Baumann's Cosmology (3)

3 The Hot Big Bang

The early universe was hot and dense. Above 10⁴ K[≈ 1 eV], stable atoms didn't exist because the average energy of particles was larger than the binding energies of atoms. The universe was therefore a plasma of free electrons and nuclei, with high-energy photons scattering between them. Above 10⁹ K, the nuclei had dissolved into their constituent protons and neutrons. The rate of interactions in the primordial plasma was very high and the universe was in a state of thermal equilibrium. The equilibrium state provides the initial conditions for the hot Big Bang. It is a very simple state in which the abundance of all the particle species are by determined by the temperature of the universe.
   As the temperature decreased, these particles dropped out of the thermal equilibrium and decoupled from the thermal bath. Non-equilibrium dynamics is required for massive particles to maintain significant abundances. Deviation from equilibrium are also crucial for understanding CMB and the formation of the light chemical elements.
   Some key events in the thermal history of the universe are listed in Table 3.1. Our story in this chapter will begin one second after the Big Bang. There were mostly free electrons (𝑒^-), positrons (𝑒⁺), photons (𝛾) and neutrinos (𝜈). A small amount were protons (𝑝⁺) and neutrons (𝑛).¹ The photons were trapped the large density of free electrons.² Neutrinos interacted with the rest of the primordial plasma through the weak nuclear interaction. The rate of these interaction soon dropped below cosmic neutrino background (C𝜈B), which still fills the universe today, but has such low energy that it is hard to detect directly. Electrons and positrons annihilated into photons shortly after neutrino decoupling. The energy of them got transferred to the photons, but not to the neutrinos. This resulted in a slight increase in the photon temperature relative to the neutrino temperature. Around three minutes later, Big Bang nucleosynthesis (BBN) took place, which fused protons and neutrons into light elements-mostly deuterium, helium and lithium. The next event, recombination occurred 370 000 years after the Big Bang. At this moment, the temperature became low enough for hydrogen atoms to form through the reaction 𝑒^- + 𝑝⁺ → 𝐻 + 𝛾. Before recombination, the strongest coupling between the photons and the rest of the plasma was through Thomson scattering, 𝑒^- + 𝛾. The sharp drop in the free electron density after recombination meant that this process became very inefficient and the photons decoupled. They have since streamed freely through the universe and are today observed as the cosmic microwave background (CMB).
   It is remarkable that this story of the hot Big Bang can now be told as a scientific fact.

* * *Let us start with so-called natural units, where the speed of light and the reduced Planck constant are set to unity
    (3.1)   𝑐 = ℏ ≡ 1.
In these units, length and time have the units, and are inverse of mass and energy. We will also introduce the reduced Planck mass
    (3.2)   𝑀_𝑃𝑙 = √(ℏ𝑐/8π𝐺) = 2.4 × 10¹⁸ 𝐺𝑒𝑉,
so that the Friedmann equation for a flat universe reads 𝐻² = 𝜌/3𝑀_𝑃𝑙². Finally, we will often set Boltzmann's constant equal to unity, 𝜅_𝛣 ≡ 1, so that temperature has units of energy. Useful conversions are
    (3.3)   𝑚_𝑝 = 1.60 × 10¹⁰ J = 1.16 × 10¹³ K = 1.78 × 10^-27 kg = (1.97 × 10^-16 J)^-1 = (6.65 × 10^-25 s)^-1,
where 𝑚_𝑝 is the proton mass. For more about the concept of natural units, refer to the beginning of Appendix C.

          3.1 Thermal Equilibrium    
The blackbody spectrum of the CMB is strong observational evidence that the early universe was in a state of thermal equilibrium.³ Moreover, on theoretical grounds, we expect the interactions of the Standard Model to have established thermal equilibrium at temperature above 100 GeV. In this section this initial state of the hot Big Bang and its subsequent evolution using the methods of thermodynamics and statistical mechanics, suitably generalized to apply to an expanding universe.

          3.1.1 Some Statistical Mechanics    
The early universe was a hot gas of weakly interacting particles. We will use a coarse-grained description of the gas using the principles of statistical mechanics. So we will characterize the properties of gas statistically. A lightning introduction to the relevant concepts of statistical mechanics and equilibrium thermodynamics will be given, if necessary, refer to some textbook on the subject.

          Distribution functions    
A key concept in statistical mechanics is the probability that a particle chose at random has a momentum 𝐩. In general, the probability function 𝑓(𝐩,𝑡) can be very complicate.⁴ However, if we wait long enough (relative to the typical interaction timescale), then the system will reach equilibrium and is characterized by a time-independent function. At this time, the gas has reached a state of maximum entropy in which the distribution function is given by either the Fermi-Dirac distribution (for fermions) or the Bose-Einstein distribution (for bosons)
    (3.4)  𝑓(𝑝,𝛵) = 1/[𝑒^{(𝐸(𝑝) - 𝜇)/𝛵} ± 1],
where the + sign for fermions and - sign for bosons. For a derivation of these functions refer to any textbook on statistical mechanics. The function in (3.4) has two parameters: temperature, 𝛵, and chemical potential, 𝜇. The latter describes the response of a system to a change in the particle number and can be positive or negative (see Section 3.1.5). The chemical potential may be temperature dependent, and since the temperature changes in an expanding universe, even the equilibrium distribution functions depend implicitly on time.

          Density of states    
To relate this microscopic description of the gas to its macroscopic properties, we must sum over all possible momentum states of the particles weighted by their probabilities. For example, the number density of particles in the gas is
    (3.5)  𝑛 = ∑_𝐩 𝑓(𝑝,𝛵).
To define this sum over state as an integral over the continuous variable 𝐩, requires the density of states. It is easiest to derive this density of states by considering the gas as a quantum system. In quantum mechanics, the momentum eigenstates of a particle in a box of side 𝐿 have a discrete spectrum. Solving the Schrödinger equation with periodic boundary conditions gives
    (3.6)  𝐩 = 𝘩/𝐿 (𝑟₁𝐱 + 𝑟₂𝐲 + 𝑟₃𝐳),
where 𝑟_𝑖 = 0, ±1, ±2, ... and 𝘩 = 4.14 × 10^-15 eVs is Planck's constant. In momentum space, the states of particle are therefore represented by a discrete set of pints, The density states in momentum space {𝐩} then 𝐿³/𝘩³ = 𝑉/𝘩³, and the states density in phase space {𝐱, 𝐩} is 1/𝘩³, If the particle has 𝑔 internal degrees of freedom (e.g. due to the intrinsic spin of elementary particles), then the density of states becomes
    (3.7)  𝑔/𝘩³ = 𝑔/(2π)³,
where we have used natural units with ℏ = 𝘩/((2π) ≡ 1.

          Densities and pressure    
Weighting each state by its probability distribution, and integrating over momentum, we obtain the number density particles
    (3.8)  𝑛(𝛵) = 𝑔/(2π)³  ∫ 𝑑³𝑝 𝑓(𝑝,𝛵).l]
The energy density and pressure of the gas are then given by
    (3.9)  𝜌(𝛵) = 𝑔/(2π)³  ∫ 𝑑³𝑝 𝑓(𝑝,𝛵) 𝐸(𝑝),
    (3.10)  𝜌(𝛵) = 𝑔/(2π)³  ∫ 𝑑³𝑝 𝑓(𝑝,𝛵) 𝑝²/3𝐸(𝑝),
where 𝐸(𝑝) = √(𝑚² + 𝑝²), if we can ignore the interaction energies between the particles.⁵. The origin of the factor in the pressure require more explanation. The volume swept out in unit time is ∣𝑣_𝑥∣𝑑𝛢 = ∣𝑝_𝑥∣²𝑑𝛢/𝐸(𝑝) which for an isotropic distribution in three direction is 𝑝²/3𝐸(𝑝). ^*[corrected in simpler way]
   Each particle species 𝑖 (with possibly distinct 𝑚_𝑖, 𝜇_𝑖, 𝛵_𝑖) has its own distribution function 𝑓_𝑖 and hence its own densities and pressure, 𝑛_𝑖, 𝜌_𝑖, 𝑃_𝑖. Species that are in thermal equilibrium share a common temperature, 𝛵_𝑖 = 𝛵. Their densities and pressures can then only differ because of difference in their masses and chemical potentials.
   At early times, the chemical potentials of all particles are much smaller than the temperature, ∣𝜇_𝑖∣ ≪ 𝛵, and can be neglected. For electrons and protons this is provable fact, for photons ii holds by definition and for neutrinos it sis likely true, but not proven.  

          3.1.2 The Primordial Plasma    
We will relate the densities and pressure of the different species in the different species in the primordial plasma to the overall temperature of the universe. Setting the chemical potential to zero, we get
    (3.11)  𝑛 = 𝑔/2π² ∫₀^∞ 𝑑𝑝 𝑝²/[exp{√(𝑝² + 𝑚²)/𝛵} ± 1],
    (3.12)  𝜌 = 𝑔/2π² ∫₀^∞ 𝑑𝑝 𝑝²√(𝑝² + 𝑚²)/[exp{√(𝑝² + 𝑚²)/𝛵} ± 1].
Defining the dimensionless variables 𝑥 ≡ 𝑚/𝛵 and 𝜉 ≡ 𝑝/𝛵, this can be written as
    (3.13)  𝑛 = 𝑔/2π² 𝛵³ 𝐼±(𝑥),    𝐼±(𝑥) ≡ ∫₀^∞ 𝑑𝜉 𝜉²/[exp√(𝜉² + 𝑥²) ± 1],
    (3.14)  𝜌 = 𝑔/2π² 𝛵³ 𝐽±(𝑥),    𝐽±(𝑥) ≡ ∫₀^∞ 𝑑𝜉 𝜉²√(𝜉² + 𝑥²)/[exp√(𝜉² + 𝑥²) ± 1].
In general, the functions 𝐼±(𝑥) and 𝐽±(𝑥) have to be elevated numerically (see Fig. 3.1), but in the relativistic and non-relativistic limits, we can determine them analytically.

          Relativistic limit    
At temperatures much larger than the particle mass, we can take limit 𝑥[≡ 𝑚/𝛵] → 0 and the integral in (3.13) reduces to
    (3.15)  𝐼±(0) = ∫₀^∞ 𝑑𝜉 𝜉²/𝑒^𝜉 ± 1.
    (3.16)  𝐼+(0) = 3/2 𝜁(3),    𝐼_(0) = 2 𝜁(3).
where we use Riemann zeta function 𝜁(3) = 1 + 1/2³ + 1/3³ + 1/4³ + ∙ ∙ ∙ ≈ 1.20205
    (3.17)  𝐼+(0)/𝐼_(0)= [3/2 𝜁(3)]/[2 𝜁(3)] = 3/4    ⇒    𝐼+(0) = 3/4 𝐼_(0).
Substituting (3.17) into (3.13), we get
    (3.18)  𝑛 = 𝜁(3)/π² 𝑔 𝛵³ { 1   bosons,   3/4   fermions. A similar computation for energy density gives
    (3.19)  𝐽±(0) = ∫₀^∞ 𝑑𝜉 𝜉³/𝑒^𝜉 ± 1.
    (3.20)  𝐽+(0) = 7/120 π⁴,    𝐽_(0) = 1/5 π⁴.
    (3.21)  𝐽+(0)/𝐽_(0) = [7/120 π⁴]/[1/5 π⁴] = 7/8    ⇒    𝐼+(0) = 7/8 𝐼_(0).
where we can use that 𝜁(4) = π⁴/90.
    (3.22)  𝜌 = π²/30 𝑔 𝛵⁴ { 1   bosons,   7/8   fermions.

Exercise 3.1   Show that 𝐽_(0) = 6𝜁(4) and 𝐽+(0) = 7/8 𝐽_(0).
[Solution]   𝐽_(0) = 7/120 π⁴/𝜁(4) = (1/15 π⁴)/(π⁴/90) = 6𝜁(4) and refer to (3.21).  ▮

   Using the observed temperature of the CMB, 𝛵₀ ≈ 2.73 𝐾, we find that the number density and energy density of photon today are       (3.24)  𝑛_𝛾,0 = 2𝜁(3)/π² 𝛵₀³  ≈ 411 photons cm^-3.  
    (3.25)  𝜌_𝛾,0 =  π²/15 𝛵₀⁴ ≈ 4.17× 10^-14 J cm^-3.  
In terms of the critical density, the photon energy density is then found in the 𝛺_𝛾𝘩² ≈ 0.26 eV m^-3.  
Finally, taking 𝑝 = 𝐸 in (3.10), we get
    (3.26)  𝑃 = 1/3 𝜌,
as expected for a gas of relativistic particles ("radiation').

          Non-relativistic limit    
At temperature below the particle mass, we take the limit 𝑥 ≫ 1 and the integral in (3.13) in the same for bosons and fermions
    (3.27)  𝐼±(𝑥) ≈ ∫₀^∞ 𝑑𝜉/[𝑒^{√(𝜉2 + 𝑥2)}].
Most of contribution to the integral comes from 𝜉 ≪ 𝑥 [𝑥 ≡ 𝑚/𝛵, mass] and we can Taylor expand the square root, [√(𝜉² + 𝑥² ≈ 𝑥 + 𝜉²/2𝑥 at 𝜉 = 0], when in the exponential to the lowest order in 𝜉,
    (3.28)  𝐼±(𝑥) ≈ ∫₀^∞ 𝑑𝜉 𝜉²/[𝑒^{{𝑥 + 𝜉2/(2𝑥)}}]
Performing the Gaussian integral, we get  
    (3.29)  𝐼±(𝑥) ≈ √(π/2) 𝑥^3/2 𝑒^-𝑥,
and, using (3.16), we find
    (3.30)  𝐼±(𝑥)/𝐼±(0) ≈ 0.5 𝑥^3/2 𝑒^-𝑥 ≪ 1.
As expected massive particles are exponentially rare at low temperatures.
   Substituting (3.29) into (3.13), we can write the density of the non-relativistic gas as a function of the temperature
    (3.31)  𝑛 = 𝑔 (𝑚𝛵/2π)^3/2 𝑒^-𝑚/𝛵
To determine the energy density in the non-relativistic limit, we can write 𝐸(𝑝) = √(𝑚² + 𝑝²) ≈ 𝑚 + 𝑝²/2𝑚, [𝐸(𝑝) = √(𝑚²𝑐⁴ + 𝑝²𝑐²), 𝑐 ≡ 1.] We then find
    (3.32)  𝜌 ≈ 𝑚𝑛 + 3/2 𝑛𝛵.
where the leading term is simply equal to the mass density.
   Finally, from (3.10), it is easy to show that the pressure of a non-relativistic gas of particles is  
    (3.33)  𝑃 = 𝑛𝛵,
which is nothing but the ideal gas law, 𝑃𝑉 = 𝑁𝑘_𝐵𝛵 (for 𝑘_𝐵 = 1). Since 𝛵 ≪ 𝑚, we have 𝑃 ≪ 𝜌, so that the gas acts like a pressureless fluid ("matter").

Exercise 3.2   Derive (3.32) and (3.33).
[Solution]    We take the limit 𝑥 ≫ 1 and the integral in (3.13) in the same for bosons and fermions   [𝑥 ≡ 𝑚/𝛵 and 𝜉 ≡ 𝑝/𝛵]
   (a)   𝑛 = 𝑔/2π² 𝛵³ 𝐼±(𝑥),    𝐼±(𝑥) ≡ ∫₀^∞ 𝑑𝜉 𝜉²/[exp√(𝜉² + 𝑥²)] ≈ √(π/2) 𝑥^3/2 𝑒^-1
   (b)   𝜌 = 𝑔/2π² 𝛵⁴ 𝐽±(𝑥),    𝐽±(𝑥) ≡ ∫₀^∞ 𝑑𝜉 𝜉²√(𝜉² + 𝑥²)/[exp√(𝜉² + 𝑥²)].  
Since in the non-relativistic limit, we can write 𝐸(𝑝) = √(𝑚² + 𝑝²) ≈ 𝑚 + 𝑝²/2𝑚
   (c)   𝜌 = 𝑔/2π² 𝛵⁴ 𝐽±(𝑥),    𝐽±(𝑥) ≡ ∫₀^∞ 𝑑𝜉 𝜉²(𝑥 + 𝜉²/2𝑥)/[exp√(𝜉² + 𝑥²) = 𝑥 𝐼±(𝑥) + 1/2𝑥 ∫₀^∞ 𝑑𝜉 𝜉⁴/[exp√(𝜉² + 𝑥²)]
                                 ≈ ∫₀^∞ 𝑑𝜉 𝜉⁴/[𝑒^{{𝑥 + 𝜉2/(2𝑥)}}] = 𝑥 𝐼±(𝑥) + 1/2𝑥 [3√(π/2) 𝑥^3/2 𝑒^-1] = 𝑥 𝐼±(𝑥) + 3/2 𝐼±(𝑥)    ⇒   𝜌 ≈ 𝑚 𝑛 + 3/2 𝑛 𝛵. ▮    

         By comparing the relativistic limit (𝛵 ≫ 𝑚) and the non-relativistic limit (𝛵 ≪ 𝑚), we see that the number density, energy density, and pressure of a particle species fall exponentially (are "Boltzmann suppressed") as the temperature drops below the mass of the particles. This can be interpreted as the annihilation of particles and antiparticles. At higher energies these annihilations also occurs, but they are balanced by by particle-antiparticle pair production. At low temperatures, the thermal energies of the particles aren't sufficient for pair production.

