Physical Cosmology 15/4/2016

Physical Cosmology 15/4/2016

Alessandro Melchiorri

alessandro.melchiorri@roma1.infn.it slides can be found here:

oberon.roma1.infn.it/alessandro/cosmo2016

T. Padmanabhan, structure formation in the universe Most of the discussion on BBN in current lectures

can be found here:

T>0.1 MeV Protons and

Neutrons are unbounded

T <0.1 MeV Protons and neutrons

are bounded to form light elements nuclei 4He, 3He, D, Li7

Big Bang Nucleosynthesis

- at T<0.1 MeV we have the formation of light elements the Big Bang Nucleosynthesis (or BBN).

- it happens at temperatures much lower than the binding energies of He4, He3, D, Li7 (all above 1 MeV) because

of the high photon/baryons ratio. Moreover, BBN starts only after D fusion.

- it predicts the correct amount of He4 in the universe (23%) and the presence of D. Both are not possible to obtain from

stellar nucleosynthesis. BBN is a pilar of the cosmological model.

- BBN does not produce heavier elements because:

- Helium 4 binding energy peak.

- low density and 3alpha process not efficient,

- it lasts 3 minutes.

- Assuming BBN we can infer from observations of He4 and D constraints on several parameters: baryon density, Neff, …

- BBN is agreement with current observations of He4 and D.

There is a disagreement with Li7.

BBN: why it starts at low T ?

The binding energies are well above 1 MeV!

Element Binding Energy

Why we need to have the T of the plasma much lower to produce these elements (T< 0.1 MeV) ?

Statistics (again)

We have found that, given a distribution function, the number density is given by:

We are now interested in the number densities of proton and neutrons when the photon plasma had a temperature T < 1 MeV. Neutrons and protons will have the same

temperature of this plasma (since they are, or they

have been, coupled to it) but since they have a mass around 1 GeV we need to compute the above equation for the non-relativistic case, when T<< m:

Statistics (again)

In this limit (T<<m) we have:

Same formula valid for bosons and fermions !

Let us define the number density of a nuclei with atomic mass A simply as:

Protons & Neutrons

For protons and neutrons we have the number densities:

Neutrons have a slightly larger mass than protons:

we can’t neglect this in the exponential but we have set in the prefactor:

Boltzmann equation

Let us suppose that two particles 1 and 2 annihilate forming two different particles 3 and 4.

The evolution of the number density of particle 1 follows the Boltzmann equation:

Boltzmann equation

Number density times aˆ3 would be constant with time without interactions (if right term is zero).

Boltzmann Equation

Rate of producing particle 1 is

proportional to distribution of particles 3 and 4

Rate of destroying particle 1 is

proportional to distribution of particles 1 and 2

Boltzmann equation

Plus (bosons) and minus (fermions) describe Bose enhancement and Pauli blocking

Boltzmann Equation

Dirac’s deltas describe momentum and energy conservation.

Boltzmann Equation

The fundamental physics of the interaction is in this matrix.

The process is reversible.

For example if we have photons scattering on electrons this

matrix would be proportional to the fine structure constant.

Boltzmann Equation

The integral on the first line are over all the possible momenta.

Since we should integrate over all possible energies but This explains the 1/2E factors in the integral over momenta, indeed:

Boltzmann Equation

We are interested in processes involving neutrons and protons (with masses approximately of 1 GeV) at temperatures

around T=0.1 MeV. We then have E/T >>1 and we can write:

Neglecting the Bose/Pauli terms and using energy conservation:

Boltzmann equation

In this limit the number density can be written as:

And we can define:

And the thermally averaged cross section:

Boltzmann Equation

We can therefore write the Boltzmann equation as:

This is a much more simple form !!!

The term on the left is of order:

The one on the right:

Is the reaction rate !

Saha Equation

If the Hubble parameter is much larger than the reaction rate we need the terms in the brackets on the right side to sum to a very small number, i.e. if we have coupling:

I.e. we have the Saha equation:

BBN at equilibrium

Let us assume that we have a reaction at equilibrium that produce a nuclei with atomic mass A out of Z protons

and A-Z neutrons.

In this case is valid the Saha equation, i.e. conservation of the chemical potential in the reaction:

BBN at equilibrium

Using the Saha equation,

we can write:

BBN at equilibrium

Using the expression for the number density of protons and electrons we have:

where we defined the binding energy of the nucleus as:

BBN at equilibrium

We have then for the number density

Let us remind that the number density of photons is given by:

We can then write the photon/baryon fraction as:

This number is very small and is conserved from BBN up to today: there are much more CMB photons than baryons in our universe !

BBN at equilibrium

Let us define as mass fraction of a nuclei of atomic mass

A its number density divided the number density of baryons, multiplied by A:

Mass fraction We can write:

BBN at equilibrium

The following formula:

Using the previous definitions of mass fraction and:

We get

with:

BBN at equilibrium

The photon/baryon appears here, to

the power of A-1 !

