A study of phantom scalar field cosmology using Lie and Noether symmetries

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by  Sourav Dutta

A study of phantom scalar field cosmology using Lie and Noether symmetries
Sourav Dutta and Subenoy Chakraborty
∗ †
Department of Mathematics, Jadavpur University, Kolkata-700032, West Bengal, India.
The paper deals with phantom scalar field cosmology in Einstein gravity. At first using Lie
symmetry, the coupling function to the kinetic term and the potential function of the scalar field
and the equation of state parameter of the matter field are determined and a simple solution is
obtained. Subsequently, Noether symmetry is imposed on the Lagrangian of the system. The
symmetryvectorisobtainedandthepotentialtakesaverygeneralformfromwhichpotentialusing
Lie Symmetry can be obtained as a particular case. Then we choose a coordinate transformation
6 (a,φ) → (u,v) such that one of the transformed variables (say u) is a cyclic for the Lagrangian.
1 Using conserved charge (corresponding to the cyclic coordinate) and the constant of motion,
0 solutions are obtained.
2

r Keywords: Noether symmetry,Lie symmetry,Phantom.
a
M
I. INTRODUCTION
5
1
The observational evidences [1–7] that we have been getting since the fag end of the last century are nicely

] accommodated in the frame work of standard cosmology by introducing an exotic matter [8–10] having large
c
negative pressure– the dark energy (DE). The natural and simplest candidate for DE is the cosmological
q
- constant [11–16]. But it is not very popular due to its extreme fine tuning and coincidence problem [17].
r As a result, dynamical DE having (negative) variable equation of state [18] are widely used to show recent
g
[ accelerated expansion. Subsequently, the idea of DE is extended further to consider matter with the equation
of state parameter w < 1, which is not ruled out by observational data, and moreover, necessary to describe

2 −
the current acceleration of the universe. Such a model is known as Phantom [19] DE model whose various
v
aspects have been studied in [20–23]. However, inspite of lot of works (for reviews see references [8–10], [14–
5
16], and [24]), still we are completely in the dark about the source of the DE, it is only characterized by its
9
7 negative pressure. Although the ΛCDM model suggests the acceleration as a final stage but there is no priori
5 reason why the acceleration of the universe will be in the final stage. On the other hand, other DE models
0 may have different fate of the universe, for example, in phantom DE model the energy density may increase
1. with time while the Hubble parameter and the curvature diverges in finite time and the final fate of the
0 universe is a big-rip singularity [20, 21, 25]. It is puzzling [26] to have a sudden feature singularity for FRW
6 cosmological models. Such singularity of pressure with finite energy density has some analogy to singularities
1 in inhomogeneous models of the universe [27, 28].
:
v
i In phantom cosmology, it has been shown recently that there is a nice duality [29] between phantom and
X
ordinary matter fields similar to superstring cosmologies [30, 31]. Finally, due to increase of energy density in
r phantom cosmology, it is possible to connect the phantom driven (low energy mevscale) dark energy phase to
a
the (high energy grand unified theories scale) inflationary era [32]. In the present work, a phantom model of
the universe is considered for which gravity is coupled to a scalar field with a scalar coupling function and a
potential. Basic geometrical symmetries (connected to the space-time) particularly, Lie point and Noether are
veryusefulinphysicalproblems. InfactNoethersymmetriesareusuallyrelatedwithsomeconservedquantities
(termed as conserved charges) which can be chosen as a selection criteria to discriminate similar physical
processesforexampleseveralDE models[33–40]. Mathematically,Liepoint/Noether symmetriesplayacrucial
role in any physical problem, as they provide first/Noether integrals which can be used to simplify a given
system of differential equation or to determine the integrability of the system. Further, Noether symmetries
are also used to examine self-consistency of phenomenological physical models. Moreover, imposing symmetry
to a physical problem, one may be able to restrict physical parameters like equation of state of a fluid [41],
potential of a scalar field etc. Recently, an explicit study of physical systems moving in a Riemannian space
and the corresponding Lie point/Noether symmetries are discussed in [42–48].
