# Physics of the hot universe!

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1 Cosmology Winter School 5/12/2011! Lecture 2:! Physics of the hot universe! Jean-Philippe UZAN! The standard cosmological models! a 0!! Eq. state! Scaling Scale factor! radiation! w=1/3! a -4! t 1/2! Matter (dust)! w=0! a -3! t 2/3!!" w=-1! a 0! e Ht! Early universe always dominated by radiation.!

2 Some numbers! Radiation will always dominate in the past. Consider the Compton diffusion between photons and electrons. The reaction rate is given by Today, it is of order 1/700 photon interact in a Hubble time At z= photon interact 80 times in a Hubble time Equilibrium! As long as we are in equilibrium, the distribution of each species can be described by its distribution function, which will be either Bose-Einstein or Fermi-Dirac. At a given spacetime point, macroscopic (fluid) quantities can be defined by integrating over the momentum space

3 Equilibrium! These integrals cannot be computed in full generality, but for physically motivated limits, they can be obtained exactly Equilibrium! At high temperature, the matter content is dominated by radiation so that with From the Friedman equations, one gets and the time/temperature relation

4 Equilibrium! Out of Equilibrium! In the description of many processes, one needs to go beyond equilibrium, which requires to write a Boltzmann equation. We introduce a distribution function f(x!,p " ) that lives in the tangent bundle (phase space) with the constraint p! p! =-m 2. In the cosmological framework, at background level, f(e,t). The boltzmann equation takes the general form: Liouville term Collision term

5 Out of Equilibrium! At the backround level, we can rewrite the Boltzmann equation as This can be integrated over p i to get where we have used that so that Out of Equilibrium! We end up with an equation of evolution for the number density For a process the collision term can be decomposed as We have 3 sources of changes for n: - dilution (due to the cosmic expansion) - creation (kl -> ij) - destruction (ij -> kl) The collision term is proportional to the densities of the interacting species Since it is often decomposed as

6 Applications! There are many applications of the Boltzmann equations: - Freeze out of an interaction and relic density [dark matter density] - Big bang nucleosynthesis [relic of light nuclei] background - Cosmic microwave background [recombination: p+e <-> H + #] - Cosmic microwave background anisotropies Propagation of photons/neutrinos in the perturbed universe. perturbation One need to describe the collision term, i.e. the microphysics. Dark matter relic! Let us consider an annihilation reaction of the form If X were at thermal equilibrium until today (T 0 <<m X ) then its relic density would be Because of the expansion, annihilation cannot keep the particle at equilibrium. Assuming equilibrium after production,

7 Dark matter relic! The conservation of energy then leads to We conclude that the collision term takes the form so that the Boltzmann equation rewrites as Dark matter relic: example! For a cold relic, decoupling occurs when the particle is NR Set The Boltzmann equation rewrites as

8 Dark matter relic: example! Before decoupling After decoupling Dark matter relic: example!

9 BBN: basics! T >> 1 Mev Nuclei cannot form because of photodissociation. The universe is mostly composed of e -, e +, #, \$, n, p g * =?? BBN: basics! T >> 1 Mev Nuclei cannot form because of photodissociation. The universe is mostly composed of e -, e +, \$, #, n, p g * = (7/8)(2x2 + 3x2x1)+2 = n and p are NR and at equilibrium via weak interactions It follows that T ~ 1 Mev weak interactions freeze out. \$ decouple & n decay T ~ 0.5 Mev electron-positron annihilation. Reheating of the \$. Light nuclei form through 2-bodies interactions. Competition between formation & photodissociation.

10 BBN: initial state! At high temperature, thermodynamics equilibrium holds (why?) so The chemical equilibrium implies so that Now, the densities of neutron and proton are given by BBN: initial state! BA is the binding energy of nuclei A We define the mass fraction of nuclei A as It follows from the expression of n A & n # (T) that with % is one of the fundamental parameter of BBN

11 BBN: entropy! If %~1, X A Z is stable as soon as T~B A since the formation of the nuclei (controlled by exp B A /T) overcomes the destruction by photodissociation (controlled by % A-1 ). If %<<1, the balance between formation and photodissociation starts only when exp B A /T ~% A-1. Thus at lower temperatures. Neglecting F(A), the temperature at which X n ~X p ~X A is For %~ , we get Note that D is very fragile so that the synthesis cannot really start before T~0.07 MeV. Deuterium bottleneck. High entropy favors free n and p. BBN: neutron abundance! At equilibrium, \$ e + n <-> p + e implies that It follows that The chemical potentials are actually very small, typically of order 10-8 to As long as T>0.1 MeV, X A ~0 so that n b =n p +n n and thus At high temperature, n p =n n =0.5, n A ~0.

12 BBN: nuclear interactions! To determine the abundances of light nuclei, one needs to solve a set of coupled Boltzmann equations. For the neutron, e.g., we have We have similar equations for all the X A and this requires the computation of the reaction rates, as functions of energy and temperature. BBN: numerical integration!

13 BBN: nuclear interactions! To determine the abundances of light nuclei, one needs to solve a set of coupled Boltzmann equations. For the neutron, e.g., we have We have similar equations for all the X A and this requires the computation of the reaction rates, as functions of energy and temperature. A good estimate can however be obtained analytically by considering the weak interaction rate between n and p at equilibrium BBN: freeze-out of the weak interaction! This allows to determine the freeze-out temperature of the weak interactions Let us assume that X n follows X n,eq until T f. Then It follows that

14 BBN: helium abundance! Light elements are formed by a series of nuclear reactions. No significant abundance can be build before Deuterium starts to be synthetized (densities are too small for more than 2-bodies reactions play any role) p+p->d+e + +\$ e is negligible (why?) so that p+n->d+# is the dominant process for the formation of D. B D being small, the synthesis starts close to T D ~0.066 MeV. This corresponds to t D ~303 s. Between t f and t D, the free neutrons decay so that BBN: helium abundance! One concludes that The neutron abundance is thus Since the binding energy of helium-4 is larger than the one of deuterium, the Boltzmann factor favorizes the production of He-4. Heavier nuclei are not formed significantly since no stable nuclei with A=5 and A=8 (which blocks n+he_4, p+he-4, He-4+He-4) and T+He-4, He-3+He-4 are slower due to Coulomb barriere. All free neutrons -> He-4

15 BBN: results!

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