          Relativistic species    
The early universe was a collection of different species and the total density 𝜌 is the sum over all contributions
    (3.34)  𝜌 = ∑_𝑖 𝑔_𝑖/2π² 𝛵_𝑖⁴ 𝐽±(𝑥_𝑖),
where the different species can have different temperatures, 𝛵_𝑖. For this complication is only relevant for neutrinos after electron-positron annihilation (see Section 3.1.4 to write ). It is common to write density in terms of the "temperature of the universe" 𝛵 (typically chosen to be photon temperature 𝛵_𝛾),
    (3.35)  𝜌 = π²/30 𝑔_*(𝛵) 𝛵⁴,
where 𝑔_* is defined as the "effective number of relativistic degree of freedom" at 𝛵, so
    (3.36)  𝑔_*(𝛵) ≡ ∑_𝑖 𝑔_𝑖 (𝛵_𝑖/𝛵)⁴ 𝐽±(𝑥_𝑖)/ 𝐽_(0).
Since the energy density of relativistic species is much greater than that of non-relativistic species, it often suffices to include only the relativistic species in (3.36). Moreover, for  𝛵_𝑖 ≫ 𝑚_𝑖, we have 𝐽±(𝑥_𝑖 ≪ 1) ≈ const and (3.36) reduces to
    (3.37)  𝑔_*(𝛵) ≡ ∑_{𝑖=𝑏𝑜} 𝑔_𝑖 (𝛵_𝑖/𝛵)⁴ + 7/8 ∑_{𝑖=𝑓𝑒} 𝑔_𝑖 (𝛵_𝑖/𝛵)⁴.
When all particles are in equilibrium at a common temperature 𝛵, determining 𝑔_*(𝛵) is then simply a counting exercise.

          Learning to count    
At early times, 𝛵 ≳ 100 GeV, all particles of the Standard Model were relativistic (see Table 3.2. To determine the corresponding value of 𝑔_*, we need to sum over the internal degree of freedom of each particle species.
   Let us start with gauge bosons. Photons have 𝑔_𝛾 = 2 corresponding to two polarizations transverse to the direction of propagation. In total massive boson of spin 𝑠 has 𝑔 = 2𝑠 + 1 polarization states. For the massive spin-1 gauge  bosons, we have 𝑔_{𝑤±, 𝑧} = 3 and hence a total 3 × 3 = 9 internal degrees of freedom. Gluon are massless and therefore contribute 𝑔_𝑔 = 2 and there are 8 of them, corresponding to the 8 generators of the group 𝑆𝑈(3),, so we get 8 × 2 = 16.
   Next, among fermions the charged leptons (𝑒^±, 𝜇^±, 𝜏^± particles) are spin-1/2 particles and therefore two spin states each.  Including a factor of 2 for antiparticles, we have 3 × 2 × 2 = 12. Similarly, each quark contributes two spin states. There are 6 favors of quark (𝑡, 𝑏, 𝑐, 𝑠, 𝑑, 𝑢) and each comes in 3 different colors. Including a factor of 2 for antiparticles, we then have 6 × 2 × 3 × 2 = 72. Lastly, we must talk about neutrinos. Neutrinos are massive spin-1/2 and they only contribute 1 internal degree of freedom.

An aside on neutrinos   For a long time, the neutrinos were thought to be massless and the gauge symmetries of the Standard Model require them massless. Massless particles travel at speed of light and their spin can be aligned or either anti-aligned with the direction of travel. Theses two options correspond to the particle having positive or negative helicity. Alternatively, we say that the particle is left-handed or right-handed. It is striking fact that only left-handed neutrinos have been observed in nature and it was long believed that right-handed neutrinos simply do not exist. This would then explain why each neutrino only contributes 1 internal degree of freedom. However, we now know that neutrinos have a small mass. Theoretically, neutrinos could have a Majorana mass and a Dirac mass. But we don't know which option is realized in nature. In case of Majorana mass, the neutrino is its own antiparticle and we then 2 spin states for each neutrino and no contribution from antiparticle. In case of Dirac mass, we get 2 spin states for each neutrinos and 2 for each antineutrino for a total of 2 + 2 =4., which is inconsistent with measurement from BBN. This means that either neutrinos are the Dirac neutrinos s of freedom of Mariorana particles or half of the degree of the Dirac neutrinos somehow decoupled in the very early universe and their energy density diluted. Models of such Dirac neutrinos exist.

   Adding up the internal degrees of freedom, we get
       𝑔_𝑏 = 28   photons (2),  𝑊^± and 𝑍 (3 × 3), gluons (8 × 2), and Higgs (1)
       𝑔_𝑓 = 90   quarks (6 × 12), charged leptons (3 × 4), and neutrinos (3 × 2),   hence
    (3.38)  𝑔_* = 𝑔_𝑏 + 7/8 𝑔_𝑓 = 106.75.
As the temperature drops, various particle species become non-relativistic and annihilate. This leads to the evolution of 𝑔_*(𝛵) shown in Fig. 3.2.
   Being the heaviest particles of the Standard Model, the top quarks annihilate first. At 𝛵 ~ 1/6 𝑚_𝑖 ~ 30 GeV,⁵ the effective number reduced to 𝑔_* =106.75 - 7/8 × 12 = 96.25. The  Higgs boson and the gauge bosons 𝑊^±, 𝑍⁰ disappear roughly at the same time next. At 𝛵 ~ 10 GeV , we have 𝑔_* = 92.25 - (1 + 3 × 3) = 86.26. Next, the bottom quark annihilate (𝑔_* = 82.25 - 7/8 × 12 = 75.75), followed by charm quark and tau leptons (𝑔_* = 75.75 - 7/8 × (12 + 4) = 75.75), Before the strange quarks have time to annihilate, matters undergo the QCD phase transition. At 𝛵 ~ 150 MeV, he quarks combine into baryons (protons, neutrinos, . . .) and mesons (pions, . . .). Although there are many different species of baryons and mesons, all except the pions (𝜋^±, 𝜋⁰) are non-relativistic below the temperature of the QCD phase transition and are therefore  Boltzmann suppressed. Thus, pions, electrons, muons, neutrinos and photons are only left in large number. The three types of pions are spin-0 bosons of 𝑔 = 3. We therefore get 𝑔_* = 2 + 3 + 7/8 (4 + 4 + 6) = 17.25. Next, electrons and positrons will annihilate. However, to understand this process we first need to talk about entropy.

          3.1.3 Entropy and Expansion History    
To describe the evolution of the universe it is useful  to track a conserved quantity. In cosmology entropy is than energy,  more informative, because it is conserved in equilibrium.

          Conservation of entropy    
We will determine the entropy of the primordial plasma from the first law of thermodynamics. The first law states that the change on the entropy (𝑆) of a system is related to the changes in its internal energy (𝑈) and volume (𝑉) as
    (3.39)  𝛵𝑑𝑆 = 𝑑𝑈 + 𝑃𝑑𝑉,
where we have assumed that any chemical potentials are small. Defining the entropy density as 𝑠 ≡ 𝑆/𝑉, we can write
    (3.40)  𝛵𝑑(𝑠𝑉) = 𝑑(𝜌𝑉) + 𝑃𝑑𝑉,   𝛵𝑠𝑑𝑉 + 𝛵𝑉𝑑𝑠 = 𝜌𝑑𝑉 + 𝑉𝑑𝜌 + 𝑃𝑑𝑉.
Since 𝑠 and 𝜌 depend only on the temperature 𝛵, and not on the volume 𝑉, this imply
    (3.41)  (𝛵𝑠 - 𝜌 - 𝑃)𝑑𝑉 + 𝑉 (𝛵 𝑑𝑠/𝑑𝛵 - 𝑑𝜌/𝑑𝛵) 𝑑𝛵 = 0.
For arbitrary variations 𝑑𝑉 and 𝑑𝛵, the two brackets have to be vanish separately. They implies
    (3.42)  𝑠 = (𝜌 + 𝑃)/𝛵,
    (3.43)  𝑑𝑠/𝑑𝛵 = 1/𝛵 𝑑𝜌/𝑑𝛵.   [also 𝑑𝑠/𝑑𝑡  = 1/𝛵 𝑑𝜌/𝑑𝑡]
Using the continuity equation [(2.106)   ῤ + 3 ȧ/𝑎 (𝜌 + 𝑃/𝑐²) = 0.] 𝑑𝜌/𝑑𝑡 = -3𝐻(𝜌 + 𝑃) = -3𝐻𝛵𝑠, (3.43) can be written in the instructive form
    (3.44)  𝑑(𝑠𝑎³)/𝑑𝑡 = 0.  
[From (3.43) 𝑑𝑠/𝑑𝑡 = 1/𝛵 𝑑𝜌/𝑑𝑡, 𝑑𝜌/𝑑𝑡 = 𝛵 𝑑𝑠/𝑑𝑡 = -3/𝑎 𝑑𝑎/𝑑𝑡 𝛵𝑠. ⇒ 𝑑𝑠/𝑑𝑡 + 3𝑠/𝑎 𝑑𝑎/𝑑𝑡 = 0, multiply 𝑎³ at the both sides, 𝑎³ 𝑑𝑠/𝑑𝑡 + 𝑠 3𝑎² 𝑑𝑎/𝑑𝑡 = 0  ⇒  𝑑(𝑠𝑎³)/𝑑𝑡 = 0.  ▮]
This means that the total entropy is conserved in equilibrium and the entropy density evolves as 𝑠 ∝𝑎³. This conservation law will be very useful for describing the expansion history of the universe.

Exercise 3.3   Including a nonzero chemical potential, the first law of thermodynamics becomes
    (3.45)  𝛵𝑑𝑆 = 𝑑𝑈 + 𝑃𝑑𝑉 - 𝜇𝑑𝑁.
   Show that the entropy density is (3.46) and it evolves as (3.47)
    (3.46-7)  𝑠 = (𝜌 + 𝑃 - 𝜇𝑛)/𝛵.   𝑑(𝑠𝑎³)/𝑑𝑡 = - 𝜇/𝛵 𝑑(𝑛𝑎³)/𝑑𝑡.
[Solution]   Given
   (a)   𝛵 𝑑(𝑠𝑉) = 𝑑(𝜌𝑉) + 𝑃𝑑𝑉 - 𝜇𝑑(𝑛𝑉),    𝛵𝑠 𝑑𝑉 + 𝛵𝑉 𝑑𝑠 = 𝜌𝑑𝑉 + 𝑉𝑑𝜌 + 𝑃𝑑𝑉 - 𝜇𝑛 𝑑𝑉 - 𝜇𝑉 𝑑𝑛.
   (b)   (𝛵𝑠 - 𝜌 - 𝑃 + 𝜇𝑛)𝑑𝑉 + 𝑉 (𝛵 𝑑𝑠/𝑑𝑡 - 𝑑𝜌/𝑑𝑡 + 𝜇 𝑑𝑛/𝑑𝑡) 𝑑𝑡 = 0.   ⇒   𝑠 = (𝜌 + 𝑃 - 𝜇𝑛)/𝛵,   𝑑𝑠/𝑑𝑡 = 1/𝛵 𝑑𝜌/𝑑𝑡 - 𝜇/𝛵 𝑑𝑛/𝑑𝑡.
Using 𝑑𝜌/𝑑𝑡 = -3𝐻(𝜌 + 𝑃), we get
   (c)   𝑑𝑠/𝑑𝑡 = -3𝐻 (𝜌 + 𝑃)/𝛵 - 𝜇/𝛵 𝑑𝑛/𝑑𝑡 = -3𝐻(𝑠 + 𝜇𝑛/𝛵) - 𝜇/𝛵 𝑑𝑛/𝑑𝑡 = -3𝐻𝑠 - 𝜇/𝛵 (𝑑𝑛/𝑑𝑡 + 3𝐻𝑛).  ⇒  𝑑𝑠/𝑑𝑡 + 3𝐻𝑠 = -𝜇/𝛵(𝑑𝑛/𝑑𝑡 + 3𝐻𝑛)
   (d)   𝑑𝑠/𝑑𝑡 + 3𝑠/𝑎 𝑑𝑎/𝑑𝑡 = -𝜇/𝛵(𝑑𝑛/𝑑𝑡 + 3𝑛/𝑎 𝑑𝑎/𝑑𝑡), multiply 𝑎³ at the both sides, 𝑎³𝑑𝑠/𝑑𝑡 + 𝑠 3𝑎² 𝑑𝑎/𝑑𝑡 = -𝜇/𝛵 (𝑎³ 𝑑𝑛/𝑑𝑡 + 𝑛 3𝑎² 𝑑𝑎/𝑑𝑡), 
which can be written as  
   (e)   𝑑(𝑠𝑎³)/𝑑𝑡 = - 𝜇/𝛵 𝑑(𝑛𝑎³)/𝑑𝑡.  ▮

   Entropy now is conserved either if the chemical potential is small, 𝜇 ≪ 𝛵, or if no particles are created or destroyed.

          Relativistic species    
Integrating (3.43), we get
    (3.48)  𝑠(𝛵) = {₀^𝛵 𝑑Ṫ/Ṫ 𝑑𝜌/𝑑Ṫ = 𝜌(𝛵)/𝛵 + {₀^𝛵 𝜌(Ṫ)/Ṫ² 𝑑Ṫ,
where we have integrated by parts and used that 𝜌/𝛵 → 0 as 𝛵 → 0. Comparing this to (3.42) , we see the second term is 𝑃/𝛵. The equation of state of plasma can then be written as     (3.49)  𝜔(𝛵) ≡ 𝑃(𝛵)/𝜌(𝛵) = {₀¹ 𝑔_*(𝑦𝛵)/𝑔_*(𝛵) 𝑦² 𝑑𝑦.  
If all particles are relativistic, then 𝑔_*(𝛵) = const and we recover the equation of state of radiation, 𝜔 =1/3.
   For a collection of different species, the total entropy density is
    (3.50)  𝑠 = ∑_𝑖 (𝜌_𝑖 + 𝑃_𝑖)/𝛵_𝑖 ≡ 2π^2/45 𝑔_*𝑆(𝛵) 𝛵³,
where we have defined 𝑔_*𝑆(𝛵) as the "effective number of degrees of freedom in entropy." At 𝛵_𝑖 ≫ 𝑚_𝑖, we have
    (3.51)  𝑔_*𝑆(𝛵) ≈ ∑_{𝑖𝑖=𝑏𝑜} 𝑔_𝑖(𝛵_𝑖/𝛵)³ + 7/8 ∑_{𝑖𝑖=𝑏𝑜} 𝑔_𝑖(𝛵_𝑖/𝛵)³.
When all species are in equilibrium at the same temperature, 𝛵_𝑖 = 𝛵,  then 𝑔_*𝑠 is simply equal to 𝑔_*. In our universe, this is the case until 𝑡 ≈ 1s. Since 𝑠 is proportional to the number density of relativistic particles, it is sometimes useful to write 𝑠 ≈ 1.8 𝑔_*𝑆(𝛵) 𝑛_𝛾, where 𝑛_𝛾 is the number density of photons. In general, 𝑔_*𝑆(𝛵) depends on temperature, so that 𝑠 and 𝑛, cannot be used interchangeably. However, after electron-positron annihilation (see below) , we have 𝑔_*𝑆 = 3.94 and hence 𝑠 ≈ 7𝑛_𝛾.
   Since 𝑠 ∝𝑎³, the number of particles in a comoving volume is proportional to the number density 𝑛, divided by the entropy density
    (3.52)  𝑁_𝑖 ≡ 𝑛_𝑖/𝑠.
If particles are neither produced nor destroyed, then 𝑛_𝑖 ∝𝑎³ and 𝑁_𝑖 is a constant. An important example of a conserved species is the total baryon number after baryogenesis, 𝑛_𝛣/𝑠 ≡ (𝑛_𝑏 - 𝑛_ḃ). A related quantity is the baryon-to photon ratio
    (3.53)  𝜂 ≡ 𝑛_𝛣/𝑛_𝛾 = 1.8 𝑔_*𝑆 𝑛_𝛣/𝑠.
After electron-positron annihilation, 𝜂 ≈ 7 𝑛_𝛣/𝑠 becomes a conserved quantity and is therefore a useful measure of the baryon content of the universe.  
   Another important consequence of entropy conservation is that
    (3.54)  𝑔_*𝑆(𝛵) 𝛵³ 𝑎³ = const     or     𝛵 ∝𝑔_*𝑆^-1/3 𝑎^-1.
Away from particle mass thresholds, 𝑔_*𝑆 is approximately constant and temperature has the expected scaling, 𝛵 ∝𝑎^-1. The factor of 𝑔_*𝑆^-1/3 accounts for the fact that a particle species becomes non-relativistic and disappears, its entropy is transferred to the other relativistic species still present in the thermal plasma, causing 𝛵 to decrease slightly more slowly than 𝑎^-1.

