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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 ¡Õ 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_{𝐹}^{2}𝛵^{2}, where 𝐺_{𝐹} ≈ 1.2 ¡¿ 10^{5} GeV^{2} is Fermi's constant. Taking the number density to be 𝑛 ≈ 𝛵^{3}, the interaction rate becomes
(3.58) 𝛤 = 𝑛𝜎∣𝑣∣ ≈ 𝐺_{𝐹}^{2}𝛵^{5}
As the temperature decrease, the interaction rate drops much more rapidly than the Hubble rate 𝐻 ≈ 𝛵^{2}/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 FermiDirac distribution (even after they become nonrelativistic at later time). In Section 2.2.1, we showed that the physical momentum of freestreaming particles scales as 𝑝 ¡ð𝑎^{1}. It's convenient to define the timeindependent combination 𝑞 ¡Õ 𝑎𝑝, so that the neutrino number density is
(3.60) 𝑛_{𝜈} ¡ð𝑎^{3} ¡ò 𝑑^{3}𝑞 1/[exp(𝑞/𝑎𝛵_{𝜈}) + 1].
After decoupling, particle number conservation requires 𝑛_{𝜈} ¡ð𝑎^{3}, which is only consistent with (3.60) if the neutrino temperature evolves as 𝛵_{𝜈} ¡ð𝑎^{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} ¡ð𝑎^{1} in the photon temperature.
Electronpositron annihilation
Shortly after the neutrinos decouple, the temperature drops below the electron mass, so that electrons and positrons can annihilate into photons.
𝑒^{+} + 𝑒^{} ¡æ 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 𝛼^{2}/𝑚_{𝑒} ~ 10^{18} s (where 𝛼 is the finestructure constant), which is much less than the age of the universe (~1 s) at the time. This means that the 𝑒^{¡¾}𝛾 plasma evolves quasiadiabatically into the 𝛾only plasma. Entropy is therefore conserved during the process. Taking 𝑔_{*𝑆}(𝑎𝛵_{𝛾})^{3} to remain constant, we find that 𝑎𝛵_{𝛾} increases after electronpositron 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 ¡ì 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 account^{9} 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.
Electronpositron 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 nonrelativistic today. It used to be believed that neutrinos were massless, in that case
(3.66) 𝜌_{𝜈} = 7/8 𝑁_{eff} (4/11)^{4/3}𝜌_{𝛾} ¢¡ 𝛺_{𝜈}𝘩^{2} ≈ 1.7 ¡¿ 10^{5} (𝑚_{𝜈} = 0). [𝘩 = 𝐻_{0}/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 ¢² 𝑚_{𝜈,𝑖} > 0.06 eV. massive neutrinos behave as radiationlike particles in the early universe (for 𝑚_{𝜈} < 0.2 eV, neutrinos are relativistic at recombination) and as matterlike particles in the late universe (see Fig. 3.4). In Problem 3.4, it will be shown that energy of massive neutrino, 𝜌_{𝜈} = ¢² 𝑚_{𝜈,𝑖}𝑛_{𝜈,𝑖}, corresponds to
(3.67) 𝛺_{𝜈}𝘩^{2} ≈ ¢² 𝑚_{𝜈,𝑖}/(94 eV). [𝘩 = 𝐻_{0}/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 ¢² 𝑚_{𝜈,𝑖} < 15 eV (using 𝘩 ≈ 0.7). Massive neutrinos also affect the latetime expansion rate which is constraint by CBM and BAO(Baryon acoustic oscillations) measurements. The current Planck constraint is ¢² 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) 𝜇_{1} + 𝜇_{2} = 𝜇_{3} + 𝜇_{4}.
• 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 particleantiparticle 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 freezeout of electrons). We will revisit these nonequilibrium 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_{𝑖}𝛵/2¥ð)^{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_{𝑒}𝑛_{𝑝})_{eq} = 𝑔_{𝐻}/𝑔_{𝑒}𝑔_{𝑝} (𝑚_{𝐻}/𝑚_{𝑒}𝑚_{𝑝} 2¥ð/𝛵)^{3/2} 𝑒^{(𝑚𝑝 + 𝑚𝑒  𝑚𝐻)/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} ¡Õ 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_{𝑒}^{2})_{eq} = (2¥ð/𝑚_{𝑒}𝛵)^{3/2} 𝑒^{𝐸𝐼/𝛵}.
