Tensor gauge boson production in high-energy collisions

NRCPS-HE-02-08 arXiv:0803.0075v3 [hep-th] 5 Mar 2008 February, 2008 Tensor Gauge Boson Production in High Energy Collisions Spyros Konitopoulos, Raffaele Fazio∗ and George Savvidy Institute of Nuclear Physics, Demokritos National Research Center Agia Paraskevi, GR-15310 Athens, Greece Abstract We calculated the leading-order cross section for the helicity two tensor gauge bosons production in fermion pair annihilation process. We compare this cross section with a similar annihilation processes in QED with two photons in the final state and with two gluons in QCD. ∗ On a leave of absence from Departamento de Fisica, Universidad Nacional de Colombia, Bogota. 1 Introduction Our intention in this article is to calculate leading-order differential cross section for the tensor gauge bosons production in the fermion pair annihilation process. The process is illustrated in Fig.1. and receives contribution from three Feynman diagrams shown in Fig.3. This diagrams are similar to the QED and QCD diagrams for the annihilation processes with two photons or two gluons in the final state. The difference between these processes is in the actual expressions for the corresponding interaction vertices. The corresponding vertices for the tensor bosons can be found through the extension of the gauge principle [9]. The extended gauge principle allows to define a gauge invariant Lagrangian L for high-rank tensor gauge fields and their cubic and quartic interaction vertices [9, 10, 11]: ′ L = LY M + L2 + L2 ... The Feynman rules for this Lagrangian can be derived from the functional integral over the fermion fields ψi , ψiµ , ... and over the gauge boson fields Aaµ , Aaµν , .... Not much is known about physical properties of such gauge field theories [1, 2, 3, 5, 6, 7, 8] and in the present article we shall ignore subtle aspects (ghosts) of functional integral quantization procedure because we limited ourselves to calculating only leading-order tree diagrams. Expanding the functional integral in perturbation theory, starting with the free Lagrangian, at g = 0, one can see that free theory contains tensor gauge bosons and fermions of different spins with cubic and quartic interaction vertices [9, 10, 11]. Explicit form of these vertices is presented in [11]. In the next section we shall present the Feynman diagrams for the given process, the expressions for the corresponding vertices, the transition matrix element and unpolarized cross section in the center-of-mass frame. In the third section we shall check that the transition amplitude is gauge invariant, that is, if we take the physical - transverse polarization - wave function for one of the tensor gauge bosons and unphysical - longitudinal polarization - for the second one, the transition amplitude vanishes. This Ward identity expresses the fact that the unphysical - longitudinal polarization states are not produced in the scattering process. In the fourth and fifth sections the squared matrix element is calculated together with traces over Dirac and isotopic matrices for unpolarized particles. In the sixth section we present the final expression for the cross section (6.25) and its comparison with the corresponding cross sections for photons and gluons in QED and QCD. 1 k = (ω , k ) 1 1 1 θ P = (E , p ) + + + P = (E- , p- ) - k = (ω , k ) 2 2 2 Figure 1: The annihilation reaction f f¯ → T T , shown in the center-of-mass frame. The p± are momenta of the fermions f f¯ and k1,2 are momenta of the tensor gauge bosons T T . 