# On the initial singularity in Kantowski-Sachs spacetime

Elham Ghorani1, and Yaghoub Heydarzade2
1Gorgan, Iran
2Department of Mathematics, Faculty of Sciences, Bilkent University, 06800
Ankara, Turkey email<EMAIL_ADDRESS><EMAIL_ADDRESS>
(August 27, 2024)

###### Abstract

The emergent universe scenario is a proposal for resolving the Big Bang
singularity problem in the standard Friedmann-Lemaitre-Robertson-Walker
cosmology. In the context of this scenario, the Universe originates from a
nonsingular static state. In the present work, considering the realization of
the emergent universe scenario, we address the possibility of having a
nonsingular Kantowski-Sachs type static state. Considering four and five
dimensional models (with and without brane), it is shown that both the
existence and stability of a nonsingular state depend on the dimensions of the
spacetime and the nature of the fluid supporting the geometry.

##  1 Introduction

Besides the great successes of standard model of cosmology based on Einstein’s
theory of general relativity (GR), there still remain several theoretical
issues. One attempt to resolve these problems was the inflationary universe
scenario [1]. Although the inflation resolves some of the problems such as
horizon problem, flatness problem and magnetic monopole problem, but it is
incapable of solving the initial big bang singularity problem. In the standard
model of cosmology, the Universe originates from an initial singular point
where all the mass, energy, and spacetime are infinitely compressed. To cure
this singularity problem, some models such as the ekpyrotic/cyclic universe
[2], the pre-big bang [3], and the emergent universe [4] have been proposed.
In the latter scenario proposed by Ellis et al [4], the Universe has no
timelike singularity, it is ever existing and stands almost static in the
infinite past. The emergent model replaces the initial big bang singularity
with a nonsingular static state, the so called Einstein static state (ESS).
Then, the Universe enters to an inflationary era and produces the same
cosmological history as in the standard model. The existence of a stable ESS
against various perturbations, such as quantum fluctuations, is a prerequisite
for a successful emergent universe scenario. The original emergent model
supposes an ESS that is characterized by Friedmann-Lemaitre-Robertson-Walker
(FLRW) metric with a perfect fluid source. In the framework of Einstein’s GR,
it was demonstrated by Eddington that an ESS is unstable versus the
homogeneous and isotropic perturbations [5]. Later studies by Gibbons [6] and
Barrow et al [7] showed that an ESS can be neutrally stable versus small
inhomogeneous vector and tensor perturbations as well as adiabatic scalar
density perturbations if the perfect fluid filling the universe has a sound
speed $c_{s}^{2}>1/5$. An approach to amend the instability problem is to
modify cosmological field equations of GR. Many works has been performed along
this line in the context of modified theories of gravity such as Einstein-
Cartan theory [8], $f(R)$ gravity [9], $f(T)$ gravity [10], brane gravity
[11], massive gravity [12] and modified Gauss-Bonnet gravity [13] among the
others.

The emergent universe scenario has been analyzed in Einstein’s GR and modified
theories only for the FLRW spacetime. One interesting question that can be
raised here is: Can a non-FLRW static state be a seed for an emergent
universe? To answer this question, one may consider Kantowski-Sachs (KS) [14]
or Bianchi type cosmological models [15] and investigate their stability
versus various perturbations. Our aim in the present work is to explore the
former. KS models are spatially homogeneous and rotation-free but possess
shear. Hence, they represent anisotropic universes. However, it is shown that
the initial anisotropies in the context of these models die away as the
universe expands from the initial singularity if there is a positive
cosmological constant, which is effectively the requirement for an inflation
scenario in the early stages of our universe [16, 17]. Therefore, the universe
could be initially anisotropic KS type, and then it passes through the
inflation and its subsequent cosmological eras toward the current homogeneous
and isotropic state. In [16], KS type anisotropic cosmological models are
classified in the presence of a nonzero cosmological constant. It is shown
that for a positive cosmological constant there exists a set of big bang
models of zero measure as well as a set of models possessing nonzero measure
with the de Sitter asymptotic. In [18], nonsingular KS type cosmological
solutions are obtained in a quantum-corrected Einstein gravity. In deriving
these analytical solutions, the authors used the analogy with the Nariai black
hole. The global structure of the KS cosmological models are studied in [19].
It has been shown that if the energy-momentum source is a perfect fluid, all
the general relativistic models are geodesically incomplete, both to the past
and to the future, and the energy density of the perfect fluid diverges at
each resulting singularity. In [20], KS models are examined versus the
classical tests of cosmology, and the results have been compared with those
belonging to FLRW spacetime in the standard model of cosmology. It is shown
that for a large class of KS models, the observations are incapable of
distinguishing between KS models and the standard model of cosmology. The
reference [21] discusses the growth of density perturbations in KS cosmologies
in the presence of a positive cosmological constant. It is found that when a
bounce occurs in the cosmic scale factor, the density gradient in the bouncing
directions experiences a local maximum at or slightly after the bounce.

The organization of this paper is as follows. In section 2, we introduce the
Kantowski-Sachs metric and its corresponding Einstein field equations in four
dimensions. We obtain the existence and stability conditions versus the scalar
perturbations for a Kantowski-Sachs static state (KSSS) for various energy-
momentum sources with generic linear and Chaplygin gas type equations of
state. The analysis in four dimensions shows that there is only one particular
fluid type that can support a nonsingular KSSS. In section 3, regarding the
result in section 2 and motivated by the fact that anisotropic models can be
exact solutions to the string effective action [22], we apply our analysis to
a braneworld generalization of the model in which a four dimensional KS
spacetime embedded in a five dimensional bulk space. The conditions for the
existence and stability of a nonsingular KSSS are also discussed in this
framework for various fluid types. In section 4, the stability analysis is
done for a 5-dimensional KS model without the brane. Finally, section 5 is
devoted to our concluding remarks.

##  2 KS geometry in four dimensions

We consider that the spacetime metric is the KS type [14]

$ds^{2}=-dt^{2}+a_{1}^{2}(t)dr^{2}+a_{2}^{2}(t)(d\theta^{2}+sin^{2}\theta
d\varphi^{2}),$ (1)

where $a_{1}(t)$ and $a_{2}(t)$ are two arbitrary functions of time and the
energy-momentum tensor supporting this geometry has the generic form
${T^{\mu}}_{\nu}=diag\left(-\rho,\,p_{r},\,p_{t},\,p_{t}\right)$ [23]. Then,
the Einstein field equations considering the above metric and energy-momentum
source read as

$\displaystyle\frac{\dot{a}_{2}^{2}}{a_{2}^{2}}+\frac{2\dot{a}_{1}\dot{a}_{2}}{a_{1}a_{2}}+\frac{1}{a_{2}^{2}}=\Lambda+k_{4}\rho,$
(2)
$\displaystyle\frac{2\ddot{a}_{2}}{a_{2}}+\frac{\dot{a}_{2}^{2}}{a_{2}^{2}}+\frac{1}{a_{2}^{2}}=\Lambda-
k_{4}p_{r},$ (3)
$\displaystyle\frac{\ddot{a}_{1}}{a_{1}}+\frac{\ddot{a}_{2}}{a_{2}}+\frac{\dot{a}_{1}\dot{a}_{2}}{a_{1}a_{2}}=\Lambda-
k_{4}p_{t},$ (4)

where $k_{4}=8\pi G$ is the four dimensional gravitational coupling constant
and $\Lambda$ is a generic (positive or negative) cosmological constant.

###  2.1 KSSS and stability analysis in four dimensions

In the following, we investigate the existence and stability of KS type static
state considering the field equations (2)-(4). To keep the generality of the
analysis, we consider two generic kinds of energy-momentum sources: $(i)$ a
fluid possessing linear equation of state, and $(ii)$ a fluid with generalized
Chaplygin gas type equations of state.

####  2.1.1 Energy-momentum source with linear equations of state

For the sake of generality, here we consider distinct equations of state for
the radial and lateral pressures $p_{r}$ and $p_{t}$, respectively, as [23,
24, 25]

$p_{r}=\omega_{r}\rho+p_{0r},\leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ p_{t}=\omega_{t}\rho+p_{0t}.$ (5)

Then, the field equations (2)-(4) become

$\displaystyle\frac{\dot{a}_{2}^{2}}{a_{2}^{2}}+\frac{2\dot{a}_{1}\dot{a}_{2}}{a_{1}a_{2}}+\frac{1}{a_{2}^{2}}=\Lambda+k_{4}\rho,$
(6)
$\displaystyle\frac{2\ddot{a}_{2}}{a_{2}}+\frac{\dot{a}_{2}^{2}}{a_{2}^{2}}+\frac{1}{a_{2}^{2}}=\Lambda-
k_{4}\omega_{r}\rho-k_{4}p_{0r},$ (7)
$\displaystyle\frac{\ddot{a}_{1}}{a_{1}}+\frac{\ddot{a}_{2}}{a_{2}}+\frac{\dot{a}_{1}\dot{a}_{2}}{a_{1}a_{2}}=\Lambda-
k_{4}\omega_{t}\rho-k_{4}p_{0t}.$ (8)

The static state corresponding to the above system of nonlinear differential
equations is defined as $\dot{a}_{1}(t)=\dot{a}_{2}(t)=\dot{\rho}(t)=0$. Then,
considering the identifications $a_{1}=a_{01}$, $a_{2}=a_{02}$ and
$\rho=\rho_{0}$ for the static state, through the field equations (6)-(8), we
obtain

$\displaystyle\frac{1}{a_{02}^{2}}=\Lambda+k_{4}\rho_{0},$ (9)
$\displaystyle\frac{1}{a_{02}^{2}}=\Lambda-
k_{4}\omega_{r}\rho_{0}-k_{4}p_{0r},$ (10) $\displaystyle 0=\Lambda-
k_{4}\omega_{t}\rho_{0}-k_{4}p_{0t}.$ (11)

From (11), the sign of $\Lambda$ depends on the sign of $\omega_{t}$ and
$p_{0t}$. Using equations (9) and (10), we have

$p_{0r}=-(1+\omega_{r})\rho_{0}=\omega_{eff}\rho_{0}.$ (12)

From (11), one gets the following relation for $p_{0t}$

$p_{0t}=-\omega_{t}\rho_{0}+\frac{\Lambda}{k_{4}}.$ (13)

The equations (12) and (13) represent the constant radial and lateral
effective equations of state of the fluid supporting the static state.