It is convenient to describe the process of recombination in terms of the free electron fraction
(3.75) 𝑋_{𝑒} ¡Õ 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)/¥ð^{2} 𝛵^{3}, [RE (3.24) 𝑛_{𝛾,0} = 2𝜁(3)/¥ð^{2} 𝛵_{0}^{3}]
where 𝜂 is the baryontophoton ratio. We can then write
(3.77) (1  𝑋_{𝑒})/𝑋_{𝑒}^{2} = 𝑛_{𝐻}/𝑛_{𝑒}^{2} 𝑛_{𝑏},
and substituting (3.74), we arrive at the socalled Saha equation
(3.78) [(1  𝑋_{𝑒})/𝑋_{𝑒}^{2}]_{eq} = 2𝜁(3)/¥ð^{2} 𝜂 (2¥ð𝛵/𝑚_{𝑒})^{}3/2 𝑒^{𝐸𝐼/𝛵}
The solution to this equation is
(3.79) 𝑋_{𝑒} = [1 ¡¾ ¡î(1 + 4𝑓)]/2𝑓,^{*} with 𝑓(𝛵,𝜂) = 2𝜁(3)/¥ð^{2} 𝜂 (2¥ð𝛵/𝑚_{𝑒})^{}3/2 𝑒^{𝐸𝐼/𝛵}, ^{*}[correcton: from [1 + ¡î(1 + 4𝑓)]/2𝑓 to [1 ¡¾ ¡î(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} ¡ì 𝐸_{𝐼} = 13.6 eV, is that there are many photons for each hydrogen atom. Even when 𝛵 < 𝐸_{𝐼}, the highenergy tail of the
𝐸_{𝐼}, the highenergy 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 ¡¿ 10^{6} has 𝐸_{𝛾} > 20 𝛵, and one in 3 ¡¿ 10^{10} has 𝐸_{𝛾} > 30 𝛵. Since there are over 10^{9} photons per baryon, rare highenergy photons are still present in sufficient numbers, unless the temperature drops far below the binding energy.
𝛵_{rec} = 𝛵_{0}(1 + 𝑧_{rec}), with 𝛵_{0} =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 matterradiation equality is at 𝑧_{rec} ≈ 3400, we conclude that recombination occurred in matterdominated 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 matterradiation 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} ¡ð 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)/¥ð^{2} 𝜂 𝑋_{𝑒}(𝛵_{dec}) 𝛵_{dec}^{3},
(3.83) 𝐻(𝛵_{dec}) = 𝐻_{0} ¡î𝛺_{𝑚} (𝛵_{dec}/𝛵_{0})^{3/2}, we get
(3.84) 𝑋_{𝑒}(𝛵_{dec}) 𝛵_{dec}^{3/2} ≈ ¥ð^{2}/2𝜁(3) 𝐻_{0}¡î𝛺_{𝑚}/𝑛𝜎_{𝛵}𝛵_{0}^{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,
(3.8586) 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 ¡æ 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} ≈ ¥ð^{2}/2𝜁(3) 𝐻_{0}¡î𝛺_{𝑚}/𝑛𝜎_{𝛵}𝛵_{0}^{3/2}. and 𝑋_{𝑒} ¡Õ 1
(b) 𝛵_{dec} = [¥ð^{2}/2𝜁(3) 𝐻_{0}¡î𝛺_{𝑚}/𝑛𝜎_{𝛵}𝛵_{0}^{3/2}]^{2/3}
To evaluate this we first write all quantities in natural units, using
(c) 𝑐 = 3 ¡¿ 10^{8} m s^{1} ¡Õ 1, ℏ𝑐 = 2 ¡¿ 10^{7} eV m ¡Õ 1
For we use 𝐻_{0} ≈ 70 km s^{1} Mpc^{1} = 70/3 ¡¿ 10^{5} (3.1 ¡¿ 10^{22} m)^{1} = 70/3 ¡¿ 10^{5} 2 ¡¿ 10^{7}/3.1 ¡¿ 10^{22} eV = 1.5 ¡¿ 10^{33} eV
The photon temperature today is 𝛵_{0} ≈ 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} m^{2} = 6.65 ¡¿ 10^{29}/(2 ¡¿ 10^{7})^{2} eV^{2} = 1.66 ¡¿ 10^{15} eV^{2}
Substituting these results, ¥ð^{2}/2𝜁(3) ≈ 4.105 and 𝛺_{𝑚} = 0.315 [RE Wikipedia LambdaCDM model] into the equation gives
(d) 𝛵_{dec} ≈ [4.105 ¡¿ (1.5 ¡¿ 10^{33} ¡¿ ¡î0.315)/{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} = 𝛵_{0}(1 + 𝑧_{dec}), [Wikipedia Cosmic micro background: 𝛵_{r} = 2.725 K ¡¿ (1 + 𝑧)]
(e) 𝑧_{dec} = 𝛵_{dec}/𝛵_{0}  1 = 9.76/0.23  1 ≈ 41. ▮
Lastscattering
The scattering of photons off electrons essentially stops at photon decoupling. To define the precise moment of lastscattering, 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 𝑡_{0} > 𝑡 is
(3.91) 𝛤_{𝛾}(𝑡) = ¡ò𝑡_{}^{𝑡0} 𝛤_{𝛾}(𝑡) 𝑑𝑡,
This probability is also called the optical depth. Taking 𝑡_{0} to be the present time, the moment of latscattering is defined by 𝜏(𝑡_{*}) ¡Õ 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 nonequilibrium analysis of recombination in Section 3.2.5.
