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김관석
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2025-03-24 15:11:12, 조회수 : 13 |
3 The Hot Big Bang: Selected Problems
3.1 Chemical potential of electrons
In this problem, we will show that the chemical potential of electrons was negligible in the early universe.
1. Consider fermions with 𝜇 ≠ 0 and 𝑚 ≪ 𝛵. Show that the difference between the number densities of particles and antiparticles is
(3.1.1a) 𝑛 - 𝑛̄ = 𝑔𝛵3/6π2 [π2 (𝜇/𝛵) + (𝜇/𝛵)3].
Note that this result is exact and not a truncated series. Hint: We may use that
(3.1.1b) ∫∞0 d𝑦 𝑦/(𝑒𝑦 + 1) = π2/12.
[Solution] Using (3.11-15), the number density of fermions in the relativistic limits is
(a) 𝑛 = 𝑔𝛵3/2π2 ∫∞0 d𝜉 𝜉2/(𝑒𝜉 - 𝛿 + 1), where 𝜉 ≡ 𝑝/𝛵 and 𝛿 ≡ 𝜇/𝛵.
The difference between the number densities of particles and antiparticles is
(b) 𝑛 - 𝑛̄ = 𝑔𝛵3/2π2 ∫∞0 d𝜉 [𝜉2//(𝑒𝜉 - 𝛿 + 1) - 𝜉2//(𝑒𝜉 + 𝛿 + 1).
Setting 𝑦 ≡ 𝜉 - 𝛿 in the first term and 𝑦 ≡ 𝜉 + 𝛿 in the second one, we get
(c) 𝑛 - 𝑛̄ = 𝑔𝛵3/2π2 [∫∞-𝛿 d𝑦 [(𝑦 + 𝛿)2/(𝑒𝑦 + 1) - ∫∞+𝛿 d𝑦 [(𝑦 - 𝛿)2/(𝑒𝑦 + 1)] ≡ 𝑔𝛵3/2π2 𝐾(𝛿). ⇒
(d) 𝐾(𝛿) = ∫∞-𝛿 d𝑦 [(𝑦 + 𝛿)2/(𝑒𝑦 + 1) + ∫+𝛿0 d𝑦 [(𝑦 - 𝛿)2/(𝑒𝑦 + 1) + ∫∞0 d𝑦 [(𝑦 + 𝛿)2 - (𝑦 - 𝛿)2]/(𝑒𝑦 + 1). ⇒
(e) 𝐾(𝛿) = ∫+𝛿0 d𝑦 [(-𝑦 + 𝛿)2/(𝑒-𝑦 + 1) + ∫+𝛿0 d𝑦 [(𝑦 - 𝛿)2/(𝑒𝑦 + 1) + 4𝛿 ∫∞0 d𝑦 𝑦/(𝑒𝑦 + 1).
We can combine the first two terms into a single integral from 0 to +𝛿
(f) (-𝑦 + 𝛿)2/(𝑒-𝑦 + 1) + (𝑦 - 𝛿)2/(𝑒𝑦 + 1) = (𝑦 - 𝛿)2/(𝑒𝑦 + 1) (𝑒𝑦 + 1) = (𝑦 - 𝛿)2. ⇒
(g) 𝐾(𝛿) = ∫+𝛿0 d𝑦 [(𝑦 - 𝛿)2 + 4𝛿 ∫∞0 d𝑦 𝑦/(𝑒𝑦 + 1) = 1/3 𝛿3 + π2/3 𝛿.
So now we can have
(h) 𝑛 - 𝑛̄ = 𝑔𝛵3/6π2 [π2 (𝜇/𝛵) + (𝜇/𝛵)3]. ▮
2. Use the above result to show that the chemical potential of electrons satisfies 𝜇𝑒/𝛵 ~ 10-9 and hence can be set to zero in the early universe.
[Solution] Since the universe has electrical neutrality 𝑛𝑒 - 𝑛̄𝑒 is to be equal to the proton density or about the baryon density and thew baryon-to-photon ratio 𝜂 (RE Appendix C.2.3) is conserved in the universe, we find
(a) (𝑛𝑒 - 𝑛̄𝑒)/𝑛𝛾 = 𝑛𝑝/𝑛𝛾 ≈ 𝑛b/𝑛𝛾 = 𝜂 ≈ 6 × 10-10.
