactified i.e. if σ spans a periodic or compact interval. But there are still unbroken Weyl/diff diffeomorphisms on the world sheet, the conformal transformations, that must be carefully taken into account during string calculations. All these symmetries are "auxiliary" much like any gauge symmetry – all the observable states and operators in the spacetime must be invariant under them (singlets).
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