          Expansion history    
As we seen in this Chapter the Friedmann equation relates the Hubble expansion rate to the energy density of the universe. At early times, the universe is dominated by the relativistic species and curvature is negligible. Hence, the Friedmann equation reads
    (3.55)  𝐻² = (1/𝑎 𝑑𝑎/𝑑𝑡)² = 𝜌/3𝑀_𝑃𝑙² ≈ π²/90 𝑔_* 𝛵/𝑀_𝑃𝑙².  
This is a single equation relating the expansion history of the universe to its temperature. We need one more equation to close system for 𝑎(𝑡) and (𝛵). Previously, we used the approximate equation of state for radiation, 𝜔 ≈ 1/3, which through determines 𝜌(𝑎). More precisely, we can substitute 𝛵 ∝𝑔_*𝑆^-1/3 𝑎^-1 into (3.55). away from mass threshold, this reproduces the result for a radiation-dominated universe, 𝑎 ∝ 𝑡^-1/2, but we see that there is a slight change in he scaling every time 𝑔_*𝑆(𝛵) changes.
   When 𝑎 ∝ 𝑡^-1/2, we have 𝐻 = 1/(2𝑡) and the Friedmann equation leads to
    (3.56)  𝛵/1MeV ≃ 1.5 𝑔_*^-1/4(1sec/𝑡)^-1/2.  
It is a useful rule of thumb that the temperature of the universe 1 second after the Big Bang was about 1 MeV (or 10 K), and evolved as 𝑡^-1/2 before that. We will show next one second after the Big Bang.

          3.1.4 Cosmic Neutrino Background
The most weekly interacting particles of the Standard Model are neutrinos. We therefore expect them to decouple first from the thermal plasma. How this produce the cosmic neutrino background (𝐶𝜈𝛣) will be shown in the following.

          Neutrino decoupling     Neutrinos were coupled to the thermal bath through weak interaction process like
    (3.57)   𝜈_𝑒 + 𝜈̄_𝑒  ⟷  𝑒⁺ + 𝑒^-,     𝑒^- + 𝜈̄_𝑒  ⟷  𝑒^- + 𝜈̄_𝑒.
The interaction rate(/particle) is 𝛤 ≡ 𝑛𝜎∣𝑣∣, where 𝑛 is the number density of the target particles, 𝜎 is the cross section, and 𝑣 is the relative velocity (which is approximately 𝑣 ≈ 𝑐 = 1). By dimensional analysis, we infer that the cross section for weak scale interaction is 𝜎 ≈ 𝐺_𝐹²𝛵², where 𝐺_𝐹 ≈ 1.2 × 10^-5 GeV^-2 is Fermi's constant. Taking the number density to be 𝑛 ≈ 𝛵³, the interaction rate becomes
    (3.58)   𝛤 = 𝑛𝜎∣𝑣∣ ≈ 𝐺_𝐹²𝛵⁵
As the temperature decrease, the interaction rate drops much more rapidly than the Hubble rate 𝐻 ≈ 𝛵²/𝑀_𝑃𝐼:
    (3.59)   𝛤/𝐻 ≈ (𝛵/1 MeV)³.
We conclude that neutrinos decouple around 1 MeV. (more accurately 0.8 MeV) After decoupling, the neutrinos more freely along geodesics and preserve the relativistic Fermi-Dirac distribution (even after they become non-relativistic at later time). In Section 2.2.1, we showed that the physical momentum of free-streaming particles scales as 𝑝 ∝𝑎^-1. It's convenient to define the time-independent combination 𝑞 ≡ 𝑎𝑝, so that the neutrino number density is
    (3.60)   𝑛_𝜈 ∝𝑎³∫ 𝑑³𝑞 1/[exp(𝑞/𝑎𝛵_𝜈) + 1].
After decoupling, particle number conservation requires 𝑛_𝜈 ∝𝑎³, which is only consistent with (3.60) if the neutrino temperature evolves as 𝛵_𝜈 ∝𝑎^-1. As long as the photon temperature 𝛵_𝛾 scales in the same way, we still have 𝛵_𝜈 =  𝛵_𝛾. However, particle annihilations will cause a deviation from 𝛵_𝛾 ∝𝑎^-1 in the photon temperature.

          Electron-positron annihilation
Shortly after the neutrinos decouple, the temperature drops below the electron mass, so that electrons and positrons can annihilate into photons.
          𝑒⁺ + 𝑒^-   →    𝛾 + 𝛾.
The energy density and entropy of electrons and positrons are transferred to the photons, but not to the decoupled neutrinos. The photons are thus "heated" (the photon temperature decreases more slowly) relative to the neutrinos (see Fig. 3.3). To quantify this effect, we consider the change in the effective number of degree of freedom in entropy. If we neglect neutrinos and other decoupled species,⁸ then we have
    (3.61)   𝑔_*𝑆 = { 2 + 7/8 × 4 = 11/2     𝛵 ≳ 𝑚_𝑒; 2     𝛵 < 𝑚_𝑒.
The annihilation of electrons and positrons occurs on a timescale of 𝛼²/𝑚_𝑒 ~ 10^-18 s (where 𝛼 is the fine-structure constant), which is much less than the age of the universe (~1 s) at the time. This means that the 𝑒^±-𝛾 plasma evolves quasi-adiabatically into the 𝛾-only plasma. Entropy is therefore conserved during the process. Taking 𝑔_*𝑆(𝑎𝛵_𝛾)³ to remain constant, we find that 𝑎𝛵_𝛾 increases after electron-positron annihilation by a factor (11/4)^1/3, while 𝑎𝛵_𝜈 remains the same. This means that after 𝑒⁺𝑒^- annihilation, the neutrino temperature is slightly lower than the photon temperature.
    (3.62)   𝛵_𝜈 = (4/11)^1/3 𝛵_𝛾.   [𝛵_𝜈≈ 0.714 𝛵_𝛾]
For 𝛵 ≪ 𝑚_𝑒, the effective number of relativistic species (in energy density and entropy) therefore is
    (3.63)   𝑔_* = 2 + 7/8 × 2 𝑁_eff (4/11)^4/3 = 3.36,
    (3.64)   𝑔_*𝑆 = 2 + 7/8 × 2 𝑁_eff (4/11) = 3.94,
where 𝑁_eff is the effective number of neutrino species in the universe. If neutrino decoupling was instantaneous then we would simply have 𝑁_eff = 3. However, neutrino decoupling was not quite complete when 𝑒⁺𝑒^- annihilation began, so some of the energy and entropy did leak to neutrinos. Taking this into account⁹ raises the effective number of neutrinos to 𝑁_eff = 3.046.¹⁰ Using this value in (3.63) explains the final value of 𝑔_*(𝛵) in Fig. 3.2.

          Electron-positron annihilation    
The relation (3.62) holds until the present. The cosmic neutrino background  therefore has a slightly lower temperature 𝛵_𝜈,0 = 1.95 K. cf. CMB 𝛵_𝜈0 = 2.73 K. The number density of neutrinos (per flavor) is
    (3.65)  𝑛_𝜈 ≈ 3/4 × 4/11 𝑛_𝛾.
Using (3.24) we see that this corresponds to 112 neutrinos cm per flavor. The present energy density of neutrinos depends on whether the neutrinos are relativistic or non-relativistic today. It used to be believed that neutrinos were massless, in that case
    (3.66)  𝜌_𝜈 = 7/8 𝑁_eff  (4/11)^4/3𝜌_𝛾     ⇒       𝛺_𝜈𝘩² ≈ 1.7 × 10^-5   (𝑚_𝜈 = 0).    [𝘩 = 𝐻₀/100 Mpc km s^-1]
However, neutrino oscillation experiments have shown that neutrinos do have a mass. The minimum sum of the neutrino masses is ∑ 𝑚_𝜈,𝑖 > 0.06 eV. massive neutrinos behave as radiation-like particles in the early universe (for 𝑚_𝜈 < 0.2 eV, neutrinos are relativistic at recombination) and as matter-like particles in the late universe (see Fig. 3.4). In Problem 3.4, it will be shown that energy of massive neutrino, 𝜌_𝜈 = ∑ 𝑚_𝜈,𝑖𝑛_𝜈,𝑖, corresponds to
    (3.67)  𝛺_𝜈𝘩² ≈ ∑ 𝑚_𝜈,𝑖/(94 eV).    [𝘩 = 𝐻₀/100 Mpc km s^-1]
By demanding that 𝛺_𝜈 <1, a cosmological upper bound can be placed on the sum of the neutrino masses ∑ 𝑚_𝜈,𝑖 < 15 eV (using 𝘩 ≈ 0.7). Massive neutrinos also affect the late-time expansion rate which is constraint by CBM and BAO(Baryon acoustic oscillations)  measurements. The current Planck constraint is ∑ 𝑚_𝜈,𝑖 < 0.13eV, which implies 𝛺_𝜈 < 0.003. Future observations promise to be sensitive enough to measure the neutrino masses.

        3.1.5 Cosmic Microwave Background  
An important event in the history of the early universe is the formation of the first atoms and the associated decoupling of photons. (see Fig. 3.5). At temperatures above about 1eV this universe still consisted of a plasma of free electrons and nuclei. Photons were tightly coupled to the electrons via Thomson scattering, which in tern strongly interacted with protons via Coulomb scattering. There was very little  neutral hydrogen. Below 0.3 eV the electons and nuclei combined to form neutral atoms and the density of free electrons decreased sharply. The photon mean free path grew rapidly and became longer than the Hubble length, 𝐻^-1. Around 0.25 eV, the photons decoupled from the matter and the universe became transparent. Today, these photons are observed as the cosmic microwave background. Key events in the formation of the CMB are summarized in Table 3.3. We will derive these facts.

          Chemical equilibrium       
During the recombination, the number of each particle species wasn't fixed, because hydrogen atoms were formed, while the number of free electrons and protons decreased. In thermodynamics we describe such a situation with chemical potential.
   Consider the generic reaction
       1 + 2  ⟷  3 + 4.
Each particle species has a chemical potential 𝜇. The second law of thermodynamics implies that particles flow to the side of the reaction where the total chemical potential is lower. Chemical equilibrium is reached when the sum of the chemical potentials each side  is equal, in which case the rates of the forward and revers reaction are equal. So
    (3.68)  𝜇₁ + 𝜇₂ = 𝜇₃ + 𝜇₄.

• There is no chemical potential for photons, because photon number is not conserved. (e.g. double Compton scattering 𝑒^- + 𝛾 ⟷ 𝑒^- + 𝛾 + 𝛾 happens in equilibrium at high temperatures.) Sometimes, this is expressed as
    (3.69)  𝜇_𝛾 = 0,
but, more accurately the concept of a chemical potential for photons doesn't exist.
• If the chemical potential of a particle 𝑋 is 𝜇𝑥, then the chemical potential of the corresponding antiparticle 𝑋̄ is
    (3.70)  𝜇𝑥̄ = -𝜇𝑥.
To see this, just consider particle-antiparticle annihilation, 𝑋 + 𝑋̄ ⟷ 𝛾 + 𝛾, and use that 𝜇_𝛾 = 0,

The equilibrium assumption will be sufficient to describe the onset of recombination, but will not capture the correct dynamics shortly thereafter (such as the freeze-out of electrons). We will revisit these non-equilibrium aspects of recombination in Section 3.2.5.

          Hydrogen recombination 
Recombination proceeds in two stages. The formation of helium atoms is followed by that of hydrogen atoms. We will assume that the universe was filled only with free electrons, protons and photons. Over 90% (by number) of the nuclei are protons, so this is reasonable approximation to reality. Moreover , helium recombination is completed before hydrogen recombination, so hat the two events can be treated separately.
   The formation of hydrogen atom occurs via the reaction
        𝑒^- + 𝑝⁺ = 𝐻 + 𝛾,
Initially, this reaction keeps the particles in equilibrium, and since 𝛵 < 𝑚_𝑖, 𝑖 = {𝑒, 𝑝, 𝐻}, we have the following equilibrium abundances
    (3.71)  𝑛_𝑖^eq = 𝑔_𝑖 (𝑚_𝑖𝛵/2π)^3/2 exp[(𝜇_𝑖 - 𝑚_𝑖)/𝛵],
where 𝜇_𝑝 + 𝜇_𝑒 = 𝜇_𝐻 (recall that 𝜇_𝛾 = 0). To remove ththe following ratioe dependence on the chemical potentials, we consider
    (3.72)  (𝑛_𝐻/𝑛_𝑒𝑛_𝑝)_eq = 𝑔_𝐻/𝑔_𝑒𝑔_𝑝 (𝑚_𝐻/𝑚_𝑒𝑚_𝑝 2π/𝛵)^3/2 𝑒^{(𝑚_𝑝 + 𝑚_𝑒 - 𝑚_𝐻)/𝛵}.
In the prefactor, we can use 𝑚_𝐻 ≈ 𝑚_𝑝, but in the exponential the small difference is crucial: it is the ionization energy of hydrogen
    (3.73)  𝐸_𝐼 ≡ 𝑚_𝑝 + 𝑚_𝑒 - 𝑚_𝐻 = 13.6 eV.
The numbers of internal degrees of freedom are 𝑔_𝑝 = 𝑔_𝑒 =2 and 𝑔_𝐻 =4.¹¹ In our knowledge the universe isn't electrically charged, so that we have 𝑛_𝑒 = 𝑛_𝑝, Equation (3.72) then becomes
    (3.74)  (𝑛_𝐻/𝑛_𝑒²)_eq = (2π/𝑚_𝑒𝛵)^3/2 𝑒^{𝐸_𝐼/𝛵}.
It is convenient to describe the process of recombination in terms of the free electron fraction
    (3.75)  𝑋_𝑒 ≡ 𝑛_𝑒/(𝑛_𝑝 + 𝑛_𝐻) = 𝑛_𝑒/(𝑛_𝑒 + 𝑛_𝐻).
A fully ionized universe corresponds 𝑋_𝑒 = 1, while a universe of only neutral atoms has 𝑋_𝑒 = 0. Our goal is to understand how 𝑋_𝑒 evolves. If we neglect the small number of helium atoms, then the denominator in (3.75) can be approximated by the baryon density
    (3.76)  𝑛_𝑏 = 𝜂 𝑛_𝛾 = 𝜂 × 2𝜁(3)/π² 𝛵³,   [RE (3.24) 𝑛_𝛾,0 = 2𝜁(3)/π² 𝛵₀³]
where 𝜂 is the baryon-to-photon ratio. We can then write
    (3.77)  (1 - 𝑋_𝑒)/𝑋_𝑒² = 𝑛_𝐻/𝑛_𝑒² 𝑛_𝑏,
and substituting (3.74), we arrive at the so-called Saha equation
    (3.78)  [(1 - 𝑋_𝑒)/𝑋_𝑒²]_eq = 2𝜁(3)/π² 𝜂 (2π𝛵/𝑚_𝑒)3/2 𝑒^{𝐸_𝐼/𝛵} The solution to this equation is
    (3.79)  𝑋_𝑒 = [-1 ± √(1 + 4𝑓)]/2𝑓,^*   with   𝑓(𝛵,𝜂) = 2𝜁(3)/π² 𝜂 (2π𝛵/𝑚_𝑒)3/2 𝑒^{𝐸_𝐼/𝛵},   ^*[correcton: from [-1 + √(1 + 4𝑓)]/2𝑓 to [-1 ± √(1 + 4𝑓)]/2𝑓]]
which is shown in Fig. 3.6 as a function of temperature (or equivalently redshift).
   Let us define the recombination temperature 𝛵_rec the temperature at which as 𝑋_𝑒 = 0/5 in (3.78).¹² For 𝜂 ≈ 6 × 10^-10, we get
    (3.80)  𝛵_rec ≈ 0.32 eV ≈ 3760 K. The reason why the recombination temperature is significantly below the binding energy of hydrogen, 𝛵_rec ≪ 𝐸_𝐼 = 13.6 eV, is that there are many photons for each hydrogen atom. Even when 𝛵 < 𝐸_𝐼, the high-energy tail of the 𝐸_𝐼, the high-energy tail of the photon distribution contains photons with energy 𝐸_𝛾 > 𝐸_𝐼, which can ionize the hydrogen atoms. Concretely, although the mean photon energy is <𝐸_𝛾> ≈ 2.7 𝛵, one in 500 photons has 𝐸_𝛾 > 10 𝛵, one in 3 × 10⁶ has 𝐸_𝛾 > 20 𝛵, and one in 3 × 10¹⁰ has 𝐸_𝛾 > 30 𝛵. Since there are over 10⁹ photons per baryon, rare high-energy photons are still present in sufficient numbers, unless the temperature drops far below the binding energy.
   𝛵_rec = 𝛵₀(1 + 𝑧_rec), with 𝛵₀ =2.73 K, gives 𝑧_rec ≈ 1380.¹³. However, we will see in Section 3.2.5 that the details of recombination are more complex and is delayed relative to the Saha prediction, with  𝑋_𝑒 = 0.5 only being reached at
    (3.81)  𝑧_rec ≈ 1270,       𝑡_rec ≈ 290 000 yrs.
Since matter-radiation equality is at 𝑧_rec ≈ 3400, we conclude that recombination occurred in matter-dominated era. Of course, it was not an instantaneous process as seen from Fig. 3.6. It took about 𝛥𝑡 ≈ 70 000 yrs (or 𝛥𝑧 ≈ 180^*) for the ionization fraction to drop from 𝑋_𝑒 = 0.9 to 𝑋_𝑒 = 0.1.  
^*[correction: from ∆𝑧 ≈ 80 to ∆𝑧 ≈ 180]