2 Tensor Gauge Bosons Production Amplitude As we already mentioned in the introduction the process is illustrated in Fig.1. Working in the center-of-mass frame, we make the following assignments: p− = (E− , p~− ), p+ = (E+ , p~+ ), k1 = (ω1 , ~k1 ), k2 = (ω2 , ~k2 ), (2.1) where p± are momenta of the fermions f ± and k1,2 momenta of the tensor gauge bosons T T . All particles are massless p2− = p2+ = k12 = k22 = 0. In the center-of-mass frame the momenta satisfy the relations p~+ = −~p− , ~k2 = −~k1 and E− = E+ = ω1 = ω2 = E. The invariant variables of the process are: s = (p+ + p− )2 = (k1 + k2 )2 = 2(p+ · p− ) = 2(k1 · k2 ) s t = (p− − k1 )2 = (p+ − k2 )2 = − (1 − cos θ) 2 s 2 2 u = (p− − k2 ) = (p+ − k1 ) = − (1 + cos θ) 2 where s = (2E)2 and θ is the scattering angle, so that the scalar products can be found in the form s (p+ · p− ) = (k1 · k2 ) = 2 s (p− · k1 ) = (p+ · k2 ) = (1 − cos θ) 4 s (p− · k2 ) = (p+ · k1 ) = (1 + cos θ). 4 2 (2.2) c, γγ’ a, αα’ q k p b, β Figure 2: The interaction vertex for vector gauge boson V and two tensor gauge bosons T - the VTT vertex - in non-Abelian tensor gauge field theory [11]. Vector gauge bosons are conventionally drawn as thin wave lines, tensor gauge bosons are thick wave lines. The Lorentz indices αά and momentum k belong to the first tensor gauge boson, the γγ́ and momentum q belong to the second tensor gauge boson, and Lorentz index β and momentum p belong to the vector gauge boson. The Feynman rules for this Lagrangian can be derived from the functional integral over the fermion fields ψi , ψ̄j , ψiµ , ψ̄jµ , ... and over the gauge boson fields Aaµ , Aaµν , .... Here Dirac indices are not shown and i, j and a are indices of the symmetry group G. In this article we shall ignore subtle aspects of functional integral quantization procedure simply because we limited ourselves to calculating only tree diagrams. Expanding the functional integral in perturbation theory, starting with the free Lagrangian, at g = 0, one can see that free theory contains a number of free fermions of different spins, each of them have equal dimension d(r) of the representation r: i, j = 1, ..., d(r) and that the number of free vector-V and tensor-T gauge bosons is equal to the number d(G) of generators of the group G: a = 1, ..., d(G) [9, 10, 11]. In momentum space the interaction vertex of vector gauge boson V with two tensor gauge bosons T - the VTT vertex - has the form† [10, 11] abc Vαabc Fαάβγγ́ , άβγγ́ (k, p, q) = −gf where Fαάβγγ́ (k, p, q) = [ηαβ (p − k)γ + ηαγ (k − q)β + ηβγ (q − p)α ]ηάγ́ − 1 − { + (p − k)γ (ηαγ́ ηάβ + ηαά ηβγ́ ) 2 † See formulas (62),(65) and (66) in [11] . 3 (2.3) + (k − q)β (ηαγ́ ηάγ + ηαά ηγγ́ ) + (q − p)α (ηάγ ηβγ́ + ηάβ ηγγ́ ) + (p − k)ά ηαβ ηγγ́ + (p − k)γ́ ηαβ ηάγ + (k − q)ά ηαγ ηβγ́ + (k − q)γ́ ηαγ ηάβ + (q − p)ά ηβγ ηαγ́ + (q − p)γ́ ηαά ηβγ }. (2.4) The Lorentz indices αά and momentum k belong to the first tensor gauge boson, the γγ́ and momentum q belong to the second tensor gauge boson, and Lorentz index β and momentum p belong to the vector gauge boson. The vertex is shown in Fig.2. Vector gauge bosons are conventionally drawn as thin wave lines, tensor gauge bosons are thick wave lines. It is convenient to write the differential cross section for our process, with tensor boson produced into a solid angle dΩ, as dσ = 1 |M|2 dΦ, 4(p+ · p− ) (2.5) where the final-state density for two massless tensor gauge bosons is dΦ = Z 1 d3 k2 d3 k1 4 (2π) δ(p + p − k − k ) = dΩ, + − 1 2 (2π)3 2ω1 (2π)3 2ω2 32π 2 so that dσ = 1 1 |M|2 dΩ. 2s 32π 2 (2.6) We shall calculate the unpolarized cross section for this reaction, to lowest order in α = g 2 /4π. The lowest-order Feynman diagrams contributing to fermion-antifermion annihilation into a pair of tensor gauge bosons are shown in Fig.3. In order g 2 , there are three diagrams. Dirac fermions ψ are conventionally drawn as thin solid lines, and RaritaSchwinger spin-vector fermions ψ µ by thick solid lines. These diagrams are similar to the QCD diagrams for fermion-antifermion annihilation into a pair of vector gauge bosons. The difference between these processes is in the actual expressions for the corresponding interaction vertices. Thus the probability amplitude of the process can be written in the form iM µανβ 1 βα i 14 g αβ b ν 1 ν b i4g t γ +γ t ta γ µ − f abc F µαρνβ γρ tc 2 }u(p− ) = (ig) v̄(p+ ){γ t 6 p− − 6 k 2 6 p− − 6 k 1 k3 2 µ a 4 k1 k2 k1 k2 k1 k2 k3 P + P- P + P + P- P- Figure 3: Diagrams contributing to fermion-anifermion annihilation to two tensor gauge bosons. Dirac fermions are conventionally drawn as thin solid lines, and Rarita-Schwinger spin-vector fermions by thick solid lines. or in equivalent form as Mµανβ = (ig)2 v̄(p+ ){γ µ ta 1 αβ g 4 6 p− − 6 k 2 tb γ ν + γ ν tb 1 αβ g 4 6 p− − 6 k 1 ta γ µ + if abc tc γρ 1 µαρνβ F }u(p− ) k32 (2.7) where u(p− ) is the wave function of spin 1/2 fermion and v(p+ ) of antifermion. The Dirac and symmetry group indices are not shown. 3 Gauge Invariance Let us check that the amplitude is gauge invariant, that is, if we take the physical - transverse polarization - wave function eT for one of the tensor gauge bosons and longitudinal polarization for the second one eL , the transition amplitude vanishes MeT eL = 0. This Ward identity expresses the fact that the unphysical - longitudinal polarization - states are not produced in the scattering process. Contracting the matrix element (2.7) with the on-shell polarization tensors of the final tensor gauge bosons e∗µα (k1 ) and e∗νβ (k2 ) we shall get (ig)2 v̄(p+ ){γ µ ta 1 αβ g 4 6 p− − 6 k 2 Mµανβ e∗µα (k1 )e∗νβ (k2 ) = tb γ ν + γ ν tb 1 αβ g 4 6 p− − 6 k 1 ta γ µ + if abc tc γρ (3.8) 1 µαρνβ F }u(p− )e∗µα (k1 )e∗νβ (k2 ). k32 Considering the last term if abc tc v̄(p+ )γρ u(p− ) 1 µαρνβ F (k1 , k3 , k2 )e∗µα (k1 )e∗νβ (k2 ) k32 5 and taking the polarization tensor e∗νβ (k2 ) to be longitudinal e∗νβ (k2 ) = k2ν ξβ + k2β ξν and the polarization tensor e∗µβ (k1 ) to be transversal, we shall get 1 µαρνβ F (k1 , k3 , k2 )e∗µα (k1 )(k2ν ξβ + k2β ξν ) = k32 1 1 1 1 = if abc tc v̄(p+ )γ ρ u(p− ) 2 [k32 e∗ρα (k1 )(ξ α − ξ α ) + k32 e∗αρ (k1 )(− ξ α + ξ α )] k3 2 2 4 if abc tc v̄(p+ )γρ u(p− ) and therefore if abc tc v̄(p+ )γρ u(p− ) 1 µαρνβ F (k1 , k3 , k2)e∗µα (k1 )(k2ν ξβ + k2β ην ) = 2 k3 1 = if abc tc v̄(p+ )γ ρ u(p− ) e∗ρα (k1 ) ξ α . 4 (3.9) Now let us consider the first two terms v̄(p+ ){γ µ ta 1 αβ g 4 6 p− − 6 k 2 tb γ ν + γ ν tb 1 αβ g 4 6 p− − 6 k 1 ta γ µ }u(p− )e∗µα (k1 )e∗νβ (k2 ). Taking again the polarization tensor to be longitudinal e∗νβ (k2 ) = k2ν ξβ + k2β ξν we shall get v̄(p+ ){γ µ ta 1 αβ g 4 6 p− − 6 k 2 tb γ ν + γ ν tb 1 αβ g 4 6 p− − 6 k 1 ta γ µ }u(p− )e∗µα (k1 )(k2ν ξβ + k2β ξν ) = 1 = v̄(p+ ){−ta tb γ µ + tb ta γ µ }u(p− )e∗µα (k1 )g αβ ξβ + 4 1 1 1 tb γ ν + γ ν tb ta γ µ }u(p− )e∗µα (k1 )g αβ k2β ξν . + v̄(p+ ){γ µ ta 4 6 p− − 6 k 2 6 p− − 6 k 1 The last term is equal to zero, e∗µα (k1 )g αβ k2β = 0, see relations (4.11), therefore we have 1 − if abc tc v̄(p+ )γ µ u(p− )e∗µα (k1 )ξ α . 4 (3.10) This term precisely cancels the contribution coming from the last term of the amplitude (3.9). Thus the cross term matrix element between transverse and longitudinal polarizations vanishes MeT eL = 0. Our intention now is to calculate physical matrix element with two transversal tensor gauge bosons MeT eT in the final state. 6 Squared Matrix Element 4 The complex conjugate of the scattering amplitude (2.7) is M ∗µανβ 2 1 αβ g 4 ν b = (−ig) ū(p− ){γ t a µ 6 p− − 6 k 2 1 αβ g 4 µ a t γ +γ t 6 p− − 6 k 1 tb γ ν − if abc tc γρ 1 µαρνβ F }v(p+ ) k32 and we can calculate now the squared matrix elements in the form ′ ′ ′ Mµανβ M∗µ α ν β ′ = (ig)2 v̄(p+ ){γ µ ta + if abc tc γρ 1 αβ g 4 6 p− − 6 k 2 1 µαρνβ F }u(p− ) k32 ′ ′ 1 αβ g 4 ′ ∗ (−ig)2 ū(p− ){γ ν tb ′ ′ ′ − if a b c tb γ ν + γ ν tb 1 αβ g 4 6 p− − 6 k 1 ′ ta γ µ + ′ ′ ′ ′ 1 αβ g 4 ′ ta γ µ + γ µ ta 6 p− − 6 k 2 ′ ′ ′ ′ ′ 1 tc γρ′ 2 F µ α ρ ν β }v(p+ ) k3 ′ ′ ′ 6 p− − 6 k 1 ′ tb γ ν − For unpolarized fermions-antifermion scattering the average over the fermion and antifermion spins is defined as follows: |M|2 = 11 X |M|2 , 2 2 spin 1/2 using completeness relations X us (p− )ūs (p− ) =6 p− , X s v s (p+ )v̄ s (p+ ) =6 p+ . s Thus averaging over spins of