Out-Of-Equilibrium and Freeze-Out: Wimps, Big-Bang Nucleosynthesis and Recombination

Home , Decoupling (cosmology), Neutrino decoupling

Chapter 5

Out-of-equilibrium and freeze-out: WIMPs, Big-bang nucleosynthesis and recombination

In the last chapter, we have studied the equilibrium thermodynamics, and calculated thermal and expansion history in the early Universe. There, as the majority of cosmic energy budget in the early time comes from relativistic particles, the equilibrium thermal history is just enough to describe the expansion history during the radiation dominated epoch. We, at the same time, have also encountered the case for neutrino decoupling where electron-neutrino scattering rate falls below the Hubble expansion rate. After the decoupling epoch, neutrino hardly interacts with cosmic thermal plasma and evolves independently. This phenomena of freeze-out and thermal decoupling is pretty generic in the expanding universe, and there are number of interactions in the Universe that undergo out-of-equilibrium: baryon-antibaryon asymmetry, (presumably) thermal decouplign of dark matter, neutrino decoupling, neutron-to-proton freezeout, big-bang nucleosynthesis, and CMB recombination, to mention a few important examples. These are the subject of this chapter. It is this out-of-equilibrium phenomena that make our Universe very rich and interesting in contrast to the majority of energy budget (at early times) which are in equilibrium. Had it stayed in a perfect equilibrium state up until now, the Universe would be a boring place characterized by one number 1 TCMB = 2.726 K. We have seen that there are residual non-relativistic particles , surviving well after the temperature fell below the rest mass of the particle. In addition, if dark matters coulped to the cosmic plasma at earlier times, their abundance is also freezed out to a ﬁnite relic density. Although minorities during the radiation dominated epoch, these non-relativistic particles eventually dominate the Universe because energy density of the equilibrium part—thermal radiation—drops more quickly (ρ T 4) compared with non-relativistic relics (ρ T 3) survive through the freeze-out; thus, matter era∝ is opened up. The other events like neutrino decoupling,∝ CMB recombination, and big-bang nucleosynthesis all provide a window for us, cosmologists, to directly look at the early history of the Universe without much of the contamination— because they are decoupled, otherwise, all memory about the initial states of the Universe would be completely erased in the name of equilibrium! General reasoning for the freeze-out goes as following with the argument that we have used before: ti < tH to maintain the equilibrium and ti > tH to decouple. Let’s think about the interaction whose n 1 n+3 cross section scales as σv T . Then, the interaction rate (Γ = ti− ) is Γ n σv T . During the radiation dominated〈 epoch,〉 ∝ the Hubble expansion rate is proportional to H' 〈T 2,〉 which ∝ means that ∝ 1This net number has to be generated by a process called baryogenesis —though, we do not have the theory that everybody agrees on.

1 2 CHAPTER 5. OUT-OF-EQUILIBRIUM AND FREEZE-OUT when n > 1, the interaction rate drops quickly than the expansion rate, and eventually hit Γ H after which the− interaction freezes out. Is n > 1 reasonable? For the interaction mediated by massive' gauge − 2 2 2 boons, e.g. weak interaction, the cross section goes as σ (gI /mI ) T , where gI is the interaction 2 ' 2 coupling constant, and 1/mI is the mass of the gauge bosons. Then, T is added to get the right dimension for the cross section. For the relativistic particles that moves with v c, we have σv T 2; thus, we have n > 1 for sure. For the scattering via massless particles such as' photons, at〈 high〉 energy∝ − 2 (T > me), Klein-Nishina cross section scales as σKN T − , but at low energy, Thomson scattering cross ∝ section is constant, σT const. ' Note that at the high energy (T ¦ MW 80 GeV), the interaction rate for both weak interaction and ' 2 2 electromagnetic interaction scales as Γ σv n α T while Hubble rate is H T /MPl, where '〈 〉 ' ' MPl ls the reduced Planck mass and α = (0.01) is the electroweak coupling constant. This reads an O 2 16 interesting conclusion that the interactions are frozen out for T ¦ α MPl 10 GeV, and they cannot maintain the thermal equilibrium during these super early times! ' In this chapter, we will reﬁne the time scale argument, which works pretty well, in a more rigorous way by using kinetic theory. On top of that, the kinetic theory allows us to calculate the decoupling and freez eout procesure more accurately. We will then move on to discuss indivisual event that we have mentioned earlier.

5.1 Kinetic theory in the FRW universe

The evolution of the phase space distribution function is described by Boltzmann equation: d f = [f ], (5.1) dλ C which looks pretty simple, and is indeed easy to understand! The left-hand side of the Boltzmann equation is the total derivative of the phase space distribution fucntion following the trajectory. That must be balanced by the change of particles (source or sink) along the pathway due to interaction, which is incoded by the collosion term in the right hand side of Eq. (5.1). If there is no source or sink, the right-hand side vanishes and the Boltzmann equation simply states the Liouville theorem that we discussed in the last chpater. One difference you might notice from the usual (classical mechanics) form of the Boltzmann equation is that we now use the afﬁne parameter λ instead of the time. It is to make the Boltzmann equation manifestly covariant; otherwise, the Boltzmann equation would depend on the choice of coordinate system. Of couse, you can relate the two derivatives by using (see, [?, ?] for more rigorous ways of writing down Boltzmann equation covariant way): d t " . (5.2) = dλ

5.1.1 Boltzmann equation in the FRW Universe

In the homogeneous isotropic FRW Universe, the particle distribution function only depends on the momentum and time f = f (p, t). The usual procedure of proceeding with Boltzmann equation is rewriting the total derivative in the left-hand-side acting on f (p, t) as d f d t ∂ f dp ∂ f , (5.3) dλ = dλ ∂ t + dλ ∂ p 5.1. KINETIC THEORY IN THE FRW UNIVERSE 3 then use geodesic equation (eq. 2.26) dp = H"p (5.4) dλ − to rewrite d f d f ∂ f ∂ f = " = " Hp . (5.5) dλ d t ∂ t − ∂ p The right hand side of the Boltzmann equation depends on the nature of the interaction. In this chapter, we will see the two body interaction (binary collosion) of 1 + 2 3 + 4 that particle 1 and particle 2 annihilate producing particle 3 and 4, or vice versa. Let us focusing↔ on particle 1. The change of f (r, p, t) from afﬁne parameter t t + δt can be written as ∼ d f f (t + δt) f (t) = δt (R¯ R)δt. (5.6) − d t ≡ − where R¯ and R are the shorthand notation of interaction rate that create (R¯) or destroy (R) particle 1 at position r and momentum p. How do we writhe [f ] = R¯ R for these binary collosions? C − Let us consider a binary collosion happening within a spatial volume d3 r at location r. The number of collosion destroying particle 1 and 2 and creating 3 and 4 within an interval δt can be written as

Rδλ = dN12dP12 34δt, (5.7) → where dN12 is the initial number of colliding pairs (p1,p2):

d4 p d4 p dN f p f p 1 2π δ(1) p2 m2 2 2π δ(1) p2 m2 d3 r. (5.8) 12 = 1( 1) 2( 2) 4 ( ) 1 1 4 ( ) 2 2 (2π) − (2π) − Here, we use the molecular chaos assumption that phase space density of particle species 1, 2 are completely uncorrelated so that expected density of pairings is simply given by multiplying the two densities. 3 3 dP12 34 is the transition rate to the final state within d p3d p4 which is related to the transition matrix →2 2 f i = 3, 4 1, 2 as |M | |〈 |M | 〉| dP12 34 =Idσ → d4 p d4 p 3 2π δ(1) p2 m2 4 2π δ(1) p2 m2 2 2π 4δ(4) p p p p = 4 ( ) 3 3 4 ( ) 4 4 f i ( ) ( 4 + 3 1 2) (2π) − (2π) − |M | − − (1 f3(p3))(1 f4(p4)), (5.9) × ± ± where I = n2(p2) v12 is the incident flux of particle 2 in the particle 1’s rest frame—in the state cat 1, 2 —, and the | in the| last line encodes influence of particle 3 and 4 to the rate due to bose enhancement| 〉 (+) and Pauli± exclusion principle ( ). Note that the expression above is manifestly covariant, as f is a scalar. When the interaction is invariant− under reflections and time reversal,

f i = p3, p4 p1, p2 = p3, p4 p1, p2 = p1, p2 p3, p4 = i f . (5.10) M |M | − − |M | − − |M | M Combining Eqs. (5.8)–(5.9), we have the expression for R as

Z d4 p Z d4 p Z d4 p R g g g 2 2π δ(1) p2 m2 3 2π δ(1) p2 m2 4 2π δ(1) p2 m2 = 2 3 4 4 ( ) 2 2 4 ( ) 3 3 4 ( ) 4 4 (2π) − (2π) − (2π) − 2 (4) f i δ (p4 + p3 p1 p2)f1(p1)f2(p2)(1 f3(p3))(1 f4(p4)). (5.11) × |M | − − ± ± 4 CHAPTER 5. OUT-OF-EQUILIBRIUM AND FREEZE-OUT

Now we can do the mass shell integration2:

Z Z (1) p 2 2 ∞ 0 1 0 2 2 2 1 1 ∞ δ (" p + m ) dp δ( )((p ) p m ) = = = d" (5.13) 2 p 2 2 − 2 | | 0 − | | − " 2 p + m 0 " | | to rewrite R as Z d3 p Z d3 p Z d3 p R g g g 2 3 4 = 2 3 4 3 3 3 (2π) 2"2 (2π) 2"3 (2π) 2"4 2 (4) f i δ (p4 + p3 p1 p2)f1(p1)f2(p2)(1 f3(p3))(1 f4(p4)). (5.14) × |M | − − ± ± 2 2 ¯ Because i f = f i , we calculate R in a similar manner, and ﬁnd the collosion term as |M | |M | g g g Z d3 p Z d3 p Z d3 p f 2 3 4 2 3 4 2π 4δ(4) p p p p 2 [ ] = 3 3 3 ( ) ( 4 + 3 1 2) f i C 2"1 (2π) 2"2 (2π) 2"3 (2π) 2"4 − − |M | f3(p3)f4(p4)(1 f1(p1))(1 f2(p2)) f1(p1)f2(p2)(1 f3(p3))(1 f4(p4)) . (5.15) × ± ± − ± ± Combining the two, we arrive at the ﬁnal form of the Boltzmann equation:

∂ f ∂ f g g g Z d3 p Z d3 p Z d3 p 1 Hp 1 2 3 4 2 3 4 2π 4δ(4) p p p p 2 1 = 3 3 3 ( ) ( 4 + 3 1 2) f i ∂ t − ∂ p1 2"1 (2π) 2"2 (2π) 2"3 (2π) 2"4 − − |M | f3(p3)f4(p4)(1 f1(p1))(1 f2(p2)) f1(p1)f2(p2)(1 f3(p3))(1 f4(p4)) . × ± ± − ± ± (5.16)

5.1.2 Equation for particle number density

To get equation for the number density, we integrate over all momenta:

Z 3 d p1 n1 = g1 f1(p1, t) (5.17) (2π)3 Then, the left hand side becomes

∂ Z d3 p Z d3 p ∂ f p , t g 1 f p , t H g 1 p 1( 1 ) 1 3 1( 1 ) 1 3 1 ∂ t (2π) − (2π) ∂ p1 dn g Z ∂ f p , t dn 1 d a3n 1 H 1 ∞ dp p3 1( 1 ) 1 3Hn ( 1). (5.18) = 2 1 1 = 1 = 3 d t − 2π 0 ∂ p1 d t − a d t The right hand side becomes more symmetric now:

Z d3 p Z d3 p Z d3 p Z d3 p Z d3 p g 1 f g 1 g 2 g 3 g 4 1 3 [ ] = 1 3 2 3 3 3 4 3 (2π) C (2π) 2"1 (2π) 2"2 (2π) 2"3 (2π) 2"4 4 (4) 2 (2π) δ (p4 + p3 p1 p2) f i × − − |M | f3(p3)f4(p4)(1 f1(p1))(1 f2(p2)) f1(p1)f2(p2)(1 f3(p3))(1 f4(p4)) . × ± ± − ± ± (5.19)

2Here, we use Z Z 1 d x 1 X 1 d xδ( )(f (x)) = d y δ( )(y) = . (5.12) d y f x zeros of f x 0( ) ( ) | | 5.2. AFTER DECOUPLING: COLLOSIONLESS BOLTZMANN EQUATION 5

Therefore, the equation for the number density becomes

1 d a3n Z d3 p Z d3 p Z d3 p Z d3 p ( 1) g 1 g 2 g 3 g 4 3 = 1 3 2 3 3 3 4 3 a d t (2π) 2"1 (2π) 2"2 (2π) 2"3 (2π) 2"4 4 (4) 2 (2π) δ (p4 + p3 p1 p2) f i × − − |M | f3(p3)f4(p4)(1 f1(p1))(1 f2(p2)) f1(p1)f2(p2)(1 f3(p3))(1 f4(p4)) . × ± ± − ± ± (5.20)

After some hard work, we just confront a lengthy equation that seems to be quite hard to integrate! Even worse, for the most general cases, we have to write something like Eq. (5.20) for all four particles and solve them at the same time. Yes, a great fun is always there to solve the integro-differential equation.

