Antisymmetric tensor fields in a generalized Randall-Sundrum scenario
aa r X i v : . [ h e p - t h ] O c t Antisymmetric tensor fields in a generalized Randall-Sundrum scenario
Ashmita Das ∗ and Soumitra SenGupta † Department of Theoretical PhysicsIndian Association for the Cultivation of Science,Kolkata - 700 032, India
Bulk antisymmetric tensor fields of different ranks have been studied in the context of a generalizedRandall-Sundrum model with a non-vanishing induced cosmological constant on the visible brane.It is shown that instead of the usual exponential suppression of the couplings of the zero modes ofthese bulk fields with the brane fermions in the original Randal-Sundrum model , here the couplingsare proportional to the brane cosmological constant. Thus in an era of large cosmological constantthese fields have significant role in physical phenomena because of their enhanced couplings withthe visible brane fermions.
Large hierarchy of mass scales between the Planck and the Tev scales results into the well-known fine tuningproblem in connection with Higgs the only scalar particle in the standard model. It has been shown that due tolarge radiative corrections the Higgs mass can not be confined within Tev unless some unnatural tuning is done orderby order in the perturbation theory. The two most successful efforts to resolve this crisis are supersymmetry[1, 2]and/or extra dimensional generalization of standard model both of which lead us to the Physics beyond standardmodel ( BSM )[3, 4]. Among various extra dimensional models, the warped geometry model proposed by Randalland Sundrum[4] has drawn special attention for the following reasons : (1) It resolves the hierarchy problem withoutintroducing any other hierarchial scale in the theory, (2) The modulus of the extra dimensional model can bestabilized [5], (3) It provides interesting new phenomenology which can be tested in the Tev scale collider experimentssay in LHC and (4) A warped solution , though not exactly same as RS model, can be found from string theorywhich as a fundamental theory predicts inevitable existence of extra dimensions [6].Randall-Sundrum scenario [4] which is defined on a 5-dimensional anti de-Sitter space-time with one spatial directionorbifolded on S /Z has the following features : • Two 3-branes namely hidden/Planck brane and visible/standard model brane are located at the two orbifoldfixed points. • The effective cosmological constant induced on the hidden and visible brane are zero i.e these are flat branes. • The brane tension of the standard model/visible brane is negative. • Without introducing any extra scale, other than the Planck scale, in the theory one can choose the braneseparation modulus r c to have a value M − P such that the desired warping can be obtained between the twobranes from Planck scale to Tev scale. • The modulus can be stabilized to the above chosen value by introducing scalar in the bulk [5] without anyfurther fine tuning.In this model it is assumed that all the standard model fields are confined on the visible brane while the gravitypropagates in the bulk. The main motivation behind this assumption has its root in string theory where the SMfields are open string modes whose end points are fixed on the brane while gravity being a closed string modecan reside in the bulk. Following this argument all the antisymmetric tensor string modes of various ranks arealso expected to propagate in the bulk. Despite having similar coupling with brane matter just as graviton modenone of these antisymmetric tensor modes so far has been detected through any experimental signature. As an ∗ Electronic address: [email protected] † Electronic address : [email protected] explanation of their invisibility it has been shown that in a warped geometry model all the antisymmetric modes oftwo or higher ranks are suppressed by successive higher powers of the exponential warp factor on the visible brane [7–9].Various phenomenological as well as cosmological implications of RS model have been discussed in severalworks[10–21]. Meanwhile RS model has been generalized [22] such that the visible 3-brane can either be de-Siter oranti-de-Sitter with positive or negative induced cosmological constant. Such models not only can resolve the gaugehierarchy problem but also may render