Yu-Xiao Liu Institute of Theoretical Physics, Lanzhou ...icts.ustc.edu.cn/chinese/seminar/transparencies/Yu... · ICTS, USTC, Hefei 2011.12.13. Content Introduction and motivation

Braneworld in f(R) gravity and critical gravity

Yu-Xiao Liu

Institute of Theoretical Physics, Lanzhou University

ICTS, USTC, Hefei

2011.12.13

Content

Introduction and motivation

f (R) thick branef (R) thick brane modelThe solutionThe stability of the solutionThe localization of gravity

Thin brane solutions in critical gravityThe case n = 4The case n 6= 4

Thick brane solutions in critical gravityThe case n = 4The case n 6= 4

Summary and open issues

Brief history of extra dimensions and braneworlds

1920’s, Kaluza and Klein, KK Theory

1980’s, Akama, Rubakov, Domain Wall Braneworld

1998, Arkani-Hamed, Dimopoulos, and Dvali, LargeExtra Dimension (ADD Braneworld Scenario)

1999, Randall and Sundrum(RS), Warped ExtraDimension (RS thin Braneworld Scenario)

1999, DeWolfe, Freedman, Gubser, and Karch,Thick Braneworld Scenario

Picture of braneworlds

4D space-time is seen as a brane (hypersurface)embedded in higher-dimensional space-time

Matter and gauge fields are confined on the brane, onlythe gravity can propagate in the bulk

Thin brane & thick brane

ds2 = e2A(y)gµνdxµdxν − dy2, e2A(y) is the warp factor.

y

Warp factor ã2 A

y

Energy density Ρ

y

Warp factor ã2 A

y

Energy density Ρ

Why braneworlds in f (R) and critical gravities?

General relativity is not a renormalizable theory.

By adding some higher order curvature terms (such asR2,R2

µν ,R2µνλσ)) to the original Einstein-Hilbert action,

we can get a renormalizable theory of gravity.

Higher-order curvature terms also appear in low-energyeffective action of string theory.

It is necessary and important to reconsider braneworldscenarios in higher-order curvature modified theories.

Why braneworlds in f (R) and critical gravities?(cont.)

f (R) gravity and critical gravity are such simple toymodels.

f (R) gravity has been applied in discussing dark energy,inflation, thermodynamics, entropic force, theholographic principle, etc..

Critical gravity [Lu and Pope, PRL 106 (2011) 181302]possess such an AdS vacuum, for which there is onlymassless tensor, and the linearized excitations havevanishing energy.

Related works about thin f (R)-brane

About thin f (R)-brane

1 In 1988, Barrow and Cotsakis found that f (R) gravityconformally equivalents to the second order Einsteingravity plus a scalar degree of freedom.

2 BC’s discovery was applied to construct thin f (R)-braneby M. Parry, S. Pichler, and D. Deeg in 2005[hep-ph/0502048]

3 For thin branes the Israel junction conditions wereinvestigated by two groups: Balcerzak etc.[arXiv:0710.3670] and Dabrowski etc. [arXiv:0711.1150]in 2007.

4 Bouhmadi-Lopez etc. discussed f (R) brane cosmology[arXiv:1001.3028] recently.

Related works about thick f (R)-brane

For thick f (R)-branes, the related works are

1 [1] V.I. Afonso, D. Bazeia, R. Menezes, and A.Yu.Petrov,“f (R)-Brane”,PLB 658(2007)71, arXiv:0710.3790.

2 [2] V. Dzhunushaliev, V. Folomeev, B. Kleihaus, and J.Kunz, “Some thick brane solutions in f (R)-gravity”,JHEP 1004(2010)130, arXiv:0912.2812.

3 [3] Y. Zhong, Y.-X. Liu and K. Yang,“Tensor perturbations of f (R)-branes,Phys. Lett. B699 (2011) 398, arXiv:1010.3478.