5.2 After decoupling: collosionless Boltzmann equation

Without diving into the equation above, we can extract an important conclusion from Eq. (5.20) about the decoupled particle species. Let’s consider a particle species, that was in thermal equilibrium state before, but decouples from thermal plazma when temperature of the Universe is T = TD, and scale factor aD. The chemical potential of the particle was µD at the time of decoupling. For the particles decoupled from the rest of the Universe, the right hand side of the Boltzmann equation vanishes, [f ] = 0), and collisionless Boltzmann equation C ∂ f ∂ f Hp = 0, (5.21) ∂ t − ∂ p governs the evolution of the phase space distribution function. This equation, indeed tells that the change of distribution function in this case is solely due to the redshift of the momentum. Let’s see this more directly by solving the equation. In principle, the phase space distribution function of the decoupled species can take any form. Let us describe this generic function by time varying temperature and chemical potential. Then, the Boltzmann equation reads f 2 µ˙ T˙ T˙ p2 a˙ µ " e(" µ)/T 0, (5.22) + + 2 − = T µ − T T " a where dot represent the time derivative. For the particles that were relativistic at decoupling, " p, and the equation reduces to ' µ˙ T˙ T˙ a˙ µ + p + = 0, (5.23) µ − T T a and the only way to satisfy this relation is to having Ta =const., and T/µ =const. that reads

aD a T(a) = TD, µ(a) = µD, (5.24) a aD 2 for a > aD. For particles that are non-relativistic " m + p /(2m) at decoupling, the equation reduces to ' µT˙ p2 T˙ p2 a˙ µ˙ T˙ p2 T˙ a˙ µ˙ + m + + (m µ) + + + 2 = 0, (5.25) − T 2m T " a ' − m µ T 2m T a 2 − to the second order in p . Again, the only way to satisfy this relation for all p is to have (m µ)/T =const., 2 and Ta =const., or − 2 aD T(a) T(a) = TD , µ(a) = m + (µD m) . (5.26) a − TD 6 CHAPTER 5. OUT-OF-EQUILIBRIUM AND FREEZE-OUT

One can easily show that, for both cases, the comoving particle number density is conserved after the decoupling, which is consistent with Eq. (5.20) which says that the comoving number density (a3n) is conserved when the right hand side vanishes. Then, from the equation we have derived in the previous chapter, the comoving entropy of the particle is saparately conserved because

3 µ 3 d(sa ) = d(na ) = 0. (5.27) − T In fact, for non-relativistic particles, this two constraits restrict the time evolution of the tempearture and the chemical potential as you have ﬁgured out in the previous homework.

5.2.1 Prelude for dark matters

What are particles completely decoupled from the thermal bath, and do not interact with photons at all after the decoupling? We call this particles dark matters. Strictly speaking, dark matters do not have to be in thermal contact with plasma at all at early times. But, there is a class of dark matter candidates which do that: WIMPs, Weakly Interacting Massive Particles. These particles had interacted with standard model particles, or among themselves via weak interaction. Almost all—except for those searching for axionic dark matters—the dark matter search programs (either direct or indirect) are looking for the signatures from WIMPs. We normally categorize dark matters into three classes, hot, warm and cold, depending on its role in the large-scale structure formation. The usual criteria is the smallest structure that the dark matter can form: cold dark matter can form structures all the way down to the sub-galactic scales, warm dark matters form structures from galactic scales, while hot dark matters can only form structures above the galactic scales. When applying this criteria to the WIMPs that decouples at early time, WIMPs become

cold dark matter if it is already non-relativistic at the time of decoupling, • warm dark matter if it is relativistic at the time of decoupling, but non-relativistic now, • hot dark matter if it has been relativistic for entire history of the Universe until now (or very • recently).

Although the critera above does not exactly coincide with the earlier deﬁnition, the division captures important physical characteristics of each category: hot dark matter ﬁrst form a huge overdensity then fragmented into smaller pieces, warm dark matter can form galactic size halos but no/small substruc- tures, cold dark matter is responsible for all the subgalactic scale minihalos. One obvious candidate for warm/hot dark matters is neutrinos: the massless neutrinos are the perfect candidate for the hot dark matter, and the massive neutrinos can be warm/hot dark matter candidates depending on the mass. There is no cold dark matter candidates in the standard model of particle physics, but going beyond the SM, there are massive right-handed neutrinos and LSP (lightest super- symmetic particles) or neutralino (electric neutral super-partners of gauge bosons) as candidates for cold dark matter. Neutralinos are mixture of superpartners of the photon, Z0 boson, and Higgs bosons (called photino, zino, higgsino, respectively), spin-1/2, and Majorana—i.e., it is its own antiparticle.

5.2.2 Massive neutrinos

In the standard model of particle physics, neutrinos are assumed to be massless particles. But, now we know that neutrinos must have mass from the neutrino oscillation observed through solar (νe from p p − 5.2. AFTER DECOUPLING: COLLOSIONLESS BOLTZMANN EQUATION 7

4 + chains —- 4p He + 2e + 2νe —-; Kamiokande, Sudbury Neutrino Observatory, etc), atmospheric → (π± µ± + νµ(ν¯µ), or π± µ± e± + νe(ν¯e) + ν¯µ(νµ); Kamiokande, Super-Kamiokande, etc), → + + → → accelerator (µ e νeν¯µ; K2K, T2K, MINOS), and reactor (ν¯e, KamLAND, Daya Bay, RENO, Double Chooz) neutrinnos.→ This clearly is one of the important clues to find out the physics beyond the standard model of particle physics. The neutrino oscillation is a result of the neutrino flavor mixing. That is, the flavor (e, µ, τ) state is not the eigenstate of the Hamiltonian (time evolution operator), and can be in general written as the linear combination of the neutrinos with mass mj: X νl L = Ujl νjL . (5.28) | 〉 j | 〉

Here, νjL stands for the mass eigenstates (massive neutrinos) and Ujl is the unitary matrix called the Pontecorvo-Maki-Nakagawa-Sakata| 〉 (PMNS) mixing matrix. From the neutrino oscillation data (if we have three massive neutrino states), we ﬁnd the mass gap between three mass states as (1 σ, 68% C.L.) range of3: −

2 2 2 5 2 ∆m21 =m2 m1 7.53 0.18 10− eV , (5.29) 2 2− 2' ± × 3 2 ∆m32 = m3 m2 2.52 0.07 10− eV (normal hierarchy : m3 > m2 > m1) | | | − | ' ± × 3 2 2.44 0.06 10− eV (inverted hierarchy : m2 > m1 > m3). (5.30) ' ± × Note that we cannot observe the absolute mass of the neutrinos from neutrino oscillation as the ﬂavor 2 changing probability only depends on the square of mass gap. ∆m21 is measured from the solar neutri- 2 nos, and ∆m32 is measured from the atmospheric neutrinos. The mass gap, however, can be translated as the lower bound of the total mass of the massive neutrinos by setting the mass of the lightest neutrino species to zero. They are X 0.067(0.106) eV < mνi, (5.31) i for normal (inverted) hierarchy spectrum of neutrino mass. The upper bound comes from the tritium beta decay4, by very accurate measurement of kinematics, that X mνi < 2 eV mνi < 6 eV. (5.33) → i

As we will discuss in the later chpaters, there might be a chance to detect the absolute mass of the neutrinos from cosmological observations and put a tighter constraint. Also, the next generation of β- spectroscopy experiments (e.g. KATRIN in Germany) are aiming at reaching a sensitivity limit m(νe) < 0.2 eV, a cosmologically relevant range.

3The numbers come from PDG http://pdg.lbl.gov 4Tritium is a radioactive material with half life 12.32 yrs. Its beta decay

3 1+ T He + e− + ν¯e (5.32) → releases only 18.6 keV of energy shared by electron and anti-neutrino, and therefore can be used as a probe of the neutrino mass. Depending on the mass of neutrinos, the high-energy tail of the emitted electron energy distribution have a cut-off at

18.6 keV mνi . This kind of experiment is hard as we have to read off the high-energy tail of the electron distribution, which is very unusual.− 8 CHAPTER 5. OUT-OF-EQUILIBRIUM AND FREEZE-OUT

But, for our purpose in this section, it is important to note that the lower limit of the mass is greater than the temperature of the neutrino

Tν = 1.946 K 0.168 meV 34(53) meV < mνi, (5.34) ' at its absolute minimun when one of the three mass state is massless (number in the paranthesis is for the inverted hierarchy). It means neutrinos are non-relativistic particles at present epoch! As we have seen in the previous chapter, cosmic background neutrinos have decoupled from the thermal plasma at

Tν dec. 1.5 MeV, and free-streaming since then. Because the neutrinos were relativistic at the time of − ' decoupling (mass upper bound (2 eV) Tν dec.), the distribution function freezes as the form, − 1 f (pD) = (5.35) e(pD µD)/TD 1 − + and all that the expansion of the Universe changes is redshifting the neutrino momentum as p = | | aD/a pD , to yield the distribution function as the same Fermi-Dirac function | | 1 f (p) = , (5.36) e(p µ)/T 1 − + with T = (aD/a)TD and µ = (aD/a)µD. What is the chemical potential of the neutrinos? As we have seen in the previous chapter, chemical potential is related to the particle-antiparticle asymmetry. With neutrino-antineutrino asymmetry, the total energy density of neutrinos is different from the standard value, and this effect allow us to bound— from BBN, for example—the neutrino chemical potential to be (at 2σ) µ 0.05 < ν < 0.07. (5.37) − Tν Note that the mixing of neutrinos leads to equilibration of asymmetries in different ﬂavors: so, we have only one number here. Therefore, we can safely neglect the chemical potential of the neutrinos. Then, the number of neutrinos is the same as the number of anti-neutrinos, and we have

gνi 3 3 ?n Tνi 3 4 3 3 nνi = γ nγ = 411 cm− 112.09 cm− (5.38) 4 g?n T 4 × 11 × ' neutrinos per species. Then, because the number density of all three species are the same, the neutrino mass density at present is X X 3 ρν = nνi mνi = 112.09 mνicm− , (5.39) i i from which we calculate the density parameter for the neutrino as P 2 ρν 2 i mνi Ωνh = h = . (5.40) ρcrit 94.02 eV

By using the known bound from the terrestrial experiments, we can ﬁnd the bound of the neutrino density parameter as 4 3 2 7.13 10− (1.13 10− ) < Ωνh < 0.0638. (5.41) × × 2 Again, parenthesis is for the inverted hierarchy. As the best-ﬁtting mass density from Planck is Ωdmh 0.12, neutrinos are just not enough to explain the observed dark matter energy density. In terms of the' energy density, it can be at most a half of the total dark matter energy budget. Looking at the opposite 5.3. BOLTZMANN EQUATION SIMPLIFIED 9

2 2 angle, one can ﬁnd an upper limit on the neutrino mass from Ωνh < Ωdmh 0.12 that is translated as '

mνi < 3.76 eV (5.42) for the mass of a neutrino species when neutrino oscillation equalize mass of three neutrino species. This cosmological bound on the neutrino mass is called Gershtein-Zeldovich bound5. This bound is just the very ﬁrst cosmological bound. The observables in the large-scale structure of the Universe such as galaxy power spectrum and angular power spectrum of CMB put more stringent upper limit on the neutrino mass. For example, from Planck alone, the 1 σ (68 % C.L.) upper bound is − X mνi < 0.403 eV. (5.43) i

A recent review [?] precents a thorough overview about the effect of massive neutrinos on cosmological observations. We will come back to this issue when we discuss the large-scale structure, but basically the constraint comes about because neutrinos are warm dark matter. For warm dark matter, perturbations cannot grow until they become non-relativistic Tν < mν, and warm dark matters generate quite different large-scale structure than cold dark matter, which is the main guest of honor in the next section.