the fermions we shall get ′ ′ ′ Mµανβ M∗µ α ν β ′ 1 αβ 1 αβ g g g4 tb γ ν + γ ν tb 4 ta γ µ + T r{6 p+ [γ µ ta 4 4 6 p− − 6 k 2 6 p− − 6 k 1 1 + if abc tc γρ 2 F µαρνβ ] k3 = ′ ′ ′ 6 p− [γ ν tb ′ ′ ′ − if a b c 1 αβ g 4 ′ ′ ′ ′ ′ ′ ta γ µ + γ µ ta 6 p− − 6 k 2 ′ ′ ′ ′ ′ 1 tc γρ′ 2 F µ α ρ ν β ]}. k3 ′ 1 αβ g 4 ′ 6 p− − 6 k 1 ′ ′ tb γ ν − Contracting the last expression with the transversal on-shell polarization tensors of the final tensor gauge bosons e∗µα (k1 ) and e∗νβ (k2 ) we shall get the probability amplitude in the form ′ ′ ′ Mµανβ M∗µ α ν β ′ e∗µα (k1 )e∗νβ (k2 )eµ′ α′ (k1 )eν ′ β ′ (k2 ) = 7 = 1 αβ 1 αβ g g g4 1 T r{6 p+ [γ µ ta 4 tb γ ν + γ ν tb 4 ta γ µ + if abc tc γρ 2 F µαρνβ ] 4 6 p− − 6 k 2 6 p− − 6 k 1 k3 ′ ν ′ 1 αβ g 4 ′ b 6 p− [γ t ′ ′ ′ a t γ ′ ′ ′ a µ µ +γ t 1 α β g 4 ′ 6 p− − 6 k 2 6 p− − 6 k 1 e∗µα (k1 )e∗νβ (k2 )eµ′ α′ (k1 )eν ′ β ′ (k2 ). ′ ′ ′ ′ ′ ′ tb γ ν − if a b c tc γρ′ 1 µ′ α′ ρ′ ν ′ β ′ ]} F k32 Using the explicit form of the vertex operator F µαρνβ (2.3), (2.4) and the orthogonality properties of the tensor gauge boson wave functions k1µ eµα (k1 ) = k1α eµα (k1 ) = k2µ eµα (k1 ) = k2α eµα (k1 ) = 0, (4.11) k2µ eµα (k2 ) = k2α eµα (k2 ) = k1µ eµα (k2 ) = k1α eµα (k2 ) = 0, where the last relations follow from the fact that ~k1 k ~k2 in the process of Fig.1, we shall get ′ ′ ′ Mµανβ M∗µ α ν = β ′ e∗µα (k1 )e∗νβ (k2 )eµ′ α′ (k1 )eν ′ β ′ (k2 ) = 1 αβ 1 αβ g g g4 1 1 tb γ ν + γ ν tb 4 ta γ µ + if abc tc γρ 2 (k2 − k1 )ρ (g µν g αβ − g µβ g να )] T r{6 p+ [γ µ ta 4 4 6 p− − 6 k 2 6 p− − 6 k 1 k3 2 ′ ′ ′ 6 p− [γ ν tb 1 αβ g 4 ′ ′ ′ ′ ′ ′ ta γ µ + γ µ ta 1 αβ g 4 6 p− − 6 k 2 6 p− − ∗ ∗ ′ ′ ′ ′ eµα (k1 )eνβ (k2 )eµ α (k1 )eν β (k2 ). ′ 6 k1 ′ ′ ′ ′ ′ ′ tb γ ν − if a b c tc γρ′ 1 1 ′ ′ ′ ′ ′ ′ ′ ′ ′ (k2 − k1 )ρ (g µ ν g α β − g µ β g ν α )]} 2 k3 2 As the next step we shall calculate the sum over transversal tensor gauge bosons polarizations. The sum over transversal polarizations of the helicity-two tensor gauge boson is [4, 11] X k1µ k̃1µ′ + k̃1µ k1µ′ k1α k̃1α′ + k̃1α k1α′ 1 )(−gαα′ + )+ e∗µα (k1 )eµ′ α′ (k1 ) = [(−gµµ′ + 2 k1 k̃1 k1 k̃1 k1α k̃1µ′ + k̃1α k1µ′ k1µ k̃1α′ + k̃1µ k1α′ )(−gαµ′ + )− k1 k̃1 k1 k̃1 k1µ′ k̃1α′ + k̃1µ′ k1α′ k1µ k̃1α + k̃1µ k1α −(−gµα + )(−gµ′ α′ + )], k1 k̃1 k1 k̃1 = (ω1 , ~k1 ) and k̃1µ = (ω1 , −~k1 ). The explicit form of the transversal polarization +(−gµα′ + where k1µ tensors, when momentum is aligned along the third axis, is given by the matrices [4, 11]   1 e1µα = √ 2  0, 0, 0, 0       0, 1, 0, 0   2   , eµα     0, 0, −1, 0      0, 0, 0, 0 8   1 =√ 2  0, 0, 0, 0       0, 0, 1, 0    .     0, 1, 0, 0      0, 0, 0, 0 From the kinematics of the process in Fig.1 it follows that ω2 = ω1 and ~k2 = −~k1 therefore k̃1µ = k2µ , k̃2µ = k1µ and the average over polarizations can be rewritten as X 1 e∗µα (k1 )eµ′ α′ (k1 ) = (Eµµ′ Eαα′ + Eµα′ Eαµ′ − Eµα Eµ′ α′ ), 2 (4.12) where Eµµ′ = −gµµ′ + k1µ k2µ′ + k2µ k1µ′ . k1 · k2 Thus the average over tensor gauge boson polarizations gives ′ ′ ′ Mµανβ M∗µ α ν = β ′ X e∗µα (k1 )e∗νβ (k2 ) X eµ′ α′ (k1 )eν ′ β ′ (k2 ) = (4.13) 1 αβ 1 αβ g g g4 1 1 tb γ ν + γ ν tb 4 ta γ µ + if abc tc 2 (6 k2 − 6 k1 )(g µν g αβ − g µβ g να )] T r{6 p+ [γ µ ta 4 4 6 p− − 6 k 2 6 p− − 6 k 1 k3 2 ′ ν ′ 6 p− [γ t ′ ′ b ′ 1 αβ g 4 ′ 6 p− − 6 k 2 ′ ′ a t γ ′ µ ′ ′ µ a +γ t 1 αβ g 4 ′ 6 p− − 6 k 1 ′ ′ ′ ′ ′ tb γ ν − if a b c tc ′ 1 1 ′ ′ ′ ′ ′ ′ ′ ′ (6 k2 − 6 k1 )(g µ ν g α β − g µ β g ν α )]} 2 k3 2 1 δ aa δ bb 1 (Eµµ′ Eαα′ + Eµα′ Eαµ′ − Eµα Eµ′ α′ ) (Eνν ′ Eββ ′ + Eνβ ′ Eβν ′ − Eνβ Eν ′ β ′ ). d(r) d(r) 2 2 In the next section we shall evaluate these traces and summation over polarizations. 5 Evaluation of Traces In order to evaluate the squared matrix element in the last expression (4.13) we have to calculate traces and then perform summation over polarizations. We shall use convenient notations for the different terms in the amplitude. The whole amplitude will be expressed as a symbolic sum of three terms M = R + L + G, exactly corresponding to the three Feynman diagrams in Fig.3, so that the squared amplitude (4.13) shall have nine terms MM∗ = (R + L + G)(R + L + G)∗ . 