d 2 Area i K i0 ν 0 (S.2) when the integral is taken over the whole space, hence the second eq. (1.12).

Transcription

1 PHY 396 K. Solutions for prolem set #. Prolem 1a: Let T µν = λ K λµ ν. Regardless of the specific form of the K λµ ν φ, φ tensor, its antisymmetry with respect to its first two indices K λµ ν K µλ ν implies µ T µν = µ λ K λµ ν = 0 S.1 and hence the first eq Furthermore, d 3 x T 0ν = i K i0 ν = d Area i K i0 ν 0 S. oundary of space when the integral is taken over the whole space, hence the second eq Prolem 1: In the Noether s formula 1.10 for the stress-energy tensor, φ a stand for the independent fields, however laeled. In the electromagnetic case, the independent fields are components of the 4 vector A λ x, hence T µν Noether EM = µ A λ ν A λ g µν L = F µλ ν A λ gµν F κλ F κλ. S.3 While the second term here is clearly oth gauge invariant and symmetric in µ ν, the first term is neither. Prolem 1c: Clearly, one can easily restore oth symmetry and gauge invariance of the electromagnetic stress-energy tensor y replacing ν A λ in eq. S.3 with F ν λ, hence eq The correction 1

2 amounts to T µν = T µν phys T µν Noether = F µλ F ν λ ν A λ = λ A ν = λ F µλ A ν A ν λ F µλ S.4 where the last term on the right hand side vanishes for the free electromagnetic field which satisfies λ F µλ = 0. Consequently, T µν phys = T µν Noether + λk λµν where K λµν = F µλ A ν = K µλν S.5 in perfect agreement with eq Prolem 1d: Let s start with the Lagrangian In component form, F i0 = F 0i = E i, F ij = ɛ ijk B k. S.6 Therefore, F i0 F i0 = F 0i F 0i = E i E i where the minus sign comes from raising one space index. Likewise, F ij F ij = +ɛ ijk B k ɛ ijl B l = +B k B k where the plus sign comes from raising two space indices at once. Thus, L = 1 4 F µν F µν = F i0 F i0 + F 0i F 0i + F ij F ij = 1 E B. S.7 Consequently, eq for the energy density gives H T 00 = F 0i F 0 i L = +E 1 E B = 1 E + B S.8 in agreement with the standard electromagnetic formulæ note the c = 1, rationalized units here. Likewise, the energy flux and the momentum density are S i T i0 = T 0i = F 0j F i j = E j +ɛ ijk B k = +ɛ ijk E j B k = E B i, S.9 in agreement with the Poynting vector S = E B again, in the c = 1, rationalized units.

3 Finally, the 3 dimensional stress tensor is T ij EM = F iλ F j λ gij L = F i0 F j 0 F ik F j k + δij L = E i E j + ɛ ikl B l ɛ jkm B m + 1 δij E B = E i E j + δ ij B l B l B i B j + 1 δij E B S.10 = E i E j B i B j + 1 δij E + B. Prolem 1e: In a sense, eq follows from eq. S.4, ut it is just as easy to derive it directly from Maxwell equations. Starting with eq. 1.14, we immediately have µ T µν EM = µ F µλ F ν λ F µλ µ F ν λ + 1 F κλ ν F κλ. S.11 Using the antisymmetry F µλ = F λµ, we rewrite the second term on the right hand side as F µλ µ F ν λ = +F µλ µ F λν = +F µλ λ F νµ = 1 F µλ λ F νµ + µ F λν = 1 F µλ ν F µλ where the last equality follows from the homogeneous Maxwell equation S.1 ɛ κλµν λ F µν = 0 λ F νµ + µ F λν + ν F µλ = 0. S.13 Consequently, the second and the third terms on the right hand side of eq. S.11 cancel each other and we are left with the first term only. Thus, µ T µν EM = µ F µλ F ν λ = Jλ F ν λ S.14 where the second equality comes from the in-homogeneous Maxwell equation µ F µλ = J λ. This proves eq. 1.16, and eq follows from that and eq Q.E.D. 3