Now, to study the stability of the static state given in Eqs. (9)-(11), we
apply the scalar perturbations to the dynamical quantities of the system
(6)-(8), i.e. the scale factors $a_{1}(t)$ and $a_{2}(t)$ and the energy
density $\rho(t),$ as

$\displaystyle a_{1}(t)\longrightarrow a_{01}\left(1+\delta a_{1}(t)\right),$
$\displaystyle a_{2}(t)\longrightarrow a_{02}\left(1+\delta a_{2}(t)\right),$
$\displaystyle\rho(t)\longrightarrow\rho_{0}\left(1+\delta\rho(t)\right).$
(14)

Hence, the perturbed field equations up to the first order take the following
form

$\displaystyle\frac{1}{a^{2}_{02}}-\frac{2\delta
a_{2}}{a^{2}_{02}}=\Lambda+k_{4}\rho_{0}+k_{4}\rho_{0}\delta\rho,$ (15)
$\displaystyle 2\delta\ddot{a}_{2}+\frac{1}{a^{2}_{02}}-\frac{2\delta
a_{2}}{a^{2}_{02}}=\Lambda-k_{4}\omega_{r}\rho_{0}\delta\rho-
k_{4}\omega_{r}\rho_{0}-k_{4}p_{0r},$ (16)
$\displaystyle\delta\ddot{a}_{1}+\delta\ddot{a}_{2}=\Lambda-
k_{4}\omega_{t}\rho_{0}\delta\rho-k_{4}\omega_{t}\rho_{0}-k_{4}p_{0t}.$ (17)

Using the constraints given by equations (9)-(11), one can reduce these
perturbation equations to

$\displaystyle\frac{2\delta a_{2}}{a^{2}_{02}}=-k_{4}\rho_{0}\delta\rho,$ (18)
$\displaystyle 2\delta\ddot{a}_{2}-\frac{2\delta
a_{2}}{a^{2}_{02}}=-k_{4}\omega_{r}\rho_{0}\delta\rho,$ (19)
$\displaystyle\delta\ddot{a}_{1}+\delta\ddot{a}_{2}=-k_{4}\omega_{t}\rho_{0}\delta\rho.$
(20)

Substituting (18) in (19) leads to

$\delta\ddot{a}_{2}+\gamma^{2}\delta a_{2}=0,$ (21)

where

$\gamma^{2}=-\frac{1+\omega_{r}}{a_{02}^{2}}=\frac{\omega_{eff}}{a_{02}^{2}}.$
(22)

Hence, one realizes that the oscillating modes for $\delta a_{2}$

$\displaystyle\delta a_{2}=C_{1}e^{i\gamma t}+C_{2}e^{-i\gamma t},$ (23)

requires the constraint

$\omega_{r}<-1.$ (24)

Similarly, using (18), (19) and (20), we obtain

$2\delta\ddot{a}_{1}=k_{4}(1+\omega_{r}-2\omega_{t})\rho_{0}\delta\rho.$ (25)

Combining (18) and (19) gives

$\delta\rho=\frac{-2\delta\ddot{a}_{2}}{k_{4}(1+\omega_{r})\rho_{0}},$ (26)

where substituting in (25) yields

$\delta\ddot{a}_{1}=\left(\frac{2\omega_{t}}{1+\omega_{r}}-1\right)\delta\ddot{a}_{2},$
(27)

which represents the relation between the radial and lateral perturbations.
Twice integration of this equation gives

$\displaystyle\delta
a_{1}=\left(\frac{2\omega_{t}}{1+\omega_{r}}-1\right)\delta
a_{2}+C_{3}t+C_{4}$ (28)

where $C_{3}$ and $C_{4}$ are integration constants. Regarding (23), the above
equation implies that the stable oscillatory modes in $\delta a_{1}$ is
subjected to the condition $C_{3}=C_{4}=0$. In this case, the amplitude of the
perturbations along the radial direction depends explicitly on the nature of
the fluid, i.e on the equations of state parameters. For the particular case
$\omega_{t}=1+\omega_{r}$, the perturbation amplitude on the radial and
lateral directions are the same.

In the following, we elaborate on two specific forms of the fluid (5) and
discuss on the stability of KSSS.

2.1.1.1 Perfect fluid

Considering a perfect fluid form for the supporting matter fields by setting
$\omega_{r}=\omega_{t}=\omega$ and $p_{0r}=p_{0t}=0$, equations (9) to (11)
for the static state reduce to

$\displaystyle\frac{1}{a_{02}^{2}}=\Lambda+k_{4}\rho_{0},$ (29)
$\displaystyle\frac{1}{a_{02}^{2}}=\Lambda-k_{4}\omega\rho_{0},$ (30)
$\displaystyle 0=\Lambda-k_{4}\omega\rho_{0}.$ (31)

Substituting $p_{0r}=0$ into Eq.(12), one gets either $w_{r}=-1$ or
$\rho_{0}=0$. The same result can be observed by comparing the equations (29)
and (30). If $\rho_{0}=0$, Eqs.(29) and (30) will be source free (because of
(31)) and cannot be held anymore for finite $a_{0}$. If
$\omega=\omega_{r}=\omega_{t}=-1$ for a perfect fluid, Eq.(31) (which demands
$\Lambda<0$) is inconsistent with Eqs.(29) and (30). In either way, one
observes that having a finite size static sate for the KS spacetime is
impossible. Then, a perfect fluid cannot support a nonsingular KSSS. Hence,
for the realization of a stable nonsingular KSSS, the modification of the
perfect fluid from is inevitable.

2.1.1.2 Anisotropic fluid

Here, we consider two possible minimal modifications of the perfect fluid form
leading to anisotropic fluids: $(1)$ a fluid having different equations of
state parameters on the radial and lateral directions, and $(2)$ a fluid
modifying the perfect fluid form with distinct constant pressures on the
radial and lateral directions.

Type 1:
In this case, considering two different equations of state parameters for the
radial and lateral directions we have

$p_{r}=\omega_{r}\rho,\leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ p_{t}=\omega_{t}\rho.$ (32)

Then, equations (9) to (11) governing the static state reduce to

$\displaystyle\frac{1}{a_{02}^{2}}=\Lambda+k_{4}\rho_{0},$ (33)
$\displaystyle\frac{1}{a_{02}^{2}}=\Lambda-k_{4}\omega_{r}\rho_{0},$ (34)
$\displaystyle 0=\Lambda-k_{4}\omega_{t}\rho_{0}.$ (35)

Satisfaction of (35) demands that $\Lambda$ and $\omega_{t}$ have the same
sign. Comparing equations (33) and (34) we obtains the constraint
$\omega_{r}=-1$ on the equation of state in the radial direction to have a
nonsingular static state. On the other hand, we have seen that the stability
of KSSS demands $\omega_{r}<-1$, i.e. equation (24). Therefore, one can
conclude that the anisotropic fluid from defined in (32) can support a static
state but this static state is not stable versus the scalar perturbations
(2.1.1).

Type 2:
We consider a fluid modifying the perfect fluid form by distinct constant
pressures on the radial and lateral directions. For this aim, we introduce the
constant pressure terms $p_{0t}$ and $p_{0r}$ to the equations of state of the
fluid with $\omega_{r}=\omega_{t}=\omega$. In this case, the equations in (5)
become

$p_{r}=\omega\rho+p_{0r},\leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ p_{t}=\omega\rho+p_{0t}.$ (36)

Then equations (12) and (13) for the static state reduce to

$\displaystyle p_{0r}=-(1+\omega)\rho_{0},$ $\displaystyle
p_{0t}=-\omega\rho_{0}+\frac{\Lambda}{k_{4}}.$ (37)

Combining them gives the following relation between the constant pressures
$p_{0r}$ and $p_{0t}$

$p_{0t}-p_{0r}=\rho_{0}+\frac{\Lambda}{k_{4}}.$ (38)

In the case $\rho_{0}+\frac{\Lambda}{k_{4}}=0$, we face an inconsistency in
the static state given by equations (9) to (11). Thus, to have a static state
we consider $\rho_{0}+\frac{\Lambda}{k_{4}}\neq 0$. In this case, $\gamma^{2}$
given by equation (22) becomes

$\displaystyle\gamma^{2}=-\frac{1+\omega}{a_{02}^{2}}.$ (39)

Therefore, the positivity condition on $\gamma^{2}$ to have a stable
nonsingular static state demands

$\omega<-1,$ (40)

which means that the fluid supporting the geometry lies in the phantom range.

For the stability of this case, the oscillatory modes of $\delta a_{2}$ is
given by equation (23) with $\gamma$ in (154). The dynamics of $\delta a_{1}$
from equation (28) reads as

$\delta a_{1}=\left(\frac{\omega-1}{\omega+1}\right)\delta
a_{2}+C_{3}t+C_{4},$ (41)

where for $C_{3}=C_{4}=0$, the oscillatory modes are possible and hence the
static state will be stable. Regarding the stability requirement (155), one
notes that although the perturbation in both directions have the same sign but
the amplitude of the perturbations on the radial direction is always greater
than the perturbations on the lateral direction.

####  2.1.2 Energy-momentum source with generalized Chaplygin gas equation of
state

Here, we consider a generalization of the Chaplygin gas type fluid [26, 27]
for the energy-momentum source possessing

$p_{r}=-\frac{\alpha_{r}}{\rho^{n}},\leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ p_{t}=-\frac{\alpha_{t}}{\rho^{m}}.$
(42)

where $0\leq m,n\leq 1$, $\alpha_{r}$ and $\alpha_{t}$ are positive constants.
If $\alpha_{r}=\alpha_{t}$ and $m=n=1$, the energy-momentum source reduces to
the generalized Chaplygin gas in [27]. The equations governing a static state
have the form

$\displaystyle\frac{1}{a_{02}^{2}}=\Lambda+k_{4}\rho_{0},$ (43)
$\displaystyle\frac{1}{a_{02}^{2}}=\Lambda+k_{4}\frac{\alpha_{r}}{\rho_{0}^{n}},$
(44) $\displaystyle 0=\Lambda+k_{4}\frac{\alpha_{t}}{\rho_{0}^{m}}.$ (45)

Satisfaction of (45) demands $\Lambda<0$ for $\alpha_{t}>0$. Then, using
(2.1.1), the perturbed field equations around the static state given by
equations (43)-(45) reduce to

$\displaystyle\frac{2\delta a_{2}}{a^{2}_{02}}=-k_{4}\rho_{0}\delta\rho,$ (46)
$\displaystyle 2\delta\ddot{a}_{2}-\frac{2\delta
a_{2}}{a^{2}_{02}}=-nk_{4}\frac{\alpha_{r}}{\rho_{0}^{n}}\delta\rho,$ (47)
$\displaystyle\delta\ddot{a}_{1}+\delta\ddot{a}_{2}=-mk_{4}\frac{\alpha_{t}}{\rho_{0}^{m}}\delta\rho.$
(48)

Combining (46) and (47), we obtain

$\delta\ddot{a}_{2}+\gamma^{2}\delta a_{2}=0,$ (49)

where

$\gamma^{2}=-\frac{n\alpha_{r}\rho_{0}^{-(n+1)}+1}{a_{02}^{2}}.$ (50)

Here, one observes that since $\alpha_{r}>0$, then $\gamma^{2}<0$ and
consequently there are no oscillatory modes for $\delta a_{2}$ and $\delta
a_{1}$.

We recapitulate our analysis in this section as follows. It is proved that in
the context of Einstein’s GR in four dimensions: $(i)$ A perfect fluid source
cannot support a finite size static KS geometry, $(ii)$ An anisotropic fluid
with linear equations of state $p_{r}=\omega_{r}\rho$ and
$p_{t}=\omega_{t}\rho\leavevmode\nobreak\ $ supports a finite size static KS
geometry but it is not stable against the scalar perturbations, $(iii)$ A
modification in the linear equation of state of the perfect fluid form as
$p_{r}=\omega\rho+p_{0r}$ and $p_{t}=\omega\rho+p_{0t}$ can support a stable
nonsingular KS type static state, and $(iv)$ A generalized Chaplygin gas fluid
having $p_{r}=-\frac{\alpha_{r}}{\rho^{n}},\leavevmode\nobreak\
p_{t}=-\frac{\alpha_{t}}{\rho^{m}}$ cannot support a stable nonsingular KS
geometry in four spacetime dimensions.

In the following section, since anisotropic models are exact solutions to the
string theory, we investigate the existence and stability conditions for a
KSSS in a five dimensional gravity theory, and show how the extra dimensional
geometric modifications affect the results in four dimensions.

##  3 KS geometry on the brane

We consider the five dimensional action [28]

$\displaystyle S=\int
d^{5}x\sqrt{-\mathcal{G}}\left(\frac{1}{2k_{5}}\mathcal{R}-\Lambda_{5}\right)+\int_{\xi=0}d^{4}x\sqrt{-g}\left(\frac{1}{k_{5}}K^{\pm}-\lambda+L_{matter}\right),$
(51)

where $k_{5}=8\pi G_{5}$, $\mathcal{G}$, $g$, $\mathcal{R}$, $\Lambda_{5}$,
$\lambda$, and $L_{matter}$ are the 5-dimensional gravitational coupling
constant, trace of the bulk space’s metric $\mathcal{G}_{AB}$, trace of the
brane’s metric $g_{\mu\nu}$, bulk space’s Ricci scalar, bulk and brane vacuum
energy, and the Lagrangian of the confined matter fields to brane,
respectively. Also, $x^{\mu}$ with $\mu=0,...,3$ represents the coordinates of
the brane while $\xi$ represents the single extra dimension orthogonal to the
brane, and $K^{\pm}$ is the extrinsic curvature on either sides of the brane.