          Photon decoupling 
At early times, photons are strongly coupled to the primordial plasma through their interaction with the free electrons
        𝑒^- + 𝛾  ⟷  𝑒^- + 𝛾,
with the interaction rate given by 𝛤_𝛾 ≈ 𝑛_𝑒 𝜎_𝛵 ≈ 2 × 10^-3 MeV^-2 is the Thomson cross section. At 𝑎 = 10^-5 (prior to matter-radiation equality), the rate of photon scattering is 𝛤_𝛾 ≈ 5.0 × 10^-6 s^-1, or three times per week. This interaction rate was much larger than the expansion rate at that time (𝐻 ≈ 2 × 10^-10 s^-1), so thah electrons and photons were in equilibrium.
   Since 𝛤_𝛾 ∝ 𝑛_𝑒, the interaction rate decrease as the density of free electrons drops during recombination. At some point, this rate becomes smaller than the expansion rate and photons decouple. We define the approximate moment of photon decoupling as 𝛤_𝛾(𝛵_dec) ≈ 𝐻(𝛵_dec).
    (3.82)  𝛤_𝛾(𝛵_dec) = 𝑛_𝑏 𝑋_𝑒(𝛵_dec) 𝜎_𝛵 = 2𝜁(3)/π² 𝜂 𝑋_𝑒(𝛵_dec) 𝛵_dec³,
    (3.83)  𝐻(𝛵_dec) = 𝐻₀ √𝛺_𝑚 (𝛵_dec/𝛵₀)^3/2,     we get
    (3.84)  𝑋_𝑒(𝛵_dec) 𝛵_dec^3/2 ≈ π²/2𝜁(3) 𝐻₀√𝛺_𝑚/𝑛𝜎_𝛵𝛵₀^3/2.
Using the Saha equation for 𝑋_𝑒(𝛵_dec) we find 𝛵_dec ≈ 0.27 eV. In the more precise treatment in Section 3.2.5, we find that decoupling occurs at a slightly lower temperature,
    (3.85-86)  𝛵_dec ≈ 0.25 eV ≈ 2970 K,     𝑧_rec ≈ 1090,     𝑡_rec ≈ 370 000 yrs.
After decoupling, the photons stream freely through the universe.
   Note that the ionization fraction decreases significantly between recombination and decoupling. 𝑋_𝑒(𝛵_rec) ≈ 5 → 𝑋_𝑒(𝛵_dec) ≈ 0.001. This show that a large degree of neutrality is necessary before the universe becomes transparent to photons.

Exercise 3.4   Imagine that he recombination did not occurs, so that 𝑋_𝑒 = 1. At what redshift would the CMB photons now decouple?
[Solution]  From definition  of photon decoupling, 𝛤_𝛾(𝛵_dec) ≈ 𝐻(𝛵_dec), so we have (3.84)
   (a)   𝑋_𝑒(𝛵_dec) 𝛵_dec^3/2 ≈ π²/2𝜁(3) 𝐻₀√𝛺_𝑚/𝑛𝜎_𝛵𝛵₀^3/2.   and    𝑋_𝑒 ≡ 1      
   (b)   𝛵_dec = [π²/2𝜁(3) 𝐻₀√𝛺_𝑚/𝑛𝜎_𝛵𝛵₀^3/2]^2/3
To evaluate this we first write all quantities in natural units, using
   (c)   𝑐 = 3 × 10⁸ m s^-1 ≡ 1,     ℏ𝑐 = 2 × 10^-7 eV m ≡ 1
For we use 𝐻₀ ≈ 70 km s^-1 Mpc^-1 = 70/3 × 10⁵ (3.1 × 10²² m)^-1 = 70/3 × 10⁵  2 × 10^-7/3.1 × 10²² eV = 1.5 × 10^-33 eV
The photon temperature today is 𝛵₀ ≈ 2.725 K or 2.348 × 10^-4 eV.
The Thomson cross section [RE Wikipedia Thomson scattering: For an electron ...] 𝜎_𝛵 = 6.65 × 10^-29 m² = 6.65 × 10^-29/(2 × 10^-7)² eV^-2 = 1.66 × 10^-15 eV^-2
Substituting these results, π²/2𝜁(3) ≈ 4.105 and 𝛺_𝑚 = 0.315 [RE Wikipedia Lambda-CDM model] into the equation gives
   (d)   𝛵_dec ≈ [4.105 × (1.5 × 10^-33 × √0.315)/{6 × 10^-10 × (1.66 × 10^-15) × (2.3 × 10^-4)^3/2}]^2/3 eV ≈ 9.76 × 10^-3 eV.
The redshift of decoupling is, using 𝛵_dec = 𝛵₀(1 + 𝑧_dec), 
   (e)   𝑧_dec = 𝛵_dec/𝛵₀ - 1 = 9.76/0.23 - 1 ≈ 41.  ▮

          Last-scattering 
The scattering of photons off electrons essentially stops at photon decoupling. To define the precise moment of last-scattering, we have to consider the probability of photon scattering. Let 𝑑𝑡 b a small time interval around the time 𝑡. The probability that a photon will scatter during this time is 𝛤_𝛾(𝑡) 𝑑𝑡, and the integrated probability between the time 𝑡 and 𝑡₀ > 𝑡 is
    (3.91)  𝛤_𝛾(𝑡) = ∫𝑡^𝑡₀ 𝛤_𝛾(𝑡) 𝑑𝑡,
This probability is also called the optical depth. Taking 𝑡₀ to be the present time, the moment of lat-scattering is defined by 𝜏(𝑡_*) ≡ 1. To a good approximation, last scattering coincides with photon decoupling, 𝑡_* ≈ 𝑡_dec. However, the optical dept is sensitive to the evolution of free electron density at the end of recombination, which isn't captured well by the equilibrium treatment of this section. A precise evaluation of 𝑡_* must therefore await our non-equilibrium analysis of recombination in Section 3.2.5.
   When we observe the CMB, we are detecting photons from this surface of last-scattering (see Fig. 3.7). Given the age of the universe, and taking into account the expansion of the universe, the distance us and the spherical last-scattering surface today is 42 billion light-years. Of course, last-scattering is a probabilistic concept-not all photons experienced their last scattering event at the same time-so there is some thickness to the last-scattering surface.

          Blackbody spectrum
The CBM is often presented as key evidence that the early universe began in a state of thermal equilibrium. Before decoupling, the number density of photons with frequency in the range 𝑓 and 𝑓 + 𝑑𝑓 is
    (3.88)  𝑛(𝑓,𝛵) 𝑑𝑓 = 2/𝑐³ 4π𝑓²/(𝑒^{𝘩𝑓/𝑘_𝛣𝛵} - 1) 𝑑𝑓,
where factors of 𝑐 and 𝑘_𝛣 were restored for clarity. This frequency distribution is called the blackbody spectrum and is characteristic of objects in thermal equilibrium.After decoupling, the photons propagate freely, with their frequencies redshifting as 𝑓(𝑡) ∝𝑎(𝑡)^-1 and number density decreasing as 𝑎(𝑡)^-3. The spectrum therefore maintains is blackbody form as long as we take the temperature to such scale as 𝛵 ∝ 𝑎(𝑡)^-1. Te early e relic radiation encodes the early equilibrium phase of the hot Big Bang.
   CMB experiments observe the so-called spectral radiation intensity, 𝐼_𝑓, which is the flux of energy per unit area per unit frequency. Let us see how this is related to the spectrum in (3.68). We first pick a specific direction and consider photons traveling in a solid angle 𝛿𝛺 around the  this direction. In a given time interval 𝛿𝑡, these photons move through a volume 𝛿𝑉 = (𝑐𝛿𝑡)³ 𝛿𝛺 and cross a cap of area 𝛿𝛢 = (𝑐𝛿𝑡)² 𝛿𝛺. The number of photons in this volume is
    (3.89)  𝛿𝑁 = 𝑛(𝑓)𝑑𝑓/4π 𝛿𝑉 = 2/𝑐³ 𝑓²𝑑𝑓/(𝑒^{𝘩𝑓/𝑘_𝛣𝛵} - 1) (𝑐𝛿𝑡)³ 𝛿𝛺,
and the number of photons crossing the surface per unit area and per unit time is
    (3.90)  𝛿𝑁/𝛿𝛢𝛿𝑡 = 2/𝑐² 𝑓²𝑑𝑓/(𝑒^{𝘩𝑓/𝑘_𝛣𝛵} - 1).
Since each photon has energy 𝘩𝑓, the flux of energy across the surface (per unit frequency) is
    (3.91)  𝐼_𝑓 = 2𝘩/𝑐² 𝑓³/(𝑒^{𝘩𝑓/𝑘_𝛣𝛵} - 1).
Figure 3.8 shows a measurement of the CMB frequency spectrum by the FIRAS instrument on the COBE satellite. What you are seeing is the most perfect blackbody ever observed in nature, proving that the early universe indeed started in a state of thermal equilibrium.

⁸ Obviously entropy is separately conserved for the thermal bath and the decoupled species.
⁹ For the precise value, we should consider that the neutrino spectrum after decoupling deviates slightly from the Fermi-Dirac distribution. The spectral distortion arise because the energy dependence of the weak interaction causes neutrinos in the high-energy tail to interact more strongly.
¹⁰ But the Planck constraint on 𝑁_eff = 2.99 ± 0.17 (≠ 3.046), so this leaves room for discovering new physics beyond the Standard Model.
¹¹ The spins of the electron and proton in a hydrogen atom can be aligned or anti-aligned, giving one singlet state and one triplet state, so 𝑔_𝐻 = 1 + 3 = 4.
¹² the choice of 𝑋_𝑒 = 0/5 looks like arbitrary. However, since 𝑋_𝑒 is exponentially sensitive to 𝛵, we don't change this criterion.
¹³ It's useful to compare this to the case of helium recombination.  This proceed in two stage: First, He²⁺ captures one 𝑒^- to create He⁺. This process occurs in equilibrium around 𝑧 ≈ 6000. Then He⁺ captures a second 𝑒^- to become a neural helium He. This is slower than predicted by Saha equilibrium and occurs around 𝑧 ≈ 2000. This means that helium recombination don't have a big effect on he hydrogen recombination and the predictions of CMB, since the universe was still optically thick after the completion of the helium recombination.

          3.2 Beyond Equilibrium    
To understand the world around us, it is crucial to understand deviations from equilibrium. For example, a massive particle species in thermal equilibrium become exponentially rare when the temperature drops below the mass of particles, the numbers of particles in a comoving volume defined by the number density  𝑛_𝑖 divided by the entropy density (3.52), 𝑁_𝑖 ≡ 𝑛_𝑖/𝑠 ~ (𝑚/𝛵)^3/2 exp(-𝑚/𝛵). In order for these particles to survive until the present time, they must drop out of thermal equilibrium before 𝑚/𝛵 becomes much larger than unity. This decoupling, and the associated freeze-out of massive particles, occurs when the interaction rate of the particles becomes smaller the expansion rate (see Fig. 3.9).
   The main tool to describe the evolution beyond equilibrium is the Boltzmann equation. In this section, the Boltzmann equation will be first introduced and then it  will be applied to three important examples: (i) the production of dark matter; (ii) the formation of the light elements during Big Bang nucleosynthesis; and (iii) the freeze-out  of electrons during recombination. Some comments about the role of non-equilibrium process in baryogenesis will also be made.

          3.2.1 The Boltzmann Equation
In the absence of interaction, the number density of a particle species 𝑖 evolves as
    (3.92)   𝑑𝑛_𝑖/𝑑𝑡 + 3 𝑎̇/𝑎 𝑛_𝑖 = 0,
which simply means that the number of particles in a fixed physical volume (𝑉 ∝𝑎³) is conserved, so that the density dilutes as 𝑛_𝑖 ∝ 𝑎³. To include the effect of interactions, we add a collision term to the right-hand side of (3.92)
    (3.93)   1/𝑎³ 𝑑(𝑛_𝑖𝑎³)/𝑑𝑡 = 𝐶_𝑖[{𝑛_𝑗}].
This is the Boltzmann equation.¹⁴ The form of the collision term depends on specific interaction. We will consider single-particle decays and two-particle scattering (or annihilations). Let us consider the following process
       1 + 2  ⟷  3 + 4,
where particle 1 interacts with particle 2 to produce particle 3 and 4, or the inverse can happen. Suppose we want to track the number density 𝑛₁ of species 1. The Boltzmann equation simply formalizes this statement
    (3.94)   1/𝑎³ 𝑑(𝑛₁,𝑎³)/𝑑𝑡 = -𝛼 𝑛₁𝑛₂ + 𝛽 𝑛₃𝑛₄.
We understand the right-hand side as follows: The first term, -𝛼𝑛₁𝑛₂ describes the distribution of particle 1, while the second term, +𝛽𝑛₃𝑛₄, account their production. The parameter 𝛼 = ⟨𝜎𝑣⟩, which denote an average over 𝑣, is the thermally-averaged cross section.¹⁵ The second parameter 𝛽 can be related to 𝛼 by noting that the collision term has to vanish in (chemical) equilibrium
    (3.95)   𝛽 = (𝑛₁𝑛₂/𝑛₃𝑛₄)_eq 𝛼,
where 𝑛_𝑖^eq are the equilibrium number densities that we calculated above. The relation in (3.95) is sometimes called detailed balance . We therefore find
    (3.96)   1/𝑎³ 𝑑(𝑛₁𝑎³)/𝑑𝑡 = -⟨𝜎𝑣⟩ [𝑛₁𝑛₂ - (𝑛₁𝑛₂/𝑛₃𝑛₄)_eq 𝑛₃𝑛₄].
It is instructive to write this in terms of the number of particles in a comoving volume, 𝑁_𝑖 ≡ 𝑛_𝑖/𝑠 ∝ 𝑛_𝑖𝑎³. This gives
    (3.97)   𝑑 ln 𝑁₁/𝑑 ln 𝑎 = -𝛤₁/𝐻 [1 - (𝑁₁𝑁₂/𝑁₃𝑁₄)_eq 𝑁₃𝑁₄/𝑁₁𝑁₂],
where 𝛤₁ ≡ 𝑛₂⟨𝜎𝑣⟩ is the interaction rate of species 1. The right-hand side of (3.97) a factor describing the interaction efficiency,
𝛤₁/𝐻, and a factor characterizing the deviation from equilibrium, [1 - ∙  ∙  ∙ ].
Derivation   𝐻 = 𝑎̇/𝑎 = 𝑑(ln 𝑎)/𝑑𝑡, 𝑛₁𝑎³ ∝𝑁_𝑖, so 1/𝑛₁𝑎³ 𝑑(𝑛₁𝑎³)/𝑑𝑡 = 𝑑(ln 𝑁_𝑖)/𝑑𝑡. By dividing the both side of (3.96) with 𝑛₁𝑑(ln 𝑎)/𝑑𝑡[= 𝐻], Cancel 𝑑𝑡's at the left-hand side, and divide by 𝑛₁𝑛₂ in the bracket of right-hand side, then [𝑛₁𝑛₂/𝑛₁𝑛₂ - (𝑛₁𝑛₂/𝑛₃𝑛₄)_eq 𝑛₃𝑛₄/𝑛₁𝑛₂] ⇒ 𝑛₁𝑛₂[1 - (𝑁₁𝑁₂/𝑁₃𝑁₄)_eq 𝑁₃𝑁₄/𝑁₁𝑁₂], 𝛤₁ ≡ 𝑛₂⟨𝜎𝑣⟩, we therefore get (3.97).  ▮
   When the interaction rate is large, 𝛤₁ ≫ 𝐻, the natural state of the system is to be equilibrium. Imagine that 𝑁₁ ≫ 𝑁₁^eq, while 𝑁_𝑖  ~ 𝑁_𝑖 ^eq ,  𝑖 = 2, 3, 4.  The right-hand side then is negative, particles of type 1 are destroyed and 𝑁₁ is reduced towards the equilibrium value 𝑁₁^eq. similarly, if 𝑁₁ ≪ 𝑁₁^eq, the right-hand side of (3.97) is positive and 𝑁₁ is driven towards 𝑁₁^eq. The same condition applies if several species deviate from their equilibrium values. As long as the interaction rates are large, the system therefore quickly relaxes to a steady state where the source in (3.97) vanishes and the particles assume their equilibrium abundances. At some point, however, the reaction rate drops below the Hubble scale, 𝛤₁ < 𝐻.  The right-hand side of (3.97) then gets suppressed and the comoving density of particles approaches a constant relic density, 𝑁₁ → const.

