Diffusion model of evolution of superthermal high-energy particles under scaling in the early Universe

0.21 MB·English

by  Yu. G. Ignatyev

Diffusion model of evolution of superthermal
high-energy particles under scaling in the early
Universe
Yu.G.Ignatyev, R.A.Ziatdinov
Kazan State PedagogicalUniversity,
1 Mezhlauk str., 1, Kazan 420021,Russia
1
0 Abstract
2
The evolution of a superthermal relic component of matter is studied
n
on thebasisofnon-equilibriummodelofUniverseandtheFokker-Planck
a
typekinetic equation offered by one of theauthors.
J

1
1 Introduction

]
c
q The diffusion equation describing cosmological evolution of superthermal par-
-
r ticles under the assumption of interactions scaling recovery in the range of su-
g perhigh energies was studied in paper [1] :
[

∂G 1 ∂ ∂G
1 = x2 +2b(τ)G , (1)
v ∂τ x2∂x ∂x
(cid:18) (cid:19)
5
7 where the dimensionless function b(τ) is determined as:
3
∞
0
. b(τ)= G(τ,x)xdx; (2)
1
Z
0 0
1
It is necessaryto solvethe equation(1) with initial andboundary conditions of
1
the form:
:
v
i G(0,x)=G(x); (3)
X
r lim G(τ,x)x3 =0, (4)
a x→∞
moreover the function G(x) must satisfy the integral conditions:
∞
G(x)x2dx=1; (5)
Z
0
∞
G(x)x3dx=1. (6)
Z
0
Asthedetailedresearcheshaveshown,thesolutionofkineticequation(1)incase
of b(τ) = 0 obtained in [1] becomes correct only at greater times of evolution,
1
therefore it becomes problematical to establish parameters relationship of this
solution with the initial distribution. In this paper we discuss the evolution of
system at small times.
It is necessary to note that the exact solution of the kinetic equation in
diffusiveapproximation(1),satisfyingthe normalizationrelations(5),(6),isan
equilibriumulatrarelativisticBolzmandistributionwithconformaltemperature
τ = 1:
3
27
f = e−3x. (7)
0
2
2 Numerical model of the initial distribution
Let’s consider the initial distribution analogous to Fermi-Dirac distribution,
given in the form of infinitely differentiable and integrable function:
A
G (x)= , (8)
0 eξx−y+1
where A,ξ,y - parameters of the initial distribution. These parameters must be
such that normalization relations (5),(6) fulfill automatically. Thus, we have
two algebraic relations on three parameters, solving which through parameter
y we find:
∞
t3dt
e(t−y)+1
Z ξ3(y)
ξ(y)= 0 ; A(y)= (9)
∞ ∞
t2dt t2dt
e(t−y)+1 e(t−y)+1
Z Z
0 0
Slow convergence of improper integrals in (9) leads to necessity of their trans-
formation to following form that is more suitable for numeric calculations:
∞
tdt
J (y)= ≡
1 e(t−y)+1
Z
0
∞ ∞ y y
dx xdx dx xdx
y + +y − . (10)
ex+1 ex+1 e−x+1 ex+1
Z Z Z Z
0 0 0 0
∞ ∞
t2dt dx
J (y)= ≡y2 +
2 e(t−y)+1 ex+1
Z Z
0 0
∞ ∞ y y y
xdx x2dx dx xdx x2dx
2y + +y2 −2y + . (11)
ex+1 ex+1 e−x+1 ex+1 ex+1
Z Z Z Z Z
0 0 0 0 0
2
∞ ∞ ∞
t3dt dx xdx
J (y)= ≡y3 +3y2
3 e(t−y)+1 ex+1 ex+1
Z Z Z
0 0 0
∞ ∞ y
x2dx x3dx dx
+3y + +y3 −
ex+1 ex+1 e−x+1
Z Z Z
0 0 0
y y y
xdx x2dx x3dx
3y2 +3y − . (12)
ex+1 ex+1 ex+1
Z Z Z
0 0 0
Using known representations of Riemann ζ - functions and Bernoulli numbers
