Physics 504, Lecture 19 April 7, L, H, canonical momenta, and T µν for E&M. 1.1 The Stress (Energy-Momentum) Tensor

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1 Last Latexed: Aprl 5, 2011 at 13:32 1 Physs 504, Leture 19 Aprl 7, 2011 Copyrght 2009 by Joel A. Shapro 1 L, H, anonal momenta, and T µν for E&M We have seen feld theory needs the Lagrangan densty LA µ, A µ / x ν,j,x ξ ) δl and the equatons of moton ome from the funtonal dervatves δa µ x ν ) δl and δ A µ x ν )). The frst s an ntegral over d3 x of δlx µ ), whh ontans δa µ x ν ) a δx x) funton. Beause we are always ntegratng over x,weused partal dervatve notaton amd treated the A µ x ν ) dependene of L as f t were a smple argument nstead of a funton. Ths gave us equatons of moton at eah pont n spae-tme. These Euler-Lagrange equatons nvolved not a total momentum but a momentum densty. For a salar feld φ j the anonal momentum densty s the µ = 0 omponent of the momentum 4-vetor urrent Π µ j x )=. µ φ j x As we have not a salar but four felds A ν, we have four 4-vetor felds α := ). A α x, t) x µ Π µ Last tme we saw that the lagrangandensty for the eletromagnet felds s L = 1 16π F µν F µν 1 J µa µ, so the anonal momentum urrents are Π µ α := A α x, t)/ x µ ) = 1 F µ α, beause, as we saw last tme, ths only nvolves the F 2 term. 504: Leture 19 Last Latexed: Aprl 5, 2011 at 13: The Stress Energy-Momentum) Tensor In dsrete mehans we defne the Hamltonan by H = P q L and then substtute for q the expresson for t n terms of P j.infelheorywe start wth the Lagrangan densty, and the felds P x) andφ x), so we get the Hamltonan densty H x) := P x) φ x) L x) = φ φ / x 0 ) x L. 0 We see that ths s naturally defned wth two Lorentz ndes, both of whh are 0, so t s one omponent of a tensor T µ ν = φ φ / x µ ) x ν δµ ν L. Ths objet goes by the names energy-momentum tensor or stressenergy tensor or anonal stress tensor. For eletromagnetsm, φ s replaed by A λ, the frst fator n the frst term s A λ / x µ ) = 1 F µ λ, the seond fator s A λ x, ν so our frst tentatve) expresson for the energy momentum tensor s T µν = 1 A λ F µλ x 1 ) ν 4 η µνf αβ F αβ. Ths tensor has some good propertes and some bad propertes. We have seen that ts 00 omponent s the hamltonan densty, whh we may nterpret as the energy densty. We mght expet, then, that T 0 ould be nterpreted as the densty of momentum, and the ntegral of t, over all spae, the th omponent of the total momentum. But we see that T 0 = 1 F 0λ A λ = 1 E j A j = 1 ) ) E B + E A. We expet the E B term from the expresson for the Poyntng vetor, but not the last term, whh s not even gauge nvarant. It s, however, n the

2 504: Leture 19 Last Latexed: Aprl 5, 2011 at 13: : Leture 19 Last Latexed: Aprl 5, 2011 at 13:32 4 absene of harges and we have not nluded the momentum of any harges) a total dervatve as then E =0,and A ) E)= E A + A E). So the ntegral of ths densty wll gve what we want. Thus we have the good property that d 3 xt 0µ = P µ, the total momentum. 1) The way T µν s defned n general guarantees that, f the Lagrangan has no explt dependene on x µ, the stream) dvergene µ T µ ν wll vansh when evaluated on felds whh obey the equatons of moton. We have µ T µ ν = ) µ ν φ + µ φ ) µ φ ) µ ν φ ν L. The dervatve n the last term s gven by the han rule so ν L = µ T µ ν = φ ν φ µ µ φ ) ν µ φ µ φ ) ) ν φ φ and the parenthess vanshes by the equatons of moton. Thus we have µ T µ ν =0. 