Table Of ContentBulk viscous FWR with time varying constants revisited
∗
J.A. Belincho´n
Dpt. of Physics ETS Architecture. UPM Madrid. Av. Juan de Herrera N-4 28040 Espan˜a.
(Dated: February 7, 2008)
We study a full causal bulk viscous cosmological model with flat FRW symmetries and where
the “constants” G,c and Λ vary. We take into account the possible effects of a c−variable into
the curvature tensor in order to outline the field equations. Using the Lie method we find the
possible forms of the “constants” G and c that make integrable the field equations as well as the
equationof statefor theviscousparameter. It isfoundthat G,cand Λfollow a powerlaw solution
2
verifyingtherelationship G/c =κ.Oncethesepossible forms havebeenobtained wecalculate the
thermodynamicalquantitiesofthemodelinordertodeterminethepossiblevaluesoftheparameters
that govern the quantities, finding that only a growing G and c are possible while Λ behaves as a
negative decreasing function.
5
0
0
I. INTRODUCTION
2
n
a In a recent paper (see [1]) we study a cosmological model with flat FRW symmetries filled by a perfect fluid and
J where the “constants”G,c andΛ were considerateas function ontime t. By differentreasonsexposedin such paper,
7 wetookintoaccountthepossibleeffectsofac variableintothecurvaturetensorinordertooutlinethefieldequations.
−
Through the Lie groupmethod we studied the possible forms of the functions G and c that make integrable the field
1 equations. In this way we were finding that G and c follow a power law solution verifying the relationship G/c2 =κ.
v
But unfortunately we were not able to determine if G and c are growing or decreasing functions on time t.
9
In order to determine if G and c are growing or decreasing functions we try to take into account thermodynamical
1
considerations, that is to say, we hope that thermodynamical restrictions help us to determine the behaviour of such
0
1 functions. For this purpose,in this paper we consider a cosmologicalmodel with flatFRW symmetries filled by a full
0 causal bulk viscous fluid and where the constants G,c and Λ are considered as functions on time t. Once we have
5 outlined the field equations (taking into account the possible effects of a c variable into the curvature tensor) we
0 rewrite them in order to obtain a second order differential equation in orde−r to apply the standard Lie procedure.
/
c The study of this ode through the Lie group method allows us to obtain the precise form of the functions G and c
q that make integrable the field equations as well as the equation of state for the bulk viscous parameter ξ.
- As we will see in section 3 the field equations only admit scaling symmetries (note that we are working with flat
r
g FRW symmetries) i.e. we are studying a self-similar model. This fact obligates that G and c follow a power law
v: solution, c= c0tK1, verifying the relationship G/c2 =κ, i.e. G and c are functions on time t but in such a way that
i this relationshipmustbe verified. We findanotherrestrictionunder this symmetry. The bulk viscosityξ,mustfollow
X the law ξ = k ρ1/2, i.e. we have found a concrete equation of state for the viscous parameter γ = 1/2. All these
γ
r results are in agreement with our previous paper ([2]).
a
Once we have found the possible forms of the functions G and c we calculate the energy density finding in a first
approach that the only physical solution imply that G and c must be a growing functions on time t since K >0.
1
Insection4wewillcalculateallthephysicalquantitiesinordertocompletethesolutionforourmodel. Inparticular
we are interested in calculating the entropy in order to obtain some restrictions for the physical parameters as K
1
and to elucidate if G and c are growing or decreasing functions. In this way we find an exact solution for Λ which
behaves as a negative decreasing function.
We end summarizing all these results in the last section.
II. THE MODEL
Following Maartens [3], we consider a Friedmann-Robertson-Walker (FRW) Universe with a line element
ds2 =c2(t)dt2 f2(t) dx2+dy2+dz2 , (1)
−
(cid:0) (cid:1)
∗Electronicaddress: [email protected]
2
filled with a bulk viscous cosmologicalfluid with the following energy-momentum tensor:
Tk =(ρ+p+Π)u uk (p+Π)δk, (2)
i i − i
whereρistheenergydensity,pthethermodynamicpressure,Πisthebulkviscouspressureandu isthefourvelocity
i
satisfying the condition u ui =1.
i
The gravitationalfield equations with variable G, c and Λ are:
1 8πG(t)
R g R= T +Λ(t)g . (3)
ik− 2 ik c4(t) ik ik
Applying the covariance divergence to the second member of equation (3) we get:
G
div Tj+δjΛ =0, (4)
c4 i i
(cid:18) (cid:19)
4c G c4δjΛ
Tj = ,j ,j Tj i ,j, (5)
i;j c − G i − 8πG
(cid:18) (cid:19)
that simplifies to:
·
Λc4 G˙ c˙
ρ˙+3(ρ+p)H +3HΠ= ρ 4ρ , (6)
−8πG − G − c
where H stands for the Hubble parameter (H =f˙/f). Therefore, our model (with FRW symmetries) is described by
the following equations:
c˙ 8πG
2H˙ 2 H +3H2 = (p+Π)+Λc2, (7)
− c − c2
8πG
3H2 = ρ+Λc2, (8)
c2
Λ˙c4 G˙ c˙
ρ˙+3(ρ+p+Π)H = ρ +4ρ , (9)
−8πG − G c
ǫ τ˙ ξ˙ T˙
τΠ˙ +Π= 3ξH τΠ 3H+ . (10)
− − 2 τ − ξ − T!
We would like to emphasize that deriving (8) and taking into account (7) it is obtained (9), that is to say, the
conservationequation(9)couldbededucedfromthefieldequationasinthestandardcasewherethe“constants”G,c
and Λ are true constants (see [1]).
In order to close the system of equations (7-10) we have to give the equation of state for p and specify T, ξ and τ.
As usual, we assume the following phenomenological(ad hoc) laws [3]:
p=ωρ, ξ =k ργ, T =D ρδ, τ =ξρ−1 =k ργ−1, (11)
γ δ γ
where 0 ω 1, and k 0, D 0 are dimensional constants, γ 0 and δ 0 (δ = ω so that 0 δ 1/2
≤ ≤ γ ≥ δ ≥ ≥ ≥ ω+1 ≤ ≤
for 0 ω 1) are numerical constants. Eqs. (11) are standard in cosmological models whereas the equation for τ is
≤ ≤
a simple procedure to ensure that the speed of viscous pulses does not exceed the speed of light. These are without
sufficient thermodynamicalmotivation, but, in absence ofbetter alternatives,we use these equations andexpect that
they will at least provide an indication of the range of possibilities. For the temperature law, we take, T = D ρδ,
δ
which is the simplest law guaranteeing positive heat capacity.
The growth of the total commoving entropy Σ over a proper time interval (t ,t) is given by [3]:
0
3 t ΠHf3
Σ(t) Σ(t )= dt, (12)
0
− −k T
B Zt0
where k is the Boltzmann’s constant.
B
Therefore, with all these assumptions and taking into account the conservation principle, i.e., div(Tj) = 0, the
i
resulting field equations are as follows:
3
c˙ 8πG
2H˙ +3H2 2 H = (p+Π)+Λc2, (13)
− c − c2
8πG
3H2 = ρ+Λc2, (14)
c2
ρ˙+3(ω+1)ρH = 3HΠ, (15)
−
Λ˙c4 G˙ c˙
+ρ 4ρ =0, (16)
8πG G − c
Π 1 ρ˙
Π˙ + = 3ρH Π 3H W , (17)
k ργ−1 − − 2 − ρ
γ (cid:18) (cid:19)
where W = 2ω+1 is a numerical constant.
ω+1
(cid:16) (cid:17)
III. LIE METHOD
The field equations (13-17) represent a system of odes with 5 unknowns. In order to integrate them and to obtain
a complete solution for the proposed model we will need to make some simplifying hypotheses (this is always a
dangerous way) or to study if the system admits symmetries. In this way, studying the possible symmetries that
admit the system, we will be able to determine the possible forms for which such system is integrable.
In order to apply the standard Lie procedure (see for example [5]-[7]) we need to rewrite eq. (13-17) by a single
one and imposing that such ode (second order ode) admits a concrete symmetry we will be able to determine the
equation of the state for the viscous parameter i.e. ξ =k ργ and the exact of the “constants” G and c for which the
γ
ode is completely integrable.
We start with the assumption Π=κρ, with κ R− (see [2] for a complete discussion of this hypothesis).The bulk
∈
viscosity evolution equation can then be rewritten in the alternative form
ρ˙
δ +k−1ρ1−γ = 3βH, (18)
ρ γ −
where β = 1 + 1 and δ = 1 W .
κ 2 − 2
Taking the derivative with respect to the time of this equation and with the use of the equation
(cid:0) (cid:1) (cid:0) (cid:1)
c˙ G(t)
H˙ = H 4πα ρ, (19)
c − c2(t)
obtained from the field equations (13) (where α=(1+ω+κ)). and taking into account eq. (15)
1 ρ˙
H = (20)
−3αρ
we obtain the following second order differential equation describing the time variation of the density of the cosmo-
logical fluid:
ρ˙2 c˙ G
ρ¨= Asρsρ˙+B ρ˙+D ρ2, (21)
ρ − c c2
where A= kγ−1, s=(1 γ), B = β and D = 12παβ.
δ − δα δ
We go next to apply all the standard procedure of Lie group analysis to this equation (see [6] for details and
notation). In this way we find that the admissible symmetries for our ode being determinate by the following system
of pdes
ρξ +ξ =0, (22)
ρρ ρ
1 c˙
2ρ2 η ξ + Asρs B ξ ρη +η =0, (23)
ρρ tρ ρ ρ
2 − − c −
(cid:18) (cid:18) (cid:19) (cid:19)
c˙ c¨ c˙2 G
2η c−2ξ 2η ρ−1+As2ηρs−1+ξ sAρs B +ξB + 3ρ2D ξ =0, (24)
tρ− tt− t t − c −c c2 − c2 ρ
(cid:18) (cid:19) (cid:18) (cid:19)
c˙ G G G˙ c˙
η +η sAρs B +D ρ2 2ξ +η 2ρ−1η D ρ2ξ 2 =0, (25)
tt t(cid:18) − c(cid:19) c2 − t ρ− − c2 G − c!
(cid:0) (cid:1)
4
Solving (22-25), we find that
1
ξ(ρ,t)=at+b, η(ρ,t)= 2aρ, s= , (26)
− ⇐⇒ 2
with the constraints, from eq. (24)
c˙2 a
c¨= c˙, (27)
c − at+2b
and from eq. (25)
G˙ c˙ G
=2 , = =κ (28)
G c ⇒ c2
being a and b numerical constants and we will assume that κ>0.
In order to solve (27), we consider the following cases.
1. Case I. Taking a=0,b=0, we get
6
c′2
c′′ = , = c(t)=K eK1t, (29)
2
c ⇒
where the (K )2 are integration constants. Therefore the equation (21) yields:
i i=1
ρ˙2 A
ρ¨= √ρρ˙+BK ρ˙+Dκρ2, (30)
1
ρ − 2
the solution obtained through invariants is:
dt dρ
= = ρ≈const. (31)
ξ η ⇒
this solution seems not to be physical.
2. Case II. Taking b=0,a=0, we get that the infinitesimal X is X =t∂ 2ρ∂ which is precisely the generator
t ρ
6 −
of the scaling symmetries. Therefore the solution will be the same than the obtained one with the dimensional
method.
With these values of a and b equation (27) yields
c′2 c′
c′′ = , = c(t)=K tK1, (32)
2
c − t ⇒
as we expected, c(t) follows a power-law solution. Therefore equation (21) yields:
ρ˙2 A K
ρ¨= √ρρ˙+B 1ρ˙+Dκρ2, (33)
ρ − 2 t
the solution obtained through invariants is:
dt dρ
= = ρ≈t−2, ρ=ρ t−2, (34)
0
ξ η ⇒
where ρ is a numerical constants that must verifies eq. (33) in such a way that
0
A A+ A2+8Dκ(1+BK ) +4Dκ(1+BK )
1 1 1
ρ = , (35)
0 2 (cid:16) p D2κ2 (cid:17)
now, since ρ must be a positive numerical constant, in a first approach,and taking into account the definition
0
of the constants A,B and D we see that ρ > 0 iff (1+BK )< 0, which imply that K > 0 since A> 0,B <
0 1 1
0,κ>0 and D <0 (since κ <0). Therefore K must be a positive numerical constant.
1
5
Once again if one insists in solving equation (33) through DA (applying the Pi-theorem) it is found that with
respecttothedimensionalbaseB= ρ,T eachquantityhasthefollowingdimensionalequation[ρ]=ρ,[t]=T
and [A]=ρ−1T−2. Therefore, we fin{d in}a trivial way that:
ρ A t
1
ρ 1 1 0 = ρ≈ . (36)
− ⇒ At2
T 0 2 1
−
3. Case III. Taking a,b=0, we get
6
′′ c′2 ac′ K1
c = , = c(t)=K2(at+b) a , (37)
c − at+b ⇒
hence equation (21) yields:
ρ˙2 A K
ρ¨= √ρρ˙+B 1 ρ˙+Dκρ2, (38)
ρ − 2 at+b
the solution obtained through invariants is:
dt = dρ = ρ≈(at+b)−2/a, (39)
ξ η ⇒
As we can see case II is a particular situation of case III.
