On the W ± and Z 0 Masses

Transcription

1 On the W ± and Z Masses Kenneth Dalton kxdalton@yahoo.com Abstract Scalar and vector fields are coupled in a gauge invariant manner, such as to form massive vector fields. In this, there is no condensate or vacuum expectation value. Transverse and longitudinal solutions are found for the W ± and Z bosons. They satisfy the nonlinear cubic wave equation. Total energy and momentum are calculated, and this determines the mass ratio m W /m Z. 1

2 1. Introduction In the electroweak theory, as it is currently understood, a uniform condensate is imposed upon the system of coupled scalar and vector fields. This gives rise to the linear wave equation for massive bosons. However, there is no experimental evidence for such a condensate, and it would be desirable to eliminate it altogether. In the theory presented here, the scalar-vector system yields the nonlinear cubic wave equation. It does so in a straightforward way, without resorting to unphysical assumptions. Vector boson solutions are found which have well-defined mass. Historically, the U(1) model of scalar electrodynamics was used to create massive photons [1]. However, it was only with the advent of electroweak theory that the weak constants, g and g, were thought to play a similar role. In this paper, the weak coupling between scalar and vector fields is shown to generate mass. The coupling terms occur in the standard electroweak Lagrangian. Transverse and longitudinal solutions are found for the W ± and Z bosons. Energy and momentum are shown to be conserved, and the total energy and momentum are calculated. The mass ratio m W /m Z emerges, during the course of this calculation. The paper begins by revisiting (and revising) the U(1) model. Most of the analysis carries over to the electroweak theory.. U(1): equations of motion The U(1) Lagrangian is [] where L = L(A) + L(Φ) + L(k) (1) L(A) = 1 4 F µνf µν () L(Φ) = g µν (D µ Φ) (D ν Φ) (3) L(k) = k 6g gµν k µ k ν (4) The constant term L(k) does not enter the field equations, but it will contribute to the energy tensor. The U(1) covariant derivatives are

3 F µν = µ A ν ν A µ (5) D µ Φ = µ Φ + iga µ Φ (6) The functional derivatives ( µ A ν ) = F µν (7) A ν = g A ν Φ Φ + ig{( ν Φ )Φ Φ ν Φ} (8) ( µ Φ ) = µ Φ + iga µ Φ (9) Φ = g A µ A µ Φ iga µ µ Φ (1) yield coupled equations of motion and µ F µν = g A ν Φ Φ + ig{( ν Φ )Φ Φ ν Φ} (11) µ µ Φ = g A µ A µ Φ ig{( µ A µ )Φ + A µ µ Φ} (1) The U(1) gauge invariance allows the transformation Φ = 1 (φ + iψ) φ (13) where φ(x) is a real function [3]. In this unitary gauge, equations (11) and (1) are simplified µ F µν + g A ν φ = (14) µ µ φ g A µ A µ φ = (15) ( µ A µ )φ + A µ µ φ = (16) Since µ ν F µν, it follows from (14) that µ (A µ φ ) = (17) 3

4 which agrees with (16). These equations admit traveling solutions, if A µ is a polarized vector. In this case, A µ (u) = ɛ µ f(u) (18) where the argument u = k µ x µ, and ɛ µ is a real polarization vector. 1 vectors satisfy Such µ A µ (u) = k µ ɛ µ df(u) du = (19) since k µ ɛ µ =. From (16), it must also be true that This condition is satisfied, if φ = φ(u) A µ µ φ = () A µ µ φ = k µ ɛ µ f(u) dφ du = (1) Polarization vectors are space-like, ɛ µ ɛ µ = 1, so that A µ A µ = f (u) () Therefore, both equations (14) and (15) will be satisfied, if f(u) = φ(u) and µ µ φ(u) + g φ 3 (u) = (3) This equation, known as the cubic wave equation, is solved by the elliptic function cn(u, 1 ) (appendix A) φ(u) = k g cn( k µx µ ) (4) The amplitude of this traveling wave is not arbitrary, but is fixed by the value of k/g. It will be shown in the following section that k is directly proportional to the mass. 1 Throughout this paper, u = k µ x µ = (k x k x ) and k = k µ k µ = (k ) k. The condition f(u) = φ(u) removes one degree of freedom from the system, leaving three. They belong to the massive vector field. The scalar field is no longer independent, since φ(u) = ɛ µ A µ (u). 4

