d 2 Area i K i0 ν 0 (S.2) d 3 x t 0ν

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1 PHY 396 K. Solutions for prolem set #. Prolem 1: Let T µν = λ K λµ ν. Regrdless of the specific form of the K λµ ν φ, φ tensor, its ntisymmetry 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 nd hence µ T µν = µ T µν Noether = hopefully = Furthermore, d 3 x T 0ν = i K i0 ν = d Are i K i0 ν 0 S. oundry of spce when the integrl is tken over the whole spce, hence P ν net d 3 x T 0ν = d 3 x t 0ν Noether. 3. Prolem 1: In the Noether s formul 1 for the stress-energy tensor, φ stnd for the independent fields, however leled. In the electromgnetic cse, the independent fields re 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 clerly oth guge invrint nd symmetric in µ ν, the first term is neither. 1

2 Prolem 1c: Clerly, one cn esily restore oth symmetry nd guge invrince of the electromgnetic stress-energy tensor y replcing ν A λ in eq. S.3 with F ν λ, hence eq. 5. The correction to the T muν Noether mounts to T µν = T µν phys T µν Noether = F µλ F ν λ + F µλ ν A λ = F µλ F ν λ ν A λ = +F µλ λ A ν = λ F µλ A ν A ν λ F µλ. S.4 On the ottom line here, the second term vnishes for the free EM field, which stisfies λ F µλ = 0. The remining first term hs the form of totl derivtive of 3-index tensor F µλ A ν which is ntisymmetric in the first two indices; in nother words, it hs the requisite form of λ K λµν. Altogether, we hve T µν phys = T µν Noether + λk λµν where K λµν = F µλ A ν = K µλν S.5 in perfect greement with eq.. Prolem 1d: Let s strt with the Lgrngin 4. 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 rising one spce index. Likewise, F ij F ij = +ɛ ijk B k ɛ ijl B l = +B k B k where the plus sign comes from rising two spce indices t 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 Now let s work out the stress-energy tensor 4 in components. For the energy density component T 00, we hve U T 00 = F 0i F 0 i L = +E 1 E B = 1 E + B S.8 in greement with the stndrd electromgnetic formulæ in the rtionlized c = 1 units.

3 Likewise, the energy flux nd the momentum density re 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 greement with the Poynting vector S = E B gin, in the rtionlized c = 1 units. Finlly, the 3 dimensionl 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 Note the nisotropy of this stress tensor. Prolem 1e: = E i E j B i B j + 1 δij E + B. In sense, eq. 7 follows from eq. S.4, ut it is just s esy to derive it directly from the Mxwell equtions. Strting with eq. 4, we immeditely hve µ T µν EM = µ F µλ F ν λ F µλ µ F ν λ + 1 F κλ ν F κλ. S.11 Let s simplify the second term here: F µλ µ F ν λ = F µλ µ F ν λ exchnging summtion indices µ λ = F λµ λ F νµ = +F µλ λ F νµ S.1 = + 1 F µλ λ F νµ 1 F µλ µ F ν λ = + 1 F µλ λ F νµ + µ F λν = 1 F µλ ν F µλ, where the lst line here follows from the homogeneous Mxwell eqution λ F νµ + µ F λν + ν F µλ = 0. S.13 Thnks to eq. S.1, the second term in eq. S.11 cncels the third term, nd we re left 3

4 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 equlity comes from the in-homogeneous Mxwell eqution µ F µλ = J λ. This proves eq. 7. Finlly, eq. 8 follows from eq. 7 nd the net stress-energy conservtion 6. Q.E.D. Prolem 1f: For ν = 0, on the LHS of eq. 7 we hve µ T µ0 = 0 T 00 + i T i0 = U t + S, S.15 while on the RHS of eq. 7 we hve J λ F 0λ = +J i F 0i = J E. S.16 Thus, eq. 7 for ν = 0 is the Poynting theorem U t + S = J E, S.17 which is the locl form of the work-energy theorem for the EM fields: The rte of the EM energy s non-conservtion is equl to the power expended y the EM forces on the electric current J. Now consider eq. 7 for ν = i = 1,, 3. On the LHS, we hve µ T µi = 0 T 0i + j T ji = Si t + j T ji, S.18 which is the locl non-conservtion of the i th component of the EM momentum. Indeed, the Poynting vector S = E B gives not only the flux of the EM energy ut lso the density of the EM momentum, while the stress-tensor T ij gives the flux of the EM momentum. 4

