LOG#109. Basic Cosmology (IV).

Today, a new post in this fascinating Cosmology thread…

bigbangnucleosynsbbpn_to_he3bbn

The Big Bang Nucleosynthesis in a nutshell

When the Universe was young, about T\sim 1MeV, the following events happen

1st. Some particles remained in thermodynamical equilibrium with the primordial plasma (photons, positrons and electrons, i.e., \gamma , e^+, e^-).

2nd. Some relativistic particles decoupled, the neutrinos! And the neutrinos are a very important particle there (the \nu “power” is even more mysterious since the discovery of the neutrino flavor oscillations).

3rd. Some non relativistic particles, the baryons, experienced a strong and subtle asymmetry. Even if we do not understand the physics behind this initial baryon asymmetry, it is important to explain the current Universe. The initial baryon asymmetry is estimated to be

\dfrac{n_b-n_{\bar{b}}}{s}\sim 10^{10}

At T\sim 1MeV, the baryon number was very large compared with the number of antibaryons. The reason or explanation of this fact is not well understood, but there are some interesting ideas for this asymmetric baryogenesis coming from leptogenesis. I will not discuss this fascinating topic today. Moreover, the fraction or ratio between the baryon density and the photon density is known to be the quantity

\eta_b\equiv \dfrac{n_b}{n_\gamma}\approx 5\mbox{.}5\cdot 10^{-10}\left(\dfrac{\Omega_b h^2}{0\mbox{.}020}\right)

In that moment, the question is: how the baryons end up? The answer is pretty simple. There are two main ideas:

1) The thermal equilibrium is kept thorugh out the whole phase in the early Universe. It means that the nuclear state will be the one with the lowest energy per baryon. That is, the most stable nucleus is iron (Fe), but, in fact,  most of the baryons end up in hydrogen (H) or Helium (He).

2) Simple nucleosynthesis is based on 2 elementary ideas and simplifications. Firstly, no element heavier than helium ^4He are produced at high ratios. Therefore, this means that protons(p), neutrons(n), the deuterium(D), the tritium (T), Helium-3 (^3He) and Helium-4 (^4He) are the main subproducts of Big Bang Nucleosynthesis in this standard scenario. Sedondly, until the Universe froze below T=0\mbox{.}1MeV, no light nuclei could form and there were only a “primordial soup” made of protons and neutrons. The neutron to proton ratio n/p IS an input for the synthesis of D, T, ^3He, and ^4He. Indeed, the 0.1 MeV temperature is relatively low considering the typical nuclear binding energy, about some MeV’s. We could have some quantities of “heavy” elements, but this effect is small, at the level of \eta_b\approx 10^{-9}, i.e., the effect of large number of photons compared to the number of baryons is comparable to the number of baryons and heavy elements beyond helium isotopes. We can loot at this effec in the deuterium (D) production:

n+p\longrightarrow D+\gamma

where the reaction is taken in the thermal equilibrium and n_\gamma=n_\gamma^0. In this equilibrium, we get

\dfrac{n_p}{n_nn_p}=\dfrac{n_D^0}{n_n^0n_p^0}=\dfrac{3}{4}\left(\dfrac{2\pi m_D}{m_nm_pT}\right)^{3/2}e^{(m_n+m_p-m_D)/T}=\dfrac{3}{4}\left(\dfrac{4\pi}{m_pT}\right)^{3/2}e^{B_p/T}

and where m_D\approx 2m_p, m_n\approx m_p and B_p is the binding energy of deuterium (D). Indeed, we get n_p\sim n_n\sim n_b=\eta_bn_\gamma and so n_\gamma\sim T^3. Moreover we also get

\dfrac{n_D}{n_b}\sim \eta_b \left(\dfrac{T}{m_p}\right)^{3/2}e^{B_p/T}

and the exponential “compensates” \eta_b and T, since the smal factor \eta_b must be chosen to be smaller than the binding energy to temperature ratio B_p/T.

By the other hand, the nuetron abundance can be also estimated. From a simple proton-neutron conversion, we obtain

p+e^- \leftrightarrow n+\nu_e

due to the weak interaction! The proton/neutron equilibrium ratio for temperatures greater than 1MeV becomes

\dfrac{n_p^0}{n_n^0}=\dfrac{e^{-m_p/T}\int dpp^2e^{-p^2/2m_pT}}{e^{-m_nT}\int dpp^2e^{-p^2/2m_nT}}=e^{Q/T}

where Q=m_n-m_p=1\mbox{.}293MeV. In fact, the exponential will not be maintained below T\approx 1 MeV and we define the neutron fraction as follows. Firstly

