\documentclass[11pt,fleqn]{article}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% template that does not use Revtex4
%%% but allows special fonts
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%
%%% Please use this template.
%%% Edit it using e.g. Notepad
%%% Ignore the header (do not change it)
%%% Process the file in the Latex site 
%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


% Page setup
\topmargin -1.5cm      
\oddsidemargin -0.04cm   
\evensidemargin -0.04cm  
\textwidth 16.59cm
\textheight 24cm 
\setlength{\parindent}{0cm} 
\setlength{\parskip}{0cm} 


% Fonts
\usepackage{latexsym}
\usepackage{amsmath} 
\usepackage{amssymb} 
\usepackage{bm}


% Math symbols I
\newcommand{\sinc}{\mbox{sinc}}
\newcommand{\const}{\mbox{const}}
\newcommand{\trc}{\mbox{trace}}
\newcommand{\intt}{\int\!\!\!\!\int }
\newcommand{\ointt}{\int\!\!\!\!\int\!\!\!\!\!\circ\ }
\newcommand{\ar}{\mathsf r}
\newcommand{\im}{\mbox{Im}}
\newcommand{\re}{\mbox{Re}}

% Math symbols II
\newcommand{\eexp}{\mbox{e}^}
\newcommand{\bra}{\left\langle}
\newcommand{\ket}{\right\rangle}

% Mass symbol
\newcommand{\mass}{\mathsf{m}} 
\newcommand{\Mass}{\mathsf{M}} 

% More math commands
\newcommand{\tbox}[1]{\mbox{\tiny #1}}
\newcommand{\bmsf}[1]{\bm{\mathsf{#1}}} 
\newcommand{\amatrix}[1]{\begin{matrix} #1 \end{matrix}} 
\newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}}

% Other commands
\newcommand{\hide}[1]{}
\newcommand{\drawline}{\begin{picture}(500,1)\line(1,0){500}\end{picture}}
\newcommand{\bitem}{$\bullet$ \ \ \ }
\newcommand{\Cn}[1]{\begin{center} #1 \end{center}}
\newcommand{\mpg}[2][1.0\hsize]{\begin{minipage}[b]{#1}{#2}\end{minipage}}
\newcommand{\Dn}{\vspace*{3mm}}

% Figures
\newcommand{\putgraph}[2][0.30\hsize]{\includegraphics[width=#1]{#2}}
\usepackage{graphicx}
\usepackage[export]{adjustbox}
\usepackage{subcaption}

% heading
\newcommand{\exnumber}[1]{\newcommand{\exnum}{#1}}
\newcommand{\heading}[1]{\begin{center} {\Large {\bf Ex\exnum:} #1} \end{center}}
\newcommand{\auname}[1]{\begin{center} {\bf Submitted by:} #1 \end{center}}


\begin{document} 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\exnumber{4442}
\heading{Chemical equilibrium for gamma == e and gamma == pi} 
\auname{Assaf Lavi, Ofir Arad}


{\bf The problem:}

\Dn
The reaction $\gamma+\gamma\leftrightarrow e^{+}+e^{-}$ occurs inside a star, where $\gamma$ is a photon and $e^\pm$ are the positron and electron (mass $m$). In general $e^\pm$ are relativistic. Assume overall charge neutrality, and that the system is in equilibrium at temperature $T$.\\

(1) Find an expression for the densities of $e^\pm$\\
(2) Find these densities in the limit $T\ll mc^2$\\
(3) Solve the same problem for the reaction $\gamma+\gamma\leftrightarrow \pi^{+}+\pi^{-}$, where $\pi$ are bosons with mass $M$\\
(4) Can the $\pi$ bosons become Bose-condensed if the temperature is sufficiently lowered? Explain the result physically.\\


\Dn\Dn


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\bf The solution:}

\Dn

(1) 
From equilibrium we deduce that $2\mu_{\gamma}=\mu_{e^+}+\mu_{e^-}$. Since photons have zero chemical potential we define $\mu\equiv\mu_{e^-}$ and as a result $\mu=-\mu_{e^+}$. The density of fermions can be calculated from the occupation function $f(\epsilon-\mu)$. For a relativistic particle, the energy is given by $\epsilon=mc^2\sqrt{1+(p/mc)^2}$, with $m$ the electron mass, and for electrons/positrons, the spin gives a factor of 2 in the occupation, so we get
\begin{equation}
\label{eq_nferm}
    n_{e^\pm}=\frac{2}{V}\sum_p\frac{1}{e^{\beta mc^2\sqrt{1+(p/mc)^2}\pm\beta\mu}+1}\ .
\end{equation}
From charge neutrality we know that these densities are equal, $n_{e^-}=n_{e^+}$. Noting that every term in the series in Eq.~(\ref{eq_nferm}) is monotonic in $\mu$, charge neutrality demands $\mu=0$.

