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\title{Nucleosynthesis}
\author{Andrii Magalich}
\chapter{Determining the conditions of nucleosynthesis}
\section{Nuclear reactions}
Generation of light chemical elements from the primordial neutrons and protons occurs through the network of nuclear reactions between nucleons and nuclei (\verb|\ref{fig:stability_line_screen}|).
\begin{figure}
\centering
\includegraphics[width=\linewidth]{images/stability_line_screen}
\caption{}
\label{fig:stability_line_screen}
\end{figure}
To determine the conditions of nucleosynthesis, we need to consider the possible processes. They constitute 3 groups by the type of interaction:
\begin{itemize}
\item Electromagnetic with photons (e.g. $n + p \leftrightarrows \ce{D} + \gamma$)
\item Weak transformations of nucleons (e.g. $n + e^+ \leftrightarrows p + \overline{\nu}_e$)
\item Other nuclear reactions
\end{itemize}
\paragraph{Reactions with photons.} These reactions directly interact with the electromagnetic plasma, so their balance is influenced by its thermodynamical state.
\paragraph{Weak reactions.} The rate of weak reactions is much smaller than the electromagnetic rate, so it is important to consider the state of Universe expansion when the nucleosynthesis takes place.
\paragraph{Regular nuclear reactions.} Nuclei are charged, so the Coulomb barrier can significantly inhibit the reactions.
On the other hand, we can classify them by the number of colliding particles:
\begin{itemize}
\item 2-particle reactions (e.g. $\ce{D} + \ce{D} \leftrightarrows p + \ce{T}$)
\item 3-particle reactions (like triple-alpha process $\ce{^4_2 He} + \ce{^4_2 He} \leftrightarrows \ce{^8_4 Be}, \ce{^8_4 Be} + \ce{^4_2 He} \leftrightarrows \ce{^12_6 C}$)
\item more complicated processes
\end{itemize}
Reactions including many particles require more time and higher densities than those that collide only 2. This means that any kind of nucleosynthesis is dominated by 2-particle processes, while 3-particle reactions can mediate generation of nuclei that cannot be created through the former.
\section{Deuterium requirement}
Generation of nucleus $\ce{^A_Z N}$ from neutrons and protons releases energy in amount equal to the binding energy of the nucleus $\Delta_{N}$. This means that the nuclei will be easily destroyed in the plasma with temperature above the binding energy scale $ T \gtrsim \Delta_N$ (as the mean photon energy is $\overline{p} \sim 3 T$).
Considering the 2-particle reactions network of the nucleosynthesis, we can immediately see that the generation of nuclei depend on the abundance of deuterium (fig. \verb|\ref{fig:BBN_reactions}|).
\begin{figure}
\centering
\includegraphics[width=\linewidth]{images/BBN_reactions}
\caption{}
\label{fig:BBN_reactions}
\end{figure}
Deuterium requirement restricts the nucleosynthesis conditions to $T < \Delta_D = 2.2 \;MeV$. The mode of deuterium generation depends on the rate of the reaction
\begin{equation}
p + n \leftrightarrow \gamma + D
\end{equation}
The cross-section of this reaction can be roughly approximated as
\begin{equation}
(\sigma v) \approx \frac{\alpha}{m_\pi^2} \approx 0.37 \cdot 10^{-6} MeV^{-2}
\end{equation}
where $m_\pi^{-1}$ corresponds to the spatial range of strong interactions and $\alpha$ is the fine structure constant related to the creation of a photon.
This leads to the following interaction rate:
\begin{equation}
\Gamma_{p(n,\gamma)D} = n_p \cdot (\sigma v) = \frac{2}{\pi^2} \eta_B \xi(3) T^3 \cdot \frac{\alpha}{m_\pi^2}.
\end{equation}
Next, comparing this interaction rate with the expansion rate of the Universe, we can determine the temperature when deuterium generation reaction can no longer support thermodynamical equilibrium.
\begin{equation}
\frac{T^2}{M_*} = H \sim \Gamma = \frac{2}{\pi^2} \eta_B \xi(3) T^3 \cdot \frac{\alpha}{m_\pi^2}
\end{equation}
\begin{equation}
\frac{1}{T_{dec}} = M_* \frac{\alpha}{m_\pi^2} \eta_B \xi(3) \frac{2}{\pi^2} = \frac{8.99 \cdot 10^{-17}}{MeV^{2}} M_* \approx \frac{1}{3 \;eV}
\end{equation}
If $T_{dec} > \Delta_D$, Deuterium will not be produced thermally and, in absence of some resonant process, will never populate the Universe significantly. In the opposite case, the nuclei will reach the nuclear statistical equilibrium density, but eventually will be driven out of equilibrium as well.