It is not enough to have B^A>T

to have X^A of order one !

We need to go to much lower temperatures ! This term is VERY

small !

BBN at equilibrium

We have X^A of order one at a temperature:

this number is 3 for Helium 4

This number is about 22 This number is of order 10 Element

Binding

Energy T^A at which X^A is of order 1

Much much lower !!!!

BBN at equilibrium

Why we have this ?

We have many more CMB photons than baryons !

Photons are distributed as a blackbody at temperature T.

Even at temperatures much lower than the binding energy, we have in the tail of the distribution many photons at energies able to destroy that element !

BBN at equilibrium

This discussion however, while it gives a physical idea of what is going on is NOT correct because BBN happens out of equilibrium !

In particular we have assumed neutrons and protons in equilibrium at the same temperature T.

This is possible thanks to the weak interaction processes:

But neutrinos decouple at T=1 MeV !! these reactions are not working a T =0.1 MeV !

Helium abundance

A rough but uselful aestimate of the final Helium 4 abundance can be obtained also in this case.

Neutrons and protons are in equilibrium thanks to the following reactions:

At equilibrium we have that:

We have then that until equilibrium holds the neutron to proton numer density decreases as:

Helium abundance

The equilibrium will end when the rate of reactions is smaller than the Hubble parameter H(T).

Computing the quantity reaction rate times neutron lifetime this a function just of the temperature.

Reaction rate n to p (it flattens because of beta decay)

Reaction rate p to n

Twice Hubble rate. To be

compared to the sum of the two reaction rates.

T=0.8 MeV

Helium abundance

Assuming for the neutron lifetime a value of 915.4 s, we have a decoupling temperature of T^D=0.8 MeV.

The neutron to proton ratio freezes at T^D=0.7 MeV :

However the beta decay still works and we have a small decrease to n/p also for later times.

From the previous calculation we had that Helium could be produced only for T< 0.28 MeV. However Helium can be produced only if D is available thanks to the reactions:

and D starts to be present only for T < 0.1 MeV…

Helium abundance

Let us assume then that BBN start at T=0.1 MeV.

At this temperature the n/p ratio is reduced thanks to the beta decay (from 1 MeV to 0.1 MeV) by a factor 0.8, to:

BBN

Assuming that at end of BBN all neutrons go to form Helium 4 (this is quite correct) we have for the

Helium mass fraction:

in agreement with the observations of primordial Helium in the Universe ! Stellar nucleosynthesis is unable to

produce such large abundance of Helium !!

Nice and very pedagogical review

BBN: accurate calculations

In order to properly compute the several abundances we need to integrate a system of Boltzmann equations.

There are several numerical codes available to do this.

In Figure we see that Helium abundance starts to be there just after D reaches enough mass fraction.

BBN: numerical calculations

BBN starts only after D is formed.

Mostly of the

neutrons goes to Helium.

Standard BBN

If we assume 3 neutrinos then BBN has just one

free parameter,

the baryon to photon ratio:

i.e. the baryon (atoms) energy density.

Predictions for the elements are in the figure in function of the baryon density.

The width of the lines are due to uncertainties in the rates.

Why we have mostly He4 ?

Helium “peak”.

If we plot nucleon binding energy versus A

we see that Helium has a

“peak”.

i.e. nuclei with larger A (6-7) are less stable.

The nuclei density is small, we don’t have any triple alpha:

Universe is expanding. BBN ends after few minutes.

Standard BBN

As we will see, observations can restrict the range in the baryon density to:

In this range we have the following approximated formulas:

What enters in the computations are several reaction rate.

Some of them are really well measured others are not.

In the estimate of D the largest error

comes from this reaction (6%)

BBN: systematics

BBN: systematics

Taken from Nollett and Holder

http://arxiv.org/pdf 1112.2683.pdf.

Few datapoint and in tension with theoretical expectations.

The LUNA400 experiment under Gran Sasso could measure the reaction rate with the better

accuracy.

BBN: Systematics

The largest systematic we have for the final Helium abundance comes from the measurement of the neutron lifetime !

Particle data group

http://pdg.lbl.gov/2014/listings/rpp2014-list-n.pdf

quotes 1.1 s error, but it comes from measurements that are scattered….

BBN:neutron lifetime

Measuring the neutron lifetime is fundamental in measuring the Fermi constant:

that enters in the weak interactions rate:

that fixes the neutron freeze-out that happens when:

BBN neutron lifetime

Lower neutron lifetime

Higher reaction rate

Freeze out happens later (smaller T) Lower density of neutrons at BBN start

Lower final Helium abundance predicted

Non-standard BBN

There is the possibility of having extra relativistic particles at BBN.

This can be parametrized by

increasing the effective neutrino number.

This increases the radiation content:

We then have larger H at BBN.

The freeze-out happens earlier (larger T).

We have more neutrons at BBN, more helium !

Non-standard BBN

We also have approximate formulas in this case if:

If we define:

We have: