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The hot, early universe
Cosmology Block Course 2014
Simon Glover & Markus Pössel
Institut für Theoretische Astrophysik/Haus der Astronomie
30 July 2014
Thermo & statistic
Back in time
Primordial nucleosynthesis
Cosmic background radiation
Contents
1 Thermodynamics & statistics in an FLRW universe
2 Going back in time to small a
3 Primordial nucleosynthesis
4 The cosmic background radiation
Simon Glover & Markus Pössel
The hot, early universe
Thermo & statistic
Back in time
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Cosmic background radiation
Thermodynamics & statistics in an FLRW universe
• Up to now, matter in our universe has not interacted
• If we get back to sufficiently small a(t) (as we must → singularity
theorems!), we cannot have had separate galaxies
• Early universe: filled with plasma, colliding particles (atoms and
photons, nucleons and nucleons) ⇒ we need a description from
thermodynamics and statistical physics!
Simon Glover & Markus Pössel
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When thermodynamics is simple and when it isn’t
Thermodynamics is simple when a system is in thermal
equilibrium, and complicated when it isn’t.
(If not in equilibrium: fluid dynamics plus reaction kinetics – can be
horribly complicated!)
In equilibrium, certain thermodynamical quantities can be
introduced, which take on constant values throughout the system.
Best-known of those: Temperature T and pressure p.
Equilibrium thermodynamics: given energy E, volume V , particle
number N , calculate T, p.
Simon Glover & Markus Pössel
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Statistical basis for thermodynamics
Thermodynamics: Macrostates specified by thermodynamic
variables like E, V, T, p, N .
Statistical mechanics: Microstates of particles (e.g. N particles
making up a gas – each has a given momentum at a given time)
Entropy as a quantity to count microstates compatible with a
macrostate:
S = k · log Ω(E, V, N).
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Systems in equilibrium
Entropy difference in terms of changing variables:
dS =
p
1
· dE + · dV
T
T
(this can be taken as definitions of T and p).
Re-write as first law of thermodynamics:
dE = T · dS − p · dV
Simon Glover & Markus Pössel
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Systems that are not in equilibrium
Second law of thermodynamics: δS ≥ 0, but never δS < 0. Entropy
cannot decrease.
Two systems in contact so that S = S1 + S2 , V = V1 + V2 ,
E = E1 + E2 :
dE1 = −dE2 ; dV1 = −dV2 so that
dS =
!
!
1
p1 p2
1
−
· dE1 +
−
· dV1 ≥ 0.
T1 T2
T1 T2
Second law means: at constant volume, dE1 < 0 if T1 > T2 . At
constant temperature, dV1 > 0 if p1 > p2 . In thermodynamics
equilibrium, dS = 0, so T1 = T2 , p1 = p2 . All as expected.
Simon Glover & Markus Pössel
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Entropy density
Define the entropy density s(T) by S(T, V) = V · s(T)
(This works because entropy is extensive!).
Then for any adiabatic change,
d(s(T)V) = Vds(T) + s(T)dV = dS(T, V) =
=
d(ρc2 V) + pdV
T
Vc2 dρ
(ρc2 + p)dV
dT +
T dT
T
This can only hold generally if the coefficients for dV are equal:
s(T) =
ρc2 + p
T
(coefficients for dT give energy conservation).
Simon Glover & Markus Pössel
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Chemical potential
Additional contribution to entropy:
dS =
X µi
p
1
· dE + · dV −
dNi
T
T
T
i
(this can be taken as the definition of the µi ). New first law:
dE = T · dS − p · dV +
Simon Glover & Markus Pössel
X µi
i
T
dNi
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Chemical potential
Within the same system, in thermal equilibrium, reactions
changing particle species 1 into 2 (and other way around), with
N1 + N2 = N =const.:
dS =
1
(µ2 − µ1 )dN1 .
T
If µ2 > µ1 , number of 1-particles increases! In full thermodynamic
equilibrium, from dS = 0, µ1 = µ2 .
Distinguish between
• thermal equilibrium (T, p constant, µi can differ from equilibrium
values)
• chemical equilibrium (µi have equilibrium values, T, p could differ)
• thermodynamic equilibrium (T, p, µi all have equilibrium values)
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Multi-particle reactions:
Particle reaction
1+2↔3+4
(z.B. H + γ ↔ p + e− , or nuclear reaction):
dN1 = dN2 = −dN3 = −dN4 , then in thermal (not necessarily
chemical!) equilibrium:
dS =
1
(µ3 + µ4 − µ1 − µ2 )dN1 ≥ 0.
