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\exnumber{4546}
\heading{Chemical equilibrium for misc reactions} 
\auname{Idan Wallerstein and Nir Kastin}
	
	
{\bf The problem:}
	
	\begin{itemize}
		\item [(a)]
		The following reaction occurs inside a star
		\[\ \gamma+ \gamma\leftrightarrow     e^{+}+e^{-}\]
		where ${\gamma}$  is a photon and ${e^{\pm}}$ are the positron and
		electron, respectively. Assume overall charge neutrality and that
		the system is in equilibrium at temperature ${T}$. Find an
		expression for the densities of ${e^{\pm}}$. (In general
		${e^{\pm}}$ with mass $m$ are relativistic). Find an explicit
		result for these densities in the limit ${k_{B}T<<mc^{2}}$. (Hint:
		no conservation law for photons).
		\item [(b)]
		Repeat (a) for the reaction
		\[\ \gamma + \gamma\leftrightarrow \pi^{+}+\pi ^{-}\]
		where ${\pi^{\pm}}$ are bosons with mass $m_{\pi}$.
		Can these bosons become Bose-condensed if the temperature is
		sufficiently lowered? What are the densities at $T=0$?
		\item[(c)] The photons in (b) are now eliminated from the system, but the
		density of the bosons is maintained. If these bosons are cooled,
		is there a temperature at which Bose condensation is possible?
		[Assume that the initial temperature satisfies $k_BT<<
		m_{\pi}c^2$]. What are the densities at $T=0$?\\
		\item[(d)] Consider initial equal densities ${\bar n}$ for $e^+$ and for $e^-$. The
		$e^{\pm}$  are allowed to equilibrate with $\pi^{\pm}$ via the reaction
		\[ e^{+}+e^{-} \leftrightarrow \pi^{+}+\pi^{-}\]
		Write equations that determine the chemical potentials at $T\neq 0$.
		Consider now $T=0$ and determine if
		Bose Einstein condensation is possible and if so what is the condition for that. Consider both $m_e>m_{\pi}$ and $m_e<m_{\pi}$ situations.
	\end{itemize}
	
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	{\bf The solution:}
	\begin{itemize}
		\item [(a)]
		The basic equation which relates the different components in a chemical reaction is the chemical equilibrium equation which states for the desired reaction that,
		\begin{equation}
		{\mu _{{+ }}} + {\mu _{{ - }}} = {\mu _\gamma } + {\mu _\gamma }
		\end{equation}
		Since the photons have zero rest mass they don't have a conservation law, which means ${\mu _\gamma } = 0$, making the equation 
		${\mu _ + } + {\mu _ - } = 0$. Moreover, charge neutrality dictates that ${n_ + }\left( {{\mu _ + }} \right) = {n_ - }\left( {{\mu _ - }} \right)$ and since both particles (electrons and positrons) have the same energy dispersion and spin, the function $n_+$ and $n_-$ will be the same with respect to their arguments, which results in having the same chemical potential ${\mu _ + } = {\mu _ - }$. Combining both equations we have $\mu={\mu _ - } = {\mu _ + } = 0$.
		In the case of relativistic fermions, $n$ as a function of $\mu$ is not analytic (not given by elementary functions). The integral expression would be
		\begin{equation}
		n_+=n_-=n =\int\limits_{m}^{\infty} {\frac{{g(\varepsilon
					)/V }}{{{e^{\beta  {(\varepsilon-\mu) }}} + 1}}d\varepsilon }=\frac{1}{\pi^2} \int\limits_{m}^{\infty} {\frac{{\varepsilon \sqrt {{\varepsilon ^2} - {m^2}} }}{{{e^{\beta  {\varepsilon }}} + 1}}d\varepsilon }
		\end{equation} 
		Where $\mu=0$ was used and the density of states was derived from the dispersion $\varepsilon  = \sqrt {{k^2} + {m^2}}$ (we use natural units $\hbar  = c = 1$): \\We have \[\frac{{dk}}{{d\varepsilon }} = \frac{d}{{d\varepsilon }}\left( {\sqrt {{\varepsilon ^2} - {m^2}} } \right) = \frac{\varepsilon }{{\sqrt {{\varepsilon ^2} - {m^2}} }}\]
		The number of states between $k, k+dk$ is
		\begin{equation}
		\frac{g\left( \varepsilon  \right) d\varepsilon}{V} =\frac{{2{d^3}k}}{{{{\left( {2\pi } \right)}^3}}} = \frac{{8\pi {k^2}dk}}{{{{\left( {2\pi } \right)}^3}}} = \frac{1}{\pi^2} \left( {{\varepsilon ^2} - {m^2}} \right)\frac{{dk}}{{d\varepsilon }}d\varepsilon  = \frac{1}{\pi^2} \left( {{\varepsilon ^2} - {m^2}} \right)\frac{\varepsilon }{{\sqrt {{\varepsilon ^2} - {m^2}} }}d\varepsilon  = \frac{1}{\pi^2} \varepsilon \sqrt {{\varepsilon ^2} - {m^2}} d\varepsilon
		\end{equation}
		Where the factor 2 in the first term is for spin degeneracy.
		