stability to the visible brane which now can be endowed with positive tension.This work was motivated by the facts that the zero cosmological constant of the visible 3-brane is not consistent withthe observed small value of the cosmological constant of our Universe and negative tension branes are intrinsically un-stable. Moreover the possibility of a de-Sitter universe by antisymmetric tensor flux compactification has been shownin the context of string inspired supergravity models [23]. Such models with flux and branes are known to have ageneric warped geometric structure. These lead us to explore the correlation between a non-vanishing 3-brane cosmo-logical constant and the antisymmetric tensor fields on the brane. From a different viewpoint the connection betweenthe cosmological constant and background space-time torsion has been studied [24]. The third rank field strengthcorresponding to the second rank anti-symmetric closed string mode namely the Kalb-Ramond field can be identifiedwith space-time torsion [25]. This field has been shown to have a highly suppressed coupling to the standard modelfields in a warped geometry model on a flat 3-brane. It is therefore important to explore whether such suppressionleading to an illusion of a vanishing torsion persists even when the space-time has non-vanishing cosmological constant.We first briefly outline the generalized RS model below.The warp factor in such a model is obtained by extremising the following the action : S = Z d x √− G ( M R −
Λ) + Z d x √− g i V i (0.1)where Λ is the bulk cosmological constant, R is the bulk (5-dimensional) Ricci scalar and V i is the tension of the i th brane ( i = hid ( vis ) for the hidden (visible) brane). It is shown that a warped geometry results from a constantcurvature brane space-time, as opposed to a flat 3-brane space-time. The generalized ansatz for the warped metric isgiven by, ds = e − A ( y ) g µν dx µ dx ν + r dy (0.2)where r corresponds to the modulus associated with the extra dimension and µ, ν stands for brane coordinate indices.As in the original RS model, the scalar mass warping is achieved through the warp factor e − A ( krπ ) = mm = 10 − n where k = q − Λ12 M ∼ Planck Mass with the bulk cosmological constant Λ is chosen to be negative. ‘ n ’ the warpfactor index must be set to 16 to achieve the desired warping and the magnitude of the induced cosmological constanton the brane in this case is non-vanishing in general and is given by =10 − N (in planckian units). A careful analysisreveals that for negative brane cosmological constant N has minimum value given by N min = 2 n leading to an upperbound on the magnitude of the cosmological constant while there is no such upper bound for the induced positivecosmological constant on the brane. Furthermore for the induced brane cosmological constant, Ω > < g µν corresponds to some de-sitter or anti de-Sitter space-time say for example dS-Schwarzschild andAdS-Schwarzschild space-times respectively [26–29].For AdS bulk i.e. Λ <
0, considering the regime for which the induced cosmological constant Ω on the visible braneis negative if one redefines ω ≡ − Ω / k ≥
0, then the following solution for the warp factor is obtained : e − A ( y ) = ω cosh (cid:16) ln ωc + ky (cid:17) (0.3)where c = 1 + √ − ω for the warp factor normalized to unity at y = 0.Similarly when the induced brane cosmological constant is positive i.e the 3-brane is de-Sitter with Ω >
0, the warpfactor turns out to be, e − A ( φ ) = ω sinh | (ln c ω − kr c φ ) | (0.4)where , c = 1 + p ω , ω = Ω3 k In order to resolve the gauge hierarchy problem the warp factor e − A ( φ ) must be equal to 10 − at φ = π and thisimplies that for both anti de-Sitter and de-Sitter branes the values of kr depend on the values of the cosmologicalconstant ω . The RS solution namely kr ∼ . ω = 0 is just one solution in theplot of solutions in figure 1. Apart from the bulk graviton various other bulk fields like scalars, gauge and fermions - N I IIA IIIB
DS ADS
FIG. 1: Graph of N versus x = krπ = 36 −
