Fermion Families from Two Layer Warped Extra Dimensions
aa r X i v : . [ h e p - ph ] A ug Preprint typeset in JHEP style - HYPER VERSION
Fermion Families from Two Layer Warped ExtraDimensions
Zhi-qiang Guo and Bo-Qiang Ma ∗ School of Physics and State Key Laboratory of Nuclear Physics and Technology, PekingUniversity, Beijing 100871, China
Abstract:
In extra dimensions, the quark and lepton mass hierarchy can be reproduced fromthe same order bulk mass parameters, and standard model fermion families can be generatedfrom one generation in the high dimensional space. We try to explain the origin of the sameorder bulk mass parameters and address the family replication puzzle simultaneously. Weshow that they correlate with each other. We construct models that families are generatedfrom extra dimensional space, and in the meantime the bulk mass parameters of same orderemerge naturally. The interesting point is that the bulk mass parameters, which are in sameorder, correspond to the eigenvalues of a Schr¨odinger-like equation. We also discuss the problemexisting in this approach.
Keywords:
Fermion Families, Warped Extra Dimensions, Mass Hierarchy. ∗ Email:[email protected] ontents
1. Introduction 12. The Setup 33. Examples of metric for finite generations 74. Further Discussions 135. Conclusions 15A. Explicit solutions for massive modes and zero mode: metric I 15B. Numerical examples for finite generations 17C. Property of hypergeometrical function 18D. Solutions for zero mode and massive modes: metric II 18E. Another metric example for orthogonality 19F. Ricci scalar curvature for metric 20
1. Introduction
In recent years, extra space dimensions have played an important role in particle physics,gravity and cosmology. Many old problems get elegant solutions in extra space dimensionalbackground with new perspectives [1, 2]. The extra space dimensions have been used toaddress the problems in flavor physics: why the masses of quarks and leptons distribute ina large range and have obvious hierarchy structure (the fermion mass hierarchy puzzle); whythe heavier generations replicate the lightest generation with the almost same properties (thefermion family replication puzzle). There have been very interesting progress that both ofthem get answers in several different approaches [3, 4, 5, 6, 7].In extra dimension, the fermion mass hierarchy has geometrical origin. It arises from thesmall overlap of wave functions in the extra space dimensions. The hierarchy mass structurecan be reproduced with the bulk mass parameters of the same order in 5-dimension warpedspace (Randall-Sundrum model) successfully (for several numerical examples, see [8]). How-ever, a new puzzle arises naturally: why the 5-dimension bulk mass parameters are of the same For definiteness, our following discussions will be based on the concise numerical example given in [9], andwe hope that our discussions can apply to other examples by some modification. – 1 –rder, or whether we can give an explanation to the origin of the same order 5-dimension bulkmass parameters. It is easy to understand that this new puzzle correlates closely with thefermion family replication problem. One bulk mass parameter stands for one fermion familyor one fermion flavor, so to explain the origin of the same order bulk mass parameter is doingthe same thing to explain the family replication problem.This new puzzle can be addressed in the approach that the three generations of standardmodel (SM) can be generated from one generation in the high dimensional space. This approachhas been adopted in several papers [3, 5], in which the 6-dimension spacetime is reduced to4-dimension spacetime directly. It is found that the SM families correspond to the zero modesof the high dimensional equation of motion. In these approaches, one missed the chance togive answers for the origin of the same order bulk mass parameters in 5-dimensions. In thepresent paper, we will adopt an alternative approach. We suggest a 6-dimension metric Ansatzof special two layer warped structure, ds = B ( z ) (cid:2) A ( y ) (cid:0) η µν dx µ dx ν + dy (cid:1) + dz (cid:3) . (1.1)As we will show below, with the help of the special structure of this two layer warped metric,we can reduce the 6-dimension (4+1)+1 spacetime to 5-dimension 4+1 spacetime at the firststep. We found that the induced 5-dimension equation of motion for fermions is similar tothat analyzed in [9], while the bulk mass parameter in this induced 5-dimension equationcorresponds to the eigenvalue of a 1-dimension Schr¨odinger-like equation, which is conformed bythe fermion wave function in the sixth space dimension. Yet, because the induced 5-dimensionequation of motion is similar to that analyzed in [9], we might expect that this induced 5-dimension equation can be applied in this model by some modification. Hence we reduce theproblem of the same order 5-dimension bulk mass parameter to the eigenvalue problem ofa second order differential equation. Because the eigenvalues of a 1-dimension Schr¨odinger-like equation are of the same order generally, we see that the same order 5-dimension massparameters emerge naturally. If we further reduce the 5-dimension spacetime to 4-dimensionspacetime, with the help of the model [9], the same order 5-dimension bulk mass parameterswill produce the hierarchy structure of the 4-dimension physical fermion mass. By this newapproach, we can address the new puzzle suggested above. We give an explanation for theorigin of the same order bulk mass parameters. Because one 5-dimension bulk mass parametercorresponds to one family, we provide an answer for the origin of the families simultaneously.However, there exist some problems in this approach, we will discuss these problems in detailand suggest methods to bypass them.According to the approach suggested above, we construct a model that the 5-dimensionbulk mass parameters of the same order emerge naturally, and hence the standard modelfamilies are generated simultaneously. In fact, we can show that the 5-dimension bulk massparameters correspond to the eigenvalues of a second order ordinary differential equation ofthe Sturm-Liouville type (similar to the Schrodinger equation in 1-dimension). Therefore,like the energy spectrum of Hydrogen atom, the bulk mass parameters or the families can begenerated from the eigenvalues of Schr¨odinger-like equation with proper boundary conditions.The details will be introduced in Sec. 2. Several examples are given in Sec. 3. We give furtherdiscussions and conclusions in Sec. 4 and Sec. 5.