Since 𝑛𝛾 = 𝜁(3)/π2 𝑔𝛾𝛵3, we get
(b) 𝜂 ≈ (𝑛𝑒 - 𝑛̄𝑒)/𝑛𝛾 = 𝑔𝑒/𝑔𝛾 π2/6𝜁(3) [𝜇𝑒/𝛵 + 1/π2 (𝜇𝑒/𝛵)3].
Therefore we find
(c) 𝜇𝑒/𝛵 ~ 10-9.
Hence the chemical potential of electrons can be set to zero in the early universe. ▮
3.2 Conservation of entropy
1. Starting from the distribution function 𝑓, and assuming vanishing chemical potential, show that the following holds in equilibrium
(3.2.1a) ∂𝑃/∂𝛵 = (𝜌 + 𝑃̄)/𝛵.
Use this result to show that entropy is conserved in equilibrium. Hint: First show that
(3.2.1b) ∂𝑓/∂𝛵 = -𝐸2/𝛵𝑝 ∂𝑓/∂𝑝.
[Solution] The Fermi-Dirac distribution or Bose-Einstein distribution function with 𝜉 = 𝐸/𝛵 = √(𝑝2 + 𝑚2) assuming vanishing chemical potential is
(a) 𝑓(𝜉) = 1/(𝑒𝜉 ± 1).
(b) ∂𝑓/∂𝛵 = ∂𝜉/∂𝛵 d𝑓/d𝜉 = -𝐸/𝛵2 d𝑓/d𝜉 = -𝐸/𝛵2 ∂𝑓/∂𝑝 ∂𝑝/∂𝜉 = -𝐸/𝛵2 𝐸𝛵/𝑝 ∂𝑓/∂𝑝 = -𝐸2/𝛵𝑝 ∂𝑓/∂𝑝.
Using (b), we can find
(c) ∂𝑃/∂𝛵 = ∂/∂𝛵 [𝑔/(2π)3 ∫ d3𝑝 𝑓(𝑝, 𝛵) 𝑝2/3𝐸] = 𝑔/2π2 ∫ d𝑝 ∂𝑓/∂𝛵 𝑝4/3𝐸 = -𝑔/2π2 ∫ d𝑝 ∂𝑓/∂𝑝 𝑝3𝐸/3𝛵.
Since for 𝑓(𝜉) ≪ 1, ∂𝑓/∂𝜉 = 𝑓(𝜉)2 - 𝑓(𝜉) ≈ -𝑓(𝜉), and ∂𝑓/∂𝑝 ≈ -𝑓(𝑝), so we get
(d) ∂𝑃/∂𝛵 ≈ 1/𝛵 𝑔/2π2 ∫ d𝑝 𝑓(𝑝)/3 ∂/∂𝑝 (𝑝3𝐸) = 1/𝛵 𝑔/2π2 ∫ d𝑝 𝑓(𝑝)/3 ∂/∂𝑝 (3𝑝2𝐸 + 𝑝3 ∂𝐸/∂𝑝) = 1/𝛵 𝑔/2π2 ∫ d𝑝 𝑓(𝑝)/3 ∂/∂𝑝 (3𝑝2𝐸 + 𝑝4/𝐸)
= 1/𝛵 𝑔/2π2 ∫ d3𝑝 [𝑓(𝑝) 𝐸 + 𝑓(𝑝) 𝑝2/3𝐸] = (𝜌 + 𝑃̄)/𝛵.
The first law of thermodynamics states that
(e) 𝛵d𝑆 = d𝑈 + 𝑃d𝑉, where 𝑆: entropy, 𝑈: the internal energy, and 𝑉: volume.