          3.2.2 Dark Matter Freeze-out
We will begin with a simple application of the Boltzmann equation (3.96): the freeze-out of massive particles. This may be how dark matter was produced in the early universe.

          Decoupling and freeze-out 
Consider a heavy fermion 𝑋 that can annihilate with its antiparticle 𝑋̄ to produce two light (essentially massless) particles,
       𝑋 + 𝑋̄  ⟷  𝓁 + 𝓁̄.
The particle 𝑋 might be the dark matter, while 𝓁 could be particles of the Standard Model.
   We will make a few simplifying assumptions: First, we will take the light particle 𝓁 to be tightly coupled to the rest of the plasma, as would be the case, for instance, if the particles were electrically charged. The light particles will then maintain their equilibrium densities throughout, 𝑛_𝓁 =  𝑛_𝓁^eq. Second, we will neglect any initial asymmetry between 𝑋 and its antiparticle 𝑋̄, so that 𝑛_𝑋 = 𝑛_𝑋̄, Finally we will assume that there are no other particle annihilations during the freeze-out of the species 𝑋. This will allow us to take 𝛵 ∝ 𝑎^-1 for the times relevant to the freeze-out.
   With these assumptions, the Boltzmann equation (3.96) for the evolution of the particle 𝑋 becomes
    (3.98)   1/𝑎³ 𝑑(𝑛_𝑋,𝑎³)/𝑑𝑡 = -⟨𝜎𝑣⟩ [𝑛_𝑋²- (𝑛_𝑋^eq)²]
Now we introduce the quantity 𝑌_𝑋 ≡ 𝑛_𝑋/𝛵³, which is proportional t the number of particles in a comoving volume, 𝑁_𝑋 ≡ 𝑛_𝑋/𝑠 ∝ 𝑌_𝑋. In fact, as long as 𝛵 ∝ 𝑎^-1, we can treat 𝑌_𝑋 and 𝑁_𝑋 interchangeably. In terms of 𝑌_𝑋, the left-hand side of (3.98) becomes 𝛵³𝑑𝑌_𝑋/𝑑𝑡. Since most of the interesting dynamics will take place when 𝛵 ~ 𝑀_𝑋, a new measure of time will be defined,
    (3.99)   𝑥 ≡  𝑀_𝑋/𝛵,   where   𝑑𝑥/𝑑𝑡 = 𝐻𝑥.
For weakly interacting particles, the decoupling occurs at very early times, during the radiation-dominated ear, where the Hubble parameter can be written as 𝐻 = 𝐻(𝑀_𝑋)/𝑥².^{[verification. needed]} [RE Dodelson p.74] Equation (3.98) then becomes the so-called Riccati equation,
    (3.100)   𝑑𝑌_𝑋/𝑑𝑡 = 𝜆/𝑥² [𝑌_𝑋² - (𝑌_𝑋^eq)²],
where we have defined the dimensionless parameter
    (3.101)   𝜆 ≡ 𝛤(𝑀_𝑋)/𝐻(𝑀_𝑋) = 𝑀_𝑋³⟨𝜎𝑣⟩/𝐻(𝑀_𝑋).
For simplicity, we will take 𝜆 to be a constant. Although there are no analytical solutions to (3.100), we can solve the Riccati equation numerically using the equilibrium abundance as an initial condition.
   Fig. 3.10 shows numerical solutions of (3.100) for different values of 𝜆. As expected, at high temperatures 𝑥 ≪ 1, we have 𝑌_𝑋 ≈ 𝑌_𝑋^eq = 3𝜁(3)/4π² ≈ 0.1, while at low temperature, 𝑥 ≫ 1, the equilibrium abundance exponentially suppressed, 𝑌_𝑋^eq ~ (𝑥/2π)^3/2𝑒^-𝑥. Eventually, the massive particles will become so rare that they will not be able to find each other fast enough to maintain the equilibrium abundance. We see that this freeze-out happens at about 𝑥_𝑓 ~ 10.
   The final relic abundance, 𝑌_𝑋^∞ ≡ 𝑌_𝑋(𝑥 = ∞), determines the freeze-out density of the massive particles. Let us estimate its magnitude as a function of the interaction strength 𝜆. Well after freeze-out, 𝑌_𝑋 will much larger than 𝑌_𝑋^eq. Thus, at late time, we can drop 𝑌_𝑋^eq from Boltzmann equation, so that
    (3.102)   𝑑𝑌_𝑋/𝑑𝑥 ≈ 𝜆𝑌_𝑋²/𝑥².
Integrating this from 𝑥 = 𝑥_𝑓 to 𝑥 = ∞, we find   [𝑑𝑌/𝑑𝑥 = 𝜆𝑌²/𝑥₂, 𝑑𝑌/𝑌² = 𝜆𝑑𝑥/𝑥₂  ⇒  -1/𝑌 = 𝜆/𝑥.]
    (3.103)   1/𝑌_𝑋^∞ - 1/𝑌_𝑋^𝑓 = 𝜆/𝑥_𝑓,  
where 𝑌_𝑋^𝑓 ≡ 𝑌_𝑋(𝑥_𝑓). Typically 𝑌_𝑋^𝑓 ≫ 𝑌_𝑋^∞, so a simple analytic approximation for relic abundance is
    (3.104)   𝑌_𝑋^∞ ≈ 𝑥_𝑓/𝜆.
We have seen that a good order-of-magnitude estimate is 𝑥_𝑓 ~ 10. Using this value, the estimate in (3.104) in excellent agreement with the numerical results in Fig. 3.10. Moreover, 𝑥_𝑓 value isn't terribly sensitive to 𝜆 value, since 𝑥_𝑓(𝜆) ∝ ∣ln 𝜆∣, where the logarithmic scaling can be understood from the decoupling condition 𝛤(𝑥_𝑓) ≈ 𝐻(𝑥_𝑓).  
   Equation (3.104) predicts that the freeze-out abundance 𝑌_𝑋^∞ decrease as the interaction rate 𝜆 increases. This makes intuitive senses: larger interaction maintain equilibrium for a longer time, deeper into the Boltzmann-suppressed regime. Since the estimate in (3.104) works very well, we will use following.

          Dark matter density
We would like to relate the freeze-out abundance of the dark matter relics to their density today. To do this we note that the number density of the particle after freeze-out decreases as 𝑛_𝑋 ∝ 𝑎^-3. We can then write the number density today as
    (3.105)   𝑛_𝑋,0 = 𝑛_𝑋,1 (𝑎₁/𝑎₀)³ = 𝑌_𝑋^∞𝛵₀³ (𝑎₁𝛵₁/𝑎₀𝛵₀)³.   [𝑌_𝑋 ≡ 𝑛_𝑋/𝛵³; 𝑛_𝑋,1 (𝑎₁/𝑎₀)³ = 𝑌_𝑋^∞𝛵₁³ (𝑎₁/𝑎₀)³]
where 𝑎₁ is an arbitrary time that is late enough for species to have reached their relic abundance, but before any additional particle annihilations have become relevant. As we have seen, such particle annihilation lead to deviations from the scaling 𝛵 ∝𝑎^-1, so that the final factor in (3.105) isn't simply unity. Using the conservation of entropy (3.54), 𝑔_*𝑆(𝑎𝛵)³ = const. we instead get
    (3.106)   𝑛_𝑋,0 = 𝑌_𝑋^∞𝛵₀³ 𝑔_*𝑆(𝛵₀)/𝑔_*𝑆(𝑀_𝑋) without momentum, 𝑝.
The normalized energy density of the particles then is, using (3.2) 𝐻² = 𝜌/3𝑀_𝑃𝑙², 𝐸 = 𝑀_𝑋 without momentum, 𝑝, (3.104) and (3.101),
    (3.107)   𝛺_𝑋 = 𝜌_𝑋,0/𝜌_crit,0 = 𝑀_𝑋𝛵₀³/3𝑀_𝑃𝑙²𝐻₀² = 𝑀_𝑋𝑛_𝑋,0/3𝑀_𝑃𝑙²𝐻₀² 𝑥_𝑓/𝜆 𝑔_*𝑆(𝛵₀)/𝑔_*𝑆(𝑀_𝑋) = 𝐻(𝑀_𝑋)/𝑀_𝑋² 𝛵₀³/3𝑀_𝑃𝑙²𝐻₀² 𝑥_𝑓/⟨𝜎𝑣⟩ 𝑔_*𝑆(𝛵₀)/𝑔_*𝑆(𝑀_𝑋),
Using (3.55), 𝐻² ≈ π²/90 𝑔_* 𝛵/𝑀_𝑃𝑙², for 𝐻(𝑀_𝑋), this becomes
    (3.108)   𝛺_𝑋 ≈ π/√90 𝛵₀³/3𝑀_𝑃𝑙²𝐻₀² 𝑥_𝑓/⟨𝜎𝑣⟩ √[𝑔_*(𝑀_𝑋)] 𝑔_*𝑆(𝛵₀)/𝑔_*𝑆(𝑀_𝑋).
Finally, we use 𝑔_*𝑆(𝑀_𝑋) = 𝑔_*(𝑀_𝑋) considering (3.51) and insert the measured value of 𝛵₀ and 𝐻₀ as well as 𝑔_*𝑆(𝛵₀) = 3.91, to get
    (3.109)   𝛺_𝑋 ~ 0.1 𝑥_𝑓/√[𝑔_*(𝑀_𝑋)] 10^-8 GeV^-2/⟨𝜎𝑣⟩
This result is insensitive to the mass of the new particles and is mostly determined by their cross section. The observed dark matter density is reproduced if        √⟨𝜎𝑣⟩ ~ 10^-4 GeV^-1 ~ 0.1 √G_𝐹.  
The fact that a thermal relic with a cross section characteristic of the weak interaction gives the right order of magnitude for the dark matter abundance today is called the WIMP miracle.¹⁶
 

          3.2.3 Baryogenesis: A Sketch* The same analysis can also be applied to annihilation of baryons 𝑏 + 𝑏̄  ⟷  𝛾 + 𝛾. Since the effective interaction between nucleons is mediated by pions, we can estimate the annihilation cross section as ⟨𝜎𝑣⟩ ~ 1/𝑚_π², with 𝑚_π ≈ 140 MeV, Substituting this into (3.109), we then find
    (3.110)   𝛺_𝑏 ≈ 10^-11,
which is much smaller than the observed value. This illustrates the need for baryogenesis to produce the abundance of baryons in the early universe. A lazy option would be to simply impose the baryon asymmetry as an initial condition. Much more satisfying would be to start with a baryon-symmetric universe and hen see the asymmetry between matter and antimatter arise dynamically.  

          Sakharov conditions
In 1967, Sakharov identified three necessary conditions that any successful theory of baryogenesis must satisfy. These Sakharov conditions are:

1. Violation of baryon number.   The interactions that generate the baryon asymmetry must violate baryon number symmetry. It should be rather obvious that a universe with vanishing baryon number, 𝛣 = 0, can only evolve into a universe with 𝛣 ≠ 0 if the interactions don't conserve the baron number.
   In the Standard Model (SM), baryon number is an "accidental symmetry," meaning that all renormalizable interactions have this symmetry because of the gauge symmetry of the SM. Despite baryon number being a symmetry of SM Lagrangian, baryon number violation occurs at the quantum level through the so-call "triangle anomaly". Nonperturbatively, this can lead "sphaleron transitions" which convert baryons to antilepton and thus change the baryon number.¹⁷ Theories beyond the SM-such as Grand Unified Theories (GUTs)- may violate baryon number even in the Lagrangian (i.e. at tree level).

2. Violation of C and CP.   Slightly less obvious is that the interactions must violate C (charge conjugation) and CP (charge conjugation combined with parity). If C was an exact symmetry, then the probability of the process 𝑖 → 𝑗 would be equal to the corresponding antiparticles, 𝑖̄ → 𝑗̄, and no net baryon number would be generated. To see why CP must be violated, we first note that CPT (with T being time reversal) is a symmetry in any relativistic quantum field theory. CP invariance is therefore equivalent to invariance under time reversal. Consider the process 𝑖(𝐩_a,𝐬_a)  → 𝑓(𝐩_a,𝐬_a), where 𝐩_a are momenta of the particles and 𝐬_a are their spins. Under time reversal, this becomes 𝑓(-𝐩_a,-𝐬_a) → 𝑖(-𝐩_a,-𝐬_a). If the time reversal is unbroken, then integrating over all momenta and summing over all spins would lead to vanishing net baryon number.
   Both C and CP are violated by the weak interaction. The SM therefore, in principle, contains the required symmetry breaking, but whether is the breaking is strong enough to explain the size of the baryon asymmetry in our universe remains to be seen.

3. Departure from equilibrium.   Finally, the process that generate the baryon asymmetry must occur out of thermal equilibrium. For non-relativistic particles and antiparticles in equilibrium, we have  
    (3.111)   𝛣 ∝ 𝑛_𝑋 - 𝑛_𝑋̄ = 2𝑔_𝑋(𝑀_𝑋𝛵)^3/2 𝑒^{-𝑀_𝑋/𝛵} sinh(𝜇_𝑋/𝛵),
where 𝜇_𝑋 is the chemical potential. By the first Sakharov condition, 𝑋 and 𝑋̄ undergo 𝛣-violating reactions, such as
        𝑋 𝑋  →  𝑋̄ 𝑋̄.
   If these reactions occur in equilibrium, then 𝜇_𝑋 = 0 and 𝛣 = 0 in (3.111). The necessary out-of-equilibrium conditions can be provided by the cosmological evolution, making baryogenesis an interesting interplay between cosmology and particle physics.