– 2 – . The Setup In this section, we introduce the setup in detail. We start with the action of a bulk Diracfermion in six dimension spacetime. The metric Ansatz for spacetime takes the form ds = B ( z ) (cid:2) A ( y ) (cid:0) η µν dx µ dx ν + dy (cid:1) + dz (cid:3) , (2.1)where η µν = diag( − , , , The bulk action for this fermion is given by the usual form, S = Z d xdydz √− g (cid:26) i (cid:2) ¯Ψ e Ma Γ a D M Ψ − D M ¯Ψ e Ma Γ a Ψ (cid:3) − i m ¯ΨΨ (cid:27) , (2.2)where e Ma is the sechsbien, and D M = ∂ M + ω abM Γ ab , Γ ab = [Γ a , Γ b ] is the covariant derivativeof spinor in curved spacetime. a and M = 0 , , , , , m to be a real number. We choose the gamma matrices representation as follows,Γ µ = γ µ γ µ ! , Γ = γ γ ! , Γ = − ! ,γ = − ! , γ i = σ i σ i ! , γ = − ! , i = 1 , , , (2.3)where µ = 0 , , , σ i are the usual Pauli matrices.By use of the metric Ansatz, the action reduces to S = S + S + S − Z d xdydz √− gi m ¯ΨΨ , (2.4) S = Z d xdydz √− g (cid:26) i B ( z ) − A ( y ) − (cid:2) ¯Ψ Γ µ ∂ µ Ψ − ∂ µ ¯ΨΓ µ Ψ (cid:3)(cid:27) ,S = Z d xdydz √− g (cid:26) i B ( z ) − A ( y ) − (cid:2) ¯Ψ Γ ∂ Ψ − ∂ ¯ΨΓ Ψ (cid:3)(cid:27) ,S = Z d xdydz √− g (cid:26) i B ( z ) − (cid:2) ¯Ψ Γ ∂ Ψ − ∂ ¯ΨΓ Ψ (cid:3)(cid:27) . Varying the action with respect to ¯Ψ, we obtain the equation of motion, A ( y ) − (cid:8) Γ µ ∂ µ Ψ + Γ (cid:0) ∂ + 2 A − A ′ (cid:1) Ψ (cid:9) + Γ (cid:18) ∂ + 52 B − ˙ B (cid:19) Ψ − m B Ψ = 0 , (2.5) The conventional way in extra dimensions is to suppose the extra dimensions as orbifolds, and that theboundary conditions are determined to be the Israel’s junction conditions [10], as adopted in [2]. However,as suggested in [11], the interval approach is more convenient in some cases, and it can produce more generalboundary conditions. In our work, we found that it is necessary to adopt the interval approach, at least whenwe deal with the boundary conditions in the sixth space dimension. Of course, it is also more convenient. Moreabout the interval approach, see [12]. – 3 –here A ′ = dA ( y ) dy , ˙ B = dB ( z ) dz , ∂ = ∂ y , ∂ = ∂ z , and the boundary term, δS bound = δS + δS , (2.6) δS = − i Z d xdz (cid:2) √− g B − A − δ ¯ΨΓ Ψ (cid:3) L ′ L ,δS = − i Z d xdy (cid:2) √− g B − δ ¯ΨΓ Ψ (cid:3) R ′ R , where we denote by [ X ] L the quantity X | L − X | . Denoting Ψ = χ χ ! , where χ and χ arefour-component Dirac spinors, we rewrite Eq. (2.5) as A ( y ) − (cid:8) γ µ ∂ µ χ + γ (cid:0) ∂ + 2 A − A ′ (cid:1) χ (cid:9) + (cid:18) ∂ + 52 B − ˙ B (cid:19) χ − m B χ = 0 , (2.7) A ( y ) − (cid:8) γ µ ∂ µ χ + γ (cid:0) ∂ + 2 A − A ′ (cid:1) χ (cid:9) + (cid:20) − (cid:18) ∂ + 52 B − ˙ B (cid:19)(cid:21) χ − m B χ = 0 . (2.8)Now we make the conventional Kluza-Klein (KK) decomposition. We expand χ and χ with spinor ψ ( x µ , y ) in 5-dimension spacetime as χ ( x µ , y, z ) = X n b F n ( z ) ψ n ( x µ , y ) , χ ( x µ , y, z ) = X n b G n ( z ) ψ n ( x µ , y ) , (2.9)in which ψ n ( x µ , y ) conforms to the Dirac equation in 5-dimension spacetime, A ( y ) − (cid:8) γ µ ∂ µ ψ n ( x µ , y ) + γ (cid:0) ∂ + 2 A − A ′ (cid:1) ψ n ( x µ , y ) (cid:9) − λ n ψ n ( x µ , y ) = 0 . (2.10)As in 6-dimension spacetime, the Dirac equation in 5-dimension spacetime requires λ n to bereal numbers. This can be verified by multiplying the two sides of Eq. (2.10) by ¯ ψ n ( x µ , y ).Here we note that it is critical that λ n must be real numbers, as it will be obvious in thefollowing discussions. With the help of Eq. (2.10), Eqs. (2.7)-(2.8) can be solved by thefollowing Ansatz, (cid:18) ddz + 52 B − ˙ B (cid:19) b F n ( z ) − mB b F n ( z ) + λ n b G n ( z ) = 0 , (2.11) (cid:18) ddz + 52 B − ˙ B (cid:19) b G n ( z ) + mB b G n ( z ) − λ n b F n ( z ) = 0 . (2.12)These equations can be simplified further by the transformations, b F n ( z ) = B ( z ) − ǫ F n ( z ) , b G n ( z ) = B ( z ) − ǫ G n ( z ) , (2.13)in which ǫ = . Then we obtain the equations (cid:18) ddz − mB (cid:19) F n ( z ) + λ n G n ( z ) = 0 , (2.14) (cid:18) ddz + mB (cid:19) G n ( z ) − λ n F n ( z ) = 0 . (2.15)– 4 –or a zero mode ( λ = 0), these bulk equations decouple and are easy to be solved. Thesolutions are given by F ( z ) = 1 √ lN exp (cid:18)Z zz mB ( ζ ) dζ (cid:19) or 0 , G ( z ) = 1 q l e N exp (cid:18) − Z zz mB ( ζ ) dζ (cid:19) or 0 . (2.16)We introduce l of the length dimension in order to make the normalization constants to bedimensionless. It will become explicit in examples in the next section. For the massive modes,we can combine the first order differential equation to obtain second order equations d dz F n ( z ) + h − m ˙ B − m B i F n ( z ) + λ n F n ( z ) = 0 , (2.17) d dz G n ( z ) + h m ˙ B − m B i G n ( z ) + λ n G n ( z ) = 0 . (2.18)Rewriting them in another form, we see that they are similar to the one dimensional Schr¨odingerequations − d dz F n ( z ) + V ( z ) F n ( z ) = λ n F n ( z ) , (2.19) − d dz G n ( z ) + e V ( z ) G n ( z ) = λ n G n ( z ) , (2.20)with potentials V ( z ) = m ˙ B + m B , e V ( z ) = − m ˙ B + m B . (2.21)In the following discussions we will illuminate that such a setup gives answers to the puzzlewe proposed in the introduction. According to the setup above, we realize an interesting factthat Eq. (2.10) is similar to the equation analyzed in the model [9] if we choose A ( y ) to be aslice of the anti-de Sitter (AdS) metric, i.e., the RS spacetime. The differences are that there isgauge field background in the model [9], and that in that work the extra dimension is adoptedto be an orbifold. However, for fermions, the equations of motion in these two cases are ofalmost similar features, and the gauge field background only makes the boundary conditionsmore involved. It is not difficult to add the gauge field background as in the model [9] to theabove setup. When we further reduce the 5-dimension spacetime to 4-dimension spacetime, wecan advance with the help of the model [9]. λ n in the above setup corresponds to the bulk massparameter in 5-dimension spacetime in the model [9]. Of course, it is obvious that one bulkmass parameter corresponds to one family in 5-dimensions. Now we can understand how thesame order bulk mass parameters or families are generated from extra space dimensions. Thebulk mass parameters are eigenvalues of Schr¨odinger-like equations (2.19)-(2.20), so generallythey should be of the same order. The eigenstates of equations (2.19)-(2.20), belonging tothe eigenvalues λ n , correspond to the generations in 5-dimensions. When we reduce furtherthe 5-dimension spacetime to the physical 4-dimensions, these generations in 5-dimensions canproduce the generations in 4-dimension spacetime. So the families in physical 4-dimensionspacetime are generated simultaneously. Of course, the eigenstates should be normalizable in– 5 –rder that we can get the effective 5-dimension action after integrating out the sixth dimen-sion. We will discuss the normalization conditions in the next section. However, a problemarises immediately from the following contradiction: On one side, the eigenvalue problem ofEqs. (2.19)-(2.20) is of the Sturm-Liouville type. The characters of the Sturm-Liouville eigen-value problem are that the number of eigenvalues is infinite and the size of eigenvalues isnon-bounded, i.e., the eigenvalue series becomes large monotonously; on the other side, as ithas been illuminated obviously in the papers [9], the larger bulk mass parameters producelighter fermions mass in 4-dimensions. Therefore, the eigenstates of Eqs. (2.19)-(2.20) produceinfinite light fermion generations. However, no lighter generations are discovered by experi-ments so far. Hence we need another mechanics to cut off the infinite series and select onlyseveral eigenstates. The left eigenstates correspond to generations in 4-dimensions.There also exists a problem about the zero mode (2.16). By the numerical examples in [9],the zero bulk mass parameter in 5-dimensions produces a very heavy fermion in 4-dimensions,and it is heavier than the SM generations. So it does not correspond to the physical generations.If the zero mode is permitted by the boundary conditions and the normalization conditions inour model, there would exist a generation that has not been discovered by experiment so far.However, the zero mode is more subtle in the example we will discussed in the next section.We will discuss this problem in more detail in that section.Now we suggest several approaches to deal with the problem about the infinite eigenvalues.