4 Prolem a: As discussed in class, Euler Lagrange equations for charged fields can e written in a manifestly covariant form as In particularly, for φ = Φ, we have D µ D µ φ φ = 0. S.15 D µ Φ = Dµ Φ, Φ = m Φ, which gives us D µ D µ Φ + m Φ = 0. S.16 Likewise, for φ = Φ we have D µ Φ = Dµ Φ, Φ = m Φ, and therefore D µ D µ Φ + m Φ = 0. S.17 As for the vector fields A ν, the Lagrangian.1 depends on µ A ν only through F µν, which gives us the usual Maxwell equation µ F µν = J ν where J ν A ν. S.18 To otain the current J ν, we notice that the covariant derivatives of the charged fields Φ and Φ depend on the gauge field: D µ Φ A ν = iqδ ν µφ, D µ Φ A ν = iqδ ν µφ. S.19 Consequently, J ν = D ν Φ iqφ = iq ΦD ν Φ Φ D ν Φ. D ν Φ iqφ S.0 Note that all derivatives on the last line here are gauge-covariant, which makes the current 4

5 J ν gauge invariant. In a non-covariant form, J ν = iqφ D ν Φ iqφd ν Φ q Φ Φ A ν. S.1 To prove the conservation of this current, we use the Leiniz rule for the covariant derivatives, D ν XY = XD ν Y + Y D ν X. This gives us µ Φ D µ Φ = D µ Φ D µ Φ = D µ Φ D µ Φ + Φ D µ D µ Φ, µ ΦD µ Φ = D µ ΦD µ Φ = D µ ΦD µ Φ + Φ D µ D µ Φ, S. and hence in light of eq. S.0 for the current, νj ν = iq D ν ΦD ν Φ + ΦD ν D ν Φ = iqφ D Φ iqφ D Φ + iq D ν Φ D ν Φ + Φ D ν D ν Φ Q.E.D. y equations of motion = iqφ m Φ iqφ m Φ = 0. S.3 Prolem : According to the Noether theorem, where T µν Noether = µ A λ ν A λ + µ Φ ν Φ + = T µν µν Noether EM + T Noether matter µ Φ ν Φ g µν L S.4 similar to the free EM fields, and T µν Noether EM = F µλ ν A λ gµν F κλ F κλ S.5 T µν Noether matter = Dµ Φ ν Φ + D µ Φ ν Φ g µν D λ Φ D λ Φ m Φ Φ. S.6 Both terms on the second line of eq. S.4 lack µ ν symmetry and gauge invariance and thus need λ K λµν corrections for some K λµν = K µλν. We would like to show that the 5

6 same K λµν = F µλ A ν we used to improve the free electromagnetic stress-energy tensor will now improve oth the T µν µν EM and T mat at the same time! Indeed, to improve the scalar fields stress-energy tensor we need T µν matter T µν µν phys matter T Noether matter = D µ Φ D ν Φ ν Φ + D µ ΦD ν Φ ν Φ = D µ Φ iqa ν Φ + D µ Φ iqa ν Φ S.7 = A ν iqφ D µ Φ iqφd µ Φ = A ν J µ, while the improvement of the EM stress-energy was worked out in prolem 1.c: T µν EM = F µλ F ν λ ν A λ = +F µλ λa ν = λ F λµ A ν + A ν J µ, S.8 cf. eq. S.4. Note that the A ν J µ term cancels etween the EM and the matter improvement terms, so the net improvement needed to symmetrize the comined stress-energy tensor is simply T µν tot T µν µν phys total T Noether total = T µν matter + T µν EM = λ F λµ A ν K λµν. S.9 Q.E.D. Prolem c: Because the fields Φx and Φ x have opposite electric charges, their product is neutral and therefore µ Φ Φ = D µ Φ Φ = D µ Φ Φ + Φ D µ Φ. Similarly, µ D µ Φ D ν Φ = D µ D µ Φ D ν Φ + D µ Φ D µ D ν Φ = m Φ D ν Φ + D µ Φ D ν D µ Φ + iqf µν Φ S.30 where we have applied the field equation D µ D µ + m Φ x = 0 to the first term on the 6

7 right hand side and used [D µ, D ν ]Φ = iqf µν Φ to expand the second term. Likewise, µ D µ Φ D ν Φ = D µ D µ Φ D ν Φ + D µ Φ D µ D ν Φ = m Φ D ν Φ + D µ Φ D ν D µ Φ iqf µν Φ S.31 and µ [ g ] µν D λ Φ D λ Φ m Φ Φ = ν D λ Φ D λ Φ + m ν Φ Φ = D ν D µ Φ D µ Φ D µ Φ D ν D µ Φ + m Φ D ν Φ + m Φ D ν Φ. S.3 Together, the left hand sides of eqs. S.30, S.31 and S.3 comprise µ T µν mat cf. eq. 7. On the other hand, comining the right hand sides of these three equations results in massive cancellation of all terms except those containing the gauge field strength tensor F µν. Therefore, µ T µν mat = D µ Φ iqf µν Φ + D µ Φ iqf µν Φ = F µν iqφ D µ Φ iqφ D µ Φ S.33 = F µν J ν in accordance with eq..7. Finally, comining this formula with eq we see that the it follows that the total stress-energy tensor.4 is conserved, µ T µν tot = µ T µν tot + µ T µν EM = 0. S.34 Q.E.D. 7