4 [4] Y.-X. Liu, Y. Zhong, Z.-H. Zhao and H.-T. Li,“Domain wall brane in squared curvature gravity”,JHEP 1106 (2011) 135,arXiv:1104.3188.

Related works about thick f (R)-brane (cont.)

5 [5] H. Liu, H. Lu and Z.-L. Wang,“f(R) Gravities, Killing Spinor Equations, ”BPS”Domain Walls and Cosmology”, arXiv:1111.6602.

In Refs. [1,2] f (R) ∝ R + Rn are assumed.

[1] PLB 658(2007)71:

Thick f (R)-brane solutions exist either for constant andvariant bulk curvatures with a background scalar field.

The curvature equals to a constant is only a special case.

For the non-constant curvature case, the solution has asingular point, which we don’t want to see in thickbraneworld models.

Related works about thick f (R)-brane (cont.)

[2] JHEP 1004(2010)130:

Some numerical thick f (R)-brane solutions exist withoutthe introducing of background scalar field.

The numerical solution can not be applied to discussmore complex problems.

There is a problem in localizing fermions on the brane.

[5] H. Liu, H. Lu and Z.-L. Wang, arXiv:1111.6602:

The Killing spinor equations are used to reduce thefourth-order differential equations of motion to the firstorder for both the domain wall and FLRW cosmologicalsolutions.

“BPS” domain walls that describe the smoothRandall-Sundrum II were obtained.

The context of the report

1 The thick f (R)-brane solution in f (R) gravity withf (R) = R + γR2 [4].

2 The analysis of the tensor perturbations of brane metricin f (R) gravity [3,4].

3 The thin and thick brane solutions in critical gravity.

Content






f (R) thick brane model

Recently, we have found an analytical thick branesolution of the action

S =

∫

d4xdy√−g

[

f (R)− 1

2gAB(∇Aφ)(∇Bφ)− V (φ)

]

, (1)

where f (R) and V (φ) are given by

f (R) =1

2κ25(R + γR2). (2)

V (φ) = λ(φ2 − v2)2 + Λ5, (3)

and the metric has the following from

ds2 = e2A(y)ηµνdxµdxν + dy2. (4)

f (R) thick brane model (cont.)

The (four-order) field equations are

f (R) + 2fR(

4A′2 + A′′)

− 6f ′RA′ − 2f ′′R = κ25(φ

′2 + 2V ), (5)

− 8fR(

A′′ + A′2)

+ 8f ′RA′ − f (R) = κ25(φ

′2 − 2V ), (6)

4A′φ′ + φ′′ =∂V

∂φ. (7)

For the case f (R) = R + γR2,

fR = 1 + 2γR , (8)

f ′R = 2γR ′, (9)

f ′′R = 2γR ′′, (10)

R ′ = −(8A′′′ + 40A′A′′), (11)

R ′′ = −(8A′′′′ + 40A′′2 + 40A′A′′′). (12)

The solution

1 The solution is 1

e2A(y) = cosh−2(ky), (13)

φ(y) = v tanh(ky). (14)

where v = 7√

3/(29κ2).

2 At the boundary of the extra dimension, i.e., y = ±∞,the bulk curvature R = −20k2, which means that thespacetime is asymptotically anti-de Sitter. Thecorresponding cosmological constant Λ = −159k2

29κ25.

3 As shown in the following figure, this solution describesa typical thick braneworld model.

1Y.-X. Liu, Y. Zhong, Z.-H. Zhao and H.-T. Li, JHEP 1106 (2011) 135

The solution

-3 -2 -1 1 2 3 y

0.2

0.4

0.6

0.8

1Warp factor ã2 A

-3 -2 -1 1 2 3 y

2

4

6

8

10Energy density Ρ

κ5 = 1, k = 1(red dotted) and k = 2(blue solid).

The stability of the solution

The analysis of a full set of gravitational fluctuations ofthe metric is a hard work.

