Radiation Energy Density

Physical Cosmology 7/4/2017

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

Radiation Energy Density

In the standard model, only two type of particles contribute to the total radiation energy density.

Photons Neutrinos

Radiation: Photons

The contribution from photons is essentially provided by the photons of the Cosmic Microwave Background.

Photons emitted by objects (galaxies, stars, gas, etc) are completely negligible since these objects forms at small redshifts (z<15) and their contribution is anyway small.

Statistics

If we consider a gas of particles at temperature T, the

probability P of having a particle with momentum between p and p+dp and position between x and x+dx is given

by the distribution function f:

We can consider for the moment an homogeneous universe (so we neglect the spatial and directional dependence). We have:

+ fermions - bosons

where m is the rest mass of the particle, mu the chemical potential, g the spin.

Statistics

we can therefore write down the following equations:

Energy Density Number Density

Pressure

Statistics

When the particles are highly relativistic and non-degenerate we have:

CMB photons

Let us apply the following two formulae to the CMB (photons, spin=2):

Using the CMB temperature of 2.73 K we get:

Entropy Density

We can define, given a gas, as entropy density the quantity:

For relativistic particles we have:

Bosons

Fermions

Entropy Conservation

It is possible to show that during the Universe evolution:

The temperature of a gas of relativistic particles during the expansion of the universe will therefore scale as:

Note that if g varies with time then the

T will not scale simply as 1/a !!

Photons increase energy by going to higher z

We can use cosmology

as an accelerator !

Strongly coupled particles

Let us consider a gas made of several kind or particles.

If the particles are relativistic the total energy density is:

If the particles are strongly interacting then they will share the same temperature and:

Strongly coupled particles

We can write down the same equations for the entropy density:

If particles are coupled and share the same temperature we have that q=g.

Plasma temperature

scales mainly as 1/a but it can vary if g changes.

(for example, one

component annihilates in photons).

Criteria for coupled particles

Let suppose that two particles interacts.

We can write down an interaction rate as:

where n is the numerical density of target particles, v the relative velocity and s the cross section.

The particles will be coupled (and will share the same temperature) if the interaction rate is faster than the

rate of expansion of the universe.

Particles are coupled, they share same T

Particles start to uncouple. They keep sharing same T until some process can change it in one of them.

Example: electrons

Free electrons (not in atoms) are coupled to photons by Thomson scattering:

Comparing with the Hubble rate (see Dodelson eq. 3.46) we get:

where Xe is the free electron fraction.

Example: electrons

The free electron fraction drops substantially after

z=1300. Most of the electrons recombine to form neutral

hydrogen.

This epoch is called epoch of recombination and we will study the process in a future lecture.

Because of recombination, the electrons decouple from the photons at redshift z=1100. Epoch of decoupling.

Example: electrons

Without recombination and fixing Xe=1 constant, the decoupling redshift would be:

i.e. a decoupling redshift around:

That is clearly in strong disagreement with current

observation of CMB anisotropies (we should not see

them in this case, we will discuss this in a future lecture).

Photons and electrons

are coupled.

They form a plasma

Photons and electrons are uncoupled.

The Universe is trasparent to

radiation

At T>0.5 MeV photons temperature is

high enough to create electron-positrons

pairs.

Neutrinos

Neutrinos are usually treated as massless particles.

This is not true: neutrinos must have a mass.

In reality what has been

measured is 2 mass differences.

Since the current CMB temperature is about 7 10ˆ-4 eV, at least

one neutrino state should be non-relativistic.

However, for the moment, we neglect the neutrino mass.

At T> 1 Mev Neutrinos are coupled to the electron-positron

plasma. They share the same temperature with it.

At T >1 Mev the universe is dominated by a radiation

component made of

photons, electrons, positrons and neutrinos all coupled and

at the same temperature.

Photons increase energy by going to higher z

We can use cosmology

as an accelerator !

At T> 1 Mev Neutrinos are coupled to the electron-positron

plasma. They share the same temperature with it.

At T >1 Mev the universe is dominated by a radiation

component made of

photons, electrons, positrons and neutrinos all coupled and

at the same temperature.

At T>0.5 MeV photons temperature is

high enough to create electron-positrons

pairs.

At T<0.5 MeV electron-positrons

pairs annihilate and disappear.

Photons temperature is not anymore high enough

to create pairs.

e+ e- pairs

annihilate heating

photons

CNB temperature

At T>1 MeV, the universe is made by electrons, photons, neutrinos and anti-neutrinos all at the same temperature.

We can write at a time 1 at T>1 Mev:

Photons

spin 1, -1 Electrons Positrons 3 neutrinos

with one spin state+

3 antineutrinos

CNB temperature

At a time T< 0.5 Mev, after electron-positron annihilation we have:

e+/e- term disappears.

Neutrino background has a different

temperature.

CNB temperature

Now we should use:

Entropy conservation Neutrino are uncoupled their temperature scales as 1/a !

CNB temperature CNB temperature

gives:

Using we have:

And, finally:

The neutrino background has a temperature

that is 0.71 lower

than the CMB temperature!

CNB temperature

If we assume that all neutrinos are still relativistic today (wrong but we assume it anyway for the moment) both CMB and CNB temperature scales as 1/a and we have today:

If relativistic,

we should have

a neutrino background surrounding us

at 1.9 K

Radiation Component

Assuming that neutrinos are relativistic today, we have that the total energy in radiation is given by:

The above formula is however not perfect.