X_n=\dfrac{n_n}{n_p+n_n}

In equilibrium, this becomes

X_n (eq)=\dfrac{1}{1+n_p^0/n_n^0}

Boltzmann equation for the process n+\nu_e\leftrightarrow p+e^- can be easily derived

a^{-3}\dfrac{d( n_na^3)}{dt}=n_n^0n_\nu^0\langle \sigma v\rangle \left( \dfrac{n_pn_e}{n_p^0n_e^0}-\dfrac{n_nn_\nu}{n_n^0n_\nu^0}\right)=n_\nu^0\langle \sigma v\rangle \left(\dfrac{n_pn_n^0}{n_p^0}-n_n\right)

and where

e^{-Q/T}=\dfrac{n_n^0}{n_p^0}

Therefore, we obtain

n_n=\left(n_n+n_p\right)X_n

and from the LHS, we calculate

a^{-3}\dfrac{d}{dt}\left[a^3(n_n+n_p)X_n\right]=a^{-3}X_n\dfrac{d}{dt}\left(a^3(n_n+n_p)\right)+\dfrac{dX_n}{dt}\left(n_n+n_p\right)

By the other hand, from the RHS

n_\nu^0\langle \sigma v\rangle \left[(n_n+n_p)(1-X_n)e^{-Q/T}-(n_n+n_p)X_n\right]

Thus, \Gamma_{np}\longrightarrow \lambda_{np} implies that

\dfrac{dX_n}{dt}=\lambda_{np}\left[(1-X_n)e^{-Q/T}-X_n\right]

If we change the variable t\longrightarrow x=Q/T, then we write

\dfrac{dX_n}{dt}=\dfrac{dX_n}{dx}\dfrac{dx}{dt}=-\dfrac{Q\dot{T}}{T^2}=-x\dfrac{\dot{T}}{T}=+x\dfrac{\dot{a}}{a}=xH

where

H=\sqrt{\dfrac{\rho_R}{3M_p^2}}=\sqrt{\dfrac{\pi^2g_\star}{90}}\dfrac{T^2}{M_p}=\sqrt{\dfrac{\pi^2g_\star}{90}}\dfrac{Q^2}{M_p}x^{-2}=H(x=1)

and

\dfrac{dX_n}{dt}=\dfrac{\lambda_{np}x}{H(x=1)}\left[e^{-x}-X_n(1+e^{-x})\right]

with

\lambda_{np}(x)=\dfrac{255}{\tau_nx^5}\left(12+6x+x^2\right)

and \tau_n is the neutron lifetime, i.e. \tau_n\approx 886\mbox{.7}s

The numerical integration of these equations provides the following qualitative sketch for X_n:

OutOfEqXcompleteAt T below 0.1MeV, the neutron decays n\longrightarrow p+e^-+\nu_e via weak interaction. It yields

X_ne^{-t/\tau_n} and $latex X_n(T_{BBN})=0\mbox{.}15\times 0\mbox{.74}=0\mbox{.}11

such as the deuterium production is done through the processes

n+p\longrightarrow D+\gamma

and it started at about T\sim 0\mbox{.}07MeV and

t=132s\left(\dfrac{0\mbox{.}1MeV}{T}\right)^2

The light element abundances

A good approximation is to consider that light element production happens instantaneously at T=T_{BBN}. Of course, the issue is…How could we determine that temperature? If we measure the abundance of deuterium abundance today, and the baryon abundance today (i.e., if we know their current densities), we can use the cosmological equations to deduce the ratio

\dfrac{n_D}{n_b}\approx \eta_b\left(\dfrac{T}{M_p}\right)^{3/2}e^{B_D/T}\sim 1

Then, we obtain from these equations and the measured densities that T_{BBN}\approx 0\mbox{.}07MeV\sim 0\mbox{.}1MeV

Moreover, since B(He)>B_D, it implies that helium-4 (^4He) production is favoured by BBN! It means that all neutron are processed inside helium-4 or hydrogen. In fact, the helium-4 abundance is known to be

X_4=\dfrac{4n(^4He)}{n_b}=2X_n(T_{BBN})\approx 0\mbox{.}22

We can compare this with an exact solution for the “yield” Y_p=0\mbox{.}2262+0\mbox{.}0135\ln (\eta_b (10^{-10}))

The observed helium-4 abundance is in good agreement with the theoretical expectations from the Standard Cosmological Model! What an awesome hit! We can also compare this with the primordial helium abundances from cosmological observations

0\mbox{.}22\sim 0\mbox{.}25

Thus, we have learned that the deuterium abundance IS a powerful probe of the baryon density!!!!

Remark: Nowadays, there is a problem with the lithium-7 abundances in stars. The origin of the discrepancy is not known, as far as I know. Then, the primordial lithium abundance is a controversial topic in modern Cosmology, so we understand BBN only as an overall picture, and some details need to be improved in the next years.

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