Since 𝑈 = 𝜌𝑉' using the result of (d), we find
(f) d𝑆 = (d𝑈 + 𝑃d𝑉)/𝛵 = 1/𝛵 [d{(𝜌 + 𝑃)𝑉} - 𝑉d𝑃] = 1/𝛵 d[(𝜌 + 𝑃)𝑉] - 𝑉/𝛵2 (𝜌 + 𝑃) d𝛵 ≈ d[(𝜌 + 𝑃) 𝑉/𝛵]. ⇒
(g) d𝑆/d𝑡 = d/d𝑡 [(𝜌 + 𝑃) 𝑉/𝛵] = 𝑉/𝛵 [d𝜌/d𝑡 + 1/𝑉 d𝑉/d𝑡 (𝜌 + 𝑃)] + 𝑉/𝛵 [d𝑃/d𝑡 - (𝜌 + 𝑃)/𝛵 d𝛵/d𝑡]
Since 𝑉 ∝ 𝑎3 d𝜌/d𝑡 + 1/𝑉 d𝑉/d𝑡 (𝜌 + 𝑃) = d𝜌/d𝑡 + 3𝑎̇/𝑎(𝜌 + 𝑃) = 0, which is the continuity equation. and using (d) again, d𝑃/d𝑡 - (𝜌 + 𝑃)/𝛵 d𝛵/d𝑡 = 0. Hence we get
(f) d𝑆/d𝑡 = 0,
which shows that entropy is conserved in equilibrium.. ▮
2. Assuming only Standard Model particles and interactions, we showed that the neutrinos and photon temperatures today are related by 𝛵𝜈, 0 = (4/11)1/3 𝛵𝛾, 0. Now imagine that after the neutrinos have decoupled from the SM particles, they remain in thermal equilibrium with a new massive spin-1 boson 𝑋 (and its antiparticles 𝑋̄). Eventually, this particle self-annihilates and transfers its energy to the neutrinos. How would this affect the neutrino temperature today?
[Solution] First considering the number of relativistic degrees of freedom before and after the annihilation of boson 𝑋, before annihilation the two bosons have 6 and the three kinds of neutrinos had 6 and after that there will be only neutrinos. So we find
(a) 𝑔*𝑆 = { 2 × 3 + 7/8 × 6 = 45/4, 𝛵 > 𝑚𝑋; 7/8 × 6 = 21/4, 𝛵 < 𝑚𝑋.
Since before annihilation the neutrino temperature 𝛵𝜈 is given by the known relation with the photon temperature 𝛵𝛾, we get
(b) 𝛵𝜈 = (4/11)1/3 𝛵𝛾.
After annihilation the neutrino temperature can be determine from the conservation of 𝑔*𝑆(𝑎𝛵𝜈)36 = const. So we get
(c) 𝛵𝜈,later = (45/21)1/3 𝛵𝜈,prev. So
(d) 𝛵𝜈,0 = (45/21)1/3 (4/11)1/3 𝛵𝛾 ≈ (60/77)1/3 𝛵𝛾,0 ≈ 0.92 × 𝛵𝛾,0.
Because 𝛵𝛾,0 = 2.73 K, we find 𝛵𝜈,0 = 2.51 K. ▮
3.4 Massive neutrinos
At least two of the three neutrino species in the Standard Model must have small masses. In this problem, we will explore the 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
(3.4.1) 𝜌𝜈 = 𝛵𝜈4/π2 ∫∞0 d𝜉 [𝜉2√(𝜉2 + 𝑚𝜈2/𝛵𝜈2]/(𝑒𝜉 + 1), where 𝛵𝜈 is the neutrino temperature.
[Solution] Since we assume that the neutrino mass is small enough and neutrino has 2 of the internal degree of freedom 𝑔, the energy density is
(a) 𝜌𝜈 = 𝑔/(2π)3 ∫ d3 𝐸(𝑝) 𝑓(𝑝, 𝛵𝜈) = 1/π2 ∫∞0 d𝑝 [𝑝2√(𝑝2 + 𝑚𝜈)]/(𝑒𝑝/𝛵𝜈 + 1)
Hence if we set 𝜉 ≡ 𝑝/𝛵𝜈, we then get
(b) 𝜌𝜈 = 𝛵𝜈4/π2 ∫∞0 d𝜉 [𝜉2√(𝜉2 + 𝑚𝜈2/ 𝛵𝜈2)]/(𝑒𝜉 + 1). ▮
2. By considering a series expansion for small 𝑚𝜈/𝛵𝜈 show that
(3.4.2) 𝜌𝜈 ≈ 𝜌𝜈0 (1 + 5/7π2 𝑚𝜈2/𝛵𝜈2),
where 𝜌𝜈0 is the energy density of massless neutrinos.