(1) An immediate proposal is that one chooses a 6-dimension spacetime in which the sixthdimension is not continuous but discrete. For example, if we discrete the finite interval to befinite points, the induced Eqs. (2.19)-(2.20) will be difference equations. The number of theireigenvalues is finite naturally. There has been a similar investigation for gravity, see [13].(2) Another bizarre proposal is that we suppose the sixth dimension to be timelike. In theAnsatz Eq. (2.1), we have chosen the sixth dimension to be spacelike. Instead we can chooseit to be timelike. This leads to the metric with two time dimensions [14]. The alternativemetric Ansatz is ds = B ( z ) (cid:2) A ( y ) (cid:0) η µν dx µ dx ν + dy (cid:1) − dz (cid:3) . (2.22)In Eq. (2.3), we let Γ = − ! , with others keeping invariant. The same procedureproduces the equations, d dz F n ( z ) − B − ˙ B ddz F n ( z ) + m F n ( z ) + h − λ n B − ˙ B − λ n i F n ( z ) = 0 , (2.23) d dz G n ( z ) − B − ˙ B ddz G n ( z ) + m G n ( z ) + h λ n B − ˙ B − λ n i G n ( z ) = 0 . (2.24)We give a simple example in which B ( z ) = constant. The solutions are F ( z ) = C exp ikz + C exp − ikz , k = p m − λ . (2.25)Here we have omitted the subscripts. If we impose the boundary conditions, F | = 0 , F | R = 0 , (2.26) For extensive investigations on two-time physics, see [15] – 6 –hen λ must conform to kR = p m − λ R = nπ, n = 1 , , , · · · . (2.27)As we emphasized above, λ and m must both be real numbers. Eq. (2.27) has solutions onlyfor finite natural number. The number of the eigenvalues depends on the size of m , hence thereare a finite number of eigenvalues.(3) However, in the following section we will adopt another approach, i.e., we can obtain afinite number of eigenstates by choosing the metric B ( z ) delicately. Although B ( z ) producesinfinite eigenstates generally, it is possible that some of them can produce only finite eigenstates.We will focus on this possibility in the following section, and give concrete examples for finitegenerations.
3. Examples of metric for finite generations
In this section, we will suggest metric examples which can produce finite generations. Wediscuss the appropriate normalization conditions for F n ( z ) , G n ( z ) at first, then we analyzeexamples in detail.We begin with the action (2.4). With the help of Eqs. (2.10), (2.11) and (2.12), the action(2.4) can be rewritten as S = Z d xdy K mn (cid:26) i A (cid:2) ¯ ψ m γ ∂ ψ n − ∂ ¯ ψ m γ ψ n + ¯ ψ m γ µ ∂ µ ψ n − ∂ µ ¯ ψ m γ µ ψ n (cid:3)(cid:27) − Z d xdy M mn A i ¯ ψ m ψ n , (3.1) K mn = Z dzB (cid:16) b F ∗ m b F n + b G ∗ m b G n (cid:17) = Z dz ( F ∗ m F n + G ∗ m G n ) , (3.2) M mn = Z dzB (cid:20)(cid:16) b F ∗ m b F n + b G ∗ m b G n (cid:17) λ n + λ ∗ m (cid:21) = Z dz (cid:20) ( F ∗ m F n + G ∗ m G n ) λ n + λ ∗ m (cid:21) . (3.3)Notice that ¯Ψ = Ψ † Γ = ( ¯ χ , ¯ χ ). Eqs. (3.1-3.3) are satisfied for all modes, including zeromodes and all massive modes.In order to get the conventional effective 5-dimensional action S eff = X n Z d xdy (cid:26) i A (cid:2) ¯ ψ n γ ∂ ψ n − ∂ ¯ ψ n γ ψ n + ¯ ψ n γ µ ∂ µ ψ n − ∂ µ ¯ ψ n γ µ ψ n (cid:3)(cid:27) − X n Z d xdyA iλ n ¯ ψ n ψ n , (3.4)we consider two cases below:Case (I): The first case is that the normalization conditions K mn = Z dz ( F ∗ m F n + G ∗ m G n ) = δ mn (3.5)– 7 –re satisfied. As in the standard Sturm-Liouville case, we can convert the normalizationconditions (3.5) to the boundary conditions. By Eqs. (2.17) and (2.18), we have (cid:2) λ n − ( λ m ) ∗ (cid:3) Z R ′ R dz ( F ∗ m F n + G ∗ m G n )= (cid:20)(cid:18) F n ddz F ∗ m − F ∗ m ddz F n (cid:19) + (cid:18) G n ddz G ∗ m − G ∗ m ddz G n (cid:19)(cid:21) | R ′ R (3.6)= ( λ n + λ ∗ m )( F ∗ m G n − G ∗ m F n ) | R ′ R . (3.7)In the last line, we have used the bulk equations (2.14) and (2.15) to simplify these expressions.We can also get (3.7) from Eqs. (2.14) and (2.15) directly. There are two types of concise choicesto make the normalization conditions satisfied,( a ) : F | R = 0 , F | R ′ = 0; (3.8)or ( b ) : F | R = G | R , F | R ′ = G | R ′ . (3.9)Then for real λ m , the orthogonality is ensured by appropriate boundary conditions. In thiscase, we can get Eq. (3.4) from Eq. (3.1) via Eq. (3.5) in a straight way.Case (II): The second case is that the normalization conditions Eq. (3.5) are not satisfied.In this case, K and M are both matrices, which means that different KK modes are mixednot just among the mass terms, but also among the kinetic terms. At the first sight, it seemsthat we can not get the the conventional effective 5-dimensional action Eq. (3.4). However, if K is positive-definite and the number of KK modes is finite , we can redefine the fermion fieldto get an action, which has the same form with that of Eq. (3.4). The difference is that theeigenvalues λ n are modified to different size. From Eq. (3.2) and Eq. (3.3), we know that K and M are both hermitian. A positive-definite hermitian matrix K can be diagonalized as K = V † Λ V = H † H, H = √ Λ V, (3.10)Λ = diag (Λ , Λ , · · · , Λ n ) , √ Λ = diag( p Λ , p Λ , · · · , p Λ n ) . In the above expressions, Λ i > , i = 1 , , · · · , n , as we have supposed that K is positive-definite. Redefine ψ n as e ψ m = H mn ψ n , (3.11)then in the new basis e ψ n , M becomes f M = ( H − ) † M H − . (3.12)After diagonalizing f M by U , we have f M = U † ∆ U, (3.13)∆ = diag( b λ , b λ , · · · , b λ n ) . We will give numerical examples to show that such conditions can be satisfied in Appendix B. – 8 –he action (3.1) can be reduced to the form like that of action (3.4) b S eff = X n Z d xdy (cid:26) i A h ¯ b ψ n γ ∂ b ψ n − ∂ ¯ b ψ n γ b ψ n + ¯ b ψ n γ µ ∂ µ b ψ n − ∂ µ ¯ b ψ n γ µ b ψ n i(cid:27) − X n Z d xdyA i b λ n ¯ b ψ n b ψ n , (3.14) b ψ m = U mn e ψ n . In the above, we have given the normalization conditions. The criteria are that we canintegrate out the sixth dimension to get an effective 5-dimension action. Now we suggest anexample that can produce finite generations. In Eqs. (2.17) and (2.18), we suppose that B ( z ) = s e ωz + ae ωz + b , s, a, b, ω > . (3.15)As in the models [2], ω can be regarded as the characteristic energy scale of the sixth dimension.We will see that it determines the size of KK modes below. The role of the dimensionlessparameters s and b will become obvious after we give the solutions of Eqs. (2.17) and (2.18).The conditions a, b > −∞ , ∞ ).Eqs. (2.17) and (2.18) can be solved by hypergeometrical functions, F ( z ) = C e − µωz ( e ωz + b ) µ − ν hypergeom (cid:18) ρ − µ + ν, − ρ − µ + ν ; 1 − µ, e ωz e ωz + b (cid:19) + C e µωz ( e ωz + b ) − µ − ν hypergeom (cid:18) ρ + µ + ν, − ρ + µ + ν ; 1 + 2 µ, e ωz e ωz + b (cid:19) , (3.16)where ρ = mω s (1 − ab ), µ = q(cid:0) mω s (cid:1) (cid:0) ab (cid:1) − (cid:0) λω (cid:1) , and ν = q(cid:0) mω s (cid:1) − (cid:0) λω (cid:1) . C and C areconstants. For the sake of simplicity, we omit the subscript n . We only display the solutionfor F ( z ) explicitly. The solution for G ( z ) can be determined by F ( z ) through Eq. (2.14)or by Eq. (2.18) directly. Now let us investigate this solution. For hypergeometrical func-tion Hypergeom( α, β ; γ, ξ ), when Re( γ − α − β ) ≤
0, it diverges at ξ = 1. In the solution(3.16), we have Re( γ − α − β ) = Re( − ν ) ≤ , (3.17)so when z → ∞ , ξ = e ωz e ωz + b →
1, a singularity happens. If we choose z to be a finiteinterval [ R, R ′ ], then the solution (3.16) is well behaved in this range. Imposing the boundaryconditions (3.8) or (3.9), we get infinite eigenvalues generally. However, we find that thefollowing choice can produce a finite number of eigenvalues. Given that z to be a semi-infinite interval [ R, ∞ ), then the solution (3.16) develops a singularity when z → ∞ . Thissingularity makes the integral in (3.5) to be divergent. In order to make the integral to befinite, the hypergeometrical series (3.16) must be cut off to be a polynomial by the requirement α = − ( n − , or β = − ( n − , n = 1 , , , · · · . In Eq. (3.16), we choose1 − ρ − µ + ν = − ( n − , n = 1 , , , · · · . (3.18) About the property of hypergeometrical function, see Appendix C. – 9 –ecause λ and m are real numbers, as we have discussed above, Eq. (3.18) might have solutionsif 1 ≤ n ≤ "r(cid:16) ab (cid:17) − − (cid:16) ab − (cid:17) mω s. (3.19)Obviously, the size of n are limited by the parameters in the metric, then only finite eigenvaluesare permitted. We see that it is important that λ and m are real numbers again. The conditions(3.18) are required by the boundary conditions when z → ∞ . This boundary condition restrictsthe solutions to the form (A.1)-(A.3). We display these solutions in Appendix A explicitly. Wesee that these solutions are determined completely up to normalization constants. For thesesolutions, F n ( z ) , G n ( z ) → x − ν n in the symbols in Appendix A, when z → ∞ . The integralin (3.5) is well defined in the interval [ R, ∞ ), if ν n > , x >
0. Because we have chosen therange of z as the interval [ R, ∞ ), we should also discuss the boundary conditions at z = R .We might want to follow the discussions in Case (I), that is, we require that the normalizationconditions (3.5) are satisfied. Because F n ( z ) , G n ( z ) → z → ∞ , the normalizationconditions (3.5) require that (cid:2) λ n − ( λ m ) ∗ (cid:3) Z R ′ R dz ( F ∗ m F n + G ∗ m G n )= − (cid:20)(cid:18) F n ddz F ∗ m − F ∗ m ddz F n (cid:19) + (cid:18) G n ddz G ∗ m − G ∗ m ddz G n (cid:19)(cid:21) | R = − ( λ n + λ ∗ m )( F ∗ m G n − G ∗ m F n ) | R = 0 . (3.20)However, it is difficult to require the solutions (A.1)-(A.3) to satisfy the conditions (3.20). From(3.18), we know that λ n are determined by the parameters mω s, ab and n . So the conditions(3.20) impose restrictions on the parameters mω s, ab and the boundary parameter R insteadof λ n . The naive numerating of parameters may mean that we can have 3 eigenstates to beorthogonal, because we have 3 parameters mω s, ab and R . Nevertheless, such choices are difficultto be implemented and it seems less natural. It is more natural to regard the parameters mω s, ab and R as the input parameters, or they should be determined by unknown physics that wedo not consider here. In the following discussions, we will not impose boundary conditionsfurther at z = R to determine the solutions, but we will simply give these parameters by handto determine the solutions. In such a choice, the normalization conditions (3.5) are not satisfied.So we should change to the Case (II), that is, K is matrix valued, and we try to diagonalizethis matrix to get the action (3.14). Before doing that, we give numerical examples to showthat only three eigenvalues are left and they are of the same order. Let ab = 9 / , mω s = 4. FromEq. (3.19), we know that only n = 1 , , λ = 3 . ω ( n = 1) , λ = 3 . ω ( n = 2) , λ = 3 . ω ( n = 3) , (3.21)which are of the same order. These massive modes together with the zero mode make K and M to be 4 × K must be positive-definite. InAppendix B, we give numerical examples to show that K is positive-definite and the modifiedeigenvalues b λ n in (3.13) are still of the same order. b λ n are given by b λ = − . ω, b λ = 4 . ω, b λ = 4 . ω, b λ = 5 . ω. (3.22)– 10 –ere we give some interpretations for our choices of the parameters. ab = 9 / , mω s = 4 arechosen to ensure that only 3 massive modes are permitted. We choose x = e ωR b = 30 as theboundary value in order to ensure that the modified b λ n are still of the same order. We foundthat small x , for example, x = 1, makes different b λ n to have big difference. Now we canunderstand the role of the parameters s and b in the metric (3.15). In order that we can trustour analysis, the condition mω < s appearsin the combination mω s . Then we can always keep mω < s despite ofthe input value of mω s . While b appears in the combination e ωz b , so it is closely related to theboundary value of z . The role of a is less obvious because it appears in a more complex way.From the above, we notice two obvious changes: (1) The massive modes are modified todifferent size, but they are still of the same order; (2) The zero mode mixes with the massivemodes. By this mixing, the zero mode gets mass of the same order with the massive modes. Astrange point is that the zero mode gets a negative mass. However, it does not form problemsin the models like [9], where only the size of the mass is relevant. In models where the signof mass is relevant, we must reconsider whether it produces problems for our model. Thisnew feature can supply a possibility to bypass the zero mode problem that we introduced inSec. 2. In the above, we get 4 massive modes from the previous 3 massive modes and 1 zeromode. They can produce 4 fermion generations in 4-dimension. This is not realistic. Theabove numerical example suggests us to start with 2 massive modes and 1 zero mode. If thezero mode gets mass of the same order with the massive modes through mixing as the abovenumerical example, we can get only 3 generations.Before giving an example about this situation, we should discuss another mass sourceabout the zero mode, that is, the zero mode can also get mass through coupling with a Higgsfield on the brane sited at z = R . Here it is appropriate to introduce the brane coupling. TheWilson line phase in [9] is not well defined because the range of z is noncompact. We do notsuggest a concrete form for this coupling. For example, it can arise from the coupling used in[6]. Here we accept the result that the zero mode gets mass λ = ǫ ω. (3.23)In the following discussions, we will consider two cases: (a) ǫ →
0, so it is negligible; (b) ǫ ∼
