14 Lecture 14: Early Universe

Transcription

1 PHYS 652: Astrophysics Lecture 14: Early Universe True science teaches us to doubt and, in ignorance, to refrain. Claude Bernard The Big Picture: Today we introduce the Boltzmann equation for annihilation as a tool for studying the early Universe. We also begin to discuss the Big Bang Nucleosynthesis BBN) during which light elements formed. The very early Universe was hot and dense, resulting in particle interactions occurring much more frequently than today. For example, while photon can today traverse the entire Universe without interacting deflection or capture), resulting in a mean-free path greater than cm, the mean-free path of a photon when the Universe was 1 second old was about the size of an atom. This resulted in a large number of interactions which kept the interacting constituents of the Universe in equilibrium. As the Universe expanded, the mean-free path of particles increased thus decreasing the rates of interactions to the point where these could no longer maintain equilibrium conditions. Different constituents of the Universe decoupled fell out of equilibrium with the rest of the Universe at different times, which determined their abundance. Falling out of equilibriulayed a vital role in: 1. the formation of the light elements during Big Bang Nucleosynthesis BBN); 2. recombination of electrons and protons into neutral hydrogen when the temperature was on the order of 1 4 ev; 3. production of dark matter in the early Universe. All three of these important phenomena are studied with the same formalism: the Boltzmann equation. Boltzmann Equation for Annihilation The Boltzmann equation generalizes the Friedmann s second equation which describes how an abundance of a specie of particles evolves with time ρ + 3ρ + P) ȧ a 0, Pρ) 0, dust approximation for matter) ρ + 3ρȧ a 0 a 3 d dt ρa 3 ) 0 a 3 d dt na 3 ) 0, 271) where n is the abundance number density) of a specie. The equation above is valid for one specie in equilibrium, and does not account for creation and annihilation of particles. The Boltzmann equation relates the rate of change in the abundance of a given particle to the difference between the rates for producing and eliminating the species. It quantifies the abundance of a specie 1 n 1 ) involved in a reaction with a specie 2 to produce a pair of species 3 and 4, 70

2 PHYS 652: Astrophysics 71 i.e., : a 3 d n1 a 3) d 3 p 1 d 3 p 2 d 3 p 3 d 3 p 4 dt 2π) 3 2E 1 2π) 3 2E 2 2π) 3 2E 3 2π) 3 2E 4 2π) 4 δ 3 p 1 + p 2 p 3 p 4 )δe 1 + E 2 E 3 E 4 ) M 2 {f 3 f 4 [1 ± f 1 ][1 ± f 2 ] f 1 f 2 [1 ± f 3 ][1 ± f 4 ]}. 272) In the absence of interactions, the right-hand side of the equation above vanishes, and the Boltzmann equation reduces to the second Friedmann s equation. From the equation above we see that: the rate of production of specie 1 is proportional to the abundance of species 3 and 4; the rate of loss of specie 1 is proportional to the abundance of species 1 and 2; the likelihood of production of a particle is higher if it is a boson than a fermion: + for Bose enhancement and - for Pauli blocking; of species 1 and 2; Dirac delta function enforce energy and momentum conservation energies are related to the momenta by E p 2 + m 2 ; 2π) 4 factor comes from replacing discrete Kronecker delta with continuous Dirac delta function; the amplitude M is determined from the physical processes taking place α, the fine structure constant for Compton scattering); to find the total number of interactions, we must integrate over all momenta; the factor 2E in the denominator arises because the phase-space integrals are four-dimensional 4-momentum) three components of spatial momenta and one of energy and confined to lie on a 3-sphere determined by E 2 p 2 + m 2. The Boltzmann equation for annihilation in the context of cosmological applications is aided by several simplifications: Scattering processes typically enforce kinetic equilibrium the scattering takes place so rapidly that the distributions of various species have the generic BE or FD forms. The only unknown then is µ, which now is a function of time. If the annihilations were to take place in equilibrium, µ would be the chemical potential, and the left- and the right-hand side would have to balance in a reaction: µ 1 + µ 2 µ 3 + µ 4. For out-of-equilibrium cases, the system is not in chemical equilibrium, which yields a differential equation for µ. In the cosmological applications we considered here, the temperatures T are smaller than the quantity E µ, which makes the term exp [E µ)/t] 1, so exp [E µ)/t] ± 1 exp [E µ)/t], yielding another simplification: f FD E) f BE E) fe) 1 e E µ)/t eµ/t e E/T. 273) 71