5.3 Boltzmann equation simpliﬁed

Now, let’s go back to the full Boltzmann equation in Eq. (5.20). To be exact, this integro-differential equations must be solved for all particle species 1, 2, 3 and 4 that are involved in the interaction. Fortunately, however, for the cases that we are interested in, Eq. (5.20) can be reduced to quite a simple form, as we make following simpliﬁcation6:

We don’t need to solve them for all four particles involved in the collosion process at all, because we • will only consider a particle/process being non-equilibrium at a time. That is, other three particle species are safely in the thermal equilibrium state. Also, for many cases the particle/process that we are interested in is maintaining kinetic equilibrium, but goes out-of-equilibrium by departure from chemical equilibrium (that is, inverse reaction of chemical equation stops happening, but still there can be frequent scattering to maintain them in the same temperature).

Because we are typically interested in the case where (" µ) ¦ T where the spin-statistics • (Fermi-Dirac and Bose-Einstein) is not important, and the distribution− function follows Maxwell- Boltzmann statistics. Therefore, we can neglect the Bose enhancement and Pauli blocking term altogether, and the phase-space distribution function becomes

("(p) µ)/T µ/T (0) f (p) e− − e f (p). (5.44) ' ≡ Using this approximation, the number density of particle species i can be written as • Z 3 Z 3 d p d p 0 0 n g f p eµi g f ( ) p eµi /T n( ), (5.45) i = i 3 i( ) = i 3 i ( ) i (2π) (2π) ≡

(0) where ni is the equilibrium number density of the species i without chemical potential.

5 Gershtein S S, Zel’dovich Ya B Pis’ma Zh. Eksp. Teor. Fiz. 4 174 (1966) [English translation JETP Lett. 4 120 (1966)] 6Here, we follow the logic in the textbooks of Kolb & Turner and Dodelson. 10 CHAPTER 5. OUT-OF-EQUILIBRIUM AND FREEZE-OUT

With these three simplification, the only unknown that we want to figure out is the chemical potential (0) µi, or equivalently, the number density ni (they are related by the known number density ni at a given temperature). Let’s see how it works. First, using Maxwell-Boltzmann distribution function, f3(p3)f4(p4)(1 f1(p1))(1 f2(p2)) f1(p1)f2(p2)(1 f3(p3))(1 f4(p4)) ± ± − ±  ±  n n n n ("3+"4)/T (µ3+µ4)/T ("1+"2)/T (µ1+µ2)/T (0) (0) 3 4 1 2 e e e e f p1 f p2   , (5.46) − − = 1 ( ) 1 ( ) (0) (0) (0) (0) ' − n3 n4 − n1 n2 and the definition of average σv as 〈 〉 4 1 Y Z d3 p σv g i 2π 4δ(4) p p p p 2 f (0) p f (0) p . (5.47) 0 0 i 3 ( ) ( 4 + 3 1 2) f i 1 ( 1) 2 ( 2) ( ) ( ) 2π 2"i 〈 〉 ≡ n1 n2 i=1 ( ) − − |M | We simplify the Boltzmann equation as   3 n n 1 d(n1a ) (0) (0) 3 4 n1n2 = σv n n   , (5.48) a3 d t 1 2 (0) (0) (0) (0) 〈 〉 n3 n4 − n1 n2 which is much simpler than before, and is indeed a simple ordinary differential equation for the number densities for a given weighted average of the interaction cross section σ and relative velocity v. Let’s digest the simplified equation. This equation contains the time scale argument that we had before. The left-hand side is of order n1/t n1H, as Hubble time is the typical time scale that the number ' density of particle has changed. The right-hand side is of order n1n2 σv n1ΓI , which is the interaction 〈 〉' rate per volume. First, when interaction rate is larger than the Hubble expansion rate ΓI H, the indivisual terms in the right hand side is much larger than the left hand side. To make the equality, the large terms must cancel each other to yield n n n n 3 4 1 2 , (5.49) (0) (0) = (0) (0) n3 n4 n1 n2 which is nothing but a condition of chemical equilibrium! You can clearly see this from Eq. (5.45). One could also derive the same equation from the detailed balance as the time derivative in the left-hand side must vanish in the equilibrium state.

At the other extreme, when the interaction rate is smaller than the expansion rate ΓI H, we can neglect the right hand side all together, and the situation is the same as no-interaction case where the 3 comoving number density n1a is conserved. Now that we have our master equation, we are ready to tackle some interesting out-of-equilibrium events. In the following sections of this chapter, we will apply this equation to three such cases: thermal decoupling of WIMPs, Big-bang nucleosynthesis and CMB recombination.

5.4 Weakly-Interacting-Massive-Particles

Because of the weak but non-zero interactions,

¯ ψ + ψ X + X¯, (5.50) 5.4. WEAKLY-INTERACTING-MASSIVE-PARTICLES 11

+ the WIMPs were in thermal equilibrium state with cosmic plasma (consists of particle X = γ, e−/e , ) at earlier times by means of pair creation/annihilation. But, at some point the interaction rate drops··· below the Hubble expansion rate to freeze out the interaction, and WIMP abundance. Let us assume that the annihilation products X are in local thermodynamical equilibrium state with other particles because of interactions other than the one we consider. This should be the case for the standard model particles which participate the electromagnetic interactions, e.g. photons, electrons, etc. Also, for simplicity, let us further assume that the chemical potential for dark matters is negligible µψ 1 so that nψ = nψ¯. This is ceratinly true for the Majorana particles, when they pair-annihilate to some particle and antiparticle pairs. For Dirac particles, this means that the number density of particles and antiparticles are the same: nψ = nψ¯. In this case, the Eq. (5.48) becomes

3 1 d(nψa ) h 2i σv n2 neq . (5.51) 3 = ψ ψ a d t − 〈 〉 − That is, we have an equation only with the equilibrium number density of the dark matter which de- termines the source term. For the Majorana particles, because we create/remove two particles per interaction, one might think that we need a symmetric factor of 2. But, that factor of 2 is cancel by that for total N identical particles, there are N(N 1)/2 different ways of pairing (to annihilate) them. − Because the comoving number density of dark matters is conserved after decoupling, it is convenient to deﬁne a new variable nψ Y , (5.52) ≡ s which takes a constant value after the decoupling. In terms of Y , we rewrite the equation as

dY 2 2 = σv s Y Yeq . (5.53) d t − 〈 〉 − Finally, we use x = mψ/T as a time variable, which is related to the cosmic time from the Friedmann equation (here, we assume that g? varies slowly): 1 45 1/2 1 m m x2 t 0.3012g 1/2 Pl 0.3012g 1/2 Pl x2 , (5.54) = = 3 2 ?− 2 = ?− 2 2H 16π GN g?(T) T ' T m ≡ 2H(m, x) 19 p where mPl = 1.2211 10− GeV = 1/ GN is the Planck mass, and we deﬁne the function H(m, x) = 1/2 2 × 2 1.6602g? (m /mPl) so that H = H(m, x)x− . Note that the x-dependence of H(m, x) comes only through the time evolution of g?(T), but we can practically ignore (tolerating % level error) that by ∼ using the approximation that g? varies only slowly. With this in mind, but not forget that we need to put x-dependency back to achieve the % accuracy, we will write H(m, x) = H(m) hereafter. Then, we trade the time derivative to ∼ d d x d H m d ( ) , (5.55) d t = d t d x = x d x then transform the Eq. (5.51) as dY x σv s 2 2 = 〈 〉 Y Yeq . (5.56) d x − H(m) − We can recast Eq. (5.56) once more to the following suggestive form 2 x dY Γψ Y = 1 , (5.57) Yeq d x − H Yeq − 12 CHAPTER 5. OUT-OF-EQUILIBRIUM AND FREEZE-OUT

eq with Γψ = nψ σv being the annihilation rate. This equation, again, tells that the ratio Γψ/H is the key 〈 〉 parameter to quantify the effectiveness of the annihilation, in a sense that when Γψ ® H, the annihilation stop being efﬁcient and the dark matter fraction Y = nψ/s freezes out.

5.4.1 Hot/warm relic dark matter

For the hot/warm dark matter WIMPs, which decouples when they are still relativistic mψ T, the decoupling procedure is exactly same as what happens to the neutrinos at T 1.5 MeV : once the decoupling temperature has reached, the particle species thermally decouples with' the relativistic particle number density at the time of decoupling. Because the number density of relativistic particles is solely determined by the temperature, as long as the decoupling happens during the time g?s =constant, the equilibrium number density are the same as the actual number density and the right hand side of Eq. (5.51) vanishes: 45ζ 3 gψn gψn Y Y ( ) 0.2777 , (5.58) = eq = 4 2π g?s ' g?s where gψn = gψ for bosons and 3/4gψ for fermion dark matters. Then, the relic abundance Y ∞ ≡ Y (x ) is also given by the Yeq at freeze-out time x f . → ∞ gψn Y = Yeq(x f ) = 0.2777 . (5.59) ∞ g?s(x f )

Using this, we also calculate the number density of hot/warm relic at present time as g ψn 3 nψ0 = s0Y = 803.4 cm− , (5.60) ∞ g?s(x f ) using 2π2 s g T 3 2893.1 cm 3, (5.61) 0 = 45 ?s0 cmb = − with T = 2.726 K. Note that the relic density fo hot/warm dark matter is very insensitive to the freezeout temperature Tf = m/x f as the dependence comes only through g?s. The energy density of hot/warm dark matter can then be written with the mass m as g ψn m 3 ρψ0 = mnψ0 = 803.4 eV/cm , (5.62) g?s(x f ) eV which corresponds to the density parameter of g 2 ψn m Ωψh = 0.07623 . (5.63) g?s(x f ) eV

Because the hot/warm dark matter can be at most the total matter density, we can ﬁnd the upper bound 2 for the mass from Ωψh ® 0.12: g?s(x f ) m ® 1.5742 eV. (5.64) gψn Also, to be counted as dark matter, the particle must be non-relativistic at present time:

1/3 1/3 af 3.909 3.909 m ¦ Tψ0 = Tψf = Tcmb = 0.2350 meV. (5.65) a0 g?s(x f ) g?s(x f ) 5.4. WEAKLY-INTERACTING-MASSIVE-PARTICLES 13

This gives us a mass bound for the hot/warm relics as 1/3 g x 4 3.909 m ?s( f ) 2.350 10− ® ® 1.5742 . (5.66) × g?s(x f ) eV gψn

As we have discussed earlier, neutrinos are the most well-known example of the warm dark matter. Neutrinos decoupled at around 1.5 MeV where g?s = 10.75, with gψn = 1.5, we reproduce the upper bound that we found earlier: X mνi < 11.28 eV. (5.67) i If hot/warm dark matter interacts with heavier gauge bosons, e.g. one in the supersymmetric models, 2 the weak interaction cross section can be modiﬁed as GF GF (MW /M) , and the decoupling tempeara- 2/3 → 4/3 ture can be increased by a factor of Tf GF− (M/MW ) . For example, the heavy gauge boson ∝ ∝ of mass M ¦ 560 TeV can lead to the decoupling temperature of Tf ¦ 200 GeV where g?s = 106.75, and the mass bound becomes (for spin-1/2 fermions)

m < 112.03 eV. (5.68)

These hypothetical dark matter particles turned to non-relativistic in the radiation dominated epoch.

5.4.2 Cold relic dark matter

For the cold dark matter that decouples when T ® mψ, the situation becomes a bit more complicated as the equilibrium number density Yeq of non-relativistic particles depends exponentially on the temperature for x = m/T 1: Z 45ζ 3 gψn 45 gψ u2 x2 Y x ( ) ∞ udup eq( ) = 4 = 4 u − 2π g?s 4π g?s x e 1 90 gψ gψ± x3/2e x 0.1447 x3/2e x , (5.69) 7/2 − − →(2π) g?s(x) ' g?s(x) using the number density of non-relativisitic particles and entropy density:

2 3/2 2 m x 2π 3 3 nψ = gψ e− , s = g?sm x− . (5.70) (2π)x 45 That is, when the temperature drops below the mass of the dark matter, the number density of dark matter particles drops drastically to reduce the interaction (annihilation) rate, which is proportional to the number density. Therefore, if interaction rate is not too strong, the annihilation interaction rate freezes out around T (1)mψ and it leaves some relic that can survive from exponentially suppressed equilibrium number density'O 7. Let us start from the annihilation cross section, which must take the form of n T n σv σ0 = σ0 x− , (5.71) 〈 〉 ≡ m 7As we have seen in the previous chapter, non-relativistic baryons and leptons are thermally coupled to plasma (therefore also have to suffer from the exponential suppression) but still have non-zero number density for those particles because they 10 have the chemical potential of order µ/T nb/s 10− from the beginning. We don’t yet have a concrete theory about how to get such a large chemical potential from' the fundamental' theory, and this is an active research area called baryogenesis. We will come back to this later in this section. 14 CHAPTER 5. OUT-OF-EQUILIBRIUM AND FREEZE-OUT

2 10−

3 10−

4 10−

s 10− /

ψ n

= 6

Y 10−

7 Yeq 10− Y(λ = 105)

8 7 10− Y(λ = 10 ) Y(λ = 109) 9 10− 100 101 102 x = m/T

5 7 Figure 5.1: Freeze-out of cold relics abundance Y = nψ/s as a function of x = m/T for λ = 10 , 10 9 and 10 (red, green, blue, respectively) along with the decoupling temperature x f = m/Tf shown as a vertical dashed lines with the same color. Black dashed line show the case for thermal equilibrium which suffers from strong—exponential— suppression for x 1.

