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Nucleosynthesis

Probably due to Geoffrey Clayton LSU

A Brief History of the Universe

The standard big bang model

we are only to do a "once over lightly" of the first 3 minutes, then we'll look at 3 items in detail
What happened to the baryons?
What happened to the electrons?
What happened to the neutrinos?

Note that our understanding of this is limited.

Assume that thermodynamics works

i.e. even though we may not know exactly how particles interact, the numbers of each are fixed via rate equations.)
eg. O₂ + 2H₂ ⇔ 2H₂O
is complex
as long as we know the energy (Q-value) given out in the process, we can calculate equilibrium.
Rates ∝ cross-sections, which can be measured.

Thermal Equilibrium

Energy/particle E = ³/₂kT for a gas.

For T >> m_ec², Number of electrons ≈ Number of photons≈ number of positrons ≈ Number of Neutrinos .....etc for all lighter mass particles.

Early Universe in detail:

A simple model:

since it is totally radiation dominated, T is a very dull function! Note scales.

10^-44s

Can't run clock back any earlier since quantum gravity dominates the early universe.

We can find "numbers" made of G, ħ(Plancks constant) and c (speed of light) with dimensions of time, mass and length.

Planck time.

\color{red}{ t_p = \left( {\frac{{G\hbar }}{{c^5 }}} \right)^{1/2} = 5.4 \times 10^{ - 44} s}

Planck length.

\color{red}{ l_P = t_p c = \left( {\frac{{G\hbar }}{{c^3 }}} \right)^{1/2} = 1.6 \times 10^{ - 35} m}

Planck mass

\color{red}{ M_p = \left( {\frac{{\hbar c}}{G}} \right)^{1/2} = 2.2 \times 10^{ - 8} kg}

Planck energy.

\color{red}{ E_P = M_p c^2 = \left( {\frac{{\hbar c^5 }}{G}} \right)^{1/2} = 1.2 \times 10^{19} GeV}

(note that if we use "natural" units c=ħ=1,

\color{red}{ M_p = E_P = \frac{1}{{t_p }} = \frac{1}{{l_p }}}

Normally we think of gravity as being "caused" by matter, but (e.g.) near a black hole, the field is so strong that the effects of the gravitational energy produce changes in the gravitational field.

For t < 10^-44s, gravitational effects of gravity >>>> stronger than matter. We have no theory here.

10^-35s
Inflation (see later): Why universe is large and flat
10^-30 s , E ≈ 10²⁰ eV
The universe was matter-antimatter symmetric: to make matter, we require CP violation. (Argument originally due to Sakharov)
a number of models give some particles
```
X₀ ⇒ p^-e⁺  (anti-matter)
X₀ ⇒ p⁺e^-  (matter) 
```
at slightly different rates.
At this stage, there were 10⁹+1 electrons (quarks) for every 10⁹ positrons (anti-quarks).

10^-15 s, E ≈ 1 TeV
If WIMPs exist, they would have been created in this era and frozen out soon after.
10^-6 s, E ≈ 1 GeV
Free quarks turn into protons and neutrons.
From here on, all the particles are the ones we would recognise today. Roughly equal numbers of
```
p, p̄, n, n̄, e⁺, e^-, ν, ν̄, γ, μ⁺ ,μ^-
```
At this temp, can create anti-protons (p^-) as readily as γ's, so >p, p̄, n, n̄ (anti-neutrons), are all in thermal equilibrium.

10^-5 s , E≈ 100 MeV
Lose p̄ and n̄, but p and n will stay in equilibrium via n + ν ⇔ p + e^-
(we normally see only n⇒ p + e^- + ν, but the density of ν's and e's was so enormous...)
μ's drop out of equilibrium and disappear.
Down to 1 proton/10⁸ γ's
Electrons and positrons still in equilibrium.
1s, E ≈ 1MeV
No longer enough ν's to keep n's in equilibrium. After this, they just decay.
```
n -> p + e^- + ν̄
```
with a half life τ = 864 s: i.e.
```
 N_n = N₀ exp(-t/τ) 
```
Electrons and positrons can no longer be created.
```
e⁺ + e^- => γ + γ
```
means that γ's now have a higher temperature than ν's as well. He could exist, but D is still unstable.