However, the problem can be simplified when one onlyconsiders the transverse and traceless (TT) part of themetric fluctuation.

Consider the tensor perturbations of the backgroundmetric:

ds2 = e2A(y)(ηµν + hµν)dxµdxν + dy2, (15)

where hµν = hµν(xµ, y).

The perturbation of the scalar field δφ(xµ, y) decouplesfrom the metric perturbations under the transverse andtraceless gauge ∂µh

µν = 0 = ηµνhµν .

The stability of the solution (cont.)

Thus, the perturbed Einstein equation is [ Y. Zhong,Y.-X. Liu and K. Yang, PLB 699 (2011) 398]

�(5)hµν =

f ′RfR

∂yhµν , (16)

For f (R) = R , this equation reduces to the KG equationfor massless spin-2 particles.

By doing the coordinate transformation dz = e−A(y)dy ,we can rewrite the metric (4) into a conformally flat one:

ds2 = e2A(y(z))(ηµνdxµdxν + dz2). (17)

Then the perturbed equation (16) reads

[

∂ 2z +

(

6∂zA+∂z fRfR

)

∂z +�(4)

]

hµν = 0. (18)


With the KK decomposition hµν = e ipxe−32A f

−1/2R εµνh(z),

the above equation is reduced to the following Schrodingerequation for h(z):

[

−∂2z + VG (z)

]

h(z) = m2h(z), (19)

where the potential is given by

VG (z) =

(

3

2A′′ +

9

4

(

A′)2)

+

(

3A′ f′R

fR− 1

4

f ′2Rf 2R

+1

2

f ′′RfR

)

. (20)

One can also factorize the Schrodinger equation (19) as

[

∂z +

(

3A′ +1

2

f ′RfR

)][

−∂z +

(

3A′ +1

2

f ′RfR

)]

h(z)

= m2h(z), (21)


which indicates that

there is no gravitational mode with m2 < 0 and thereforethe solution for our model is stable.

The zero mode (m2 = 0)

h0(z) = N0e32A(y(z))f

1/2R (R(z)) (22)

is the ground state of the Schrodinger equation (19).

In f (R) gravity, the zero mode depends not only on theform of the warp factor, but also on the form of f (R).

The localization of gravity

For our model, the warp factor e2A(y) = cosh−2(ky)becomes e2A(y(z)) = (1 + k2z2)−1 in z coordinate, and thepotential takes the form

VG (z) =15k2

(

−14 + 37k2z2 + 28k4z4 + 4k6z6)

4 (5 + 7k2z2 + 2k4z4)2. (23)

When |z | → ∞, VG (z) → 154z2

. The corresponding zero modecan be normalized as

h0(z) =

√

k

8

√5 + 2k2z2

(1 + k2z2)5/4. (24)

The localization of gravity (cont.)

-1.5 -1 -0.5 0.5 1 1.5z

-30

-20

-10

10

VG

-1.5 -1 -0.5 0.5 1 1.5z0.250.5

0.751

1.251.5

h0

The existence of the zero mode indicates that thegravity can be localized on the brane.

The zero mode gives the Newtonian potential U ∝ 1rfelt

by the massive objects on the brane.

In addition to the zero mode, there is a series ofcontinues massive KK states, which would give acorrection to the Newtonian potential.

Content






Thin brane solutions in critical gravity

First, we consider the thin braneworld with co-dimensionone generated in n = (p + 1)-dimensional critical gravity,where n ≥ 3. The action is

S = Sg + Sb, (25)

where the gravity part Sg and the brane part Sb are given by

Sg =1

2κ2

∫

dnx√−g

[

R − (n − 2)Λ0 + αR2 + βRMNRMN

]

, (26)

Sb =

∫

dn−1x

√

−g (b) V0. (27)

Here g(b)µν is the induced metric on the brane, and V0 is the

brane tension.

Thin brane solutions in critical gravity (cont.)