Variation of the action (51) with respect to the bulk metric
$\mathcal{G}_{AB}$ gives the $5D$ Einstein field equations

${}^{(5)}G_{ab}=k_{5}{}^{(5)}T_{ab},$
$\displaystyle{}^{(5)}T_{ab}=-\Lambda_{5}\mathcal{G}_{ab}+\delta(\xi)\left(-\lambda\mathcal{G}_{ab}+T_{ab}^{matter}\right),$
(52)

where $a,b=0,...,4$. In order to keep the generality of our analysis, similar
to the four dimensional case, we consider the energy-momentum tensor
$T^{matter}_{\mu\nu}$ corresponding to $L_{matter}$ on the brane as
$T^{\mu}_{\nu}=diag\left(-\rho,\,p_{r},\,p_{t},\,p_{t}\right)$ [23].
Considering the bulk space metric in the form

$ds^{2}=\left(N_{a}N_{b}+g_{ab}\right)dx^{a}dx^{b},$ (53)

where $N^{a}$ represents the unit normal vector to the hypersurface
$\xi=constant$ and $g_{ab}$ is the induced metric on this hypersurface, the
induced field equations on the brane take the following form [29]

$G_{\mu\nu}=-\Lambda
g_{\mu\nu}+k_{4}T_{\mu\nu}+k_{5}^{2}S_{\mu\nu}-E_{\mu\nu},$ (54)

in which

$\displaystyle
S_{\mu\nu}=\frac{1}{12}TT_{\mu\nu}-\frac{1}{4}T_{\mu}^{\alpha}T_{\alpha\nu}+\frac{1}{24}g_{\mu\nu}(3T^{\alpha\beta}T_{\alpha\beta}-T^{2}).$

Here $\Lambda=k_{5}(\Lambda_{5}+k_{5}\lambda^{2}/6)$ is the effective
cosmological constant on the brane, and $E_{ab}=C_{abcd}N^{a}N^{b}$ where
$C_{abcd}$ represents the 5-dimensional Weyl tensor of the bulk space.

Considering the brane cosmology with zero Weyl tensor and a generic energy-
momentum tensor possessing the form we mentioned above, the Einstein field
equations for the metric (1) on the brane read as

$\displaystyle\frac{\dot{a}_{2}^{2}}{a_{2}^{2}}+\frac{2\dot{a}_{1}\dot{a}_{2}}{a_{1}a_{2}}+\frac{1}{a_{2}^{2}}=\Lambda+k_{4}\rho+\frac{1}{12}k_{5}^{2}(\rho^{2}-p_{r}^{2}-p_{t}^{2}+2p_{r}p_{t}),$
(55)
$\displaystyle\frac{2\ddot{a}_{2}}{a_{2}}+\frac{\dot{a}_{2}^{2}}{a_{2}^{2}}+\frac{1}{a_{2}^{2}}=\Lambda-
k_{4}p_{r}-\frac{1}{12}k_{5}^{2}(\rho^{2}-p_{r}^{2}+p_{t}^{2}+2\rho p_{t}),$
(56)
$\displaystyle\frac{\ddot{a}_{1}}{a_{1}}+\frac{\ddot{a}_{2}}{a_{2}}+\frac{\dot{a}_{1}\dot{a}_{2}}{a_{1}a_{2}}=\Lambda-
k_{4}p_{t}-\frac{1}{12}k_{5}^{2}(\rho^{2}+p_{r}^{2}+\rho p_{r}+\rho
p_{t}-p_{r}p_{t}).$ (57)

###  3.1 KSSS and stability analysis on the brane

In the following, we investigate the existence and stability of KS type static
state on the 4D brane considering the field equations (55)-(57). For the sake
of generality of the analysis, we study two generic kinds of energy-momentum
sources: $(i)$ the sources possessing linear equation of state, and $(ii)$ the
sources with generalized Chaplygin gas type equation of state.

####  3.1.1 Energy-momentum source with linear equations of state

We consider a general equation of state with the form given in (5), and
introduce the parameters

$\displaystyle
A=\frac{1}{12}k_{5}^{2}(1-\omega_{r}^{2}-\omega_{t}^{2}+2\omega_{r}\omega_{t}),$
$\displaystyle
B=k_{4}+\frac{1}{6}k_{5}^{2}\left(\omega_{r}(p_{0t}-p_{0r})+\omega_{t}(p_{0r}-p_{0t})\right),$
$\displaystyle C=\frac{1}{12}k_{5}^{2}(2p_{0r}p_{0t}-p_{0r}^{2}-p_{0t}^{2}),$
$\displaystyle
D=\frac{1}{12}k_{5}^{2}(1-\omega_{r}^{2}+\omega_{t}^{2}+2\omega_{t}),$
$\displaystyle
E=\frac{1}{6}k_{5}^{2}(\omega_{t}p_{0t}-\omega_{r}p_{0r}+p_{0t})+k_{4}\omega_{r},$
$\displaystyle F=\frac{1}{12}k_{5}^{2}(p_{0r}^{2}-p_{0t}^{2})-k_{4}p_{0r},$
$\displaystyle
G=\frac{1}{12}k_{5}^{2}(1+\omega_{r}^{2}+\omega_{r}+\omega_{t}-\omega_{r}\omega_{t}),$
$\displaystyle
H=k_{4}\omega_{t}+\frac{1}{12}k_{5}^{2}\left(\omega_{r}(2p_{0r}-p_{0t})-\omega_{t}p_{0r}+p_{0r}+p_{0t}\right),$
$\displaystyle I=\frac{1}{12}k_{5}^{2}(p_{0r}p_{0t}-p_{0r}^{2})-k_{4}p_{0t}.$
(58)

Hence the field equations (55)-(57) reduce to the following forms

$\displaystyle\frac{\dot{a}_{2}^{2}}{a_{2}^{2}}+\frac{2\dot{a}_{1}\dot{a}_{2}}{a_{1}a_{2}}+\frac{1}{a_{2}^{2}}=\Lambda+A\rho^{2}+B\rho+C,$
(59)
$\displaystyle\frac{2\ddot{a}_{2}}{a_{2}}+\frac{\dot{a}_{2}^{2}}{a_{2}^{2}}+\frac{1}{a_{2}^{2}}=\Lambda-D\rho^{2}-E\rho+F,$
(60)
$\displaystyle\frac{\ddot{a}_{1}}{a_{1}}+\frac{\ddot{a}_{2}}{a_{2}}+\frac{\dot{a}_{1}\dot{a}_{2}}{a_{1}a_{2}}=\Lambda-G\rho^{2}-H\rho+I.$
(61)

For a static state defined by
$\dot{a}_{1}=\dot{a}_{2}=\ddot{a}_{1}=\ddot{a}_{2}=0$, let $a_{1}=a_{01}$,
$a_{2}=a_{02}$ and $\rho=\rho_{0}$. Then, the field equations (59)-(61) give

$\displaystyle\frac{1}{a_{02}^{2}}=\Lambda+A\rho_{0}^{2}+B\rho_{0}+C,$ (62)
$\displaystyle\frac{1}{a_{02}^{2}}=\Lambda-D\rho_{0}^{2}-E\rho_{0}+F,$ (63)
$\displaystyle 0=\Lambda-G\rho_{0}^{2}-H\rho_{0}+I.$ (64)

The Using equations (62) and (63), we have the constraint equation for
$\rho_{0}$

$(A+D)\rho_{0}^{2}+(B+E)\rho_{0}+C-F=0,$ (65)

Now, to study the stability of static state defined in (62)-(64), we consider
the scalar perturbations in the form of equations (2.1.1). Keeping up to the
first order perturbation terms, equations (59) to (61) give

$\displaystyle\frac{1}{a^{2}_{02}}-\frac{2\delta
a_{2}}{a^{2}_{02}}=\Lambda+A\rho_{0}^{2}(1+2\delta\rho)+B\rho_{0}(1+\delta\rho)+C,$
(66) $\displaystyle 2\delta\ddot{a}_{2}+\frac{1}{a^{2}_{02}}-\frac{2\delta
a_{2}}{a^{2}_{02}}=\Lambda-D\rho_{0}^{2}(1+2\delta\rho)-E\rho_{0}(1+\delta\rho)+F,$
(67)
$\displaystyle\delta\ddot{a}_{1}+\delta\ddot{a}_{2}=\Lambda-G\rho_{0}^{2}(1+2\delta\rho)-H\rho_{0}(1+\delta\rho)+I.$
(68)

Using the constraints given by equations (62) to (64), one can reduce above
equations to

$\displaystyle-\frac{2\delta
a_{2}}{a^{2}_{02}}=(2A\rho_{0}^{2}+B\rho_{0})\delta\rho,$ (69) $\displaystyle
2\delta\ddot{a}_{2}-\frac{2\delta
a_{2}}{a^{2}_{02}}=(-2D\rho_{0}^{2}-E\rho_{0})\delta\rho,$ (70)
$\displaystyle\delta\ddot{a}_{1}+\delta\ddot{a}_{2}=(-2G\rho_{0}^{2}-H\rho_{0})\delta\rho.$
(71)

From equation (69), we have

$\delta\rho=-\frac{2}{a^{2}_{02}(2A\rho_{0}^{2}+B\rho_{0})}\delta a_{2},$ (72)

where substituting in equation (70) leads to

$\delta\ddot{a}_{2}+\alpha^{2}\delta a_{2}=0,$ (73)

where

$\alpha^{2}=-\frac{1}{a_{02}^{2}}\left(1+\frac{2D\rho_{0}+E}{2A\rho_{0}+B}\right).$
(74)

Hence, the oscillating modes for $\delta a_{2}$ requires

$1+\frac{2D\rho_{0}+E}{2A\rho_{0}+B}<0.$ (75)

The solution to equation (73) with the condition (75) is

$\displaystyle\delta a_{2}=C_{1}e^{i\alpha t}+C_{2}e^{-i\alpha t},$ (76)

which shows the oscillatory behavior of $\delta a_{2}$.
On the other hand, combining equations (69), (70) and (71) gives

$2\delta\ddot{a}_{1}=((2A+2D-4G)\rho_{0}^{2}+(B+E-2H)\rho_{0})\delta\rho,$
(77)

and subtracting (70) from (69) leads to

$2\delta\ddot{a}_{2}=-((2A+2D)\rho_{0}^{2}+(B+E)\rho_{0})\delta\rho.$ (78)

Hence, using (78) and (77) we obtain

$\delta\ddot{a}_{1}=\beta\delta\ddot{a}_{2},$ (79)

where $\beta$ is defined as

$\beta=\frac{4G\rho_{0}+2H}{(2A+2D)\rho_{0}+B+H}-1.$ (80)

Twice integration of equation (79) gives

$\displaystyle\delta a_{1}=\beta\delta a_{2}+C_{3}t+C_{4}.$ (81)

This shows that $\delta a_{1}$ can also have an stable oscillatory mode if
$C_{3}=C_{4}=0$, and

$2(A+D)\rho_{0}+B+H\neq 0.$ (82)

Hence, both the conditions in (75) and (82) should be satisfied to have a
stable static state. This result holds for a fluid with the general form of
equation of state (5). In Figures 1, the existence and stability conditions
(i.e. equations (64), (65), (75) and (82)) for a static state are plotted for
some typical values of the parameters. The presence of some $\rho_{0}>0$
ranges represents the satisfaction of the constraints that means a stable
static state exists for the given values of the parameters. The intersection
of $Y=0$ and $\Phi=0$ lies in the range $\omega_{r}<-1$ that guarantees a
stable static state. When $p_{0r}=0.1$ and $p_{0t}=0.2$, $\omega_{t}$ should
be also negative, but for $p_{0r}=0.1$, $p_{0t}=-0.1$, and small amounts of
$\rho_{0}>0$ , the constraints for having a stable static state can be
satisfied for a positive value of $\omega_{t}$. It is also clear from figure 1
that, in this setup, changing the value of $p_{0t}$ did not affect the result
of equation (65), but have a significant effect on equation (64) for small
values of $\rho_{0}>0$.