3 mins,

⁸

D becomes stable, n + p -> d + γ (Note: this doesn't occur in stars. There are no free n's.)
Then all remaining n's become cooked to He, via.

d + d -> ³He + n,  d + ³He -> ⁴He + p

I.e., the universe has to drop to T ≈ 10⁹K before n's can make heavier nuclei.
Now doing this properly:

Baryons

Neutrons and protons are almost exactly in balance, except that n is slightly heavier. m_n= 939.6 MeV
m_p = 938.3 MeV

so ΔE = 1.29 MeV

This means that there is a slight preference to have p's: Boltzmann gives

\color{red}{ \frac{{n_n }}{{n_p }} \simeq e^{ - \frac{{\delta E}}{{kT}}} ,\delta E = \left( {m_n - m_p } \right)c^2 }

ν's,n's stay in equilibrium via n + ν ⇔ p + e^-

Similar argument to γ freezeout, but X- sect is weak:

\color{red}{ \sigma \sim G_F^2 E^2 \approx 10^{ - 47} \left( {\frac{{kT}}{{1MeV}}} \right)^2 {\rm{m}}^{\rm{2}} }

so rate
\color{red}{ \Gamma = n_\nu c\sigma _w \propto t^{ - 5/2} }

Freezeout occurs (as before) when \color{red}{\frac{1}{\Gamma } \ge \frac{1}{H}} Complicated, but E_freezeout ≈ .8 MeV when t ≈ 1s. Ratio is

\color{red}{ \frac{{n_n }}{{n_p }} \simeq e^{ - \frac{{\delta E}}{{kT}}} = e^{ - \frac{{1.29}}{{.8}}} \approx .2}

Neutrons then decay:

\color{red}{ n_n \left( t \right) = n_n \left( 0 \right)e^{ - t/\tau } ,n_p \left( t \right) = n_p \left( 0 \right) + n_n \left( 0 \right)\left( {1 - e^{ - t/\tau } } \right)}

creating protons.

until temp drops enough for deuteron to be formed (reaction is very fast, so don't need to worry about freezeout). Saha equation gives ratio as

\color{red}{ \frac{{n_D }}{{n_n }} = 6.5\eta \left( {\frac{{kT}}{{m_n c^2 }}} \right)^{3/2} \exp ^{B_D /kT} }

Very strong function of T (and hence t) so turns all n's into D's

Again:

After this heavy nuclei form vary quickly: principles are the same but equations are messy

Amount of ⁴He depends crucially on how long the universe takes to cool.

Time taken depends on the density of γ's and ν's, so we can run rate equations to get predictions...

Note that all the action takes place at around 3 mins.

However, this is not unique: it assumes a certain density of protons.

(Roughly, the more protons, the more neutrons so the more He.)

Hence we can run these equations for different numbers of protons.

We have several predictions out of this:
Proportion of He ≈ .1 ⇒.3, whatever we do.
Can measure ⁴He, D, ³He, and with great difficulty ⁷Li, and we know the number of γ's now.
And the result is.......... All the measurements are consistent and predict a number for the density of protons ≈ .1 m^-3.

Ω_B ≈ .03

consistent with WMAP (acually, this result preceded WMAP)
Note totally inconsistent with Ω_B =1.
Note the value Li ≈ 10^-10 H is a minimum: very hard to fiddle this either way.
Combining this with dark matter measurements tells us that we need
- Ω_b≈ 0.03
- Ω_CDM≈ 0.26

Radiation

We've argued that early universe is rad-dominated, but what does this mean? More precisely: The momentum distribution of particles in thermal equilibrium

\color{red}{ f(\vec p) = \frac{1}{{e^{(E - \mu )/kT} \pm 1}}}

(+ is F-D, - is B-E). Assume μ = 0 (# of particles = # of anti-particles) and energies are high

\color{red}{ E = \sqrt {p^2 c^2 + m^2 c^4 } \Rightarrow pc}

Then energy/number densities are

\color{red}{ \begin{array}{l} u = \frac{g}{{(2\pi )^3 }}\int {\frac{{4\pi p^2 dp \cdot p}}{{e^{pc/kT} \pm 1}}} \\ n = \frac{g}{{(2\pi )^3 }}\int {\frac{{4\pi p^2 dp}}{{e^{pc/kT} \pm 1}}} \\ \end{array}}

can be solved exactly:

\color{red}{ \begin{array}{l} n_\nu = \frac{{3\varsigma (3)}}{{2\pi ^2 }}T_\nu ^3 = \frac{3}{4}n_\gamma ,\zeta (3) = 1.202.. \\ u_\nu = \frac{{7\pi ^2 }}{{120}}T_\nu ^4 = \frac{7}{8}u_\gamma \\ \end{array}}

Hence energy density in any relativistic gas is

\color{red}{ {\rm{Q = }}\frac{{{\rm{2}}\sigma }}{c}{\rm{(g}}_{\rm{b}} {\rm{ + }}\frac{{\rm{7}}}{{\rm{8}}}{\rm{g}}_{\rm{f}} {\rm{)a}}^{\rm{3}} {\rm{ T}}^{\rm{4}} }

where

g_B number of bosonic spin states (2 for γ's: would expect 3, but longitudinal mode (m=0) doesn't exist)
g_f number of fermionic spin states (2 for electrons, 1 for ν's, since they are only left-handed
7/8 comes about because one is integrating an F-D distribution instead of a B-E
Better \color{red}{\varepsilon = \frac{{\pi ^2 }}{{30}}g_{eff} T^4 }