1/2 for non-relativistic dark matter particles (T ® m) where v T . Here, n depends on the decay channel: n = 0 for s-wave, n = 1 for p-wave, etc. Then, we〈 rewrite〉 ∝ the Boltzmann equation Eq. (5.56) as dY n 2 2 2 = λx− − Y Yeq , (5.72) d x − − with x σv s σ s m g λ 0 ( ) 0.2642 ?s mm σ (5.73) H〈 m〉 = H m g Pl 0 ≡ ( ) x=1 ( ) ' p ? T=m a constant depending on the mass of dark matter, and g?s, pg?, and σv measured at T = m. 〈 〉 We can then integrate the equation for a given value of λ and m. As an example, let us consider the dark 5 9 matter with gψ = 2, mψ ¦ 100 GeV (this makes g?s g? 106.75), and λ = 10 ∼ . We show the result ' ' of numerically integrating Eq. (5.72) in Fig. 5.1. As we expected earlier, for x < x f , the abundance Y closely follows the equilibrium values, while it freezes out after the decoupling and asymptotes to the constant value Y . As increasing λ, decoupling temperature decreases to reduce the ﬁnal relic abundance. ∞

Let’s estimate the freezeout temperature and the abundance Y ﬁrst. Freeze-out happens at Γψ(x f ) = ∞ H(x f ). Transforming the equation to the form of Eq. (5.57),

2 x dY n 1 Y = λYeq x− − 1 (5.74) Yeq d x − Yeq − 5.4. WEAKLY-INTERACTING-MASSIVE-PARTICLES 15

we ﬁnd the criteria of x f as

Γψ x 1 λY x x n 1 H ( f ) = = eq( f ) −f − gψ gψ n+1/2 x f n+1/2 x f =0.1447λ x−f e− = 0.03823 mmPlσ0 x−f e− . (5.75) g?s pg?

Note that here we drop all the time dependence of g?i’s as the time scale that is relevant here is usually much smaller than the time scale that g?i are varying (which is of order tens of Hubble time). We take the logarithm of both side, gψ 1 ln 0.03823 mmPlσ0 = n ln x f + x f , (5.76) pg? − 2 which can be iteratively solved to yield gψ 1 gψ x f = ln 0.03823 mmPlσ0 n ln ln 0.03823 mmPlσ0 + . (5.77) pg? − − 2 pg? ···

After the freeze-out, the actual number density is much bigger than the equilibrium number density Y Yeq and the differential equation becomes dY n 2 2 = λx− − Y , (5.78) d x − which can be integrated as 1 1 λ n 1 = x−f − . (5.79) Yf − Y − n + 1 ∞ Typically Yf is signiﬁcantly larger than Y , as the number density further decays even after the freeze- out time, then we ﬁnd ∞

3.785 n 1 x n+1 1/2 n 1 ( + ) f 3.785(n + 1) g /g?s x f Y + x n+1 ? (5.80) = λ f 1/2 = mm σv ∞ ' g?s/g? mmPlσ0 Pl f 〈 〉 where σv f refers to the value at freezeout. Note that the relic density is larger for smaller cross- section.〈 It is〉 because the smaller cross section means earlier decoupling, which saves the dark matters out of the equilibrium density (which is exponentially suppressed) earlier. Dark matters with larger annihilation cross-section, on the other hand, decouple later and yield smaller relic abundance. The estimate of Y is directly related to the dark matter abundance and number density: ∞ 1/2 n+1 n 1 g /g?s x 4 ( + ) ? f 3 nψ0 =s0Y = 1.095 10 cm− , (5.81) ∞ × mmPlσ0 1/2 n+1 n 1 g /g?s x 4 ( + ) ? f 1 3 ρψ0 =mnψ0 = 1.095 10 eV/cm , (5.82) × mPlσ0 eV that corresponds to the density parameter of

n 1 g1/2/g x 2 9 ( + ) ? ?s f 1 Ωψ0h = 1.039 10 GeV− . (5.83) × mPl σv 〈 〉 16 CHAPTER 5. OUT-OF-EQUILIBRIUM AND FREEZE-OUT

Fourth generation massive neutrinos

As an example of the cold relic, let us consider the fourth generation massive neutrinos with m MeV which is non-relativistic at decoupling (x f ¦ 1). The annihilation cross-section depends on the nature of particle: for the Dirac type neutrinos, annihilations process is dominated by s-wave (n = 0) with 2 2 2 10 m 2 σ0 GF m = 1.36044 10− GeV− , (5.84) ' × GeV 9 3 then λ = 3.400 10 mGeV, and the freezeout happens at (using gψ = 2, g? 60) × ' 1 x f 16.613 + 3 ln mGeV + ln (16.613 + 3 ln mGeV) 18.02 + 3 ln mGeV, (5.85) ' 2 ' which gives the WIMP number density 18.02 3 ln m 18.02 3 ln m n 7 + GeV 3 n + GeV ψ0 8.5096 10− 3 cm− 3.744 p0 3 , (5.86) ' × mGeV ' mGeV Here, we use the proton density of Y 2 p ρcritΩb 7 Ωbh 3 np0 = 1 = 2.273 10− cm− , (5.87) − 2 mp × 0.023 with Yp = 0.24. The density parameter for the cold relic is 18.02 3 ln m 1.455 1 h2 + GeV m Ωψ0 0.08074 2 2 1 + ln GeV , (5.88) ' mGeV ' mGeV 6 because we have the same amount of anti-particles, we have 2.91 h2 Ωψψ¯0 2 . (5.89) ' mGeV The relic density in this case gets smaller for more massive particles, just because cross-section becomes bigger for the larger mass. 2 From the measured dark matter density parameter Ωdmh < 0.12, we have a ﬁrm lower bound of the right-handed neutrinos as m ¦ 4.9 GeV. (5.90) This bound is called Lee-Weinberg bound [?], although it was discovered many others (all published in the same year, 1997) including P.Hut, K. Sato and, of course, Y. Zel’dovich.

WIMP miracle

2 To get the observed cold dark matter fraction of Ωcdmh 0.12, we need ' n 1 g1/2/g x 2 9 ( + ) ? ?s f 1 Ωψ0h = 1.039 10 GeV− 0.12, (5.91) × mPl σv ' 〈 〉 or, equivalently,

10 1/2 2 27 1/2 3 σv 7.0906 10− (n + 1)g? /g?s x f GeV− 8.2824 10− (n + 1)g? /g?s x f cm /s. (5.92) 〈 〉' × ' × 5.5. BIG-BANG NUCLEOSYNTHESIS 17

2 2 8 2 For a weak scale mass m 100 GeV, σ0 αW /m 10− GeV− , then x f 25. Having g?s g? 106.75, we need (for s-wave' annihilate), ' ' ' ' '

26 3 9 2 σv 2.004 10− cm /s 1.713 10− GeV− , (5.93) 〈 〉' × ' × 8 2 which is remarkably close to the cross section at the weak-scale σ0 10− GeV− . ' That is, if there is new physics at the electroweak scale that involves the introduction of a new stable neutral particle, then the particle must have a relic density in the ballpark of that required to explain the cold dark matter density. It’s truely a remarkable coincidence as there is no a priori reason that a constant in the elementary particle physics (the weak scale) has anything to do with the dark matter density in the unit of critical density, which is set by the expansion rate of the Universe. Completely different scale, and completely different physics! This is called a WIMP miracle, and this coincidence led a number of theorists and experimentalists to take the idea of WIMP dark matter very seriously!

5.4.3 Need for baryogenesis

Another application of this calculation is the freeze-out of baryon-antibaryon. Consider the Universe begins without any baryon asymmetry. With the exactly same logic that we study in this section, when annihilation rate drops below the Hubble expansion rate (happens after baryons turns non- relativistic), baryon-antibaryon pair annihilation freezes out and we have a ﬁnite relic density of baryons 2 26 2 2 and antibaryons. The annihilation cross section is σann v 1 fm = 10− cm 25.683 GeV− , with 〈 〉' 19 ' m 1 GeV, it makes g? 60 and we ﬁnd x f ln(0.03823(2/p60)(10 )(25)) 42, then Tf 23 MeV. ' ' ' ' ' That is, 10 decades increase in the cross section increases x f by a factor of 2. But, that means that the 20 relic abundance decrease by twice longer time in the exponent to yield Y = nb/s = n¯b/s 6.5 10− ! This, of course is 9 orders of magnitude too small compared to the observed∞ baryon-to-photon' × ratio. Therefore, there must have been an initial baryon asymmetry, and the abundance of antibaryons should be entirely neglibible today.

5.5 Big-bang Nucleosynthesis

The next case of the freeze-out that we consider is Big-Bang Nucleosynthesis. The observed Helium abundance from the metal-poor part of the Universe suggests that the primordial Helium abundance is about 24% of total baryonic mass: Yp = 4nHe/nb 0.24. ' What does this number implies about the formation of Helium nucleus? Let us suppose that all the Helium nuclei in the Universe is formed in the core of the stars. Becase the binind energy of a Helium 5 nucleus is 28.3 MeV, 7.1 MeV ( 1.14 10− erg) of energy is realeased per baryon (mb mp 1.67 24 ' × ' ' × 10− g) when Helium nucleus is formed. If we assumes that this process has been converted a quarter 17 of baryons to Heliums during the past ten billion years (the Hubble time, tH 4.41 10 s), then the average baryonic mass-to-light ratio must be ' ×

M M 1.67 10 24 g g M 4 − 0.258 0.5 . (5.94) L = E/t /4 1.14 10 5erg×/ 4.41 1017 s erg/s L ( H ) ' − ( ) ' ' × × This is far much smaller than the observed value of M/L ¦ 20M /L . Therefore, only a small fraction of Helium can form inside the stars, and majority of Helium nuclei must form at early time as the cosmological event called Big-bang nucleosynthesis (BBN)! 18 CHAPTER 5. OUT-OF-EQUILIBRIUM AND FREEZE-OUT

By the way, where are the fusion energy of BBN now? Because the number density of neutron is about 9 10− of the photon number density, and BBN happens at around 0.1 MeV scale, the fusion energy released from BBN is only a small fraction of the total radiation energy. At this early time, free-free emission and double-compton scattering is so efficient that any energy release can be very efficiently thermalized to absorb and digest the fusion energy as a blackbody radiation field that we observe today as Cosmic Microwave Background. BBN happens around T 0.1 MeV which happens a few minutes after the Big-bang. As end products, BBN forms the light nuclei' such as D, T, 3He, 4He, 7Li, etc, whose primordial abundance can be measured from the observations such as QSO absorption lines (D), 3.46 cm spin-flip transition (3He) as well as the abundance in the most metal poor environments (4He and 7Li). The dominant processes of forming those elements is the two-body interaction of protons and neutrons and their compound nuclei like Deutrium and Tritium:

n + p D + γ → 3 D + D He + n → 3 D + p He → D + D T + p 3 → He + n T + p 3 → 4 He + D He + p → 4 T + D He + n 3 4 → 7 He + He Be 4 → 7 He + T Li 7 → 4 4 Li + p He + He 7 → 7 Be + n Li + p. (5.95) →

Using these interactions [?] find out analytic approximation to the final light element abundance from BBN (see also [?] for the fixed point approach). General BBN calculation follows three steps:

[1] n/p-freeze out, [2] Deutrium bottle-neck, [3] onset of light-element formation.