2) Note that the tensor we have defned so far, T µν s not symmetr under µ ν, whh s a problem when t omes to defnng the angular momentum. If T 0j s truly the momentum densty, we would expet ɛ jk x j T 0k d 3 x to be the angular momentum, but n fat that only works f T µν s symmetr. So we have good propertes 1) and 2), but we have the problems that T µν s not gauge nvarant, s not symmetr, and dffers from Poyntng loally. Can we add somethng to T whh wll fx these problems wthout messng up the good propertes? Note that f we add ψ µν to T µν, wth the requrement that ψ µν = ψ µν, the extra pee wll hange 2) by µ ψ µν =0,. e. unhanged, beause the dervatves are symmetr whle ψ s antsymmetr. Furthermore, the ntegral over spae of T 0µ gets an adon of d 3 x ψ 0ν = d 3 x j ψ j0ν = S n jψ j0ν 0 where the surfae S goes to nfnty, where we an assume all our felds go to zero 1. Thus addng ψ µν preserves all the good propertes. So onsder ψ µν = A ν F µ /, and addng 1 A ν F µ )= 1 A ν ) F µ beause F µ = 0 n the absene of a soure J µ. But ths s just what we need to add to T µν to make Θ µν = T µν + 1 F µ A ν = 1 F µ F ν 1 ) 4 ηµν F αβ F αβ. Ths expresson has all the good propertes and s also gauge nvarant and symmetr. Furthermore, Θ 0 = 1 F 0j F j = 1 E jɛ jk B k = 1 E B), the orret momentum densty or energy flux, as gven by Poyntng. 1.2 Ambgutes n the Aton The aton we used for the eletromagnet feld by tself depends only on F µν, that s, on the eletr and magnet felds, but the nteraton term wth urrents, A nt = 1/) d 4 xj µ A µ depends on the 4-vetor potental, whh, as we know, s not unquely defned, beause the physal felds E and B are unhanged by a gauge transformaton A µ A µ = A µ + µ Λ, that s, we ould add a pee to the aton of 1/) d 4 xj µ µ Λ. Ths would, however, make no dfferene to the equatons of moton, beause the d 4 xj µ µ Λ= n µ J µ Λ d 4 xλ µ J µ, S where S s a hypersurfae surroundng the four dmensonal regon we are onsderng, whh means a hypersurfae at nfnty. We an provde some exuse for lamng ether J µ or Λ vanshes at nfnty, so the hyper)surfae term an be dsarded, and the other term nvolves the dvergene of the urrent, whh we know has to be zero by onservaton of harge. 1 We often take a avaler atttude about suh arguments, but you should keep a small reservaton n the bak of your head that under some rumstanes there may be anomoles that make t mpossble to assure that these terms an be gnored

3 504: Leture 19 Last Latexed: Aprl 5, 2011 at 13: : Leture 19 Last Latexed: Aprl 5, 2011 at 13:32 6 Ths nvarane s a general feature of lagrangan mehans. Beause t s only the varaton of the lagrangan, and not the lagrangan, that matters physally, and the felds are vared only nsde the regon and not on the surfae, any hange n the lagrangan densty by a dvergene, or of the lagrangan by a total tme dervatve, s rrelevant to the physs. 1.3 Θ µν n the presene of urrents We saw that the energy-momentum tensor of the eletromagnet feld an be onsdered to be Θ µν = 1 F µ F ν 1 ) 4 ηµν F αβ F αβ, and n the absene of any soures, t s onserved, µ Θ µν = 0. What happens f there are soures? Now µ Θ µν = µ F µ F ν + 1 ) 4 ηµν F αβ F αβ = µ F µ ) F ν = J F ν + F µ µ F ν F αβ α F ν β F αβ ν F αβ β F ν α + ν F αβ ) = J F ν F αβ η ν α F β + β F α + F αβ ) = J F ν, as the term n parentheses s zero by the homogeneous Maxwell equatons. Thus the total 4-momentum of the eletromagnet feld P ν = 1 d 3 xθ 0ν x), EM s not onserved, but rather dp ν EM = 1 d d 3 x Θ 0ν x) = d 3 x 0 Θ 0ν x) = 1 d 3 xj x)f ν x) 1 d 3 x Θ ν = 1 d 3 xj x)f ν x), as the ntegral of a dvergene an be thrown away as a surfae term at nfnty. Consder a harged partle of mass m, harge q at pont x t). Its mehanal 4-momentum hanges by dp ν ) Ths partle orresponds to a 4-urrent = 1 dp ) ν γ = 1 q dτ γ F ν x )U. J =, J)=q δ 3 x x ),q u δ 3 x x )=q γ 1 U δ 3 x x ). Pluggng ths nto our expresson for the hange n the momentum of the eletromagnet feld, we have dp ν EM = q d 3 xf ν x)γ 1 U δ 3 x x )= q F ν γ x )U, and the total momentum, P ν EM + P) ν s onserved. 