aa r X i v : . [ phy s i c s . g e n - ph ] M a r Forces on a Clifford bundle jason hansonApril 1, 2019
Abstract
In [3], the Clifford bundle over spacetime was used as a geometricframework for obtaining coupled Dirac and Einstein equations. Otherforces may be incorporated using minimal coupling. Here the fun-damental forces that are allowed within this framework are explicitlyenumerated.
In a previous article [3], we used the Clifford bundle Cl ∗ M over a spacetimemanifold M as a geometric framework for incorporating both the Dirac andEinstein equations. We also indicated how other forces can be introducedusing minimal coupling. That is, by using a connection on the Clifford bundleof the form ∇ α = ∂ α + ˆΓ α + θ α . Here ˆΓ α is the metric–compatible connectionon M extended to Cl ∗ M , and θ α is a collection of four 16 ×
16 matrices thatencode the force. These matrices must satisfy certain constraints in order forthe variational principle to yield the Dirac equation γ α ∇ α ψ = mψ . Such acollection θ = { θ , θ , θ , θ } is a tensor, which we call a force tensor .In this article, we enumerate the distinct types of forces that are allowedby minimal coupling. This is done by considering the action of the Lorentzgroup on the space of all possible force tensors V . Under this action, V isa representation of the Lorentz group. And as such, it can be decomposedinto irreducible subrepresentations. Each irreducible subrepresentation cor-responds to a distinct force. Article summary.
In the remainder of this section, we review the relevantconstructions used in geometric framework introduced in [3], as well as the1ecessary constraints on a force tensor θ . In section 2, we find a basis forthe space of all possible force tensors V , and in section 3 we write down theirreducible subspaces. Each irreducible subspace is identified with a typeof fundamental force field on spacetime M : scalar, vector, anti–symmetrictensor, symmetric tensor, and Fierz tensor. In section 4, we use the curvatureof the connection ∇ α to obtain field equations for each of the fundamentalforces. Let M be a Lorentz manifold with metric g . Let x = x α , with α = 0 , , , M , and let e α . = ∂/∂x α denote the correspondingbasis vectors of the tangent bundle of M at x . In this basis, g αβ denotes thecomponents of g , and g αβ denotes the components of g − .The Clifford bundle Cl ∗ M is formed by taking the Clifford algebra ofeach fiber of the tangent bundle. That is, we enforce the algebra relation e α e β + e β e α = 2 g αβ on each fiber. We may choose e ∅ . = 1 , e , e , e , e , e , e , e , e , e , e , e , e , e , e , e as basis vectors for the fibers of Cl ∗ M . Here e I . = e α e α · · · e α k for the multi–index I = α α · · · α k . We denote the length of I by | I | ; i.e., | I | = k . A field ψ is a section M → Cl ∗ M , and we write ψ = ψ I e I , where I = ∅ , , , . . . , γ µ is defined as left multiplication by the tangentvector e µ . That is, γ µ ψ . = e µ ψ . Moreover, we set γ α . = g αβ γ β .The spacetime metric g is extended to a metric ˆ g on Cl ∗ M by the ruleˆ g ( ψ, φ ) . = − h ψ † φ + φ † ψ i ∅ . Here ψ † is the linear extension of e † α α ··· α k . =( − k e α k ··· α ··· α , and h ψ i ∅ is linear projection onto the e ∅ component of ψ .One shows that γ Tα ˆ g + ˆ gγ α = 0.A transformation A on the tangent bundle of M extends to a transfor-mation ˆ A on Cl ∗ M via the rule ˆ A e α α ··· α k . = ( A e α )( A e α ) · · · ( A e α k ). Inparticular, suppose that B is a change of basis for the tangent bundle of M : e ′ α = B − e α . This extends to an action ˆ B on Cl ∗ M . Under the extendedchange of basis, the gamma matrices transform as γ ′ α = ( B − ) βα ˆ Bγ β ˆ B − and γ ′ α = B αβ ˆ Bγ β ˆ B − . Whereas the extended metric transforms as ˆ g ′ =ˆ B − T ˆ g ˆ B − .The transformation extension rule in the previous paragraph applies toLie groups. That is, if G is a Lie group acting locally on the tangent bundle,2hen we can extend the action of G to Cl ∗ M . On the other hand, theLie algebra g of G extends to a Lie algebra action via the rule ˆ a e α ··· α k = P ki =1 e α · · · e α i − ( a e α i ) e α i +1 . . . e α k for all a ∈ g . We use minimal coupling to model forces. That is, we assume there is aconnection ∇ on Cl ∗ M , where ∇ α ψ = ∂ α ψ + C α ψ for some collection of four16 ×
16 matrices C α . However, we need to ensure that we obtain the Diracequation Dψ + µψ = 0 , where Dψ . = γ α ∇ α ψ, by varying the (partial) Lagrangian density L K = ( ψ T ˆ gDψ + µψ T ˆ gψ ) ω, ω . = p − det( g ) (1)with respect to ψ . As observed in [3], the conditions C Tα ˆ g + ˆ gC α = ∂ α ˆ g and [ γ α , C β ] = ∂ β γ α + Γ αβǫ γ ǫ , (2)where Γ αβǫ are the Christoffel symbols for the metric connection on M , aresufficient to guarantee this.The metric connection on M extends to a connection on Cl ∗ M via theLeibniz rule. Moreover, the extended metric connection matrices ˆΓ α satisfyboth conditions in equation (2). So to incorporate forces other than gravity,we look for connection matrices of the form C α = ˆΓ α + θ α , with θ Tα ˆ g + ˆ gθ α = 0 and [ γ α , θ β ] = 0 . (3)We call any collection θ α of 16 ×
16 matrices that satisfy (3) a force tensor .We write θ ∗ to denote the collection { θ , θ , θ , θ } .By general principles, under a local change of basis B for the tangentbundle of M , the connection matrices C α transform as C ′ α = ( B − ) βα ( − ∂ β ˆ B + ˆ BC β ) ˆ B − . (4)The extended metric connection ˆΓ α will satisfy this equation, so a force tensormust satisfy the transformation rule θ ′ α = ( B − ) βα ˆ Bθ β ˆ B − (5)3oreover, a force tensor remains a force tensor under a transformation. Thatis, the equations in (3) are satisfied by the transformed force tensor: θ ′ Tα ˆ g ′ +ˆ g ′ θ ′ α = 0 and [ γ ′ α , θ ′ β ] = 0.We remark that a force tensor is, in general, not the matrix of an actualconnection. It does not transform as a connection matrix is required to,equation (4). However in the case of Minkowski space M , where the extendedmetric connection is trivial ˆΓ α = 0, a force tensor is indeed the matrix of aconnection on the Clifford algebra Cl ( M ). The set V ( g ) of all force tensors is necessarily a (real) vector space. We willdetermine a basis in the case of the Minkowski metric g = η , where η . = diag( − , , , . Note that ˆ η is also diagonal. The following fact, which we verify at the endof this section, facilitates the computation. Fact.
Let N be a × matrix acting on Cl ( M ) as a vector space. Then N commutes with gamma matrices if and only if for all multi–indices I we have N e I = e I · N e ∅ . Moreover for such N , N T ˆ η + ˆ ηN = 0 if and only if N e ∅ lies in the subspace of Cl ( M ) spanned by the basis vectors e I with | I | = 1 , . Observe that from the first statement, any matrix N that commutes withgamma matrices is completely determined by its effect on e ∅ .Let us introduce the following notation. For a multi–index I , { I } de-notes the 16 ×