Figure 1:  The left plots show the intersecting surfaces corresponding to
constraints (64) and (65) for having a static state. $Y=0$ represents equation
(64) and $\Phi=0$ represents equation (65). The right plots show the
intersection of surfaces for the existence conditions in the solution space of
stability conditions, i.e equations (75) and (82). The curves in the right
plots satisfy all the conditions (64),(65),(75) and (82) to have a totally
stable static state. We used the typical values of $p_{0r}=0.1$ and
$p_{0t}=0.2$ for the first line and $p_{0r}=0.1$ and $p_{0t}=-0.1$ for the
second line. We also assumed $k_{4}=1$, $k_{5}=1.5$, $\Lambda=10^{-5}$.

In the following, we consider some special types of fluids and discuss under
what conditions these fluids can support a nonsingular KSSS on the brane.

3.1.1.1 Perfect fluid

In this case, we consider perfect fluid which means
$\omega_{r}=\omega_{t}=\omega$ and $p_{0r}=p_{0t}=0$. Then, the coefficients A
to I defined in equations (3.1.1) reduce to

$\displaystyle A=\frac{1}{12}k_{5}^{2},$ $\displaystyle B=k_{4},$
$\displaystyle D=G=\frac{1}{12}k_{5}^{2}(1+2\omega),$ $\displaystyle
E=H=k_{4}\omega,$ $\displaystyle C=F=I=0.$ (83)

For the the static state given by equations (62) to (64), one obtains

$\displaystyle\frac{1}{a_{02}^{2}}=\Lambda+\frac{1}{12}k_{5}^{2}\rho_{0}^{2}+k_{4}\rho_{0},$
(84)
$\displaystyle\frac{1}{a_{02}^{2}}=\Lambda-\frac{1}{12}k_{5}^{2}(1+2\omega)\rho_{0}^{2}-k_{4}\omega\rho_{0},$
(85) $\displaystyle
0=\Lambda-\frac{1}{12}k_{5}^{2}(1+2\omega)\rho_{0}^{2}-k_{4}\omega\rho_{0}.$
(86)

Similar to four dimensions, these equations lead to

$\frac{1}{a_{02}^{2}}=0,$ (87)

which means that a perfect fluid cannot support a KSSS even in the presence of
higher dimensional modifications.

3.1.1.2 Anisotropic fluid

In four dimensions, we showed that a stable static state can exist only for an
anisotropic fluid of Type 2. Here, we show that in the presence of higher
dimensional modifications to the field equations, both Type 1 and Type 2
fluids can support a stable nonsingular KSSS.

Type 1:
In this case we consider equations of state in (32). Hence, the coefficients A
to I defined by equations (3.1.1) reduce to

$\displaystyle A$ $\displaystyle=$
$\displaystyle\frac{1}{12}k_{5}^{2}(1-\omega_{r}^{2}-\omega_{t}^{2}+2\omega_{r}\omega_{t}),$
$\displaystyle B$ $\displaystyle=$ $\displaystyle k_{4},$ $\displaystyle D$
$\displaystyle=$
$\displaystyle\frac{1}{12}k_{5}^{2}(1-\omega_{r}^{2}+\omega_{t}^{2}+2\omega_{t}),$
$\displaystyle E$ $\displaystyle=$ $\displaystyle k_{4}\omega_{r},$
$\displaystyle G$ $\displaystyle=$
$\displaystyle\frac{1}{12}k_{5}^{2}(1+\omega_{r}^{2}+\omega_{r}+\omega_{t}-\omega_{r}\omega_{t}),$
$\displaystyle H$ $\displaystyle=$ $\displaystyle k_{4}\omega_{t},$
$\displaystyle C$ $\displaystyle=$ $\displaystyle F=I=0.$ (88)

Then the constraints (64) and (65) for the existence of a static state,
respectively, become

$\Lambda-\left(\frac{1}{12}k_{5}^{2}(1+\omega_{r}^{2}+\omega_{r}+\omega_{t}-\omega_{r}\omega_{t})\right)\rho_{0}^{2}-k_{4}\omega_{t}\rho_{0}=0,$
(89)

and

$\frac{1}{6}k_{5}^{2}(1-\omega_{r}^{2}+\omega_{t}+\omega_{r}\omega_{t})\rho_{0}^{2}+k_{4}(1+\omega_{r})\rho_{0}=0.$
(90)

Also the conditions (75) and (82) to have an oscillatory mode, respectively,
become

$\frac{\frac{1}{6}k_{5}^{2}(1-\omega_{r}^{2}+\omega_{t}^{2}+2\omega_{t})\rho_{0}+k_{4}\omega_{r}}{\frac{1}{6}k_{5}^{2}(1-\omega_{r}^{2}-\omega_{t}^{2}+2\omega_{r}\omega_{t})\rho_{0}+k_{4}}<-1,$
(91)

and

$\rho_{0}\neq-\frac{3k_{4}}{k_{5}^{2}\left(1+\frac{\omega_{r}(\omega_{t}-\omega_{r})}{1+\omega_{t}}\right)}.$
(92)

Thus, in contrast to four dimensions, the anisotropic fluid of Type 1 can
support a stable nonsingular ESU on the brane under the conditions (89)-(92).
In Figure 2, these conditions are plotted for some typical values of the
parameters. We see from the figure that the constraints can be satisfied for
$\rho_{0}>0$. This means a stable static state do exist for the given values
of the parameters. In this case, it is obvious that to have a stable static
state we need $\omega_{r}<-1$, but $\omega_{t}$ can have any value in the
given range of $-1.5<\omega_{t}<1$.

Figure 2:  The first plot shows the intersecting surfaces corresponding to
constraints (89) and (90) for having a static state. $Y=0$ and $\Phi=0$
represent equations (89) and (90), respectively. The second plot shows the
intersection of surfaces for the existence conditions in the solution space of
stability conditions, i.e equations (91) and (92). The value of the parameters
in blue in the second plot satisfy all the conditions (89),(90),(91) and (92)
to have a totally stable static state. The used typical values and ranges are:
$k_{4}=1$, $k_{5}=1.5$, $p_{0r}=0$, $p_{0t}=0$, $\Lambda=10^{-5}$,
$-1.5<\omega_{r},\leavevmode\nobreak\ \omega_{t}<1$, $0.0001<\rho_{0}<5$.

Type 2:
Now we consider anisotropic fluid given in (36). Thus, the coefficients in
(3.1.1) become

$\displaystyle A=\frac{1}{12}k_{5}^{2},$ $\displaystyle B=k_{4},$
$\displaystyle C=\frac{1}{12}k_{5}^{2}(2p_{0r}p_{0t}-p_{0r}^{2}-p_{0t}^{2}),$
$\displaystyle D=G=\frac{1}{12}k_{5}^{2}(1+2\omega),$ $\displaystyle
E=\frac{1}{6}k_{5}^{2}(\omega p_{0t}-\omega p_{0r}+p_{0t})+k_{4}\omega,$
$\displaystyle F=\frac{1}{12}k_{5}^{2}(p_{0r}^{2}-p_{0t}^{2})-k_{4}p_{0r},$
$\displaystyle H=k_{4}\omega+\frac{1}{12}k_{5}^{2}(\omega p_{0r}-\omega
p_{0t}+p_{0r}+p_{0t}),$ $\displaystyle
I=\frac{1}{12}k_{5}^{2}(p_{0r}p_{0t}-p_{0r}^{2})-k_{4}p_{0t}.$ (93)

Then the constraints (64) and (65) obtained for the existence of a static
state, respectively, become

$\displaystyle\Lambda-\frac{1}{12}k_{5}^{2}(1+2\omega)\rho_{0}^{2}-\left(k_{4}\omega+\frac{1}{12}k_{5}^{2}(\omega
p_{0r}-\omega
p_{0t}+p_{0r}+p_{0t})\right)\rho_{0}+\frac{1}{12}k_{5}^{2}(p_{0r}p_{0t}-p_{0r}^{2})-k_{4}p_{0t}=0,$
(94)

and

$\displaystyle\frac{1}{6}k_{5}^{2}(1+\omega)\rho_{0}^{2}+\left(k_{4}(1+\omega)+\frac{1}{6}k_{5}^{2}(\omega
p_{0t}-\omega
p_{0r}+p_{0t})\right)\rho_{0}+\frac{1}{6}k_{5}^{2}(p_{0r}p_{0t}-p_{0r}^{2})+k_{4}p_{0r}=0.$
(95)

In this case, the conditions (75) and (82) governing the stability of the the
static state become

$\frac{\frac{1}{6}k_{5}^{2}(1+2\omega)\rho_{0}+\frac{1}{6}k_{5}^{2}(\omega
p_{0t}-\omega
p_{0r}+p_{0t})+k_{4}\omega}{\frac{1}{6}k_{5}^{2}\rho_{0}+k_{4}}<-1,$ (96)

and

$\rho_{0}\neq-\frac{3k_{4}}{k_{5}^{2}}-\frac{1}{4}\left(p_{0r}+p_{0t}(\frac{1-\omega}{1+\omega})\right).$
(97)

Figure 3 shows the possibility of satisfying the conditions (94) to (97).
Similar to GR, considering $\omega<-1$, we can have a stable static state for
the given value of parameters.

Figure 3:  The figures show the contours for constraints (94) and (95) in the
solution space of equations (96) and (97). $Y=0$ represents equation (94) and
$\Phi=0$ represents equation (95). The used typical values and ranges are:
$k_{4}=1$, $k_{5}=1.5$, $\Lambda=10^{-5}$, $-1<p_{0r},\leavevmode\nobreak\
p_{0t}<1$, and $\omega=-\frac{3}{2}$. Here, the surfaces $Y=0$ and $\Phi=0$
intersect on a line that means the satisfaction of all the conditions
(94),(95),(96) and (97).