9 The first contribution can be evaluated in the following way: (GG∗ )µανβ ′ ′ ′ µ α ν β 1 1 g4 T r{6 p+ if abc tc 2 (g µν g αβ − g µβ g να )(6 k2 − 6 k1 ) 2 4d (r) k3 2 ′ ′ ′ 1 ′ ′ ′ 1 ′ ′ ′ ′ ′ ′ ′ ′ 6 p− (−i)f a b c tc 2 (g µ ν g α β − g µ β g ν α )(6 k2 − 6 k1 )}δ aa δ bb = k3 2 4 ′ ′ T r{6 p (6 k − 6 k ) 6 p (6 k − 6 k )} g + 2 1 − 2 1 = 2 tr(f abc f abc tc tc ) 4d (r) (2k1 k2 )(2k1 k2 ) 1 1 ′ ′ ′ ′ ′ ′ ′ ′ (g µν g αβ − g µβ g να )(g µ ν g α β − g µ β g ν α ). 2 2 ′ = We can calculate traces over the symmetry group indices using formulas from the Appendix A: N(N 2 − 1) 2 and then the traces of gamma matrices using relation from the Appendix B: ′ ′ tr(f abc f abc tc tc ) = d(r)C2(r)C2 (G) = d(G)C(r)C2(G) = g4 p+ · (k2 − k1 ) p− · (k2 − k1 ) + p+ · p− k1 · k2 d(r)C2(r)C2 (G)8 2 4d (r) (2k1 k2 )(2k1 k2 ) 1 1 ′ ′ ′ ′ ′ ′ ′ ′ (g µν g αβ − g µβ g να)(g µ ν g α β − g µ β g ν α ) = 2 2 1 1 ′ ′ ′ ′ g4 ′ ′ ′ ′ = 2 d(r)C2(r)C2 (G)2 sin2 θ(g µν g αβ − g µβ g να )(g µ ν g α β − g µ β g ν α ). 4d (r) 2 2 Now it is easy to calculate summation over tensor gauge boson polarizations using expression (4.12) and the corresponding scalar products (2.2) GG∗ = g4 d(r)C2(r)C2 (G) sin2 θ. 4d2 (r) (5.14) The next contribution in (4.13) can be evaluated as follows: 1 αβ g4 µ a 4g = 2 T r{6 p+ [γ t (LG ) tb γ ν ] 4d (r) 6 p− − 6 k 2 ′ ′ ′ 1 ′ ′ ′ 1 ′ ′ ′ ′ ′ ′ ′ ′ 6 p− [−if a b c tc 2 (g µ ν g α β − g µ β g ν α )(6 k2 − 6 k1 )]}δ aa δ bb = k3 2 4 g T r{6 p+ γ µ (6 p− − 6 k2 )γ ν 6 p− (6 k2 − 6 k1 )} 1 αβ µ′ ν ′ α′ β ′ 1 µ′ β ′ ν ′ α′ = −i 2 tr(f abc ta tb tc ) g (g g − g g ), 4d (r) (−2p− k2 )(2k1 k2 ) 4 2 ′ ′ ′ ∗ µανβ µ α ν β ′ and then using traces from the Appendix A and the Appendix B we shall get g 4 d(r)C2 (r)C2 (G) 4{g µν [p+ · p− k1 · k2 + p+ · (k2 − k1 ) p− · k2 + p+ · k2 p− · (k2 − k1 )] + 4d2 (r) 2 + k1 · k2 (pµ+ pν− − pµ− pν+ ) + 2p− · (k2 − k1 )pµ+ pν− + 2p+ · (k2 − k1 )pµ− pν− } 1 αβ µ′ ν ′ α′ β ′ 1 µ′ β ′ ν ′ α′ 1 g (g g − g g ). (−2p− k2 )(2k1 k2 ) 4 2 10 Using again expression (4.12) and scalar products (2.2) we can sum over the polarizations of tensor gauge bosons: LG∗ = 1 g4 d(r)C2(r)C2 (G)(− sin2 θ). 2 4d (r) 4 (5.15) The third contribution is 1 αβ g4 ν b 4g ta γ µ ] T r{6 p [γ t + 4d2 (r) 6 p− − 6 k 1 ′ 1 ′ ′ ′ ′ ′ 1 ′ ′ ′ ′ ′ ′ ′ ′ 6 p− [−if a b c tc 2 (g µ ν g α β − g µ β g ν α )(6 k2 − 6 k1 )]}δ aa δ bb = k3 2 4 g T r{6 p+ γ ν (6 p− − 6 k1 )γ µ 6 p− (6 k2 − 6 k1 )} 1 αβ µ′ ν ′ α′ β ′ 1 µ′ β ′ ν ′ α′ = −i 2 tr(f abc tb ta tc ) g (g g − g g ) 4d (r) (−2p− k1 )(2k1 k2 ) 4 2 (RG∗ )µανβ ′ ′ ′ µ α ν β ′ = and can be evaluated in the similar way: d(r)C2(r)C2 (G) g4 (− )4{g νµ[−p+ · p− k1 · k2 + p+ · (k2 − k1 ) p− · k1 + p+ · k1 p− · (k2 − k1 )] + 2 4d (r) 2 1 + k1 · k2 (pν− pµ+ − pν+ pµ− ) + 2p− · (k2 − k1 )pν+ pµ− + 2p+ · (k2 − k1 )pν− pµ− } (−2p− k1 )(2k1 k2 ) 1 αβ µ′ ν ′ α′ β ′ 1 µ′ β ′ ν ′α′ g (g g − g g ), 4 2 so that after summation over polarization we shall get: (RG∗ ) = 1 2 g4 d(r)C (r)C (G)(− sin θ). 2 2 4d2(r) 4 (5.16) As one can get convinced, the next two terms GL∗ and GR∗ give similar contributions: g4 1 d(r)C2(r)C2 (G)(− sin2 θ), 2 4d (r) 4 (5.17) g4 1 GR = 2 d(r)C2(r)C2 (G)(− sin2 θ). 4d (r) 4 (5.18) GL∗ = ∗ The sixth contribution is ′ ′ ′ ′ ∗ µανβ µ α ν β (LL ) ′ ′ 1 αβ 1 α β ′ ′ ′ ′ ′ ′ g g g4 = 2 T r{6 p+ [γ µ ta 4 tb γ ν ] 6 p− [γ ν tb 4 ta γ µ ]}δ aa δ bb = 4d (r) 6 p− − 6 k 2 6 p− − 6 k 2 ′ ′ g4 T r{6 p+ γ µ (6 p− − 6 k2 )γ ν 6 p− γ ν (6 p− − 6 k2 )γ µ } 1 αβ 1 α′ β ′ = 2 tr(ta tb tb ta ) g g 4d (r) (2p− k2 )(2p− k2 ) 4 4 11 and involves trace of eight gamma matrices. It can be performed using expression presented in the Appendix B: ′ ′ ′ ′ g4 µ µ µ µ ν ν µµ d(r)C (r)C (r){16p p [p p + p p − p · p g ]+ 2 2 + − − + + − − − 4d2 (r) ′ ′ ′ ′ ′ + 8pν− [(p+ · k2 pµ− + p− · k2 pµ+ )g µν − (p+ · k2 pµ− − p− · k2 