T
In equilibrium,
µ3 + µ4 = µ1 + µ2
(more generally: one such relation for each conserved quantum
number: baryon number, lepton number, . . . )
Simon Glover & Markus Pössel
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Particles in thermal equilibrium
Grand-canonical example: E, V, N given — what is the equilibrium
state? (Sum over quantum states, treat bosons and fermions
differently).
Number density in momentum space cell d3 p = dpx · dpy · dpz :
n(px , py , pz ) =
with E(p) =
p
mc2 + (pc)2 .
g
d3 p
exp ([E(p) − µ]/kT) ∓ 1 h3
Integrate up to get total particle number density!
Simon Glover & Markus Pössel
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Highly relativistic particles 1/2
kT > mc2 , kT > µ, E ≈ pc:
Equipped with these formulae, it is straightforward to show that
8π
ζ(3) g(kT)3 ·
n=
(ch)3
(
1
bosons
3/4 fermions
wIth ζ(3) = 1.2020569031. The density for highly relativistic
particles is
4π5
g(kT)4 ·
ρc =
15(ch)3
2
(
1
bosons
7/8 fermions
For bosons, this is Bose-Einstein statistics, for fermions,
Fermi-Dirac statistics.
Simon Glover & Markus Pössel
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Highly relativistic particles 2/2
Pressure:
p=
1 2
ρc (as for radiation!)
3
Entropy density:
4 ρc2
16π5
s(T) =
=
gk(kT)3 ·
3 T
45(ch)3
(
1
bosons
7/8 fermions
Note that the chemical potential ν features nowhere in here – for
highly relativistic particles, lots of particle-antiparticle pairs flying
around, the chemical potential can be neglected!
Simon Glover & Markus Pössel
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Photons
Thermal photon gas: g = 2 (two polarizations), bosons, m = 0,
E = hν, in thermal equilibrium µ = 0:
Number density of photons with frequencies between ν and ν + dν
is
8πν2 /c2
dν
exp(hν/kT) − 1
Simon Glover & Markus Pössel
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Photons
8πh
ρc2 = 3
c
Z∞
0
ν3
dν
exp(hν/kT) − 1
From the bosonic energy distribution, it follows that the number
n(m) of photons with energies greater or equal to m · kT is
n
n(m) =
2ζ(3)
Z∞
m
x2 dx
.
exp(x) − 1
For instance, 10−9 of the photons have energies > 26kT , while
10−10 have energies > 29kT .
Simon Glover & Markus Pössel
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Non-relativistic particles
For non-relativistic particles, mc2 kT and
E(p) ≈ mc2 +
p2
.
2m
The number density is
g
mc2 − µ
n = 3 (2π m kT)3/2 exp −
kT
h
!
even when not in chemical equilibrium, the energy density is
3
ρc = n · mc + kT
2
2
2
and the pressure
!
p = n kT.
These last expressions are as expected.
Simon Glover & Markus Pössel
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Particle interactions and time scales
How many particle interactions (collisions) in a given situation? For
non-relativistic particles in thermal (not necessarily chemical!)
equilibrium, number densities n1 and n2 , reduced mass
µ = m1 m2 /(m1 + m2 ), the collision rate density C is
C = n1 n2 huσ(E)iu ,
where the averaging is over a Maxwell-Boltzmann distribution for
the relative velocity u, and σ(E) is the cross section (= collision
probability), with E(u) the (velocity-dependent!) energy,
!
µ̄ 3/2 Z∞
−µ̄u2
exp
σ(E) u3 du.
huσ(E)iu = 4π
2π kT
2kT
0
Simon Glover & Markus Pössel
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Particle interactions and time scales
If σ(E) is independent, or weakly dependent on E, the integral
becomes
s
huσiu = σhuiu . = σ
8kT
πµ̄
If one of the particle species is photons, we will approximate the
collision rate by
C = n1 n2 σc,
scaling, if necessary, with the fraction of photons with an energy
larger than the reaction we’re interested in.
Simon Glover & Markus Pössel
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Particle interactions and time scales
The number of reactions per particle of species 1 is
Γ=
C
= n2 huσ(E)iu ,
n1
which has physical dimension 1/time.