5 3. U(1): energy and momentum The energy tensor for the vector field is T µν (A) = F µη F η ν g µνf ηρ F ηρ (5) with energy and momentum densities T (A) = { 1 } F 1 + F + F3 + F3 + F31 + F1 (6) T i (A) = F 1 F i1 + F F i + F 3 F i3 (7) For the real scalar field T µν (φ) = µ φ ν φ + g A µ A ν φ g µν L(φ) (8) with energy and momentum densities T (φ) = 1 { ( φ) + ( φ) + g [(A ) + A ]φ } (9) T i (φ) = φ i φ + g A A i φ (3) The constant term L(k) contributes so that T µν (k) = k 3g (k µk ν 1 g µνk ) (31) T (k) = k 6g (k + k ) (3) T i (k) = k 3g k k i (33) The following formulas occur repeatedly in the calculations and are placed here for reference (app. A): φ = k g cn (u) (34) ( dφ) k { = 1 cn 4 (u) } du g (35) 5

6 3.1 Transverse polarization For plane waves moving along the x 3 -axis, k µ = (k, k 3 ). The linear polarization vectors are ɛ µ 1 = 1 ɛ µ = 1 (36) If the polarization is along x 1, then A µ = ɛ µ 1 φ(u) has the single component A 1 = φ(u). In this case, the energy contributions are T (A) = 1 ( A 1 ) + 1 ( 3A 1 ) = k + k3 ( dφ) du = k k + k { 3 1 cn 4 (u) } (37) g T (φ) = 1 { ( φ) + ( 3 φ) + g A 1φ } = k { k + k3 {1 cn 4 (u)} + (k g k 3)cn 4 (u) } (38) T (k) = k k + k3 g 3 Similarly, the momenta are (39) Sum terms, in order to obtain T 3 (A) = A 1 3 A 1 = k g k k 3 {1 cn 4 (u)} (4) T 3 (φ) = φ 3 φ = k g k k 3 {1 cn 4 (u)} (41) T 3 (k) = k g 3 k k 3 (4) T = k g { 3 (k + k 3) k 3 cn 4 (u) } (43) T 3 = k g k k 3 { 4 3 cn4 (u) } (44) 6

7 These expressions obey the energy conservation law, ν T ν =. It can be shown that the momentum is conserved as well, ν T 3ν =. The calculation of total energy and momentum follows the introduction of a volume element l 3 dv/v. In order to arrive at the correct quantum expressions, the integrals must be independent of the ratio k /g. This factor is eliminated from (43) and (44) by setting The integrals are l 3 V 3π g dv = K k k V dv (45) E = 3π K = 3π K cp 3 = 3π K = 3π K g T k k dv V hc { k V 3 (k + k3) k3 cn 4 (u) } dv (46) g T 3 dv k k V hc k V k k 3{ 4 3 cn4 (u) } dv (47) The elliptic function cn(u, 1 ) admits the integral (app. A) cn 4 (u) du = u 3 + sn(u)cn(u)dn(u) + constant (48) 3 Over one period, (or any integral number of periods) 1 4K cn 4 (u) du = 1 (49) 4K 3 sn() = sn(4k) =. Therefore, the energy and momentum are given by E = π K hck = hω (5) cp 3 = π K hck3 = hc π (51) λ 3. Longitudinal polarization In this case, A µ (u) = ɛ µ φ(u) is expressed in terms of the zero-helicity vector 7