Notably, this reaction stays in the equilibrium almost until the recombination, when most nuclear reactions cease, inhibited by the Coulomb barrier. Due to this we assume that the nucleosynthesis takes place somewhere in between recombination and $T = \Delta_D$, with Deuterium production reaction in equilibrium.
Notice that the upper bound on the nucleosynthesis does not depend on the details of the Universe evolution, while the lower one can shift, depending on the Hubble rate. Higher Hubble rate means interaction freeze-out at higher temperature.
(For conditions when BBN does not happen due to the lack of Deuterium, see App. \verb|\ref{App:no-bbn-hubble}|)
\section{Equilibrium theory}
\subsection{Nuclear statistical equilibrium}
As nuclei are charged non-relativistic particles, their densities will be governed by Boltzmann statistics in kinetic equilibrium:
\begin{equation}
n_A = g_A \left( \frac{m_A T}{2 \pi} \right)^\frac{3}{2} e^\frac{\mu_A -m_A}{T}
\end{equation}
Electromagnetic processes are sufficiently fast during all the nucleosynthesis such that the only changing parameter in the formula above is the chemical potential of the specie $\mu_A$.
In the case of sufficiently fast interactions between nuclei, chemical potentials $\mu_A$ will also enter the chemical equilibrium:
\begin{equation}
\mu_A = Z \mu_p + (A-Z) \mu_n
\end{equation}
This condition ensures that for any nuclear reaction conserving the total number of protons and neutrons, the following will hold:
\begin{equation}
\sum_{in} \mu_i = \sum_{out} \mu_j
\end{equation}
As this applies also to protons and neutrons as well, the chemical potential of any nucleus can be expressed as
\begin{align}
e^\frac{\mu_A}{T} &= e^\frac{Z \mu_p + (A-Z) \mu_n}{T}
\\ &= n_p^Z n_n^{A-Z} \left( \frac{2 \pi}{m_n T} \right) ^\frac{3A}{2} 2^{-A} e^\frac{Z m_p + (A-Z) m_n}{T}
\end{align}
where we neglect the difference in the masses of protons, neutrons and mean nucleon mass in the nuclei $\frac{m_A}{A}$ (while the same difference is important in the exponential).
From this we get the total nuclei densities in the nuclear statistical equilibrium:
\begin{equation}
n_A = g_A A^\frac32 2^{-A} \left(\frac{2 \pi}{m_n T}\right)^\frac{3(A-1)}{2} n_p^Z n_n^{A-Z} e^\frac{B_A}{T}
\end{equation}
where $B_A = Z m_p + (A-Z) m_n - m_A$ is the binding energy of the nucleus.
If all nuclear reactions were in chemical equilibrium, the final abundances of the nuclei would be defined by the formula above and the amount of chemical elements beside the Hydrogen would be negligible.