We compare this with the Hubble parameter
H(t) =
ȧ
a
which is the ratio of the change of a to a itself, and thus a measure
for the time it takes a to change significantly.
Simon Glover & Markus Pössel
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Local equlibrium vs. freeze-out
1
Γ H for reactions that establish thermal equilibrium: Local
Thermal Equilibrium (LTE): Adiabatic (=isentropic) change from one
temperature-dependent equilibrium to the next
2
H Γ: Freeze-out – particle concentrations remain constant (or
change because of decay, or alternative reactions). Temperature
decouples.
Simon Glover & Markus Pössel
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For small x = a/a0 : Radiation dominates!
Density/present density
1021
1014
107
100
Dust
Dark energy
10−7
10−14 −6
10
Radiation
10−5
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10−1 100
Relative scale factor a/a0
10−4
10−3
10−2
101
102
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Radiation dominates
For early times, assume that the only significant contribution
comes from radiation:
q p
a = a0 2 Ωr0 H0 t
so with Ωr0 = 5 · 10−5 and H0 = 2.18 · 10−18 /s,
−10
a = a0 · (1.76 · 10
Hubble parameter goes as
H(t) =
Simon Glover & Markus Pössel
)
r
t
.
1s
1
.
2t
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For the phase directly following radiation dominance
For later times assume the only significant contribution comes from
the matter density,
a = a0
3p
Ωm0 H0 t
2
!2/3
,
so with Ωm0 = 0.317 and H0 = 2.18 · 10−18 /s,
a = a0 · (1.5 · 10−12 )
Hubble parameter goes as
H(t) =
Simon Glover & Markus Pössel
t 2/3
.
1s
2
.
3t
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How does this compare to the exact treatment?
10-1
Relative scale factor x
10-2
10-3
10-4
10-5
10-6 9
10
Full evolution
Radiation only
Matter only
1010
1011
1012
1013
Cosmic time [s]
1014
1015
1016
. . . works except for t = 8 · 1010 − 2 · 1014 s = 6k – 300k years.
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How does the energy distribution evolve?
At some time t1 , scale factor value a(t1 ), in some volume V1 , let the
photon number between ν1 and ν1 + dν1 be
V1
8π(ν1 )2 /c3
dν1 .
exp(hν1 /kT1 ) − 1
At some later time t2 , the same photons are now spread out over a
volume V2 = (x21 )3 with x21 = a(t2 )/a(t1 ). They have been
redshifted to ν2 = ν1 /x21 , and their new frequency interval is
dν2 = dν1 /x21 .
Simon Glover & Markus Pössel
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How does the energy distribution evolve?
We can re-write the new number density in terms of the new
frequency and interval values ν2 and dν2 ; the x21 mostly cancel,
which gives a new number density in the frequency interval
ν2 . . . ν2 + dν2 at time t2 that is
8π(ν2 )2 /c3
dν2 .
exp(hν2 x2 1/kT1 ) − 1
This corresponds to the number of photons we would expect in the
given frequency range for thermal radiation with temperature
T2 = T1 · a(t1 )/a(t2 ),
which for a(t2 ) > a(t1 ) corresponds to lower temperature.
Temperature scales as ∼ 1/a(t)! Radiation remains Planckian!
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Photon number baryon number
Now it becomes important that the number of photons is so much
larger than the baryon number (as you will estimate in the
exercise):
η=
nB
≈ 6 · 10−10 .
nγ
Everything that’s going on will take place in a photon bath! Even
absorption reactions hardly matter – they will change the bath by
at most 10−9 !
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Photon energy over time
Temperature effects in the early universe
Photon energy [eV]
1010
2.7 kT (average boson energy)
109
27 kT (a billionth of all photons)
8
10
7
10
106
105
104
103
102
101
100
10-1
10-2
10-3
10-4 -5
10 10-3 10-1 101 103 105 107 109 1011 1013 1015 1017
Cosmic time [s]
The hot, early universe
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Photon energy over time
Temperature effects in the early universe
Photon energy [eV]
1010
2.7 kT (average boson energy)
109
27 kT (a billionth of all photons)
8
10
7
10
106
105
Pu ionization: 120 keV
104
103
102
101
Hydrogen ionization: 13.6 eV
100
10-1
10-2
10-3
10-4 -5
10 10-3 10-1 101 103 105 107 109 1011 1013 1015 1017
Cosmic time [s]
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Photon energy over time
Temperature effects in the early universe
Photon energy [eV]
1010
109
108
Ni-56 binding energy: 8.8 MeV/nucleon
107
106
Electron pair production: 1.2 MeV
105
104
103
102
101
100
10-1
10-2
2.7 kT (average boson energy)
10-3
27 kT (a billionth of all photons)
10-4 -5
-3
10 10 10-1 101 103 105 107 109 1011 1013 1015 1017
Cosmic time [s]
The hot, early universe
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Reaction numbers?