8 Prolem 3a: Unitary transform.9 of the Φ a fields into each other leaves the ilinear comination a Φ aφ a invariant which immediately leads to the invariance of the scalar potential in the Lagrangian 1.8. Indeed, a unitary matrix U satisfies U U = 1 i.e., a U a U ac = δ ac and hence a Φ a Φ a = a =,c Φ U a U ac Φ c Φ Φ c U a U ac = δ c a c = Φ Φ. S.35 Similarly, the kinetic part of the Lagrangian would e invariant provided the derivatives D µ Φ a and D µ Φ a transform into each other just line the fields itself, D µ Φ a U a D µ Φ & D µ Φ a D µ Φ U a = a D µ Φ ad µ Φ a invariant. S.36 Thus, all we need to prove is the assumptions on these line or equivalently, that the covariant derivatives D µ commute with the unitary transform.9. For a gloal symmetry, the U matrix is x-independent, thus µ U a = 0 which makes the symmetry transform commute with the ordinary derivatives µ. To commute with the covariant derivatives D µ = µ + ia µ x ˆQ, S.37 the symmetry transform should also commute with the electric charge operator Q. In other words, it should not mix fields having different electric charges. For the prolem at hand, all the Φ a fields have the same charge +q while all the conjugate fields Φ a have the opposite charge q. Thus, a symmetry may mix a Φ a field with any other Φ fields as in eq..9 ut not with any of the Φ conjugate fields. Likewise, it may mix a Φ a with the other Φ ut now with any of the Φ. 8

9 Prolem 3: Consider an infinitesimal unitary symmetry U = 1 + iɛt + Oɛ for some hermitian matrix T = T. The generator T of this symmetry acts on the fields as δφ a = ɛtφ a = ɛ it a Φ, δφ a = ɛtφ a = ɛ it a Φ S.38 so the Noether current of this symmetry is J µ T = a µ Φ a TΦ a + a µ Φ a TΦ a = a D µ Φ a it a Φ + a D µ Φ a i T a = T a Φ = a T a id µ Φ Φ a id µ Φ a Φ S.39 = T a J µ a a tr T J µ. where J µ is the hermitian matrix of currents J µ a = iφ ad µ Φ iφ Dµ Φ a = J µ a..10 Note: a different hermitian matrix T would generate a different infinitesimal unitary symmetry U = 1 + iɛt, which in term would have a different Noether current J µ T = tr T J µ. However, all such currents are linear cominations of the same N currents J µ a, and there is ovious linear etween a particular hermitian generator T and the corresponding Noether current J T. Consequently, the whole matrix J µ = J µ a act as a matrix of Noether currents of the UN symmetry. 9

10 Prolem 3c: Since the product Φ a D µ Φ is electrically neutral, we have µ Φ a D µ Φ = D µ Φ a D µ Φ = D µ Φ a D µ Φ + Φa Dµ D µ Φ, S.40 and likewise µ Φ Dµ Φ a = D µ Φ Dµ Φ a = D µ Φ D µ Φ a + Φ Dµ D µ Φ a. S.41 Consequently µ Φ a D µ Φ Φ Dµ Φ a = Φ a Dµ D µ Φ Φ Dµ D µ Φ a S.4 while the remaining terms in eqs. S.40 and S.41 cancel each other, which means µ J µ a = iφ a Dµ D µ Φ iφ Dµ D µ Φ a. S.43 Moreover, when the scalar fields oey their equations of motion D µ D µ Φ a D µ D µ Φ a = V = Φ a m + λ Φ, Φ a = V Φ = Φ a m + λ Φ, a S.44 the right hand sides of eqs. S.43 vanish altogether and the Noether currents.10 are conserved: µ J µ a = iφ a Dµ D µ Φ iφ Dµ D µ Φ a = iφ a Φ m + λ Φ Q.E.D. = 0. iφ Φ a m + λ Φ S.45 10

