Table Of ContentDynamic model of spherical perturbations in the
Friedmann universe. II. retarding solutions for
the ultrarelativistic equation of state
Yu.G. Ignatyev, N. Elmakhi
Tatar State Humanitarity PedagocicalUniversity
1 Mezhlauk St., Kazan 420021,Russia
1
1 Abstract
0
ExactlinearretardingsphericallysymmetricsolutionsofEinsteinequa-
2
tions linearized around Friedmann background for the ultrarelativistic
n
equation of state are obtained and investigated. Uniqueness of the so-
a lutions in theC1 class is proved.
J
7
1 General solution of evolutionary equation in
]
c the form of power series and private cases.
q
-
r
1.1 Equations of spherical perturbations model
g
[
AsitisshowninRef.[1],sphericallysymmetricperturbationsofFriedmannmet-
1 rics are described by one scalar function δν(r,η) connected with perturbation
v
of the metric tensor component g in the isotropic coordinates (r,η) with the
4 44
relation
4
5
δg =a2(η)δν, (1)
1 44
.
1 at that the representation turns out to be convenient
0
1 Φ(r,η) Ψ(r,η) µ(η)
1 δν =2 =2 − , (2)
ar ar
:
v
i it allows to factor the solution into particlelike mode
X
r 2µ(η)
a δνp = , (3)
− ar
equivalenttoNewtonpotentialofthepointmassµ(t),andthenonsingularmode
2Ψ(r,η)
δν = , (4)
0
ar
the “potential” of which Ψ(r,η) and its first derivative by a radial variable at
the beginning of the coordinates satisfy the relations
∂Ψ(r,η)
lim Ψ(r,η) =]0, lim r =0. (5)
r 0| | r 0(cid:12) ∂r (cid:12)
→ → (cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
1
By a constant coefficient of barotrope κ the mass of a particle source µ(η) and
the nonsingular mode potential Ψ(r,η) satisfy the evolutionary equations (75)
and (84) [1]:
2 6(1+κ) µ
µ¨+ µ˙ =0, (6)
η − (1+3κ)2η2
2 6(1+κ) Ψ
Ψ¨ + Ψ˙ κΨ =0. (7)
′′
η − (1+3κ)2η2 −
In the previous paper [1] an exact solution of the evolutionary equation (6) for
the particlelike source mass was also obtained
2 3(1+κ)
µ=µ+η1+3κ +µ η− 1+3κ , (1+κ)=0; (8)
− 6
µ
µ=µ+η+ −, (1+κ)=0. (9)
η
In[1]it wasnotedthat the peculiarity inthe solution(8)by 1+3κ=0is coor-
dinate disappearing by coming from the temporal coordinate η to the physical
time t
(1+κ) 1+3κ
t= a(η)dη; t=c¯η31+3κ; η =C¯ t3(1+κ). (10)
1
∫ ⇒
Letusrecall,thenonsingularpartoftheenergydensityperturbationδε(r,η)
and the radial velocity of medium v(r,η) are determined with help of the po-
tential function Ψ(r,η) by the relations
δε 1 a˙
= 3 Φ˙ Ψ , (11)
′′
ε −4πra3ε (cid:18) a − (cid:19)
0 0
1 ∂ Φ˙
(1+κ)v = , (12)
−4πra3ε ∂r r
0
where ε (η) is a nonperturbed energy density of Friedmann Universe?
0
6(1+κ) 2 6κ
ε0 η− 1+3κ ; a η1+3κ; ε0a3 η−1+3κ. (13)
∼ ∼ ∼
In[1]bythevariablesseparationmethodageneralsolutionoftheevolution-
ary equation(7) satisfying the conditions (5) in form of anintegral from Bessel
functions was also obtained. In this paper we shall obtain more convenient
solutions in form of power series.