γ	e^-	e⁺	{\nu _e }	{\bar \nu _e }	{{\bar \nu _\mu }}	{{ \nu _\mu }}	{{ \nu _\tau }}	{{\bar \nu _\tau }}	g_eff	{{ \nu _X }}
2	7/4	7/4	7/8	7/8	7/8	7/8	7/8	7/8	2+35/4	x

\color{red}{g_{eff} = g_{BE} + \frac{7}{8}g_{FD} }

Note that g_eff will change with time: e.g. as kT → 100 keV, electrons become non-rel, and drop out of this formula. So although all these lead to universe expanding at t^1/2 coefficient is different

Can use this to count unseen massless particles. \color{red}{T^2 = \frac{{2.42}}{{t\sqrt {g_{eff} } }}}

e.g. if we add extra ν's Subir Sarkar

gives \color{red}{\left| {N_\nu - 3} \right| < 1.6} i.e. at most there is 1 more massless particle)
OPAL finds \color{red}{\left| {N_\nu - 3} \right| < .01}
Note that this measures something different: OPAL measures only particles that couple to the Z, cosmology measures all rel. particles that were in thermal equilibrium (not axions)

Neutrinos

Also ν's freeze out early (but they stay at same temp T as γ's: T_ν = T_γ ), but e's stay in equilm.

e⁺ + e^- ⇔ γ + γ

becomes

e⁺ + e^- ⇒ γ + γ

i.e. electrons "freeze out" and create γ's, so they are "reheated".

Before reheating: entropy of electron-photon ideal gas
\color{red}{ S_{before} = \frac{{11}}{3}\sigma _b (T_\gamma )^3 V}
Afterwards photons
\color{red}{ S_{after} = \frac{4}{3}\sigma _b (T_\gamma )^3 V}
(11/3 = (4+7)/3!).
Process must be adiabatic, so γ's heat up and
\color{red}{ T_\nu = T_\gamma \left( {\frac{4}{{11}}} \right)^{1/3} \Rightarrow T_{\nu 0} {\rm{ = 1}}{\rm{.945 K}}}
(most accurately known unobservable number in physics!).

Since ν's were formed in the BB and are about as numerous as γ's, we can ask what mass they must have to give Ω = 1?

ρ_c≈ 5×10^-27 kg m^-3 ≈ 5 GeV m^-3,  N_ν ≈ N_γ ≈ 10⁹ m^-3

so we'd need

m_ν ≈ 5 eV ≈ 9×10^-36 kg (≈ 10^-5 m_e)

which we can live with.

(note this means
\color{red}{ \bar \varepsilon _0 = kT_0 \sim 10^{ - 5} eV < < m_\nu c^2 }
so they were relativistic at the time of the BB, but are very non-rel now)

It's more complicated (you were surprised?) SInce there are three kinds of ν's, the correct result is that

\color{red}{ \begin{array}{l} \Omega _\nu \le 1 \Rightarrow \sum {m_\nu } \le 45eV \\ \Omega _\nu \le 0.26 \Rightarrow \sum {m_\nu } \le 15eV \\ \end{array}}

However, it is not possible for all of dark matter to be ν's: we need a fraction of CDM particles.
e.g. Suppose we make all of the galaactic halo out of ν's: need to have \color{red}{v \le v_{escape} } (obviously!).
Would then have a degererate gas of ν's:, so standard F-D stats gives number density
\color{red}{ n_{\nu ,\max } = \frac{{p_{\nu ,\max }^3 }}{{3\pi ^2 }} = \frac{{m_\nu ^3 v_{esc}^3 }}{{3\pi ^2 }}}
and energy
\color{red}{ u{}_{\nu ,\max } = n_{\nu ,\max } m_\nu = \frac{{m_\nu ^4 v_{esc}^3 }}{{3\pi ^2 }} = u_{halo} \approx 0.3GeVm^{ - 3} }
which has to be the required density.
Hence need to have \color{red}{m_\nu \ge 40eV}
which gives us a problem!

How will we see the BB ν's?

I wish I knew!
e.g. \color{red}{\sigma _{\nu N} \approx G_F^2 m_\nu ^2 /\pi \cong 10^{ - 56} \left( {m_\nu /eV} \right)^2 cm^2 } but energy transfer to target nucleus \color{red}{\delta E \sim \frac{{m_\nu }}{{m_A }}kT_0 \sim 10^{ - 6} eV}
various ideas for coherent scattering: e.g Cavendish type expt with polarised target ⇒ forces 10^-11 x current technology
collect your Nobel prize!
We now understand the "small" components" of the universe: what's left?