(e.g., see [6]):
∞
xn−1
dx=(1−21−n)Γ(n)ζ(n); (13)
ex+1
Z
0
∞
x2n−1 22n−1−1
dx= π2nB (14)
ex+1 2n n
Z
0
and introducing dimensionless functions:
y
xn
S(n,y)= dx, (n=0,1,2,...), (15)
e−x+1
Z
0
(S(0,y)=ln(1+ey)/2) we get expressions for (10)-(12):
π2
J (y)=yln2+ +yS(0,y)−S(1,y); (16)
1
12
π2 3
J (y)=y2ln2+y + ζ(3)+y2S(0,y)−
2
6 2
2yS(1,y)+S(2,y); (17)
π2 9 7
J (y)=y3ln2+y2 +y ζ(3)+ π4+
3
4 2 120
y3S(0,y)−3y2S(1,y)+3yS(2,y)−S(3,y). (18)
Thus, by means of introduced functions (10)-(12) we find expression for b(τ):
∞
xdx
b (y)=A(y) ≡
0 eξ(y)x−y+1
Z
0
∞
A(y) xdx J (y)J (y)
3 1
= . (19)
ξ2(y) ex−y+1 J2(y)
Z 2
0
3
Presentation of (9) by means of S(n,y) and Riemann ζ - functions makes nu-
meric computations simpler. The results of integration are shown on Figure
1,2 and we see that ξ(y) is monotone increasing, A(y) is monotone decreasing
function.
3.32
3.3
3.28
3.26
3.24
3.22
3.2
3.18
3.16
0 0.2 0.4 0.6 0.8 1
y
Figure 1: Plot of ξ(y) function.
16
14
12
10
8
6
4
2
0 2 4 6 8 10
y
Figure 2: Plot of A(y) function.
4
As a result, normalized initial distribution function is defined by one arbitrary
parameter, y:
A(y)
G (x,y)= , (20)
0 eξ(y)x−y+1
which controlsthe the degree of non-equilibrium of initial distribution (8), Fig-
ure 3.
12
10
8
6
4
2
0
0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
x
Figure 3: Normalized initial distribution G (x,y) according to parameter y:
0
top-down: y =1, y =3, y =10. The initial distribution (8) is shown by dotted
line.
3 Diffusion equation and integral conservation
laws
Asknown(see,e.g. [5]),thestrictconsequencesfromgeneralrelativistickinetic
theoryincaseofelasticcollisionsareintegralconservationlawsofparticlestotal
number and their energy. The solution of diffusion equation (1), obtained on
the basis of general relativistic kinetic equations, also must satisfy these laws.
Therefore it is necessary to verify this fact.
Multiplying by x2 the both sides of equation (1) and integrating received
expression in parts on all interval of impulsive variable x, we get:
∞
∞
d ∂G
x2G(x,τ)dx= x2 +2b(τ)G . (21)
dτ ∂x
Z (cid:20) (cid:18) (cid:19)(cid:21)(cid:12)0
0 (cid:12)
(cid:12)
(cid:12)
5
Assuming later on:
∂G
limx2G; limx2 =0; (22)
x→0 x→0 ∂x
∂G
lim x2G=0; lim x2 =0; (23)
x→∞ x→∞ ∂x
we obtain from (21):
∞
x2G(x,τ)dx=Const. (24)
Z
0
According to (5) the constant at the right side of (24) equals to 1.
Multiplying nowbyx3 the bothsidesof(1)andintegratingreceivedexpres-
sion in parts on all interval, we get:
∞
∞
d ∂G
x3G(x,τ)dx = x3 +2b(τ)G −
dτ ∂x
Z (cid:20) (cid:18) (cid:19)(cid:21)(cid:12)0
0 (cid:12)
(cid:12)
∞ (cid:12)
∂G
− x2 +2b(τ)G dx (25)
∂x
Z (cid:18) (cid:19)
0
Considering (22), (24), (4) and once more integrating received expression in
parts, we find:
∞
d ∞
x3G(x,τ)dx =−x2G +
dτ 0
Z
0 (cid:12)
∞ ∞ (cid:12)
+2 xGdx−2b(τ) x2Gdx. (26)
Z Z
0 0
Taking into account relations (2),(5), and also (22),(24), we finally obtain from
(26):
∞
x3G(x,τ)dx=Const. (27)
Z
0
According to (6) the constant at the right side of (27) equals to 1.