3 PHYS 652: Astrophysics 72 so that This also means that exp [ E µ)/t] f 1, so that 1 ± f 1 1. These approximations cause the last line of the Boltzmann equation [eq. 272)] to simplify to f 3 f 4 [1 ± f 1 ][1 ± f 2 ] f 1 f 2 [1 ± f 3 ] [1 ± f 4 ] f 3 f 4 f 1 f 2 e µ 3+µ 4 )/T e E 3+E 4 )/T e µ 1+µ 2 )/T e E 1+E 2 )/T e E 1+E 2 )/T [ e µ 3+µ 4 )/T e µ 1+µ 2 )/T ]. 274) We have also used the conservation of energy here E 1 +E 2 E 3 +E 4. This now constitutes a integrodifferential equation for µ i. It is, however, convenient to directly solve for the number densities n i by relating the two via n i g i where g i is the degeneracy of the species. d 3 p 2π) 3 f i g i e µ i/t It is useful to define the equilibrium number density i g i d 3 p 2π) 3 e E i/t e µ i/t n i i : d 3 p 2π) 3e E i/t, 275) ) 3/2 g mi T i 2π e m i /T m i T, T g 3 i π 2 m i T, i so that the last line of the Boltzmann equation now becomes [ e E 1+E 2 )/T [ e µ 3+µ 4 )/T e µ 1+µ 2 )/T ] e E 1+E 2 )/T 276), 277) 3 n0) 4 n 1n 2 1 n0) 2 ]. 278) After defining the thermally averaged cross section as 1 d 3 p 1 d 3 p 2 d 3 p 3 d 3 p 4 σv 1 n0) 2π) 3 2E 1 2π) 3 2E 2 2π) 3 2E 3 2π) 3 e E 1+E 2 )/T 2E 4 2 2π) 4 δ 3 p 1 + p 2 p 3 p 4 )δe 1 + E 2 E 3 E 4 ) M 2, 279) the Boltzmann equation simplifies to a 3 d dt n1 a 3) 1 n0) 2 σv [ 3 n0) 4 n 1n 2 1 n0) 2 ]. 280) This is a simple first order differential equation for the number density n i. Although some of the details will be application-dependent i.e., dependent on which particles are interacting), we will use this to treat three different reactions: 1. neutron-proton ratio: n + ν e p + e, n + e + p + ν e, 281) 72

4 PHYS 652: Astrophysics recombination: e + p H + γ 282) 3. dark matter production: X + X l + l. 283) Saha equation. The left-hand side of the Boltzmann equation given in 280) is of the order of Hn 1 since a 3 d dt n1 a 3) ṅ 1 + 3ȧ a n 1 Hn 1 ), while the right-hand side is of order n 1 n 2 σv. Therefore, if the reaction rate is much larger than the expansion rate: n 2 σv H, then the terms on the right-hand side will be much larger than the terms on the left-hand side. In order for the equality to be preserved, the terms in the brackets on the right-hand side should cancel each other out be extremely close to each other). This yields the Saha equation: 3 n0) 4 n 1n 2 1 n0) ) Big Bang Nucleosynthesis BBN) As the temperature of the early Universe cools to 1 MeV, the cosmic plasma consists of: Relativistic particles in equilibrium: photons, electrons and positrons. These interact among themselves via electromagnetic interaction e + e γγ. The abundances of these constituents are given by Fermi-Dirac and Bose-Einstein statistics. Decoupled relativistic particles: neutrinos. At temperatures above 1 MeV, the rate of interactions such as νe νe which keeps neutrinos coupled to the rest of the plasma drops below the rate of expansion of the Universe. Therefore, neutrinos have the same temperature as the other relativistic particles, and hence are roughly as abundant, but they do not couple to them. Nonrelativistic particles: baryons. If the number of baryons and antibaryons was completely symmetric, they would completely annihilate away by 1 MeV. However, there was an initial asymmetry between baryons and antibaryons n b n b 10 10, 285) s throughout the early history of the Universe, until the antibaryons were annihilated away at about T 1 MeV. The resulting ration between baryons and photons is given in terms of the present-day baryon content of the Universe Ω b and the current Hubble rate h as η b n b n γ ρ b n γ ρ crω b n γ 1.87h g cm 3 Ω b g 411cm Ω b h Ωb h 2 ), 286)