1.4 Equaton of Moton for A µ We saw that the equatons of moton for the 4-vetor potental are σ F σµ = σ σ A µ µ σ A σ = J µ. If we had an equaton that told us to nfore the Lorenz onon σ A σ =0, we ould drop the seond term and have the equaton σ σ A µ = J µ, whh has as ts solutons a partular soluton gven n terms of the Green s funton for the wave equaton, together wth an arbtrary soluton of the homogeneous equaton σ σ A µ = 0. Let us dsuss ths homogeneous soluton frst. Wth the Lorenz onon mposed, the soluton s smply k A µ k + e k x ω k t + A µ k e k x+ω k t ), where ω = k. Ths soluton s onstraned by the Lorenz onon to have ωa 0 k ± k A k ± = 0. These are the solutons for an eletromagnet wave n empty spae.

4 504: Leture 19 Last Latexed: Aprl 5, 2011 at 13: : Leture 19 Last Latexed: Aprl 5, 2011 at 13:32 8 But f we don t mpose the ad-ho Lorenz onon, the equatons σ σ A µ µ σ A σ =0 are not enough to determne the evoluton of A µ x, t) as a funton of tme, even f we ompletely spefy ntal onons on A µ x, 0) and ts tme dervatve at t = 0. Ths s most easly seen n the Fourer transformed equaton k σ k σ A µ k µ k σ A σ =0, whh, though t looks lke four equatons for A, s atually only three, beause f we ontrat wth k µ we get k σ k σ k µ A µ k µ k µ k σ A σ =k 2 k 2 )k A =0, whh s not a onstrant on A but an dentty. In other words, the equaton only determnes the omponents of A transverse to k. Ths s yet another ndaton of the gauge nvarane, the statement that a gauge-nvarant aton prnple annot determne the evoluton of a gaugevarant feld beause no equaton wll determne the gauge transformaton Λ. We an, however, adopt the Lorenz gauge onon and ask what the equaton that determnes A µ n that gauge s. So we turn to the nhomogeneous equaton A µ = β β A µ = J µ, wth the soluton A µ x) = d 4 x Dx, x )J µ x ), where Dx, x ) s a Green s funton for D Alembert s equaton xdx, x )=δ 4 x x ). We are nterested n solvng ths n all of spaetme, wthout boundares at fnte dstanes, so the equaton s translaton nvarant and D must be a funton only of the dfferene, Dx, x )=Dx x )=Dz). The equaton may be solved by Fourer transform, Dz) = 1 d 4 k µ Dk µ )e kµzµ. 2π) 4 As δ 4 z µ )= 1 2π) 4 d 4 ke kµzµ, the soluton for the Green s funton s Dk µ )= 1 k 2, and Dzµ )= 1 2π) 4 d 4 k e kµzµ k 2. Now ths looks very smlar to the Green s funton for the Laplae equaton, exept that the 1/k 2 s muh more dangerous here, as t vanshes whenever k0 2 = k 2, and not just at one pont n a three dmensonal spae. For Laplae s equaton the ll-defned pont n the ntegraton was just a sgn that a potental satsfyng Laplae s equaton ould have an arbtrary onstant and frst dervatve, the solutons of the homogeneous equaton. Here too the lldetermned part of D represents the homogeneous solutons, but ths s now an nfnte dmensonal spae of solutons, all free eletro magnet waves. The ll-defned ntegral through the sngular pont an be larfed by wrtng the Green s funton frst as Dz) = 1 d 3 ke k z e k0z0 dk 2π) 4 0 Γ k0 2 k. 2 We may make a well defned Green s funton by spefyng that the ontour Γ should not go rght r k 0 