You will calculate some reaction numbers (photon colliding with
atom, or with nucleus) in the exercises today. Collision rates will
not be a problem — as long as the photons carry sufficient energy
to trigger a reaction (ionization, splitting a nucleus...)!
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The big picture
Radiation era
Matter era
Big bang
10−33 s
Inflation
1e-6 s
quark confinement
1 s to 3 min
light elements
Simon Glover & Markus Pössel
Now
13.8e9. a
1e8 a
galaxies
380,000 a
CMB
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Cosmic background radiation
Where do we begin?
In this lecture, we trace temperature back until we have a sea of
single nucleons (protons and neutrons). We leave earlier phases
(inflation etc.) for later, namely for the last lecture.
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Primordial nucleosynthesis
As we have seen, at about 1s of cosmic time, sufficient photon
energy to tear apart the most stable nuclei (Ni-56).
Three-particle reactions much too uncommon (will occur in stars,
but not here!), so nuclei have to be built from two-particle reactions.
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Reaction network
012 January 10
Coc et al.
(α,n)
(α,γ)
11
12
C
(β-)
(p,γ)
C
(t,γ)
10
11
B
B
12
B
(n,γ)
X
7
9
Be
Be
(β+)
6
3
7
Li
Li
(d,γ)
4
He
8
Li
He
(t,n)
(d,n)
(t,p)
1
H
2
3
H
X
(p,α)
H
(d,p)
ated S-factor. The solid line
(n,p)
(d,nα)
n
line journal.)
(n,α)
Figure 22. Reduced network displaying the important reactions for 4 He, D,
3
7
6
9
10,11
He, and Li (blue), Li (green), Be (pink),
B (cyan), and CNO (black)
11 B=
Key: There isproduction.
no stable
with
! (Fig.
from
but 5
follows
a different
pathCoc 2012)
Note thatelement
CNO production
is via A
11 B formation through the late-time 11 C decay.
(A color version of this figure is available in the online journal.)
than primordial
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When can nucleosynthesis start?
All nucleosynthesis starts with deuterium production. Binding
energy of Deuterium: 2.2MeV .
Nothing happens until the photon energy 27kT (this or more
carried by 6 · 10−10 of all photons!) goes below 2.2 MeV , which is
(k = 8.6 · 10−5 eV/K ) at
or
so
TD = 9.5 · 108 K
a/a0 = T0 /TD = 3 · 10−9
t = 290 s.
. . . at which time, all neutrons are quickly built into 4 He; most stable
configuration, gap at A = 5! But how many neutrons do we have in
the first place?
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How many neutrons do we have to start with?
Weak interactions between protons and neutrons:
n + ν e ↔ p + e−
Reaction rate for these weak interactions is:
−47
σw = 10
2
m
kT
1 MeV
!2
Estimating the reaction rate (similar to exercise), this works only
while there is still pair production, and stops at
t ≈ 1s, kT ≈ 0.8 MeV.
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How many neutrons do we have to start with?
n + νe ↔ p + e−
in equilibrium above or about kT = 1 MeV means that
µn + µνe = µp + µe− ⇒ µn = µp
(where, as mentioned above, we have neglected µ for highly
relativistic, pair-produced particles. But the non-relativistic number
density was
n=
so
Simon Glover & Markus Pössel
!
mc2 − µ
g
3/2
(2π
m
kT)
exp
−
,
kT
h3
nn
mn
=
np
mp
!3/2


 (mn − mp )c2 

 .
exp −
kT
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How many neutrons do we have to start with?
mn
nn
=
np
mp
!3/2


 (mn − mp )c2 

 .
exp −
kT
Inserting kT = 0.8 MeV and (mn − mp )c2 = 1.293 MeV , neglecting
the pre-factor:
nn
1
= 0.198 ≈
np
5
at t = 1s.
Problem: Neutrons decay, with half-life 611 s! Between t = 1 s and
t = 290 s, the number ratio has dropped from 1 neutron to 5
protons to
1
(1/2)290 s/611 s ≈ 0.72 neutrons per proton ≈ .