[Solution] 𝜌𝜈0 set 𝜌𝜈 with 𝑚𝜈 = 0, that is
(a) 𝜌𝜈0 ≡ 𝛵𝜈4/π2 ∫∞0 d𝜉 [𝜉3/(𝑒𝜉 + 1) = 7π2/120 𝛵𝜈4.
We can expand, for 𝜉 = ∞, √(𝜉2 + 𝑚𝜈2/𝛵𝜈2) ≈ 𝜉 + 1/2𝜉 𝑚𝜈2/𝛵𝜈2 and so we find
(b) 𝜌𝜈 ≈ 𝛵𝜈4/π2 ∫∞0 d𝜉 [𝜉2/(𝑒𝜉 + 1) [𝜉 + 1/2𝜉 𝑚𝜈2/𝛵𝜈2] = 𝜌𝜈0 + 𝛵𝜈2/π2 ∫∞0 d𝜉 𝜉/(𝑒𝜉 + 1) = 𝜌𝜈0 + 𝛵𝜈2𝑚𝜈2/24 = 𝜌𝜈0 (1 + 5/7π2 𝑚𝜈2/𝛵𝜈2). ▮
3. If 𝜌𝜈 is significantly larger than 𝜌𝜈0 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] The case that 𝜌𝜈 is significantly larger than 𝜌𝜈0 at recombination is, according to (b) in the previous problem and the relation between neutrino and photon temperature (3.62),
(a) 𝑚𝜈 > √(7π2/5) 𝛵𝜈, recombination = √(7π2/5) (4/11)3 𝛵𝛾, recombination ≈ 2.65 𝛵𝛾, recombination.
According to Table 3.3, 𝛵𝛾, recombination ≈ 0.29 eV, so we can find that the smallest mass of the neutrinos affects the CMB anisotropies is
(b) 𝑚𝜈 ≳ 0.77 eV. ▮
4. Estimate the redshift at which the neutrinos become non-relativistic.
[Solution] When their temperature falls below the mass. the neutrinos become non-relativistic, 𝛵𝜈, non-rel= ~ 𝑚𝜈. Since 𝛵𝜈 ∝ 1/𝑎 = 1 + 𝑧, we find
(a) 𝛵𝜈, non-rel/𝛵𝜈, 0 = 1 + 𝑧non-rel.
Since 𝛵𝜈, 0 = (4/11)1/3 𝛵𝛾, 0 and According to Appendix C.2.3 𝛵𝛾, 0 ≈ 0.235 meV, so we get 𝛵𝜈, 0 ≈ 0.17 meV
(b) 𝑧non-rel ≈ 𝑚𝜈/0.17 meV - 1. ▮
5. Compute the number density of these neutrinos today.
[Solution] Referring to (3.13-19), we know that the number density of the neutrinos at decoupling is
(a) 𝑛𝜈 = 𝑔𝜈 3𝜁(3)/4π2 𝛵3𝜈, decouple.
Since after decoupling, the neutrinos preserve the relativistic Fermi-Dirac distribution, so we have
(b) 𝑛𝜈, 0 = 𝑔𝜈 3𝜁(3)/4π2 𝛵3𝜈, 0.
And the photon number density toady is
(c) 𝑛𝛾, 0 = 𝑔𝛾 𝜁(3)/4π2 𝛵3𝛾, 0 ≈ 410 cm-3.
Hence, using (3.62), we can compute the number density of these neutrinos today, so
(d) 𝑛𝜈, 0 = 3/4 𝑔𝜈/𝑔𝛾 (𝛵𝜈, 0/𝛵𝛾, 0)3 𝑛𝛾, 0 ≈ 3/4 × 2/2 × 4/11 × 410 photons cm-3 ≈ 112 neutrinos cm-3. ▮
6. Show that their contribution to the energy density in he universe today is,
(3.4.6) 𝛺𝜈𝘩2 ≈ 𝑚𝑣/94𝑒𝑉.