16 matrix that commutes with gamma matrices and such that { I } e ∅ = e I . Moreover, we let { β, I } ∗ denote the collection of matrices with { β, I } α = 0 if α = β, and { β, I } α = { I } if α = β. I.e., { β, I } α = δ αβ { I } . It should be pointed out that even though { α } e ∅ = e α and γ α e ∅ = e α , necessarily { α } 6 = γ α .From the above fact, we see that the space of force tensors V ( η ) is a realvector space with basis given by { β, I } ∗ with I = 0 , , , , , , , , ,
23 and β = 0 , , , . (6)In particular, V ( η ) has dimension 40.4n the remainder of this section, we verify the previously claimed proper-ties of the matrix 16 ×
16 matrix N . The first statement is readily seen fromthe definition of the gamma matrices. For the second statement, let I = α · · · α k be a multi–index. Set γ I . = γ α · · · γ α k , and γ † I . = ( − k γ α k · · · γ α ,so that γ I e ∅ = e I and γ † I e ∅ = e † I . Moreover, ˆ ηγ † I = γ TI ˆ η . Viewing e I as acolumn vector, we have e T ∅ ˆ ηN e I = e T ∅ ˆ ηN γ I e ∅ = e T ∅ ˆ ηγ I N e ∅ . The symmetry of ˆ η implies that the quantity on the right is equal to e T ∅ N T γ TI ˆ η e ∅ = e T ∅ N T ˆ ηγ † I e ∅ = e T ∅ N T ˆ η e † I . It follows that ( ⋆ ) e T ∅ (ˆ ηN + N T ˆ η ) e I = e T ∅ N T ˆ η ( e † I + e I ) for any multi–index I . Observe that e I + e † I = 0 if | I | = 1 , e I if | I | = 0 , , N T ˆ η + ˆ ηN = 0, ( ⋆ ) and the fact that ˆ η is diagonal imply that N e ∅ can only lie in the subspace spanned by those e I with | I | = 1 , J and K be any multi–indices. Now e † J e K reduces to amultiple of a basis vector, say r e I . So, e TJ N T ˆ η e K = e T ∅ γ TJ N T ˆ η e K = e T ∅ N T γ TJ ˆ η e K = e T ∅ N T ˆ ηγ † J e K = r e T ∅ N T ˆ η e I . Similarly, one computes e TJ ˆ ηN e K = r e T ∅ N T ˆ η e † I , so that e TJ ( N T ˆ η + ˆ ηN ) e K = r e T ∅ N T ˆ η ( e I + e † I ). Identity ( ⋆ ) thus implies that N T ˆ η + ˆ ηN = 0 if N e ∅ liesin the stated subspace. A Lorentz transformation Λ defines a change of basis for the tangent bundleof M . We can extend this to a change of basis ˆΛ for Cl ∗ M . In this way theLorentz group O ( g ) acts on the space of force tensors V ( g ). Indeed, fromequation (5), (Λ · θ ) α . = (Λ − ) βα ˆΛ θ β ˆΛ − for any force tensor θ ∗ . The corresponding Lorentz algebra action is then( L · θ ) α = − L βα θ β + ˆ Lθ α − θ α ˆ L = [ ˆ L, θ α ] − L βα θ β (7)for L ∈ so ( g ). 5e say that a force tensor θ ∗ is irreducible if its orbit under the O ( g )action, or equivalently under the so ( g ) action, is an irreducible real represen-tation of O ( g ). The irreducible sub–representations of V ( g ) may be computedusing standard Lie algebra techniques, such as in [2]. We will present theresults for the case g = η . We explicitly compute how so ( η ) affects each of the basic force tensors { β, I } ∗ in equation (6). As a vector space, so ( η ) is six–dimensional, and we may takeas basis the six 4 × s , s , s , a , a , a s k . = E k + E k and a jk . = E jk − E kj where E αβ is the 4 × α, β )–entry is unity, and all otherentries are zero. In particular, we have s k e = e k , s k e k = e , a jk e j = − e k , a jk e k = e j (8)and s k e l = 0 and a jk e l = 0 in all other cases.The Lie algebra action of so ( η ) on M extends to a Lie algebra action on Cl ( M ). For example,ˆ s e = ( s e )( e ) + ( e )( s e ) = ( e )( e ) + ( e )(0) = − e e = − e Now from equation (7), the action of L ∈ so ( η ) on { β, I } is( L · { β, I } ) α = [ ˆ L, { β, I } α ] − L να { β, I } ν = δ βα [ ˆ L, { I } ] − L βα { I } Figure 1 gives a listing of values for [ ˆ L, { I } ] for our chosen basis matrices L of so ( η ). For instance,ˆ s { } e J = ˆ s e J e = (ˆ s e J )( e ) + ( e J )(ˆ s e )= { } ˆ s e J − e J e = { } ˆ s e J − { } e J So that [ˆ s , { } ] = −{ } . Figure 1 and equation (8) can then be used tocompute the effect of the so ( η ) action on the basis elements of V ( η ). E.g.,( a · { , } ) = 0 − ( a ) { } = 0( a · { , } ) = 0 − ( a ) { } = 0( a · { , } ) = [ˆ a , { } ] − ( a ) { } = −{ } − −{ } ( a · { , } ) = 0 − ( a ) { } = −{ } I }\ L s s s a a a { } { } { } { } { } { } −{ } −{ } { } { } { } −{ }{ } { } { } { }{ } −{ } −{ } −{ } −{ } { } { } −{ } { } −{ }{ } { } { } { } { }{ } { } −{ } { } −{ }{ } { } −{ } −{ } { }{ } { } −{ } { } −{ } L, { I } ].Thus, ( a · { } ) ∗ = (0 , , −{ } , −{ } ) = −{ , } ∗ − { , } ∗ . Viewing the space of force tensors V ( g ) as a Lie algebra representation of so ( g ), we decompose it into irreducible summands. Schematically, the de-composition is V ( g ) = ⊕ ⊕ ′ ⊕ ⊕ ⊕ (9)where each summand has the indicated dimension. The decomposition is overthe reals. Further decomposition of the summands and can be achievedover the complex numbers. Indeed when complex coefficients are allowed,each of these two summands decomposes into two conjugate summands ofcomplex dimensions equal to half the real dimension: = ⊕ ¯ and = ⊕ ¯ .The 4–dimensional summands and ′ are (real) isomorphic. In fact, we willsee that and ′ are both isomorphic to the standard 4–vector representationof the Lorentz group. In the classification of representations of the Lorentzgroup, this representation is denoted ( , ). The summand is isomorphicto (1 , ⊕ (0 , isomorphic to (1 , isomorphic to ( , ) ⊕ ( , ).7 .3 One–dimensional connection The summand is the one–dimensional subspace spanned by the single forcetensor u = { , } ∗ + { , } ∗ + { , } ∗ + { , } ∗ . Indeed, one computes that L · u = 0 for all generators L = s , s , s , a , a , a of so ( η ), and hence for all L ∈ so ( η ). That is, so ( η ) acts trivially on u . Any force tensor in can thus be written in the form U ∗ = φu (10)for a scalar field φ . The field φ is necessarily unaffected by a change of basis;that is, its transformation rule is φ ′ = φ . The individual component matricesof U ∗ are U = φ { } , U = φ { } , U = φ { } , U = φ { } There two four–dimensional summands in (9). We will see that the twosummands are isomorphic. As a consequence of this, if we set
V . = ⊕ ′ ,then the decomposition of V into irreducible summands is not unique: wecan find irreducible subspaces V , V of V such that V = V ⊕ V , but V isequal to neither nor ′ . For the moment, we will choose summands thatare comparatively easily to write down. Specifically, set V . = R { v , v , v , v } and V . = R { v ′ , v ′ , v ′ , v ′ } where v . = −{ , } ∗ − { , } ∗ − { , } ∗ v ′ . = −{ , } ∗ + { , } ∗ − { , } ∗ v . = { , } ∗ − { , } ∗ − { , } ∗ v ′ . = { , } ∗ + { , } ∗ − { , } ∗ v . = { , } ∗ + { , } ∗ − { , } ∗ v ′ . = −{ , } ∗ − { , } ∗ + { , } ∗ v . = { , } ∗ + { , } ∗ + { , } ∗ v ′ . = { , } ∗ + { , } ∗ − { , } ∗ Observe that as matrices, { }{ } = { } . In this way, we can write v ′ α = J v α , so that V = J V , where J . = { } .In general, for any angle ζ we obtain an irreducible summand of V bychoosing basis vectorscos ζ v α + sin ζ v ′ α = cos ζ v α + sin ζ J v α = e ζJ v α e ζJ = cos ζ I + sin ζ J is the matrix exponential (note that J = − I ).Let us set ζ . = e ζJ V = R { e ζJ v , e ζJ v , e ζJ v , e ζJ v } . (11)For different values of ζ , we obtain different summands of V . Moreover for ζ , ζ ′ ∈ [0 , π ) with ζ = ζ ′ , V = ζ ⊕ ζ ′ .One computes the extended Lie algebra action of so ( η ) on the basis vec-tors of V to beˆ s k v = v k , ˆ s k v k = v , ˆ a jk v j = − v k , ˆ a jk v k = v j for j, k = 1 , , j = k (with all other actions trivial). Moreover, thematrix J commutes with the extended Lie algebra action, so that the basisvectors of ζ in (11) transform in the exact same way as those of V . It followsthat ζ is indeed stable under the so ( η ) action, and that all summands of thisform are isomorphic. Furthermore, the basis vectors of ζ transform exactlylike the basis vectors e α of M under Lorentz transformations. Therefore, anyforce tensor in ζ can be identified with a 4–vector field. I.e., we have theforce tensor F ∗ = A α e ζJ v α (12)for any 4–vector field A α . Explicitly, we have F = e ζJ ( A { } + A { } + A { } ) F = e ζJ ( − A ( { } + A { } + A { } ) F = e ζJ ( − A { } − A { } + A { } ) F = e ζJ ( − A { } − A { } − A { } ) The summand in (9) is spanned by the force tensors v . = { , } ∗ − { , } ∗ v . = { , } ∗ − { , } ∗ v . = { , } ∗ − { , } ∗ v . = { , } ∗ + { , } ∗ v . = −{ , } ∗ − { , } ∗ v . = { , } ∗ + { , } ∗ (13)The action of so ( η ) on these basic force tensors is given in table in figure 2.This verifies that is indeed stable under the so ( η ) action. In fact, this isexactly the same action as that of so ( η ) on Λ M , the vector space spanned9 s s a a a v − v − v − v − v v v − v v − v v v v v v v v − v v − v v v − v − v v v v − v v − v so ( η ) on the basic force tensors of .by the 2–forms e α ∧ e β . = e α ⊗ e β − e β ⊗ e α . Hence we may identify anyconnection from with an anti–symmetric tensor field W αβ . That is, wemay write H ∗ = W αβ v αβ (14)for some anti–symmetric tensor field W αβ , provided we define the basic con-nections v αβ to satisfy v αβ = 0 if α ≥ β . Explicitly, H = W { } − W { } + W { } H = − W { } + W { } + W { } H = W { } − W { } − W { } H = − W { } + W { } + W { } A basis for summand in (9) is given by the following basic force tensors. u . = −{ , } ∗ + { , } ∗ u . = { , } ∗ − { , } ∗ u . = { , } ∗ − { , } ∗ u . = { , } ∗ − { , } ∗ u . = { , } ∗ − { , } ∗ u . = { , } ∗ + { , } ∗ u . = { , } ∗ + { , } ∗ u . = { , } ∗ − { , } ∗ u . = { , } ∗ + { , } ∗ (15)The action of so ( η ) on these is summarized in the table in figure 3, fromwhich we see that the above collection of force tensors is stable under the so ( η ) action. 10e show that the basic force tensors in (15) form a basis for the space ofsymmetric traceless 4 × M . Indeed, suppose S αβ is such that S βα = S αβ and S αα = η αβ S αβ = 0. The latter condition isequivalent to S = S − S − S . Therefore, we may write S αβ e α ⊗ e β = S ( e ⊗ e + e ⊗ e ) + S ( e ⊗ e + e ⊗ e )+ S ( e ⊗ e + e ⊗ e ) + S ( e ⊗ e + e ⊗ e )+ S ( e ⊗ e − e ⊗ e ) + S ( e ⊗ e + e ⊗ e )+ S ( e ⊗ e + e ⊗ e ) + S ( e ⊗ e − e ⊗ e )+ S ( e ⊗ e + e ⊗ e )which defines a basis for the space of symmetric traceless matrices. Onecomputes that the so ( η ) action on this basis is exactly the same as the actionon the corresponding basic force tensors in equation (15).In sum, if we define u αβ = 0 for α > β and u = 0, then a force tensorin can be written in the form