pµ+ )g µ ν + (p+ · k2 pν− − p− · k2 pν+ )g µµ ] ′ ′ ′ ′ ′ ′ ′ ′ + 8pν− [(p+ · k2 pµ− + p− · k2 pµ+ )g µ ν − (p+ · k2 pµ− − p− · k2 pµ+ )g µν + (p+ · k2 pν− − p− · k2 pν+ )g µµ ] ′ ′ ′ ′ ′ ′ 1 αβ 1 α′ β ′ 1 g g , + 8p+ · k2 p− · k2 [g µν g µ ν − g µν g µ ν + g µµ g νν ]} (2p− k2 )(2p− k2 ) 4 4 and after summation over polarizations we shall get LL∗ = g4 1 d(r)C2(r)C2 (r) sin2 θ. 2 4d (r) 4 (5.19) The seventh contribution is identical with the sixth one and gives RR∗ = g4 1 2 d(r)C (r)C (r) sin θ. 2 2 4d2 (r) 4 (5.20) The eighth contribution is ′ (LR∗ )µανβ ′ ′ ′ µ α ν β ′ 1 αβ ′ ′ ′ 1 gα β ′ ′ ′ g g4 tb γ ν ] 6 p− [γ µ ta 4 tb γ ν ]}δ aa δ bb = = 2 T r{6 p+ [γ µ ta 4 4d (r) 6 p− − 6 k 2 6 p− − 6 k 1 ′ ′ ′ g4 T r{6 p+ γ µ (6 p− − 6 k2 )γ ν 6 p− γ µ (6 p− − 6 k1 )γ ν } 1 αβ 1 α′ β ′ = 2 tr(ta tb ta tb ) g g 4d (r) (2p− k2 )(2p− k1 ) 4 4 and gives ′ ′ ′ ′ g4 1 µν νµ µν νµ + d(r)C (r)(C (r) − C (G)){−4k · k p · p g g + 4p · k p · k g g 2 2 2 1 2 + − + 2 − 1 4d2 (r) 2 ′ ′ ′ ′ ′ ′ ′ ′ + 4p+ · k1 p− · k2 g µν g νµ − 4k1 · k2 p+ · p− g µν g µ ν + + 4p+ · k2 p− · k1 g µν g µ ν + 4p+ · k1 p− · k2 g µν g µ ν + ′ ′ ′ ′ + 4k1 · k2 p+ · p− g µµ g νν − 4p+ · k2 p− · k1 g µµ g νν − ′ ′ ′ ′ ′ ′ ′ ′ − 4p+ · k1 p− · k2 g µµ g νν − 4k1 · k2 g νν pµ+ pµ− + 4k1 · k2 g µ ν pν+ pµ− + 4k1 · k2 g νµ pν+ pµ− − ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ − 4k1 · k2 g νν pµ+ pµ− + 8p− · k2 g νν pµ+ pµ− + 4k1 · k2 g µν pν+ pµ− − 8p− · k2 g µν pν+ pµ− − ′ ′ ′ ′ − 4k1 · k2 g µν pν+ pµ− + 8p− · k2 g µν pν+ pµ− − 8p+ · k2 g νν pµ− pµ− − 4k1 · k2 g µ ν pµ+ pν− + ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ + 8p− · k1 g µ ν pµ+ pν− + 4k1 · k2 g µν pµ+ pν− − 8p− · k1 g µν pµ+ pν− − 4k1 · k2 g µµ pν+ pν− + ′ ′ ′ ′ ′ + 8p− · k1 g µµ pν+ pν− + 8p+ · k1 g µ ν pµ− pν− + 8p+ · k1 g µν pµ− pν− + 8p+ · k2 g µν pµ− pν− − ′ ′ ′ ′ ′ ′ ′ ′ − 16p+ · p− g µν pµ− pν− + 16pν+ pµ− pµ− pν− + 4k1 · k2 g νµ pµ+ pν− + 4k1 · k2 g µν pµ+ pν− − 12 ′ ′ ′ ′ ′ ′ ′ ′ − 4k1 · k2 g µµ pν+ pν− + 8p+ · k2 g µν pµ− pν− − 8p+ · k1 g µµ pν− pν− + 16pµ+ pµ− pν− pν− } 1 αβ 1 α′ β ′ 1 g g . (2p− k2 )(2p− k1 ) 4 4 After summation over polarizations we shall get LR∗ = 1 1 g4 d(r)C2(r)(C2 (r) − C2 (G)) (− sin2 θ) 2 4d (r) 2 4 (5.21) and for the ninth contribution we shall get identically RL∗ = g4 1 1 d(r)C2 (r)(C2 (r) − C2 (G)) (− sin2 θ). 2 4d (r) 2 4 (5.22) We are now in a position to calculate the total contribution to the squared matrix element (4.13). Putting together all pieces of the squared matrix element (5.14), (5.15), (5.16), (5.17), (5.18), (5.19), (5.20), (5.21), (5.22), we finally obtain g4 1 MM = 2 d(r)C2 (r) C2 (G) sin2 θ. 4d (r) 4 ∗ 6 (5.23) Cross Section We can calculate now the leading-order differential cross section for the tensor gauge bosons production in the annihilation process. This process, as we discussed in the introduction, receives contribution from three Feynman diagrams shown in Fig.3 and for the unpolarized fermion pairs the squared matrix element was presented above (5.23). Plugging everything into our general cross-section formula (2.6) yields the differential cross section in the centerof-mass frame: 1 1 1 g4 d(r)C2 (r) C2 (G) sin2 θ dΩ = 2 4d (r) 4 2s 32π 2 g 2 1 C2 (r)C2 (G) = ( )2 sin2 θ dΩ = 4π s 64d(r) α2 C2 (r)C2 (G) sin2 θ dΩ, = s 64d(r) dσ = where α= (6.24) g2 . 4π Thus the unpolarized cross section is dσ = α2 C2 (r)C2(G) sin2 θ dΩ, s 64d(r) 13 (6.25) where for the SU(N) group we have C2 (r)C2 (G) 64d(r) = (N 2 −1) . 