# Big Bang Nucleosynthesis

By

Dr Lim Zhi Han

●

Published 2022-10-09

One of the success of the Big Bang theory is the model of primordial nucleo-synthesis which correctly predicts the abundance of helium and other light elements in the Universe. In this section we will outline the various nuclear reactions in the first three (and a half) minutes of the Big Bang. We shall begin our discussions at 10^{12} \text{ K} or t \sim 10^{−4} \text{ s}, for any earlier and hotter, the physics involved will be out of the scope of this course. Our Universe just below 10^{12} \text{ K} was a soup of particles comprising of photons (\gamma), electrons (e^−), positrons (e^+), neutrinos and their anti-particles (\nu_e, \bar \nu_e), protons (p^+) and neutrons (n). Apart from the massless photons, the speed of the particles follows the Maxwell-Boltzmann distribution¹,

\begin{equation}
f(v)=4\pi (\frac{m}{2 \pi kT})^{3/2}v^2\exp(\frac{-mv^2}{2kT}).
\end{equation}

The characteristic thermal energy at a temperature T, derived by evaluating the most probable speed in the distribution, is kT . At 10^{12} \text{ K}, the characteristic energy

\begin{equation}
kT \approx \text{86 MeV}
\end{equation}

is much larger than the energy difference between the proton and neutron

\begin{equation}
(m_p-m_n)c^2=-1.293.
\end{equation}

Because of the excessive amount of energy available, protons and neutrons constantly transforms from one to the other via the following equations

\begin{equation}
n+e^+ \rightleftharpoons p^+ + \bar \nu_e
\end{equation}

\begin{equation}
\space n+\nu_e \rightleftharpoons p^+ + e^-
\end{equation}

The ratio of the number density of neutrons and protons is given by the Boltzmann factor

\begin{equation}
\frac{n_n}{n_p}=\exp[(m_p-m_n)c^2/kT]=0.985.
\end{equation}

This ratio was to drop as the temperature of the Universe falls.

At T = 10^{10} \text{ K}, n_n/n_p = 0.223. At this temperature, the rates of reactions of n \leftrightarrow p^+ drops drastically for two reasons. Firstly the energy of neutrinos had fell for them to become much less interacting and unable to initiate reactions in Eqs. (8.14) and (8.15). Secondly, as the temperature continues to drop from 10^{10} \text{ K}, energetic photons that are able to fuel the positron–electron pair production 2 \gamma \leftrightarrow e^+ + e^− are depleting significantly (see Ex. below). This means that more electron and positrons annihilate each other without being replaced. As a result the ratio of 223 neutrons to 1000 protons was almost frozen.
Almost, but not entirely. The n_n/n_p ratio now falls gradually due to a slow beta decay process

\begin{equation}
n \rightarrow p^+ + e^- + \bar \nu_e
\end{equation}

with a decay half-life of \tau_{1/2} = 617s.

The proton p^+ is also known as the hydrogen ^1 \text{H} nucleus. The next stable nucleus to form is ^4_2 \text{He}, the second lightest element in the Periodic Table. The formation of ^4_2 \text{He} could not possibly be the product of two protons and two neutrons colliding simultaneously since such events will be too rare. The efficient reactions that produce ^4_2 \text{He} include

\begin{equation}
\begin{align*}
p^+ + n \space & \rightleftharpoons \space ^2_1 \text{H} + \gamma \\
^2_1 \text{H} + ^2_1 \text{H} \space  &  \rightleftharpoons \space ^3_1 \text{H} + p^+ \\
^3_1 \text{H} + ^2_1 \text{H} \space  & \rightleftharpoons \space ^4_2 \text{He} + n
\end{align*}
\end{equation}

and

\begin{equation}
\begin{align*}
p^+ + n \space & \rightleftharpoons \space ^2_1 \text{H} + \gamma \\
^2_1 \text{H} + ^2_1 \text{H} \space  &  \rightleftharpoons \space ^3_2 \text{He} + n \\
^3_2 \text{He} + ^2_1 \text{H} \space  & \rightleftharpoons \space ^4_2 \text{He} + p^+
\end{align*}
\end{equation}

Notice that in both the reaction schemes above, deuterium ^2_1 \text{H} needs to be first produced. Above 10^9 \text{ K}, the energetic radiation would have dissociated any ^2_1 \text{H} nucleus formed, preventing and subsequent reactions. This is know as the deuterium bottleneck. Just below the 10^9 \text{ K}, the deuterium bottleneck is cleared and almost all the neutrons we had started with before 10^9 \text{ K} were cooked into helium ^4_2 \text{He}. Our sample of 164 \space n and 1000 \space p^+ becomes 82 \space ^4_2 \text{He} and 836 \space p^+ (or ^1 \text{H}). The mass fraction of ^4_2 \text{He} in the Universe is thus estimated to be 4(82)/[836 + 4(82)] = 0.28. This is consistent with observations which finds the primordial abundance of ^4_2 \text{He} to be 24\%. This is important as it was shown in separate studies that stella nucleo-synthesis could not have produced so much helium. The fact that Big Bang nucleo-synthesis quantitatively accounts for the helium abundance scores the triumph against the steady state theory, which relied on the stars to produce all the elements.

While the simple account presented above gives a reasonably good estimate to the helium-4 abundance, detailed numerical simulations were made to give a more accurate quantitative abundance of helium-4 and other light elements (see figures 2 and 3 in Wagoner, 1973). The abundance of light elements other than helium-4 is strongly dependent on the baryon² number density. This can be qualitatively understood easily for deuterium. Other than reactions with itself as shown in Eqs. (8.18) and (8.19), deuterium also reacts with protons through

\begin{equation}
^2_1 \text{H} + p \rightarrow \space ^3_2 \text{He} + \gamma
\end{equation}

If the baryon number density is high, the reaction is fast, pushing more deuterium to be converted to ^3_2 \text{He} (which subsequently converts to ^3_2 \text{He}), resulting in a low deuterium abundance. Conversely a low baryon density will result in more deuterium left over. Astronomical observations of the deuterium abundance is hence of great importance as it gives information on the baryon number density. Current observations finds the relative abundance of the light elements to be

\begin{align}
^1_1 \text{H} \quad &:& ^2_1 \text{H} \quad &: &^3_2 \text{He} \quad &: &^4_2 \text{He} \quad &: &^7_3 \text{Li} \nonumber \\
= 0.75  \quad &:& 10^{-5} \quad &: &10^{-5}  \quad &: &0.25  \quad &: \space &10^{-9}
\end{align}