The line-element is assumed as

ds2 = gMNdxMdxN = e2A(y)ηµνdx

µdxν + dy2, (28)

where e2A is the warp factor with the normalized conditione2A(0) = 1 on the brane located at y = 0. We introduce theZ2 symmetry by setting A(y) = A(−y).

The equations of motion are given by

GMN + αE(1)MN + βE

(2)MN = κ2TMN − 1

2(n − 2)Λ0gMN , (29)

where Tµν = −V0 gµνδ(y), Tnn = 0,

E(1)MN = 2R

(

RMN − 1

4R gMN

)

+ 2gMN�R − 2∇M∇NR ,

E(2)MN = 2RPQ(RMPNQ − 1

4RPQ gMN) +�(RMN +

1

2R gMN)−∇M∇NR .


The junction conditions are determined by

∫ 0+

0−dy

[

Rµν −1

2R gµν + αE (1)

µν + βE (2)µν

]

= −κ2V0 gµν(y = 0).(30)

At y 6= 0, the explicit forms of the above EOMs are

(p − 1)(Λ0 + pA′2 + 2A′′)− p2(p − 3)[(p + 1)α + β]A′4

− 2p[

(6p2 − 4p − 6)α+ (p2 + 3p − 6)β]

A′2A′′

−[

4pα+ (p + 1)β](

2A(4) + 4pA′A(3) + 3pA′′2)

= 0, (31)

(p − 1)(

Λ0 + pA′2)

− p2(p − 3)((p + 1)α+ β)A′4

+ p[

4pα+ (p + 1)β](

A′′2 − 2A′A(3) − 2pA′2A′′)

= 0. (32)


It is very difficult to find the solution of thin brane forarbitrary α and β for the above four-order differentialequations and the junction conditions (30).

However, at the critical point 4pα+ (p + 1)β = 0, theEOMs are reduced toµµµ

4Λ0 + 4pA′2 + p(p − 1)(p − 3)βA′4 = 0, (33)[

2 + (p − 1)(p − 3)βA′2]

A′′ = 0, (34)

and the junction condition is

∫ 0+

0−dy

[

2 + (p − 1)(p − 3)βA′2]

A′′ = − 2κ2

p − 1V0. (35)

It can be seen from Eqs. (33) and (34) that thecurvature-squared modifications in the 4D critical gravity(p = 3 or n = 4) has no effect on the brane solution.


In the following, we give the solutions of the abovebrane equations for n = 4 and n 6= 4, respectively.

The case n = 4

For this case (p = 3), the solution is

A(y) = −√

−Λ0

3|y |, (36)

V0 =4

κ2

√

−Λ0

3(> 0) . (37)

Thus, we get a brane with positive tension and the warpfactor exponentially falling from the brane to infinity. Thebrane is embedded in a 4D AdS spacetime.

This is nothing but the RS solution in four dimensions.But the perturbation structure of the system is very differentfrom the RS one.

The case n 6= 4

For the case n 6= 4, we first give the solutioncorresponding to β = 0:

A(RS)(y) = −√

−Λ0

p|y|, (38)

V(RS)0 =

2(p − 1)

κ2

√

−Λ0

p(> 0) , (39)

which describes the RS positive tension brane embedded inan AdS spacetime.

Note that the tension brane is positive.

The case n 6= 4 (cont.)

In critical gravity (β 6= 0), we have two solutions:

A±(y) = −

√

2[

− p ±√

p[p − (p − 1)(p − 3)βΛ0]]

p(p − 1)(p − 3)β|y|, (40)

V0± = ±p − 1

pκ2

√

−8Λ0p[p − (p − 1)(p − 3)βΛ0]

p ±√

p[p − (p − 1)(p − 3)βΛ0], (41)

which correspond to positive and negative tension branes,respectively.

For the positive tension brane solution, the constrainconditions for the parameters are Λ0 < 0 and β > 0, or Λ0 < 0and p

(p−1)(p−3)Λ0< β < 0.