Use the lower bound on the sum of the neutrino masses from oscillation experiments, ∑ 𝑚𝑣 > 0.06 eV, to derive a lower bound on the total neutrino density. How does this compare to the cosmological bound 𝛺𝜈𝘩2 < 0.001?
[Solution] Since the neutrinos become non-relativistic when 𝛵𝜈, 0 ~ 𝑚𝜈, the energy density is the same as the mass density
(a) 𝜌𝜈, 0 = 𝑚𝜈 𝑛𝜈, 0 = 3/11 𝑛𝛾, 0
Referring (3.24-25) we can get
(b) 𝛺𝜈𝘩2 = 𝜌𝜈, 0/𝜌𝛾, 0 𝛺𝛾𝘩2 = 3/11 𝑚𝜈 𝑛𝜈, 0/𝜌𝛾, 0 𝛺𝛾𝘩2 = 3/11 2𝜁(3)/𝜋2 𝛵03 15/𝜋2 1/𝛵04 𝛺𝛾𝘩2.
According to Appendix C.2.3, 𝛵𝛾, 0 ≈ 0.235 meV and 𝛺𝛾𝘩2 ≈ 2.473 × 10-5, so we find
(c) 𝛺𝜈𝘩2 ≈ 𝑚𝜈/94eV.
Since the lower bound on the sum of the neutrino masses from oscillation experiments, ∑ 𝑚𝑣 > 0.06 eV, we find 𝛺𝜈𝘩2 > 0.0006. The result is close to the cosmological bound 𝛺𝜈𝘩2 < 0.001. ▮
7. Discuss qualitatively why a much larger mass would still be compatible with the standard cosmology.
[Solution] If at recombination 𝑚𝜈 was much larger than 𝛵𝜈, then the neutrinos were non-relativistic at that time. So we can try to measure the matter density today, 𝑚𝜈 𝑛𝜈, 0. ▮
3.7 Baryon symmetry
Consider particles and antiparticles with mass 𝑚 and number density 𝑛(𝑡) and 𝑛̄(𝑡). The Boltzmann equation for 𝑛(𝑡) is
(3.7) d𝑛/d𝑡 = -3𝑛 𝑎̇/𝑎 - 𝑛𝑛̄(𝑡)⟨𝜎𝑣⟩ + 𝑃(𝑡).
Describe the physical significance of each term appearing in this equation
1. Show that (𝑛 - 𝑛̄)𝑎3 is a constant.
[Solution] We can change (3.7) into
(a) - 𝑛𝑛̄(𝑡)⟨𝜎𝑣⟩ + 𝑃(𝑡) = d𝑛/d𝑡 + 3𝑛 𝑎̇/𝑎 = 1/𝑎3 d(𝑛𝑎3)/d𝑡
By substituting 𝑛 by 𝑛̄, we get
(b) - 𝑛𝑛̄(𝑡)⟨𝜎𝑣⟩ + 𝑃(𝑡) = 1/𝑎3 d(𝑛̄𝑎3)/d𝑡
Subtracting (b) from (a), we find
(c) 1/𝑎3 d[(𝑛 - 𝑛̄)𝑎3]/d𝑡 = 0. ⇒ (𝑛 - 𝑛̄)𝑎3 = const. ▮
2. Assuming initial particle-antiparticle symmetry, show that the Boltzmann equation can be written
(3.7.2) 1/𝑎3 d(𝑛𝑎3)/d𝑡 = -⟨𝜎𝑣⟩(𝑛2 - 𝑛2eq),
where 𝑛eq denotes the equilibrium number density.