8 ɛ µ () = 1 k k 3 k (5) A µ has two components (A, A 3 ) = (k 3, k )k 1 φ(u). The contributions to the energy density are T (A) = 1 ( A 3 3 A ) = k k k { 3 1 cn 4 (u) } (53) g T (φ) = 1 { ( φ) + ( 3 φ) + g (A + A 3)φ } = k k + k { cn 4 (u) } (54) g T (k) = k k + k3 g 3 while those for the momentum are (55) Sum terms to find T 3 (A) = (56) T 3 (φ) = φ 3 φ + g A A 3 φ = k g k k 3 { 1 + cn 4 (u) } (57) T 3 (k) = k g 3 k k 3 (58) T = k g { 3 k 1 3 k 3 + k 3 cn 4 (u) } (59) T 3 = k g k k 3 { cn4 (u) } (6) Here, too, the energy and momentum are conserved, ν T ν = ν T 3ν =. The integrals are 8

9 E = 3π K = 3π K = π cp 3 = 3π K g T k k dv V hc { k V 3 k 1 3 k 3 + k3cn 4 (u) } dv K hck = hω (61) g T 3 dv = 3π K k k V hc k V = π K hck3 = hc π λ These calculations establish the equality k k 3{ cn4 (u) } dv E p = ( π ) k K showing that k is directly proportional to the mass m = (6) (63) π K k (64) 4. U(1) SU() L : field equations; energy tensors The Lagrangian is given by L = L(W ) + L(B) + L(Φ) + L(k) + L(leptons) (65) The lepton term contains the electroweak interaction and will be set aside. The focus is upon the coupling between vector and scalar fields. The covariant derivatives are [4] G µν (W ) = µ (W i νt i ) ν (W i µt i ) + g Σ ijk ɛ ijk W i µw j ν T k (66) F µν (B) = µ B ν ν B µ (67) D µ Φ = { µ + ig T i Wµ i + ig Y B µ} Φ = { µ + ig τ i W µ i + ig 1 B } ( φ + ) µ φ (68) 9

10 The final term in G µν (W ) yields third- and fourth-order vector boson interactions and will also be ignored [5]. This leaves L(W ) + L(B) = 1 Tr G µν(w )G µν (W ) 1 4 F µν(b)f µν (B) = 4{ 1 Fµν (A)F µν (A) + F µν (Z)F µν (Z) +F µν (W 1 )F µν (W 1 ) + F µν (W )F µν (W ) } (69) where W 3 µ = B µ = g A µ + g + g g A µ g + g g g + g g g + g Z µ = sin θa µ + cos θz µ (7) Z µ = cos θa µ sin θz µ (71) The charged bosons W µ ± real fields Wµ 1 and Wµ. The expansion of = (W 1 µ iw µ)/ are expressed in terms of the L(Φ) = g µν (D µ Φ) D ν Φ) (7) is carried out in appendix B, where the functional derivatives are also found. Before writing the equations of motion, the scalar field is transformed to unitary gauge ( ) φ + Φ = φ ( ) φ/ where φ(x) is real. This greatly simplifies the functional derivatives (73) ( µ A ν ) = F µν (A) ( µ Z ν ) = F µν (Z) ( µ W ν ) = F µν (W ) = A ν (74) = 1 Z ν 4 (g + g )Z ν φ (75) = 1 W ν 4 g W ν φ (76) 1

11 where W ν is either Wν 1 are or W ν. The equations of motion for the vector fields µ F µν (A) = (77) µ F µν (Z) = 1 4 (g + g )Z ν φ (78) µ F µν (W ) = 1 4 g W ν φ (79) Since µ ν F µν, these equations yield For the scalar field, ν (Z ν φ ) = ν (W ν φ ) = (8) ( µ φ ) = µ φ i g + g Z µ φ (81) φ = i g + g Z µ µ φ + g W µ W µ + φ (g + g )Z µ Z µ φ (8) ( µ φ + ) = i gw +µ φ (83) φ + = i gw +µ µ φ + (ea µ g sin θz µ ) 1 gw + µ φ (84) The equations of motion are (φ = φ/ ) µ µ φ 1 4 g W 1µ Wµ 1 φ 1 4 g W µ Wµ φ 1 4 (g + g )Z µ Z µ φ = (85) and (ea µ g sin θz µ ) g W µ + φ = (86) where (8) has been used. The vector fields couple individually with φ, so that (86) will be satisfied. The energy tensors are much the same as for U(1). Each vector field A µ, Wµ, 1 Wµ, Z µ contributes 11

