Lagrangian. µ = 0 0 E x E y E z 1 E x 0 B z B y 2 E y B z 0 B x 3 E z B y B x 0. field tensor. ν =

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1 Lagrangian L = 1 4 F µνf µν j µ A µ where F µν = µ A ν ν A µ = F νµ. F µν = ν = µ = 0 0 E x E y E z 1 E x 0 B z B y 2 E y B z 0 B x 3 E z B y B x 0 field tensor. Note that F µν = g µρ F ρσ g σν for g µν = diag(1, 1, 1, 1), so F 0i = F 0i and F ij = F ij for i, j 1, 2, 3 so 1 4 F µνf µν = 1 2 (E 2 B 2 ). C. DeTar (U Utah) University of Utah December 5, / 14

2 Vector potential We have A µ = (Φ, A) and j µ = (ρ, J) and E = Φ t A B = A so S = dtl(t) and L = d 3 xl = [ 1 d 3 x ( Φ + t 2 12 ) ] ( A)2 Φρ + A J We might want to use Φ and A as the field coordinates. C. DeTar (U Utah) University of Utah December 5, / 14

3 Gauge invariance But note gauge invariance. The change of variables A A χ Φ Φ + t χ Leaves F µν unchanged. The action changes as S = d 4 x ( 14 ) F µνf µν j µ A µ ( d 4 x 1 ) 4 F µνf µν j µ A µ j µ µ χ = S + d 4 x ( µ j µ )χ S is invariant for a conserved current: µ j µ = ρ/ t + J = 0. C. DeTar (U Utah) University of Utah December 5, / 14

4 Euler-Lagrange Equations δl t δ A/ t δl δa = 0 δl t δ Φ/ t δl δφ = 0 i.e. ( Φ + t ) t A + ( A) = J ( Φ t ) A = ρ E t + B = J E = ρ The other two Maxwell s equations follow from the expressions relating E and B to A and Φ. C. DeTar (U Utah) University of Utah December 5, / 14

5 Gauss Law as a Constraint Canonical momenta: p A = δl δ A/ t = E p Φ = δl δ Φ/ t = 0, so Gauss Law is a constraint equation, not an equation of motion, and Φ can t be a canonical variable. Take divergence of the Maxwell-Ampère law t E + B = t E = J = ρ t so once we satisfy Gauss Law at one time, it is satisfied at all times. C. DeTar (U Utah) University of Utah December 5, / 14

6 Inconsistencies If we try to use all three spatial components of the vector potential as canonical coordinates we have an inconsistency: [A i (x), p A,j (x )] = iδ ij δ 3 (x x ) wrong! since p A = E and the divergence with respect to x gives [A i (x), ρ(x )] = 0 = i δ 3 (x x ). x i This difficulty can be traced to gauge invariance. Not all information about A is determined by the Euler-Lagrange equations. C. DeTar (U Utah) University of Utah December 5, / 14

7 Gauge Fixing A way out is to fix the gauge. Choices (infinitely many) A 0 = 0 Axial gauge A = 0 Coulomb gauge µ A µ = 0 Landau, Lorentz, or Feynman gauge C. DeTar (U Utah) University of Utah December 5, / 14

8 Coulomb Gauge In Coulomb gauge A = 0. The Lagrangian becomes L = d 3 x [ 1 2 ( ) ] A 2 1 t 2 ( A)2 + J A 1 2 Φρ where now 2 Φ = ρ, so we solve for Φ and eliminate it. Since A = 0 we change the canonical commutator to [A i (x), p A,j (x )] = iδij tr (x x ) δij tr (x x ) = ( δ ij i j 2 d 3 k (2π) 3 eik (x x ) ( δ ij k ) ik j k 2 ) δ 3 (x x ) C. DeTar (U Utah) University of Utah December 5, / 14

9 Photons The result is that there are only two states of polarization: A(x, t) = p A (x, t) = where ω = k and and d 3 k 2ω(2π) 3 d 3 k 2ω(2π) 3 2 λ=1 ɛ(k, λ) [a(k, λ)e ik x + a (k, λ)e ik x] 2 ɛ(k, λ)iω [a (k, λ)e ik x a(k, λ)e ik x] λ=1 [a(k, λ), a (k, λ )] = δ λ,λ δ 3 (k k ) k ɛ(k, λ) = 0 A = 0 2 λ=1 ɛ i (k, λ)ɛ j (k, λ) = δ ij k ik j k 2. C. DeTar (U Utah) University of Utah December 5, / 14

10 Propagator The Feynman propagator in Coulomb gauge is ig ij (x, x ) = 0 T [A i (x)a j (x )] 0 d 4 k e ik (x x ) ( = i (2π) 4 k 2 δ ij k ) ik j + iɛ k 2. This is not Lorentz covariant because Coulomb gauge is not. C. DeTar (U Utah) University of Utah December 5, / 14

11 Covariant gauges L = 1 4 F µνf µν j µ A µ 1 2 λ( µ A µ ) 2 The last term is a gauge fixing term. With λ = 1 we get Feynman gauge. The Euler-Lagrange equation is 2 A µ + (1 λ) µ ( A) = j µ and propagator is ig µν (x, x ) = i d 4 k e ik (x x ) ( (2π) 4 k 2 + iɛ g µν + 1 λ λ ) k µ k ν k 2 The Feynman rules vary with the gauge. Physical results do not. C. DeTar (U Utah) University of Utah December 5, / 14

12 Exercise (1) Show that the Euler-Lagrange equations for the electromagnetic field interacting with a conserved current j µ without gauge fixing reproduce the Maxwell-Ampère law and Gauss law. That is, fill in the details of the derivation sketched in the lecture notes. Exercise (2) (a) Under a gauge transformation A µ A µ = A µ + µ χ, show that F µν = µ A ν ν A µ is invariant. (b) Find χ such that the transformed field is in the gauge A 0 = 0 (That is, find χ in terms of the original A µ.) (c) Once in the gauge A 0 = 0, a restricted class of gauge transformations is still possible. What is the restriction on χ? C. DeTar (U Utah) University of Utah December 5, / 14

13 Exercise: Longitudinal and transverse fields Exercise (3) A vector field can be decomposed into a longitudinal and a transverse part: E(x) = E (x) + E (x) where the longitudinal part has no curl and the transverse part, no divergence: E = 0 E = 0 (a) Find the natural decomposition of the electric field in Coulomb gauge, i.e. express E and E in terms of Φ and A. (b) Write the E and B fields in terms of the plane wave expansion. (Don t neglect the c-number E contribution to E. C. DeTar (U Utah) University of Utah December 5, / 14

14 Wave equation: Exercise (4) Starting from the Lagrangian in Coulomb gauge, show that the Euler-Lagrange equations can be expressed as ( ) 2 t 2 2 A = J. Coulomb gauge propagator: Exercise (5) From the plane wave expansion for the vector potential, derive the expression for the Coulomb gauge photon propagator G ij (x, x ) = d 4 k e ik (x x ) ( (2π) 4 k 2 δ ij k ) ik j + iɛ k 2 C. DeTar (U Utah) University of Utah December 5, / 14