5 PHYS 652: Astrophysics 74 where we have used n γ 411 cm 3 Homework set #2) and the critical density computed on top of the page 21 of the notes: ρ cr 1.87h g cm 3. Therefore, there are orders of magnitude more relativistic particles than baryons at about T 1 MeV. The goal of these next few lectures is to determine how the baryons arrange themselves. If the equilibrium was maintained throughout the expansion, the final state of baryons would only be dictated by energetics all baryons would end up in iron, the element with the highest binding energy. However, nuclear reactions are too slow to keep the Universe in equilibrium as its temperature drops. Therefore, the reactions do not lead up to iron, but stop at light elements when the Universe becomes sparse enough to keep the further reactions from taking place. In order to understand what happens to the baryons, we need to solve a set of coupled Boltzmann differential equations [eq. 272)] for all reactions which are taking place. This indeed is a daunting task, which is greatly ameliorated by two simplifications: 1. No elements heavier than helium are produced at appreciable levels with the exception of lithium at one part in ). Therefore, the only nuclei that need to be traced are hydrogen H) and helium He), and their isotopes: deuterium 2 H or D), 2. The physics separates rather neatly into two parts since no light nuclei form above T 0.1 MeV only free protons and neutrons exist. This means that we first have to solve for neutron/proton abundance, and then use that result as input for the formation of nucleons of light elements. These simplifications rely on the physical fact that, at high temperatures comparable to binding energies, whenever a nucleus is formed in a reaction, it is destroyed by a collision with a highenergy photon. This can be quantified by the Saha equation [eq. 284)]. Let us consider binding of a neutron and proton into a nucleus of deuterium: Photons have n γ γ, the Saha equation becomes 3 n0) 4 n + p D + γ. 287) n 1 n 2 1 n0) 2 D n0) γ n p n Dn γ n0) n n n p n 3n 4 n0) n 1 n 2 n D n n n p 3 n0) 4 1 n0) 2 n0) D n p 288) We are considering how this reaction takes place when the temperature of the Universe is on the order of the binding energy of deuterium, which is B D 2.22 MeV. The masses of protons and neutrons are MeV and m n MeV, and the mass of deuterium is m D +m n B D MeV, which means that we use the m i T regime of eq. 276), to obtain note: g D 3 because of 3 spin states of D, and g p 2 and g n 2 because of their spin states): n D n n n p ) 3/2 g md T D 2π e m D /T ) 3/2 ) e m n/t mpt 3/2 g p e /T g mnt n 2π 2π g ) D T 3/2 ) 3/2 md e m D m n )/T g n g p 2π m n 3 ) 3/2 2πmD e BD/T, 289) 4 m n T 74

6 PHYS 652: Astrophysics 75 because B D m n + m D. If we approximate m D 2 and m n which is valid to within 0.15%), the equation above becomes n D 3 ) 4π 3/2 e B D/T n n n p 4 T 290) Because both neutron and proton density are proportional to the baryon density n b, the equation above further simplifies into n D n D 3 ) 4π 3/2 e B D/T n n n p n b n b 4 T ) n D 3 4π 3/2 ) n b e BD/T 3 4π 3/2 η b n γ e B D/T n b 4 T 4 T 3 3 ) 4π 3/2 η b 4 2T π 2 e BD/T 12 ) T 3/2 T π 1/2η b e B D/T n ) D T 3/2 η b e BD/T. 291) n b As long as B D /T is not too large and we are doing this analysis in the regime B D T), the prefactor dominates. Not only is T, and hence T/ 1, but the baryon-to-photon ratio η b is extremely small [see eq. 286)], so the right-hand side of the equation above vanishes. This means that the density of deuterium nuclei also vanishes. Small baryon-to-photon ratio thus inhibits nuclei production until the temperature drops well beneath the nuclear binding energy T B D ). This is why at temperatures T > 0.1 MeV virtually all baryons are in the form of neutrons and protons. Around this temperature, the production of deuterium and helium starts, but the reaction rates are too low to produce heavier elements. Not having a stable isotope with mass number 5 means that heavier elements cannot be produced via reaction 4 H + p X. 292) The heavier elements are formed in stars triple alpha process): 4 He + 4 He + 4 He 12 C, 293) but that is only much later. The early Universe is too sparse for these reactions to take place, i.e. for three helium nuclei to find one another on relevant timescales. 75