          Out-of-equilibrium decay
A simple toy model for baryogenesis will be sketched: the out-of-equilibrium decay a massive particle. The decay rate of particles is 𝛤_𝑋, and most of the decay occurs when the age of the universe is of the order of the particles's lifetime, 𝑡 ~ 𝐻^-1 ~ 𝛤_𝑋^-1. If the decay violates baron number, then a baryon asymmetry will be generated.
   For concreteness, we assume that the 𝑋 particle has two decay channels 𝑎 and 𝑏, with baryon numbers 𝛣_𝑎 and 𝛣_𝑏. Similarly, the antiparticle 𝑋̄ has the decay channels 𝑎̄ and 𝑏̄ with baryon numbers -𝛣_𝑎 and -𝛣_𝑏. The branching ratios for the decay are
    (3.112)   𝑟 ≡ 𝛤(𝑋 → 𝑎)/𝛤_𝑋,     𝑟̄ = 𝛤(𝑋̄ → 𝑎̄)/𝛤_𝑋,     1 - 𝑟 ≡ 𝛤(𝑋 → 𝑏)/𝛤_𝑋,      1 - 𝑟̄ = 𝛤(𝑋̄ → 𝑏̄)/𝛤_𝑋,
where we have used that the total decay rates of 𝑋 and 𝑋̄ are equal (because of CPT symmetry and unitarity). The net baryon number produced in the decays of 𝑋 and 𝑋̄ then is
    (3.113)   ∆𝛣 ≡ 𝛣_𝑋 + 𝛣_𝑋̄ = [𝑟𝛣_𝑎 + (1 - 𝑟)𝛣_𝑏] + [1 - 𝑟̄𝛣_𝑎 - (1 - 𝑟̄)𝛣_𝑏] = (𝑟 - 𝑟̄)(𝛣_𝑎 - 𝛣_𝑏).
If the baryon number isn't violated in the individual decays, then 𝛣_𝑎 = 𝛣_𝑏 = 0 and hence ∆𝛣 = 0. Similarly, if C and CP are preserved, then 𝑟 = 𝑟̄ and ∆𝛣 = 0. This illustrates the necessity of Sakharov conditions for baryogenesis.
   The net baryon number density produced by the decays is 𝑛_𝛣 = ∆𝛣 𝑛_𝑋. Using that 𝑛_𝑋 ~ 𝑛_𝛾 at the time of the decay, we get
    (3.114)   𝜂 = 𝑛_𝛣/𝑛_𝛾 ~ ∆𝛣,
which relates the cosmologically observed baryon-to-photon ratio to the amount of baryon number violation. Hence, we have assumed that the entropy release in the decays is negligible, which is not always a good approximation. Finally, a more accurate computation would also involve explicit integration of the Boltzmann equation.

          Models of baryogenesis
There are many other scenarios. Some popular models of baryogenesis will be described very briefly. Refer to the suggested reference for more detail.

• Gut baryogenesis   Grand Unified Theories unifying the strong and weak interactions generally contain baryon number-violating interactions. In the effective theory of the Standard Model, these interaction appear as higher-dimensional operators suppressed by a large scale and are often constrained by a the absence of proton decay. The baryon asymmetry then be created by the out-of-equilibrium decay of a heavy particle, as sketched above. The non-observation of proton decay puts a lower bound on the mass of the decaying particle, which can be in tension with the reheating temperature after inflation.
(Refer to some papers [12-18] for more detail.)

• Electroweak baryogenesis   Within the SM, at first sight, all the Sakharnov conditions could be satisfied: (i) sphalerons provide baryon number violation, (ii) weak interactions (EW) violate C and CP, and (iii) the electroweak phase transition provide out-of-equilibrium conditions. However, on closer inspection, The EW phase transition is not strongly first order (as required because sphaleron transition would wash out the baryon asymmetry in a second-order phase transition) and the size of the CP violation is too small. Therefore new physics is required to change the nature of the EW phase transition and provide new sources of CP violation. Nevertheless, it remains an attractive option. (Refer to reviews [11, 26-28].
   During a first-order phase transition, bubbles of the broken phase (with nonzero Higgs vacuum expectation value, ⟨𝐻⟩ ≠ 0) nucleate within the surrounding plasma in the unbroken phase (with ⟨𝐻⟩ = 0); see Fig. 3.11. These bubbles expand and coalesce until only the broken phase remains. If the underlying theory contain CP violation, then the scattering of particles with the bubble walls can generate CP asymmetries  in the particle number densities. These asymmetries diffuse into the symmetric phase leading to a biasing of electroweak sphaleron transitions which producing a net baryon symmetry. Some of the excess baryons are then transferred to the broken phase where the rate of sphaleron transition transitions is strongly suppressed, so that the baryon asymmetry is preserved.

• Leptogenesis   Another interesting possibility is to generate via leptogenesis.{29] Imagine a theory beyond the SM with an extra heavy (but sterile) neutrino (like in the "seesaw mechanism" for neutrino masses). If the decay of this heavy neutrino violates CP, then it could produce more antileptons than leptons. These antileptons are then transformed into baryon by sphaleron transitions. The end result is an excess of baryons over antibaryons.¹⁸    

       3.2.4 Big Bang Nucleogenesis
As our next example, we will use the Boltzmann equation to follow the evolution of baryonic matter in the early universe. since baryon number is conserved, the total number of nucleons stays constant. However, weak nuclear reactions convert neutrons and protons into each other and strong nuclear reactions will build heavier nuclei from them. In this section, how the light elements-hydrogen, helium and lithium-were synthesized in the Big Bang will be shown. But all of the complicated detail of BBN won't be given.¹⁹ Instead our goal is a theoretical understanding of a single number, the ratio of the (mass) density of helium to hydrogen, This number is one of the key predictions of the Big Bang theory.
    (3.115)   𝑌_𝑃 ≡ 𝑛_⁴𝐻𝑒/𝑛_𝐻 ~ 1/4.   ^*[Correction: 4𝑛_𝐻𝑒 → 𝑛_⁴𝐻𝑒]
   Let us briefly sketch the main sequence of events during BBN. At early times, the baryonic matter was mostly in the form of protons and neutrons, which were coupled to each other by 𝛽-decay and inverse 𝛽-decay:
    (3.116)  𝑛 + 𝜈_𝑒  ⟷  𝑝⁺ + 𝑒^-,     𝑛 + 𝑒⁺ ⟷  𝑝⁺ + 𝜈̄_𝑒,
where 𝜈_𝑒 and 𝜈̄_𝑒 are electron neutrinos and antineytrinos. Initially, the relative abundance of the protons and neutrons were determined by equilibrium thermodynamics. Around 1 MeeV the reaction in (3.116) became inefficient and the neutrinos decoupled. We will determine the freeze-out abundance by solving the relevant Boltzmann equation. Free neutrons decay until neutrons and protons combined into deuterium²⁰
    (3.117)  𝑛 + 𝑝⁺   ⟷   𝐷 + 𝛾.
This occurred at a temperature around 0.1 MeV. Finally, the deuterium nuclei fused into helium
    (3.118)   𝐷 + 𝑝⁺   ⟷   ³𝐻 + 𝛾.     𝐷 + ³𝐻   ⟷   ⁴𝐻 + 𝑝⁺.
In this way, essentially all of the primordial neutrons got converted  into helium. The formation of heavier elements was very inefficient, which is why most protons survived and became the hydrogen in the universe.

          Equilibrium
About 𝛵 ≈ 0.1 MeV, only free protons and neutrons existed, while other light nuclei hadn't been formed yet. We can therefore first solve for the neutron-to-proton ratio and the use this abundance as an input for the synthesis of deuterium, helium, etc.
   Consider the reactions in (3.116). Let us assume that the chemical potentials of the electrons and neutrinos are negligibly small, so that 𝜇_𝑛 = 𝜇_𝑝. Using (3.71) for the equilibrium number densities 𝑛_𝑖^eq, we then have
    (3.119)  (𝑛_𝑛/𝑛_𝑝)^eq = (𝑚_𝑛/𝑚_𝑝)^3/2 𝑒^{-(𝑚_𝑛 - 𝑚_𝑝)/𝛵}.
The small difference between the proton and neutron masses can be ignored in the prefactor, but has to be kept in the exponential. Hence, we find
    (3.120)  (𝑛_𝑛/𝑛_𝑝)^eq = 𝑒^-𝑄/𝛵,
where 𝑄 ≡ 𝑚_𝑛 - 𝑚_𝑝 = 1.30 MeV. For 𝛵 ≫ 1 MeV, there are as many neutron as protons, while 𝛵 < 1 MeV, the neutron fraction gets smaller. If the weak interactions would operate efficiently enough to maintain equilibrium indefinitely, then the neutron abundance would drop to zero. But in the real world , the weak interactions are not so efficient.

          Neutron freeze-out
Since essentially all the neutrons eventually become incorporated into ⁴𝐻, the primordial ratio of neutrons protons is particularly important. For convenience we define
    (3.121)   𝑋_𝑛 ≡ 𝑛_𝑛/(𝑛_𝑛 + 𝑛_𝑝),
and, using (3.120), we get
    (3.122)   𝑋_𝑛^eq(𝛵) = 𝑒^-𝑄/𝛵/(1 + 𝑒^-𝑄/𝛵).
follow this equilibrium abundance until weak interaction processes such as (3.116) effectively shut off around 𝛵 ~ 1MeV (roughly at the same time as neutrino decoupling).²¹ To follow this evolution in detail, we must solve the Boltzmann equation (3.96), with 1 = neutron, 3 = proton, and 2, 4 = leptons (𝑛_𝓁 = 𝑛_𝓁^eq):
    (3.123)   1/𝑎³ 𝑑(𝑛_𝑛𝑎³)/𝑑𝑡 = -𝛤_𝑛 [𝑛_𝑛 - (𝑛_𝑛/𝑛_𝑝)_eq𝑛_𝑝],  
where 𝛤_𝑛 = 𝑛_𝓁⟨𝜎𝑣⟩ is the rate for neutron /proton conversion. Substituting (3.121) and (3.122) into (3.123), we find
    (3.124)   𝑑𝑋_𝑛/𝑑𝑡 = -𝛤_𝑛 [𝑋_𝑛 - (1 - 𝑋_𝑛) 𝑒^-𝑄/𝛵].  [Dodelson p.66, the total density (𝑛_𝑛 + 𝑛_𝑝) times 𝑎³ (is preserved and) can be taken outside the derivative]
   The key input from article physics is the rate 𝛤_𝑛. We have from (3.116)
    (3.125)   𝛤_𝑛 = 𝑛_𝜈⟨𝜎(𝑛𝑣_𝑒 → 𝑝⁺𝑒^-)𝑣⟩ + 𝑛_𝑒⟨𝜎(𝑛𝑒⁺ → 𝑝⁺𝜈̄^𝑒)𝑣⟩,
where the neutrino and positron densities are
    (3.126)   𝑛_𝑒 = 2𝑛_𝜈 = 3𝜁(3)/2π² 𝛵³.
Moreover, the relevant thermally-averaged cross sections are
    (3.127)   ⟨𝜎(𝑛𝑣_𝑒 → 𝑝⁺𝑒^-)𝑣⟩ = 510π²/{3𝜁(3) 𝜏_𝑛} (12𝛵 + 6𝑄𝛵 + 𝑄²)/𝑄³,      ⟨𝜎(𝑛𝑒⁺ → 𝑝⁺𝜈̄^𝑒)𝑣⟩ = 255π²/{3𝜁(3) 𝜏_𝑛} (12𝛵 + 6𝑄𝛵 + 𝑄²)/𝑄³,
where 𝜏_𝑛 = 886.7±0.8 s is the neutron lifetime. The total neutron/proton conversion rate then is
    (3.128)   𝛤_𝑛(𝑥) = 255/𝜏_𝑛 (12 + 3𝑥 + 𝑥²)/𝑥⁵,
where 𝑥 ≡ 𝑄/𝛵,
    (3.129)   𝐻(𝑥) = √(𝜌/3𝑀_𝑝𝑙²) = π/3 √(𝑔_*/10) 𝑄²/𝑀_𝑝𝑙 1/𝑥²
where 𝑔_* = 10.75 and 𝐻 = 1.13 s^-1. [Dodelson pp.66-67. 𝐻 = √(8π𝐺𝜌)/3, 𝜌 = π²/30 𝛵² [𝛴_𝑖=bosons 𝑔_𝑖 + 7/8 𝛴_{𝑖=fermions} 𝑔_𝑖] (𝑖 relativistic) ≡ 𝑔_* π²/30 𝛵²,  𝑔_* = 2(𝑔_𝛾) + 7/8[6(𝑔_𝜈 + 2(𝑔_𝑒) + 2(𝑔_𝑝) ≃ 10.75, 𝐻(𝑥 = 1) = √(4π³𝐺𝑄⁴/45) × √10.75 = 1.13 sec^-1.] This shows s that the conversion time 𝛤_𝑛^-1 = 0.60 is comparable to the age of the universe 𝑡 = (2𝐻)^-1 = 0.74 s. [RE (3.55-56)] Since 𝛤_𝑛^-1 ∝ 𝛵^-3 ∝ 𝑡^3/2, the interaction time eventually becomes longer than the age of the universe and the neutron- to-proton ration approaches a constant.  
   We can solve the Boltzmann equation (3.124) numerically. It is useful to choose an evolution variable that focuses on the dynamics at 𝛵 ~ 𝑄. Using 𝑥 = 𝑄/𝛵, [i.e. 𝑥 =1], we can the left-hand side of (3.124) as
    (3.130)   𝑑𝑋_𝑛/𝑑𝑡 = 𝑑𝑥/𝑑𝑡 𝑑𝑋_𝑛/𝑑𝑥 = -𝑥/𝛵 𝑑𝛵/𝑑𝑡 𝑑𝑋_𝑛/𝑑𝑥 = 𝑥𝐻 𝑑𝑋_𝑛/𝑑𝑥,   where we have 𝛵 ∝𝑎^-1 in the last equality. [Since 𝛵 ∝𝑎^-1 from (3.54), 𝛵 ∝ 𝑔_*𝑆^-1/3𝑎^-1, then we have -1/𝛵 𝑑𝛵/𝑑𝑡 = 𝐻.] So we get
    (3.131)   𝑑𝑋_𝑛/𝑑𝑥 = 𝛤_𝑛(𝑥)/𝐻(𝑥 = 1) 𝑥[𝑒^-𝑥 - 𝑋_𝑛(1 + 𝑒^-𝑥)].
The numerical solution of this equation is shown in Fig. 3.12. We see that the freeze-out of neutron abundance is       (3.132)   𝑋_𝑛^∞ ≡ 𝑋_𝑛(𝑥 = ∞) = 0.15.
A rough estimate of the freeze-out abundance can also be obtained by evaluating the equilibrium abundance (3.122) at the time of neutrino decoupling, 𝛵_𝑓 ~ 0.8 MeV, which gives 𝑋_𝑛^∞ ~ 1/6.

          Neutron Decay
At temperature below 0.2 MeV (or 𝑡 ≳ 100 s) the finite lifetime of neutron becomes important. To include this neutron decay in our computation, we simply multiply the freeze-out abundance (3.132) by an exponential decay factor
    (3.133)  𝑋_𝑛(𝑡) = 𝑋_𝑛^∞ 𝑒^{-𝑡/𝜏_𝑛} ≈ 0.15 𝑒^{-𝑡/887 s}.
It is now a race against time. If the onset of nucleosynthesis takes significantly longer than the neutron lifetime, than the number of neutrons becomes exponentially small and not much fusion will occur. Fortunately, this is not the case in our universe. 

¹⁴ Strictly speaking, it's the "integrated Boltzmann equation." A more fundamental Boltzmann equation for the evolution of the distribution function 𝑓_𝑖 will be introduced in Appendix B.
¹⁵ Cross sections are the fundamental observables in particle physics. They describes the probability of a certain scattering process. We will simply use dimensional analysis to estimate the few cross sections. They may depend on the relative velocity 𝑣 of particles 1 and 2.
¹⁶ WIMP stands for weakly interacting massive particle. WIMPs used to be a very popular dark matter candidate because of the result in (3.109). However, recent direct detection experiments have put pressure on the WIMP scenario. Alternative dark matter candidate, such as non-thermal axions have increase in popularity.
¹⁷ Sphalerons are nonpertubative solution of the elctroweak field equation, corresponding to the collective excitation of 𝑊, 𝑍 and Higgs fields. They rather spectacular objects. The typical size of a sphaleron is 10^-17 m, so that its volume is a 1000 000 times smaller than that of proton, while the mass is around 10 TeV, or 10 000 times the mass of a proton. Sphalerons are therfore 10 billion times denser than proton.
¹⁸ Interest in leptogenesis was recently revived by hints of confirmed CP violation for the ordinary neutrinos by the T2K experiment [Ref. 3-30]. If confirmed, this would make a plausible that CP violation also play an imortant role for sterile neutrinos (if they exist).
¹⁹ The detailed predictions of BBN can be computed numerically using the code PRIMAT. Thanks to Cyril Pitrou for making the code open to public.
²⁰ The reaction is very similar to the reaction 𝑒^- + 𝑝⁺ → 𝐻 + 𝛾 in Section 3.1.5. Our analysis will therefore be virtually identical to the Saha analysis of recombination.
²¹ It is fortunate that 𝛵_𝑓 ~ 𝑄. This seams to be a coincidence: 𝑄 is determined by the strong and electromagnetic interaction, while 𝛵_𝑓 is by the weak interaction. Imagine a 𝛵_𝑓 ≪ 𝑄.