40, for n = 16 and for both positive and negative brane cosmological constant. Thecurve in region-I corresponds to positive cosmological constant on the brane, whereas the curve in regions-II & III representsnegative cosmological constant on the brane. fields have been considered in different work to obtain their massless as well as massive KK towers on the visiblebrane [30, 31]. However in the context of string theory where many higher rank anti-symmetric tensor fields appearas closed string modes [32], the study of bulk fields have been widened to include these fields also. One such fieldnamely the second rank antisymmetric tensor field ( called Kalb-Ramond field )[33] with third rank antisymmetricfield strength can be viewed upon as the torsion field in the background space-time. The apparent torsion-freeuniverse implies that such field, if exists, must be heavily suppressed on the visible brane. It was then shown thata Randall-Sundrum warped geometry model can indeed explain such a suppression of this field on the visible branecompared to the graviton through the large exponential warping which appears in the space-time metric. Inspiredby this result the authors of [7] then extended this calculation for even higher rank antisymmetric tensor fields whichinevitably appear in string-based models. It turned out that all such higher rank fields are even more suppressed byhigher powers of the warp factors [8] and Randall-Sundrum model thus can explain the apparent invisibility of allthese antisymmetric tensor fields in our universe.Here, we propose to re-examine this feature in the context of generalized RS model described earlier. Ourprime concern is to find out possible modifications in the projections of the antisymmetric tensor fields on the(3+1)dimensional brane due to the inclusion of cosmological constant induced on the 3-brane.We organize our work as follows : • To determine the massless and the massive KK-modes of the two-form KR field and also of the higher rankantisymmetric tensor field for anti-deSitter visible brane • To repeat the same calculation for the deSitter visible brane. • We also determine the coupling of KR field as well as the higher rank antisymmetric tensor fields to the fermionfields localized on the visible brane for both de-Sitter and anti de-sitter branes to look for their possinle presencethrough interactions with brane fermions.We consider space-time with torsion in a generalized Randall-Sundrum scenario,that is with a cosmological constantΩ induced on the visible brane. We recall from our previous discussions that the warp metric is, ds = e − A ( φ ) g µν dx µ dx ν + r c d φ (0.5)where , e − A ( φ ) = ω cosh(ln ωc + kr c φ )and , c = 1 + p − ω , ω = − Ω3 k Modes of the Antisymmetric tensor fields in anti de-Sitter 3-brane
We begin with the second rank antisymmetric tensor field ( namely the Kalb-Ramond field) with a third rank tensorfield strength. As mentioned earlier we can identify the background space-time torsion with the rank 3 antisymmetricfield strength tensor H MNL corresponding to the second rank anti-symmetric Kalb-Ramond tensor field B MN whichare related as H MNL = ∂ [ M B NL ] . The KR gauge invariance allows us to set B µ = 0.Using the explicit form of generalized RS metric and keeping B µ = 0 the action is given as, S H = Z d x Z dφr c e A ( φ ) [ η µα η νβ η λγ H µνλ H αβγ − r c e − A ( φ ) η µα η νβ B µν ∂ φ B αβ ] (0.6)Using the Kaluza-Klein decomposition for the KR field: B µν ( x, φ ) = ∞ X n =0 B nµν ( x ) χ n ( φ ) √ r c (0.7)the effective four dimensional action becomes S H = Z d x ∞ X n =0 [ η µα η νβ η λγ H nµνλ H nαβγ + 3 m n η µα η νβ B nµν B nαβ ] (0.8)provided − r c d χ n dφ = m n χ n e A ( φ ) (0.9)along with the orthonormality condition Z e A ( φ ) χ m ( φ ) χ n ( φ ) dφ = δ mn (0.10)In terms of z n = m n k e A ( φ ) equation (0.9) can be recast in the form z n d χ n dz n + z n dχ n dz n − a z n (1 − a z n ) dχ n dz n + z n χ n (1 − a z n ) = 0 (0.11)where, a = k ω m n Keeping the leading order terms we get,[ z n d dz n + z n ddz n + z n (1 + a z n )] χ n = 0 (0.12)The solution of the above equation can be written as, χ n ( φ ) = 1 N n [ J ( z n ) + α n Y ( z n ) + ω ξ n ] (0.13)From continuity condition at φ = 0 we obtain , α n ≃ x n e − A ( π ) where x n = z n ( π ). This implies α n <<