The motivation of the present work is to study the phantom scalar field cosmology in the perspective of
[email protected]
∗
[email protected]
†
2
symmetry. Our aim is two fold:
(a) Specifying the coupling function and the scalar potential from geometric principles.
(b)Cosmological solutions using the symmetries.
Geometric approach have been applied to symmetries in the minisuperspace approach and they are related
with basic geometric properties of the minisuperace, hence there are geometric selection rules and that have
been studied in different gravitationaltheories [54, 55, 56]
The paper is organized as follows: Section II deals with basic equations for the phantom scalar field model
whileLiepointsymmetryhasbeenimposedinSectionIII.SectionIVdealswiththeNoethersymmetryapproach
and the corresponding solutions. At the end, symmetry of the work is presented in section V.
II. BASIC EQUATION AND AUTONOMOUS SYSTEM
The action of a scalar field φ coupled to gravitywith coupling parameter,a function of the scalar field φ has
the form [49, 54] :
R 1
A= d4x√ g[ λ(φ)gµν(∆ φ)(∆ φ) V(φ)]+S , (1)
Z − 2k2 − 2 µ ν − m
where k2 =8πG is the gravitational coupling, λ(φ) the coupling parameter is an arbitrary function of φ, V(φ)
is the potential of the scalar field and S is the action of the matter field which is chosen as the cold dark
m
matter (DM) in the form of perfect fluid with constant equation of state. For flat FRW model, if we vary the
action with respect to the metric variable then we have the Einstein equations :
a˙2 λ(φ)
3 =κ2 ρ + φ˙2+V(φ) , (2)
a2 (cid:18) m 2 (cid:19)
a¨ κ2
2 = (3γ 2)ρ +2(λ(φ)φ˙2 V(φ)) , (3)
a2 − 3 h − m − i
with ρ = 1λ(φ)φ˙2+V(φ), and p = 1λ(φ)φ˙2 V(φ), the energy density and thermodynamic pressure of the
φ 2 φ 2 −
scalarfield. AssumingthereisnointeractionbetweenthescalarfieldandcoldDM,thentheenergyconservation
relations take the form :
ρ˙ +3γρ =0, i.e., ρ =ρ a 3γ, (4)
m m m 0 −
and
ρ˙ +3H(ρ +p )=0. (5)
φ φ φ
As a result, the evolution of the scalar field is given by
1 ∂V
λ(φ)φ¨+ λ(φ)φ˙2+3Hλ(φ)φ˙ + =0. (6)
′
2 ∂φ
Due to non-linearity in nature and complicated in form, the evolution equations (i.e., eq. (3) and eq. (6)) have
been studied [49] through qualitative analysis. Using the auxiliary variables
x= √λ(φ)φ˙, y = √V(φ), z = √6,
√ǫH √3H φ
the evolution equations can be written as autonomous system
3
x = x(x2 y2 1) α (z)y2,
′ 1
2 − − −
3
3
y′ =y α1(z)x+ (x2 y2+1) ,
(cid:20) 2 − (cid:21)
and
xz2
z′ = ,
− λ (z)
1
p
where α (z) = α(φ) = 3 V′(φ) , λ (z) = λ(φ), and stands for differentiation with respect to
1 q2V(φ)√λ(φ) 1 ′′ ′
N =ln a.
Note that the auxiliary variables are not independent rather they are connected by the relation
x2+y2+Ω =1.
m
In Ref [49] the potential V(φ) and the coupling function λ(φ) has been chosen in an ad hoc manner as
exponential or power law form. From the stability analysis it is generally, concluded that although there are
late time attractors with phantom scalar field but they are not quantum mechanically stable.