          Deuterium
At this point, the universe is mostly protons and neutrons. The first nucleus to form via reaction in (3.117), 𝑛 + 𝑝⁺ ⟷ 𝐷 + 𝛾, is deuterium.²²  Only when deuterium is available can helium be produced. Since deuterium is formed directly from neutrinos and protons it can follow its equilibrium abundance as long as enough free neutrons are available. However, since the deuterium binding energy is rather small (see Table 3.4), it takes a while for the its abundance to become large. Although heavier nuclei have larger binding energies and hence would have larger equilibrium abundance, they cannot be formed until sufficient deuterium has become available. This is the deuterium bottleneck.²³ To get a rough estimate for the time of nucleosynthesis, we determine the temperature 𝛵_nuc when the deuterium fraction in equilibrium would be of order one, (𝑛_𝐷/𝑛_𝑝)_eq ~ 1.
   Deuterium is produced by the reaction in (3.117). Since 𝜇_𝛾 = 0, where we have 𝜇_𝑛 + 𝜇_𝑝 = 𝜇_𝐷. To remove  the dependence of on the chemical potential, we consider
    (3.134)   (𝑛_𝐷/𝑛_𝑛𝑛_𝑝)_eq = 3/4 (𝑚_𝐷/𝑚_𝑛𝑚_𝑝 2π/𝛵)^3/2 𝑒^{-(𝑚_𝐷 - 𝑚_𝑛 - 𝑚_𝑝)/𝛵},   [RE (3.71) 𝑛_𝑖^eq = 𝑔_𝑖 (𝑚_𝑖𝛵/2π)^3/2 exp[(𝜇_𝑖 - 𝑚_𝑖)/𝛵.]
where we have used (3.71) with 𝑔_𝐷 = 3 and 𝑔_𝑝 = 𝑔_𝑛 = 2. [𝑔_𝐷/𝑔_𝑛𝑔_𝑝 = 3/4)] In the prefactor 𝑚_𝐷 can be set equal to 2𝑚_𝑛 ≈ 2𝑚_𝑝 ≈ 1.9 GeV, but in the exponential the small deference between  𝑚_𝑛 + 𝑚_𝑝 and 𝑚_𝐷 is crucial: it is the binding energy of deuterium
    (3.135)   𝛣_𝐷 ≡ 𝑚_𝑛 + 𝑚_𝑝 - 𝑚_𝐷 = 22.2 MeV.
    (3.136)   (𝑛_𝐷/𝑛_𝑝)_eq = 3/4 𝑛_𝑛^eq (4π/𝑚_𝑝𝛵)^3/2 𝑒^{𝛣_𝐷/𝛵}.
To get an order-of-magnitude estimate, we approximate the neutron density by the baryon density and write this in terms of the photon temperature and the baryon-to-photon ratio,   [RE (3.76) 𝑛_𝑏 = 𝜂 𝑛_𝛾 = 𝜂 × 2𝜁(3)/π² 𝛵³,]
    (3.137)   𝑛_𝑛 ~ 𝑛_𝑏 = 𝜂 𝑛_𝛾 = 𝜂 × 2𝜁(3)/π² 𝛵³.
    (3.138)   (𝑛_𝐷/𝑛_𝑝)_eq ≈ 𝜂 (𝛵/𝑚_𝑝)^3/2 𝑒^{𝛣_𝐷/𝛵}.   [RE 3/4 2𝜁(3)/π² 𝛵³ (4π/𝑚_𝑝𝛵)^3/2 ≈ 8.14 (𝛵/𝑚_𝑝)^3/2]
As in the case of recombination, the smallness of baryon-to-photon ratio 𝜂 inhibits the production of deuterium until the 𝛵 drops well below binding energy 𝛣_𝐷. The 𝛵 has to drop enough so that 𝑒^{𝛣_𝐷/𝛵} can compete with 𝜂 ≈ 6 × 10^-10. We find (𝑛_𝐷/𝑛_𝑝)_eq ~ 1 at 𝛵_nuc ~ 0.06 MeV, which via (3.56), [When 𝑎 ∝ 𝑡^-1/2, we have 𝐻 = 1/(2𝑡) and the Friedmann equation leads to 𝛵/1MeV ≃ 1.5 𝑔_*^-1/4(1sec/𝑡)^-1/2], with 𝑔_* = 3.38 [Dodelson p.67 neutron day (𝑛 → 𝑝 + 𝑒 + 𝜈) and deuterium production (𝑛 + 𝑝  → 𝐷 + 𝛾). When neutrons decay electrons and positrons have annihilated, so 𝑔_* = (𝑔_result/𝑔_𝑛) = [2(𝛾) + 7/8 {3(𝐷) + 𝑓_d × 2(𝜈)}]/ [7/8 2(𝑛)] = 2.64 + 𝑓_d ≈ 3.38, with decay factor 𝑓_d = 0.74.], translates into
    (3.139)    𝑡_nuc = 120 s (0.1 MeV/𝛵_nuc)² ~ 330 s.
We see from Fig. 3.13 that a better estimate would be 𝑛_𝐷^eq(𝛵_nuc) ≈ 10^-3𝑛_𝑝^eq(𝛵_nuc), which gives 𝛵_nuc ≈ 0.07 MeV and 𝑡_nuc ≈ 250 s. Notice that 𝑡_nuc ≪ 𝜏_𝑛, so the result for 𝑋_𝑛(𝑡_nuc) from (3.133) won't be very sensitive to the estimate for 𝑡_nuc. Once the fusion of deuterium has started, the subsequent production of helium is very rapid.

          Helium
Since the binding energy is larger than that of deuterium, the Boltzmann factor 𝑒^𝛣/𝛵 favors helium over deuterium. Fog. 3.13 shows that helium is produced almost immediately after deuterium. The reaction proceeds in two steps: First, helium-3, ³𝐻𝑒, are formed via
       𝐷 + 𝑝⁺  ⟷   ³𝐻𝑒 + 𝛾,     𝐷 + 𝐷  ⟷   ³𝐻 + 𝑝⁺,     𝐷 + 𝐷  ⟷   ³𝐻𝑒 + 𝑛.
The rate of 𝐷-𝐷 fusion is much faster, but there are more protons than deuterium. Next, helium-4, ⁴𝐻𝑒, is produced through the following chain of reactions
       ³𝐻 + 𝑝⁺  ⟷   ³𝐻 + 𝑛,     ³𝐻 + 𝐷  ⟷   ⁴𝐻𝑒 + 𝑛,     ³𝐻 + 𝐷  ⟷   ⁴𝐻𝑒 + 𝑝⁺.
Virtually, all the neutrons at 𝑡 ~ 𝑡_nuc are processed into ⁴𝐻𝑒 by these reactions. Substituting 𝑡_nuc ≈ 250 s into (3.133), [𝑋_𝑛(𝑡) = 𝑋_𝑛^∞ 𝑒^{-𝑡/𝜏_𝑛} ≈ 0.15 𝑒^{-𝑡/887 s}], we get
    (3.140)    𝑋_𝑛(𝑡_nuc) ≈ 0.11.    
Since two neutrons go into one nucleus of ⁴𝐻𝑒, the final helium abundance is equal to half of the neutron abundance at 𝑡_nuc, so that 𝑛_𝐻𝑒 = 1/2 𝑛_𝑛(𝑡_nuc. Hence, the mass fraction of helium is
    (3.141)    𝑌_𝑃 = 𝑛_⁴𝐻𝑒/𝑛_𝐻 = 𝑌_𝑃 = 𝑛_⁴𝐻𝑒/𝑛_𝑝 = 2𝑋_𝑛(𝑡_nuc/(1 - 𝑋_𝑛(𝑡_nuc) ~ 0.25, [= 1/4]
as we wish to show. A more accurate calculation gives [Ref.3-31]
    (3.142)    𝑌_𝑃 = = 0.2262 + 0.0135 ln(𝜂/10^-10,
where the logarithmic dependence on 𝜂 can be trace back to the effect of baryon-to-photon ratio on the time of nucleosynthesis in (1.139). Fig. 3.14 shows the predicted helium mass fraction as a function of 𝜂. The abundance of ⁴𝐻𝑒 increases with increasing 𝜂 because nucleosynthesis starts earlier for larger baryon density.
   In making the prediction (3.141), we have taken the effective number of relativistic species, 𝑔_*, to be fixed at the value by the Standard Model. It is also interesting to let the value of 𝑔_* float and use measurement to put constraints on any new physics contributing to 𝑔_* at BBN. The dependence on 𝑔_* is through 𝐻 ∝ √𝑔_* 𝛵², so that increasing the value of 𝑔_* leads to a larger expansion rate (at the same 𝛵) which implies an earlier neutron freeze-out, 𝛵_𝑓 ∝ 𝑔_*^1/6.^{[verification. needed]4}𝐻𝑒.

          Other light elements
To determine the abundances of other light elements, the coupled Boltzmann equations have to be solved numerically. A qualitative discussion of the most important reactions will just be provided. Wee see in Fig. 3.13 that most of deuterium is processed into ³𝐻𝑒 and ⁴𝐻𝑒. (Tritium is unstable and decay into ³𝐻𝑒.) The final abundance of 𝐷 and ³𝐻𝑒, as a function of 𝜂, are shown in Fig. 3.14. Since both of them are burnt by fusion, their abundance decrease as 𝜂 increases.
   Nucleosynthesis is a very inefficient process, so very few nuclei beyond helium are formed. Small amount of beryllium and lithium are created by the reactions
       ³𝐻𝑒 + ⁴𝐻𝑒  ⟷  ⁷𝛣𝑒 + 𝛾,     ³𝐻𝑒 + ⁴𝐻𝑒  ⟷  ⁷𝐿𝑖 + 𝛾,
  and can be converted via     ⁷𝛣𝑒 + 𝑛  ⟷  ⁷𝐿𝑖 + 𝑝⁺.
Lithium can be destroyed by capturing a proton
       ⁷𝐿𝑖 + 𝑝⁺  ⟷  ⁴𝐻𝑒 + ⁴𝐻𝑒.
This last reaction reduces the amount of lithium relative to beryllium. However, once the temperature gets low enough, beryllium can decay into lithium by capturing an electron
       ⁷𝛣𝑒 + 𝑒^-  ⟷  ⁷𝐿𝑖 + 𝜈_𝑒,
The predicted amount of lithium is shown in Fig. 3.14. At low 𝜂, ⁷𝐿𝑖 is destroyed by protons with an efficiency. On the other hand, its precursor ⁷𝛣𝑒 is produced more efficiently as 𝜂 increases. This explains the valley in the curve for ⁷𝐿𝑖.
   Almost no nuclei with mass number 𝛢 > 7 are produced in the Big Bang. The basic reason is that there are no stable nuclei with 𝛢 = 8 that can be formed fast enough to sustain the chain reaction. The half-life of ⁸𝛣𝑒 is 10^-12 s.The merger of three helium-4 nuclei into carbon-12 is too slow. Similarly the capture of neutrons and protons into ⁸𝐿𝑖 and ⁹𝛣𝑒 is very inefficient.

          Observations
To test the predictions of Big Bang nucleosynthesis, the element abundances must be measured in the region where very little post-processing of the primordial gas has taken to place. In particular, nuclear fusion inside of stars changes the abundances. The measurements cited below are those suggested by the Particle Data Group [Ref. 3-33]. The fact that we find good quantitative agreement with observations remains one of the great triumphs of the Big Bang model.

•Hellium-4 can be measured from the light of ionized gas clouds, because the strength of some emission lines depends on the amount of helium. We have to correct for the fact that ⁴𝐻𝑒 is also produced in stars. One way to do this is to correlate the measured helium abundance with the abundances of heavier elements, such as nitrogen and oxygen. The larger the amount of oxygen, the more helium has been created. Extrapolating the measurements to zero oxygen gives an improved estimate of the primordial helium abundance. Using this approach, the measured abundance of primordial helium-4 is found to be [34-36].
    (3.143)    𝑌_𝑃 = 0.245 ± 0.003.
We see from Fig. 3,14 that this is consistent with the prediction from BBN, given the CMB measurement of the baryon density.

•Deuterium is weakly bound and therefore easily destroyed in the late universe. BBN is the only source of significant deuterium in the universe. The best way to determine the primordial (unprocessed) value of the deuterium abundance is to measure the spectra of high-redshift quasars. These spectra contain absorption lines due to gas clouds along the line-of-sight, and in particular, the Lyman-𝛼 line is a sensitive probe of the amount of deuterium (see e.g. [37]). A weighted mean of several such measurements is 𝐷/𝐻 = (25.47±0.25) × 10^-6 [33]. In Fig. 3.14 this agrees precisely with the prediction from BBN.

•Helium-3 can be created and destroyed in stars.In abundance is therefore hard to measure and interpret. Given these uncertainties, helium-3 is usually not used as a cosmological probe,

•Lithium is mostly destroyed by stellar nucleosynthesis. The best estimate of its primordial abundance follows from the measurement of metal-poor stars in the Galactic halo. Averageing the measurements of [38-41] gives 𝐿𝑖/𝐻 = (1.6±0.3) × 10^-10. As we see from Fig. 3.14, the measured lithium abundance is significantly smaller than all the predicted value. It is unclear whether this lithium problem is due to systematic errors in the interpretation\n of the measurements or signals the need for new physics during BBN.

          3.2.5 Recombination Revisited^*
Finally, we will use the Boltzmann equation to provide a more accurate treatment of recombination. It's important to determine the evolution of free-ectron  density accurately because it affects the prediction for the CMB anisotropy (see Chapter 7). We assumed previously that thermal equilibrium holds throughout the process. In reality, this assumption breaks down rather quickly and non-equilibrium effects become important. As we will see, recombination directly to the ground state of the hydrogen atom is very inefficient and a more complicated path needs to be considered for  the era of precision cosmology.

          Non-equilibrium recombination
As thr free-electron fraction drops during recombination, the interaction rate also decreases and can fall below the expansion rate of the universe. When this happens, we must use Boltzmann equation to follow the evolution. Applying (3.96(), 1/𝑎³ 𝑑(𝑛₁𝑎³)/𝑑𝑡 = -⟨𝜎𝑣⟩ [𝑛₁𝑛₂ - (𝑛₁𝑛₂/𝑛₃𝑛₄)_eq 𝑛₃𝑛₄], to the reaction 𝑒^- + 𝑝⁺ ⟷ 𝐻 + 𝛾, we get
    (3.144)   1/𝑎³ 𝑑(𝑛_𝑒𝑎³)/𝑑𝑡 = -⟨𝜎𝑣⟩ [𝑛_𝑒² - (𝑛_𝑒²/𝑛_𝐻)_eq𝑛_𝐻],  
where we have used that 𝑛_𝑒 = 𝑛_𝑝 and 𝑛_𝛾 = 𝑛_𝛾^eq. For the factor of  (𝑛_𝑒²/𝑛_𝐻)eq we can use (3.74), (𝑛_𝐻/𝑛_𝑒²)_eq = (2π/𝑚_𝑒𝛵)^3/2 𝑒^{𝐸_𝐼/𝛵}. Moreover, the electron and hydrogen densities can be written as, 𝑛_𝑒 = 𝑋_𝑒𝑛_𝑏 and 𝑛_𝐻 = (1 - 𝑋_𝑒) 𝑛_𝑏.
Using that 𝑛_𝑏𝑎³ = const, we then get
    (3.145)   𝑑𝑋_𝑒/𝑑𝑡 = [𝛽(𝛵) (1 - 𝑋_𝑒) - 𝛼(𝛵)𝑛_𝑏𝑋_𝑒²],
where we have introduced
    (3.146-7)   𝛼(𝛵) ≡ ⟨𝜎𝑣⟩,      𝛽(𝛵) ≡ ⟨𝜎𝑣⟩ (𝑚_𝑏𝛵/2π)^3/2 𝑒^{-𝐸_𝐼/𝛵}.
The parameter 𝛼 characterizes the recombination rate, while 𝛽 is associated with the ionization rate. When 𝛼 is large, right-hand side of (3.145) must vanish and the evolution of 𝑋_𝑒 is given by Saha equilibrium. The function 𝛼(𝛵) depends on the precise way that the electrons get captured into the ground state, it releases photons with energy larger than 13.6 eV. These photons will then quickly ionized other atoms, leading to no net recombination.