It is easy to see from the Chemical reactions in Eq. (5.95) that Deutrium is the key element of this process: Deutriums are needed to form Tritium and Helium-3 that are required to form Helium-4. Therefore, if there is not enough Deutriums in the Universe, the reaction rate will not be enough to form other nuclei. It is only after the Deutrium abundance becomes large enough that the Deutrium bottle-neck is opened to form elements like T, 3He and 4He. The abundance of Deutrium is sensitive to the number density of protons and neutrons from which the Deutrium is formed. Therefore, we shall start our study on BBN from neutrons and protons that set up the initial condition. In this section, we discuss each item in a greater detail. 5.5. BIG-BANG NUCLEOSYNTHESIS 19

5.5.1 Setting up the initial condition: neutron freeze-out

Even without any two-body interactions, neutrons can β-decay into protons:

n p + e− + ν¯e, (5.96) → with lifetime of τn = 880.3 s. The mass of neutron is mn 939.5654 MeV and of proton is mp ' ' 938.2720 MeV, therefore the β-decay releases Q mn mp 1.2933 MeV. But, in the energy scale ≡ − ' that we are in here, T 1 0.5 MeV, age of the Universe is t ® 10 second after the Big-Bang and the β-decay of neutrons is' not∼ yet important. At these high temperatures, the following weak interactions are dominant channel to convert neutrons to protons and vice versa:

n + νe p + e− + n + e p + ν¯e, (5.97) whose combined interaction rate is given8 by

5 2 255 T Q Q 1 λn p = + 6 + 12 s− , (5.98) → τn Q T T

1 1 which is about 5.5 s− at T = Q = 1.2933 MeV, faster than the expansion rate (H 1.133 s− ) at T = Q. Such a fast interaction rate leads the neutron to proton ratio to be the equilibrium' value:

(eq) 3/2 nn nn mn m m /T µ µ /T Q/T µ µ /T Q/T e ( n p) e( n p) e e( e νe ) e . (5.99) = eq = − − − − − − np ( ) mp T¦Q np ' ' Here, I use the chemical equilbrium condition . This means that number density of µn µe = µp µνe neutrons are the same as the number density of protons− at high− temperature T Q. Like the previous section of dark matter, let us deﬁne the dimensionless quantity Xn as

nn Xn = , (5.100) nn + np and use Q x (5.101) = T as the time variable. The equilibrium condition becomes

1 X x 1 X eq . (5.102) n( ® ) = n x ' e + 1 We the apply Eq. (5.48) to the case at hand to ﬁnd out the evolution equation for neutron number density: 3 (eq) ! 1 d(a nn) nn Q/T = λn p nn np = λn p nn e np , (5.103) a3 d t (eq) − − → − np − → −

8 Because the temperature is greater than the mass of electron T > me 0.511 MeV, one can safely assume that the number density of non-relativistic particles (neutrons and protons) are much' smaller than the relativistic particles (electrons n n n n and neutrinos): n, p e, νe . 20 CHAPTER 5. OUT-OF-EQUILIBRIUM AND FREEZE-OUT

time t sec 100 101 102 100 ) p n + n

n 1 ( 10− / n n = n X

eq Xn

Xn(T )

Xn(T)+ β-decay 2 10− 0 1 10 10− T[MeV]

Figure 5.2: Freeze-out of neutron fraction (Xn) to total baryon density.

P where λn p = n σv is the total interaction rate that sums over the two reactions, and we use the detailed balance→ to〈 ﬁx〉 the ratio between the source term and the sink term of the neutrons. Using nn = (nn + np)Xn, we rewrite the equation in terms of Xn,

λ x λ x dXn n p x n p x eq = → Xn e− (1 Xn) = → 1 + e− Xn Xn . (5.104) d x − H(Q) − − − H(Q) − Here, we convert the time derivative to the x derivative by using Eq. (5.55) with

1/2 2 16 1 H(Q) = 1.6602g? (Q /mPl) 7.4561 10− MeV 1.133 s− (5.105) ' × ' 2 is the Hubble parameter at T = Q: H(x) = H(Q)x− . So, we end up with an ordinary differential equation Eq. (5.104) describing the evolution of the neutron fraction Xn, that we can solve with an initial condition of Eq. (5.102). We show the result of numerical integration in Fig. 5.2 The asymptotic value that we ﬁnd numerically is Xn∞ 0.147617 when we ignore the β-decay that follows the freeze- out that further reduce the neutron fraction' as:

t/τn t/τn Xn(t > t f ) Xn∞e− 0.147617e− . (5.106) ' '

Let us estimate the freeze-out temperature of the neutrons. The story goes pretty much similar to what we have discussed earlier for the dark matter. As temperature drops, the interaction rate reduces sharply so that it eventually falls below the Hubble expansion rate when λn p = H, or →

255 2 2 5 x f + 6x f + 12 = 1.133x−f , (5.107) τn x f 5.5. BIG-BANG NUCLEOSYNTHESIS 21 which yields Q x f = 1.9056 Tf = = 0.6787 MeV. (5.108) → x f Indeed, the freeze-out happens before electron-positron pair annihilation.

Now we estimate the freeze-out fraction Xn∞ = Xn(x ) without the β-decay. To do so, let’s ﬁnd eq→ ∞ the formal solution of Eq. (5.104). Let f (x) = Xn Xn , then Eq. (5.104) becomes − λ x x d f (x) n p x e + → (1 + e− )f (x) = . (5.109) d x H(Q) (ex + 1)2

Homogeneous solution is

Z x λ x 0 n p( 0) x f ( ) x exp d x → 1 e 0 , (5.110) ( ) 0 x H x ( + − ) ∝ − 0 ( 0)

0 then general solution may be written as f (x) = g(x)f ( )(x) where g(x) satisﬁes

x d g(x) 1 e = . (5.111) d x f (0)(x) (ex + 1)2

Integrating once more, we ﬁnd

Z x x˜ Z x λ x 0 e n p( 0) x f x f ( ) x g x d x˜ exp d x → 1 e 0 , (5.112) ( ) = ( ) ( ) = ex˜ 1 2 0 x H x ( + − ) ( + ) − x˜ 0 ( 0) and Z x x˜ Z x λ x eq e n p( 0) x X x X x d x˜ exp d x → 1 e 0 . (5.113) n( ) = n ( ) + ex˜ 1 2 0 x H x ( + − ) 0 ( + ) − x˜ 0 ( 0) eq Note that we apply the boundary condition Xn(x 0) Xn at the last equation. At freeze-out, eq → → Xn Xn and we estimate it by setting x : → ∞ Z x˜ Z λ x ∞ e ∞ n p( 0) x X d x˜ exp d x → 1 e 0 . (5.114) n∞ = ex˜ 1 2 0 x H x ( + − ) 0 ( + ) − x˜ 0 ( 0)

The integral in the bracket can be done analytically to read:

Z λ x ∞ n p( 0) x 1 4 x˜ 1 3 4 d x → 1 e 0 0.25567 e . (5.115) 0 x H x ( + − ) = x˜2 + x˜3 − + x˜ + x˜2 + x˜3 x˜ 0 ( 0)

Then, the freeze-out neutron fraction is

Xn∞ = 0.147617, (5.116) as we found in the numerical solution earlier! Wait a minute. Didn’t I intend to ‘estimate’ the result? Well, if there is a full analytical solution, why not? 22 CHAPTER 5. OUT-OF-EQUILIBRIUM AND FREEZE-OUT

5.5.2 Deuterium bottleneck

After neutron fraction freezes out, Deutrium (mD 1875.612 MeV) must form ﬁrst via ' n + p D + γ. (5.117)

But, if tempearture of the Universe is high enough so that there are enough photons to destroy the Deuteriums ("(γ) > BD = mn +mp mD 2.225 MeV), the reverse process dominates. As we see earlier, no further interactions will be proceeded− ' until we have enough Deuterium. This is called Deuterium bottleneck. Let us consider the equilibrium abundance of Deuterium. In the last chapter, we derived the number density of non-relativistic particles

3/2 mi T n T g e (mi µi )/T , (5.118) i( ) = i 2π − − then ni eµi /T emi /T (5.119) = 3/2 gi(mi T/2π) In chemical equilibrium, µn + µp = µD, which yields 1 nn 1 np 1 nD emn/T emp/T emD/T , (5.120) 3/2 3/2 = 3/2 2 (mn T/2π) 2 (mp T/2π) 3 (mD T/2π) or the mass fraction of Deuterium (XD)

3/2 2n 3 nnnp 2π m D D (mn+mp mD)/T XD = = e − nn + np 2 nn + np T mnmp 3/2 3 2π m D BD/T = XnXpnb e . (5.121) 2 T mnmp

Using the baryon-to-photon ratio parameter9 n n η 1010 b 1010 b 4.105 1010n T 3, (5.123) 10 = 2ζ 3 b − nγ ( ) 3 ≡ π2 T ' × the Deutrium fraction becomes

3/2 m 11 3/2 D BD/T XD = 3.654 10− η10XnXp(2πT) e × mnmp

14 3/2 2.225/TMeV 5.647 10− η10XnXp TMeVe . (5.124) ' ×

This fraction is quite small earlier. For example, at TMeV = BD 2.225 (> Q), Xn 0.359 and Xp 13 2' ' ' 0.641 (their equilibrium fraction), then X D 7.369 10− (Ωbh /0.023) 1. We show the equilibrium ' × 9Baryon-to-photon ratio is constant over time, and it takes the value of

5 2 2 10 nb 10 ρcritΩb 10 1.1232 10− Ωbh Ωbh η10 10 = 10 10 × = 6.286 (5.122) ≡ nγ mpnγ ' 411 0.023 5.5. BIG-BANG NUCLEOSYNTHESIS 23 fraction of neutrons, protons and Deuteriums in Fig. 5.3 as a thin cyan dashed line (exact calculation is shown as the thick cyan line). As shown in Fig. 5.3, the thermal equilibrium approximation does a fair job to explain the full result, although it shows some deviation at earlier time when Xn abundance freezes out (because we assumes the full thermal equilibrium), and later time when nucleosynthesis starts. Anyway, for both approximation and exact solution, it is clear that it takes some time for the deuterium abundance X D to catch up with the neutron and proton abundance even after the temperature of the Universe drops below T ® BD 2.225 MeV. For example, let us estimate the time it takes to achieve ' XD 0.01 from Eq. (5.124). For this estimate, we ignore the neutron decay and use Xn = Xn∞ and ' Xp = 1 Xn∞ to have the equation for the temperature as − T 3/2 13 1% BD/T1% X D 1.4824 10− e = 0.01, (5.125) ' × BD or B 3 B D ln D 24.93, (5.126) T1% − 2 T1% ' then, we ﬁnd BD 3 24.93 + ln(24.93) + 29.76, (5.127) T1% ' 2 ···' so that T1% BD/29.76 0.0748 MeV. Although Deuterium abundance in the exact solution does not ' ' quite reach that value (maximum abundance is about XD 0.0036 at T 0.0729 MeV), it is close enough to the temperature when the bottleneck is opened (that' we will estimate' shortly).

This delay of Deuterium formation comes about because the baryon-to-photon ratio η = nb/nγ is so small. That is, because we have about a billion photons per every baryon, there still exist photons at Wein tail of the Planck curve—the photons energetic enough to destroy the Deuterium nucleus—even when the mean photon energy is below the binding energy of the Deuterium. Therefore, we have wait until the Universe is cool enough so that photons with " > BD disappear. This happens at T ® BD/30!

From Eq. (5.124) (using Xn = Xn∞ 0.1476), we calculate the equilibrium temperature (TMeV) abun- ' dance (X D) relation as, ! B 3 B X X D ln D ln D ln D 29.54, (5.128) = 3/2 + TMeV 2 TMeV 14 η10/6.286 − 5.647 10 η10XnXpBD ' ( ) × − then using the iteration, BD X D 3 X D 29.54 + ln + ln 29.54 + ln , (5.129) TMeV ' (η10/6.286) 2 (η10/6.286) we find the temperature (in MeV) corresponding to the Deuterium concentration XD: 1 X D X D TMeV− = 13.28 + 0.449 ln + 0.674 ln 29.54 + ln . (5.130) (η10/6.286) (η10/6.286) This equation is the key to estimating the final Helium abundance. Then, when does the Deuterium bottleneck open? Let us estimate T (open). The Deuterium bottleneck opens up when following interactions destroying Deuterium are efficient to form heavier elements:

3 D + D He + n → D + D T + p (5.131) → 24 CHAPTER 5. OUT-OF-EQUILIBRIUM AND FREEZE-OUT

The interaction rate of these reactions between T = 0.06 MeV and T = 0.09 MeV is measured to be 17 3 1 σv DD 3Hen = (1.3 2.2) 10− cm s− 〈 〉 → − × 17 3 1 σv DD Tp = (1.2 2) 10− cm s− . (5.132) 〈 〉 → − × Then, the change of number density of Deuterium due to these two interactions are

3 1 d(a nD) 2 2 σv 3 σv n σv n . (5.133) 3 = DD Hen + DD Tp D D a d t − 〈 〉 → 〈 〉 → ≡ − 〈 〉 In terms of the adundance, XD = 2nD/nb, the equation can be written as

dXD 1 2 = σv XDnb, (5.134) d t −2 〈 〉 and the change due to the interactions are visible when

1 2 σv XDnb t XD, (5.135) 2 〈 〉 ' or 1 1/2 open 2 η g ( ) 6 1 10 − ? X D = 2.2697 10− TMeV− (XD) . (5.136) ' σv nb t × 6.286 3.363 〈 〉 Here, we use the time-temperature relation t 2.420/ g T 2 s, and n η / 4.105 1010 T 3 = p ? MeV− b = 10 ( ) 21 3 3 × ' 3.171 10 η10 TMeV cm− (Eq. (5.123)). Plugging the temperature-concentration relation Eq. (5.130) into Eq.× (5.136), we have

open ( ) 1 1/2 open X η g ( ) 5 6 D 10 − ? XD 3.014 10− + 1.019 10− ln , (5.137) ' × × (η10/6.286) 6.286 3.363 whose solution is, again using iteration,

1 1/2 5 1/2 open η g 3.012 10 g X ( ) 3.014 10 5 10 − ? 1 0.03381 ln − ? D − + × 2 ' × 6.286 3.363 (η10/6.286) 3.363 η 1 g 1/2 h η g i 5 10 − ? 10 ? 1.953 10− 1 0.1043 ln + 0.02609 ln . ' × 6.286 3.363 − 6.286 3.363 (5.138)

2 Therefore, for our reference cosmology, Ωbh = 0.023, g? = 3.363, the Deuterium bottleneck opens (open) 5 when the concentration becomes XD 1.800 10− which happens at T 0.0967 MeV. Note that ' × ' one can calculate the concentration XD at the bottle neck opening for given g? and η10 from Eq. (5.138).