III. LIE POINT SYMMETRY AND CONSEQUENCES
Suppose we have a system of second order ordinary differential equations of the form
x¨i =wi(t,xj,x˙j). (7)
Now a vector field X =ξ ∂ +ηi ∂ in the augmented space (t,xi) is the generator of a Lie point symmetry of
∂t ∂xi
the above system of ordinary differential equations if the following relation hold [44] :
X[2] x¨i w(t,xi,x˙i) =0, (8)
−
(cid:0) (cid:1)
where X[2] stands for the second prolongation of X and has the expression
x[2] =ξ∂ +ηi∂ +(η˙i x˙iξ˙)∂x˙i+(η¨i x˙iξ¨ 2x¨iξ˙)∂x¨i. (9)
t i
− − −
Equivalently, the lie point symmetry condition can be written as
X[1],A =λ(xi)A, (10)
h i
where,
x[1] =ξ∂ +ηi∂ +(η˙i x˙iξ˙)∂x˙i, (11)
t i
−
is the first prolongation vector of X and the Hamiltonian vector field A has the expression
A=∂ +x˙i∂ +wi(t,xj,x˙j)∂ . (12)
t xi x˙i
In the present phantom cosmology, we have 3D augmented space (t,a,φ) and the evolution equations(2), (3)
and (6) can be written as
a˙2 κ2
a¨= a γE a 3γ +λ(φ)φ˙2 w (t,a,a˙,φ,φ˙), (13)
a − 2 (cid:16) m − (cid:17)≡ 1
and
1λ(φ) 1 dV
φ¨= ′ φ˙2 3Hφ˙ w (t,a,a˙,φ,φ˙). (14)
−2 λ(φ) − − λ(φ) dφ ≡ 2
4
So the infinitesimal generator of the point transformation corresponding to equations (13) and (14) can be
written as
∂ ∂ ∂ ∂ ∂
X =ξ∂t +η1∂a +η2∂φ +η1′∂a˙ +η2′ ∂φ˙, (15)
where ξ =ξ(t,a,φ) and η =η (t,a,φ),i=1,2. Using the total derivative operator as
i i
d ∂ ∂ ∂
= +a˙ +φ˙ , (16)
dt ∂t ∂a ∂φ
one can write
dη dξ dη dξ
η1′ = dt1 −a˙ dt and η2′ = dt2 −φ˙dt. (17)
Now, the lie point symmetry condition corresponding to equation (13) can be written as [44]
Xw1 =η1′′, (18)
and that for equation (14) is
Xw2 =η2′′, (19)
with,
dη dξ dη dξ
η = 1′ a¨ = 1′ w ,
1′′ dt − dt dt − 1dt
dη dξ dη dξ
η = 2′ φ¨ = 2′ w . (20)
2′′ dt − dt dt − 2dt
Solving the overdetermined system of equations corresponding to the symmetry conditions (18) and (19), the
component of the symmetries takes the form:
ξ =αt+β, η =αa, η =αφ,
1 2
with
λ µ
λ= 0 , V(φ)=V 0, (21)
φ2 0− φ2
and the equation of state parameter γ takes values 0 and 2. Here, α,β,λ ,V and µ are arbitraryconstants.
3 0 0 0
Hence the symmetry vector is
∂ ∂
X =(αt+β)∂ +αa +αφ . (22)
t
∂a ∂φ
The characteristic equation corresponding to this symmetry is given by
dt da dφ
= = , (23)
αt+β αa αφ
which gives an one parameter family of functions
β
a=t+ =φ. (24)
α
Now using the above one parameterfamily of functions to the field equations(2), (3), and (6) the constants are
related by the relations:
2
λ = = µ , V = ρ , when γ =0,
0 κ2 − 0 0 − 0
2 3 2
µ = ρ = λ , V =0, when γ = . (25)
0 3(cid:18) 0− κ2(cid:19) − 0 0 3
5
IV. THE NOETHER SYMMETRY APPROACH
In Noether symmetry approach, some conserved quantities corresponding to a dynamical system can be
determined from the invariance of the corresponding Lagrangian under the lie derivative along an appropriate
vector field. Thus due to this symmetry constraints are imposed on dynamics and as a result it is possible
to solve the equations of motion. In general, for a point-like canonical Lagrangian L which depends on the
variables qα(xi) and on their derivatives ∂ qα(xi), the Euler-Lagrangeequations are
j
∂L ∂L
∂ = . (26)
k(cid:18)∂∂ qα(cid:19) ∂qα
j
Now contracting with some unknown functions λα(qβ), we have
∂L ∂L
λα ∂ =0, (27)
(cid:20) k(cid:18)∂∂ qα(cid:19)− ∂qα(cid:21)
j
which on simplification can be written as
∂L
L=∂ λα . (28)
LX k(cid:18) ∂∂ qα(cid:19)
j
Here stands for the Lie derivative along the vector field
X
L
∂ ∂
X =λα +(∂ λα) . (29)
∂qα j ∂∂ qα
j
ThusasaconsequenceofNoethertheorem(whichstatesthat L=0),wecansaythattheLagrangianLis
X
L
invariant along the vector field X which is also called the generator of symmetry. Consequently, we can define
the charge [51]
∂L
Ji =λα , (30)
∂∂ qα
i
which is conserved as
∂ Ji =0, (31)
i
The energy function associated with Lagrangianis
∂
E = Lq˙α L, (32)
∂q˙α −
which is the total energy (T +V) of the system and is a constants of motion [35].