2
1.2 Class C∞ general solution in the perturbation area;
κ = 0, 1+κ = 0
6 6
As it was shown in the previous paper [1], the singular part of the potential
function δν(r,η) is uniquely extracted by the representation (2) in which the
potential function Ψ(r,η) is nonsingular at the beginning of the coordinates,
that is satisfies the relations (5) in consequence of which, in particular
Ψ(0,η)=0. (14)
Further supposing in the final neighborhood r = 0;r [0,r ) in which the
0
∈
metrics perturbation is localized, the potential function Ψ(r,η) belongs to the
C class,letusrepresentthesolutionoftheevolutionaryequation(6)satisfying
∞
the conditions (5) in the form of power series of the radial variable r
∞
Ψ(r,η)= Ψ (η)rn. (15)
n
nX=1
Let us underline the expansion (15) does not consist of a member with a zero
power rin consequence of the relation (14). Substituting the function Ψ(r,η))
in the form of the Eq. (15) into the evolutionary equation (6) and setting the
coefficientsateachpowerrintheobtainedequationequaltozero,wegetachain
of linking equations
Ψ˙ 6(1+κ) Ψ
Ψ =0; Ψ¨ +2 2m+1 2m+1
2m 2m+1 η − (1+3κ)2 η2
=κ(2m+3)(2m+2)Ψ ; m=0, , (κ=0, 1). (16)
2m+3
∞ 6 −
Thus, by κ = 0 the general solution of the evolutionary equation (6) for the
6
potentialfunctionΨ(r,η) isaseriesbyoddpowersofthe radialvariabler, that
isbyκ=0, 1thepotentialfunctionΨ(r,η)oftheC classisanoddfunction
∞
6 −
of the radial variable r
∞
Ψ(r,η)= Ψ (η)r2p+1. (17)
2p+1
Xp=0
For the private particlesolutions respondingthe specific physicalconditions
the series can be broken at any odd n =N >3. In this case supposing for the
last member of the series
Ψ (η)=0; n>N =2p+1, (p=1,2...), (18)
n
we obtain from the Eq. (16) a closed equation
Ψ˙ 6(1+κ) Ψ
Ψ¨ +2 2p+1 2p+1 =0. (19)
2p+1 η − (1+3κ)2 η2
3
As far as this equation does not differ from the evolutionary equation for the
particlelikesourcemassatall,itsgeneralsolutionwillcoincidewiththesolution
(8) accurate within reterming
2 3(1+κ)
Ψ2p+1 =C+pη1+3κ +Cpη− 1+3κ , (20)
−
where Cp and Cp are some constants.
+
Substituting t−his solutioninto the nextto lastequationofthe chain(16)we
shall get the equation for determining Ψ :
2p 1
−
Ψ˙ 6(1+κ) Ψ
Ψ¨2p−1+2 2ηp−1 − (1+3κ)2 2ηp2−1
2 3(1+κ)
=κ(2p+3)(2p+2)(cid:20)C+pη1+3κ +C−pη− 1+3κ (cid:21). (21)
In consequence on this equation linearity its general solution is a sum of the
general solution of the corresponding homogeneous equation Ψ0 and a pri-
2p 1
vate solution of the inhomogeneous one Ψ1 . But the general so−lution of the
2p 1
homogeneousone coincides with the already−mentioned abovesolution (8), and
the private solution can be introduced in the form of a sum of two solutions
corresponding to the two members of the right part (19). Therefore, evidently
the corresponding private solution has the form
2 3(1+κ)
Ψ12p 1 =Ap 1η1+3κ+2+Bp 1η− 1+3κ +2. (22)
− − −
Thus, we find
κ(1+3κ) κ(1+3κ)
A = (2p+1)Cp; B = (2p+1)Cp. (23)
p−1 2(7+9κ) + p−1 − 6(1 κ) −
−
Substituting the obtained solutions into the previous equations let us repeat
the analogues calculations/ Thus we pointed out the algorithm of constructing
a general solution of an evolutionary equation (7) reduced to repeating differ-
entiation operations. This general solution corresponding to the highest power
N =(2p+1)ofthe radialvariableconsistsof2N arbitraryconstantsappearing
every time by solving the corresponding homogeneous differential equations.