N ∗ = S αβ u αβ for some symmetric traceless tensor field S αβ ; i.e., such that S αβ = S βα and S αα = 0. Explicitly, N = − S { } − S { } − S { } − S { } N = S { } + S { } + S { } + S { } N = S { } + S { } + S { } + S { } N = S { } + S { } + S { } + S { } s s a a a u u u u u u u u + 2 u u u − u − u u u u + 2 u u u − u u u u u u u u u − u − u − u − u u u u u − u − u − u u u u − u u u u u − u u − u − u u u u u u u Figure 3: The action of so ( η ) on the basic force tensors of . The sixteen–dimensional summand in equation 9 is spanned by the fol-lowing force tensors: u . = { , } ∗ − { , } ∗ u . = { , } ∗ − { , } ∗ u . = { , } ∗ − { , } ∗ u . = { , } ∗ − { , } ∗ u . = { , } ∗ + { , } ∗ u . = { , } ∗ + { , } ∗ u . = { , } ∗ − { , } ∗ u . = { , } ∗ + { , } ∗ u . = { , } ∗ + { , } ∗ u . = { , } ∗ − { , } ∗ u . = { , } ∗ + { , } ∗ u . = { , } ∗ + { , } ∗ u . = { , } ∗ − { , } ∗ u . = { , } ∗ − { , } ∗ u . = { , } ∗ + { , } ∗ u . = { , } ∗ + { , } ∗ (16)The action of so ( η ) on these force tensors is given by the table in figure 4.We will identify the coefficients of a force tensor in with a Fierz tensor.Such tensors are described briefly in [1].Consider the subspace F of the tensor product space M ⊗ M ⊗ M consistingof tensors F αρσ of the form F αρσ = − F ασρ , η αρ F αρσ = 0 , ǫ αρστ F αρσ = 0 . (17)12hese three constraints allow us to write F αρσ e α ⊗ e ρ ⊗ e σ = F ( f + f ) + F ( f + f ) + F ( f + f )+ F ( f + f ) + F ( f − f ) + F ( f − f )+ F ( f + f ) + F ( f − f ) + F ( f + f )+ F ( f + f ) + F ( f − f ) + F ( f − f )+ F ( f − f ) + F ( f − f ) + F ( f − f )+ F ( f − f )where f αρσ . = e α ⊗ e ρ ⊗ e σ − e α ⊗ e σ ⊗ e ρ . One computes that the action of so ( η ) on the basis of F thus defined is exactly as the table in figure 4. Thatis, the basis element f + f transforms in the same way as the basic forcetensor u , f + f transforms exactly as u , and so on.It follows that a general force tensor in can be expressed in the form X ∗ = F αρσ u αρσ , where F αρσ satisfies equation (17), and provided we set u αρσ = 0 for indicesthat do not match those of the stated basis elements. E.g., u = 0, u = 0,et cetera. Moreover, we have X = − F { } − F { } − F { } − F { } − F { } − F { } X = F { } + F { } + F { } + F { } + F { } + F { } X = F { } + F { } + F { } + F { } + F { } + F { } X = F { } + F { } + F { } + F { } + F { } + F { } Each distinct irreducible summand in (9) corresponds to a distinct funda-mental force. That is, there are five fundamental forces (not including grav-ity) predicted by this model. In this section, for each fundamental force wecompute the the potential energy term under the assumption L V . = ω tr (Ω αβ Ω αβ ) , s s a a a u u u u − u − u − u u u u u − u − u u u u u u − u − u − u u u u u − u − u u u − u − u − u − u − u − u − u u u u u − u u + 2 u u + u − u − u u u u u u u u − u u u − u − u u u − u − u u u u − u − u u − u − u − u u − u u u u u − u u − u − u u − u − u − u − u − u − u u u u − u − u u u u − u u u u + u u u + u u + 2 u u + u u u − u − u u u − u u + 2 u u − u u u − u − u − u u u + u u − u u − u − u u − u u Figure 4: The action of so ( η ) on the basic force tensors of .where Ω αβ = ∂ α C β − ∂ β C α + C α C β − C β C α is the curvature matrix of thetotal connection C α = ˆΓ α + θ α , with θ ∗ the force tensor. We also computethe force field equations obtained by varying the Lagrangian density L = L K + τ L V , (18)with L K as in equation (1) and τ a constant, with