Due to the presence of the Noether symmetry, evolution equations are reduced and hence the dynamics
can be solved exactly. Further, if the symmetry (i.e., conserved quantity) have some physical meaning, then
Noether symmetry approach can be used to select reliable models [37].
In the present cosmologicalmodel with phantom scalar field, the point like Lagrangiantakes the form
κ2
L=3aa˙2+κ2ρ0a3(1−γ) a3λ(φ)φ˙2+κ2a3V(φ). (33)
− 2
Here, the infinitesimal generator of the Noether symmetry( We recall that we consider a special case of
Noether’s theorem. The applicationof the completes Noether’s theoremin Scalarfield cosmologycanbe found
6
in [54] ) is
∂ ∂ ∂ ∂
X =α +β +α˙ +β˙ , (34)
∂a ∂φ ∂a˙ ∂φ˙
where α=α(a,φ), β =β(a,φ), α˙ = ∂αa˙ + ∂αφ˙, and β˙ = ∂βa˙ + ∂βφ˙.
∂a ∂φ ∂a ∂φ
Now, the existence of Noether symmetry of the above Lagrangian demands L = 0 and hence α,β will
X
L
satisfy the following partial differential equations :
∂α
α+2a =0, (35)
∂a
∂α ∂β
6 κ2a2λ(φ) =0, (36)
∂φ − ∂a
∂β
3αλ+βaλ +2aλ =0, (37)
′
∂φ
and
3αρ +3αV(φ)+βaV (φ)=0, for γ =0,
0 ′
3αV(φ)+βaV′(φ)=0, for γ =1. (38)
In order to solve this coupled partial differential equation, we use the method of separation of variable as
α=A (a)B (φ) , β =A (a)B (φ). (39)
1 1 2 2
At first we choose λ(φ)= λ0 from the Lie point symmetry and the possible solutions are the following:
φ2
Case I: λ >0
0
A
α= 0cosh(p ln φ+b ), (40)
√a 1
β =−κ42Aλ0 pa−23φsinh(p ln φ+b1), (41)
0
V sinh2(p ln φ+b ), for γ =1
V = 0 1 . (42)
(cid:26)V0sinh2(p ln φ+b1) ρ0, for γ =0(cid:27)
−
Where, p2 = 3κ2 λ , A is arbitrary.
8 | 0| 0
Case II: λ <0
0
A
α= 0cos(p ln φ+b ), (43)
√a 1
β =−κ42Aλ0 pa−23φsin(p ln φ+b1), (44)
0
| |
V sin2(p ln φ+b ), for γ =1
V = 0 1 . (45)
(cid:26)V0sin2(p ln φ+b1) ρ0, for γ =0(cid:27)
−
7
It should be noted here that if we choose λ=λ , a constant then the infinitesimal generator of the Noether
0
symmetry coincides with [38].