####  3.1.2 Energy-momentum source with generalized Chaplygin gas equation of
state

For an energy-momentum source with a generalized Chaplygin equation of state
of the form (42), the Einstein field equations on the brane will be

$\displaystyle\frac{\dot{a}_{2}^{2}}{a_{2}^{2}}+\frac{2\dot{a}_{1}\dot{a}_{2}}{a_{1}a_{2}}+\frac{1}{a_{2}^{2}}=\Lambda+k_{4}\rho+\frac{1}{12}k_{5}^{2}\left(\rho^{2}-\frac{\alpha_{r}^{2}}{\rho^{2n}}-\frac{\alpha_{t}^{2}}{\rho^{2m}}+2\frac{\alpha_{r}\alpha_{t}}{\rho^{m+n}}\right),$
(98)
$\displaystyle\frac{2\ddot{a}_{2}}{a_{2}}+\frac{\dot{a}_{2}^{2}}{a_{2}^{2}}+\frac{1}{a_{2}^{2}}=\Lambda+k_{4}\frac{\alpha_{r}}{\rho^{n}}-\frac{1}{12}k_{5}^{2}\left(\rho^{2}-\frac{\alpha_{r}^{2}}{\rho^{2n}}+\frac{\alpha_{t}^{2}}{\rho^{2m}}-2\frac{\alpha_{t}}{\rho^{m-1}}\right),$
(99)
$\displaystyle\frac{\ddot{a}_{1}}{a_{1}}+\frac{\ddot{a}_{2}}{a_{2}}+\frac{\dot{a}_{1}\dot{a}_{2}}{a_{1}a_{2}}=\Lambda+k_{4}\frac{\alpha_{t}}{\rho^{m}}-\frac{1}{12}k_{5}^{2}\left(\rho^{2}+\frac{\alpha_{r}^{2}}{\rho^{2n}}-\frac{\alpha_{r}}{\rho^{n-1}}-\frac{\alpha_{t}}{\rho^{m-1}}-\frac{\alpha_{r}\alpha_{t}}{\rho^{m+n}}\right).$
(100)

The corresponding static state is given by the following equations

$\displaystyle\frac{1}{a_{02}^{2}}=\Lambda+k_{4}\rho_{0}+\frac{1}{12}k_{5}^{2}\left(\rho_{0}^{2}-\frac{\alpha_{r}^{2}}{\rho_{0}^{2n}}-\frac{\alpha_{t}^{2}}{\rho_{0}^{2m}}+2\frac{\alpha_{r}\alpha_{t}}{\rho_{0}^{m+n}}\right),$
(101)
$\displaystyle\frac{1}{a_{02}^{2}}=\Lambda+k_{4}\frac{\alpha_{r}}{\rho_{0}^{n}}-\frac{1}{12}k_{5}^{2}\left(\rho_{0}^{2}-\frac{\alpha_{r}^{2}}{\rho_{0}^{2n}}+\frac{\alpha_{t}^{2}}{\rho_{0}^{2m}}-2\frac{\alpha_{t}}{\rho_{0}^{m-1}}\right),$
(102) $\displaystyle
0=\Lambda+k_{4}\frac{\alpha_{t}}{\rho_{0}^{m}}-\frac{1}{12}k_{5}^{2}\left(\rho_{0}^{2}+\frac{\alpha_{r}^{2}}{\rho_{0}^{2n}}-\frac{\alpha_{r}}{\rho_{0}^{n-1}}-\frac{\alpha_{t}}{\rho_{0}^{m-1}}-\frac{\alpha_{r}\alpha_{t}}{\rho_{0}^{m+n}}\right).$
(103)

Combining equations (101) and (102) we get

$k_{4}\left(\rho_{0}-\frac{\alpha_{r}}{\rho_{0}^{n}}\right)+\frac{1}{6}k_{5}^{2}\left(\rho_{0}^{2}-\frac{\alpha_{r}^{2}}{\rho_{0}^{2n}}+\frac{\alpha_{r}\alpha_{t}}{\rho_{0}^{m+n}}-\frac{\alpha_{t}}{\rho_{0}^{m-1}}\right)=0$
(104)

Thus to have a static state the constrains (103) and (104) should be
satisfied. Then, the perturbed field equations versus the scalar perturbations
(2.1.1) take the forms

$\displaystyle\frac{1}{a^{2}_{02}}-\frac{2\delta a_{2}}{a^{2}_{02}}$
$\displaystyle=$ $\displaystyle\Lambda+k_{4}\rho_{0}(1+\delta\rho)$ (105)
$\displaystyle+$
$\displaystyle\frac{k_{5}^{2}}{12}\left(\rho_{0}^{2}(1+2\delta\rho)-\frac{\alpha_{r}^{2}}{\rho_{0}^{2n}}(1-2n\delta\rho)-\frac{\alpha_{t}^{2}}{\rho_{0}^{2m}}(1-2m\delta\rho)+2\frac{\alpha_{r}\alpha_{t}}{\rho_{0}^{m+n}}(1-(n+m)\delta\rho)\right),$
$\displaystyle 2\delta\ddot{a}_{2}+\frac{1}{a^{2}_{02}}-\frac{2\delta
a_{2}}{a^{2}_{02}}$ $\displaystyle=$
$\displaystyle\Lambda+k_{4}\frac{\alpha_{r}}{\rho_{0}^{n}}(1-n\delta\rho)$
(106) $\displaystyle-$
$\displaystyle\frac{k_{5}^{2}}{12}\left(\rho_{0}^{2}(1+2\delta\rho)-\frac{\alpha_{r}^{2}}{\rho_{0}^{2n}}(1-2n\delta\rho)+\frac{\alpha_{t}^{2}}{\rho_{0}^{2m}}(1-2m\delta\rho)-\frac{2\alpha_{t}}{\rho_{0}^{m-1}}(1-(m-1)\delta\rho)\right),$
$\displaystyle\delta\ddot{a}_{1}+\delta\ddot{a}_{2}$ $\displaystyle=$
$\displaystyle\Lambda+k_{4}\frac{\alpha_{t}}{\rho_{0}^{m}}(1-m\delta\rho)$
(107) $\displaystyle-$
$\displaystyle\frac{k_{5}^{2}}{12}\left(\rho_{0}^{2}(1+2\delta\rho)+\frac{\alpha_{r}^{2}}{\rho_{0}^{2n}}(1-2n\delta\rho)-\frac{\alpha_{r}}{\rho_{0}^{n-1}}(1-(n-1)\delta\rho\right)$
$\displaystyle-$
$\displaystyle\frac{\alpha_{t}}{\rho_{0}^{m-1}}(1-(m-1)\delta\rho)-\frac{\alpha_{r}\alpha_{t}}{\rho_{0}^{m+n}}\left(1-(n+m)\delta\rho\right)).$

Using equations (101) to (103) and defining

$\displaystyle
A=k_{4}\rho_{0}+\frac{k_{5}^{2}}{6}\left(\rho_{0}^{2}+n\frac{\alpha_{r}^{2}}{\rho_{0}^{2n}}+m\frac{\alpha_{t}^{2}}{\rho_{0}^{2m}}-(n+m)\frac{\alpha_{r}\alpha_{t}}{\rho_{0}^{m+n}}\right),$
(108) $\displaystyle
B=-nk_{4}\frac{\alpha_{r}}{\rho_{0}^{n}}-\frac{k_{5}^{2}}{6}\left(\rho_{0}^{2}+n\frac{\alpha_{r}^{2}}{\rho_{0}^{2n}}-m\frac{\alpha_{t}^{2}}{\rho_{0}^{2m}}+(m-1)\frac{\alpha_{t}}{\rho_{0}^{m-1}}\right),$
(109) $\displaystyle
C=-mk_{4}\frac{\alpha_{t}}{\rho_{0}^{m}}-\frac{k_{5}^{2}}{12}\left(2\rho_{0}^{2}-2n\frac{\alpha_{r}^{2}}{\rho_{0}^{2n}}+(n-1)\frac{\alpha_{r}}{\rho_{0}^{n-1}}+(m-1)\frac{\alpha_{t}}{\rho_{0}^{m-1}}+(n+m)\frac{\alpha_{r}\alpha_{t}}{\rho_{0}^{m+n}}\right).$
(110)

The perturbed equations (105) to (107) reduce to

$\displaystyle-\frac{2\delta a_{2}}{a^{2}_{02}}=A\delta\rho,$ (111)
$\displaystyle 2\delta\ddot{a}_{2}-\frac{2\delta
a_{2}}{a^{2}_{02}}=B\delta\rho,$ (112)
$\displaystyle\delta\ddot{a}_{1}+\delta\ddot{a}_{2}=C\delta\rho.$ (113)

Combining (111) and (112) we obtain

$\delta\ddot{a}_{2}+\left(\frac{B}{A}-1\right)\frac{1}{a_{02}^{2}}\delta
a_{2}=0.$ (114)

Then an oscillatory mode for $\delta a_{2}$ is subjected to the condition

$\frac{B}{A}>1.$ (115)

Combining equations (111)-(113), we also obtain

$-2\delta\ddot{a}_{1}=(-2C+B-A)\delta\rho.$ (116)

Using (111) and (112)

$2\delta\ddot{a}_{2}=(B-A)\delta\rho,$ (117)

Combining it with (116) we get

$\delta\ddot{a}_{1}=\left(\frac{2C}{B-A}-1\right)\delta\ddot{a}_{2}.$ (118)

Then, the dynamics of $\delta a_{1}$ reads as

$\delta a_{1}=\left(\frac{2C}{B-A}-1\right)\delta a_{2}+\alpha t+\beta.$ (119)

where $\alpha,\beta$ are constants and for $\alpha=\beta=0$ and $B\neq A$ the
oscillatory modes are possible and the static state will be stable. Figure 4
shows the possibility of satisfying the constraints (100) and (104) in the
solution space of equation (115) for the given values of parameters. It is
seen from the figure that we need $n,m>1$ and very small amount of $\rho_{0}$,
which corresponds to a large radial and lateral pressure, to have a stable
static state. However, for $\rho_{0}<0.004$, $m$ can be less than 1.

Figure 4:  This figure shows the surfaces which satisfy equations (100) and
(104) in the solution space of equation (115). $Q=0$ and $W=0$ represent
equations (100) and (104), respectively, and their overlap shows the region in
which we have a stable static state. The used typical values are: $k_{4}=1$,
$k_{5}=1.5$, $\alpha_{r}=\frac{1}{2}$, $\alpha_{t}=2$, $\Lambda=10^{-5}$.

Hence, the following is the summary of the analysis of the existence and
stability of a KSSS on a brane: $(i)$ A perfect fluid cannot support a finite
size static KS geometry even in the presence of higher dimensional
modifications, $(ii)$ In contrast to GR in four dimensions, an anisotropic
fluid with $p_{r}=\omega_{r}\rho$ and
$p_{t}=\omega_{t}\rho\leavevmode\nobreak\ $ supports a stable finite size
static KS geometry, and $(iii)$ A modification of the perfect fluid form as
$p_{r}=\omega\rho+p_{0r}$ and $p_{t}=\omega\rho+p_{0t}$ can also support a
stable nonsingular KS type static state, and $(iv)$ In contrast to the case in
four dimensions, a stable nonsingular KS geometry can be supported by a
generalized Chaplygin gas source of energy-momentum.

##  4 KS geometry in five dimensions

In this section, we consider a five-dimensional KS type metric

$ds^{2}=-dt^{2}+a_{1}^{2}(t)dr^{2}+a_{2}^{2}(t)(d\theta_{1}^{2}+sin^{2}\theta_{1}d\theta_{2}^{2}+sin^{2}\theta_{1}sin^{2}\theta_{2}d\theta_{3}^{2}),$
(120)

where $a_{1}(t)$ and $a_{2}(t)$ are two arbitrary functions of time and the
energy-momentum tensor supporting this geometry has the generic form
${T^{\mu}}_{\nu}=diag\left(-\rho,\,p_{r},\,p_{t},\,p_{t},\,p_{t}\right)$.
Then, the Einstein field equations are

$\displaystyle
3(\frac{\dot{a}_{2}^{2}}{a_{2}^{2}}+\frac{\dot{a}_{1}\dot{a}_{2}}{a_{1}a_{2}}+\frac{1}{a_{2}^{2}})=\Lambda+k_{5}\rho,$
(121) $\displaystyle
3(\frac{\ddot{a}_{2}}{a_{2}}+\frac{\dot{a}_{2}^{2}}{a_{2}^{2}}+\frac{1}{a_{2}^{2}})=\Lambda-
k_{5}p_{r},$ (122)
$\displaystyle\frac{\ddot{a}_{1}}{a_{1}}+\frac{2\ddot{a}_{2}}{a_{2}}+\frac{2\dot{a}_{1}\dot{a}_{2}}{a_{1}a_{2}}+\frac{\dot{a}_{2}^{2}}{a_{2}^{2}}+\frac{1}{a_{2}^{2}}=\Lambda-
k_{5}p_{t}.$ (123)

###  4.1 KSSS and stability analysis in five dimensions

In the following, we study the existence and stability of five-dimensional KS
type static state considering the field equations (121)-(123). Similar to
previous sections, we consider two generic kinds of energy-momentum sources:
$(i)$ a fluid possessing linear equation of state, and $(ii)$ a fluid with
generalized Chaplygin gas type equations of state.