For the negative tension case, the constrain conditionsare β < 0 and Λ0 > 0, or β < 0 and p

(p−1)(p−3)β < Λ0 < 0.


Next, we study the limits of the solutions (40) under thecondition of βΛ0 → 0.

For the negative tension brane solution A−(y), the limitis divergent.

While, for the case of positive tension, A+(y) and V0+

can be expanded as

A+(y) = −√

−Λ0

p

[

1 +(p − 1)(p − 3)

8pβΛ0 +O(βΛ0)

2

]

|y |, (42)

V0+ =2(p − 1)

κ2

√

−Λ0

p

[

1− 3(p − 1)(p − 3)

8pβΛ0 +O(βΛ0)

2

]

. (43)

So, when β → 0 but keep Λ0 as a constant, the abovepositive tension brane solution (40)-(41) can be reduced tothe RS one (38)-(39), while the negative one cannot.


It is interesting to note that, when Λ0 and β satisfy thefollowing relation

Λ0 =p

(p − 1)(p − 3)β, (44)

the brane tension vanishes and the warp factor is simplified as

A(y) = −√

−2Λ0

p|y |. (45)

Obviously, such solution could not appear in standardEinstein gravity theory. While, in critical gravity theory,although the naked brane tension is zero, but with the

identification of κ2T(eff)µν ≡ −αE

(1)µν − βE

(2)µν , there will be a

positive effective brane tension coming from the contributionof the curvature-squared terms.

Content






Thick brane solution in critical gravity

Next, we consider the thick brane generated by a scalarfield in n-dimensional critical gravity. The action reads as

S = Sg + Sm, (46)

where Sg is given by (26) and the matter part is

Sm =

∫

dnx√−g

[

− 1

2gMN∂Mφ∂Nφ− V (φ)

]

. (47)

The line-element is also assumed as (28) and the scalar fieldφ = φ(y) for static branes.

Thick brane solution in critical gravity (cont.)

The EOMs for general α and β are four-order, while theyare reduced to the following two-order ones in critical case:

[

(p − 1)(p − 3)βA′2 − 2]

A′′ =2κ2

p − 1φ′2, (48)

p(p − 1)(p − 3)βA′4 + 4pA′2 + 4Λ0 =8κ2

p − 1

(1

2φ′2 − V

)

,(49)

φ′′ + pA′φ′ = Vφ, (50)

where Vφ ≡ dVdφ .

Note that eq. (50) can be derived from eqs. (49) and(48). Hence, the above three equations are not independent.

We can solve the above second-order differentialequations by the superpotential method.


Introducing the superpotential function W (φ), theEOMs (49)-(50) can be solved by the first-order equations:

A′ = − κ2

p − 1W , (51)

φ′ =(

1− hW 2)

Wφ, (52)

V =1

2

(

1− hW 2)2

W 2φ − p − 1

2κ2Λ0

− pκ2

2(p − 1)W 2

(

1− 1

2hW 2

)

, (53)

where h = − p−32(p−1)βκ

4.Again, the parameter β has no effect on the Einstein

equations in four-dimensional case.


The energy density ρ(y) of the system is given by

ρ(y) = −TMN uMuN = −T 00 =

1

2φ′2 + V , (54)

where uM = (E−A, 0, 0, 0, 0).For brane solutions, we require that the energy density

on the boundaries of the extra dimension y should vanish:

ρ(|y | → ∞) → 0, (55)

with which the naked cosmological constant Λ0 will bedetermined.

The case n = 4

For n = 4, the EOMs read

A′ = −κ2

2W , (56)

φ′ = Wφ, (57)

V = −3

4κ2W 2 +

1

2W 2

φ − Λ0

κ2. (58)

In order for the scalar to get a kink solution, the potentialV (φ) should at least has two finite vacua. And the usual φ4

potential is a natural choice.However, with the superpotential method, the φ4

potential derived from the superpotential W (φ) = aφ+ bφ2

can not support kink solution for the scalar.