We assumed that the neutrino decoupling is before electron-positron annihilation and instantaneous.

In reality is not instantaneous and some decoupling still happens during e-e+ annihilation.

We keep the formula but we write it now as:

With:

Massive neutrinos

After electron-positron annihilation the neutrino number density is given by

The neutrino number density must satisfy:

Today

This is true even if neutrino are non relativistic today ! Neutrino number per

species. (Multiply by 6 to have the total)

Massive neutrinos

If neutrinos are non relativistic today they contribute to matter (not radiation!) with an energy density given by

where mⁱ are the 3 mass eigenstates.

An useful formula is given by:

Energy density

in NON relativistic neutrinos.

Brief history of neutrino

discoveries

The Beta Decay

Heavy atom decays to lighter atom, emitting electron.

Why is the spectrum continuous? Why is electron energy not equal to difference between two atomic energies?

A brief history of the Neutrino

1930. Wolfgang Pauli proposed a “desperate remedy” to save the law of energy conservation in nuclear beta decay by introducing a new neutral particle with spin ½ dubbed the “neutron”.

1932. James Chadwick discovered what we now call the neutron, but it was clear that this particle was too heavy to be the “neutron”.

1933. Enrico Fermi formulated the first theory of nuclear beta decay and invented a new name: the neutrino (a little neutral one in italian).

1956. Clyde Cowan and Fred Reines first detected antineutrinos emitted from a nuclear reactor at Savannah River in South Carolina.

1957. Goldhaber, Grodzins and Sunyar measured the “handedness” of neutrinos in a ingenious experiment at the Brookhaven National Lab.

Neutrino are always left-handed and therefore particle physicist concluded that they had to be massless.

Standard Neutrinos

⬥ 3 neutrinos, corresponding to 3 families of leptons

⬥ Electron, muon, and tau neutrinos

⬥ They are massless because we see only left-handed neutrinos.

⬥ If not they are not necessarily mass eigenstates

(Pontecorvo, 1958): one species can “oscillate” into another

iHt

0 e

ν

ν ≠

1) Sun

2) Cosmic Rays hitting the

atmosphere

Two Obvious Sources of neutrinos

A brief history of neutrino discoveries

1968. Ray Davis and colleagues get first radiochemical solar neutrino results using cleaning fluid in the Homestake Mine in North Dakota, leading to the observed deficit known as the “solar neutrino problem”.

1985. The “atmospheric neutrino anomaly” is observed at IMB and Kamiokande.

1998. The Super-Kamiokande experiment reports finding oscillations for atmospheric neutrinos and, thus, mass muon neutrinos.

2001. SNO announces observations of neutral currents from solar

neutrinos providing convincing evidence that neutrino oscillations are the cause of the solar neutrino problem. By combining with

SuperKamiokande the SNO collaboration determined how many muon neutrinos or tau neutrinos were incident at the Japanese detector.

SuperKamiokande

SNO

Normal hierarchy Inverted hierarchy

If neutrino masses are hierarchical then oscillation experiments do not give information on the absolute value of neutrino masses

Moreover neutrino masses can also be degenerate

c atmospheri 3

2

1

, m , m m

m >> δ

SOLAR ν KAMLAND ATMO. ν

K2K

1 2

3

m m

m > > m

> m

> m

Laboratory bounds on neutrino mass

2 / 1 2 2

⎟⎠

⎜ ⎞

⎝

= ⎛

∑

i

i

ei m

U m_β

Experiments sensitive to absolute neutrino mass scale : Tritium beta decay:

2 2

2 2

2 . 3 3

. 2

0 . 3 2

. 1

eV m

eV m

±

−

=

±

−

=

β

β (Mainz)

(Troitsk)

) 2

( 8

.

1 σ

β

eV

m <

Best fit gives a negative mass !!!

KATRIN experiment (future)

∑

=

i

i ei

m U

m

Bounds on neutrino mass

Experiments sensitive to absolute neutrino mass scale : Neutrinoless double beta decay (only if neutrino are

majorana particles!):

Neutrinoless doule beta decay processes have been searched in many experiments with different isotopes, yielding negative results.

Some years ago, members of the Heidelberg-Moscow experiment have claimed the detection of a 0ν2β signal from the ⁷⁶Ge isotope.

If the claimed signal is entirely due to a light Majorana neutrino masses then we have the constraint:

) 3

( 0

. 2 17

.

0 eV < m

< eV σ

Constraints from Cuore-0

Constraints from double-beta decay

are plagued by unknowns in the nuclear matrix.

Current Cosmological Constraints

We will see that cosmology produce extremely tight bounds on the sum of the 3 neutrino mass eigenstates.

- These bounds are “model dependent”. Maybe the cosmological model will turn out to be wrong.

- If the cosmological model is OK, both beta and double beta decay experiments will not reach enough sensitivity to detect the neutrino mass.

Current constraint (CMB+BAO):

Allowed by

cosmology 95% c.l. n.h. .

i.h.

Let us now go back at temperatures just below 0.5 MeV.

Neutrinos are relativistic and their energy density is related to photons.

No matter what

is the neutrino mass, at

0.5 MeV > T > 1 KeV the neutrino energy

density is given by:

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

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:

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 !