####  4.1.1 Energy-momentum source with linear equations of state

By considering a general equation of state with the form given in (5), the
field equations (121)-(123) become

$\displaystyle
3(\frac{\dot{a}_{2}^{2}}{a_{2}^{2}}+\frac{\dot{a}_{1}\dot{a}_{2}}{a_{1}a_{2}}+\frac{1}{a_{2}^{2}})=\Lambda+k_{5}\rho,$
(124) $\displaystyle
3(\frac{\ddot{a}_{2}}{a_{2}}+\frac{\dot{a}_{2}^{2}}{a_{2}^{2}}+\frac{1}{a_{2}^{2}})=\Lambda-
k_{5}\omega_{r}\rho-k_{5}p_{0r},$ (125)
$\displaystyle\frac{\ddot{a}_{1}}{a_{1}}+\frac{2\ddot{a}_{2}}{a_{2}}+\frac{2\dot{a}_{1}\dot{a}_{2}}{a_{1}a_{2}}+\frac{\dot{a}_{2}^{2}}{a_{2}^{2}}+\frac{1}{a_{2}^{2}}=\Lambda-
k_{5}\omega_{t}\rho-k_{5}p_{0t}.$ (126)

Then, the corresponding static state is given by the following equations

$\displaystyle\frac{3}{a_{02}^{2}}=\Lambda+k_{5}\rho_{0},$ (127)
$\displaystyle\frac{3}{a_{02}^{2}}=\Lambda-
k_{5}\omega_{r}\rho_{0}-k_{5}p_{0r},$ (128)
$\displaystyle\frac{1}{a_{02}^{2}}=\Lambda-
k_{5}\omega_{t}\rho_{0}-k_{5}p_{0t}.$ (129)

From (127) and (128) we get

$p_{0r}=-(1+\omega_{r})\rho_{0}=\omega_{eff}\rho_{0},$ (130)

and combining (127) and (129) leads to

$p_{0t}=\frac{2}{3}\frac{\Lambda}{k_{5}}-(\omega_{t}+\frac{1}{3})\rho_{0}.$
(131)

To study the stability of the static state given by (127)-(129), we consider
the scalar perturbations in the form of (2.1.1) and keep up to the first order
perturbation terms. Then equations (124)-(126) give

$\displaystyle 3(\frac{1}{a^{2}_{02}}-\frac{2\delta
a_{2}}{a^{2}_{02}})=\Lambda+k_{5}\rho_{0}+k_{5}\rho_{0}\delta\rho,$ (132)
$\displaystyle 3(\delta\ddot{a}_{2}+\frac{1}{a^{2}_{02}}-\frac{2\delta
a_{2}}{a^{2}_{02}})=\Lambda-k_{5}\omega_{r}\rho_{0}\delta\rho-
k_{5}\omega_{r}\rho_{0}-k_{5}p_{0r},$ (133)
$\displaystyle\delta\ddot{a}_{1}+2\delta\ddot{a}_{2}+\frac{1}{a^{2}_{02}}-\frac{2\delta
a_{2}}{a^{2}_{02}}=\Lambda-k_{5}\omega_{t}\rho_{0}\delta\rho-
k_{5}\omega_{t}\rho_{0}-k_{5}p_{0t}.$ (134)

Using the static state defined in (127)-(129), the above equations reduce to

$\displaystyle-6\frac{\delta a_{2}}{a^{2}_{02}}=k_{5}\rho_{0}\delta\rho,$
(135) $\displaystyle 3\delta\ddot{a}_{2}-6\frac{\delta
a_{2}}{a^{2}_{02}}=-k_{5}\omega_{r}\rho_{0}\delta\rho,$ (136)
$\displaystyle\delta\ddot{a}_{1}+2\delta\ddot{a}_{2}-\frac{2\delta
a_{2}}{a^{2}_{02}}=-k_{5}\omega_{t}\rho_{0}\delta\rho.$ (137)

Substituting (135) in (136) leads to

$\delta\ddot{a}_{2}+\gamma^{2}\delta a_{2}=0,$ (138)

where

$\gamma^{2}=-\frac{2(1+\omega_{r})}{a_{02}^{2}}=\frac{2\omega_{eff}}{a_{02}^{2}}.$
(139)

Hence, the oscillating modes for $\delta a_{2}$

$\displaystyle\delta a_{2}=C_{1}e^{i\gamma t}+C_{2}e^{-i\gamma t},$ (140)

requires the constraint

$\omega_{r}<-1.$ (141)

Similarly, using (135), (136) and (137), we obtain

$\delta\ddot{a}_{1}=k_{5}(\frac{2}{3}\omega_{r}-\omega_{t}+\frac{1}{3})\rho_{0}\delta\rho.$
(142)

Combining (135) and (136) gives

$\delta\rho=\frac{-3\delta\ddot{a}_{2}}{k_{5}(1+\omega_{r})\rho_{0}},$ (143)

where substituting in (142) yields

$\delta\ddot{a}_{1}=\left(\frac{3\omega_{t}-\omega_{r}}{1+\omega_{r}}-1\right)\delta\ddot{a}_{2},$
(144)

Twice integration of this equation gives

$\displaystyle\delta
a_{1}=\left(\frac{3\omega_{t}-\omega_{r}}{1+\omega_{r}}-1\right)\delta
a_{2}+C_{3}t+C_{4}.$ (145)

Similar to 4D case, the stable oscillatory modes in $\delta a_{1}$ is
subjected to the condition $C3=C4=0$. Here, for $3\omega_{t}=3\omega_{r}+2$
the perturbation amplitude on the radial and lateral directions are the same.

In the following, we consider two specific forms of the fluid (5) and discuss
on the stability of KSSS.

4.1.1.1 Perfect fluid

To study perfect fluid, we set $\omega_{r}=\omega_{t}=\omega$ and
$p_{0r}=p_{0t}$. Then, equations (127)-(129) for the static state reduce to

$\displaystyle\frac{3}{a_{02}^{2}}=\Lambda+k_{5}\rho_{0},$ (146)
$\displaystyle\frac{3}{a_{02}^{2}}=\Lambda-k_{5}\omega\rho_{0},$ (147)
$\displaystyle\frac{1}{a_{02}^{2}}=\Lambda-k_{5}\omega\rho_{0}.$ (148)

Comparing (146) and (147) leads to $\omega=-1$. Similar to the case in four
dimensions, (148) is not consistent with (147) regardless of $\omega$ values
meaning that having a static state is not possible for a perfect fluid in five
dimensions.

4.1.1.2 Anisotropic fluid
Here, similar to previous sections, we consider two modifications of the
perfect fluid.

Type 1:

In this case, by considering two different equations of state parameters for
the radial and lateral directions in the form of (32), the equations
(127)-(129) governing the static state become

$\displaystyle\frac{3}{a_{02}^{2}}=\Lambda+k_{5}\rho_{0},$ (149)
$\displaystyle\frac{3}{a_{02}^{2}}=\Lambda-k_{5}\omega_{r}\rho_{0},$ (150)
$\displaystyle\frac{1}{a_{02}^{2}}=\Lambda-k_{5}\omega_{t}\rho_{0}.$ (151)

comparing equations (149) and (150) we obtains the constraint $\omega_{r}=-1$
which is not allowed based on equation (141). Then a Type 1 fluid fails to
support a stable static sate.

Type 2:
Now we consider anisotropic fluid with the form of (36). Then the equations
(130) and (131) for $p_{0r}$ and $p_{0t}$ become

$p_{0r}=-(1+\omega)\rho_{0},$ (152)
$p_{0t}=\frac{2}{3}\frac{\Lambda}{k_{5}}-(\omega+\frac{1}{3})\rho_{0}.$ (153)

In this case, $\gamma^{2}$ given by equation (139) becomes

$\displaystyle\gamma^{2}=-\frac{3(1+\omega)}{a_{02}^{2}},$ (154)

Therefore, the positivity condition on $\gamma^{2}$ to have a stable
nonsingular static state demands

$\omega<-1,$ (155)

which means that the fluid supporting the geometry lies in the phantom range.
Also, equation (144) becomes

$\delta\ddot{a}_{1}=\left(\frac{\omega-1}{\omega+1}\right)\delta\ddot{a}_{2},$
(156)

The dynamics of $\delta a_{1}$ reads as

$\delta a_{1}=\left(\frac{\omega-1}{\omega+1}\right)\delta
a_{2}+C_{3}t+C_{4}.$ (157)

Then, similar to four dimension, for $C3=C4=0$ the oscillatory modes are
possible and hence the static state will be stable.

###  4.2 Energy-momentum source with generalized Chaplygin gas equation of
state

Using energy-momentum source with a generalized Chaplygin equation of state of
the form (42), the Einstein field equations in five dimensions will be

$\displaystyle
3(\frac{\dot{a}_{2}^{2}}{a_{2}^{2}}+\frac{\dot{a}_{1}\dot{a}_{2}}{a_{1}a_{2}}+\frac{1}{a_{2}^{2}})=\Lambda+k_{5}\rho,$
(158) $\displaystyle
3(\frac{\ddot{a}_{2}}{a_{2}}+\frac{\dot{a}_{2}^{2}}{a_{2}^{2}}+\frac{1}{a_{2}^{2}})=\Lambda+k_{5}\frac{\alpha_{r}}{\rho^{n}},$
(159)
$\displaystyle\frac{\ddot{a}_{1}}{a_{1}}+\frac{2\ddot{a}_{2}}{a_{2}}+\frac{2\dot{a}_{1}\dot{a}_{2}}{a_{1}a_{2}}+\frac{\dot{a}_{2}^{2}}{a_{2}^{2}}+\frac{1}{a_{2}^{2}}=\Lambda+k_{5}\frac{\alpha_{t}}{\rho^{m}}.$
(160)

Then, the corresponding static state is

$\displaystyle\frac{3}{a_{02}^{2}}=\Lambda+k_{5}\rho_{0},$ (161)
$\displaystyle\frac{3}{a_{02}^{2}}=\Lambda+k_{5}\frac{\alpha_{r}}{\rho_{0}^{n}},$
(162)
$\displaystyle\frac{1}{a_{02}^{2}}=\Lambda+k_{5}\frac{\alpha_{t}}{\rho_{0}^{m}}.$
(163)

applying the perturbations given by (2.1.1) to the field equations (158)-(160)
leads to

$\displaystyle 3(\frac{1}{a^{2}_{02}}-\frac{2\delta
a_{2}}{a^{2}_{02}})=\Lambda+k_{5}\rho_{0}+k_{5}\rho_{0}\delta\rho,$ (164)
$\displaystyle 3(\delta\ddot{a}_{2}+\frac{1}{a^{2}_{02}}-\frac{2\delta
a_{2}}{a^{2}_{02}})=\Lambda+\frac{k_{5}\alpha_{r}}{\rho_{0}^{n}}(1-n\delta\rho),$
(165)
$\displaystyle\delta\ddot{a}_{1}+2\delta\ddot{a}_{2}+\frac{1}{a^{2}_{02}}-\frac{2\delta
a_{2}}{a^{2}_{02}}=\Lambda+\frac{k_{5}\alpha_{t}}{\rho_{0}^{m}}(1-m\delta\rho).$
(166)

Considering equations (161)-(163), the above equations reduce to

$\displaystyle\frac{6\delta a_{2}}{a^{2}_{02}}=-k_{5}\rho_{0}\delta\rho,$
(167) $\displaystyle 3\delta\ddot{a}_{2}-\frac{6\delta
a_{2}}{a^{2}_{02}}=-nk_{5}\frac{\alpha_{r}}{\rho_{0}^{n}}\delta\rho,$ (168)
$\displaystyle\delta\ddot{a}_{1}+2\delta\ddot{a}_{2}-\frac{2\delta
a_{2}}{a^{2}_{02}}=-mk_{5}\frac{\alpha_{t}}{\rho_{0}^{m}}\delta\rho.$ (169)

Combining (167) and (168) one gets

$\delta\ddot{a}_{2}+\gamma^{2}\delta a_{2}=0,$ (170)

where

$\gamma^{2}=-\frac{2(n\alpha_{r}\rho_{0}^{-(n+1)}+1)}{a_{02}^{2}}.$ (171)

Thus, $\gamma^{2}$ is always negative and consequently there are no
oscillatory modes for $\delta a_{1}$ and $\delta a_{2}$.