5.5.3 BBN: after bottleneck is opened

In the previous section, we show that the Deuterium bottleneck opens up at T 0.0967 MeV when (open) 5 ' XD 1.8 10− . Then, the things happens so rapidly and the equilibrium abundance of the Deu- terium' reaches× about 1% of the total baryon mass shortly after (when T 0.075 MeV). Around that time, the Deuterium abundance peaks because the deuterium production rate' (n + p D + γ) and the 3 consumption rate (D + D He + n and D + D T + p) are equal around that time: → → → λ X X 4 2 pn p n 4 10− 2 10 , (5.139) λDDXD ' X D 5.5. BIG-BANG NUCLEOSYNTHESIS 25

time t 1s 1m 5m 15m 60m 100 1 10− 2 10− 3 10− n 4 p 10−

i D

X 5 10− T 6 3 10− He 4 7 He 10− 7Li 8 mass fraction 10− 7Be 9 10− β-decay 10 Eq.(5.124) 10− 11 10− 12 10− 13 10− 0 1 2 10 10− 10− temperature T [MeV]

Figure 5.3: Evolution of the mass fraction of various nuclides relevant in the Big-bang nucleosynthesis process: neutron (n), proton (p), Deuterium (D), Triton (T), 3-Helium (3He), 4-Helium (4He) 7-Lithium (7Li), 7-Belilium (7Be) from about 1 sec after Big-bang to one hour. We use the publically available 2 version (v4.1) of Kawano code with the standard cosmology: Nν = 3, Ωbh = 0.023. 26 CHAPTER 5. OUT-OF-EQUILIBRIUM AND FREEZE-OUT

3 where λ σv nb is the interaction rate normalized to total baryon, and λpn/λDD 10− at T 0.07 0.08 MeV.≡ Fig.〈 〉 5.3 also shows that the nucleosyinthesis ends around this time so that' the abundance' of− Triton, 3-Helium, and 4-Helium saturate and freeze out. Because the binding energy per nuclide of 4- Helium nuclei is highest among the light elements, 4-Helium is the energetically favored among all the light element formed during BBN. Therefore, once the nucleosynthesis starts, essentially all neutrons contribute to form 4-Helium to yield

Yp 2Xn(TBBN), (5.140) ' where TBBN is the temperature at the beginning of the nucleosynthesis. One indication of the beginning of the nucleosynthesis is when the neutron concentration drops below the level that the β-decay predicts, or the Deuterium concnetration peaks (XD 0.01), which happens at '

TBBN 0.075 MeV, (5.141) ' the cosmic time corresponding to this temperature is

tBBN 225.6 sec (5.142) ' after the Big-band, and the helium abundance is

225.6/880.3 Yp 2 0.147617e− 0.23, (5.143) ' × ' which is close to the magic number! The numerical integration gives Yp 0.2474. ' The final abundance of light elements, in particular Helium and Deuterium, depends sensitively on the assumed value of the baryon number density (η10) as shown in Fig. 5.4 which is the result of the full numerical calculation. When increasing the baryon density but fix photon number density, the deuterium can form earlier because there are less photons per Deuterium that can break the Deuterium bound state. Then, there are more Deuterium available to start the nucleosynthesis earlier. That means, more neutrons (in Deu- terium) are available (simply because they avoid β-decay) to form Helium. This increases the final

Helium abundance. However, Yp may not exceed the maximum value Yp = 2Xn∞ 0.2952 unless BBN happens before the neutron-to-proton ratio freezed out. In the other extreme case,' if the baryon density is too low, the Deuterium formation process (n + p D + γ) had freezed out before the nucleosynthesis → began (then neutrons β-decay into protons). Then, Xn does not drop below XD and only fraction of neutrons end up processed into light elements. Therefore, the ﬁnal Helium fraction is much less than the standard value.

The relic Deuterium abundance traces the freeze-out concentration. For higher η10, once Xn drops below XD, essentially all neutrons go into the Deuterium, then XD is controlled by its depletion via D+D channel and D + p channel. Schematically, the rate of change of Deuterium concentration is given by

dXD 2 η10 XD + RXpXD , (5.144) dT ∝ 5 where R λpD/λDD 10− is relative ratio between the interaction rate of DD-channel and Dp- ∝ ' f channel. For small η10 < 10, the freeze-out concentration XD is larger than RXp, then the ﬁrst term f 1 (DD-channel) dominates to have XD η−10 dependence. On the other hand, when η10 > 10, the Dp- ∝f channel dominates and we have ln XD η10 dependence on the baryon density. That is the reason why the relic Deuterium abundance is a∝ sensitive probe of the baryon density. 5.6. HYDROGEN RECOMBINATION AND CMB 27

Nucleosynthesis calculation can also be altered if we have different fundamental parameters/constants. As an exameple, we plot the dependence of Yp on the additional relativistic degrees of freedom (param- eterized in terms of effective number of neutrino species Nν) from Nν = 2 to Nν = 5. The effect here is changing g? from the standard value 3.363 to g? = 2.908, 3.817, 4.271 for Nν = 2, 4, 5, respectively. 1/2 Then, because the cosmic time depends t g?− for a given temperature, adding (subtracting) relativistic degrees of freedom decreases (increases)∝ time for neutron-to-protonration freeze-out and the f beginning of BBN. Therefore, for larger (smaller) Nν, A) the freeze-out concentration of neutrons (Xn ) increases (decreases) and B) more (less) neutrons are available to form Helium nucleus that increases (decreases) Yp. This qualitative argument is consistent with what we see in the top panel of Fig. 5.4. You will quantify the change of Helium concentration under the change of various fundamental parameters in the homework.

5.6 Hydrogen recombination and CMB

After the Big-bang nucleosynthesis, besides the neutrinos that have been free-streaming after T 1.5 MeV, the thermal bath of the Universe is ﬁlled with Helium nuclei, Hydrogen nuclei (protons),' electrons, and photons. Next major event is the recombination of Helium that the Helium nuclei captures one electron to form a He+ ion, then a He+ ion captures another electron to form a He atom. + The ionization energy of He and He are, respectively, IHe+ = 54.4 eV and IHe = 24.62 eV, greater than the ionization of Hydrogen, IH = 13.6eV and this process must happen earlier than the hydrogen recombination. You will have some fun with Helium recombination in the homework.

5.6.1 Recombination in equilibrium

Our main interest in this section is the recombination of Hydrogen, the process that the free electrons and protons combines to form hydrogen atoms in the graound (1s) state:

p + e H1s + γ (13.6 eV). (5.145) At early times when the photon temperature is high, the reaction should be in an equilibrium due to the frequent interaction between protons and electrons. In this case, we can use chemical equilibrium condition µe + µp = µH, (5.146) or ne np nH eme/T emp/T emH/T , (5.147) 3/2 3/2 = 3/2 2(me T/2π) 2(mp T/2π) 4(mH T/2π) then 3/2 3/2 nenp mpme T m T (mH mp me)/T e IH/T = e − − e− . (5.148) nH mH 2π ' 2π Let us deﬁne the ionization fraction as

ne np Xe = , (5.149) ≡ ne + nH np + nH then the equation becomes X 2 1 m T 3/2 e e IH/T e− , (5.150) 1 Xe ' np + nH 2π − 28 CHAPTER 5. OUT-OF-EQUILIBRIUM AND FREEZE-OUT

0.30 0.25 He 4 X 0.20 Nν = 2 0.15 Nν = 3 0.10 Nν = 4

ﬁnal abundance 0.05 Nν = 5 0.00

1 4 10− He

2 10−

3 10− D

4 10− i X 5 10− 3He

6 10− ﬁnal abundance

7 10−

8 10− 7Li

9 10−

10 10−

11 10− 2 1 0 1 2 3 10− 10− 10 10 10 10 10 η10 = 10 nb/nγ

Figure 5.4: The mass fraction of various nuclides relevant in the Big-bang nucleosynthesis 5.6. HYDROGEN RECOMBINATION AND CMB 29 which is called Saha equation. Because we assume Heliums are completely neutral, 3 10 8 T 3 nH,tot np + nH = 1 Yp nb 0.76 10− η10nγ 3.124 10− η10 cm , (5.151) ≡ − ' × ' × 2.726 K the right hand side becomes

3/2 3/2 1 me T IH e IH/T 2.498 1016η 1 e IH/T , (5.152) 10 2 3 − −10 − (1 Yp)10 η102ζ(3)/π T 2π ' × T − − and the Saha equation becoems 2 Xe IH 3 IH = exp 37.76 + ln ln η10 . (5.153) 1 Xe − T 2 T − − 30.474 13 When T IH 157, 820 K, the right hand side of the Saha equation is e 1.72 10 , extremely ' ' ' × large, which dictates Xe 1. The ionization fraction Xe decreases and the neutral fraction is of order unity when the right hand' side of the Saha equation is order unity; that happens when IH 3 IH IH 3 IH 37.76 + ln ln η10 0 ln 37.76 ln η10 = 35.922, (5.154) − T 2 T − ' → T − 2 T ' − and the temperature can be calculated from (again, using iteration)

IH 3 = 35.922 + ln(35.922) 41.294, (5.155) T 2 ' (eq) which gives T 3822 K at which Xe 0.618. Again, the reason why we have to wait long after the temperature' reaches the ionization temperature' until the formation of neutral Hydrogen is because the baryon-to-photon ratio is small. Even though the mean photon temperature is small, we have large number of photons at the tail of the Planck distribution that can ionize all the hydrogen atom. Below this temperature, the equilibrium number density drops rapidly. The equilibrium ionization fraction (eq) becomes half (Xe = 0.5) at Trec 3737 K, (5.156) ' but the ionization fraction at, say, T 3000 K is already (using that Xe 1), ' (eq) 15.19 7 Xe (T = 3000 K) e− 2.53 10− . (5.157) ' ' × The redshift at the recombination is zrec = Trec/Tcmb 1 1370, and the cosmic time is trec 252, 000 yrs from the Big-bang. − ' '

5.6.2 Non-equilibrium process

The equilibrium calculation above is not what happens in our Universe, in particular after the ionization fraction drops signiﬁcantly. Instead, the sharp drop of the number density of electrons and protons freezes out the recombination process, and we have to use full kinetic equation. Yes, as you can easily suspect at this point, we will solve the freeze-out process by using the Boltzmann equation for the electron number density:

3 (eq) (eq) ! 1 d(a ne) ne np = σv nH nenp . (5.158) a3 d t (eq) 〈 〉 nH − 30 CHAPTER 5. OUT-OF-EQUILIBRIUM AND FREEZE-OUT

You might temp to use the Saha equilibrium ratio in Eq. (5.148) into the equation and solve for it, but that is NOT the way. We need to include more physics here. First of all, the recombination to the ground state will not read a net change of the ionization fraction. It is because the recombination photon with E > 13.6 eV will be emitted in this process. This photon, has a photo-ionization cross section of (at E = 13.6 eV) 29πα a πa2 6.30 10 18cm2, (5.159) ν0 = 4 0 − 3e ' × where α = 1/137 is the fine structure constant, and a0 = 0.5292Å is the Bohr radius. The total 2 number density of hydrogens and protons at recombination temperatre T 3500 is (Ωbh = 0.023) 3 ' nH,tot 416 cm− , which gives the mean-free-time of ' 1 λ 12720 1 X 1 s. (5.160) γ = n a c = ( e)− H ν0 − Therefore, the mean-free time for the ionizing photon is much smaller than the Hubble time ( 105 yrs) 9 ∼ as long as the neutron fraction 1 Xe > 10− (which is almost always true for our case of interest). This means that the recombination− photon emitted from the ground-state recombintaion always ionize another neutral hydrogen to yield no net change. Therefore, the recombination must be done first to higher-n state then cascade down to the ground state. This process is called case-B (on-the-spot) approximation. The case-B recombination rate coefficient is 4.309 10 13 T 0.6166 v − 4− 3 αB σ B = × 0.5300 cm /s, (5.161) ≡ 〈 〉 1 + 0.6703T4 4 where T4 T/(10 K). ≡ But, even after excluding the recombination to the ground state, recombination to the higher-n level then cascade down will produce Lyman-series photons that have a large cross sections for captured by a Hydrogen atom. These photons excite adjacent Hydrogen atoms to the high-energy states so that it can be easily photoionized again. Therefore, both of them are not the main channel through which cosmic recombination has happened. The two main routes to the recombinations are A) two-photon decay and B) redshifted Lyman-alpha photons. The details of the recombination calculation is presented in [?,?] using the three-level approximation (that we will follow in this section later). This calculation is refined later (including effectively 300 atomic levels and improved treatment of Helium recombination—same issue is there for Helium that we have to use case-B approximation) by [?] which reads to the public code RecFAST. More recently, further refinement has been done for the calculation to less than 1% accuracy by heroic personals like Jens Chluba, Chris Hirata, etc, and that effort reads public code HyRec and CosmoRec. The codes calculate a detailed physical processes including other two-photon decays, recombination to higher levels (including some 500 shells), feedback from other Lymann lines, etc. You can find them in the following URL 10. As for the discussion here, we will use Peebles’s prescription.