5 Physiclly, this non-conservtion is due to mechnicl forces etween the EM fields nd the currents, so we should hve S i t + j T ji = f i, S.19 where f is the density of the net EM forces on the chrges nd the currents, f = ρe + J B. S.0 And indeed, on the RHS of eq. 7 for ν = i we hve J λ F iλ = J 0 F i0 + J j F ij = ρe i ɛ ijk J k B k = ρe + J B i = f i, S.1 in perfect greement with eq. S.19. In other words, the ν = 1,, 3 components of eq. 7 give the locl form of the momentum-impulse theorem for the EM field. Prolem : As discussed in clss, Euler Lgrnge equtions for the chrged fields cn e written in mnifestly covrint form s In prticulrly, for φ = Φ, we hve D µ D µ φ φ = 0. S. D µ Φ = Dµ Φ, Φ = m Φ, which gives us D µ D µ Φ + m Φ = 0. S.3 Likewise, for φ = Φ we hve D µ Φ = Dµ Φ, Φ = m Φ, nd therefore D µ D µ Φ + m Φ = 0. S.4 As for the vector fields A ν, the Lgrngin 9 depends on µ A ν only through F µν, which 5

6 gives us the usul Mxwell eqution µ F µν = J ν where J ν A ν. S.5 To otin the current J ν, we notice tht the covrint derivtives of the chrged fields Φ nd Φ depend on the guge field: D µ Φ A ν = iqδ ν µφ, D µ Φ A ν = iqδ ν µφ. S.6 Consequently, J ν = D ν Φ iqφ D ν Φ iqφ = iqφ D ν Φ + iqφ D ν Φ. S.7 Note tht ll derivtives on the lst line here re guge-covrint, which mkes the current J ν guge invrint. In the non-covrint form, J ν = iqφ D ν Φ iqφd ν Φ q Φ Φ A ν. S.8 To prove the conservtion of this current, we use the Leiniz rule for the covrint derivtives, 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.9 nd hence in light of eq. S.7 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 equtions of motion = iqφ m Φ iqφ m Φ = 0. S.30 6

7 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 mtter µ Φ ν Φ g µν L S.31 similr to the free EM fields, nd T µν Noether EM = F µλ ν A λ gµν F κλ F κλ S.3 T µν Noether mtter = Dµ Φ ν Φ + D µ Φ ν Φ g µν D λ Φ D λ Φ m Φ Φ. S.33 Both terms on the second line of eq. S.31 lck µ ν symmetry nd guge invrince nd thus need λ K λµν corrections for some K λµν = K µλν. We would like to show tht the sme K λµν = F µλ A ν we used to improve the free electromgnetic stress-energy tensor will now improve oth the T µν µν EM nd T mt t the sme time! Indeed, to improve the sclr fields stress-energy tensor we need T µν mtter T µν µν phys mtter T Noether mtter = 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 = A ν iqφ D µ Φ iqφd µ Φ S.34 = A ν J µ, while the improvement of the EM stress-energy ws worked out in prolem 1.c: T µν EM = F µλ F ν λ ν A λ = +F µλ λ A ν = λ F λµ A ν + A ν J µ, S.35 cf. eq. S.4. Note tht the A ν J µ term cncels etween the EM nd the mtter improvement terms, so the net improvement needed to symmetrize the comined stress-energy tensor is 7

8 simply Q.E.D. T µν tot T µν µν phys totl T Noether totl = T µν mtter + T µν EM = λ F λµ A ν K λµν. S.36 Prolem c: Becuse the fields Φx nd Φ x hve opposite electric chrges, their product is neutrl nd therefore µ Φ Φ = D µ Φ Φ = D µ Φ Φ + Φ D µ Φ. Similrly, 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.37 where we hve pplied the field eqution D µ D µ + m Φ x = 0 to the first term on the right hnd side nd used [D µ, D ν ]Φ = iqf µν Φ to expnd 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.38 nd 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.39 Together, the left hnd sides of eqs. S.37, S.38 nd S.39 comprise µ T µν mt cf. eq. 7. On the other hnd, totlling up the right hnd sides of these three equtions results in mssive cncelltion of ll terms except those contining the guge field strength tensor F µν. Therefore, µ T µν mt = D µ Φ iqf µν Φ + D µ Φ iqf µν Φ + mssive cncelltion = F µν iqφ D µ Φ iqφ D µ Φ S.40 = F µν J µ = +F νλ J λ in ccordnce with eq

