The evolution of matter and radiation in the universe
Dynamics of a homogeneous universe
Evolution of scale factor (Friedman with k = 0):
ȧ2
2
2
−4
−3
= H = H0 Ωr x + Ωmx + Ωv ,
2
a
where
a
,
x=
a0
Ωr ≈ 0.8 × 10−4,
Ωm = 0.26,
Evolution of energy density:
d(εa3)
da3
+p
= 0.
dt
dt
Ωv = 0.74.
Radiation in thermal equilibrium:
ε
p=
→ εa4 = constant.
3
Stefan-Boltzmann law
ε = αT 4;
→ in radiation-dominated universe:
1
1
T ∝ ∝√ .
a
t
Non-relativistic matter in thermal equilibrium at T > 0
Boyle-Gay-Lussac:
3
3
2
εa = nmc + nkB T,
2
pa3 = nkB T ;
→ in a matter-dominated universe:
3
dT
nkB T da3
nkB
+
=0
2
dt
a3 dt
and
1
1
T ∝ 2 ∝ 4/3 .
a
t
Decoupling
Radiation/relativistic gas → Planck distribution
8πν 2dν
nT (ν)dν = hν/k T
.
B −1
e
After decoupling:
a → a0
⇒
νa
0
ν = 0.
a
Then
nT (ν)dν =
!3
0
a
a
8πν 0 2dν 0
0
0
ehν /kB T − 1
= n0T 0 (ν 0)dν 0,
with new temperature such that
ν
ν0
= 0
T
T
⇔
aT = a0T 0.
Similarly for non-relativistic gas with initial distribution
2
n(p) d3p = Ce−p /2mkB T d3p,
after expansion a → a0
p0 =
pa
a0
⇒
a2 T = a0 2 T 0 .
Conclusion:
· after decoupling relativistic/non-relativistic gases behave as if
in equilibrium at a temperature T which scales as for a radiation/matter dominated universe.
· non-relativistic matter cools much faster than massless particles
(radiation).
CMB
CMB temperature:
T = 2.725 ± 0.001 K
Photon and energy density:
2ζ(3) kT 3
9 m−3 ,
nγ =
=
0.41
×
10
π2
~c
π 2 (kB T )4
−3 .
=
0.260
MeV
m
εγ =
15 (~c)3
→
Ωγ = 0.46 × 10−4
Baryons
Galaxy counts, star counts,
nucleosynthesis:
→ ΩB ≈ 0.04
nB0 ≈ 0.2 m−3
Present baryon-to-photon ratio:
ηB ≡
nB0
≈ 0.4 - 0.6 × 10−9
nγ0
Electrons: ne ≈ nB → Ωe ≈ 2 × 10−5.
Decoupling of thermal photons
Ionisation energy of hydrogen: Eion = 13.6 eV.
Large photon density →
· only a fraction ∼ 10−10 of photons with energy ∼ Eion;
· after H-formation atoms permanently in excited state.
Optically thick e-p-γ plasma for temperatures kTdec > 0.27 eV;
hence decoupling of photons takes place at
Tdec ≈ 3130 K
⇒
adec
T0
xdec =
=
≈ 0.87 × 10−3,
a0
Tdec
xdec
xdx
1
√
≈ 3.5 × 105 yr.
tdec =
H0 0
Ωr + Ωmx
Z
Neutrinos
At energies kB T > 0.8 MeV neutrinos interacted with electrons
and baryons via pair creation or elastic scattering, e.g.:
ν + ν̄ ↔ e− + e+ ↔ γ + γ,
n + ν ↔ p+ + e−
→ thermal equilibrium with baryons, electrons and photons.
Decoupling happens when the interaction rate of neutrinos is
smaller than the expansion rate of the universe
s
Γν = σν |v|n < H =
8πGαef f
T2
3
Condition satisfied at kB T = 0.8 MeV when
a
xν dec = ν dec = 0.2 × 10−9,
tν dec ∼ 1 sec.
a0
Cosmic neutrino background
· Cosmic ν-background unobserved;
· After decoupling a thermal background of relativistic neutrinos
is expected with an effective temperature at late times
Tν = 0.71 Tγ .
If ν’s still relativistic now (mν ≤ 0.17 meV), this temperature is
Tν 0 = 1.95 K. For 3 stable relativistic ν-species
nν = 3 ×
εν = 3 ×
Tν
3
×
4
Tγ
7
Tν
×
8
Tγ
→
!3
× nγ
→
0.33 × 109 m−3,
!4
× εγ
→
0.17 MeV m−3,
Ων(massless) ≈ 0.30 × 10−4.
Massive neutrinos
Neutrinos are relativistic up to temperatures kB Tν ∼ mν c2
At this time the photon temperature is
mν c2
.