12 while the scalar Lagrangian T µν = F µη F η ν g µνf ηρ F ηρ (87) L(φ) = 1 gµν{ µ φ ν φ g (W 1 µw 1 ν + W µw ν )φ (g + g )Z µ Z ν φ } (88) gives T µν (φ) = µ φ ν φ g (W 1 µw 1 ν + W µw ν )φ (g + g )Z µ Z ν φ g µν L(φ) The constant term T µν (k) is treated in the following section. (89) 5. The W ± and Z bosons The coupling between W ± and φ is described by (79, 85) µ F µν (W ) g W ν φ = (9) µ µ φ 1 4 g W µ W µ φ = (91) where µ W µ = W µ µ φ =. These equations are identical to those of U(1), (14) and (15), with the replacement g g/. The solutions of the cubic wave equation (3) are polarized, W µ (u) = ɛ µ φ(u), where φ(u) = k W g Similarly, the coupling between Z and φ is given by cn( k µ x µ ) (9) µ F µν (Z) (g + g )Z ν φ = (93) µ µ φ 1 4 (g + g )Z µ Z µ φ = (94) The solutions are Z µ (u) = ɛ µ φ(u), where 1

13 φ(u) = k Z g + g cn( k µx µ ) (95) There are eleven degrees of freedom in the system, belonging to the massless photon and the three massive bosons. The scalar field is not independent, i.e., φ(u) = ɛ µ W µ (u) or φ(u) = ɛ µ Z µ (u). The expressions for T and T 3 are exactly the same as those in U(1), apart from the new coupling constants. They give rise to factors k W /g and k Z/(g + g ). Before integration can occur, the volume element must eliminate these two factors. This is possible, if they are equal k W g = k Z g + g (96) Only then will the volume element be uniquely defined (compare (45)) l 3 V 3π g dv = 8K kw k V 3π (g + g ) dv = dv (97) 8K kzk V Integration may now go forward as in U(1). 3 The integrals for W ± and Z, including transverse and longitudinal fields, all assume the form E = l3 V cp 3 = l3 V T dv = hω (98) T 3 dv = hc π λ (99) 6. Concluding remarks The coupling theory is made tractable with the choice of unitary gauge. It eliminates the scalar currents, leaving direct nonlinear coupling between the 3 Moreover, the constant terms are identical T µν (k) = 4 kw 3 g (k µk ν 1 g µνk ) k Z = 4 3 g + g (k µk ν 1 g µνk ) 13

14 real scalar and vector fields. The equations of motion yield both transverse and longitudinal solutions, indicating the presence of mass. In the U(1) model, scalar-vector coupling generates a single massive vector field, while in the electroweak theory there are three, the W ± and Z. They satisfy relation (96), yielding the mass ratio m W m Z = k W k Z = g g + g (1) In linear field theory, it is customary to include the integration volume V as part of the plane wave expansion. For example, the expression ( hc/k V ) 1/ appears in each boson term. This is possible because the energy tensor T µν is second-order in the field. However, the non-linear tensor (89) contains both second- and fourth-order terms, making such a procedure untenable. Instead, the volume is introduced as a factor, l 3 /V, during the integration of T µν. Finally, the constant term in the energy tensor T µν (k) has been introduced out of necessity. Without it, the energy and momentum would not satisfy the relation c p = E v, as they must; the work would simply be incomplete. Appendix A. Elliptic functions [6, 7] The elliptic functions sn(u, m), cn(u, m), and dn(u, m) satisfy Their derivatives are sn (u, m) + cn (u, m) = 1 (11) dn(u, m) = 1 m sn (u, m) (1) In particular, d sn(u, m) du d cn(u, m) du d dn(u, m) du = cn(u, m)dn(u, m) (13) = sn(u, m)dn(u, m) (14) = m sn(u, m)cn(u, m) (15) 14