1.3 Case N=3
Letusdemonstratethetasksolutioninthesimplestbutasitturnsoutthemost
important case when N = 3 (p = 1) and the series (15) consists of only two
members corresponding to the valuesp = 0,1. In this case from Eqs. (15)-(23)
we find
2 3(1+κ)
Ψ3 =C+1η1+3κ +C1η− 1+3κ ; (24)
−
4
2 3(1+κ)
Ψ1 =C+0η1+3κ +C0η− 1+3κ +
−
3κ(1+3κ) (2+3κ) κ(1+3κ) 1 3κ
2(7+9κ) C+1η2 1+3κ − 2(1 κ) C−1η−1+−3κ. (25)
−
Let us study now the specific cases κ = 0 and 1+κ = 0 which are out of the
general solution (15).
1.4 Nonrelativistic matter κ = 0
In this case we get from (8) the law of mass evolution of a particlelike source
(see Ref. [4])
µ=C+η2+C η−3. (26)
−
The equation (7) for the potential function Ψ takes the form
Ψ˙
Ψ¨ +2 6Ψ=0, κ=0, (27)
η −
whence we find
Ψ=φ (r)η2+φ (r)η 3, (28)
+ −
−
where φ (r) are arbitrary functions r. In this case the mass of a particlelike
±
source evolves according to the law
µ=µ η2+µ η 3. (29)
+ −
−
1.5 Inflationary case κ+1 = 0
In this case the equation for the field function F is elliptic
Φ˙
Φ¨ +2Φ +Φ =0, (κ+1=0), (30)
′′
η
and the radial velocity of perturbation is not determined by the equation (12)
which in this case gives
∂ Φ
=0, 1+κ=0. (31)
∂η (cid:18)r(cid:19)
Integrating (31) we find:
Φ=φ(r)+ξ(η)r, (32)
whereφ(r)andξ(η)aresomearbitraryfunctionsoftheirarguments. Substitut-
ing this solution into the equation (31) and dividing the variables we find the
function Φ(r)
C C
Φ=C 2r3+r C + 3 +C η , (33)
1 2 1
− 2 (cid:18) η (cid:19)
where C , C and C are arbitrary constants.
1 2 3
5
2 Retarding spherical perturbations in ultrarel-
ativistic universe
2.1 Boundary conditions for the retarding solutions
By κ > 0 The spherically symmetric perturbations of Friedmann metrics are
described by the hyperbolic equation, by κ < 0 - by elliptic one. Peculiarities
of the hyperbolic equations (7) are convergentand nonconvergentwaves
r √κη =Const, (34)
∓
spreading at the sound velocity
v =√κ. (35)
s
Let us study the task for spherically symmetric solutions of linearized Einstein
equationsagainstthebackgroundofFriedmannmetricswithzeroboardingcon-
ditions for the potential function Φ(r,η) at the sound horizoncorresponding to
the causality principle
Σ: r =r +√κ(η η ). (36)
0 0
−
In terms of the introduced functions Φ(r,η) and Ψ(r,η these conditions can be
written in the form
Φ(r,η) =0; Ψ(r,η) =µ(η); (37)
|r=r0+√κ(η η0) ⇔ |r=r0+√κ(η η0)
− −
Φ(r,η) =0; Ψ(r,η) =0. (38)
′ |r=r0+√κ(η η0) ⇔ ′ |rr0+√κ(η η0)
− −
In this case in consequence of the equations for perturbations out of the
boundary of the sound horizon the perturbations of energy density, pressure
and velocity must automatically turn into zero:1
δε(r,η) =0; v(r,η) =0. (39)
|r>r0+√κ(η η0) |r>r0+√κ(η η0)
− −
An important private case of the boundary conditions (37) and (38) are the
conditions at “the zero sound horizon “
Σ : r =√κη. (40)
0
In this case instead of (37) and (38) we have
Ψ(r,η) =µ(η); Ψ(r,η) =0. (41)
|r√κη ′ |r√κη
1 Attheveryboundaryofthesoundfronttheenergydensityperturbationsandtheveloc-
itiesmaypracticallyhavefinalbreaks.
