in general, this is complicated — but we only care about it forημGμa= 0— thenthe first term on the RHS vanishes and the change inημGμais independent ofG— just a constant — infinite but uninteresting— NO GHOSTSthis is called “temporal gauge” ifημημ>0, it is called “axial gauge” ifημημ<0,and it is called “light-cone gauge” ifημημ= 0— so you see that there are manygauges in which gauge fixing works very simply — you just impose a conditiononGμto break the symmetry and calculate — for gauge invariant quantities, theresults are independent of which gauge you chose21
Δ(G) = Πx,aflflflfl[dημGμΩa][dΩ]flflflflημGμΩa=0because we are evaluating atημGμa= 0we only need infinitesimaltransformationsΩ = 1 +idωaTaGμ=TaGμa→ΩGμΩ-1-iΩ∂μΩ-1→dGμa=i[dωaTa, TbGμb]-Ta∂μdωa=-Ta∂μdωa-i[TbGμb, dωaTa]=-Dμ(Tadωa)d(ημGμa) =ημdGμa=i[dωaTa, TbημGμb]-Taημ∂μdωain general, this is complicated — but we only care about it forημGμa= 0— thenthe first term on the RHS vanishes and the change inημGμais independent ofG— just a constant — infinite but uninteresting— NO GHOSTSthis is called “temporal gauge” ifημημ>0, it is called “axial gauge” ifημημ<0,and it is called “light-cone gauge” ifημημ= 0— so you see that there are manygauges in which gauge fixing works very simply — you just impose a conditiononGμto break the symmetry and calculate — for gauge invariant quantities, theresults are independent of which gauge you chose22
funny propagator — look at oneGμaand drop thea— assumeημημ6= 0L=-14‡∂μGν-∂νGμ·‡∂μGν-∂νGμ·+KημGμ+sμGμ+· · ·Kequationof motionημGμ= 0Gνequationof motion∂μ∂L∂μGν=∂L∂Gν-∂μ‡∂μGν-∂νGμ·=Kην+sν0 =∂ν‡Kην+sν·⇒K=-(∂s)(η∂)(η∂)(∂G) =-ην∂μ‡∂μGν-∂νGμ·=ην‡Kην+sν·=-(ηη)(∂s)(η∂)+ (ηs)(∂G) =(η∂)(ηs)-(ηη)(∂s)(η∂)2-(∂∂)Gν=sν-ην(∂s)(η∂)-∂ν(η∂)(ηs)-(ηη)(∂s)(η∂)2=--gνμ+∂νημ+ην∂μ(η∂)-(ηη)∂ν∂μ(η∂)2¶sμ23
another example of gauge fixing functions —f(G, φ) = Πx,aδ‡∂μGμa(x)·I call this “Landau gauge” (also called “Lorenz gauge” or “Lorentz gauge”)Ka∂μGμaso that the equation of motion forKais∂μGμa= 0what isZf(GΩ, φΩ) [dΩ]?Δ(G, φ)-1=Zf(GΩ, φΩ) [dΩ] =ZΠx,aδ‡∂μGμΩa(x)·[dΩ]=ZΠx,aδ‡∂μGμΩa(x)·1`flflflfl[d∂μGμΩa][dΩ]flflflfl¶[d∂μGμΩa]= Πx,aˆ1,flflflfl[d∂μGμΩa][dΩ]flflflfl∂μGμΩa=0!