[Solution] According to (3.92) the number of the particles in a fixed volume 𝑉 ∝ 𝑎3 is conserved and assuming initial particle-antiparticle symmetry in equilibrium, we find
(a) 1/𝑎3 d(𝑛eq𝑎3)/d𝑡 = 0. ⇒ 𝑃(𝑡) = 𝑛2eq⟨𝜎𝑣⟩
Hence we have
(b) 1/𝑎3 d(𝑛𝑎3)/d𝑡 = -⟨𝜎𝑣⟩(𝑛2 - 𝑛2eq). ▮
3. Taking ⟨𝜎𝑣⟩ to be a constant, derive the late-time value of 𝑌 = 𝑛/𝛵3 in terms of the freeze-out temperature 𝛵𝑓.
[Solution] Consider the dark matter possibly produced in the early universe. Assuming there are no other particle annihilation during the dark matter freeze out, we may take 𝛵 ∝ 𝑎-1. So we have
(a) 𝑌 = 𝑛/𝛵3 = 𝑛𝑎3/𝛵03 ⇒ d𝑌/d𝑡 = 1/𝛵03 d(𝑛𝑎3)/d𝑡 = 𝑎3/𝛵0 1/𝑎3 d(𝑛𝑎3)/d𝑡 = - ⟨𝜎𝑣⟩/𝛵3 (𝑛2 - 𝑛2eq) = - 𝛵3⟨𝜎𝑣⟩(𝑌2 - 𝑌2eq).
As in (3.99), we introduce 𝑥 ≡ 𝑀/𝛵, where d𝑥/d𝑡 = 𝐻𝑥 and we get
(b) d𝑌/d𝑡 = d𝑌/d𝑥 d𝑥/d𝑡 = - 1/𝐻𝑥 𝛵3⟨𝜎𝑣⟩(𝑌2 - 𝑌2eq).
Since, as in (3.100), for weakly interacting particles the decoupling occurs at very early times, during the radiation-dominated era, where we can write 𝐻(𝛵) = 𝐻(𝑀)/𝑥2. Then it becomes so-called Riccati equation
(c) d𝑌/d𝑥 = -𝑀3/𝐻(𝑀) ⟨𝜎𝑣⟩/𝑥2 (𝑌2 - 𝑌2eq) = -𝜆/𝑥2 (𝑌2 - 𝑌2eq), where 𝜆 ≡ 𝑀3⟨𝜎𝑣⟩/𝐻(𝑀).
After freeze-out 𝑌 ≫ 𝑌eq, so that
(d) d𝑌/d𝑥 ≈ -𝜆/𝑥2 𝑌2. ⇒ -d𝑌/𝑌2 = 𝜆 d𝑥/𝑥2
Integrating this from 𝑥 = 𝑥𝑓 to 𝑥 = ∞, we find
(e) 1/𝑌∞ - 1/𝑌𝑓 ≈ 𝜆/𝑥𝑓, where 𝑌𝑓 ≡ 𝑌(𝑥𝑓).
Typically 𝑌𝑓 ≫ 𝑌∞, so we finally get
(f) 𝑌∞ ≈ 𝑥𝑓/𝜆 = 1/𝛵𝑓 𝐻(𝑀)/⟨𝜎𝑣⟩𝑀2. ▮
4. If there was a speed-up in the expansion rate of the universe caused by the addition of extra relativistic species, what would happen to the abundance of the surviving massive particles and why?
[Solution] According to (e) in the previous problem, 𝑌 ≡ 𝑛/𝛵3 and 𝑌∞ ∝𝐻(𝑀), so the speed-up in the expansion rate of the universe would cause the greater abundance of the surviving massive particles and at the same time the faster expansion would make 𝛵𝑓 higher, i.e. earlier in period then the the surviving massive particles number density would increase more rapidly. ▮
Now apply this to proton-anti-proton annihilation. We may use that ⟨𝜎𝑣⟩ ≈ 100 GeV-2 and 𝑚𝑝 ≈ 1 GeV.
5. Estimate the proton-to-photon ratio, 𝑛𝑝/𝑛𝛾. How does our result compare to observations?
[Solution] We can write the photon density today 𝑛𝑝,0 = 𝑌∞𝛵03 and the photon density is given by (3.23) 𝑛𝛾,0 = 2𝜁(3)/π2 𝛵03 in the relativistic equilibrium. So we have
(a) 𝑛𝑝,0/𝑛𝛾,0 = π2/2𝜁(3) 𝑌∞ = π2/2𝜁(3) 1/𝛵𝑓 𝐻(𝑚𝑝)/⟨𝜎𝑣⟩𝑚𝑝2.