          The effective three-level atom
To avoid the instantaneous reionization of the hydrogen atoms, recombination must first to an excited state, which then decays to the ground state. The photon created during this multi-step recombination have lower energy and are therefore less likely to ionize the hydrogen atoms. The details were worked out by Peebles in 1968 [43] (see also Zel'dovich, Kurt and Sunyaev [43]).
   Peebles first argued That the hydrogen atom can be treated as an effective three-level atom. The three relevant states are the ground  state (1𝑠), the excited states (mostly 2𝑠 and 2𝑝), and the continuum states of ionized hydrogen (see Fig. 3.15). The excited states are in thermal equilibrium with each other since radiative excitations and decays are very fast. Peebles therefore considered all excited states as a single entity.Since the direct recombination to the ground state  is very inefficient, the rate of recombination will be determined  by the rate of decayof the first excited state. A standard computation in quantum field theory gives
    (3.148)   𝛼(𝛵) ≈ 9.8 𝛼²/𝑚_𝑒² (𝐸_𝐼/𝛵)^1/2 ln (𝐸_𝐼/𝛵),
where 𝛼 ≈ 1/137 is the fine-structure constant.
   But, this is still bot the answer. When an atom in the first excited state decays to the ground state, it produces a Lyman-𝛼 photon. This photon has a large probability to be absorbed by a near by atom. which can then be ionized.To achieve a significant level of recombination we must avoid these resonant excitation. There are two way out:

• First, there is a small probability that the 2𝑠 state will decay to the 1𝑠 state through the emission of two photons. These photons then don't have enough energy to excite the atoms in the ground state back to the first excited state. The rate for this two-photon decay is
    (3.149)   𝛬_2𝜆 = 8.227 s^-1.
About 57% of all natural hydrogen is formed via this route.

• Second, as the universe expands, the energy of the Lyman-𝛼 photons that are created by the 2𝑝 → 1𝑠 transition is redshifted. This moves these photons off resonance. If a photon avoids being reabsorbed for a sufficiently long time, then this effect allows them to escape. There is a small probability that this will happen. The rate of recombination via this resonance escape is [44].²⁴
    (3.150)   𝛬_𝛼 = 8π/𝜆_𝑎³𝑛_1𝑠 𝐻,
where 𝜆_𝑎 ≡ 2π𝐸_𝛼^-1 = 8π/(3𝐸_𝐼) is the wavelength of a Lyman--𝛼 photon and 𝑛_1𝑠  ≈ (1 - 𝑋_𝑒)𝑛_𝑏 is the abundance of hydrogen atoms in the 1𝑠 state. Substituting 𝑛_𝑏 = 𝜂 𝑛_𝛾, the recombination rate becomes
    (3.151)   𝛬_𝛼 = 27/[128𝜁(3)] 𝐻(𝛵)/[(1 - 𝑋_𝑒) 𝜂 (𝛵/𝐸_𝐼)³].
About 43% of all hydrogen atoms are formed in this way.

   Although the details of the details of the analysis are rather complex, the final answer is easy to state and interpret.The main new parameter is the effective branching ratio
    (3.152)   𝐶_𝑟(𝛵) ≡ (𝛬_2𝜆 + 𝛬_𝛼)/(𝛬_2𝜆 + 𝛬_𝛼 + 𝛽_𝛼),
where 𝛽_𝛼 ≡ 𝛽𝑒^{3𝐸_𝐼/4𝛵} is the ionization rate of the 𝑛 = 2 state. 𝐶_𝑟 will be called Peebles factor. It describe the probability that an atom in the first excited state reaches the ground state through either of the two ways described above before being ionized. When 𝐶_𝑟 is much smaller than unity, recombination is suppressed. The evolution of 𝐶_𝑟(𝑧) is shown in Fig. 3.16. We see that 𝐶_𝑟 ≪ 1 for 𝑧 > 900, which implies that recombination will be delayed.
   The branching ratio 𝐶_𝑟 appears as a multiplicative factor in the evolution equation (3.145), leading to the so-called Peebles equation
    (3.153)   𝑑𝑋_𝑒/𝑑𝑡 = 𝐶_𝑟 [𝛽(𝛵) (1 - 𝑋_𝑒) - 𝛼(𝛵)𝑛_𝑏𝑋_𝑒²].
This equation can be integrated numerically to obtain the evolution of free-electron fraction.

          Electron freeze-out
It will be convenient to use redshift instead of timeas the indipendent variable during the evolution. Using [(2.63) 𝑎 = (1 + 𝑧)^-1, 𝐻 = 1/𝑎 𝑑𝑎/𝑑𝑡 = (1 + 𝑧) 𝑑 (1 + 𝑧)^-1/𝑑𝑡 = (1 + 𝑧) × -(1 + 𝑧)^-2 𝑑𝑧/𝑑𝑡 = -(1 + 𝑧)^-1 𝑑𝑧/𝑑𝑡.  ⇒  𝑑𝑧/𝑑𝑡 = -(1 + 𝑧) 𝐻. ▮)  
    (3.154)   𝑑𝑋_𝑒/𝑑𝑡 = 𝑑𝑋_𝑒/𝑑𝑧 𝑑𝑧/𝑑𝑡 = -𝑑𝑋_𝑒/𝑑𝑧 𝐻(1 + 𝑧),
the Peebles equation (3.153) can then be written as
    (3.155)   𝑑𝑋_𝑒/𝑑𝑧 = -𝐶_𝑟(𝑧)/𝐻(𝑧) 𝛼(𝑧)/(1 + 𝑧) [(1 - 𝑋_𝑒) (𝑚_𝑏𝛵/2π)^3/2 𝑒^{-𝐸_𝐼/𝛵} - 𝑋_𝑒² 𝜂 2𝜁(3)/π² 𝛵³],
where 𝛵 = 0.235 meV (1 + 𝑧). The evolution of the Hubble parameter is
    (3.156)   𝐻(𝑧) = √𝛺_𝑚 𝐻₀ (1 + 𝑧)^3/2 [1 + (1 + 𝑧)/(1 + 𝑧_eq)]^1/2,
with 𝐻₀ ≈ 1.5 × 10^-33 eV. Fig. 3.17  shows the evolution of the free-electron fraction as a function of redshift. We see that recombination is indeed delayed relative to the Saha expectation. In particular, 𝑋_𝑒 = 0.5 at 𝑧 = 1270 (compared to 𝑧 = 1380 for Saha.) Unlike the Saha prediction the electron density freeze out at a non-zero value about 2 × 10^-4.

          Decoupling and last-scattering
In Section 3.1.5, we defined last-scattering as the moment when the optical depth, (𝑡_* ≡ 1. The probability that a photon did not scatter off an electron in the redshift interval [𝑧₀, 𝑧₁] is,
    (3.157)   𝑃(𝑧₀, 𝑧₁) = 𝑒^{-𝜏(𝑧₀,𝑧₁)})
where 𝜏 is the optical depth
    (3.158)   𝜏(𝑧₀, 𝑧₁) ≡ ∫_𝑡1^𝑡₀ 𝑑𝑡 𝜎_𝛵 𝑛_𝑒(𝑡) = ∫_𝑧0^𝑧₁ 𝑑𝑧 𝜎_𝛵 𝑛_𝑏𝑋_𝑒(𝑧)/[𝐻(1 + 𝑧)].
The probability that a photon scattered for the last time in the interval [𝑧₁, 𝑧₁ + 𝑑𝑧₁] then is
    (3.159)   𝑃(𝑧₀, 𝑧₁) - 𝑃(𝑧₀, 𝑧₁ + 𝑑𝑧₁) ≡ 𝑔(𝑧₀, 𝑧₁) 𝑑𝑧₁, where 𝑔(𝑧₀, 𝑧₁) is the visibility function. We will be interested in 𝑔(𝑧) ≡ (0, 𝑧), which which is the probability that a photon observed today scattered last at redshift 𝑧. Using (3.157) and (3.159), we have²⁵
    (3.160)   𝑔(𝑧) = -𝑑/𝑑𝑧 𝑒^{-𝜏(0,𝑧)} = 𝑑𝜏/𝑑𝑧 𝑒^-𝜏 Note that 𝜏(0, 0) = 0 and 𝜏(0, ∞) = ∞, so that the visibility function satisfies
    (3.161)   ∫₀^∞ 𝑑𝑧 𝑔(𝑧) = -𝑒^-𝜏∣₀^∞ = 1.
as required for a probability density. At early times, 𝜏 is large and the visibility function is exponentially suppressed. After combination, 𝑑𝜏/𝑑𝑧 is small because the density of free electrons small. This means that 𝑔(𝑧) will be pealed at the moment of decoupling. Indeed, this is what find in Fig. 3.18. The maximum of the visibility function is at 𝑧_* = 1080, which we take as our definition of the moment of last-scattering. The function has a width of 𝛥𝑧_* = 80, which characterizes the finite width of the last-scattering surface.

          Precision cosmology
It is remarkable how much of the comp;ex physics of recombination was understood by Peebles and others in the 1960s. For the analysis of modern CMB experiments, however, the above analysis is still not enough.The most imortant development of refined analysis of recombination will be sketched, since they made the modern era of precision cosmology possible.
   The three-level atom assumes that all excited staes are equilibrium with the 𝑛 = 2 staes. This assumption breaks down during the later stages of precise multi-level atom was first studied by Seager, Sasselov and Scott in 1999 [46]. It was found that accounting for the additional excited states leads to an increase in the rate of recombination in the late times (see Fig. 3.19). To follow the evolution of each state, a large number of coupled Boltzmann equation has to be solved. Fortunately, the dynamics can be mimicked by solving it with an an enhanced recombination coefficient 𝛼 → 1.14 𝛼. This approach became the basis of the recombination code RECFAST.
   While RECFAST was precise unbiased analysis of the enough for the analysis of the WMAP data, an unbiased analysis of the Planck data required konwing the recombination history to better than 0.1%. many new effects were identified that change the predictions at the 1% level and therefore needed to to be included in the analysis:

• Seager et al. [46] assumed that states with same energy, but different angular momentum quantum number 𝓁, were in equilibrium. This assumption breaks down at the percent level [48]. As in the original analysis of Seager et al., solving for the abundances of all the excited states at every time-step is numerically very expensive. An elegant solution to this problem was found by Ali-Haimoud and Hirata [49]. They realized only the populations of a few excited states had to be tracked, provided one uses precomputed effective states of hydrogen. This effective multi-level atom is the basis of the recombination codes HYREC [47] and COSMOREC [50].

• At the precision of the Planck experiment, the rates for these processes have to be modeled more accurately than we have done. For example, stimulated two-photon emission cannot be ignored and changes the effective two-photon decay rate 𝛬_2𝛾 at the percent level [51]. Similarly, the absorption of non-thermal photons (produced in previous transitions) must be accounted for [52]. More generally, the time dependent radiative transfer problem must be solved to high accuracy; see [53-57] for further relevant corrections. Finally, it was realized that two-photon decay from higher excited states, 𝑛𝑠 and 𝑛𝑑, become relevant at this level of accuracy [58-60].

          3.3 Summary 
In this chapter, we have studied the thermal history of the universe. At early times, the rates of particle interactions, 𝛤, were much larger than the expansion rate, 𝐻, so that all particles were in equilibrium at a common temperature, 𝛵. The energy density was dominated by relativistic species
    (3.162)   𝜌 = π ²/30 𝑔_*𝛵⁴,
where 𝑔_* is the effective number of relativistic degrees of freedom (which is the sum of the internal degrees of freedom for each particle weighted by a factor of 7/8 for fermions). The pressure of the primordial plasma was 𝑃 = 𝜌/3, so that it behaved like radiation.    An important quantity is the entropy density     (3.163)   𝑠 = (𝜌 + 𝑃)/𝛵 = 2π^2/45 𝑔_*𝑆𝛵³ where 𝑔_*𝑆 is the effective number of relativistic degrees of freedom in entropy which for most of the universe's history was equal to 𝑔_*. The conversation of entropy then implies 𝑔_*𝑆𝛵³ ∝ 𝑎³ . We used this to relate the temperature of the cosmic neutrino background to that of the cosmic microwave background 𝛵_𝑣  = (4/11)^1/3𝛵_𝛾.
   To study non-equilibrium effects, we introduced the Boltzmann equation. For processes of the form 1 + 2 ⟷ 3 + 4, the Boltzmann equation for the species 1 can be written as
    (3.164)   𝑑 ln 𝑁₁/𝑑 ln 𝑎 = -𝛤₁/𝐻 [1 - (𝑁₁𝑁₂/𝑁₃𝑁₄)_eq 𝑁₃𝑁₄/𝑁₁𝑁₂],
where 𝑁_𝑖 = 𝑛_𝑖/𝑠 is proportional to the number of particles in comoving volume. As long as the interaction rate is larger than the expansion rate, 𝛤₁ ≫ 𝐻, the particle abundance tracks its equilibrium value. Once the interaction rate drops below the expansion rate, however, the particles drop out of thermal equilibrium and decouple from the thermal bath. by integrating the Boltzmann equation, we can follow the non-equilibrium evolution of the particle abundances. We presented three examples: (i) dark matter freeze-out (ii) Big Bang nucleosynthesis; and (iii) recombination.
   As a simple example of dark matter production, we considered the reaction 𝑋 + 𝑋̄ ⟷ 𝓁 + 𝓁̄, We used the Boltzmann equation to follow the evolution of the density of dark matter particles 𝑋. The resulting dark matter density today depends inversely on the annihilation cross section.
   We then studied nucleosynthesis. Initially, the most abundant nuclei were neutrinos and protons, kept in equilibrium through reactions of the form
       𝑛 + 𝜈_𝑒  ⟷  𝑝⁺ + 𝑒^-.
Around 1 MeV, these reactions become inefficient and the neutrons decoupled. We derived the freeze-out abundance by solving the Boltzmann equation. At 0.1 MeV, neutrinos and protons fused into deuterium, which then combined with protns into helium-3 and helium-4:
       𝑛 + 𝑝⁺  →  𝐷 + 𝛾,     𝐷 + 𝑝⁺  →  ³𝐻𝑒 + 𝛾,     𝐷 + ³𝐻𝑒  →  ⁴𝐻𝑒 + 𝑝⁺.
The resulting mass fraction of helium-4 is 𝑌_𝑃 ≡ 𝑛_⁴𝐻𝑒/𝑛_𝐻 ≈ 0.25, with a weak dependence on the baryon density of the universe.
   Finally, we investigate the recombination reaction
      𝑒^- + 𝑝⁺   →  𝐻 + 𝛾.
We used the Boltzmann equation to follow the evolution of the free-electron fraction, 𝑋_𝑒 ≡ 𝑛_𝑒/𝑛_𝑏. We took into account that recombination directly to the ground state is very inefficient and instead proceeds via two-photon decay and resonance escape from the first excited state. Defining recombination as the time when 𝑋_𝑒 = 0.5, we found
    (3.165)   𝑧_rec ≈ 1270,     𝑡_rec ≈ 290 000 yrs.
As the number of free electron dropped, the reaction 𝑒^- + 𝛾 ⟷ 𝑒^- + 𝛾 became inefficient and the photons decoupled. This happened at
    (3.166)  𝑧_dec ≈ 1090,     𝑡_dec ≈ 370 000 yrs.
The photons from this era are observed today as the CMB.

²² The fusion of two protons is inefficient because the nuclei have to overcome the Coulomb repulsion before the nuclear force can take over. The fusion of two neutrons produces a very unstable nucleus.
²³ Even in equilibrium the production of helium  would rather late, at 𝛵 ~ 0.3 MeV. The deuterium bottleneck leads to a further delay of this equilibrium expectation.
²⁴ This rate is a product of rate of the decay 2𝑝 → 1𝑠 + 𝛾 and the probability that the Lyman-𝛼 photon escapes to infinity without exciting another hydrogen atom. We cna find a more detailed discussion on Peebles' orginal paper {42] or in Weinberger's book [45]
²⁵ In Chaper 7, we will define the visibility function in terms of conformal time, 𝑔(𝑧) ≡ (𝑒^-𝜏)' = -𝜏' 𝑒^-𝜏, where ' ≡ ∂_𝜂. In that case, its peak is at 𝑧_* = 1090.