The case n = 4 (cont.)

Hence, we turn to use another superpotential

W (φ) = a(

φ− φ3

3v20

)

, which corresponds to the φ6 model:

V (φ) = − a2κ2

12v40

(

φ2 − v20)2 [

φ2 − 2(3κ−2 + 2v20 )]

(59)

with the two vacua are at φ± = ±v0 (the extreme points ofthe superpotential W (φ)).

The solution is found to be

φ(y) = v0 tanh(ky), (60)

e2A(y) =[

cosh(ky)]− 2

3κ2v2

0 exp(

− 1

6κ2v20 tanh

2(ky))

, (61)

Λ0 = −1

3a2v20κ

4 < 0 , (62)

where k = a/v0.


The scalar curvature and the energy density are

R(y) =1

3a2κ2

[

κ2v20 sech6(ky) + 3(κ2v20 + 3)sech4(ky)− 4κ2v20

]

,(63)

ρ(y) = a2[

κ2v2012

(

3 + sech2(ky))

+ 1

]

sech4(ky). (64)

The energy density vanishes at the boundaries:

ρ(|y | → ∞) ∝ e−4k|y | → 0. (65)

And the bulk curvature R(|y | → ∞) → −43a

2v20κ4 = 4Λ0,

which means that the spacetime is asymptotically AdS. Thecorresponding cosmological constant is just the naked one:Λ = Λ0.


As shown in the following figure, the above solutiondescribes a typical thick braneworld embedded in an AdSspacetime.

-6 -4 -2 2 4 6y

-10

-7.5

-5

-2.5

2.5

5

R

-6 -4 -2 2 4 6y

1

2

3

4

5

Ρ

Figure: The shapes of the scalar curvature R(y) and the energy densityρ(y) for the case n = 4.

The case n 6= 4

For general n 6= 4 (n ≥ 5 and n = 3) and β 6= 0, h 6= 0 andso we have chance to get the usual φ4 potential by settingW = aφ. The potential is

V (φ) = b(φ2 − v20 )2, (66)

where

b =p − 3

8(p − 1)2[

(p − 3)a2κ2β2 − pβ]

a4κ6, (67)

v20 = − 2(p − 1)

(p − 3)a2κ4β. (68)

When (p − 3)β > 0, the above V (φ) (66) is not a usualφ4 potential with two degenerate vacua since v20 < 0. Suchpotential does not support a thick brane solution.


So we consider the case of (p − 3)β < 0, namely,

p > 3 and β < 0, or

p = 2 and β > 0,

for which v20 > 0, b > 0, so the potential V (φ) = b(φ2 − v20 )2

has two vacua at φ± = ±v0. The solution is

φ(y) = v0 tanh(ky), (69)

e2A(y) =[

cosh(ky)]− 2

p−1κ2v2

0 , (70)

Λ0 =p

(p − 1)(p − 3)β< 0 , (71)

where k = a/v0 =√

− (p−3)2(p−1)κ

4β. This solution stands for a

thick flat brane with the energy density given by

ρ(y) =1

2v20

(

k2 + 2bv20)

sech4(ky). (72)


On the boundaries |y | → ∞,

A(|y | → ∞) → −√

−2Λ0

p|y |. (73)

So the asymptotic solution (73) with the relation (71) (Λ0 =

p(p−1)(p−3)β ) is in accord with the thin brane solution

(44)-(45) given in previous section.From the asymptotic solution (73), we have

RMN(|y | → ∞) → 2Λ0gMN = ΛgMN .Therefore, the thick flat brane is embedded in an AdS

spacetime with the effective cosmological constant Λ = 2Λ0.