128N This cross section should be compared with the analogous annihilation cross sections in QED and QCD. Indeed let us compare this result with the electron-positron annihilation into two photons. The e+ e− → γγ annihilation cross section [12] in the high-energy limit is dσγγ = α2 1 + cos2 θ dΩ s sin2 θ (6.26) except very small angles of order m/E. Angular dependence of cross section is such that at θ = π/2 it has a minimum and then increases for small angles [13, 14]. The quark pair annihilation cross section into two gluons q q̄ → gg in the leading order of the strong coupling αs is dσgg = αs2 C2 (r)C2 (r) 1 + cos2 θ C2 (G) [ (1 + cos2 θ)]dΩ − 2 s d(r) 4C2 (r) sin θ (6.27) and also has minimum at θ = π/2 and increases for small scattering angles [15]. The production cross section of tensor gauge bosons (6.25) shows dramatically different behavior with its maximum at θ = π/2 and decrease for small angles. One of the authors (R.F.) is indebted to the Demokritos National Research Center for kind hospitality. The work of (R.F.) was supported by ENRAGE (European Network on Random Geometry), Marie Curie Research Training Network, contract MRTN-CT-2004005616. The work of (G.S.) was partially supported by the EEC Grant no. MRTN-CT2004-005616. 7 Appendix A The gauge group matrices ta form a representation r of the Lie group G. The matrices ta obey the algebra [ta , tb ] = if abc tc , where the structure constants f abc are totally antisymmetric. The invariant operators C(r) and C2 (r) are defined by the equations ta tb = C2 (r)1, and satisfy the relation C(r) = tr(ta tb ) = C(r)δ ab d(r) C2 (r), d(G) where d(r) is the dimension of the representation r. By convention the i and a are indices of the symmetry group G. A number of fermions ψ i is equal to the dimension d(r) of the 14 representation r: i = 1, ..., d(r). The number of gauge bosons Aa is equal to the number d(G) of generators of the group G: a = 1, ..., d(G). For the fundamental N and adjoint G representations of the SU(N) groups we have C(N) = 1/2, C2 (N) = N2 − 1 , 2N C(G) = 1/2 = C2 (G) = N. The traces over symmetry group indices now can be evaluated: 1 N(N 2 − 1) 1 , − itr(f abc ta tb tc ) = d(r)C2(r)C2 (G) = d(G)C(r)C2(G) = 2 2 4 ′ ′ N(N 2 − 1) abc abc c c tr(f f t t ) = d(r)C2(r)C2 (G) = d(G)C(r)C2(G) = , 2 (N 2 − 1)2 , tr(ta tb tb ta ) = d(r)C2(r)C2 (r) = d(G)C(r)C2(r) = 4N 1 (N 2 − 1) tr(ta tb ta tb ) = d(G)C(r)C2(r)(C2 (r) − C2 (G)) = − . 2 4N 8 Appendix B In this appendix we shall perform calculation of traces which appear in the squared matrix element (4.13). The traces under consideration have terms proportional to the momentum of the tensor gauge bosons k1µ and k2ν . These terms can be ignored, because after contraction with the corresponding transverse wave functions of the tensor gauge bosons eµα (k1 ) and eνβ (k2 ) they give zero contribution. Therefore we shall calculate the traces up to the longitudinal terms which are proportional to the vectors k1µ and k2ν . They are GG ∼ T r{6 p+ (6 k2 − 6 k1 ) 6 p− (6 k2 − 6 k1 )} = 8[p+ · (k2 − k1 ) p− · (k2 − k1 ) + p+ · p− k1 · k2 ], (LG∗ )µν ∼ T r{6 p+ γ µ (6 p− − 6 k2 )γ ν 6 p− (6 k2 − 6 k1 )} = = 4{g µν [p+ · p− k1 · k2 + p+ · (k2 − k1 ) p− · k2 + p+ · k2 p− · (k2 − k1 )] + + k1 · k2 (pµ+ pν− − pµ− pν+ ) + 2p− · (k2 − k1 )pµ+ pν− + 2p+ · (k2 − k1 )pµ− pν− }, (RG∗ )µν ∼ T r{6 p+ γ ν (6 p− − 6 k1 )γ µ 6 p− (6 k2 − 6 k1 )} = = 4{g νµ[−p+ · p− k1 · k2 + p+ · (k2 − k1 ) p− · k1 + p+ · k1 p− · (k2 − k1 )] + + k1 · k2 (pν− pµ+ − pν+ pµ− ) + 2p− · (k2 − k1 )pν+ pµ− + 2p+ · (k2 − k1 )pν− pµ− }, 15 (GL∗ )µν ∼ T r{6 p+ (6 k2 − 6 k1 ) 6 p− γ ν (6 p− − 6 k2 )γ µ } = = 4{g µν [p+ · p− k1 · k2 + p+ · (k2 − k1 ) p− · k2 + p+ · k2 p− · (k2 − k1 )] + + k1 · k2 (pµ+ pν− − pµ− pν+ ) + 2p− · (k2 − k1 )pµ+ pν− + 2p+ · (k2 − k1 )pµ− pν− }, (GR∗ )µν ∼ T r{6 p+ (6 k2 − 6 k1 ) 6 p− γ µ (6 p− − 6 k1 )γ ν } = = 4{g µν [−p+ · p− k1 · k2 + p+ · (k2 − k1 ) p− · k1 + p+ · k1 p− · (k2 − k1 )] + + k1 · k2 (pµ+ pν− − pµ− pν+ ) + 2p− · (k2 − k1 )pν+ pµ− + 2p+ · (k2 − k1 )pµ− pν− }, ′ ′ ′ ′ (LL∗ )µνµ ν ∼ T r{6 p+ γ µ (6 p− − 6 k2 )γ ν 6 p− γ ν (6 p− − 6 k2 )γ µ } = ′ ′ ′ ′ = {16pν− pν− [pµ− pµ+ + pµ+ pµ− − p+ · p− g µµ ] + ′ ′ ′ ′ ′ + 8pν− [(p+ · k2 pµ− + p− · k2 pµ+ )g µν − (p+ · k2 pµ− − p− · k2 pµ+ )g µ ν + (p+ · k2 pν− − p− · k2 pν+ )g µµ ] ′ ′ ′ ′ ′ ′ ′ ′ + 8pν− [(p+ · k2 pµ− + p− · k2 pµ+ )g µ ν − (p+ · k2 pµ− − p− · k2 pµ+ )g µν + (p+ · k2 pν− − p− · k2 pν+ )g µµ ] ′ ′ ′ ′ ′ ′ + 8p+ · k2 p− · k2 [g µν g µ ν − g µν g µ ν + g µµ g νν ]}, ′ ′ ′ ′ (RR∗ )µνµ ν ∼ T r{6 p+ γ ν (6 p− − 6 k1 )γ µ 6 p− γ µ (6 p− − 6 k1 )γ ν } = ′ ′ ′ ′ = {16pµ− pµ− [pν+ pν− + pν− pν+ − p+ · p− g νν ] + ′ ′ ′ ′ ′ + 8pµ− [(p+ · k1 pν− + p− · k1 pν+ )g νµ − (p+ · k1 pν− − p− · k1 pν+ )g ν µ + (p+ · k1 pµ− − p− · k1 pµ+ )g νν ] + 8pµ− [(p+ · k1 pν− + p− · k1 pν+ )g ν ′ ′ ′ ′ ′ ′ µ ′ ′ ′ ′ ′ ′ − (p+ · k1 pν− − p− · k1 pν+ )g νµ + (p+ · k1 pµ− − p− · k1 pµ+ )g νν ] ′ ′ + 8p+ · k1 p− · k1 [g νµ g µ ν − g νµ g µν + g νν g µµ ]}, ′ ′ ′ ′ (LR∗ )µνµ ν ∼ T r{6 p+ γ µ (6 p− − 6 k2 )γ ν 6 p− γ µ (6 p− − 6 k1 )γ ν } = ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ = −4k1 · k2 p+ · p− g µν g νµ + 4p+ · k2 p− · k1 g µν g νµ + 4p+ · k1 p− · k2 g µν g νµ − 4k1 · k2 p+ · p− g µν g µ ν + 4p+ · k2 p− · k1 g µν g µ ν + 4p+ · k1 p− · k2 g µν g µ ν + 16 ′ ′ ′ ′ 4k1 · k2 p+ · p− g µµ g νν − 4p+ · k2 p− · k1 g µµ g νν − ′ ′ ′ ′ ′ ′ 4p+ · k1 p− · k2 g µµ g νν − 4k1 · k2 g νν pµ+ pµ− + ′ ′ 4k1 · k2 g µ ν pν+ pµ− + 4k1 · k2 g νµ pν+ pµ− − ′ ′ ′ ′ ′ ′ ′ ′ 4k1 · k2 g νν pµ+ pµ− + 8p− · k2 g νν pµ+ pµ− + 4k1 · k2 g µν pν+ pµ− − 8p− · k2 g µν pν+ pµ− − ′ ′ ′ ′ 4k1 · k2 g µν pν+ pµ− + 8p− · k2 g µν pν+ pµ− − ′ ′ ′ ′ 8p+ · k2 g νν pµ− pµ− − 4k1 · k2 g µ ν pµ+ pν− + ′ ′ ′ ′ ′ ′ 8p− · k1 g µ ν pµ+ pν− + 4k1 · k2 g µν pµ+ pν− − ′ ′ 8p− · k1 g µν pµ+ pν− − 4k1 · k2 g µµ pν+ pν− + ′ ′ ′ ′ ′ ′ 8p− · k1 g µµ pν+ pν− + 8p+ · k1 g µ ν pµ− pν− + ′ ′ 8p+ · k1 g µν pµ− pν− + 8p+ · k2 g µν pµ− pν− − ′ ′ ′ ′ 16p+ · p− g µν pµ− pν− + 16pν+ pµ− pµ− pν− + ′ ′ ′ ′ ′ ′ 4k1 · k2 g νµ pµ+ pν− + 4k1 · k2 g µν pµ+ pν− − ′ ′ 4k1 · k2 g µµ pν+ pν− + 8p+ · k2 g µν pµ− pν− − ′ ′ ′ ′ 8p+ · k1 g µµ pν− pν− + 16pµ+ pµ− pν− pν− , ′ ′ ′ ′ (RL∗ )µνµ ν ∼ T r{6 p+ γ ν (6 p− − 6 k1 )γ µ 6 p− γ ν (6 p− − 6 k2 )γ µ } = ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ = −4k1 · k2 p+ · p− g µν g νµ + 4p+ · k2 p− · k1 g µν g νµ + 4p+ · k1 p− · k2 g µν g νµ − 4k1 · k2 p+ · p− g µν g µ ν + 4p+ · k2 p− · k1 g µν g µ ν + 4p+ · k1 p− · k2 g µν g µ ν + ′ ′ ′ ′ 4k1 · k2 p+ · p− g µµ g νν − 4p+ · k2 p− · k1 g µµ g νν − ′ ′ ′ ′ 4p+ · k1 p− · k2 g µµ g νν − 4k1 · k2 g νν pµ+ pµ− + ′ ′ ′ ′ 8p− · k2 g νν pµ+ pµ− − 4k1 · k2 g µ ν pν+ pµ− + ′ ′ ′ ′ 8p− · k2 g µ ν pν+ pµ− + 4k1 · k2 g νµ pν+ pµ− − ′ ′ ′ ′ 8p− · k2 g νµ pν+ pµ− − 4k1 · k2 g νν pµ+ pµ− + ′ ′ ′ ′ 4k1 · k2 g µν pν+ pµ− + 4k1 · k2 g µν pν+ pµ− − ′ ′ ′ ′ 8p+ · k2 g νν pµ− pµ− + 4k1 · k2 g µ ν pµ+ pν− + ′ ′ ′ ′ 4k1 · k2 g µν pµ+ pν− − 4k1 · k2 g µµ pν+ pν− + 17 ′ ′ ′ ′ 8p+ · k2 g µ ν pµ− pν− + 4k1 · k2 g νµ pµ+ pν− − ′ ′ ′ ′ 8p− · k1 g νµ pµ+ pν− − 4k1 · k2 g µν pµ+ pν− + ′ ′ ′ ′ ′ ′ ′ ′ ′ 8p− · k1 g µν pµ+ pν− − 4k1 · k2 g µµ pν+ pν− + ′ 8p− · k1 g µµ pν+ pν− + 8p+ · k1 g νµ pµ− pν− + ′ ′ 8p+ · k2 g νµ pµ− pν− − 16p+ · p− g νµ pµ− pν− + ′ ′ ′ ′ 8p+ · k1 g µν pµ− pν− + 16pν+ pµ− pµ− pν− − ′ ′ ′ ′ 8p+ · k1 g µµ pν− pν− + 16pµ+ pµ− pν− pν− . (8.28) All these traces have been calculated with the use of the Mathematica program [16]. References [1] M. Fierz. Über die relativistische Theorie kräftefreier Teilchen mit beliebigem Spin, Helv. Phys. Acta. 12 (1939) 3. [2] M. Fierz and W. Pauli. On Relativistic Wave Equations for Particles of Arbitrary Spin in an Electromagnetic Field, Proc. Roy. Soc. A173 (1939) 211. [3] J.Schwinger, Particles, Sourses, and Fields (Addison-Wesley, Reading, MA, 1970) [4] H. van Dam and M. J. G. Veltman, “Massive And Massless Yang-Mills And Gravitational Fields,” Nucl. Phys. B 22 (1970) 397. [5] C.Fronsdal, Massless fields with integer spin, Phys.Rev. D18 (1978) 3624 [6] A. K. Bengtsson, I. Bengtsson and L. Brink, Cubic Interaction Terms For Arbitrary Spin, Nucl. Phys. B 227 (1983) 31. [7] E. Witten, Noncommutative Geometry And String Field Theory, Nucl. Phys. B 268 (1986) 253. [8] R. R. Metsaev, “Cubic interaction vertices for fermionic and bosonic arbitrary spin fields,” arXiv:0712.3526 [hep-th]. 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