It is enough in order to G(x) function’s degree of magnitude satisfy the
following strong inequalities for realization of relations (22),(24),(4):
1 1
G(x)|x→0 < x, G(x)|x→∞ < x3. (28)
We also note that realization of the energy conservation law is provided with
the presence of term with b(τ) coefficient at the right side of diffusion equation
(1). If this term is missing, the process of energy transmitting at small times is
not considered.
6
4 Expansion of diffusion equation by smallness
of τ
This approximationmatch up the early stages of universe evolution, when par-
ticle interactions are inessential:
τ ≪1. (29)
Then the term at the left side of diffusion equation is main and taking into
account the initial distribution (3) we have:
∂G
=0,⇒G(x,τ)=G (x). (30)
0
∂τ
Expanding the right side of equation (1), substituting expression (30) as the
initialdistributionandintegratingintimevariableweobtainthefirstcorrection.
Sequentially iterating this procedure we get the recurring formula for finding
higher approximations:1
1 ∂ ∂G
G = x2 k+
k+1 x2∂x ∂x
Z (cid:20)
k ∞
2 Gi Gk−ixdx dτ, (31)
i=0(cid:18) Z0 (cid:19)#
X
Consideringthatcorrectionofk-orderisproportionaltoτk andintegrating(31),
we get:
τk+1 1 ∂ ∂g
G = x2 k+
k+1 k+1x2∂x ∂x
(cid:20)
k ∞
2 gi gk−ixdx . (32)
i=0(cid:18) Z0 (cid:19)#
X
Differentiating in expression (32), we finally get:
τk+1 ∂2g 2∂g
k k
G = + +
k+1 k+1 ∂x2 x ∂x
(cid:26)
k
∂g 2g
i i
2 + bk−i . (33)
∂x x
i=0(cid:20)(cid:18) (cid:19) (cid:21))
X
We also find the recurring formula for determining function b(τ) in equation
(1). Considering (32) we obtain according to (2):
τk+1 ∞ dx ∂ ∂g
b = x2 k+
k+1
k+1 x ∂x ∂x
Z0 (cid:20)
1Hereandfurther,ifnotmentionedespecially,k∈N.
7
k ∞
2 gi gk−ixdx . (34)
i=0(cid:18) Z0 (cid:19)#
X
Integrating in parts an integral in (34) and supposing henceforward:
∂g
k
limxg =0; limx =0; (35)
i
x→0 x→0 ∂x
∂g
k
lim xg =0; lim x =0. (36)
i
x→∞ x→∞ ∂x
we get finally:
b =
k+1
τk+1 ∞ ∂g k
k
+2 (gibk−i) dx. (37)
k+1 ∂x
Z0 " i=0 #
X
Nowweprovethatthe correctionstonormalizedinitialdistribution(8)cannot
change the total number density and energy density at every step of iterations.
Multiplying by x2 the both sides of (32) and integrating on all interval, we
get:
∞
τk+1 ∂g
x2G dx= x2 k+
k+1
k+1 ∂x
Z (cid:26) (cid:20)
0
∞
k ∞
2 gi gk−ixdx . (38)
Xi=0(cid:18) Z0 (cid:19)#)(cid:12)(cid:12)0
(cid:12)
Supposing henceforward: (cid:12)
(cid:12)
∂g
limx2g =0; limx2 k =0; (39)
i
x→0 x→0 ∂x
∂g
lim x2g =0; lim x2 k =0. (40)
i
x→∞ x→∞ ∂x
we obtain from (38):
∞
x2G dx=0. (41)
k+1
Z
0
Multiplying by x3 the both sides of (32) and integrating on all interval, we get:
∞
x3G dx=
k+1
Z
0
∞
τk+1 ∂g k ∞
= x3 k +2 gi gk−ixdx −
k+1 ( " ∂x Xi=0(cid:18) Z0 (cid:19)#)(cid:12)(cid:12)0
(cid:12)
(cid:12)
(cid:12)
8
∞
τk+1 ∂g k ∞
x2 k +2 gi gk−ixdx dx. (42)
k+1 ∂x
Z " i=0(cid:18) Z0 (cid:19)#
0 X
Considering (39), (40), (27) and once more integrating received expression in
parts, we find from (42):
∞ ∞
τk+1 ∞
x3G dx= −x2g +2 xg dx−
k+1 k+1 k 0 k
Z0  (cid:12) Z0
(cid:12)
∞ k ∞ 
−2 x2 gi gk−ixdx dx . (43)