To demonstrate, let's consider the nuclei in equilibrium before nucleosynthesis at the temperature of $1 MeV$ with $\eta_B(1 MeV) = 10^{-9}$:
\begin{equation}
n_\gamma = \frac{2}{\pi^2} \xi(3) T^3 \approx 0.2435 MeV^3
\end{equation}
\begin{equation}
\frac{n_n}{n_p} \approx e^{-\frac{m_n - m_p}{T}} = 0.2744
\end{equation}
\begin{align}
n_B &= n_p \left(1 + \frac{n_n}{n_p}\right) = \eta_B n_\gamma
% \\ n_p &= \frac{\eta_B n_\gamma}{1 + (\frac{n_n}{n_p})}
% \\ n_n &= \frac{\eta_B n_\gamma}{1 + (\frac{n_n}{n_p})^{-1}}
\end{align}
\begin{multline}
n_A = g_A A^\frac32 2^{-A} \left(\frac{2 \pi}{m_n T}\right)^{\frac32(A-1)} \\ \left(\frac{n_n}{n_p}\right)^{A-Z} \left(\frac{\eta_B n_\gamma}{1 + \frac{n_n}{n_p}}\right)^{A} e^\frac{B_A}{T}
\end{multline}
\begin{figure}
\begin{center}
$\begin{tabu}{|c|c|c|c|}
\hline \ce{^A Z} & g_A & B_A, MeV & n_A(1 MeV), MeV^3 \\
\hline p & 2 & & 1.9 \cdot 10^{-10} \\
\hline n & 2 & & 5.2 \cdot 10^{-11} \\
\hline \ce{^2 H} & 3 & 2.22 & 1.1 \cdot 10^{-22} \\
\hline \ce{^3 H} & 2 & 6.92 & 2.1 \cdot 10^{-34} \\
\hline \ce{^3 He} & 2 & 7.72 & 1.7 \cdot 10^{-33} \\
\hline \ce{^4 He} & 1 & 28.3 & 1.6 \cdot 10^{-38} \\
\hline \ce{^12 C} & 1 & 92.2 & 1.5 \cdot 10^{-118} \\
\hline
\end{tabu} $
\end{center}
\caption{Nuclear Statistical Equilibrium (NSE) abundances of the chemical elements}
\end{figure}
The abundances of elements at the later times can be found from the system of algebraic equations:
\begin{align}
\begin{cases}
n_B &= n_n + n_p + \sum_A n_A \\
\frac{n_n}{n_p} &= e^\frac{-\Delta m}{T} \\
n_A &= g_A A^\frac32 2^{-A} \left(\frac{2 \pi}{m_n T}\right)^{\frac32 (A-1)} n_p^Z n_n^{A-Z} e^\frac{B_A}{T}
\end{cases}
\end{align}
Or, in terms of mass fractions $X_A = \frac{n_A A}{n_B}$:
\begin{align}
\begin{cases}
1 &= X_n + X_p + \sum_A X_A \\
X_n &= e^{-\frac{m_n-m_p}{T}} X_p \\
X_A &= g_A \left( \zeta(3)^{A-1} \pi^{\frac12 (1-A)} 2^{\frac12 (3A-5)} \right) A^\frac52 (T/m_n)^{\frac32 (A-1)} \\
& \cdot \eta^{A-1} X_p^Z X_n^{A-Z} e^\frac{B_A}{T}
\end{cases}
\end{align}
\section{Kinetic approach}
Because of the expansion, equilibrium conditions for all particles cannot be reached during the whole period of nucleosynthesis. More fundamental, kinetic theory is required to describe the dynamics of elements production in the partially equilibrated system.
For derivation of kinetic equations, see App. \verb|\ref{App:boltzmann-equation-in-expanding-universe}|.
\subsection{Nucleosynthesis as a kinetic system}
As the system is in partial equilibrium, one has to write Boltzmann equation only for some of the quantities. For example, electromagnetic interactions are sufficiently fast at all times, so all electrically charged particles are considered to be in dynamical (but not necessarily chemical) equilibrium. These particles have thermal distribution with temperature and chemical potential. Fast interactions equilibrate individual temperatures of te species with the temperature of radiation (photons). This fact greatly reduces the system of kinetic equations and number of quantities to find solutions for.
Additionally, at all times relevant for the nucleosynthesis, nuclei are non-relativistic and can be described by number density instead of the distribution function.
Then, the following system of equation arises:
\begin{align}
\frac{\partial f_{\nu_i}(t, y)}{\partial t} + 3 H &= I_{coll}\{f, n_n\} \\
\frac{\partial n_n(t)}{\partial t} + 3 H &= I_{coll}\{f, n\} \\
\frac{\partial n_X(t)}{\partial t} + 3 H &= I_{coll}\{f, n\}
\end{align}
These equations describe the evolution of neutrinos $\nu_i$, neutrons $n$ and nuclei $X = D, T, He, Li, \dots$. To close the system of equations, one has to supply the Friedmann equation and condition of energy conservation:
\begin{align}
H^2 &= \frac{8 \pi G \rho}{3} \\
\frac{d \rho}{d t} &= -3 H (\rho + P)
\end{align}
These equations contain the following independent variables: distributions functions of the relativistic particles that departed from equilibrium $f_\nu$, number densities of the neutrons and nuclei $n_n,X$, temperature $T$ and scale factor $a$.
To be complete, this system also requires the presently measured baryon-to-photon ratio and CMB temperature to define the overall baryon charge in the Universe. Starting from the recombination-time value, baryon-to-photon ratio can be traced back through electron-positron annihilation (and other transitions that produce entropy) to the value before neutron decoupling.