6
2.2 Solutions with zero boundary conditions at zero sound
Letusstudysomeprivatesolutionsintheformofpowerseriesbyradialvariable
satisfying the boundary conditions (41). Such perturbations can be generated
by the metrics fluctuations at zero moment of time and concentrated at this
moment of time in zero volume. Coming to the most convenientradialvariable
̺
1
r =√κρ, (= ρ), (42)
√3
in the case of κ = 1/3 let us write down the field equations (6) and (7). The
dot denotes derivatieves by anew temporal variable
2 2µ
µ¨+ µ˙ =0, (43)
η − η2
2 2Ψ
Ψ¨ + Ψ˙ Ψ =0. (44)
′′
η − η2 −
then solving
µ=µ+η+µ η−2, (45)
−
where µ and µ are arbitrary constants.
+
−
N=3
As far as the Taylor-seriesexpansionof the function Ψshouldnot consist of the
zero power r by definition and therefore all the even powers in this expansion
automatically vanish, so setting further
Ψ(ρ,η)=Ψ (η)r+Ψ (η)r3 (46)
1 3
and dividing the variables in the equation (44) we get an equation for the
functionsΨ (η):
i
Ψ¨ +2Ψ˙ 2Ψ1 =6Ψ ; (47)
1 η 1− η2 3
Ψ¨ +2Ψ˙ 2Ψ3 =0. (48)
3 η 3− η2
In compliance with (45) we find from (48)
Ψ3 =C3+η+C3−η−2, (49)
whereC3 aresomeconstants? Substitutingthesolution(49)intotherightpart
of the eq±uation (47) and defining private solutions of the obtained equation we
shall find its general solution
3
Ψ1 =C1+η+C1−η−2+ 5C3+η3−3C3−. (50)
7
Thus,
3
Ψ=(C1+η+C1−η−2+ 5C3+η3−3C3−)r+(C3+η+C3−η−2)r3. (51)
Further calculating Ψ(ρ,η) Ψ(η,η) , substituting the result into the
r=η
| ≡
boundarycondition(41)andsettingequalthecoefficientsinthe obtainedequa-
tion by simultaneous powers η we shall get a set of equations for the constants
Ci±,µ
±
η 2 0 =µ ;
−
η−1 (cid:12)(cid:12) C1− =0−;
η (cid:12)(cid:12) −2C3− =µ+; (52)
η2 (cid:12) C+ =0;
(cid:12) 1
η4 (cid:12) 8C+ =0.
(cid:12) 5 3
(cid:12)
(cid:12)
Thus, there are only two nonzero constants included into the expression for ν:
µ+ and C3− =−1/2µ+. So, finally
3 1 ρ3
Ψ(ρ,η)= µ ρ µ , rη; µ η, ρ>η (53)
(cid:26)2 + − 2 +η2 + ⇒
3 1 ρ3
Φ(ρ,η)= µ ρ µ χ(η ρ),
(cid:18)2 + − 2 +η2(cid:19) − −
It is the obtained previously retarding solution with zero boundary conditions
at zerosound horizon[2] andχ(z)is the Heavyside function. At that supposing
the following for the ultrarelativistic equation of state
a(η)=η; (κ=1/3), (54)
we obtain
µ ρ2 µ
+ +
ν(ρ,η)= 3 µ 2 χ(η ρ), (55)
(cid:18) η − +η3 − ρ (cid:19) −
It is easy to see that in this case at the surface of zero sound front ρ = η the
boundary conditions (41) are fulfilled identically.