In n(6= 4)-dimensional critical gravity without matterfields, there are two disconnect AdS vacua with thecosmological constants determined by Λ0 = Λ− (p−1)(p−3)β

4p Λ2

2.In the case here with matter field, because of the

relation Λ0 =p

(p−1)(p−3)β , which is caused by the condition

(55) ρ(|y | → ∞) → 0, the two asymptotic AdS vacua becomethe same one with the cosmological constant Λ = 2Λ0.

2H. Lu and C. N. Pope, Critical Gravity in Four Dimensions, Phys. Rev.Lett. 106 (2011) 181302 [arXiv:1101.1971[hep-th]].

The tensor fluctuation of the metric

The tensor perturbation of the background metric is

ds2 = e2A(y)(ηµν + h̄µν)dxµdxν + dy2. (74)

The fluctuation equations in TT gauge are given by

G(L)MN +

1

2(n − 2)Λ0e

2Ah̄µν + αE(1)(L)MN + βE

(2)(L)MN = 0, (75)

where the fluctuation of G(1)µν is

G (L)µν = −1

2

[

�(4)h̄µν + e2Ah̄′′µν + 4A′e2Ah̄′µν + e2A(4A′2 + A′′)h̄µν

]

,(76)

The tensor fluctuation of the metric (cont.)

and the fluctuations of E(1)µν and E

(2)µν read

E (1)(L)µν = 4(2A′′ + 5A′2)�(4)h̄µν + 4(2A′′ + 5A′2)e2Ah̄′′µν

+8e2A(A′′′ + 9A′A′′ + 10A′3)h̄′µν

−8e2A(2A′′′′ + 16A′A′′′ + 12A′′2 + 37A′2A′′ + 5A′4)h̄µν ,

E (2)(L)µν = −1

2e−2A

�(4)

�(4)h̄µν −�

(4)h̄′′µν

−2A′�

(4)h̄′µν + 2(A′′ + 3A′2)�(4)h̄µν

−1

2e2Ah̄′′′′µν − 4e2AA′h̄′′′µν − 4e2AA′2h̄′′µν

+e2A(40A′3 + 16A′A′′ + 2A′′′)h̄′µν

−e2A(5A′′′′ + 40A′A′′′ + 30A′′2 − 72A′4 + 20A′′A′2)h̄µν .

It is unclear whether the tensor perturbation of the branemetric is stable.

Summary

For f (R) gravity theory:

In the frame of fourth-order f (R) gravity theory, we havefound a smooth thick brane solution.

The brane is generated by a background scalar field withthe usual φ4 potential.

The tensor perturbation of the metric solution is stable.

The 4D massless graviton can be localized on suchbrane.

By introducing the Yukawa coupling between thefermion and the background scalar field, the fermionscan be trapped on the brane.

Summary

For critical gravity theory:

The curvature-squared modifications in the 4D criticalgravity has no effect on the brane solution. So the branesolutions are same with the RS ones in four dimensions

In other dimensions, the thin brane solutions are alsosimilar to the RS ones. But the brane tension can bepositive, negative or vanishing in some conditions.

The thick branes are also similar to the RS ones. Theyare embedded in AdS space-times with effectivecosmological constant Λ = Λ0 and Λ = 2Λ0 for n = 4 andn 6= 4, respectively.

The fluctuation equations of the brane solution are verydifferent for the critical gravity (four-order) and thestandard Einstein gravity (two-order).

Open issues

There are some questions should be addressed:

It is unclear whether the scalar perturbations of thebrane metric are stable in F (R) gravity with matterfields.

It is unclear whether the tensor and scalar perturbationsof the brane metric are stable in critical gravity.

It is not known whether the four-dimensional Newton’sLaw can be recovered on the brane in critical gravity.

The reduced Einstein equations on the branes may beapplied to cosmology.

Thanks!

Documents

Yu-Xiao Liu Institute of Theoretical Physics, Lanzhou ...icts.ustc.edu.cn/chinese/seminar/transparencies/Yu... · ICTS, USTC, Hefei 2011.12.13. Content Introduction and motivation