The summary of the result obtained in this section is as follows. The analysis
here reveals that the existence and stability conditions for a four and a five
dimensional KS geometries without a brane are similar in some manners. More
specifically, (i) finite size static KS geometry does not exist for a perfect
fluid source, (ii) an anisotropic type 1 fluid cannot support a static state,
but an anisotropic type 2 fluid supports a stable nonsingular KS type static
state, and (iii) a generalized Chaplygin gas fluid cannot support a stable
nonsingular KS geometry in a 5-dimensional model without brane. The results of
the analysis are different than the case when a four dimensional brane is
embedded in a five (or higher) dimensional Ricci flat bulk space.
Specifically, a stable nonsingular KS geometry can be supported by both the
generalized Chaplygin gas fluid and an anisotropic fluid in a brane model. One
interpretation of the differences in these two 5-dimensional models (with and
without brane) is that in a braneworld scenario, the matter fields are
confined to the brane and have no way to propagate along the extra
dimension(s). Due to this confinement, matter fields have one less degree of
freedom in comparison to the case where they are distributed in a five
dimensional space. The confinement of the matter fields to the brane affects
the local extrinsic curvature and dynamics of the brane within its bulk space.
This induces a modification to the Einstein’s field equations on the brane.
This modification provides a geometrical interpretation for dark energy as the
manifestation of the local extrinsic shape of the brane, see for instances
[30]. In our study, this modification provides the possibility of the
existence and stability of an anisotropic KS type static state for a wider
range of fluid types on the brane in comparison to the four and five
dimensional models without brane.

##  5  Conclusion

In the present work, the possibility of having a nonsingular KS type spacetime
as a seed for an emergent universe is investigated. It is discussed that the
existence and stability of the nonsingular KSSS depend on the dimensions of
the spacetime and the nature of the fluid supporting the geometry. In
particular, it is found that:

  * •

In the context of GR in four dimensions:

$(i)$

A perfect fluid cannot support a finite size static KS geometry.

$(ii)$

An anisotropic fluid with equations of states $p_{r}=\omega_{r}\rho$ and
$p_{t}=\omega_{t}\rho\leavevmode\nobreak\ $ can support a finite size static
KS geometry but it is not stable against the scalar perturbations.

$(iii)$

A modification of the perfect fluid form possessing equations of state
$p_{r}=\omega\rho+p_{0r}$ and $p_{t}=\omega\rho+p_{0t}$ can support a stable
nonsingular KS type static state.

$(iv)$

A generalized Chaplygin gas fluid with the equations of state
$p_{r}=-\frac{\alpha_{r}}{\rho^{n}}$ and $p_{t}=-\frac{\alpha_{t}}{\rho^{m}}$
cannot support a stable nonsingular KS geometry.

  * •

In the context of a five dimensional braneworld scenario:

$(i)$

A perfect fluid cannot support a finite size static KS geometry even in the
presence of higher dimensional modifications.

$(ii)$

In contrast to the four dimensional case, an anisotropic fluid having
equations of state $p_{r}=\omega_{r}\rho$ and
$p_{t}=\omega_{t}\rho\leavevmode\nobreak\ $ supports a stable finite size
static KS geometry.

$(iii)$

A fluid having the equations of state $p_{r}=\omega\rho+p_{0r}$ and
$p_{t}=\omega\rho+p_{0t}$ can support a stable nonsingular KS type static
state.

$(iv)$

In contrast to the case in four dimensions, a stable nonsingular KS geometry
can be supported by a generalized Chaplygin gas fluid.

  * •

In the context of a five dimensional model without brane, the results of the
analysis for the existence and stability conditions are similar to the four
dimensional model addressed above.

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# Generalized Linear Models of $T_{90}$-$T_{50}$ relation to classify GRBs

Sourav Dutta<EMAIL_ADDRESS>Sunanda  Reetanjali Moharana
<EMAIL_ADDRESS>Manish Kumar  Department of Physics, Indian Institute of
Technology Jodhpur, Karwar 342037, India.

###### Abstract

Gamma-ray bursts (GRBs) can be classified with their linearly dependent
parameters alongside the standard $T_{90}$ distribution. The Generalized
linear mixture model(GLM) identifies the number of linear dependencies in a
two-parameter space. Classically, GRBs are classified into two classes by the
presence of bimodality in the histogram of T90. However, additional classes
and sub-classes of GRBs are fascinating topics to explore. In this work, we
investigate the GRBs classes in the $T_{90}{-}T_{50}$ plane using the
Generalized Linear Models(GLM) for Fermi GBM and BATSE catalogs. This study
shows five linear features for the Fermi GBM catalog and four linear features
for the BATSE catalog, directing towards the possibility of more than two GRB
classes.

††preprint: APS/123-QED

##  I Introduction

Gamma-ray bursts (GRBs) can last ten milliseconds to several hundred seconds
in the gamma-ray sky. Using the distribution of the period $T_{90}$, refers to
the duration at which 90% of fluence is collected on the detector, reference
[1] noted binary behavior with 222 events in the 1B BATSE catalog. Several
attempts are performed with multiple observed GRB parameters like fluence,
hardness ratio (higher energy flux compared to lower energies) [2, 3, 4, 5, 6]
to identify the number of classes in GRBs. All the attempts till now confirm
the existence of two classes, notably known as short GRBs (SGRBs), that last a
few milliseconds to 2 sec and long GRBs (LGRBs) with $T_{90}>$ 2 sec,
significantly. However, the exiting of the third class of GRB has long been
debated and is being attempted with the univariate and bivariate analyses of
GRBs with Gaussian distribution. For example, [7, 8] claimed the existence of
the third class in BATSE data with less than a $0.5\%$ rejection probability.
Analog evidence was found for BeppoSaX [9] and Fermi-GBM GRB catalogs [5].
Further claims of three class preferences by GRBs are studied with maximum
likelihood and $\chi^{2}$ methods [7, 5]. A detailed study exploring the
possibility of a third class of GRBs has been reviewed in [10].

The two classes of GRBs further have different astrophysical origins, hence
establishing the classification. LGRBs are associated with Supernovae (SNe)
explosions [11, 12]. In contrast, the SGRBs originated during compact star
mergers, either neutron star-neutron star (NS-NS) [13] or NS-blackhole (BH)
mergers due to their observational association with macronovae/kilonovae [14]
sGRB:130603B, [15], 060614 [16] and 050709 [17]. The advanced Laser
Interferometer Gravitational-Wave Observatory (LIGO) and Virgo gravitational
radiation observatories detected the first binary neutron star merger on 17
August 2017. The gravitational wave (GW) known as GW170817 revealed a merger
of two low-mass compact objects, whose mass range clearly puts them as neutron
stars rather than black hole candidates [18].

Not long back [19], used a machine learning algorithm, Gaussian mixed model
(GMM) to the $T_{90}$ distribution, for Fermi GMB, Beppo, Swift/BAT data and
CGRO)/BATSE to look into GRB classification. Their result confirmed two
population components to describe the CGRO/BATSE and Fermi/GBM GRBs in the
observer frame and for the Swift/BAT bursts in the rest frame. Additionally,
trimodal distribution explains the Swift GRBs in the observer frame. However,
this result may be affected as it is not complete data. A search for different
classes with other observed significant parameters is relevant to establish
the robustness of bimodal behavior.

Here, we investigate the classification of GRBs with the GMM algorithm for a
single parameter, like T50, T90, and Hardness-ratio (HR) distributions with
larger data sets for Fermi-GBM. Importantly, we investigated the linearly
correlated GRB parameters with the Generalized Mixture model (GMM) algorithm.
Specifically, the $T_{90}$ vs $T_{50}$ plane collected at the detector, where
T50 and T90 refer to the time to accumulate 50% and 90% of the burst fluence,
starting at the 25% and 5% of fluence level, respectively. From the definition
of $T_{90}$ and $T_{50}$, one can expect linear relation between these
parameters. Hence, we use the GLM algorithm for Fermi GBM and BATSE catalogs
to study the correlation coefficient in $T_{90}$-$T_{50}$ plane. We also check
for multi-linear relations of $T_{90}$ and $T_{50}$ with the hardness ratio
(HR).

This paper is organized as follows: In section II, we describe the data
selection method for Fermi GBM and BATSE catalog. Section III describes the
GMM method to verify bimodality in the histogram plot of $T_{90}$ and the
selection method used for GMM analysis. Section IV is dedicated to the GLM
method to study the correlation coefficient in $T_{90}$-$T_{50}$, $T_{90}$-HR
and $T_{50}$-HR. Section V discusses the possible outcome of the study.

##  II Data set of GRBs

We consider two major GRB catalogs for our analysis, $Fermi$ GBM, and BATSE
4B. The number of GRBs in the $Fermi$ GBM catalog is 3440 collected from 14th
July, 2008 to 19th January,
2023111https://heasarc.gsfc.nasa.gov/W3Browse/fermi/fermigbrst.html. However,
the GRB 090626707, have no record for T90 and T50 values. Hence, we excluded
this in the T90 distribution study. 2299 valid samples have been considered
for investigating the duration and spectral hardness relationship. The BATSE4B
current catalog contains 1637 entries (GRBs from 1991 April 21 to 1996 August
14)222https://heasarc.gsfc.nasa.gov/W3Browse/cgro/batse4b.html. We selected
1233 samples for the study of T90 distribution and 1179 samples for
investigating the duration and spectral hardness relationship.

We collect the HR data for Fermi-GBM following[19], where HR is the ratio
between the flux from 100 keV to 300 keV to 25 keV to 50 keV, for Comptonised
model fit to the spectrum. The HR for BATSE 4B is obtained as the ratio of
fluences (flux over the entire burst duration) in the energy range 100-300 keV
to 50-100 keV.

##  III  Analysis Methods

We first searched for the classification of GRBs in the scattering plot of T50
vs T90. We first used the Gaussian Mixture model (GMM) following [19]. We then
performed another known maximum likelihood classification method, the
Generalized Linear Models (GLM) [22] to search for different classes of GRBs
we studied GLM on the scattering distribution of T50 vs T90, T50 vs HR, and
T90 vs HR.

###  III.1 Gaussian Mixture Model

We use the machine learning python package scikit-learn333https://scikit-
learn.org/stable/ to use the GMM method in T90 distribution as performed in
[19] with at least three times more data. This method helps to avoid the
classic histogram method for classifying GRBs. GMM is a mixture of the
weighted sum of different Gaussian distributions that describe the number of
classes in the data. Assuming k number of Gaussian component
$(C_{i},i=1,2....,k)$ the ${i}^{th}$ Gaussian distribution for $x$ number of
data is,

$N(x|\mu_{i},\Sigma_{i})=\frac{1}{2\pi}\frac{1}{\sqrt{\Sigma_{i}}}exp\\{-\frac{1}{2}(x-\mu_{i})^{T}\Sigma_{i}^{-1}(x-\mu_{i})\\}$
(1)

with $\mu_{i}$ mean and $\Sigma_{i}$ covariance. The probability distribution
function for complete data $X=x_{j}(j=1,2…..,N)$

$P(X|\omega,\mu,\Sigma)=\sum_{j=1}^{N}\left(\sum_{i=1}^{k}\omega_{i}\;N(x_{j}|\mu_{i},\Sigma)\right),$
(2)

where $w_{i}$ is the weight of $i^{th}$ Gaussian distribution.

The GMM method uses an iterative algorithm alternating between Expectation and
Maximization steps (E and M steps) to estimate the parameters of $i^{th}$
number of Gaussian distributions as explained in [19].