When we observe the CMB, we are detecting photons from this surface of lastscattering (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 lastscattering surface today is 42 billion lightyears. Of course, lastscattering is a probabilistic conceptnot all photons experienced their last scattering event at the same timeso there is some thickness to the lastscattering 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/𝑐^{3} 4¥ð𝑓^{2}/(𝑒^{𝘩𝑓/𝑘𝛣𝛵}  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) ¡ð𝑎(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 ¡ð 𝑎(𝑡)^{1}. Te early e relic radiation encodes the early equilibrium phase of the hot Big Bang.
CMB experiments observe the socalled 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/4¥ð 𝛿𝑉 = 2/𝑐^{3} 𝑓^{2}𝑑𝑓/(𝑒^{𝘩𝑓/𝑘𝛣𝛵}  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/𝑐^{2} 𝑓^{2}𝑑𝑓/(𝑒^{𝘩𝑓/𝑘𝛣𝛵}  1).
Since each photon has energy 𝘩𝑓, the flux of energy across the surface (per unit frequency) is
(3.91) 𝐼_{𝑓} = 2𝘩/𝑐^{2} 𝑓^{3}/(𝑒^{𝘩𝑓/𝑘𝛣𝛵}  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 FermiDirac distribution. The spectral distortion arise because the energy dependence of the weak interaction causes neutrinos in the highenergy tail to interact more strongly.
^{10} But the Planck constraint on 𝑁_{eff} = 2.99 ¡¾ 0.17 (¡Á 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 antialigned, 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, He^{2+} 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.
Problem 3.4 Massive neutrinos
At least two of the three neutrino species in the Standard Model must have small masses. We will explore the cosmological consequences of this neutrino mass.
1. Let us assume that the neutrino mass is small enough, so that the neutrinos are relativistic at decoupling. Show that the energy density after decoupling is
(a) 𝜌_{𝜈} = 𝛵_{𝜈}^{4}/¥ð^{2} ¡ò_{0}^{¡Ä} 𝑑𝜉 𝜉^{2} ¡î(𝜉^{2} + 𝑚_{𝜈}^{2}/𝛵_{𝜈}^{2})/(𝑒^{𝜉} + 1),
where 𝛵_{𝜈} is the neutrino temperature.
[Solution] Since the neutrinos are relativistic when they decouple and The contribution of a neutrino flavor with mass 𝑚_{𝜈} to the energy density is
(b) 𝜌_{𝜈} = 𝑔/(2¥ð)^{3} ¡ò 𝑑^{3}𝑝 𝑓(𝑝,𝛵) 𝐸(𝑝), 𝑔_{𝜈} = 2; 𝜌_{𝜈} = 2/2¥ð^{2} ¡ò_{0}^{¡Ä} 𝑑𝑝 𝑝^{2}¡î(𝑝^{2} + 𝑚^{2})/[𝑒^{{¡î(𝑝2 + 𝑚2)/𝛵}} ¡¾ 1]. + sign for fermion, 𝑚_{𝜈}^{2}/𝛵_{𝜈} ≈ 0;
(c) 𝜌_{𝜈} = 1/¥ð^{2} ¡ò_{0}^{¡Ä} 𝑑𝑝 𝑝^{2}¡î(𝑝^{2} + 𝑚^{2})/𝑒^{𝑝/𝛵𝜈} + 1]. 𝜉 ¡Õ 𝑝/𝛵_{𝜈}; 𝜌_{𝜈} = 1/¥ð^{2} ¡ò_{0}^{¡Ä} 𝑑(𝜉𝛵_{𝜈}) (𝜉𝛵_{𝜈})^{2}¡î[(𝜉𝛵_{𝜈})^{2} + 𝑚^{2}]/(𝑒^{𝜉} + 1) ¢¡
(d) 𝜌_{𝜈} = 𝛵_{𝜈}^{4}/¥ð^{2} ¡ò_{0}^{¡Ä} 𝑑𝜉 𝜉^{2} ¡î(𝜉^{2} + 𝑚_{𝜈}^{2}/𝛵_{𝜈}^{2})/(𝑒^{𝜉} + 1). ▮
2. By considering a series expansion for small 𝑚_{𝜈}/𝛵_{𝜈} show that
(e) 𝜌_{𝜈} ≈ 𝜌_{𝜈0} (1 + 5/7¥ð^{2} 𝑚_{𝜈}^{2}/𝛵_{𝜈}^{2}),
where 𝜌_{𝜈0} is the energy density of massless neutrinos.