N=5
Nowletusstudy the fifthdegreemultinomialasa solution. Inthis caseinstead
of the relations (44) and (45) we have
Ψ(ρ,η)=Ψ (η)ρ+Ψ (η)ρ3+Ψ (η)ρ5; (56)
1 3 5
Ψ¨ +2Ψ˙ 2Ψ1 =6Ψ ; (57)
1 η 1− η2 3
Ψ¨ +2Ψ˙ 2Ψ3 =20Ψ ; (58)
3 η 3− η2 5
8
Ψ¨ +2Ψ˙ 2Ψ5 =0. (59)
5 η 5− η2
Similarly to the previous case we have
Ψ5 =C5+η+C5−η−2. (60)
Substituting (60)intotherightpartoftheequation(58),solvingitsimilarly
to the previous case and substituting the obtained solution into the equation
(57) we get finally
Ψ3 =C3+η+C3−η−2+2C5+η3−10C5−; (61)
3 3
Ψ1 =C1+η+C1−η−2+ 5C3+η3−3C3−+ 7C5+η5−15C5−η2; (62)
and thus
3 3
Ψ(ρ,η)=(C1+η+C1−η−2+ 5C3+η3−3C3−+ 7C5+η5−15C5−η2)r+
+(C3+η+C3−η−2+2C5+η3−10C5−)r3+(C5+η+C5−η−2)r5. (63)
Substituting theequation(63)intotheboundaryconditions(41)andequat-
ing the coefficients under equal η degrees we get equations for the constants
Ci±,µ
±
η 2 0 = µ ;
−
η−1 C1− = 0;−
η −2C3− = µ+;
η2 C+ = 0; (64)
1
η3 −24C5− = 0;
η4 8C+ = 0;
5 3
η6 24C+ = 0.
7 5
Thus,
C5± =0, (65)
and the rest constants values coincide with the obtained ones above. It means
that the solution coincides with the obtained one above. It is easy to show,
adding of any new members of the series does not change the situation.
Thus, we have proved the theorem:
Theorem 1. The only spherically-symmetric solution of the C1 class of
linearized around Friedmann space-plane solution to the Einstein equations for
an ideal ultrarelativistic fluid, corresponding to the zero boundary conditions
at the zero sound horizon (41), is the solution (53) (it is equivalent to (55)).
9
2.3 Investigation of the retarding solution
From the formula (55) one can immediately see continuity of not only the first
radialderivativesbutalsothefirsttemporalonesoftheδν metricsperturbation
and the potential functions
∂δν(ρ,η)
=0; (66)
∂ρ (cid:12)
(cid:12)ρ=η
(cid:12)
∂δν(ρ,η)(cid:12)
=0; (67)
∂η (cid:12)
(cid:12)ρ=η
(cid:12)
In Fig.1 and 2 g(cid:12)raphs of temporal evolution of the potential functions Φ(ρ,η)
and δν(ρ,η) are shown
−
4 10
8
3
6
−Φ(ρ,τ) 2 −δν(ρ,τ)
4
1
2
0 1 2 3 4 0 1 2 3 4
ρ
Ρ
Fig.1. Evolution of the potential Fig 2. Evolution of the metrics per-
function Φ(ρ,η). Bottom-up η = turbation δν(ρ,η). Bottom-up η =
− −
1;2;3;4. Along the abscissa axis the 1;2;3;4. Along the abscissa axis the
radial variable values ρ = r√3 are radial variable values ρ = r√3 are
laid off. laid off.
The second radial derivatives of the potential functions and metrics have a
final break at the sound horizon. In consequence of this the energy density
perturbationhasafinalbreakatthesoundhorizonalso. Calculatingtherelative
energy density of the spherical perturbations according to the formula (11) we
find
δε 3√3µ ρ3 ρ
+
= +3 1 . (68)
ε − 4πρ (cid:18)η3 η − (cid:19)
0
The jump of the relative energy density at the sound horizon is
δε 9√3µ
+
∆= = (69)
ε (cid:12) − 4πη
0(cid:12)ρ=η
(cid:12)
(cid:12)
10