Now we shall assume the potential in the form of exponential function, i.e., V = V eµφ where V and µ are
0 0
the constants. Using the set of 1st order Partialdifferential equations (35)–(38) we can determine the function
λ(φ) and the infinitesimal generator of the Noether symmetry as follows
When γ =0
1
α = α0a−21 (cid:20)eµφ(cid:18)1+ Vρ0e−µφ(cid:19)− Vρ0ln(cid:18)eµφ+ Vρ0(cid:19)(cid:21)2 ,
0 0 0
1
β = −β0a−23 (cid:18)1+ Vρ0e−µφ(cid:19)(cid:20)eµφ(cid:18)1+ Vρ0e−µφ(cid:19)− Vρ0ln(cid:18)eµφ+ Vρ0(cid:19)(cid:21)2 ,
0 0 0 0
eµφ
λ = λ . (46)
0 1+ ρ0e µφ 2 eµφ 1+ ρ0e µφ ρ0ln eµφ+ ρ0 21
(cid:16) V0 − (cid:17) h (cid:16) V0 − (cid:17)− V0 (cid:16) V0(cid:17)i
When γ =1
α = α0a−21 b0+c0eµφ 21 ,
β = β0a−2(cid:2)3 b0+c0eµ(cid:3)φ 21 ,
−
eµ(cid:2)φ (cid:3)
λ = λ , (47)
0[b +c eµφ]
0 0
where, α ,β ,λ ,b ,c are constants.
0 0 0 0 0
Hence the infinitesimal generator of the Noether symmetry is completely determined.
In the next step, we shall determine a change of variables in the augmented space, i.e., (a,φ) (u,v) such
→
that one of the variables become cyclic, i.e., the transformed Lagrangian reduced the dynamical system and
then it can be solved.
The explicitformofthetwoconstantofmotionnamelyconservedcharge(see eq. (30))andconservedenergy
(see eq. (32)) are given by:
∂L ∂L
Jκ =α +β , (48)
∂a˙ ∂φ˙
and
∂L ∂L
E =a˙ +φ˙ L. (49)
∂a˙ ∂φ˙ −
Further introducing the carton one form [35] namely
∂L ∂L
θ = da+ dφ. (50)
L ∂a˙ ∂φ˙
The conserved charge can be written as
Jκ =i θ , (51)
X L
where, i θ indicates contractions between the vector field X and the differential form θ.
X L
So by a point transformation (a,φ) (u,v) the vector field X becomes
→
∂ ∂ d d d d
X =(i du) +(i dv) + (i du) + (i dv) . (52)
X x X x
∂u ∂v (cid:18)dt (cid:19)du˙ (cid:18)dt (cid:19)dv˙
e
8
Note that X is still the lift of a vector field defined on the space of positions. As X is a symmetry so we
choose a coordinate transformation such that
e
i du=1 , i dv =0, (53)
X X
so that X = ∂ and ∂L =0. Thus u is a cyclic coordinate and the dynamics can be reduced [38].
∂u ∂u
e
Hence by solving the equations (53) a and φ can be obtained as
for: λ >0
0
2
3A 3 1
a= 0 u2 v2 3 ,
(cid:18) 2 (cid:19) −
(cid:0) (cid:1)
1 v
φ=exp tanh 1( ) b . (54)
(cid:20)p(cid:16) − u − 1(cid:17)(cid:21)
for: λ <0
0
2
3A 3 1
a= 0 u2+v2 3 ,
(cid:18) 2 (cid:19)
(cid:0) (cid:1)
1 v
φ=exp tan 1( ) b . (55)
(cid:20)p(cid:16) − u − 1(cid:17)(cid:21)
The transformed Lagrangiantakes the form:
3A2(u˙2 v˙2)+ 9κ2A2V v2+κ2ρ for λ >0,γ =1
0 − 4 0 0 0 0
 3A2(u˙2 v˙2)+ 9κ2A2V v2 for λ >0,γ =0
L= 3A200(u˙2+−v˙2)+ 944κ2A020V00v2+κ2ρ0 for λ00<0,γ =1, (56)
3A2(u˙2+v˙2)+ 9κ2A2V v2 for λ <0,γ =0
0 4 0 0 0
 
with u as a cyclic coordinate.
Now the conserved charge gives
∂L
J = =6A2u˙ =J , (57)
u ∂u˙ 0 0
which integrates to give, u(t)=u t+u .
