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.So in order for this expression to vanish, the derivative must be zero.That is,⎛ ∂Lμ⎞∂∂νϕ −δ Lμ⎝⎜ ∂[∂νμϕ ]⎠⎟ =This expression is so important that we give this quantity its own name.It turns out this is the energy-momentum tensor.We write this as∂LT μμ(2.21)ν =ν∂ ϕ − δν L∂[∂μϕ]Hence the conservation relation expressed by zero total divergence is∂ μμ νT = 0 (2.22)Notice that∂LT 0 =ϕ − L = H (2.23)∂ ϕThat is, T 0 is nothing other than the Hamiltonian density: it’s the energy density and the 0equation ∂T0 refl ects conservation of energy.The components of momentum=density are given by T 0 where i runs over the spatial indices.The components of imomentum of the fi eld are found by integrating each of these terms over space, that is, P = d x Tii∫ 3 0 (2.24)38Quantum Field Theory Demystifi edConserved CurrentsNow let’s go through the process used in the last section again to see how Noether’s theorem can be applied to derive a conserved current and an associated conservedcharge.We let the fi eld vary by a small amountϕ → ϕ + δϕ (2.25)We then start from the premise that under this variation, the Lagrangian does notchange.The variation in the Lagrangian due to Eq.(2.25) will be of the formL → L + δL (2.26)So what we mean is thatδL = 0 (2.27)Now following the usual procedure the variation of the Lagrangian due to a variation in the fi eld will beδL∂L∂L =δϕ +∂μ δϕ()ϕ∂∂(∂μϕ)Once again, from the Euler-Lagrange equations [Eq.(2.14)] we can write∂L⎛ ∂L ⎞= ∂∂ϕμ ⎝⎜ ∂[∂μϕ]⎠⎟Therefore, we have⎛L∂⎞⎛ ∂⎞δL∂LL = ∂δϕμ∂μ( ) = ∂μ⎝⎜ ∂[∂μϕ δϕδϕ]⎠⎟+ ∂[∂ ]⎝⎜ ∂[∂ ]μϕ⎠⎟μϕSince we are operating under the premise that the variation in the fi eld does not change the Lagrangian [Eq.(2.27)], this leads us to the result⎛ ∂L⎞∂δϕμ0 (2.28)⎝⎜ ∂[∂μϕ] ⎠⎟ =CHAPTER 2 Lagrangian Field Theory39We call the quantity in the parentheses a conserved current.In analogy with electrodynamics we denote it with the letter J and write∂LJ μ =δϕ∂(2.29)[∂μϕ]The conservation law Eq.(2.28) then can be written as∂ μμ J= 0 (2.30)This is the central result of Noether’s theorem:• For every continuous symmetry of the Lagrangian—that is, a variationin the fi eld that leaves form of the Lagrangian unchanged—there is aconserved current whose form can be derived from the Lagrangian usingEq.(2.29).There is a conserved charge associated with each conserved current that results from a symmetry of the Lagrangian.This is found by integrating the time component of J: Q = d x J∫ 3 0 (2.31)We see that the translation symmetry in spacetime worked out in the previous section that led to the energy-momentum tensor is a special case of Noether’s theorem, with the conserved “charges” being energy and momentum.The Electromagnetic FieldThe Maxwell or electromagnetic fi eld tensor is given by⎛− E− E− Exyz ⎞⎜ E− BB ⎟xzyF μνμ Aνν Aμ= ∂− ∂= ⎜⎟ (2.32)EB⎜− Byzx ⎟⎜ E⎝− BB0 ⎟zyx⎠The Aμ is the usual vector potential, but this is a 4-vector whose time component is the scalar potential and whose spatial component is the usual vector potential usedto write down the magnetic fi eld, that is, Aμ = ψ( , A).It can be shown that F μν40Quantum Field Theory Demystifi edsatisfi es or leads to Maxwell’s equations.Note that F μν is antisymmetric; the sign fl ips when the indices are interchanged.That is,F μνFνμ= −(2.33)Without sources, the homogeneous Maxwell’s equations can be written in terms ofthe electromagnetic fi eld tensor as∂λ μν + ∂ν λμ + ∂μ νλFFF(2.34)Meanwhile, the inhomogeneous Maxwell’s equations can be written as∂ μννμ F− J = 0 (2.35)where Jν are the current densities.It is an instructive exercise to derive Maxwell’s equations using a variational procedure so that you can learn how to work withtensors of higher order, that is, vector fi elds.In the next example we derive Eq.(2.35) from a Lagrangian.EXAMPLE 2.5Show that the Lagrangian L = − 1 F F μν J μ A leads to the inhomogeneous 4μν−μMaxwell’s equations [Eq.(2.35)] if the potential A is varied, leaving the current mdensity constant.SOLUTIONWe begin as usual by writing the action as an integral of the Lagrangian density.The action in this case is⎛ 1μνμ⎞S = d x − F Fμν− J A∫ 4 ⎝⎜μ4⎠⎟The variation we will compute is δ μA.We have11δ⎛⎞S = d x −δμνμνμ(μνF ) F−μνFδ( F ) − J δ μA∫ 4 ⎝⎜ 44⎠⎟Let’s consider the fi rst term.Using the defi nition of the electromagnetic fi eld tensor Eq.(2.32), we can write F= ∂ A − ∂ Aμνμ νν μ and so we have− 1 (δ F ) μνμνμν F= − 1 (∂μδ νA − ∂νδ A )μ F44CHAPTER 2 Lagrangian Field Theory41Now we integrate by parts, transferring the derivatives from the δ νA terms to F μν.This allows us to write− 1 (δ F ) μνμνμν F= − 1 (∂μδ νA − ∂νδ A )μ F44= 1 (∂ μνμνμ F δ νA − ∂ F δ A )νμ4But, repeated indices are dummy indices [ Pobierz całość w formacie PDF ]