1, so it is comparable with the massive modes. We will give numerical examples aboutthese two cases respectively. According to the same spirit with the above example, we let ab = 8 / , mω s = 3 to ensure that only 2 massive modes are permitted. These massive modesare given by λ = 2 . ω ( n = 1) , λ = 2 . ω ( n = 2) . (3.24)We still choose x = e ωR b = 30, then by the same procedure with that in Appendix B, we canget the modified b λ n as b λ = − . ω, b λ = 3 . ω, b λ = 3 . ω, for ǫ = 0; (3.25) b λ = 1 . ω, b λ = 2 . ω, b λ = 3 . ω, for ǫ = 2 . (3.26)Here we only give the results. The details are similar to that in Appendix B. The zero modegets mass of the same order with the massive modes in both cases. The difference is that the– 11 –ign of the zero mode mass is opposite in these two cases. In both cases, with the help of thezero mode mixing with massive modes or the zero mode coupling with Higgs field, we maysuggest a possibility to bypass the zero mode problem in Sec. 2. In each case, we may get just3 generations. In models where the sign of mass is relevant, we can check which case may berealistic.In the above example, the orthogonal conditions (3.5) are not satisfied, because we choosea special metric and a special range for z . This choice induces mixing between differentmodes. In the following, we try to construct an example, which can ensure that the orthogonalconditions (3.5) are satisfied. This example can be constructed by changing the metric (3.15)to the following form, B ( z ) = s e ωz − ae ωz − b , s, a, b, ω > . (3.27)Because a, b >
0, this metric develops singularity at the point z = log aω . This may make thismetric unrealistic. However, we find that it can satisfy the orthogonal conditions (3.5) justbecause it has such special structure. Here we let aside the problem of singularity, and focus onhow it can satisfy the orthogonal conditions. The solutions can be given by hypergeometricalfunctions yet, b F ( z ) = C e − µωz ( e ωz − b ) ρ hypergeom (cid:18) ρ − µ − ν, ρ − µ + ν ; 1 − µ, e ωz b (cid:19) (3.28)+ C e µωz ( e ωz − b ) ρ hypergeom (cid:18) ρ + µ − ν, ρ + µ + ν ; 1 + 2 µ, e ωz b (cid:19) , where ρ, µ and ν keep the same form with that in Eqs. (3.16). By use of the property of thehypergeometrical function, we see that if Re(1 − ρ ) ≤
0, a singularity happens at z = R = log bω .Then if we choose the range of z to be ( −∞ , R ], like the above example, the boundaryconditions at z = R = log bω impose the conditions like (3.18). We should also make thesolutions well behaved when z → −∞ . These two requirements can be satisfied by the followingcondition ρ + µ − ν = − n, n = 0 , , , · · · . (3.29)For ab <
1, Eq. (3.29) has solutions and n is limited by0 ≤ n ≤ "r − (cid:16) ab (cid:17) − (cid:16) − ab (cid:17) mω s. (3.30)We give these solutions in Appendix D explicitly. We suppose that mω s = 10 , ab = , in orderthat only 3 massive modes are permitted. Besides these massive modes, Eq. (3.29) also has azero mode solution. These solutions are given by λ = 0 , λ = 3 . ω, λ = 4 . ω, λ = 4 . ω. (3.31) See Appendix F. – 12 –rom Appendix D, we know that when z → −∞ , then x → F n ( z ) , G n ( z ) →
0; while at z = R = log bω , x = 1, then F n ( z ) , G n ( z ) = 0. So in this example, the boundary conditions(3.8) are satisfied. Then the orthogonal conditions (3.5) are ensured. There are no mixingamong different modes. So in this example, the zero mode can only become massive throughcoupling with Higgs field. If the zero mode can get mass comparable to the massive modes,we can adjust the parameters mω s and ab to make that only 2 massive modes are permitted.Then we can get just 3 generations. If this zero mode gets small mass, a generation heavierthan the SM generations is produced. We should check whether it is allowed by experiments.If it is objected by experiments, it will make a problem for our model.In Appendix E, we suggest another example, in which the orthogonal conditions (3.5) aresatisfied. Like the metric (3.27), there exists a singularity in the range of z we considered.In order to avoid the singularity, we may change the range of z . For example, for the metric(3.27), we can choose the range of z to be [ z , R ], where z > log aω . In this new range, thesingularity of the metric (3.27) at log aω is avoided. But the orthogonal conditions (3.5) will benot satisfied. We have not found a metric which satisfies the requirements: 1) it can producefinite generations; 2) it ensures that the orthogonal conditions (3.5) are satisfied, and is wellbehaved in the range of z .
4. Further Discussions
In this section, we compare the approach adopted in [3, 5] with our setup in Sec. 2 at first. Inthe approach adopted in [3, 5], the authors reduced the 6-dimension spacetime to 4-dimensionspacetime directly. Distinct from this approach, we reduce two layer warped 6-dimension(4+1)+1 spacetime to the 5-dimension 4+1 spacetime which is still warped at the first step,then we reduce the 5-dimension spacetime to the physical 4-dimension spacetime. Their dif-ferences are the different ways that one treats the zero mode and massive modes.(1) For the zero mode: In the approach adopted in [3, 5], the authors reduced the 6-dimension spacetime to 4-dimensions directly, and got zero modes in 4-dimension spacetime,so these zero modes correspond to the standard model (SM) generations. These zero modesget mass through coupling to the Higgs field. In the present approach, we reduce the 6-dimension spacetime to 5-dimensions at first, so we get a zero mode in 5-dimension spacetime.When one further reduces the 5-dimension spacetime to the physical 4-dimension spacetime,the zero mode in 5-dimension spacetime can produce a very heavy fermion in 4-dimensions,as illuminated obviously in [9]. It is very heavy, hence it does not correspond to the SMgeneration. However, as we discussed in Sec. 3, the zero mode can also become massive, thenit can produce SM generation if it can get large mass. But if it gets small mass, then itproduces a new generation objected by experimental data. This can cause a problem for ourmodel.(2) For the massive modes: In the approach adopted in [3, 5], the authors got the massivemodes in 4-dimension spacetime. The massive modes are heavy Kluza-Klein (KK) particles.They do not correspond to the SM fermions. However, in the present approach, we reduce the6-dimension spacetime to 5-dimensions at first, so we get the massive modes in 5-dimensionspacetime again. These massive modes are KK states in 5-dimension spacetime. When one– 13 –educes further the 5-dimension spacetime to the physical 4-dimension spacetime, these massivemodes in 5-dimension spacetime can produce massive fermions in 4-dimension spacetime, whichare light and correspond to the SM fermions, as it is obvious in [9].These differences provide a new chance that we can address some extra issues in the presentapproach: we give an explanation for the origin of the same order bulk mass parameters, andgive an answer for the fermion generation puzzle in the meantime. Note that it is the specialmetric Ansatz (2.1) to supply such an explanation. The two layer structure of the metricenables one to reduce the 6-dimension spacetime to 5-dimension spacetime. The metric Ansatzfor A ( y ) can be the AdS metric, A ( y ) = 1 ky . (4.1)By this choice, the five dimensional Dirac equation (2.10) is similar to that analyzed in models[9]. Besides, we want to address another two issues:(1) Whether the metric (2.1) can be the background solutions of Einstein equations? Themetric Ansatz (2.1) has been analyzed in [16] in high derivative gravity with matter sources andin [17] with a negative bulk cosmological constant. Their solutions are not the metric whichwe suggested in Sec. 3. Here we consider