\subsection{Illustration: Deuterium production model}
The simplest model of nucleosynthesis to be solved contains photons, neutrons, protons and Deuterium nuclei involved in the reaction $n+p \leftrightarrow D+\gamma$ while neutrons are allowed to decay $n\rightarrow p$:
\begin{align*}
\partial_t n_n + 3 H(t) n_n &= -\Gamma_n n_n - (\partial_t n_D + 3 H n_D) \\
\partial_t n_p + 3 H(t) n_p &= +\Gamma_n n_p - (\partial_t n_D + 3 H n_D) \\
\partial_t n_D + 3 H(t) n_D &= \frac{\alpha}{m_\pi^2} \left(\sqrt{\frac{3T}{m}} n_n n_p - e^{-\frac{\Delta}{T}} n_D n_\gamma\right) \\
T(t) &= \sqrt{\frac{M_{pl}}{2 t}} \left( \frac{8 \pi}{3} \frac{\pi^3}{30} g(t) \right)^{-\frac14}\\
H(t) &= \frac{1}{2t}
\end{align*}
The expansion of the Universe can be taken here as whatever background we wish. Particularly useful control parameter here is the effective number of relativistic degrees of freedom $g(t)$. Consider 3 cases:
\begin{itemize}
\item constant $g(t) = 7.25 = 2 + \frac78 \cdot 2 \cdot 3 $ (standard cosmology after electron-positron annihilation)
\includegraphics[width=\linewidth]{images/D_model_g725}
$X_D(\infty) = 0.26675$
\item constant $g(t) = 9 = 2 + \frac78 \cdot 2 \cdot (3+1) $ (additional neutrino-like particle)
\includegraphics[width=\linewidth]{images/D_model_g900}
$X_D(\infty) = 0.26877$
\item varying $g(t) = 7.25 + 10 e^{-\left(\frac{t-10s}{5s}\right)^2} $ (model background of some transition in the plasma consistent with observations)
\includegraphics[width=\linewidth]{images/D_model_gvar}
$X_D(\infty) = 0.19567$
\end{itemize}
Hence, changes in expansion directly influence the observables of nucleosynthesis.
\appendix
\chapter{No-BBN Hubble rate}
\label{App:no-bbn-hubble}
One can calculate the expansion rate that excludes the possibility of the nucleosynthesis altogether (when Deuterium is not produced thermally):
\begin{equation}
M_* = \frac{m_\pi^2}{\alpha} \frac{\pi^2}{2 \eta_B \xi(3) T} \approx \frac{10^{16} MeV^2}{T} = \frac{M_{pl}}{1.66 \sqrt{g_*}}
\end{equation}
\begin{equation}
H = \frac{T^3}{10^{16} MeV^2}
\end{equation}
This corresponds to a ridiculous energy density of the Universe at $\sim MeV$ temperatures.
\begin{equation}
g_* = \left(\frac{M_{pl}}{1.66 M_*} \right)^2 \approx 0.5 \cdot 10^{12} \left(\frac{T}{MeV}\right)^2
\end{equation}
\chapter{Boltzmann equation in expanding Universe}
\label{App:boltzmann-equation-in-expanding-universe}
In general, Boltzmann equation for the distribution function is defined as
\begin{equation}
\frac{d f(t, x, p)}{d t} = I_{coll}\{f\}
\end{equation}
where $I_{coll}$ is the collision term, describing the processes that can lead to the non-conservation of phase density (Liouville's theorem).
At early times, the Universe is isotropic and homogeneous, such that the distribution function does not depend on the spatial coordinates:
\begin{equation}
\frac{d f(t, p)}{d t} = I_{coll}\{f\}
\end{equation}
Furthermore, if the particle in question is non-relativistic, one can average over momenta as well to obtain the equation on the number density:
\begin{equation}
\frac{d n(t)}{d t} = I_{coll}\{n\}
\end{equation}
In expanding Universe, change of variables allows to disentangle the momentum coordinate from the time dependence:
\begin{equation}
y = p a(t)
\end{equation}
\begin{equation}
\frac{\partial f(t, y)}{\partial t} + 3 H = I_{coll}\{f\}
\end{equation}
And, correspondingly
\begin{equation}
\frac{\partial n(t)}{\partial t} + 3 H = I_{coll}\{n\}
\end{equation}
The term $3H$ describes the natural decrease in the phase density due to the expansion of the Universe.

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