During the radiation era, as (3.55), the Friedmann equation reads
(b) 𝐻2 = 𝜌/3𝑀𝑝𝑙2 ≈ π2/90 𝑔*𝛵4, ⇒ 𝐻 ≈ 𝛵2/𝑀𝑝𝑙,
where we assume that before electrons and positrons annihilate 𝑔* = 10.75. Then we get
(c) 𝑛𝑝,0/𝑛𝛾,0 = π2/2𝜁(3) 1/𝛵𝑓⟨𝜎𝑣⟩.
According to (3.104) the decoupling condition is 𝛤(𝑥𝑓) ≈ 𝐻(𝑥𝑓) with 𝑥𝑓 ≡ 𝑚𝑝/𝛵𝑓 and 𝐻 = 𝐻(𝑚𝑝)/𝑥𝑓2, where 𝛤 = 𝑛𝑝⟨𝜎𝑣⟩ = 𝑛̄𝑝⟨𝜎𝑣⟩. Since proton is the non-relativistic, 𝑛𝑝,0 ≈ 𝑛𝑝,,eq, so we get
(d) 𝑛𝑝,0 ≈ 𝑛𝑝,eq = 𝑔 (𝑚𝑝𝛵/2π)3/2 𝑒-𝑚𝑝/𝛵 = 2𝑚3𝑝/(2π)3/2 𝑥-3/2𝑒-𝑥.
Since 𝛤(𝑥𝑓) ≈ 𝐻(𝑥𝑓), we find
(e) 2𝑚3𝑝⟨𝜎𝑣⟩/(2π)3/2 𝑥𝑓-3/2𝑒-𝑥𝑓 = 𝐻(𝑚𝑝)𝑥𝑓-2.
Assuming 𝑚𝑝 ≈ 𝛵𝑓 we get 𝐻(𝑚𝑝) ≈ 𝑚𝑝2/𝑀𝑝𝑙, so
(f) 𝑒𝑥𝑓 = 2/(2π)3/2 𝑚𝑝𝑀𝑝𝑙⟨𝜎𝑣⟩𝑥𝑓1/2 ⇒ 𝑥𝑓 = ln [2/(2π)3/2 𝑚𝑝𝑀𝑝𝑙⟨𝜎𝑣⟩] + 1/2 ln 𝑥𝑓.
Since 𝑚𝑝 = 1 GeV and ⟨𝜎𝑣⟩ = 100 GeV-2 and if we ignore the second term of the right side, 1/2 ln 𝑥𝑓, then we can get the first approximation of 𝑥𝑓. Next we substitute it the second term and find the second approximation which
(g) 𝑥𝑓 ≈ 44.7. ⇒ 𝑥𝑓 ≈ 46.6.
(h) 𝛵𝑓 = 𝑚𝑝/𝑥𝑓 = 1 GeV/46.6 ≈ 20 MeV.
Finally we can estimate the proton-to-photon ratio, so that
(i) 𝑛𝑝,0/𝑛𝛾,0 = π2/2𝜁(3) 1/𝛵𝑓⟨𝜎𝑣⟩ ≈ 10-18.
Referring to Appendix C.2.3, the observed baryon-to-photon ratio, 𝜂 ≈ 6 × 10-10. So our result is too much far from the observation. ▮
3.8 Big Bang nucleosynthesis
BBN is sensitive to the physical conditions in the first three minutes of the hot Big Bang. In this problem, we will explore how this can be used to probe various types of physics beyond the Standard Model. We will be asked to assess how changes to the physics during BBN would affect the amount of helium being produced,
1. Let us approximate the freeze-out temperature of the neutrons, 𝛵𝑓, by the decoupling temperature of the neutrinos. Write down an expression for 𝛵𝑓 in terms of the number of relativistic species, 𝑔*, Newton's constant 𝐺, and Fermi's constant, 𝐺𝐹. Explain how a change in 𝛵𝑓 affects the neutrino-to-proton ratio and the final helium abundance.