a minimum coupled scalar-gravity system in order toinvestigate whether the metric in Sec. 3 can be realized. This is a simple and convenient way.The action is given by S = Z d xdydz √− g { M R } + Z d xdydz √− g (cid:26) g MN ∇ M φ ∇ N φ + V ( φ ) (cid:27) , (4.2)in which V ( φ ) is the potential term for scalar field. Supposing that the metric Ansatz (2.1)and that φ only depends on z , we get the following equations,4 B − B zz + 2 B − B z + 3 A − A yy = 14 M (cid:20) − B (cid:18) B − φ z + V ( φ ) (cid:19)(cid:21) , (4.3)4 B − B zz + 2 B − B z + 6 A − A y = 14 M (cid:20) − B (cid:18) B − φ z + V ( φ ) (cid:19)(cid:21) , (4.4)10 B − B z + 4 A − A yy + 2 A − A y = 14 M (cid:20) φ z − B (cid:18) B − φ z + V ( φ ) (cid:19)(cid:21) , (4.5) B − φ zz + 4 B − B z φ z − dV ( φ ) dφ = 0 , (4.6)in which A y = dAdy , B z = dBdz , φ z = dφdz . Obviously, A ( y ) should be of the form A ( y ) = ky + c , inwhich k, c are constant. A minimum coupled scalar-gravity coupled system has been analyzedin [18] in five dimensions. The result is that there always exists appropriate form of V ( φ )to ensure that the metric have solutions, and V ( φ ) and the metric can be expressed with asuperpotential. The similar result applies to the above system, that is, for any B ( z ), thereexists appropriate V ( φ ), which makes Eqs. (4.3)-(4.6) satisfied. However, the present systemis more complex, and it is difficult to express the solutions with a superpotential. Of course,in order to get the solution of the metric adopted in Sec. 3, the boundary conditions must beadopted appropriately. – 14 –2) Whether the examples used in Sec. 3 are acceptable physically? The metric (3.15),(3.27) and (E.1) are similar to that analyzed in [19], that is, they are both asymptotically flat.They are also both noncompact, and both have infinite volume [20]. Especially for the metric(3.27) and the metric (E.1), there is singularity [21] in the range of z that we choose. So it needsfurther work to investigate whether they are acceptable physically. The similar problem existsfor the metric (2.22), in which the sixth space dimension is timelike. As being emphasizedin [22], the violations of casuality and probability give stringent restrictions on the timelikedimension. It also needs further work to investigate whether it is acceptable physically. Wehave not found a metric, which is finite volumed, as in the models [2], and can produce finitegenerations simultaneously.
5. Conclusions
Now we summarize the main points in our work. In this paper, we try to explain the ori-gin of the same order bulk mass parameters, and give answers to the generation replicationpuzzle simultaneously. The fermion masses are of hierarchy structure in 4-dimension space-time. It seems that it is difficult to interpret them as the eigenvalues of a Schr¨odinger-likeequation. However, the hierarchy structure can be reproduced with the bulk mass parametersin 5-dimension spacetime. The 5-dimension mass parameters are in the same order, as havebeen shown in many papers [6, 7, 8]. This interesting feature supplies a chance to interpretthe 5-dimension mass parameters, which are of the same order, as the the eigenvalues of aSchr¨odinger-like equation. Supposing that the six dimension spacetime metric has special twolayer (4+1)+1 structure, we can reduce the 6-dimension spacetime to 5-dimension spacetime atthe first step. We find that the bulk mass parameters are the eigenvalues of a Schr¨odinger-likeequation. Hence the same order mass parameters emerge naturally. However, the problem isthat the number of eigenvalues is infinite generally, which leads to infinite light generations. Wesuggest several approaches to deal with this problem. Obviously, this problem arises from thefact that in the conventional Kluza-Klein (KK) decomposition, one gets infinite KK particlesgenerally. However, as in the example given by Madore [23], in the noncommutative geomet-rical background, and by the choice of the internal structure, the modification of KK theorygives rise to finite spectrum of particles. Therefore, it is possible to overcome the difficultiesin our work by use of the noncommutative geometry. It will require modifying the frameworkin Sec. 2. We hope we can address these issues in the future.
Acknowledgement
This work is partially supported by National Natural Science Foun-dation of China (Nos. 10721063, 10575003, 10528510), by the Key Grant Project of ChineseMinistry of Education (No. 305001), by the Research Fund for the Doctoral Program of HigherEducation (China).
A. Explicit solutions for massive modes and zero mode: metric I
In this appendix, we give the solutions of Eq. (3.16) for massive modes under the conditions(3.18) explicitly. The solutions should be well behaved in the range [ R, ∞ ). Let x = e ωz b , the– 15 –olutions are given by F ( x ) = √ ω √ N x − µ ( x + 1) µ − ν , (A.1) F ( x ) = √ ω √ N x − µ ( x + 1) µ − ν (cid:20) − α γ xx + 1 (cid:21) , (A.2) F ( x ) = √ ω √ N x − µ ( x + 1) µ − ν " − α γ xx + 1 + α ( α + 1) γ ( γ + 1) (cid:18) xx + 1 (cid:19) , (A.3)...in which γ n = 1 − µ n , n = 1 , , , · · · (A.4) α n = ρ − µ n + ν n , ρ = mω s (cid:16) − ab (cid:17) , (A.5) ν n = s(cid:16) mω s (cid:17) − (cid:18) λ n ω (cid:19) , (A.6) µ n = s(cid:16) mω s (cid:17) (cid:16) ab (cid:17) − (cid:18) λ n ω (cid:19) . (A.7)We have dropped the hypergeometrical function after the coefficient C in (3.16), because itis divergent under the conditions (3.18) when z → ∞ . The solutions for G n ( z ) can be gottenfrom (2.14) as G n ( x ) = ωλ n (cid:20) mω s x + ab x + 1 F n ( x ) − x ddx F n ( x ) (cid:21) . (A.8)The zero mode solution is given by F ( x ) = 0 , G ( x ) = √ ω √ N x − mω s ab ( x + 1) mω s ( ab − . (A.9)When z → ∞ , x = e ωz b → ∞ F n ( x ) → x − ν n , G n ( x ) → ωλ n (cid:16) mω s + ν n (cid:17) x − ν n , (A.10) F ( x ) = 0 , G ( x ) → x − mω s . (A.11)They are all well behaved when ν n >
0. In terms of x , the integral in Eq. (3.2) can be rewrittenas K mn = Z dz ( F ∗ m F n + G ∗ m G n ) = 1 ω Z dxx ( F ∗ m F n + G ∗ m G n ) . (A.12)They are also well behaved in the range [ R, ∞ ) when ν n > x >
0. In the numericalexamples we give in Sec. 3, the conditions ν n > x > . Numerical examples for finite generations In order to get numerical results, we need to input the parameters mω s, ab and R . From thesolutions in Appendix A, we know that it is enough to input value for x = e ωR b .Let mω s = 4 , ab = 9 / n = 1 , , x = e ωR b = 30. We normalize the solutions for massive modes and zeromode according to (3.5) as Z dz ( F ∗ n F n + G ∗ n G n ) = 1 , n = 0 , , , . (B.1)These conditions determine the normalization constants. Then the matrix K is determined tobe K = . . . . . . . . . . . . . (B.2)The indices of K are determined according to (3.2) as K mn = Z dz ( F ∗ m F n + G ∗ m G n ) , m, n = 0 , , , . (B.3) K can be diagonalized as K = V T Λ V, Λ = diag(3 . , . , . , . , (B.4) V = − . − . − . − . − . − . . . − . . . − . . − . . − . , where V T means the transpose of V . This example shows that K is positive-definite, as weexpected. λ n is given by (3.21) in Sec. 3. M is determined by (3.3) as M mn = Z dz (cid:20) ( F ∗ m F n + G ∗ m G n ) λ n + λ ∗ m (cid:21) , m, n = 0 , , , . (B.5)Given the parameters, M is determined to be M = .