Z0 Xi=0(cid:18) Z0 (cid:19) 
After an obvious simplifications 
∞ ∞
τk+1 ∞
x3G dx= −x2g +2 xg dx−
k+1 k+1 k 0 k
Z0  (cid:12) Z0
(cid:12)
∞  ∞
−2 xg dx x2g dx−
k 0
Z Z
0 0
∞
k ∞
−2 gk−ixdx x2gidx , (44)
 
Xi=1 Z0 Z0 
 
and taking into account expressions (5), (39), (40), (41), we get from (44) at
last:
∞
x3G dx=0. (45)
k+1
Z
0
Thus, we certain that the distribution function iterations of each step don’t
changethetotalnumberofparticlesandenergy,thisisusefultoolforcalculation
correctness. It follows from (45) that the correctionof any order is alternating-
sign on the interval [0,+∞). Hence, small time τ approximation is completely
equivalent to expansion of exact function G(x,τ) to Taylor series by τ powers.
4.1 The first order approximation
As a first approximation according to recurring formula (32) we have:
G =τg ; (46)
1 1
1 ∂ ∂G
g = x2 0 +2b G . (47)
1 x2∂x ∂x 0 0
(cid:18) (cid:19)
9
Substituting expression (47) into (2) and integrating in parts, also considering
that function G (x,y), defined by expression (8), with its derivatives approach
0
to 0 at x→∞ faster than any of exponential functions and at x→0 has finite
derivatives, we obtain:
∞
dx ∂ ∂G
0
b =τ +2b G (x) =
1 0 0
x ∂x ∂x
Z (cid:18) (cid:19)
0
∞
∂G
0
=τ x +2b G (x) +
0 0
∂x
(cid:20) (cid:18) (cid:19)(cid:21)(cid:12)0
(cid:12)
∞ (cid:12)
∂G (cid:12)
0
+τ +2b G (x) dx=
0 0
∂x
Z (cid:18) (cid:19)
0
∞
=−τG (0)+τ2b G dx⇒
0 0 0
Z
0
∞
A(y)ey
b (τ)=−τ +τ2b G dx.
1 ey+1 0 0
Z
0
Substituting here the initial distribution (8), we find:
∞
A(y)
G dx= ln(1+ey), (48)
0
a(y)
Z
0
therefore
2b ln(1+ey) ey
0
b (τ)=τA(y) − . (49)
1 a(y) ey+1
(cid:20) (cid:21)
Then, substituting the function G (x,y) into (47), we determine an explicit
0
form of linear approximation to the initial distribution:
10