7
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Helium fraction
With 2 neutron per 14 protons (1/7), you can make 1 4 He plus 12
protons. Mass ratio between He and total mass is
Y=
4
= 25%
16
. . . this is the fairly robust main prediction for big bang
nucleosynthesis!
Simon Glover & Markus Pössel
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Time evolution
The Astrophysical Journal, 744:158 (18pp), 2012 January 10
2
2
ΩBh =WMAP
ΩBh =WMAP
10
1
10
10
10
10
3
10
4
1
p
-1
4
n
10
He
-3
2
10
H
-5
3
3
H
10
He
-7
10
-1
-3
-5
Mass fraction
Mass fraction
10
2
10
10
-14
12
10
-15
3
13
C
C
14
N
10
-16
13
-7
10
15
-17
N
O
16
10
10
-9
7
10
Li
7
Be
-11
10
6
10
-13
-9
-11
10
10
10
10
-13
10
11
10
B
10
-17
-19
8
Be
10
10
10
10
10
10
12
Be
10
10
10
10
O
O
-22
-21
10
-23
10
-23
16
N
14
O
12
N
-24
B
-25
2
18
-21
17
9
Li
13
B
C
-19
Li
B
C
-17
10
B
8
-23
N
-20
-15
10
9
-21
O
15
-19
10
C
10
10
-18
14
Li
11
-15
10
3
10
-25
Time (s)
Figure 12. Standard big bang nucleosynthesis production of H, He, Li, Be, and
Simon Glover & Markus
Pössel
B isotopes
as a function of time, for the baryon density taken from WMAP7.
(A color version of this figure is available in the online journal.)
10
-25
4
10
Fig. from Coc 2012
3
Time
Figure 13. Standard big bang
nucleosynthesis
production of C, N, and
The
hot, early universe
as a function of time. (Note the different time and abundance range
to Figure 12.)
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Cosmic background radiation
-2
ΩBh
2
0.26
4
0.24
He
0.22 -3
10
10
3
He/H, D/H
Mass fraction
10
WMAP 2011
Comparing with observations
10
D
-4
-5
3
He
10
10
-6
7
-9
Li
-10
1
Simon Glover & Markus Pössel
WMAP
7
Li/H
10
10
η×10
10
Image from Coc 2012
The hot, early universe
Figure 3. Abundances of 4 He (mass fraction), D, 3 He and 7 Li (by number relative to H)
Thermo & statistic
Back in time
Primordial nucleosynthesis
Cosmic background radiation
The cosmic background radiation
How do we get from the plasma state (hydrogen and helium nuclei,
electrons, photons) to an atomic, transparent universe?
Reaction:
H + γ ↔= p + e−
As you will estimate in the exercises (at least the LHS), lots and
lots of collisions – system will be in equilibrium!
In thermodynamic equilibrium, thermodynamic potentials add up:
µH + µγ = µp + µe− .
. . . but µγ in chemical equilibrium is zero!
Simon Glover & Markus Pössel
The hot, early universe
Thermo & statistic
Back in time
Primordial nucleosynthesis
Cosmic background radiation
Photon chemical potential
But there are atomic reactions where either one or two photons
can be produced:
A→B+γ
versus
A → B∗ + γ, B∗ → A + γ.
This means
µA = µB + µγ ,
but also
µA = µB∗ + µγ = µB + 2µγ
⇒ this can only hold if µγ = 0!
Simon Glover & Markus Pössel
The hot, early universe
Thermo & statistic
Back in time
Primordial nucleosynthesis
Cosmic background radiation
Equilibrium state for ionization
Since µγ = 0, in equilibrium for the reaction H + γ ↔= p + e− ,
µH = µp + µe− .
Our particles are all non-relativistic, so
!
mc2 − µ
g
3/2
exp −
,
n = 3 (2π m kT)
kT
h
and with the ge = gp = 2 (spin ±1/2) and gH = 1 + 3 = 4 (spin 0
plus spin 1), so
np ne (2πme kT)3/2 mp
=
nH
mH
h3
!3/2
B
exp −
kT
where B = (mp + me − mH )c2 = 13.6 eV is the binding energy.