###  III.2 Selection Method

To choose the number of classes, we use the Bayesian information criterion
(BIC), also known as the Schwarz[24] criterion from a finite set of Gaussian
distribution models. The model with the smallest value of BIC ($BIC_{min}$) is
the preferred model. BIC is defined as:

$BIC=p\ln N-2\ln P_{max}$ (3)

where Pmax, p, N are the maximum likelihood (ML) achieved by the models, the
number of parameters of the model, and the sample size, respectively.

To compare the two or more models, BIC difference ($\Delta_{i}$) is calculated
[25, 6].

$\Delta_{i}=BIC_{i}-BIC_{min}$ (4)

If $0<\Delta_{i}<2$, then $i^{th}$ model and the minimum BIC model are
accepted, and when $2<\Delta_{i}<6$, $i^{th}$ model becomes a considerable
model. For another case where $\Delta_{i}>6$, indicates strong evidence
against the $i^{th}$ model, following $\Delta_{i}>10$ shows robust proof
against it.

###  III.3 GMM analysis for $T_{90}$ distribution

We study the feature of T90 distribution of both the Fermi GBM and BATSE 4B
catalogs using the GMM algorithm.

####  III.3.1 Fermi GBM

Fig. 1 shows the variation of BIC values for the number of components and the
distribution. This study shows the T90 distribution of GRBs fitted with a two-
component mixture, namely, Short Gamma Ray Bursts (SGRBs) and Long Gamma-Ray
Bursts (LGRBs).

Figure 1: GMM analysis for T90 distribution of Fermi GRB catalog. Left: BIC
variations. Right: Distribution plot for two components with the intersection
at 3.7 sec.

In the presented dataset, the mean duration of SGRBs is 0.965 sec, with a
weight of 0.23 and a standard deviation of 0.6. Similarly, for LGRBs, the mean
duration is 28.47 sec, with a weight of 0.77 and a standard deviation of 0.45.
The value of BICmin is 7400.4.

####  III.3.2 BATSE 4B

Fig. 2 shows the variation of BIC values for the number of components and
distribution. This study shows the T90 distribution of GRBs fitted with the
two-component mixture.

Figure 2: GMM analysis for T90 distribution of BATSE 4B catalog. Left: BIC
variations. Right: Distribution plot for two components with the intersection
at 4 sec.

In the presented dataset, for SGRBs, the mean duration was 0.57 sec.), with a
weight of 0.32 and a standard deviation of 0.58. Similarly, for LGRBs, the
mean duration is 32.15 sec, with a weight of 0.68 and a standard deviation of
0.44. The value of BICmin is 1233.

The above study has been performed, with nearly one-third of data by other
groups for the T90 distribution, and found the existence of two GRB
classes[19]. We also studied 3 the BIC variation of T50 distribution for both
the BATSE 4B and Fermi-GBM datasets. The mean distribution of SGRBs is 0.23
sec, with a weight of 0.31 for the BATSE 4B dataset, and 0.37 sec, with a
weight of 0.23 for Fermi-GBM dataset, while that of LGRBs is 11.41 sec, with a
weight of 0.68 for BATSE 4B dataset, and 10.53 sec, with a weight of 0.76 for
Fermi-GBM dataset. This study again supports two classes of GRBs.

This classification should be validated with the co-dependence of two
parameters. Rather than looking for clusters in the scattering plots of two
parameters, we checked for the linear dependency of the parameters. One would
expect a linear relation between T50 and T90, T90-HR and T50-HR. Hence we use
the generalized linear regression mixture (GLM) testing for these parameter
spaces to check the number of distinct linear features present.

Figure 3: GMM analysis for T50 distribution for both BATSE 4B (top) and Fermi-
GBM (bottom) datasets

##  IV Generalized Linear Model

Generalized Linear Models (GLM) are powerful, decisive tools that offer
flexible and computationally attractive models for large datasets. Model
choice is made important due to this resulting flexibility and complexity. GLM
has been used in a wide range of areas from longitudinal data analysis[26] to
penalized spline smoothing[27] and functional data analysis[28]. Considering
finite mixture models with k components of the form:

$\displaystyle h(y|x,\psi)=\sum_{i=1}^{k}\pi_{i}f(y|x,\theta_{i})$
$\displaystyle\pi_{i}\geq 0,\sum_{i=1}^{k}\pi_{i}=1,$ (5)

where y is a dependent variable with conditional density h, x is a vector of
independent variables, $\pi_{i}$ is the prior probability of component i,
$\theta_{i}$ is the component-specific parameter vector of the density
function f, and $\psi=(\pi_{1},...,\pi_{k},\theta_{1},...,\theta_{k})$ is a
vector for all the parameters. The posterior probability for the class j with
the observation (x,y) is given by,

$P(j|x,y,\psi)=\frac{\pi_{j}f(y|x,\theta_{j})}{\sum_{k}\pi_{k}f(y|x,\theta_{i})}.$
(6)

The posterior probabilities are used for data classification by assigning each
observation to that class with the maximum value of posterior probability. The
log-likelihood of a sample of N observations
${(x_{1},y_{1}),...,(x_{N},y_{N})}$ is given by

$\displaystyle\log L$ $\displaystyle=$ $\displaystyle\sum_{n=1}^{N}\log
h(y_{n}|x_{n},\psi)$ (7) $\displaystyle=$
$\displaystyle\sum_{n=1}^{N}\sum_{i=1}^{k}\pi_{i}f(y_{n}|x_{n},\theta_{i}).$

This value can not be maximized directly. The most popular method for maximum
likelihood estimation for the parameter vector $\psi$ is through the iterative
EM algorithm.

Estimate(E) the posterior class probabilities corresponding to each
observation and derivation of the prior class probabilities,

$\widehat{\pi_{i}}=\frac{\sum_{n=1}^{N}P(i|x_{n},y_{n},\psi)}{N}.$ (8)

Equation 6 is used to calculate the posterior probabilities.

Maximize(M) the log-likelihood for each component separately using the
posterior probabilities as weights

$max\sum_{n=1}^{N}\widehat{p_{ni}}\log f(y_{n}|x_{n},\theta_{i}),$ (9)

where, $\widehat{p_{ni}}$ is given by equation 10.

$\widehat{p_{ni}}=\sum_{n=1}^{N}P(i|x_{n},y_{n},\psi)$ (10)

The E and M steps are repeated until the likelihood improvement falls under
the pre-specified threshold.

###  IV.1 GLM analysis

We have used the ”flexmix” model, available in R [22], for GLM analysis. The
”flexmix” model is a general framework for finite mixtures of regression
models that implements the EM algorithm. The E-step and all data handling are
provided, while the user can supply the M-step to define new models easily. We
have used the default generalized linear models (glm) method for clustering
the following parameter space. The randomness arises in the BIC values due to
random assignment to classes with probabilities $\widehat{p_{ni}}$. To control
this randomness, the determination of BICmin was performed $10^{5}$ times for
each parameter space. The results of the same are shown in fig. 7 and
tabulated in tables 1.

####  IV.1.1 T90-T50

The correlation coefficient, $\rho$ from linear regression (lm) for T90-T50
parameter space is 0.96 for the Fermi-GBM dataset. The error bars for T90 and
T50 are significantly large. Hence we also used orthogonal distance regression
(ODR) methods to calculate the slope of the linear line, $m_{ODR}$, and found
its value 1.05. Figure 4 shows the results. However, with this analysis, we
can not tell the possibility of the existence of multilinear dependence of the
parameters.

Figure 4: $\text{T}_{50}-\text{T}_{90}$ correlation with linear regression.

Our analysis through the GLM is shown in fig. 5 for one run. It shows the
variation of BIC values for the number of components of the T90-T50 parameter
space corresponding to Fermi GBM dataset for one case. In this case, we find
$\Delta_{4}=10.8$. Hence the total number of classes is five. On the right
side of the fig 5 we show the five groups along with their fitted linear lines
for this case.

Figure 5: GLM analysis for T90 vs T50 distribution of Fermi GRB catalog. Left:
Sample plot for BIC variation for different components. Right: The five
classes with their linear fit slopes as, 0.95,1.2,1.18,0.99 and 0.82

The GLM study for the T90-T50 parameter space with the BATSE 4B dataset is
shown in the figure. 6. In this case $\rho$, is 0.97. In this case, we find
$\Delta_{3}=3.97$. Hence the total number of classes is four. We also used
orthogonal distance regression (ODR) methods to calculate the slope of the
linear line, $m_{ODR}$, and found its value of 1.08.

Figure 6: GLM analysis for T90 \- T50 distribution of BATSE 4B catalog. Left:
Sample plot for BIC variation for different components. Right: The four groups
with linear slopes, 1.07,0.92,1.3 and 0.8

To check the randomness in the BICmin, we performed the GLM analysis for
$10^{5}$ times for T90-T50, shown in fig 7 with slant strike histogram. We
list the details of the occurrence of different classes for both Fermi-GBM and
BATSE 4B data in table 1.

Figure 7: Distribution of BICmin values for each parameter space Table 1: Distribution of component numbers for Fermi-GBM and BATSE 4B GRB events x-axis | y-axis | Number of components | % of occurrence
---|---|---|---
$Fermi$-GBM
T90 | T50 | 4 | 17.05
|  | 5 | 77.00
|  | 6 | 5.35
T90 | HR | 3 | 0.02
|  | 4 | 53.38
|  | 5 | 46.20
T50 | HR | 3 | 0.04
|  | 4 | 62.07
|  | 6 | 0.15
BATSE 4B
T90 | T50 | 3 | 3.12
|  | 4 | 96.88
T90 | HR | 2 | 100.00
T50 | HR | 2 | 100.00

####  IV.1.2 T50-HR & T90-HR

We study the linear dependence of T90-HR & T50-HR to further check the
different GRB classes. The correlation coefficient, $\rho$ for T90-HR
parameter space is -0.32 for the Fermi GBM dataset. Our analysis through the
GLM is shown in fig. 8. It shows the variation of BIC values for the number of
components of the T90-HR parameter space corresponding to Fermi GBM dataset
for one case suggesting the possibility of more than 3 linear fits to the
data. In contrast, the T90-HR BATSE 4B dataset suggests bilinear dependency.

The correlation coefficient,$\rho$ for T50-HR parameter space is -0.31 for the
Fermi GBM dataset. Our analysis through the GLM is shown in fig. 8 right-
bottom. It shows the variation of BIC values for the number of components of
the T50-HR parameter space corresponding to Fermi GBM dataset for one case. In
this case, we find $\Delta_{5}=0.7$, suggesting the possibility of four
different linear dependencies. The GLM study for T50-HR parameter space with
BATSE 4B dataset is shown in fig. 8 left-bottom. In this case $\rho$, is
-0.38. In this case, we find $\Delta_{3}=19.6$, again suggesting bilinear
behavior.

Figure 8: Sample plot for BIC variation for different components for T90 vs HR
(top) and T50 vs HR (bottom) distributions

##  V Discussion & Outlook

The binary classification, long and short, of GRBs is well established.
However, the possibility of further classes and sub-classes is still there.
Several attempts have been made to find the third class of GRB using different
statistical methods. Following [19], we study the classification with GMM
technique using EM-algorithm not only for T90 distribution but also for T50
distribution. This study showed bimodal behavior in both distributions, even
with a larger data set than in [19]. However, the bimodal result can be
further established by studying the dependency of different parameters on each
other. Motivated by this, we have applied the Generalized Linear Models (GLM)
to check the linear dependency of T90-T50, T90-HR, and T50-HR for both the
Fermi GBM and BATSE 4B GRB events. This study indicates the possibility of
more than two classes of GRBs, specifically in the linear relation of T90-T50.
Importantly the technique can group the GRB events with a slower increase of
$T_{50}$ than $T_{90}$.

###### Acknowledgements.

The authors acknowledge Science and Engineering Research Board (SERB) Grant
No. SRG/2020/001932.

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