1. Also at φ = π we get, J ( x n ) = π x n e − A ( π ) .The differential equation for ξ n now becomes, (cid:20) z n d dz n + z n ddz n + z n (cid:21) ξ n + k m n z n J ( z n ) = 0 (0.14)To the leading order J ( z n ) = and the above differential equation becomes (cid:20) z n d dz n + z n ddz n + z n (cid:21) ξ n + 12 k m n z n = 0 (0.15)We therefore obtain the solution for ξ n as, ξ n = 14 k m n [2 πz n J ( z n ) Y ( z n ) − πz n J ( z n ) J ( z n ) Y ( z n ) − πz n J ( z n ) Y ( z n ) + πz n J ( z n ) Y ( z n ) − πz n J ( z n ) Y ( z n )+ πz n J ( z n ) Y ( z n ) + 4 πz n J ( z n ) Y ( z n )] (0.16)Using the orthonormality condition (0.10) and performing the numerical integration we find, N n = 1 √ kr c km n (cid:20) . − k ω m n . (cid:21) (0.17)The final solution for the massive modes turns out to be, χ n ( z n ) = p kr c m n k (cid:20) . − k ω m n . (cid:21) − [ J ( z n ) + ω ξ n ] (0.18)We now turn our attention to massless mode. The differential equation for the massless mode is,1 r c d χ n dφ = 0 (0.19)The solution of the above equation is, χ ( φ ) = c φ + c (0.20)Applying the continuity condition we find c = 0.Hence, we get χ = c . Now using orthonormality condition, Z π e A ( φ ) c dφ = 1 (0.21)We obtain c = 2 kr c e − kr c π h ω (1 + e kr c π ) i . Plugging in the solution of c we finally arrive at the expression for χ as, χ = √ p kr c e − kr c π (cid:20) ω e kr c π ) (cid:21) (0.22)This the solution for the massless KR mode on the brane.It may be observed that both the massless and the massive KR mode depend on the induced brane cosmologicalconstant ω . For ω = 0 i.e when the warped geometry corresponds to RS model, we retrieve the result that the KRfield is heavily suppressed on the visible brane as obtained in [7].Let us now generalize the above analysis and consider the bulk antisymmetric tensor field of higher rank i.e a rank-3tensor, X MNA , with the corresponding field strength tensor Y MNAB . S = Z d x √− GY MNAB Y MNAB (0.23)G is the determinant of the 5 dimensional metric.In general one should be able to write down a rank-( n + 1) antisym-metric tensor field strength tensor as, Y a a ....a n +1 = ∂ [ a n +1 X a a ....a n ] (0.24)Using the explicit form of the generalized RS metric and using the gauge fixing condition i.e X µνy = 0 one obtains, S x = Z d x Z dφ [ e A ( φ ) η µλ η νρ η αγ η βδ Y µναβ Y λργδ + 4 e A ( φ ) r c η µλ η αδ η νρ ∂ φ X µνα ∂ φ X λργ ] (0.25)Considering the KK decomposition of the field X, X µνα ( x, φ ) = ∞ X n =0 X nµνα ( x ) χ n ( φ ) √ r c (0.26)an effective action can be obtained for the projection X µνα on the visible brain, S x = Z d x X n [ η µλ η νρ η αγ η βδ Y nµναβ Y nλργδ + 4 m n η µλ η νρ η αδ X nµνα X nλργ ] (0.27)where m n is defined through the relation, − r c ddφ ( e A ( φ ) ddφ χ n ) = m n χ n e A ( φ ) (0.28) χ n satisfies the orthonormality condition, Z e A ( φ ) χ m ( φ ) χ n ( φ ) dφ = δ mn (0.29)Introducing f n = e A ( φ ) χ n equation (0.28) can be recast in the form, (cid:20) z n d f n dz n + z n df n dz n + f n z n (1 − a z n ) − f n + f n a z n (1 − a z n ) − z n a df n dz n (cid:21) = 0 (0.30)where, z n = m n k e A ( φ ) and a = k ω m n . Ignoring the last term in comparison to the term containing z n we find, (cid:20) z n d dz n + z n ddz n + z n (1 − a z n ) − a z n (1 − a z n ) (cid:21) f n = 0 (0.31)The solution of the above equation can be written as, f n = 1 N n [ J ( z n ) + α n Y ( z n ) + ω ξ n ] χ n = e − A ( φ ) f n = e − A ( φ ) N n (cid:2) J ( z n ) + α n Y ( z n ) + ω ξ n (cid:3) (0.32)We can reduce the equation (0.31) into the differential equation for ξ n as, z n d ξ n dz n + z n dξ n dz n + [ z n (1 + a ) − ξ n = 0 (0.33)The solution for ξ n turns out to be, ξ n = α n Y ( p a z n ) − J ( − p a z n ) χ n = e − A ( φ ) N n [ J ( z n ) + α n Y ( z n ) + ω { α n Y ( p a z n ) − J ( − p a z n ) } ] (0.34)The desired mass value of M n on the visible brane should be of the order of the Tev scale ( << k ). Using the continuitycondition at φ = 0 and noting that e kr c π >> α n = − J h m n k (cid:16) − ω (cid:17)i + ω m n k Y (cid:2) m n k (cid:0) − ω (cid:1)(cid:3) − k ω πm n (cid:18) − ω (cid:18) − k m n (cid:19)(cid:19) (0.35)Estimating the order of α n , we get α n <<
1. We therefore can write, χ n = e − A ( φ ) N n h J ( z n ) − ω J ( − p a z n ) i (0.36)Again from the continuity condition at φ = π we find, J ( x n ) + ω J ( p a x n ) = 0 (0.37)Here, x n = z n ( π ) = m n k e A ( π ) . Once again from orthonormality condition (0.29) ,and performing the integrationnumerically , we can have the expression for N n as, N n = km n √ kr c (cid:2) .