[Solution] For a massless neutrinos, the energy density formula derived above with 𝑚_{𝜈} = 0,
(f) 𝜌_{𝜈0} = 𝛵_{𝜈}^{4}/¥ð^{2} ¡ò_{0}^{¡Ä} 𝑑𝜉 𝜉^{3}/(𝑒^{𝜉} + 1) = 𝛵_{𝜈}^{4}/¥ð^{2} 7¥ð^{4}/120 = 7¥ð^{2}/120 𝛵_{𝜈}^{4}.
Since 𝑚_{𝜈}/𝛵_{𝜈} is small and using Taylor expansion ¡î(𝜉^{2} + 𝑚_{𝜈}^{2}/𝛵_{𝜈}^{2}) ≈ 𝜉 + 1/2𝜉 (𝑚_{𝜈}/𝛵_{𝜈})^{2}
(g) 𝜌_{𝜈} ≈ 𝛵_{𝜈}^{4}/¥ð^{2} ¡ò_{0}^{¡Ä} 𝑑𝜉 𝜉^{2}/(𝑒^{𝜉} + 1) [𝜉 + 1/2𝜉 (𝑚_{𝜈}/𝛵_{𝜈})^{2}] = 𝜌_{𝜈0} + 𝛵_{𝜈}^{2}𝑚_{𝜈}^{2}/2¥ð^{2} ¡ò_{0}^{¡Ä} 𝑑𝜉 𝜉/(𝑒^{𝜉} + 1) = 𝜌_{𝜈0} + 𝛵_{𝜈}^{2}𝑚_{𝜈}^{2}/2¥ð^{2} ¥ð^{2}/12,
(h) 𝜌_{𝜈} ≈ 𝜌_{𝜈0} + 𝛵_{𝜈}^{2}𝑚_{𝜈}^{2}/24 = 𝜌_{𝜈0}(1 + 5/7¥ð^{2} 𝑚_{𝜈}^{2}/𝛵_{𝜈}^{2}). ▮
3. If 𝜌_{𝜈} is significantly larger than 𝜌_{𝜈0} at recombination, then the mass of the neutrinos affects the CMB anisotropies. What is the smallest neutrino mass that is observable in the CMB?
[Solution] This implies that 5/7¥ð^{2} 𝑚_{𝜈}^{2}/𝛵_{𝜈}^{2} > 0, so
(i) 𝑚_{𝜈} > ¡î(7¥ð^{2}/5) 𝛵_{𝜈,rec} = ¡î(7¥ð^{2}/5) (4/11)^{2}1/3 𝛵_{𝛾,rec}
where we have used the relation between neutrino and photon temperatures after electronpositron annihilation. 𝛵_{𝛾,rec} ≈ 0.32 eV, so
(j) 𝑚_{𝜈} > ¡î(7¥ð^{2}/5) (4/11)^{2}1/3 ¡¿ 0.32 eV ≈ 0.85 eV. ▮
If the neutrino mass is larger than the present photon temperature 𝛵_{𝛾,0} ≈ 0.235 meV, then these neutrinos will be nonrelativistic today.