Tγ = 1.40 Tν = 1.40
kB
The corresponding scale factor is
Tγ0
kB Tγ0
a
=
= 0.71
a0
Tγ
mν c2
After that neutrinos are non-relativistic and
kB Tν0 =
a
a0
!2
kB Tγ0 2
2
m
c
kB Tν = 0.5
ν
mν c2
0.029 meV2
=
.
mν c2
For mν c2 = 10 meV:
Tν0 = 0.034 K
Neutrino number densities are fixed at decoupling, when they are
still very relativistic → nν0 = 0.33 × 109 m−3.
Then with mν as above, for three species:
εν0 = 3.3 MeV m−3
⇒
Ων(massive) = 0.6 × 10−3 Ωm,
and the cross-over to non-relativistic behaviour takes place at
a
≈ 0.02
⇔
z ≈ 50.
a0
Photon vs. neutrino temperature
After neutrino decoupling, the neutrino temperature and the
photon temperature are given by
a T = adec Tdec.
Subsequently electron-positron annihilation heats the photon gas:
e− + e+ → γ + γ.
As a result the photon density and temperature are raised compared to the neutrino density and temperature. How much?
Computation based on thermodynamics of relativistic gases
First law of thermodynamics
T dS = dU + pdV
⇒
T d(sV ) = d(εV ) + pdV
with
S
s(T ) = ,
V
→
U
ε(T ) = ,
V
Ts = ε + p = T
dp
.
dT
Radiation:
p=
N.B.:
ε
3
αF =
⇒
7
αB .
8
ε = αT 4,
s=
4α 3
T .
3
During e+ e−-annihilation entropy remains constant:
⇒
s+ = s−
3 = α T 3.
α+T+
− −
Before decoupling 3 components in thermodynamic equilibrium:
(γ, e−, e+)
7
11
→ α+ = 1 + 2 ×
αB =
αB ;
8
4
after decoupling 1 component left: γ
→
α− = αB .
Therefore
T−
11 1/3
=
= 1.40.
T+
4
Primordial nucleosynthesis
With the known baryon-to-photon ratio ηB spontaneous synthesis of light nuclei: D, 3He, 4He, from free nucleons was possible
at photon and baryon temperatures between 100 and 10 keV.
4 He is by far the most stable of the light nuclei
⇒ practically all neutrons (> 99%) end up in α-particles.
Then the mass of 4He as a fraction of total baryon mass is
Y (4He) =
4nα
2nn
2(nn/np)
=
=
.
nn + np
nn + np
1 + (nn/np)
Observed: Y (4He) = 0.24
⇒
nn/np = 0.13.
Initial condition determined by thermal equilibrium
!
2
nn
= e−∆mc /kB T0 = 0.2
np 0
at kB T0 = 0.8 MeV.
Neutron life time τ = 886 sec
nn
nn0 e−∆t/τ
=
np 1 np0 + nn0(1 − e−∆t/τ )
!
→
= 0.13
for
∆t = 328 sec.
→ kB T1 ≈ 0.09 MeV.
Very sensitive to ηB .
Formation of neutral hydrogen
In a plasma of hydrogen atoms, protons, electrons and photons,
we can calculate the degree of ionisation as a function of temperature:
np
nH
X=
,
1−X =
np + nH
np + nH
Use charge neutrality and chemical equilibrium:
ne = np,
µ H = µp + µe .
In units c = kB = ~ = 1, non-relativistic equilibrium densities are:
miT 3/2
µi − mi
ni = gi
exp
2π
T
→
nH
1−X
g
=
= H
X
np
gp
≈
mH
mp
!3/2
µH − mH − µp + mp
exp
T
gH
µe − me + B
exp
gp
T
gH
2π 3/2
ne eB/T
=
gegp meT
Now ge = gp = 2, gH = 4, and
ne = np = 0.76 XηB nγ = 0.76 X × ηB
→
2ζ(3) 3
T
2
π
T 3/2 B/T
1−X
= 2.9 ηB
e
≡ S(T ),
X2
me
where the ionization energy is B = 13.6 eV.
η = 0.6 x 10−9
B
1.0
X
0.5
0
0.20
0.30
0.40
0.50
T (eV)
q
X(T ) =
1 + 4S(T ) − 1
2S(T )
.
Photon decoupling temperature
Interaction rate of photons and electrons:
8πα2~2
Γγ = c σT ne =
ne
3m2
c
e
Compare with expansion rate of the universe
c σT ne
c σT np H0
Γγ
=
=
H
H
H0 H
c σT
x2
=
× 0.76 X(T ) × ηB nγ √
H0
Ωr + Ωmx
where
x=
Tγ 0
a
2.35
=
× 10−4,
=
a0
Tγ
T (eV)
nγ = nγ 0 ×
Tγ
Tγ 0
!3
= nγ 0 ×
T (eV)
2.35
!3
× 1012 m−3,
and we have a dimensionless factor
c σT nγ 0
= 0.34 × 109.
H0
Then
T
Γγ
= 0.0016 X(T ) ×
H
T0
Γγ
<1
H
⇔
!3
X(T ) < 0.01
ηB
x2
√
.
×
−9
Ωr + Ωmx
0.6 × 10
⇔
T < 0.27 eV = 3130 K.