136 + ω . (cid:3) (0.38) χ n = e − A ( φ ) m n k p kr c (cid:2) .
136 + ω . (cid:3) − h J ( z n ) − J ( − p a z n ) i (0.39)For massless mode the differential equation becomes,1 r c ddφ (cid:18) e A ( φ ) ddφ χ n (cid:19) = 0 (0.40)Solving the above differential equation we derive the solution for massless mode; χ = c (cid:20) ω e kr c φ kr c − e − kr c φ kr c + ω φ (cid:21) + c (0.41)Applying continuity condition, we find ( just as in previous case ), c = 0 and χ = c . This leads to χ = c . Finallyapplying orthonormality condition, Z π e A ( φ ) c dφ = 1 (0.42)we arrive at c = 4 kr c e − kr c π (cid:0) ω e kr c π (cid:1) which on substitution yields the final expression for χ as, χ = 2 p kr c e − kr c π (cid:18) ω e kr c π (cid:19) (0.43)The massless as well as massive modes thus depend on the induced brane cosmological constant.The masses for the various order KK modes as well as massless mode can be determined from zeros of the Besselfunction in the continuity condition at φ = π . It is interesting to observe ( say from Equ.0.37) that these values allare in the Tev range and do not change significantly for a very wide range of the values of the cosmological constant ω say over a range of 1 ≤ ω ≥ − . We present below the the masses of various modes of both KR and higherrank antisymmetric tensor field (see Table I). n m torn ( for KR field )( T eV ) 3.75 7.015 10.173 13.323 m n ( for higher rank tensor field )( T eV ) 5.135 8.417 11.619 14.796TABLE I: Table of mass modes on anti de-Sitter 3-brane
Coupling with brane fermions
Let us consider the interaction of both massless and the massive modes of the antisymmetric tensor fields withspin- fermions on the visible brane. Starting from the 5-dimensional action and remembering that the fermion andits interactions are confined to the brane at φ = π we evaluate the coupling of the KR field strength with the branefermions. The fermion action is given as, S ψ = i Z d x Z dφ [ detV ] ψ [ γ c v µa ( ∂ µ − i G LN σ ab v νa ∂ µ v λb δ Nν δ Lλ − G AD σ ab v βa v δb Γ AMB δ Mµ δ Bβ δ Dδ )] ψδ ( φ − π ) (0.44)where G MN is given by , G MN = v aM v bN η ab (0.45)and the vierbein v aµ is, v = 1; v aµ = e − A ( φ ) δ aµ ; detV = e − A ( φ ) (0.46)a,b etc. being the tangent space indices.Integrating out the compact dimension and using the fact that the fermion field on the brane is consistently renor-malized as ψ → e A ( π ) / ψ , one obtains the effective 4-dimensional fermion KR interaction as, L ψψH = iψγ µ σ νλ [ 1 M P e kr c π (cid:26) ω (cid:0) e kr c π (cid:1)(cid:27) H µνλ +(1 . m n M P k (cid:18) k ω m n . (cid:19) (cid:0) J ( x n ) + ω ξ n (cid:1) ∞ X n =1 H nµνλ ] ψ (0.47)where , H nµνλ = ∂ [ µ B nνλ ] Substituting the leading order approximation of ξ n from eq.(0.16) in the eq.(0.47)We obtain, L ψψH = iψγ µ σ νλ [ 1 M P e kr c π (cid:26) ω (cid:0) e kr c π (cid:1)(cid:27) H µνλ +(1 . m n M P k (cid:18) k ω m n . (cid:19) (cid:18) J ( x n ) − ω π ke Aπ m n J ( x n ) J ( x n ) Y ( x n ) (cid:19) ∞ X n =1 H nµνλ ] ψ (0.48)The leading order coupling of massless KR field to the brane fermion now becomes ∼ M p e − krπ + ωM p whereas theleading order coupling of massive KR fields to the brane fermion is ∼ e A ( π ) M p + ω e A ( π ) M p .In the limit ω = 0 we retrieve the expressions of the coupling as obtained in the flat brane scenario. Though thecorrections to the couplings indeed depend on the brane cosmological constant but due to the upper-bound ( ∼ − ) on the