Simon Glover & Markus Pössel
The hot, early universe
Thermo & statistic
Back in time
Primordial nucleosynthesis
Cosmic background radiation
Equilibrium state for ionization
Charge neutrality means ne = np . Define the ionization fraction
xe ≡
ne
,
ne + nH
so that
xe2
ne
(2πme kT)3/2 mp
ne
=
=
=
1 − xe nH (ne + nH ) nH nb
mH
nB h3
!3/2
B
exp −
kT
with nb the baryon number density. But the number density is
related to the modern value, and the present (CMB) temperature
T0 , as
nb (t) = nb0
Simon Glover & Markus Pössel
a0
a(t)
!3
= nb0
T
T0
!3
.
The hot, early universe
Thermo & statistic
Back in time
Primordial nucleosynthesis
Cosmic background radiation
Equilibrium state for ionization
Inserting this into the equation, neglecting the mp /mH term,
xe2
1 − xe
=
1
2πme k
T
nb0 h3
21
= 8.78 · 10
Simon Glover & Markus Pössel
!3/2
1K
T
B
exp −
kT
!3/2
1.6 · 105 K
exp −
T
!
The hot, early universe
Thermo & statistic
Back in time
Primordial nucleosynthesis
Cosmic background radiation
Ionization fraction by temperature
31
1−
x to
or
44)
the
on-
ination
Ionisation fraction as a function of
Simon Glover & Markus Pössel
Graph: M. Bartelmann
The hot, early universe
Thermo & statistic
Back in time
Primordial nucleosynthesis
Cosmic background radiation
Recombination at what redshift?
From the previous graph, Trec ≈ 0.3 eV ≈ 3500 K .
By scaling behaviour of T :
T0
1
a(trec )
=
= 7.8 · 10−4 =
a0
Trec
1+z
so
z ≈ 1280.
Using the “matter only” approximation
a = a0
we get
3p
Ωm0 H0 t
2
!2/3
trec = 376, 000 a.
Simon Glover & Markus Pössel
The hot, early universe
Thermo & statistic
Back in time
Primordial nucleosynthesis
Cosmic background radiation
Precision CMB: COBE-FIRAS (Mather et al.)
400
Data from Fixsen et al. 1996
Intensity in MJy/sr
350
Best Planck fit: T = 2.728 K
300
Range shown: spectrum ±3 σ
250
200
150
100
50
0
0
5
10
Frequency in 1/cm
15
20
Data from Fixsen et al. 1996 via http://lambda.gsfc.nasa.gov
Simon Glover & Markus Pössel
The hot, early universe
Thermo & statistic
Back in time
Primordial nucleosynthesis
Cosmic background radiation
Precision CMB: COBE-FIRAS (Mather et al.)
400
Data from Fixsen et al. 1996
Intensity in MJy/sr
350
Best Planck fit: T = 2.728 K
300
Range shown: spectrum ±100 σ
250
200
150
100
50
0
0
5
10
Frequency in 1/cm
15
20
Data from Fixsen et al. 1996 via http://lambda.gsfc.nasa.gov
Simon Glover & Markus Pössel
The hot, early universe
Thermo & statistic
Back in time
Primordial nucleosynthesis
Cosmic background radiation
Precision CMB: COBE-FIRAS (Mather et al.)
400
Data from Fixsen et al. 1996
Intensity in MJy/sr
350
Best Planck fit: T = 2.728 K
300
Range shown: spectrum ±500 σ
250
200
150
100
50
0
0
5
10
Frequency in 1/cm
15
20
Data from Fixsen et al. 1996 via http://lambda.gsfc.nasa.gov
Simon Glover & Markus Pössel
The hot, early universe
Thermo & statistic
Back in time
Primordial nucleosynthesis
Cosmic background radiation
The big picture
Radiation era
Matter era
Big bang
10−33 s
Inflation
1e-6 s
quark confinement
1 s to 3 min
light elements
Simon Glover & Markus Pössel
Now
13.8e9. a
1e8 a
galaxies
380,000 a
CMB
The hot, early universe
Thermo & statistic
Back in time
Primordial nucleosynthesis
Cosmic background radiation
Literature
Little, Andrew J.: An Introduction to Modern Cosmology. Wiley
2003 [brief and basic]
Dodelson, Scott: Modern Cosmology. Academic Press 2003.
[more advanced]
Weinberg, Steven: Cosmology. Oxford University Press 2008
[advanced]
Weinberg, Steven: Gravitation and Cosmology. Wiley & Sons 1972
[advanced and detailed]
Simon Glover & Markus Pössel
The hot, early universe