0 1
Similarly, the conservedenergy in terms of u,v gives
9
E =3A2(u˙2 ǫv˙2) κ2A2V v2 κ2ρ γ =0, (58)
l 0 − − 4 0 0 − 0
where ǫ= 1 corresponding to λ >or<0 and γ =0 or 1.
0
±
Now using the solution for u, the solution of v takes the form
√u2 γ2m2
 √m02−l u2 sin[l(t−t0)] for ǫ=+1 
v = √u20l−−lγ20mc2ocso[ls(ht[−l(tt0−)]t0)] ffoorr ǫu0=>+m10 , (59)
√m2 u2
 l− 0sinh[l(t−t0)] for m>u0 
9
Figure 1: representsa¨(t) against t for Figure 2: shows thevariation of Ω against t
d
λ0>0, γ =1 with for λ0 >0, γ =1 with
A0=1, u0 =1, l2 =3, m2= 31. A0 =1, u0 =1, l2 =3, m2= 13.
where γ =0 or 1 and l2 = 3κ2V0, m2 = κ2ρ0.
4 3A20
Hence from (54) or (55) the explicit form of the scale factor and the phantom scalar field are given by
A. for λ > 0, γ =1
0
and a(t)=(cid:0)(cid:0)33AA2200(cid:1)(cid:1)3322 hhuu2020tt22−− ((um202−ll−22mu220))scoins22((lltt))ii1331 , (60)
exp(cid:20)p1(tanh−1[√ulu20−0tm2sin(lt)]−b1)(cid:21) 
φ(t)= , (61)
exp(cid:20)p1(tanh−1[√mlu20−tu20cos(lt)]−b1)(cid:21) 
accordingas u > m or u < m. Inthese solutions we have chosenu =0=t without any loss of generality.
0 0 1 0
B. for λ > 0, γ =0
0
2 1
3A u 3 1 3
a(t)= 0 0 t2 sin2(lt) ,
(cid:18) 2 (cid:19) (cid:20) − l2 (cid:21)
1 sin(lt)
φ(t)= exp (tanh 1[ b ) . (62)
(cid:20)p − (lt) − 1 (cid:21)
C. for λ < 0, γ =1
0
a(t)= (cid:0)33AA2200(cid:1)3322 huu2020tt22++ ((um202−ll−22mu220))csoinshh22((lltt))i1313 , (63)
and  (cid:0) (cid:1) h i 
exp(cid:20)p1(tan−1[√ulu20−0tm2cosh(lt)]−b1)(cid:21) 
φ(t)= , (64)
exp(cid:20)p1(tan−1[√mlu20−tu20sinh(lt)]−b1)(cid:21) 
 
10
Figure 3: shows thegraphical representation of
wφ against t for λ0 >0, γ =1 with
λ0=8, b1=1, p2=3, V0 =4.
Figure 4: corresponds the graphical
Figure 5: representsΩ against t for
representation of a¨(t) against t for d
λ0 >0, γ =0 with
A0=1,λ0u>0 =0,1,γl=2 =0 w3,ithm2= 31. A0 =1, u0 =1, l2 =3, m2= 31.
according as u > m or u < m.
0 0
D. for λ < 0, γ =0
0
2 1
3A u 3 1 3
a(t)= 0 0 t2 cosh2(lt) ,
(cid:18) 2 (cid:19) (cid:20) − l2 (cid:21)
1 cosh(lt)
φ(t)= exp (tanh 1[ b ) . (65)
(cid:20)p − (lt) − 1 (cid:21)
Ifwenowimposetheinitialbig-bangsingularityatt=0,i.e.,a(0)=0thenwehavethefollowingrestrictions:
(i) u >m for λ >0 , (II) u <m for λ <0 and case D is not possible. Thus for λ >0 it is possible to have
0 0 0 0 0
both γ = 0 and 1 while for λ <0 only the dark matter in the form of dust is permissible. It should be noted
0
thatduetocomplicatedformofthesymmetrycondition(46),itisnotpossibletofindthecosmologicalsolution
correspondingto it. Further, if we setthe presenttime t =1,i.e., we use the time scalein whichthe ageif the
0
universe (not known in principle) is chosen as unit of time and setting a =a(t )=1 we have
0 0