magnitude of the induced brane cosmological constant in the anti- de-Sitter brane, both these correctionsare vanishingly small on the visible brane. This implies that the brane cosmological constant on the anti-de Sitterbrane does not modify the result significantly from that in the flat 3-brane case and the massless mode again hasextremely weak coupling whereas the massive modes have inverse Tev coupling. Thus the massless KR mode whichcan be identified with background space-time torsion still remains invisible in an anti-de Sitter warped geometry model.Next we take up the coupling of higher rank field strength tensor to the brane fermion located on the Ads brane.Here we consider the rank-3 antisymmetric tensor field X MNA with the corresponding rank four field strength tensor Y MNAB . Proceeding similarly as in case of the KR field, the final expression for the coupling becomes, L ψψY = iψγ µ Σ νλβ [ e A ( π ) M p e kr c π (cid:26) ω e kr c π (cid:27) Y µνλβ + m n M p k (2 . − . ω )( J ( x n ) − ω J ( − p a x n )) ∞ X n =1 Y nµνλβ ] ψ (0.49)where , a = k ω m n It is evident fom the above expression that in absence of the cosmological constant i.e for a flat brane in RS scenariothe couplings for both the massless and massive modes are heavily suppressed. The leading order correction to thecoupling term for the massless higher rank tensor field to the brane fermion is now ∼ e A ( π ) ωM p e krcπ , while that for themassive higher rank tensor fields can be written as ∼ m n ω M p k . Both these corrections once again are very tiny due tothe upper-bound of 10 − on the value of the brane cosmological constant in the anti de-Sitter case.Thus we conclude that for anti de-Sitter brane their is not much change in the scenario of the presence for theantisymmetric tensor fields on the anti de-Sitter brane from that in flat brane. All the massless and massive modesare heavily suppressed except the massive modes for the rank two KR fields which has an inverse Tev coupling withthe brane fermions.We now shift our attention to the de-Sitter 3-brane solution in the generalized RS model described earlier andexamine the presence of various rank antisymmetric tensor fields on the visible brane. Our result reveals a drasticchange of scenario for the de-Sitter brane from that in an anti de-Sitter brane. Modes of the Antisymmetric tensor fields in de-Sitter 3-brane
Considering the induced brane cosmological constant on the visible 3-brane to be positive i.e Ω > e − A ( φ ) = ω sinh | (ln c ω − kr c φ ) | (0.50)where , c = 1 + p ω , ω = Ω3 k Unlike the Ads scenario, the induced cosmological constant in this case has no bound and the warp factor beingdifferent from the Ads scenario, the perturbed solution ξ n changes. Repeating the same procedure as has been donefor Ads brane , we can recast the equation (0.9) for χ n in terms of z n = m n k e A ( φ ) as, z n d χ n dz n + z n dχ n dz n − a z n (1 + a z n ) dχ n dz n + z n χ n (1 + a z n ) = 0 (0.51)where, a = k ω m n As the third term is small compared to z n , we obtain[ z n d dz n + z n ddz n + z n (1 − a z n )] χ n = 0 (0.52)The solution of the above equation can be written as, χ n ( φ ) = 1 N n [ J ( z n ) + α n Y ( z n ) + ω ξ n ] (0.53)From continuity condition at φ = 0 we obtain , α n ≃ x n e − A ( π ) where we have used x n = z n ( π ).This implies that α n << φ = π we get, J ( x n ) = π x n e − A ( π ) ξ n now becomes, (cid:20) z n d dz n + z n ddz n + z n (cid:21) ξ n − k m n z n J ( z n ) = 0 (0.54)To the leading order J ( z n ) = the above differential equation becomes (cid:20) z n d dz n + z n ddz n + z n (cid:21) ξ n − k m n z n = 0 (0.55)The solution for ξ n is given as, ξ n = 14 k m n [ − πz n J ( z n ) Y ( z n ) + 4 πz n J ( z n ) J ( z n ) Y ( z n ) + 2 πz n J ( z n ) Y ( z n ) − πz n J ( z n ) Y ( z n ) + 4 πz n J ( z n ) Y ( z n ) − πz n J ( z n ) Y ( z n ) − πz n J ( z n ) Y ( z n )] (0.56)Using the orthonormality condition (0.10) and doing the numerical integration we finally arrive at the expression for N n as, N n = km n √ kr c (cid:20) .
714 + k ω m n . (cid:21) (0.57)The final solution for the massive modes turns out to be, χ n ( z n ) = p kr c m n k (cid:20) .
714 + k ω m n . (cid:21) − [ J ( z n ) + ω ξ n ] (0.58)To examine the presence of the massless mode of the antisymmetric tensor field on the visible brane , we turn ourattention to the differential equation for the massless mode which is now given as,1 r c d χ n dφ = 0 (0.59)The solution for the massless mode is obtained as, χ ( φ ) = c φ + c (0.60)Here applying continuity condition once again we get, c = 0. Use of the orthonormality condition yields , c = | kre − kr c π [1 − ω e kr c π )] | (0.61)Putting the solution of c in the final expression for χ , we find, χ = √ p kr c e − kr c π | (cid:20) − ω e kr c π ) (cid:21) | (0.62)This is the solution for the massless KR field on a de-Sitter 3-brane.Let us now turn our attention to the bulk antisymmetric tensor field of higher rank i.e a rank-3 tensor, X MNA ,with the corresponding field strength tensor Y MNAB . Following the procedure described so far and introducing f n = e A ( φ ) χ n , the equation (0.28) in terms of z n = m n k e A ( φ ) can be recast as, (cid:20) z n d f n dz n + z n df n dz n + f n z n (1 + a z n ) − f n − f n a z n (1 + a z n ) (cid:21) = 0 (0.63)where, a = k ω m n Now we can reduce the equation (0.63) in terms of the perturbed solution ξ n z n d ξ n dz n + z n dξ n dz n + [ z n (1 − a ) − ξ n = 0 (0.64)1The solution for ξ n turns out to be, ξ n = α n Y ( p − a z n ) − J ( − p − a z n ) χ n = e − A ( φ ) N n [ J ( z n ) + α n Y ( z n ) + ω { α n Y ( p − a z n ) − J ( − p − a z n ) } ] (0.65)Applying the continuity condition at φ = 0 we again arrive at α n <<
1. This leads to, χ n = e − A ( φ ) N n h J ( z n ) − ω J ( − p − a z n ) i (0.66)The continuity condition at φ = π yields, J ( x n ) + p − a ω J ( p − a x n ) = 0 (0.67)Using equation (0.29) i.e the orthonormality condition and performing the numerical integration we get the expressionfor N n , N n = (0 . km n √ kr c (0.68) χ n = e − A ( φ ) m n k p kr c (2 . h J ( z n ) − J ( − p − a z n ) } i (0.69)This is the solution for the massive modes on the de-Sitter brane.For the massless mode the solution is, χ = c (cid:20) ω e kr c φ kr c − e − kr c φ kr c − ω φ (cid:21) + c (0.70)Applying continuity condition we again find, c = 0 and χ = c Furthermore the orthonormality condition yields, c = | e − kr c π [1 − ω e kr c π ] | (0.71)Putting the solution of c ,we get the final expression for χ χ = 2 p kr c e − kr c π | (cid:18) − ω e kr c π (cid:19) | (0.72)Just as in anti de-Sitter case here also we determine various masses from the continuity condition at φ = π andestimating the zeros of the Besel function. These are depicted in Table II. n m torn ( for KR field )( T eV ) 3.726 6.996 10.17 13.27 m n ( for higher rank tensor field )( T eV ) 5.106 8.418 11.55 14.79TABLE II: Table of mass modes on de-Sitter 3-brane