Throughthe symmetryanalysisofeq. (21)wehavebeenableofdeterminingthe equationofstatethatmustfollow
the viscous parameter ξ, obtaining that γ =1/2, as well as, that under the scaling symmetry c and G must follow a
power law solution in such a way that they must verify the relationship G/c2 =κ.
IV. A COMPLETE SOLUTION
Once we know the behaviour of the quantities ρ,G and c, in this section, we are going to determine the rest of the
quantities i.e. the scale factor and mainly the behaviour of the entropy with the hope that this quantity help us to
determine the possible values of the constant K . For this purpose we begin considering eq. (15)
1
ρ˙+3(ω+1+κ)ρH =0, (40)
which trivially leads us to the well known relationship between the energy density ρ and the scale factor f
ρ=A f−3(ω+1+κ) or ρ=A f−3α, (41)
ω ω
where α=(ω+1+κ). Now, taking into account the eq. (17) and simplifying it, we obtain,
W 1 1 √3ρ
1 ρ˙ = , (42)
− 2 − κα − 2α − k
(cid:18) (cid:19) γ
thereby obtaining ρ=ρ(t). On simplifying further, we obtain,
ρ˙ 1
= = ρ=ρ t−2, (43)
0
√3ρ −Kkγ ⇒
where
W 1 1
K = 1 , ρ =(2Kk )2. (44)
− 2 − κα − 2α 0 γ
(cid:18) (cid:19)
From equation (41) we obtain:
A 1/3α
f = ωt2 , i.e. f t3(ω+21+κ). (45)
ρ ∝
(cid:18) 0 (cid:19)
6
An important observational quantity is the deceleration parameter q = d 1 1. The sign of the deceleration
dt H −
parameter indicates whether the model inflates or not. The positive sign of q correspondsto “standard”decelerating
(cid:0) (cid:1)
models whereas the negative sign indicates inflation. In our model, the deceleration parameter behaves as:
3α
q = 1+ . (46)
− 2
In this way we find that for ω =1, κ = 4/3 is a critical value since q =0, therefore and under these considerations,
q <0 if κ < 4/3 and q >0 if κ > 4/−3.
− −
We proceed with the calculation of the other physical quantities under the condition γ =1/2:
ξ =k ργ k ρ t−2 γ ≈k2t−1, (47)
γ ∝ γ 0 γ
T =D ρδ =D(cid:0) ρ t−2(cid:1) δ ≈t−2 with δ = ω , (48)
δ δ 0
ω+1
τ =ξρ−1 =k (cid:0)ρ t−2 (cid:1)−1/2, i.e. τ =(2K)−1t (49)
γ 0
We see from τ = (2K)−1t that this result is in(cid:0) agree(cid:1)ment with the theoretical result obtained in [3]. For viscous
expansion to be non-thermalizing, we should have τ < t, or otherwise the basic interaction rate for viscous effects
should be sufficiently rapid to restore the equilibrium as the fluid expands.
The commoving entropy is
Σ(t)= 2κΣ0 tx2δ+α2−3dx=(ω+1)Σ0t−κα (50)
− α
Z
where Σ = ρ01−δ−1/αA1ω/α and we find that κ > (ω+1) i.e. κ ( 2,0]. We notice that the parameter κ weakly
perturbs0the perDfeδcktBfluid FRW Universe. When−κ = 0, the com∈mo−ving entropy assumes a constant value and we
recover the perfect fluid case.
Finally, we will use equation (16) to obtain the behaviour of the cosmological “constant” Λ. Since we know the
behaviour of the constants G, c and ρ i.e.
G=κc2, c(t)=K tK1, and ρ=ρ t−2, (51)
2 0
then eq. (16) yields
Λ˙c2 c˙
2 =0, (52)
8πκρ − c
and substituting eq. (51) into eq. (52) it is obtained the following ODE
K ρ K
Λ˙ =16πκρ 1t−2K1−3 = Λ= 8πκ 0 1 t−2(K1+1) (53)
0K2 ⇒ − K2K +1
2 2 1
with K > 1.