402 1 . . . √ . . . . √ . . . . √ . (B.6)– 17 –ollowing the procedure in Sec. 3, we can get f M as f M = . − . − . . − . . − . . − . − . . . . . . . = U T ∆ U,U = . . − . . − . . . . . − . . . − . − . − . . , (B.7)∆ = diag(5 . , . , . , − . . C. Property of hypergeometrical function
We cite a theorem [24] about the hypergeometrical function F ( α, β ; γ, ξ ).Theorem 2 . . F ( α, β ; γ, ξ ) with | ξ | = 1 converges absolutely if Re( γ − α − β ) >
0. The series converges conditionally if ξ = e iθ = 1 and 0 ≥ Re( γ − α − β ) > − γ − α − β ) ≤ − D. Solutions for zero mode and massive modes: metric II
For the metric (3.27), the solutions for massive modes well behaved in the range ( −∞ , R ] aregiven by F ( x ) = √ ω √ N x µ (1 − x ) ρ (cid:20) − β γ x (cid:21) , (D.1) F ( x ) = √ ω √ N x µ (1 − x ) ρ (cid:20) − β γ x + β ( β + 1) γ ( γ + 1) x (cid:21) , (D.2)...in which γ n = 1 + 2 µ n , n = 1 , , , · · · (D.3) β n = ρ + µ n + ν n , (D.4) ν n = s(cid:16) mω s (cid:17) − (cid:18) λ n ω (cid:19) , (D.5) µ n = s(cid:16) mω s (cid:17) (cid:16) ab (cid:17) − (cid:18) λ n ω (cid:19) . (D.6)In the above expressions, we have defined x = e ωz b . ρ is defined as in Sec. 3. λ n are determinedby (3.29), and n is limited by (3.30). We have dropped the hypergeometrical function after– 18 –he coefficient C in (3.28), because it is divergent when z → −∞ . The solutions for G n ( x )are determined by G n ( x ) = ωλ n (cid:20) mω s x − ab x − F n ( x ) − x ddx F n ( x ) (cid:21) . (D.7)The solutions (D.1)-(D.7) are well behaved when µ n > , ρ ≥
1. They are satisfied in ournumerical example in Sec. 3.We find that Eq. (3.29) also has a well behaved zero mode solution. This zero modesolution is given by F ( x ) = √ ω √ N x mω s (1 − x ) ρ , G ( x ) = 0 . (D.8)It is consistent with what we get from (2.16). E. Another metric example for orthogonality
In this appendix, we suggest another metric which ensures that the the orthogonal conditions(3.5) are satisfied. This metric is given by B ( z ) = s ωz − aωz − b , s, a, b, ω > . (E.1)We suppose a, b > a > b here. The conditions a, b < a < b work well also. Butthe conditions a < , b > γ = 2(1 + ν ) , ν = mω s ( a − b ) is notinteger, the solutions are given by F ( z ) = C e − µ ( ωz − b ) ( ωz − b ) ν hypergeom ( α ; γ, µ ( ωz − b ))+ C e − µ ( ωz − b ) ( ωz − b ) ν +1 − γ hypergeom ( α + 1 − γ ; 2 − γ, µ ( ωz − b )) , (E.2)in which µ = q(cid:0) mω s (cid:1) − (cid:0) λω (cid:1) and α = 1 + ν − mω s νµ . We choose the range of z to be [ R, ∞ ),where R = bω . The confluent hypergeometrical function F ( α ; γ, ξ ) ∼ e ξ when ξ → ∞ . In orderto make the solutions well behaved when z → ∞ , the confluent hypergeometrical functionmust be cut off to be a polynomial by the requirement α = 1 + ν − mω s νµ = − ( n − , n = 1 , , · · · . (E.3)Then λ n are determined to be λ n = " mω s (cid:18) − ν ( ν + n ) (cid:19) ω. (E.4)By (E.4), we know that " mω s (2 ν + 1) ν + 1 ω ≤ λ ≤ (cid:16) mω s (cid:17) ω. (E.5)– 19 –e should also require that the solutions are well behaved at z = R = bω . This conditionrequires the solutions further to be F n ( z ) = √ ω √ N e − µ ( x − b ) ( x − b ) ν hypergeom ( − n ; γ, µ ( x − b )) . (E.6)Here we define x = ωz , and G n ( z ) are determined by G n ( x ) = ωλ n (cid:20) mω s x − ax − b F n ( x ) − ddx F n ( x ) (cid:21) . (E.7)There is also a well behaved zero mode solution, which is given by F ( x ) = 0 , G ( x ) = √ ω √ N e − s mω x ( x − b ) ν . (E.8)We have supposed a > b . So when z → ∞ , x → ∞ , F n ( z ) , G n ( z ) →
0; while F n ( z ) , G n ( z ) = 0at z = R, x = b if ν ≥
2. So the orthogonal conditions (3.8) can be satisfied.According to the analysis in [9], if we let λ = ms to be the lightest generation of SM,there exist infinite heavier generations corresponding to λ < ms , in which λ = 0 is the heaviestgeneration. However, there exists a problem in this case: the lighter generations approximateto be continuous, which conflicts with the experimental fact. Therefore, special boundaryconditions must be adopted to remove the reductant generations. We hope that there are onlyfinite generations left with the help of the special boundary conditions. But it seems that it isdifficult to impose such boundaries naturally. F. Ricci scalar curvature for metric
Ricci scalar curvature for metric Ansatz (2.1) is given by R = − (cid:2) B − B zz + B − B z ) + (8 A − A yy + 4 A − A y ) B − (cid:3) , (F.1)in which A y = dAdy , B z = dBdz . From the second term in (F.1), we know that there is singularityfor the metric (3.27) at z = log aω and the metric (E.1) at z = aω . At the point z = log bω and z = bω ,the metric (3.27) and (E.1) are not well defined respectively, but the Ricci scalar for them arewell defined. So it needs further work to determine whether they are true singularities. References [1] N. Arkani-Hamed, S. Dimopoulos, G. R. Dvali, Phys. Lett. B 429 (1998) 263; I. Antoniadis,N. Arkani-Hamed, S. Dimopoulos, G. R. Dvali, Phys. Lett. B 436 (1998) 257.[2] L. Randall, R. Sundrum, Phys. Rev. Lett. 83 (1999) 4690; ibid. 83 (1999) 3370.[3] M. V. Libanov, S. V. Troitsky, Nucl. Phys. B 599 (2001) 319; J.-M. Fr`ere, M. V. Libanov,S. V. Troitsky, Phys. Lett. B 512 (2001) 169.[4] B. A. Dobrescu, E. Poppitz, Phys. Rev. Lett. 87 (2001) 031801.[5] A. Nernov, Phys. Rev. D 65 (2002) 044004; S. Aguilar, D. Singleton, Phys. Rev. D 73 (2006)085007; M. Gogberashvili, P. Midodashvili, D. Singleton, J. High Energy Phys. 0708 (2007) 033. – 20 –