[Solution] When interaction rate, 𝛤𝜈 is similar to Hubble scale 𝐻, neutrinos decouple, using (3.58) and (3.55), we get
(a) 𝛤𝜈 ≈ 𝐺2𝐹𝛵5dec ~ 𝐻 ≈ √(π2/90 𝑔*𝛵4dec/𝑀2𝑝𝑙) = √(8π3/90 𝐺𝑔*𝛵4dec). ⇒
(b) 𝛵dec ∝ 𝑔*1/6𝐺1/6𝐺-2/3𝐹.
Since the freeze-out temperature of the neutrons 𝛵𝑛_𝑓 is similar to the decoupling temperature of neutrinos 𝛵𝜈_dec, using (3.120) we get
(c) 𝑛𝑛/𝑛𝑝 ≈ 𝑒-𝑄/𝛵𝑛_𝑓, where 𝑄 ≡ 𝑚𝑛 - 𝑚𝑝 = 1.30 MeV.
For below 𝛵𝑛_𝑓 ~ 1 MeV, the neutron fraction gets smaller and essentially all the neutrons would incorporated into 4He. This process would increase the final helium abundance. ▮
2. Describe how a change in the baryon-to-photon ratio 𝜂 would affect the amount of helium and deuterium produced by BBN.
[Solution] According to (3.138), the smallness of 𝜂 inhibits the production of deuterium until the temperature drops below the binding energy of deuterium 𝛣𝐷.
(a) (𝑛𝐷/𝑛𝑝)eq ≈ 𝜂 (𝛵/𝑚𝑝)3/2 𝑒𝛣𝐷/𝛵, where 𝛣𝐷 = 2.2 MeV.
Once the formation of deuterium starts, 3𝐻, 4𝐻 and 4𝐻 can also begin. So referring Fig. 3.14, we find that raising 𝜂 leads to a small increase in 4𝐻 abundance, a small decrease in 3𝐻 abundance and a rapid decrease in deuterium abundance. ▮
3. Discuss the effect of the following suppositions on the production of helium:
• 𝑔*: There were more relativistic species during BBN than we expected.
• 𝐺𝐹: The weak interaction strength was smaller than we thought.
• 𝐺: Newton's constant was larger during BBN.
• 𝑄: The neutron-proton mass difference was larger than supposed.
• 𝜏𝑛: The neutron lifetime was shorter than assumed.
• 𝜇𝜈: There were many more neutrinos than antineutrinos.
[Solution] Using (b) in Problem 3.8-1, we find
• 𝑔*: Since 𝛵dec ∝ 𝑔*1/6, more relativistic species during BBN raises the 𝛵𝑓 and hence leads to a increase in helium abundance.
• 𝐺𝐹: Since 𝛵dec ∝ 𝐺-2/3𝐹, the smaller weak interaction strength raises the 𝛵𝑓 and hence leads to a increase in helium abundance.
• 𝐺: Since 𝛵dec ∝ 𝐺1/6, Newton's constant was larger during BBN raises the 𝛵𝑓 and hence leads to a increase in helium abundance.
Using (c) in Problem 3.8-1, 𝑛𝑛/𝑛𝑝 ≈ 𝑒-𝑄/𝛵𝑛_𝑓 we find
• 𝑄: The larger neutron-proton mass difference would decrease 𝑛𝑛/𝑛𝑝 and hence decrease the final helium abundance.
Using (3.133), the neutron fraction 𝑋𝑛 = 0.15 𝑒-𝑡/𝜏𝑛 we find
• 𝜏𝑛: The neutron lifetime was shorter than assumed would decrease 𝑛𝑛/𝑛𝑝 and hence decrease the final helium abundance.
• 𝜇𝜈: At early times the baryonic matter was mostly in the form of protons and neutrons. which were coupled
(a) 𝑛 + 𝜈𝑒 ⟷ 𝑝 + 𝑒-, 𝑛 + 𝑒+ ⟷ 𝑝 + 𝜈̄𝑒.
So if there were many more neutrinos than antineutrinos, then more neutrinos could convert more protons and hence this would decrease the final helium abundance. ▮
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