respect to the force fieldcomponents. Again we assume that g = η , the Minkowski metric, so that C α = θ α .With the exception of the scalar field, all computations were performedwith the assistance of a symbolic algebra package. The potential energy of the force tensor U ∗ in equation (10), associated withthe one–dimensional irreducible summand , is found to be L V = 96 ( ∂ α φ )( ∂ α φ ) − φ Variation of the Lagrangian density (18) with respect to the scalar field φ then gives the field equation ∂ α ∂ α φ + 16 φ = τ ψ T ˆ ηγ α { α } ψ (19)for some constant τ . 14 .2 Vector field The potential energy term in the Lagrangian for the force tensor F ∗ on ζ given in equation (12) is L V = − A α A α ) cos 4 ζ + 256( A α A α ∂ β A β − A α A β ∂ α A β ) cos 3 ζ − ∂ α A β )( ∂ α A β ) + ( ∂ α A α ) ] cos 2 ζ + 32 ǫ αβρσ ( ∂ α A β )( ∂ ρ A σ ) sin 2 ζ Here ǫ αβρσ is the totally antisymmetric tensor of rank four. The field equationfor the four–vector field A α is then( ∂ β ∂ β A α + ∂ α ∂ β A β ) cos 2 ζ − A β ∂ α A β − A α ∂ β A β ) cos 3 ζ − A β A β ) A α cos 4 ζ = − τ ψ T ˆ ηγ β e ζJ ( v α ) β where v α is the basic force tensor from section 3.4, and ( v α ) β is the β –component matrix of v α In this case, the potential energy term of the Lagrangian for the connection H ∗ in (14) is L V = − W αβ W αβ ) + 4( ǫ αβρσ W αβ W ρσ ) − ∂ α W ρσ )( ∂ α W ρσ ) − ∂ β W αβ )( ∂ σ W ασ )The field equation for the antisymmetric tensor field W αβ is ∂ µ ∂ µ W αβ − ∂ α ∂ σ W σβ + ∂ β ∂ σ W σα − W ρσ W ρσ ) W αβ + ( ǫ µνρσ W µν W ρσ ) ǫ αβκλ W κλ = − τ ψ T ˆ ηγ σ ( v αβ − v βα ) σ ψ where ( v αβ ) σ is the σ –component matrix of the basic force tensor v αβ inequation (13). For a symmetric traceless tensor field S αβ , the potential energy term for theLagrangian is L V = − S αβ S αβ ) + 256 det( S ) + 32( ∂ α S βρ )( ∂ α S βρ ) − ∂ α S βρ )( ∂ β S αρ )15here det( S ) is the determinant of the field S αβ as a 4 × S αβ , we needto take into the constraint S = S − S − S . The result is ξ + ξ = χ , ξ = χ , ξ = χ ,ξ = χ , ξ − ξ = χ , ξ = χ ,ξ = χ , ξ − ξ = χ , ξ = χ where ξ αβ . = ∂ µ ∂ µ S αβ − ∂ α ∂ µ S µβ − ∂ β ∂ µ S µα + 2( S ρσ S ρσ ) S αβ − S ) αβ + cof( S ) βα ] χ αβ . = τ ψ T ˆ ηγ ρ ( u αβ + u βα ) ρ ψ Here, cof( S ) is the cofactor matrix of S , and u αβ are the basic force tensorsin equation (15). In this case, the potential energy term of the Lagrangian is given by L V = − F αρσ F αρσ ) + 2( ǫ ρσµν F αρσ F αµν ) + 16( ∂ β F αρσ )( ∂ α F βρσ ) − ∂ β F αρσ )( ∂ β F αρσ ) + 128( ∂ β F αρσ ) F αρµ F βσµ where F αρσ is a Fierz tensor; i.e., a tensor that satisfies equation (17). Con-strained variation of the total Lagrangian yields ξ + ξ = χ , ξ + ξ = χ , ξ + ξ = χ , ξ + ξ = χ ,ξ − ξ = χ , ξ − ξ = χ , ξ + ξ = χ , ξ − ξ = χ ,ξ + ξ = χ , ξ + ξ = χ , ξ − ξ = χ , ξ − ξ = χ ,ξ − ξ = χ , ξ − ξ = χ , ξ − ξ = χ , ξ − ξ = χ , where ξ αρσ . = ∂ β ∂ β F αρσ − ∂ α ∂ β F βρσ + 4 F βνσ ∂ β F αρν − F βνρ ∂ β F ασν + 2 F κλσ ∂ α F κλρ − F κλρ ∂ α F κλσ + 2 F ασκ ∂ β F βρκ − F αρκ ∂ β F βσκ − ( F βµν F βµν ) F αρσ + ( ǫ κλµν F βκλ F β µν ) ǫ ρσθφ F αθφ χ αρσ . = − τ ψ T ˆ γ ν ( u αρσ − u ασρ ) ν ψ Here, u αρσ are the basic force tensors in equation (16).16 eferences [1] H. I. Arcos, C. S. O. Mayor, G. Ot´alora, and J. G. Pereira, Spin–2 fieldsand helicity,
Found. Phys. (2012) 42: 1339–1349.[2] William Fulton and Joe Harris,
Representation Theory: A First Course,
Graduate Texts in Mathematics, Springer, New York, 1991.[3] jason hanson,