Coupling with brane fermions
We now consider the coupling of torsion as well as the higher rank antisymmetric tensor field to the matter fieldson the de-Sitter visible brane.2The warp factor for the de-Sitter case is e − A ( φ ) = ω sinh(ln c ω − kr c φ )where , c = 1 + p ω Performing the same calculation as in the case of Ads brane for KR field , we get the expression for the coupling oftorsion to the fermion ,residing on the visible brane as, L ψψH = iψγ µ σ νλ [ 1 M P e kr c π (cid:26) − ω (cid:0) e kr c π (cid:1)(cid:27) H µνλ + m n M P k m n ωk . (cid:0) J ( x n ) + ω ξ n (cid:1) ∞ X n =1 H nµνλ ] ψ (0.73)Substituting the leading order approximation of ξ n from the eq.(0.56) in the eq.(0.73) L ψψH = iψγ µ σ νλ [ 1 M P e kr c π (cid:26) − ω (cid:0) e kr c π (cid:1)(cid:27) H µνλ + m n M P k ω . (cid:18) J ( x n ) + ω π ke Aπ m n J ( x n ) J ( x n ) Y ( x n (cid:19) ∞ X n =1 H nµνλ ] ψ (0.74)In this case the leading order coupling of massless KR field to the brane fermion is ∼ M p e krcπ and the leading ordercoupling of massive KR field to the brane fermion can be written as ∼ m n ωM p k e A ( π ) .It may be observed from fig.(1) that the value of ω rises very steeply with the decrease in the value of kr from thecorresponding RS value. It is given by the relation [22] e − krπ = 10 − [1 + √ ω ](1 + √ ω ) (0.75)Due to the decrease in the value of kr with increase in the value of the induced positive brane cosmological constantthere will be a region where both the above couplings ( for massless and massive modes with brane fermions ) becomestrong and can be comparable or larger than that of the gravity mode with the brane fermions.Proceeding similarly the coupling of the higher rank tensor field with the brane localized fermion can be determined.From the coupling term, L ψψY = iψγ µ Σ νλβ [ e A ( π ) M p e kr c π (cid:26) − ω e kr c π (cid:27) Y µνλβ +(2 . m n M p k ( J ( x n ) − ω J ( − p − a x n )) ∞ X n =1 Y nµνλβ ] ψ (0.76)where , a = k ω m n we can easily derive the leading order correction to the coupling term for the massless higher rank tensor field to thebrane fermion as ∼ e A ( π ) M p e krcπ For large ω , the exponential factor in the denominator can be small ( due to decrease in the value of kr where asthat in the numerator is ∼ . This leads to enhanced coupling for the massless modes.The leading order coupling term for the massive higher rank tensor fields however can be written as, ∼ m n ω M p k , whichis suppressed by an additional factor of k in the denominator and therefore is heavily suppressed. Conclusions
Bulk antisymmetric tensor fields of two and higher rank have been studied in a generalized Randall-Sundrum modelwith an induced cosmological constant on the visible 3-brane. We have considered both de-Sitter and anti de-Sitter3non flat 3-branes with an appropriate warp factor which can resolve the gauge hierarchy problem in connection withthe Higgs mass. The massless modes of the bulk antisymmetric fields which have vanishingly small coupling withfermion matter field on the visible brane in an usual RS scenario now acquires much larger coupling due to the presenceof non-vanishing cosmological constant on the 3-brane. It is shown that due to the constraints on the magnitude ofthe cosmological constant in an anti de-Sitter 3-brane in the generalized warped model these couplings continue to besmall. 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