6] T. Gherghetta, A. Pomarol, Nucl. Phys. B 586 (2000) 141; Y. Grossman, M. Neubert, Phys. Lett.B 474 (2000) 361; N. Arkani-Hamed, M. Schmaltz, Phys. Rev. D 61 (2000) 033005; D. E. Kaplan,T. M. Tait, J. High Energy Phys. 11 (2001) 051.[7] C. Cs´aki, C. Grojean, J. Hubisz, Y. Shirman, J. Terning, Phys. Rev. D 70 (2004) 015012;C. Cs´aki, hep-ph/0510275; Y. Hosotani, S. Noda, Y. Sakamura, S. Shimasaki, Phys. Rev. D 73(2006) 096006.[8] Y. Hosotani, S. Noda, Y. Sakamura, S. Shimasaki, Phys. Rev. D 73 (2006) 096006; C. Cs´aki,C. Grojean, J. Hubisz, Y. Shirman, J. Terning, Phys. Rev. D 70 (2004) 015012; S. Chang,C. S. Kim, M. Yamaguchi, Phys. Rev. D 73 (2006) 033002.[9] Y. Hosotani, S. Noda, Y. Sakamura, S. Shimasaki, Phys. Rev. D 73 (2006) 096006.[10] W. Israel, Nuovo Cimento. B 44 (1966) 1; Erratum-ibid. B 48 (1967) 463.[11] C. Cs´aki, hep-ph/0510275.[12] C. Cs´aki, C. Grojean, J. Hubisz, Y. Shirman, J. Terning, Phys. Rev. D 70 (2004) 015012;M. Carena, J. Lykken, and M. Park, Phys. Rev. D 72 (2005) 084017.[13] C. Deffayet, J. Mouradb, Phys. Lett. B 589 (2004) 48.[14] J. M. Overduin, P. S. Wesson, Phys. Rep. 283 (1997) 303; M. Gogberashvili, Phys. Lett. B 484(2000) 124; Z. Berezhiani, M. Chaichian, A. B. Kobakhidze and Z. H. Yu, Phys. Lett. B 517(2001) 387.[15] I. Bars, Phys. Rev. D 74 (2006) 085019.[16] H. Collins, B. Holdom, Phys. Rev. D 64 (2001) 064003.[17] D. Choudhury, S. SenGupta, Phys. Rev. D 76 (2007) 064030.[18] O. DeWolfe, D. Z. Freedman, S. S. Gubser, A. Karch, Phys. Rev. D 62 (2000) 046008; C. Cs´aki,J. Erlich, T. J. Hollowood, Y. Shirman, Nucl. Phys. B 581 (2000) 309.[19] R. Gregory, V. A. Rubakov, S. M. Sibiryakov, Phys. Rev. Lett. 84 (2000) 5928; C. Cs´aki,J. Erlich, T. J. Hollowood1, Phys. Rev. Lett. 84 (2000) 5932.[20] G. Dvali, G. Gabadadze, M. Porrati, Phys. Lett. B 485 (2000) 208; G. Dvali, G. Gabadadze,Phys. Rev. D 63 (2001) 065007.[21] M. Gell-Mann, B. Zwiebach, Nucl. Phys. B 260 (1985) 569; A. G. Cohen, D. B. Kaplan, Phys.Lett. B 470 (1999) 52.[22] G. Dvali, G. Gabadadze, G. Senjanovi´c, hep-ph/9910207; F. Yndurain, Phys. Lett. B 256 (1991)15; M. Chaichian, A. B. Kobakhidze, Phys. Lett. B 488 (2000) 117.[23] J. Madore, Phys. Rev. D 41 (1990) 3709.[24] G. E. Andrews, R. Askey, R. Roy, Special Functions, Cambridge University Press, CambridgeU.K. (2000).6] T. Gherghetta, A. Pomarol, Nucl. Phys. B 586 (2000) 141; Y. Grossman, M. Neubert, Phys. Lett.B 474 (2000) 361; N. Arkani-Hamed, M. Schmaltz, Phys. Rev. D 61 (2000) 033005; D. E. Kaplan,T. M. Tait, J. High Energy Phys. 11 (2001) 051.[7] C. Cs´aki, C. Grojean, J. Hubisz, Y. Shirman, J. Terning, Phys. Rev. D 70 (2004) 015012;C. Cs´aki, hep-ph/0510275; Y. Hosotani, S. Noda, Y. Sakamura, S. Shimasaki, Phys. Rev. D 73(2006) 096006.[8] Y. Hosotani, S. Noda, Y. Sakamura, S. Shimasaki, Phys. Rev. D 73 (2006) 096006; C. Cs´aki,C. Grojean, J. Hubisz, Y. Shirman, J. Terning, Phys. Rev. D 70 (2004) 015012; S. Chang,C. S. Kim, M. Yamaguchi, Phys. Rev. D 73 (2006) 033002.[9] Y. Hosotani, S. Noda, Y. Sakamura, S. Shimasaki, Phys. Rev. D 73 (2006) 096006.[10] W. Israel, Nuovo Cimento. B 44 (1966) 1; Erratum-ibid. B 48 (1967) 463.[11] C. Cs´aki, hep-ph/0510275.[12] C. Cs´aki, C. Grojean, J. Hubisz, Y. Shirman, J. Terning, Phys. Rev. D 70 (2004) 015012;M. Carena, J. Lykken, and M. Park, Phys. Rev. D 72 (2005) 084017.[13] C. Deffayet, J. Mouradb, Phys. Lett. B 589 (2004) 48.[14] J. M. Overduin, P. S. Wesson, Phys. Rep. 283 (1997) 303; M. Gogberashvili, Phys. Lett. B 484(2000) 124; Z. Berezhiani, M. Chaichian, A. B. Kobakhidze and Z. H. Yu, Phys. Lett. B 517(2001) 387.[15] I. Bars, Phys. Rev. D 74 (2006) 085019.[16] H. Collins, B. Holdom, Phys. Rev. D 64 (2001) 064003.[17] D. Choudhury, S. SenGupta, Phys. Rev. D 76 (2007) 064030.[18] O. DeWolfe, D. Z. Freedman, S. S. Gubser, A. Karch, Phys. Rev. D 62 (2000) 046008; C. Cs´aki,J. Erlich, T. J. Hollowood, Y. Shirman, Nucl. Phys. B 581 (2000) 309.[19] R. Gregory, V. A. Rubakov, S. M. Sibiryakov, Phys. Rev. Lett. 84 (2000) 5928; C. Cs´aki,J. Erlich, T. J. Hollowood1, Phys. Rev. Lett. 84 (2000) 5932.[20] G. Dvali, G. Gabadadze, M. Porrati, Phys. Lett. B 485 (2000) 208; G. Dvali, G. Gabadadze,Phys. Rev. D 63 (2001) 065007.[21] M. Gell-Mann, B. Zwiebach, Nucl. Phys. B 260 (1985) 569; A. G. Cohen, D. B. Kaplan, Phys.Lett. B 470 (1999) 52.[22] G. Dvali, G. Gabadadze, G. Senjanovi´c, hep-ph/9910207; F. Yndurain, Phys. Lett. B 256 (1991)15; M. Chaichian, A. B. Kobakhidze, Phys. Lett. B 488 (2000) 117.[23] J. Madore, Phys. Rev. D 41 (1990) 3709.[24] G. E. Andrews, R. Askey, R. Roy, Special Functions, Cambridge University Press, CambridgeU.K. (2000).