		The limit $T\ll m$ is now taken, leading to a Boltzmann approximation to the occupation function since $e^{\beta\varepsilon}\gg1$. Thus, the integral would be:
		\begin{equation}\begin{array}{l}
		{n_ + } = {n_ - } = n = \frac{1}{{{\pi ^2}}}\int\limits_m^\infty  {\varepsilon \sqrt {{\varepsilon ^2} - {m^2}} {e^{ - \beta \varepsilon }}d\varepsilon }  = T{\left( {\frac{m}{\pi }} \right)^2}{K_2}\left( {\frac{m}{T}} \right) \approx \\
		\approx {e^{ - \beta m}}\frac{{{m^3}}}{{{\pi ^2}}}\sqrt {\frac{\pi }{2}} {\left( {\frac{1}{{\beta m}}} \right)^{3/2}} =  = 2\frac{{{{\left( {2m} \right)}^{3/2}}}}{{{{\left( {2\pi } \right)}^2}}}\Gamma \left( {\frac{3}{2}} \right){T^{3/2}}{e^{ - \beta m}}
		\end{array}\end{equation}
		Where $K_2(x)$ is the modified Bessel function of the second kind. \\ 
		
		We can see that in the limit $T=0$ the density of the electrons/positrons is $0$, thus all the particles annihilate to make a photon gas. 
		
		\item [(b)]
		For the boson gas the integral for the number of particles with relativistic energy would be:
		\begin{equation}\begin{array}{l}
		n_ + ^{ > 0} = n_ - ^{ > 0} = {n^{ > 0}} = \frac{1}{{2{\pi ^2}}}\int\limits_{}^{} {\frac{{\varepsilon \sqrt {{\varepsilon ^2} - {m^2}} }}{{{e^{\beta \left( {\varepsilon  - \mu } \right)}} - 1}}d\varepsilon }  \approx \frac{1}{{2{\pi ^2}}}\int\limits_m^\infty  {\varepsilon \sqrt {{\varepsilon ^2} - {m^2}} {e^{ - \beta \varepsilon }}d\varepsilon }  = \\
		= \frac{T}{2}{\left( {\frac{m}{\pi }} \right)^2}{K_2}\left( {\frac{m}{T}} \right) \approx \frac{{{{\left( {2m} \right)}^{3/2}}}}{{{{\left( {2\pi } \right)}^2}}}\Gamma \left( {\frac{3}{2}} \right){T^{3/2}}{e^{ - \beta m}}
		\end{array}\end{equation} 
		Where:
		\\ \qquad	1. Like item (a) the chemical potentials obey  $\mu={\mu _ - } = {\mu _ + } = 0$ since $\mu_\gamma=0$ .
		\\	\qquad 2. The same limit as in item (a) - $T\ll m$ is taken.
		\\	\qquad 3. $g(\varepsilon)$ is obtained exactly the same as in item (a) except for the spin degeneracy, meaning: \[\frac{g\left( \varepsilon  \right)d\varepsilon}{V} = \frac{1}{2 \pi^2} \varepsilon \sqrt {{\varepsilon ^2} - {m^2}} d\varepsilon\]
		
		Thus, we can see that at the limit $T\to 0$ the number of excited bosons goes to zero. In addition since the chemical potential is always zero, it  is lower than the minimum energy $\mu=0<m=\varepsilon_{min}$ forbidding a condensation to appear. Hence, taking the limit of $T\to0$ we get that all the pions have been annihilated to make a photon gas.
		
