An Extended Transformation to Reach Continuous Gauge Invariance for Gauge Fields with Static Masses

The paper provides an extended local gauge transformation. By means of it, the Lagrangian of gauge fields with masses can keep unchanged in the continuous transformations. In this way, the Higgs mechanism becomes unnecessary and we do not need to suppose the existence of the Higgs particles. PACS numbers: 1110, 1130 According to the Yang-Mills theory, in order to keep the Lagrangian of a system unchanged under the local gauge transformation, the transformation rules of the field φ and its covariant differentiation should be defined as φ(x) → φ(x) = exp[iθ(x)T]φ(x) (1) Dμ(x)φ(x) → D ′ μ(x)φ (x) = exp[−iθ(x)T ]Dμ(x)φ(x) (2) in which Dμ(x) = ∂μ +Aμ(x) Aμ(x) = −igA i μ(x)T i (3) From Eq.(2), we can get the transformation rule of gauge under the infinitesimal transformation Aiμ(x) → A ′i μ(x) = A i μ(x) + f ijkθj(x)Akμ(x)− 1 g ∂μθ (x) (4) The intensity of the gauge field is defined as F i μν(x) = ∂μA i ν(x)− ∂νA i μ(x) + gf ijkAjμ(x)A k ν(x) (5) The Lagrangian of the free gauge field with zero mass is L0 = − 1 4 F i μνF i μν (6) It is invariable under the gauge transformation. But the Lagrangian of the free gauges fields with static masses (suppose their masses are the same for simplification) is L1 = − 1 4 F i μνF iμν − 1 2 m2AiμA i μ (7) It can not keep unchanged under the transformation. However, the gauge particles just as W and Z have masses. In order to solve this problem, in the current theory, the Higgs mechanism has to be introduced. By means of the concept of spontaneous symmetry breaking of vacuum, the gauge particles obtain their masses. The current gauge theory has achieved great success but has also two basic faults. The first is that the Higgs particles can not be found up to now. The second is that the theory can not keep unchanged again after the Higgs mechanism is introduced, that is to say, the invariance is not thorough. Besides, there exist some other problems, for example, too many parameters have to be introduced and the reason of spontaneous symmetry breaking of vacuum is also unclear and so on. It is proved below that we can solve these problems by introducing a supplement function B μ(x). Let Bμ(x) = B i μ(x)T , similar to Eq.(2), we define the extended transformation as Dμ(x)φ(x) → D ′′ μ(x)φ (x) = exp[−iφ(x)T ][Dμ(x) +Bμ(x)]φ(x) (8) Here Dμ(x) = ∂μ +A ′′i μ (x). It can be calculated from Eq.(8) that by the extended transformation, the transformation rule of Aμ(x) becomes A μ (x) = A i μ(x) +G i μ(x) (9) in which Giμ(x) = f ijkθj(x)Akμ − 1 g ∂μθ (x) +B μ(x) + f θ(x)B μ(x) (10) The transformation means that we let Aiμ(x) → A i μ(x) +B i μ(x) (11) in Eq.(4) again, or let F i μν(x) → F i μν(x) +K i μν(x) (12) in Eq.(5). Here K μν = ∂μB i ν − ∂νB i μ + gf ijk(AjμB k ν +A j νB k μ +B j μB k ν ) (13) Considering the fact that F i μνF i μν is invariable in the current gauge transformation with (F i μνF i μν) ′ = F i μνF i μν (14) Putting Eq.(12) into the two sides of Eq.(14), we have (F i μνF i μν) ′′ = F i μνF i μν +Q1 (15) In the formula Q1 = 2F i μνK i μν +K i μνK i μν (16) Putting Eq.(11) and Eq.(15) into Eq.(7), we get the extended Lagrangian L 1 = L1 − 1 4 Q1 − 1 2 Q2 (17) in which Q2 = m 2(2AiμG i μ +G i μG i μ) (18) We have L 1 = L1 as long as let Q1 + 2Q2 = 0 (19) In this way, the Lagrangian is invariable under the extended gauge transformation. It is obvious that the invariance of continuous transformations can also be kept in this way. Now let us discuss how to determine the forms of the functions B μ(x) from Eq.(19). Suppose i = 1.2 . . .N , the number of B μ(x) is 4N . But they only satisfy one equation, that is to say, only one of them is independent. The other 4N-1 B μ(x) can be chosen arbitrarily. As the simplest form, if we take B μ(x) = B 1 1 (x) = B(x) μ = 1 i = 1 B μ(x) = 0 μ 6= 1 i 6= 1 (20) Eq.(19) can be written as