15 d cn(u, m) du = cn(u)dn (u) + m sn (u)cn(u) = (m 1) cn(u) m cn 3 (u) (16) If the parameter m = 1, then d cn(u, 1 ) du = cn 3 (u, 1 ) (17) The cubic wave equation (3) yields the ordinary differential equation k µ k µ d φ(u) + g φ 3 (u) = (18) du Substitute φ(u) = a cn(u, 1 ) to find k a cn 3 (u, 1 ) + g a 3 cn 3 (u, 1 ) = (19) This equation is satisfied, if a = k /g. The solution of (18) is φ(u) = k g cn(u, 1 ) = k g cn( k µx µ ) (11) is The elliptic functions, with m = 1, satisfy two useful identities. The first ( d cn(u) ) = sn (u)dn (u) = 1 du {1 cn4 (u)} (111) The second identity is d du {sn(u)cn(u)dn(u)} = 1 {3 cn4 (u) 1} (11) and it follows that cn 4 (u) du = u 3 + sn(u)cn(u)dn(u) + constant (113) 3 The period of an elliptic function is 4K. (If m = 1, then K = ) For motion along the x 3 -axis, cn( k µ x µ ) = cn(k 3 x 3 k x ) = cn 4K ( x 3 λ t T ) (114) 15

16 It follows that ck = 4K T = K π πf = K π ω (115) k 3 = 4K λ = K π π λ (116) and Appendix B. Scalar Lagrangian B.1 Lagrangian L(Φ) = g µν (D µ Φ) (D ν Φ) = µ φ + µ φ + + µ φ µ φ E = hω = π K hck (117) cp 3 = hc π λ = π K hck3 (118) + { [ea µ + 1 (g cos θ g sin θ)z µ ] + 1 g W +µ W µ +[ea µ g sin θ Z µ ] { g W + µ φ + φ + g W µ φ φ +} + { 1 g W µ W + µ (g + g )Z µ Z µ } φ φ } φ + φ + i[ea µ + 1 (g cos θ g sin θ)z µ ](φ + µ φ + µ φ + φ + ) i gw µ (φ µ φ + µ φ φ + ) i gw +µ (φ + µ φ µ φ + φ ) + i g + g Z µ (φ µ φ µ φ φ ) (119) B. Functional derivatives ( µ φ ) = µ φ + i gw µ φ + i g + g Z µ φ (1) 16

17 φ = i gw µ µ φ + + i g + g Z µ µ φ +[ea µ g sin θ Z µ ] 1 gwµ φ g W µ W + µ φ (g + g )Z µ Z µ φ (11) and ( µ φ + ) = µ φ + + i[ea µ + 1 (g cos θ g sin θ)z µ ]φ + + i gw +µ φ (1) φ = + i[eaµ + 1 (g cos θ g sin θ)z µ ] µ φ g W +µ Wµ φ + +[ea µ + 1 (g cos θ g sin θ)z µ ] φ + +[ea µ g sin θ Z µ ] 1 gw µ + φ i gw +µ µ φ (13) References 1. I. Aitchison and A. Hey, Gauge Theories in Particle Physics: A Practical Introduction, (Hilger, Bristol, 198) p. 1.. A.O. Barut, Electrodynamics and Classical Theory of Fields and Particles, (Macmillan, New York, 1964) p F. Mandl and G. Shaw, Quantum Field Theory, (Wiley, New York, 1984) p. 9, F. Halzen and A. Martin, Quarks and Leptons, (Wiley, New York, 1984) p W. Greiner and B. Müller, Gauge Theory of Weak Interactions, 3rd ed., (Springer, Berlin, ) p J. Mathews and R. Walker, Mathematical Methods of Physics, nd ed., (Addison-Wesley, New York, 1969) p M. Abramowitz and I. Stegun, Handbook of Mathematical Functions, (Dover, New York, 1965) p Also, 17