9 Finlly, comining this formul with eq. 7 we see tht the it follows tht the totl stress-energy tensor 1 is conserved, µ T µν tot = µ T µν tot + µ T µν EM = 0. S.41 Q.E.D. Prolem 3: Unitry trnsform 17 of the Φ fields into ech other leves the iliner comintion Φ Φ invrint which immeditely leds to the invrince of the sclr potentil in the Lgrngin 16. Indeed, unitry mtrix U stisfies U U = 1 i.e., U U c = δc nd hence Φ Φ = =,c Φ U U cφ c c Φ Φc U U c = δc = Φ Φ. S.4 Similrly, the kinetic prt of the Lgrngin would e invrint provided the derivtives D µ Φ nd D µ Φ trnsform into ech other just like the fields itself, D µ Φ U D µφ & D µ Φ D µ Φ U D µ Φ D µ Φ S.43 D µ Φ D µ Φ : invrint. Thus, ll we need to prove is the ssumptions on the top line here, or equivlently, tht the covrint derivtives D µ commute with the unitry trnsform 17. For glol symmetry, the U mtrix is x-independent, thus µ U = 0 which mkes the symmetry trnsform commute with the ordinry derivtives µ. To commute with the covrint derivtives D µ = µ + ia µ x ˆQ, S.44 the symmetry trnsform should lso commute with the electric chrge opertor Q. In other words, it should not mix fields hving different electric chrges. 9

10 For the prolem t hnd, ll the Φ fields hve the sme chrge +q while ll the conjugte fields Φ hve the opposite chrge q. Thus, symmetry my mix Φ field with ny other Φ fields s in eq. 17 ut not with ny of the Φ conjugte fields. Likewise, it my mix Φ with the other Φ ut now with ny of the Φ. Prolem 3: Consider n infinitesiml unitry symmetry U = 1 + iɛt + Oɛ for some hermitin mtrix T = T. The genertor T of this symmetry cts on the fields s δφ = ɛtφ = ɛ it Φ, δφ = ɛtφ = ɛ iφ T, S.45 so the Noether current of this symmetry is J µ T = µ Φ TΦ + µ Φ TΦ = D µ Φ it Φ + D µ Φ iφ T = T iφ D µ Φ iφ D µ Φ S.46 = T J µ tr T J µ. where J µ is the hermitin mtrix of currents J µ = iφ D µ Φ iφ D µ Φ = J µ. 18 Note: different hermitin mtrix T would generte different infinitesiml unitry symmetry U = 1 + iɛt, which in term would hve different Noether current J µ T = tr T J µ. However, ll such currents re liner comintions of the sme N currents J µ, nd there is ovious liner reltion etween prticulr hermitin genertor T nd the corresponding Noether current J T. Consequently, the whole mtrix J µ = J µ cts s mtrix of Noether currents of the UN symmetry. 10

11 Prolem 3c: Since the product Φ D µ Φ is electriclly neutrl, we hve µ Φ D µ Φ = D µ Φ D µ Φ = D µ Φ D µ Φ + Φ D µ D µ Φ, S.47 nd likewise µ Φ Dµ Φ = D µ Φ Dµ Φ = D µ Φ D µ Φ + Φ Dµ D µ Φ. S.48 Consequently µ Φ D µ Φ Φ Dµ Φ = Φ D µ D µ Φ Φ Dµ D µ Φ S.49 while the remining terms in eqs. S.47 nd S.48 cncel ech other, which mens µ J µ = iφ D µ D µ Φ iφ Dµ D µ Φ. S.50 Moreover, when the sclr fields oey their equtions of motion D µ D µ Φ D µ D µ Φ = V Φ = Φ m + λ Φ, = V Φ = Φ m + λ Φ, S.51 the right hnd sides of eqs. S.50 vnish ltogether nd the Noether currents 18 re conserved: µ J µ Q.E.D. = iφ D µ D µ Φ iφ Dµ D µ Φ = iφ Φ m + λ Φ = 0. iφ Φ m + λ Φ S.5 11