We can also write the densities as an integral, using $\frac{(2\pi)^3}{V}$ as the spacing in momentum space,
\begin{equation}
\label{eq_nint}
    n_{e}=\frac{2}{(2\pi)^3}\int_0^{\infty}\frac{1}{e^{\beta mc^2\sqrt{1+(p/mc)^2}}+1}d^3p\ .
\end{equation}
%%%%%%%%%%%%%%
\Dn
(2)
In the limit $\beta mc^2\gg1$, the main contributions in the integral in Eq.~(\ref{eq_nint}) come from small $p$. Thus we can approximate the particles energies as non-relativistic and use the Boltzmann approximation to yield
\begin{align}
    &\epsilon\approx mc^2+\frac{p^2}{2m}\ ,\\
    &n_{e}\simeq\frac{2}{(2\pi)^3}\int_0^{\infty}\frac{1}{e^{\beta( mc^2+p^2/2m)}+1}d^3p\simeq\frac{2 e^{-\beta mc^2}}{(2\pi)^3}\int_0^{\infty}e^{-\frac{\beta p^2}{2m}}d^3p=\frac{2e^{-\beta mc^2}}{\lambda_T^3}\nonumber\ ,
\end{align}
where $\lambda_T\equiv \left(\frac{m}{2\pi \beta}\right)^{-1/2}$. This result is similar to what we would expect from a classical ideal gas.
% %%%%%%%%%%%%%%%

% \Dn

% (2)
% In the limit $\epsilon_f<<T<<mc^2$ we can estimate the particles energy to be non relativistic, and the occupation number as Boltzmann distribution. More explicitly:
% \begin{equation}
% \epsilon\approx mc^2+\frac{p^2}{2m}    
% \end{equation}
% \begin{equation}
%     n_{e^-}=2e^{\beta\mu}\sum_p e^{-\beta(mc^2+\frac{p^2}{2m})}\ ,\\
%     n_{e^+}=2e^{-\beta\mu}\sum_p e^{-\beta(mc^2+\frac{p^2}{2m})}
% \end{equation}
% Since the densities are equal, an immediate conclusion is that $\mu=0$ which is self-consistent with the assumption that $\mu\beta\approx\epsilon_f\beta<<1$.

% Thus we can get an expression, which is similar to classical ideal gas,
% \begin{equation}
%     n_{e^-}=n_{e^+}=\frac{2 e^{-\beta mc^2}}{ \lambda^3_T} \ ,\\ \lambda_T\equiv \sqrt{\frac{2\pi \beta}{m}}
% \end{equation}


%%%%%%%%%%%%%%
\Dn

(3)
Now we have $\pi^{\pm}$ bosons, with mass $M$, so the occupation function changes to Bose-Einstein, and the densities become
\begin{equation}
\label{eq_nbos}
    n_{\pi^\pm}= \frac{1}{V}\sum_p\frac{1}{e^{\beta Mc^2\sqrt{1+(p/Mc)^2}\pm\beta\mu}-1}\ ,
\end{equation}
where $\mu\equiv \mu_{\pi^-}=-\mu_{\pi^+}$. For bosons, the chemical potential must be smaller than the ground state energy, which in our case demands $-Mc^2<\mu<Mc^2$. In that region, similarly to the fermionic case, each term in the series in Eq.~(\ref{eq_nbos}) is monotonic in $\mu$, so charge neutrality assures us that $\mu=0$ here too.

Writing the densities as integrals, we have
\begin{align}
    n_{\pi}=\frac{1}{(2\pi)^3} \int\frac{1}{e^{\beta Mc^2\sqrt{1+(p/Mc)^2}}-1}d^3p\ .
\end{align}
In the $T\ll Mc^2$ limit, the same logic applies to justify the Boltzmann approximation, and we have
\begin{equation}
\label{eq_npi}
    n_{\pi}\simeq \frac{e^{-\beta Mc^2}}{(2\pi)^3}\int_0^{\infty}e^{-\frac{\beta p^2}{2M}}d^3p=\frac{e^{-\beta M c^2}}{\lambda_T^3}\ .
    \end{equation}
%%%%%%%%%%%%%
\Dn

(4)
Since we showed that charge neutrality demands $\mu=0$, no condensation is possible. In more detail - to have condensation, the chemical potential must tend to the ground state energy, which in our case would mean $\mu\to \pm Mc^2$, but this is impossible and would mean a net charge in the system.

To understand what happens physically, we examine Eq.~(\ref{eq_npi}),

\begin{equation}
    n_{\pi}\simeq \frac{e^{-\beta M c^2}}{\lambda_T^3}\propto \frac{e^{-Mc^2/T}}{T^{3/2}}
\end{equation}
so when we lower the temperature, the densities become smaller and smaller, i.e. there are less particles instead of the particles being in condensation. 
%%%%%%%%%%%%%

\Dn

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{document} 