11 Prolem 3d: N complex fields Φ a x are equivalent to N real fields φ a,α x for a = 1,..., N and α = 1, : Φ a x = φ a1x + iφ a x, Φ ax = φ a1x iφ a x. S.46 The scalar potential part of the Lagrangian 8 is a function of Φ aφ a a = a φ a1 + φ a = 1 φ a,α a,α S.47 so it is invariant under any SON rotation of the N real fields. For electrically neutral fields, the net kinetic part of the Lagrangian L kinetic neutral = a µ Φ a µ Φ a = a,α 1 µ φ a,α S.48 would also e invariant under any SON symmetry as long as the transformations are gloal, i.e., x independent, ut for the electrically charged fields the situation is more complicated. Indeed, for a charged field Φ, the gauge-covariant kinetic term D µ Φ D µ Φ comprises oth the kinetic terms per se ut also the coupling of the charged field to the EM fields A µ : D µ Φ D µ Φ = µ Φ iqa µ Φ µ Φ + iqa µ Φ = µ Φ µ Φ + iqa µ Φ µ Φ Φ µ Φ + q A µ A µ Φ Φ. S.49 In terms of the two real fields φ α comprising the complex field Φ, this formula ecomes D µ Φ D µ Φ = 1 µφ µφ + qa µ φ µ φ 1 φ 1 µ φ + 1 q A µ A µ φ φ. S.50 Consequently, for the N real fields comprising the N complex fields of similar electric charge 11

12 we have L kinetic charged = D µ Φ ad µ Φ a a = 1 a,α µ φ a,α qa µ a where E αβ is a antisymmetric matrix E αβ φ a,α µ φ a,β + 1 q A µ A µ α,β a,α φ a,α S.51 E = , that is E 11 = E = 0, E 1 = +1, E 1 = 1. S.5 Clearly, the first and the third sums on the second line of eq. S.51 are invariant under any SUN rotation of the N real fields, ut the second sum is more finicky it is invariant only under rotations that preserve the E matrix, or rather the N N antisymmetric matrix 0N N +1 N N E = E 1 N N = S.53 1 N N 0 N N Thus, due to the coupling of the N charged fields to the EM fields A µ, the Lagrangian.8 does not have a gloal SON symmetry; instead, the symmetries are limited to the sugroup of SON that preserves the E matrix. That sugroup happens to e equivalent to UN. This maye hard to see in the language of N real fields, ut it ecomes ovious once we go ack to complex fields that have definite electric charges +q for all the Φ a and q for all the Φ a. In this asis, good symmetries should commute with the electric charge operators, so they should not mix the Φ and Φ fields with each other. On the other hand, any complex-linear mixing of the Φ a with each other and of the Φ a with each other is OK. As we saw in part a the symmetry group satisfying this condition in addition to preserving the a Φ a is the unitary group UN. 1

13 Prolem 3e: Suppose an infinitesimal SON transform.11 were a symmetry of the theory. Then the Noether current for this symmetry would e J µ C = a µ Φ a = Dµ Φ a C a Φ = Ca Φ Dµ Φ a + C a Φ Dµ Φ a a, Ca Iµ a + C a Iµ a = 1 a + a µ Φ a = Dµ Φ a C a Φ = 1 tr C I µ + 1 tr CI µ S.54 where I µ a is the complex antisymmetric matrix of currents I µ a = Φ D µ Φ a Φ a D µ Φ = I µ a Iµ a. S.55 These current are electrically charged the I µ a q so let s take their covariant divergences have charge +q while the Iµ a have charge D µ I µ a = µi µ a + iqa µi µ a and D µ I µ a = µ I µ a iqa µ I µ a. S.56 Using the Leiniz rules for the D µ, we have D µ I µ a = D µφ D µ Φ a + Φ D µ D µ Φ a a = Φ D µ D µ Φ a Φ a D µ D µ Φ y equations of motion = Φ m λ Φ Φ a Φ a m λ Φ Φ S.57 = 0, and likewise D µ I µ a = 0. In light of eqs. S.56 this means that the ordinary non-covariant divergences of these currents do not vanish; instead µ I µ a = iqa µi µ a, µ I µ a = +iqa µ I µ a, S.58 and the corresponding charges Q a = d 3 xi 0 a and Q a are NOT conserved. 13

14 For a particular generator C of SON/UN we have Noether current J µ C = 1 tr C I µ + 1 tr CI µ = Re trc I µ S.59 with a divergence µ J µ C = Re trc µ I µ = qa µ Im tr C I µ. S.60 For generic fields, this divergence 0 and the corresponding charge Q C is NOT conserved. 14