Cosmology Winter School 5/12/2011 !

Physics of the hot universe!

Jean-Philippe UZAN!

Lecture 2:!

The standard cosmological models !

Eq. state! Scaling ! Scale factor!

radiation! w=1/3 ! a-4! t1/2!

Matter (dust)! w=0! a-3! t2/3!

!" w=-1! a0! eHt!

a0!

Early universe always dominated by radiation.!

Some numbers!

Consider the Compton diffusion between photons and electrons. The reaction rate is given by

Radiation will always dominate in the past.

Today, it is of order

1/700 photon interact in a Hubble time

At z=1000

1 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

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

Equilibrium!

Out of Equilibrium!

In the description of many processes, one needs to go beyond equilibrium, which requires to write a Boltzmann equation.

Liouville term Collision term

We introduce a distribution function f(x!,p") that lives in the tangent bundle (phase space) with the constraint p!p!=-m2.

In the cosmological framework, at background level, f(E,t).

The boltzmann equation takes the general form:

Out of Equilibrium!

At the backround level, we can rewrite the Boltzmann equation as

This can be integrated over pi 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

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]

- Cosmic microwave background [recombination: p+e <-> H + #]

- Cosmic microwave background anisotropies Propagation of photons/neutrinos in the perturbed universe.

background

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 (T0<<mX) then its relic density would be

Because of the expansion, annihilation cannot keep the particle at equilibrium.

Assuming equilibrium after production,

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

Dark matter relic: example!

Before decoupling

After decoupling

Dark matter relic: example!

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 = 10.75 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.

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 nA & n#(T) that

with

% is one of the fundamental parameter of BBN

BBN: entropy!

If %~1, XAZ is stable as soon as T~BA since the formation of the nuclei

(controlled by exp BA/T) overcomes the destruction by photodissociation (controlled by %A-1).

If %<<1, the balance between formation and photodissociation starts only when exp BA/T ~%A-1. Thus at lower temperatures.

Neglecting F(A), the temperature at which Xn~Xp~XA is

For %~5. 10-10, 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 10-10.

As long as T>0.1 MeV, XA~0 so that nb=np+nn and thus

At high temperature, np=nn=0.5, nA~0.

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 XA and this requires the computation of the reaction rates, as functions of energy and temperature.

BBN: numerical integration!

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 XA 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 Xn follows Xn,eq until Tf. Then

It follows that

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.

BD being small, the synthesis starts close to TD~0.066 MeV. This corresponds to tD~303 s.

Between tf and tD, 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

BBN: results!