12 Prolem 3d: N complex fields Φ x re equivlent to N rel fields φ,α x for = 1,..., N nd α = 1, : Φ x = φ 1x + iφ x, Φ x = φ 1x iφ x. S.53 The sclr potentil prt of the Lgrngin 16 is function of Φ Φ = φ 1 + φ = 1,α φ,α S.54 so it is invrint under ny SON rottion of the N rel fields. For electriclly neutrl fields, the net kinetic prt of the Lgrngin L kinetic neutrl = µ Φ µ Φ =,α 1 µ φ,α S.55 would lso e invrint under ny SON symmetry s long s the trnsformtions re glol, i.e., x independent, ut for the electriclly chrged fields the sitution is more complicted. Indeed, for chrged field Φ, the guge-covrint kinetic term D µ Φ D µ Φ comprises oth the kinetic terms per se ut lso the coupling of the chrged 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.56 In terms of the two rel fields φ α comprising the complex field Φ, this formul 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.57 Consequently, for the N rel fields comprising the N complex fields of similr electric chrge 1

13 we hve L kinetic chrged = D µ Φ D µ Φ = 1,α µ φ,α qa µ where E αβ is ntisymmetric mtrix E αβ φ,α µ φ,β + 1 q A µ A µ α,β,α φ,α S.58 E = , tht is E 11 = E = 0, E 1 = +1, E 1 = 1. S.59 Clerly, the first nd the third sums on the second line of eq. S.58 re invrint under ny SUN rottion of the N rel fields, ut the second sum is more finicky it is invrint only under rottions tht preserve the E mtrix, or rther the N N ntisymmetric mtrix 0N N +1 N N E = E 1 N N = S.60 1 N N 0 N N Thus, due to the coupling of the N chrged fields to the EM fields A µ, the Lgrngin 16 does not hve glol SON symmetry; insted, the symmetries re limited to the sugroup of SON tht preserves the E mtrix. Tht sugroup hppens to e equivlent to UN. This mye hrd to see in the lnguge of N rel fields, ut it ecomes ovious once we go ck to complex fields tht hve definite electric chrges +q for ll the Φ nd q for ll the Φ. In this sis, good symmetries should commute with the electric chrge opertors, so they should not mix the Φ nd Φ fields with ech other. On the other hnd, ny complex-liner mixing of the Φ with ech other nd of the Φ with ech other is OK. As we sw in prt, the symmetry group stisfying this condition in ddition to preserving the Φ is the unitry group UN. 13

14 Prolem 3e: Suppose n infinitesiml SON trnsform 19 were symmetry of the theory. Then the Noether current for this symmetry would e J µ C = µ Φ = Dµ Φ C Φ = C Φ D µ Φ + C Φ Dµ Φ, C Iµ + C I µ = 1 + µ Φ = Dµ Φ C Φ = 1 tr C I µ + 1 tr CI µ S.61 where I µ is the complex ntisymmetric mtrix of currents I µ = Φ D µ Φ Φ D µ Φ = I µ S.6 nd I µ is similr ntisymmetric mtrix of the complex conjugte currents I µ = Φ Dµ Φ Φ D µ Φ = I µ. S.63 These current re electriclly chrged the I µ hve chrge +q while the I µ q so let s tke their covrint divergences hve chrge D µ I µ = µ I µ + iqa µ I µ nd D µ I µ = µ I µ iqa µ I µ. S.64 Using the Leiniz rules for the D µ, we hve D µ I µ = D µ Φ D µ Φ + Φ D µ D µ Φ = Φ D µ D µ Φ Φ D µ D µ Φ y equtions of motion = Φ m λ Φ Φ Φ m λ Φ Φ S.65 = 0, nd likewise D µ I µ = 0. In light of eqs. S.64 this mens tht the ordinry non-covrint 14

15 divergences of these currents do not vnish; insted µ I µ = iqa µ I µ 0, µ I µ = +iqa µ I µ 0. Therefore, the corresponding chrges Q = Q = d 3 xi 0 nd Q = Q conserved. S.66 re not 15