1
−
We would like to emphasize that for K > 0 then Λ < 0 and for K < 0 then Λ > 0 but unfortunately we do not
1 1
able to fix a better value for the constant K . Finally it is observed that
1
1
Λ≈ . (54)
c2t2
as it is expected in this self-similar solution.
As we have seen, with the followed way, we have not been able to determine a better value for the constant K
1
than the obtainedone in eq. (35). for this reasonwe try a new tentative whichconsists in consideringthe effects of a
c variable into the scale factor in such a way that when we calculate the entropy such effect may by reflected better
−
than in our previous way.
A. A new tentative.
In order to take into account the effects of a c variable into the scale factor we will determine it from eq. (14)
−
since we already know the behaviour of G,c,ρ and Λ. Therefore, after a simplification eq. (14) yields.
K
3H2 =8πκρ 1 1 t−2, (55)
0
− (K +1)
(cid:18) 1 (cid:19)
7
therefore
1/2
8πκρ 1
f =K tυ / υ = 0 , (56)
f
3 K +1
(cid:18) 1 (cid:19)
finding again that K > 1.
1
−
In this way the entropy formula (12) yields:
Σ(t)=Σ0t4Kπ1κ+ρ01+2δ−2, (57)
whereΣ0 =−κ√6ρ0KkBf32π√κρ(0π+κ(ρK0(1K+11+)(1δ)−)1).Ifwemake2πκρ0 =1(a freehypothesis),Σ(t)≈Σ0t2(cid:16)K11+1−ω+11(cid:17) wefind
therefore K > ω, but we are not able to find a better value (or limit) for K .
1 1
−
Or taking into account the field eq. (13)
K K
2H˙ +3H2 2 1H = 8πκρ (ω+κ)+ 1 t−2 (58)
0
− t − K +1
(cid:18) 1 (cid:19)
which is a Riccati ode and which particular solution (scaling solution) is:
a
H = = f =K ta (59)
f
t ⇒
where
(K +1)2 (K +1)4 3A α(K +1)2+K2+K
1 ± 1 − 1 1 1
a= r (60)
(cid:16) (cid:17)
3(K +1)
1
finding that K > 1, with α=(ω+κ), A=8πκρ and imposing the condition.
1 0
−
(K +1)4 3A α(K +1)2+K2+K 0 (61)
1 − 1 1 1 ≥
(cid:16) (cid:17)
(note that α may take negative values) therefore the entropy behaves as:
Σ≈t−2+2δ+(K1+1)2+√(K1+1)(4K−13+A1()α(K1+1)2+K12+K1) (62)
As we can see following these ways we have only been able to determine that K > 1 or that
1
−
2
ψ(K )> (63)
1
ω+1
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
but as ψ(K1) depends on the factors like A and α it is very(cid:12)difficu(cid:12)lt to determine a good value for this constant.
V. CONCLUSIONS.
In this paper we have studied a flat FRW cosmological model filled with a bulk viscous fluid and where the
“constants” G,c and Λ are considered as functions on time t. In order to outline the field equations we have taken
into account the possible effects of a c variable into the curvature tensor. In this way we have been able to deduce
−
the general conservation principle from the field equations obtaining an equation that coincides with the obtained
one from the divergence of the right hand of the field equation i.e. div GTj+δjΛ =0. Using the Lie group tactic
c4 i i
we have been able to determine the equation of state for the bulk visc(cid:16)ous paramete(cid:17)r, ξ = kγργ, as well as the exact
form for the “constants” G and c which make integrable the field equations.
Wehaveobtainedthatourmodelisself-similari.e. admitsascalingsymmetryiffγ =1/2andcfollowsapowerlaw
solution i.e. c =c0tK1, while G and c must verify the relationship G/c2 =κ. Under these results we have calculated
the behaviour of the energy density finding, in a first approach, that the physical solution is only possible if K >0,
1
that is to say, G and c are growing functions on time t. We have also calculated the rest of the main quantities but
8
underphysicalconsiderationswehaveonlybeenabletodeterminethatK > 1.Thelatterresultdoesnotcontradict
1
−
our previous result K >0.
1
Therefore we conclude that G and c are growing functions while Λ is a negative decreasing function.
Acknowledgement. I wouldwishto expressmy gratitude to JavierAceves forhis translationinto Englishofthis
paper.
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