4. Estimate the redshift at which the neutrinos become nonrelativistic.
[Solution] Neutrinos become nonrelativistic when their temperature falls below their mass at 𝛵_{𝜈,nr} ~ 𝑚_{𝜈}. Recall that after decoupling theit temperature evolves 𝛵_{𝜈} ¡ð 𝑎^{1} and 1 + 𝑧 = 𝑎^{1}, so
(k) 𝛵_{𝜈,nr}/𝛵_{𝜈,0} = 1 + 𝑧_{nr} ¢¡ 𝑧_{nr} = 𝛵_{𝜈,nr}/𝛵_{𝜈,0}  1 ≈ 𝛵_{𝜈,nr}/1.95 K  1 ≈ 𝑚_{𝜈}/(4/11)^{1/3}𝛵_{𝛾,0} ≈ (𝑚_{𝜈}/0.714 ¡¿ 0.2348 meV) 1 ≈ (𝑚_{𝜈}/0.17 meV) 1. ▮
5. Compute the number density of these neutrinos today.
[Solution] When the neutrinos decoupled, they were in relativistic limit, so we can use equation 𝑛 = 3/4 𝜁(3)/¥ð^{2} 𝑔 𝛵^{3} (for femion), therefore
(l) 𝑛_{𝜈} = 3𝜁(3)/4¥ð^{2} 𝑔_{𝜈} 𝛵_{dec}^{3}.
Since neutrinos are decoupled and temperature redshifts simply as 𝛵_{𝜈} ¡ð 𝑎_{1} (see Fig. 3.3), so we can write
(m) 𝑛_{𝜈,0} = 3𝜁(3)/4¥ð^{2} 𝑔_{𝜈} 𝛵_{𝜈,0}^{3},
where we allow the neutrinos to evolve on geodesics. However, we have the photon number density today which is given by relativistic thermodynamics
(n) 𝑛_{𝛾,0} = 𝜁(3)/¥ð^{2} 𝑔_{𝛾} 𝛵_{𝛾,0}^{3},
Now we can compute the ratio of the two number density using 𝛵_{𝜈} = (4/11)^{1/3} 𝛵_{𝛾},
(o) 𝑛_{𝜈,0}/𝑛_{𝛾,0} = 3/4 𝑔_{𝜈}/𝑔_{𝛾} (𝛵_{𝜈,0}/𝛵_{𝛾,0})^{3} = 3/4 ¡¿ 2/2 ¡¿ 4/11 = 3/11. ▮
6. Show that their contribution to the energy density in the universe today is
(p) 𝛺_{𝜈}𝘩^{2} ≈ 𝑚_{𝜈}/94 EV.
Use the lower bound on the sum of the neutrino masses from oscillation experiments, ¢² 𝑚_{𝜈} > 0.06 eV, to derive bound on the total neutrino density. How does this compare to the cosmological bound 𝛺_{𝜈}𝘩^{2} < 0.001 ?
[Solution] Since the neutrinos are nonrelativistic today, their energy density is the same as their mass density, so
(q) 𝜌_{𝜈,0} = 𝑚_{𝜈}𝑛_{𝜈,0} = 3/11 𝑚_{𝜈}𝑛_{𝛾,0}.
The fractional energy in such a neutrino is, using 𝛵_{𝛾,0} ≈ 0.2348 meV and 𝛺_{𝛾,0}𝘩^{2} ≈ 2.47 ¡¿ 10^{5},
(r) 𝛺_{𝜈}𝘩^{2} = 𝜌_{𝜈,0}/𝜌_{𝛾,0} 𝛺_{𝛾}𝘩^{2} = 3/11 𝑚_{𝜈}𝑛_{𝛾,0}/𝜌_{𝛾,0} = 3/11 𝑚_{𝜈} (2𝜁(3)/¥ð^{2} 𝛵_{𝛾,0}^{3})/(¥ð^{2}/15 𝛵_{𝛾,0}^{4}) = 90𝜁(3)/11¥ð^{2} 𝑚_{𝜈}/0.2348 meV 2.47 ¡¿ 10^{5} ≈ 𝑚_{𝜈}/94eV. ▮
If ¢² 𝑚_{𝜈} > 0.06 eV, we get 𝛺_{𝜈}𝘩^{2} > 0.0006, which is not far from the cosmological upper bound, 𝛺_{𝜈}𝘩^{2} < 0.001. ▮
7. Discuss qualitatively why a much larger mass would still be compatible with the standard cosmology.
[Solution] If 𝑚_{𝜈} ¡í 𝛵_{dec}, then the neutrinos were already nonrelativistic at decoupling. As we can see in Fig. 3.4, at late time the energy density of massive neutrinos start to dominate over photons. Now we can measure the matter density of neutrinos without any constraint. ▮ 