along the real axs of k 0, but rather around the poles at k 0 = ± k. Three suh ontours are shown. As the ntegrand s analyt exept at the ponts k 0 = ± k, theontours may be deformed so that they beome the real k 0 axs beyond the regon wth the poles. F The retarded r), advaned a), and Feynman F ) ontours for defnng the Green s funton. Consder frst the Green s funton as gven by the ontour r. If the soure ats at tme 0, and f we evaluate Dz) at a tme after that, wth z 0 > 0, the ontour Γ may be losed by takng a large semrle n the lower half omplex plane, where e k0z0 = e Im k 0 z 0 0, so ths semrle makes k no ontrbuton to the ntegral but does allow us to evaluate t as 2π tmes the sum of the two resdues. The mnus s due to our rlng these resdues lokwse rather than ounterlokwse, and the resdues are Res k 0= k e k0z0 k 0 + k )k 0 k ) + Res k 0= k e k0z0 k 0 + k )k 0 k ) a

5 504: Leture 19 Last Latexed: Aprl 5, 2011 at 13: : Leture 19 Last Latexed: Aprl 5, 2011 at 13:32 10 = e k z 0 2 k + e k z 0 2 k = sn k z 0 ). k On the other hand, f z 0 < 0 we may lose the ontour n the upper half plane, as the semrle ontrbuton now vanshes there, and we have enrled no sngulartes and the Cauhy-Goursat theorem tells us t vanshes. Thus D r z) = Θz0 ) 2π) 3 d 3 ke k z sn k z 0 ). k Choosng the North pole along z usng spheral oordnates, ths beomes D r z) = Θz0 ) k 2 dk dθ sn θe kr os θ snkz0 ) 2π) 2 0 k = Θz0 ) dk snkr)snkz 0 ), 2π 2 R 0 where R = z. The Green s funton s alled the retarded Green s funton beause the effets our only after the soure ats. It s also alled the ausal Green s funton beause ths s how thngs ought to be, though t would be perfetly onsstent to use the ontour a and the advaned Green s funton to ask what onfguraton of nomng waves ould be magally made to dsappear by nteratng wth a gven soure J µ. The thrd ontour shown n the fgure gves the Feynman propagator, whh s used n quantum feld theory. But we need not dsuss that here. The expresson for D r z) an be further smplfed by wrtng snkr)snkz 0 ) = 1 [ oskr z 0 )) oskr + z 0 )) ] 2 = 1 [ e z 0 R)k e z0+r)k + e z0 R) k) + e z0 R) k)] 4 so D r z) = Θz0 ) dk [ e z0 R)k e z0+r)k] 8π 2 R = Θz0 ) R [δz 0 R) δz 0 + R)] = Θz0 ) R δz 0 R), where the seond δ was dropped beause both z 0 and R are postve. So the Green s funton only ontrbutes when the soure and effet are separated by a lghtlke path, wth z 0 = z. So how do we desrbe the feld when we know what the soures are throughout spae-tme? We an use any of the Green s funtons to get the nhomogeneous ontrbuton, and then allow for an arbtrary soluton of the homogeneous equaton. Thus we an wrte A µ = A µ n x)+ d 4 x D r x x )J µ x ) = A µ out x)+ d 4 x D a x x )J µ x ). If the soures are onfned to some fnte regon of spae-tme, there wll be no ontrbuton from D r at tmes earler than the frst soure, and A µ n x) desrbes the felds before that tme. Also after the last tme that the soure nfluenes thngs, the feld wll be gven by A µ out x) alone. Of ourse the soure may be persstent, for example f there s a net harge, but we may often onsder that the effet of the soure s onfned to the hange from A µ n x) toaµ out x) and defne the radaton feld to be A µ rad x) =Aµ out x) Aµ n x) = d 4 x Dx x )J µ x ), where Dz) :=D r z) D a z). The expresson we wrote earler for the urrent densty of a pont harge, J = q γ 1 U δ 3 x x ) an be wrtten n ths four-dmensonal language as J x µ )=q δt x 0 /)γ 1 U δ 3 x x t)) = q dτδ 4 x µ x µ τ))u, where τ measures proper tme along the path of the partle.