Peebles’s recombination : three-level approximation

Here, we model the cosmic recombination history by using the three-level approximation: ground state with fraction x1, n = 2 state with fraction x2, and ionization state with xp = xe. We assume that the 10HyRec: http://www.sns.ias.edu/ yacine/hyrec/hyrec.html Cosmological Recombination Project: http://www.cita.utoronto.ca/ jchluba/Science_Jens/Recombination/Welcome.html ∼ 5.6. HYDROGEN RECOMBINATION AND CMB 31 higher n 2 energy level is populated according to the equilibrium ratio: ≥ x g n` n` (En` E2)/T = e− − . (5.162) x2 g2

Because the net decay into 1s state is very slow, we can write the rate of change of n = 2 level from the competition between case-B recombination and photo-ionization:

1 d n a3 ( 2 ) α n n βn . (5.163) 3 = B e p 2 a d t − where we can ﬁgure out β from the detailed balance as

(eq) (eq) 3/2 3/2 αB ne np ge gp me T αB me T β α e IH /4T e IH /4T . (5.164) = eq = B − = − ( ) g2 2π 4 2π n2

2 6 Here, we use g2 = 8 electron : 2(1s ) + 6(2p ) 2(proton spin) = 16, then the equation becomes × d x 1 m T 3/2 2 2 e IH /4T = αB nH,tot xe e− x2 . (5.165) d t n>2 − 4 2π

On top of the recombination and photoionization to n = 2, we have to include the two-photon decay and the redshifted Lyman-alpha lines that give the net transition from n = 2 to n = 1 state. Two photon decay is a forbidden (2s 1s) dipole transition with the transition rate → 1 Λ2γ = 8.2206 s− . (5.166) Neither of the two photons have enough energy to excite the ground-state hydrogen; thus, this process reads to the net production of the ground state hydrogen atom. Rate of change of the x2 fraction due to the two-photon decay is thus

d x2 h x2 i = Λ2γ β2γ x1 , (5.167) d t 2γ − 4 − where β2γ is the detailed-balance correction that makes sure that the total rate does not change in the equilibrium state. That is, from the Saha equation for H(n = 1) + γ(3/4IH) H(n = 2), we find n g 2 2 3IH/4T 3IH/4iT = e− = 4e− (5.168) n1 g1 that reads 3IH/4T β2γ = e− , (5.169) and d x h x i 2 2 3IH/4T = Λ2γ x1e− . (5.170) d t 2γ − 4 − Finally, we consider the redshfit of the Lyman-alpha line. The decay rate for the Lyman-alpha transition 8 1 (2p 1s) is ALyα 6.25 10 s− . Let us first calculate the optical depth of the Lyman-alpha line. Because→ the line width' is narrow× we estimate the cross section as

σ(ν) = σ0δ(ν νLyα), (5.171) − 32 CHAPTER 5. OUT-OF-EQUILIBRIUM AND FREEZE-OUT where 2 2 σ0 = 3π ALyαλ−Lyα, (5.172) is the integral constant11. Then, the optical depth of Lyman-alpha photons is

Z Z σ ν τ n σd t n x ( ) dν, (5.177) = 1 = H,tot 1 ν˙ | | then we use the cosmological redshift ν˙ = Hν to calculate the optical depth as − 2 nH,tot x1 3π ALyαnH,tot x1 τ σ . (5.178) = H 0 = 3 νLyα HνLyα This is called the Sobolev optical depth that plays a key role in the line formation in expanding media. During recombination, the optical depth is typocally of order 108, and Lyman-alpha photons are immediately re-absorbed. The escape probability12 of the Lyman-alpha∼ photons is then

τ 1 e− 1 8 P = − 10− (5.182) τ ' τ ' which makes the ﬁnal rate ALyα P 1. Then, the change of the x2 due to Lyman-alpha emission is ' A Hν3 d x2 3 Lyα 3I /4T Lyα 3I /4T x 4x e H x 4x e H (5.183) = 2 1 − = 2 2 1 − d t Lyα −4 τ − −4π nH,tot x1 −

11This constant can be found from the detailed balance as following. Put a Hydrogen atom in a black-body, then in equilibrium we have Z ∞ dnγ ALyα(1 + f (νLyα))n2p = n1s σ(ν)dν. (5.173) 0 dν Then, using 2 1 dnγ ν f ν , f ν , (5.174) ( ) = eν/T 1 dν = π2 ( ) we ﬁnd − ν /T 2 e Lyα νLyα ALyα n2p n1sσ0 , (5.175) νLyα/T = 2 νLyα/T 1 e 1 π (e − ) or − 2 2 π n2p π νLyα/T σ0 = 2 ALyαe = 3 2 ALyα. (5.176) νLyα n1s νLyα

12 The escaping probability of a line profile P(ν)dν emitted at time t can be written as Z ∞ τ(ν,t) P(t) = dν (ν)e− , (5.179) P −∞ τ(ν) which is the profile-weighted e− , the probability that a photon emitted at frequency ν will escape. When the photon is emittedi to an expanding media, the optical depth is Z Z 0 Z 0 Z ν ∞ σ(ν0) n1s τ(ν, t) = d t0n1sσ(νa(t)/a(t0)) = dν0n1s σ0 dν0δ(ν0 νLyα) = τ dν0δ(ν0 νLyα). (5.180) ν˙ Hν − t − ν 0 ' ν − − 0 − with ν0 = νa(t)/a(t0) being the redshfited frequency. For the final equality, we assume that the emitted photons will be re- absorved in a time much less than the Hubble time. Then, ν˙ Hν. Putting all pieces together, with ν δ ν ν we 0 ( ) = ( Lyα) calculate the escaping probability as ' − P −

τ Z ν Z τ ∞ τ R dν δ ν ν 1 x 1 e 0 0 ( 0 Lyα) P(t) = dνδ(ν νLyα)e− − = d xe− = − . (5.181) − τ 0 τ −∞ 5.6. HYDROGEN RECOMBINATION AND CMB 33

Combining all three contributions, we ﬁnd

d x x 2 2 2 3IH/4T = αB nH,tot xe β x2 Λ2γ + Λα x1e− . (5.184) d t − − 4 − Here, we deﬁne 3 HνLyα Λ . (5.185) α 2 ≡ π nH,tot x1 We assumes that the excited Hydrogen atoms are in equilibrium (it makes sense as they are short-lived), then set x˙2 = 0 to ﬁnd 2 3IH/4T αB nH,tot xe + Λ2γ + Λα x1e− x2 = 4 . (5.186) Λα + Λ2γ + 4β Feeding this into the net rate of loss of electron (note that two-photon decay and Lyman-α re-absorption do not invlove electrons), and using that x2 1: α n X 2 Λ Λ x e 3IH/4T dXe 2 2 B H,tot e + 2γ + α 1 − = αB nH,totXe + β x2 = αB nH,totXe + 4β d t − − Λα + Λ2γ + 4β Λα Λ2γ + 2 3IH/4T = αB nH,totXe 4β x1e− − Λα + Λ2γ + 4β − 3/2 Λα Λ2γ m T + 2 e IH/T = αB nH,totXe e− (1 Xe) (5.187) − Λα + Λ2γ + 4β − 2π − with

Hν3 3/2 Lyα me T Λ 8.225 s 1, Λ , 4β α e IH/4T (5.188) 2γ = − α = 2 = B − π nH,tot(1 Xe) 2π − This equation is called Peebles equation! Note that the terms in the bracket satisfy the detailed balance because of the Saha equation Eq. (5.148). Had we ignored the delay of the recombination due to the inefﬁciency of two-photon decay and redshifted Lyman-alpha channels, we could have ended up with the equation without the prefactor, simply from the detailed balance and the case-B approximation. The prefactor can be interpreted as the branching raio of two-photon decay and Lyman-alpha escape to total of two-photon decay, Lyman-alpha escape and photoionization from n = 2. time-temperature relation during recombination

To solve the Peebles equation numerically, we need to supplement the time-temperature relation. The recombination happens close to the matter-radiation equality, the two main player here is the radiation density and the matter density. As usual, we start from the Friedmann equation (let’s ignore the mass of neutrinos for this calculation): 2 3H = 8πG (ρm + ρR) . (5.189)

It is convenient to deﬁne the new time variable y = a/aeq, where aeq = ΩR/ΩM is the scale factor at equality, to rewrite the Friedmann equation as

1 d y 2 ρ(eq) 1 3H2 3 4πG 1 , (5.190) = y d t = y3 + y 34 CHAPTER 5. OUT-OF-EQUILIBRIUM AND FREEZE-OUT

(eq) (eq) (eq) where ρ = 2ρM = 2ρR is the total energy density at equality. Then, we can solve the differential equation to have 1/2 (eq) § ª 4πGρ − 2 3/2 1/2 t = (1 + y) + 2 2 (1 + y) . (5.191) 3 3 − Then, now let’s work out some numbers. From g? = 3.363 and Tcmb = 2.726 K, the radiation energy density now is 2 π 4 34 3 ρR = g? Tcmb = 7.8177 10− g/cm (5.192) 30 × and the matter density now is

29 2 3 ρM = ΩMρcrit = 1.8788 10− (ΩMh ) g/cm . (5.193) × This gives the scale factor at equality as

5 2 1 aeq = 4.161 10− (ΩMh )− , (5.194) × the redshift at equality as 2 1 + zeq = 1/aeq 24033 ΩMh , (5.195) ' Then, the photon temperature at equality is

eq 2 2 Tγ = Tcmb/aeq 65513 ΩMh K 5.6456 ΩMh eV, (5.196) ' ' and the energy density at equality is

(eq) 3 16 2 4 3 ρ = 2ρM/aeq = 5.2158 10− (ΩMh ) g/cm , (5.197) × which makes the Hubble parameter at equality as v v t8πG t16πG Ç H(eq) ρ(eq) Ω ρ a 3 H 2Ω a 3 = 3 = 3 M crit eq− = 0 M eq− 11 22 1 =1.7075 10− ΩMh s− (5.198) × The time equation Eq. (5.191) is then becomes 2.625 kyr §2 ª t 1 y 3/2 2 2 1 y 1/2 , (5.199) = 2 2 ( + ) + ( + ) (ΩMh ) 3 − and temperature is T (eq) γ 65513 2 5.6456 2 Tγ(y) = ΩMh K ΩMh eV. (5.200) y ' y ' y Finally, the redshift is 2 1 1 24033 ΩMh 1 + z = = = , (5.201) a yaeq y and the Hubble parameter is v v H(eq) t1 y 1.7075 10 11 Ω h2 2 t1 y H y + − ( M ) + s 1 (5.202) ( ) = y2 2 = × y2 2 −

For example, at the recombination temperature, Trec = 3737 K, the redshift is zrec = 1370, which gives 2 (for ΩMh = 0.143), yrec = 2.507. This gives the age of the Universe at recombination as trec = 14 1 1 252.41 kyr, and the Hubble parameter at the time as H(yrec) = 7.357 10− s− = (430.74 kyr)− . × 5.6. HYDROGEN RECOMBINATION AND CMB 35

Result: recombination history

Using y = a/aeq as a time variable, we can rewrite the left-hand side of the Peebles equation as dX dX d y dX e e e yH y . (5.203) d t = d y d t = d y ( )

Then, the Peebles equation becomes

3/2 dX Λα Λ2γ α m T e + B 2 e IH/T = nH,totXe e− (1 Xe) . (5.204) d y − Λα + Λ2γ + 4β yH(y) − 2π − We listed all parameters for convenience:

Ω h2 T 3 4.309 10 13 T 0.6166 n 7 b 3 − 4− 3 H,tot = 1.964 10− cm , αB = × 0.5300 cm /s, × 0.023 2.726 K 1 + 0.6703T4 Hν3 3/2 Lyα me T Λ 8.2206 s 1, Λ , 4β α e IH/4T . (5.205) 2γ = − α = 2 = B − π nH,tot(1 Xe) 2π − 3 One cautionary note. Because we use ħh = 1 in our natural unit, we need to multiply νLyα by (2π) :

3 8πHνLyα 8πH Λ , (5.206) α = n 1 X = 3 H,tot( e) nH,totλLyα(1 Xe) − − where λLyα = 1216Å. In other words, for νLyα, what we really meant was ωLyα = 2πνLyα. We show the result of the nmerical calculation in Fig. 5.5. As can be seen in Fig. 5.5, the recombination proceeds quite slowly compared to the other non-equilibrium processes. For the previous two cases, the the non-equilibrium solutions follow the equilibrium solution quite well before the freezeout, then freezeout happens to yield the deviation. But, in the case of cosmic recombination, the non-equilibrium solution already diviates from the equilibrium solution qhite early in the process and the ionization fraction gradually decreases toward the residual value. Of course, the reason is because the recombination process is delayed due to the slowness of the relevant channels—two-photon decay and redshifted Lymann-alpha photons (with a rate of a couple happening in a second)—compared to the normal radiative processes (A2p 1s 0.6 billion in a second) would yield the thermal equilibrium. But, why isn’t the recombination delayed→ ' by the same factor? It is because the time scale that set the equilibrium density to drop is the rate at which the number density of ionizing photon (Eγ > 13.6 eV) decreases due to the cosmological redshift, which is much slower than the radiative decay time scale. Now, let us estimate the freezeout redshift and the residual electron number density from the usual method. The interaction rate for the case-B recombination is

Γep = neαB = nH,totXeαB. (5.207)

As the decoupling happens when Xe 1, we approximate it as 7.891 3 15.782 Xe(T4) exp 17.96 + ln , (5.208) ' − T4 4 T4 and 3 3 nH,tot 9695.36T4 cm− , (5.209) ' 36 CHAPTER 5. OUT-OF-EQUILIBRIUM AND FREEZE-OUT

100 Saha equation (equilibrium) Peebles Peebles (fudge=1.14)

1 HyRec Full calculation 10− zrec = 1370 e

X 2 10−

3 10−

4 10− 103 102 redshift z

Figure 5.5: Recombination history calculated from Saha equation (Eq. (5.148)), Peebles equation (Eq. (5.204)), Peebles equation with a fudge factor 1.14 to αB, and the full calculation from HyRec 2 2 code. We use ΩMh = 0.143, Ωbh = 0.023. 5.6. HYDROGEN RECOMBINATION AND CMB 37 to estimate the temperature at which the recombination rate is equal to the Hubble rate as

1.6334 3.1133T4 q e 7.891/T4 H 13 T T 3 Γep = 0.53 − = = 2.8131 10− (0.9368 + 4) 4 (5.210) 1.49187 + T4 × to ﬁnd out f Trec 2714 K, (5.211) ' with the ionization fraction f 4 Xe 3.129 10− . (5.212) ' × 4 Surprise! surprise! The ionization fraction from the full calculation at z 200 is 3.2696 10− ! It’s not too different from our crude guesstimation!! ' × The residual free electron plays very important role in the primordial chemistry, as it is the main catalyst of H− ion that is needed to form a molecular hydrogen, which is an important coolant for forming the ﬁrst generation of stars and galaxies.

5.6.3 Decoupling of photon

Because the ionization fraction drops in the course of the Hydrogen recombination, the light can ﬁnally decoupled from the electrons13! When does it happens? The cross-section for light scattering off the electron at this energy scale is given by the Thomson scattering cross section:

8π 8π e4 σ r2 6.6524 10 25 cm2. (5.213) T = 0 = 2 − 3 3 me ' × Then, the mean free path of the photon is 1 λ = . (5.214) neσT The mean free path is the mean path length that the photon can travel without being scattered once. Because the volume swiped by a incident particle with vertical (to the moving direction) cross section, σ, after moving a distance x is V = xσ, the expected mean number of electrons inside of that volume is neσT x. As the mean number of electrons is the same as expected number of scattering, the distance 1 at which one scattering is expected on average is λ = (neσT)− . With the same logic, the probability of photon being scattered in the path length between x and x + δx is δx P(x x + δx) = , (5.215) ∼ λ(x) which gives the scattering rate of

P x x δx Γ ( + ) n σ c (5.216) scat. = ∼δx/c = e T that we have been using so far. The decoupling happens when the scattering rate is the same as the expansion rate: neσTc = XenH,totσTc = H(T), (5.217)

13Note that we ignore the photon-proton scattering because the cross section is far much smaller, suppressed by the inverse- 7 mass-square ratio ( 10− ), for this process. ∼ 38 CHAPTER 5. OUT-OF-EQUILIBRIUM AND FREEZE-OUT or

13q 3 H T 2.8131 10 (0.9368 + T4)T4 X T ( ) − e( ) = = 3 × 25 10 nH,totσTc 9695.36T 6.6524 10 2.9979 10 4 ( − ) ( ) v × × × × u0.9368 T 3t + 4 =1.4549 10− 3 (5.218) × T4

Using the equilibrium ionization fraction, I ﬁnd that the decoupling happens at

(est) (est) Tdec = 3086 K, zdec = 1131, (5.219)

(est) with Xe,dec = 0.0095. The result from the full calculation is

Tdec = 2461 K, zdec = 901.82, (5.220) and Xe,dec = 0.013. What we have calculated above was the time at which the scattering rate is the same as the Hubble rate. The more rigorous way of deﬁning the decoupling is through the visibility function of photon. The visibility function of photon is the probability distribution of photon’s last scattering redshift. That is, what we have to calculate is the probability that the photon had last-scattered at zs, but has been freely streaming to the observer at z = 0 ever since (ignoring for the reionization for a moment). Let’s go back to the path-length argument. The probability that the photon has not scattered until x, and does not scatter inbetween the path length x and x + δx is δx Pno(x + δx) = Pno(x) 1 . (5.221) − λ(x) The equation above can be tranlated into the differential equation:

d ln Pno(x) 1 = = neσT, (5.222) d x −λ(x) − from which we calculate the probability of no scattering from z = 0 to all the way to the redshift zs (path length x) as τ(x) Pno(x) = e− (5.223) where Z x τ(x) = ne(x0)σTd x0 (5.224) 0 is the optical depth of the photon. Probability that the photon had scattered at least one times between x and now is P>1(x) = 1 Pno(x), (5.225) − which is called opacity. The probability that the last-scattering happens further away than x + δx is

Pls(> x + δx) = Pno(x + δx), (5.226) and the probability that the last-scattering happens closer than x is

Pls(< x) = 1 Pno(x). (5.227) − 5.6. HYDROGEN RECOMBINATION AND CMB 39

Combining the two, the probability that the last-scattering happens between x and x + δx is

Pls(x)d x = Pls(< x + d x) Pls(< x) = Pno(x + δx) + Pno(x), (5.228) − − so that dP x dτ no( ) τ(x) τ(x) Pls(x) = = e− = neσTe− . (5.229) − d x d x The probability distribution function Pls(x) of the last-scattering is the desired visibility function. More often, the visibility function is defined as the last-scattering redshift instead of the optical path length x (physical distance that each photon travels). That is, Z z dτ σTnH,tot(z)Xe(z) nH,tot(z )Xe(z ) g z e τ(z) exp σ dz 0 0 . (5.230) ( ) = dz − = 1 z H z T 0 1 z H z ( + ) ( ) − 0 ( + 0) ( 0) We show the visibility function in Fig. 5.6. The visibility function peaks at around z 1077, which means that the most of photons were last-scattered off electrons at that redshift. The isotropic' sphere around us with radius χ(z = 1077) therefore forms a last-scattering surface of photon after which the photons free-stream without being scattered. The photons at higher redshift underwent frequent scattering so that the primordial fluctuations on scales smaller than diffusion scale got equalized and erased. At redshift z > 2 106, the double Compton scattering and Bremsstrahlung process is efficient to generate new entropy (photon× number) out of the erased fluctuations, but below that redshift the photon number density is conserved. Conserved number density, but changing energy (from the small scale fluctuation) will induce the spectral distortion of the 5 observed CMB spectrum. Currently, spectral distortion is limited to less than 10− level, but the limit should be improved by future CMB mission with many frequency channels such as PIXIE.

5.6.4 Thermal decouplig of matter from radiation : Compton heating

We have discussed the photon’s last-scattering off the electron in the previous section, and show that most of photons have decoupled from electron at z 1100. Did electrons set free about the same time? The answer is no! It is because there are so many photons' per an electron. The simplest way of looking at this problem is to calculate the mean free path. For photon, mean free 1 1 path is given by λγ = (neσT)− and for electron it is λe = (nγσT)− , therefore, λ γ ne nb 11 = = (1 Yp) Xe = 7.6 10− η10Xe. (5.231) λe nγ − nγ × That is, the electron mean-free path is MUCH smaller compared to the photon’s mean-free path. Well, photons do NOT care for one scattering per ten billion, but electrons certainly do. This frequent scattering of electron off the photon result energy exchange between electrons and pho- 2 tons. In a adiabatic expansion, the temperature of electron must scale as Te 1/a . But, as a result of the tight thermal coupling due to energy exchange (via Compton scattering),∝ the electron temperature scales as the same way as the photon temperature Te 1/a! ∝ Let’s calculate when does that thermal coupling stop. The average energy transfer per Compton scattering (Compton heating) is 4 k T ργ ∆" β 2 " 4 B e , (5.232) b = γ = 2 3 mec nγ 40 CHAPTER 5. OUT-OF-EQUILIBRIUM AND FREEZE-OUT

0.0045 visibility function 0.0040

0.0035

0.0030

0.0025 ) z ( g 0.0020

0.0015

0.0010

0.0005

0.0000 500 1000 1500 2000 2500 3000 redshift z

Figure 5.6: The visibility function of the Cosmic Microwave Background photons. Most of photons were 2 2 last-scattered off electrons at z 1077 for ΩMh = 0143, Ωbh = 0.023. Note that the wing at the lower redshift side is broader due to the' delayed recombination. 5.6. HYDROGEN RECOMBINATION AND CMB 41

2 where we use the equipartition theorem 3/2kB T = 1/2mv , and ργ = nγ "γ . Note that the energy that electron gains from photon is shared by all baryons. Then, the rate of change of baryonic energy density is given by dE k T b n n σ c∆" 4n σ cρ B e , (5.233) = e γ T b = e T γ 2 d t mec from which we can read off the time scale for the themal energy exchange:

1 dE ργ k T t 1 b 4n σ c B e . (5.234) Compton− = = e T 2 Eb d t Eb mec

Using Eb = 3/2(nb + ne)kB Te, we ﬁnd

8 X ργσTc t 1 e . (5.235) Compton− = 1 2 3 (1 Yp) + Xe mec − − Let’s work out for the numbers (in the natural units).

π2 ρ T 4 13216.5 1 z 4 cm 4 γ = 15 γ = ( + ) − 2 10 1 mec = 2.5897 10 cm− , (5.236) × from which we ﬁnd 20 4 1 2.714 10− Xe(1 + z) tCompton− = × . (5.237) 1.3158 + Xe We have to equate this with Hubble rate at the matter domination:

18 3/2 H(z) = 1.22546 10− (1 + z) , (5.238) × 4 with the freezeout value Xe 3 10− , we ﬁnd that the baryons thermally decouple from photon at ' × 2/5 1.316 + Xe (1 + z) = 4.6 131.72, (5.239) Xe ' or zdec 130! The actual calculation using Xe(z) gives zdec 134, which agree pretty well with our estimation!' The temperature of the baryons were the same as the' temperature of the photons before that redshift. After the decoupling redshift, temperature of baryons drops well below the photon temperature (with different power-law index). That is, CMB has been inﬂuencing the evolution of the Universe until that redshift.