		\item[(c)]
		When the photons are eliminated, there is no chemical equilibrium and the pions regain their maximum chemical potential $\mu_{max}=\varepsilon_{min}=m_{\pi}$, and the gas becomes a regular three-dimensional ideal gas which can condensate. Quantitatively for $\mu\to \varepsilon_{min}=m_{\pi}$, 
		\begin{equation}{n_ +^{>0} } = {n_ -^{>0} } =n^{>0} = \frac{1}{2\pi^2}\int\limits_{m}^{\infty} {\frac{{\varepsilon \sqrt {{\varepsilon ^2} - {m^2}} }}{{{e^{\beta \left( {\varepsilon  - m } \right)}} - 1}}d\varepsilon }
		\end{equation}
		Specifically at the limit $T\to 0$ The integrand is exactly $0$, so there are no excited particles and all the pions occupy the ground state leading to $n_+=n_-=n=\frac{N}{V}$ where $\frac{N}{V}$ is the density of the bosons which was maintained after the elimination of the photons from the system.
		
		\item[(d)]
		Since the initial densities are equal the reaction enforces
		\begin{equation}{n_{e + }} = {n_{e - }} \qquad {n_{\pi  + }} = {n_{\pi  - }} \qquad {n_{e + }} + {n_{\pi  + }} = \bar n\end{equation}
		and for the chemical potentials themselves:
		${\mu _{e + }} + {\mu _{e - }} = {\mu _{\pi+ }} + {\mu _{\pi  - }}$.
		At $T=0$ we would have ${\varepsilon _{F + }} + {\varepsilon _{F - }} = {\mu_{\pi  + }} + {\mu_{\pi  - }}$. Again, the dispersion for $e^+$ and $\pi^+$ is the same as $e^-$ and $\pi^-$ respectively so the relation is simplified to ${\varepsilon _F} = {\mu_\pi }$\\.
		The density of the fermions is given by:
		\begin{equation}
		{n_e} = \frac{1}{{{\pi ^2}}}\int\limits_{{m_e}}^\infty  {\frac{{\varepsilon \sqrt {{\varepsilon ^2} - m_e^2} }}{{{e^{\beta \left( {\varepsilon  - {\mu _e}} \right)}} + 1}}d\varepsilon }  \to \frac{1}{{{\pi ^2}}}\int\limits_{{m_e}}^\infty  {\Theta \left( {{\varepsilon _F} - \varepsilon } \right)\varepsilon \sqrt {{\varepsilon ^2} - m_e^2} d\varepsilon }  = \frac{1}{{{\pi ^2}}}\int\limits_{{m_e}}^{{\varepsilon _F}} {\varepsilon \sqrt {{\varepsilon ^2} - m_e^2} d\varepsilon }  = \frac{1}{{3{\pi ^2}}}{\left( {\varepsilon _F^2 - m_e^2} \right)^{3/2}}
		\end{equation}
		Where the $T\to0$ limit was taken.\\
		From chemical equilibrium and the fact that $\varepsilon_F=\mu_\pi\le\varepsilon_{min}=m_\pi$ it is derived that:
		\begin{equation}
		{n_e} = \frac{1}{{3{\pi ^2}}}{\left( {\varepsilon _F^2 - m_e^2} \right)^{3/2}} =\frac{1}{{3{\pi ^2}}}{\left( {\mu _\pi ^2 - m_e^2} \right)^{3/2}} \le \frac{1}{{3{\pi ^2}}}{\left( {m_\pi ^2 - m_e^2} \right)^{3/2}}
		\end{equation} 
		Separation of cases:\\
		1. If $m_\pi<m_e$ then $\varepsilon_F<m_e$ leading to $n_e=0$, and $n_\pi=\bar n-n_e=\bar n$. Thus, the case of $T=0$ yields in a pure pion gas in the ground state as a BEC, note the left scenario in Figure 1.\\
		2. If $m_\pi>m_e$ then $n_e$ is finite, thus the electrons occupy the energy levels between $m_e$ to $\varepsilon_F$. As can be seen from the inequality above, if $m_\pi>\varepsilon_F(\bar n)$ then $n_e=\bar n$ and $n_\pi=0$ as it is shown in the middle scenario in Figure 1. Else, $m_e<m_\pi<\varepsilon_F(\bar n)$ leading to occupation of the electrons up to the energy of $m_\pi$, meaning $n_e=\frac{1}{{3{\pi ^2}}}{\left( {m_\pi ^2 - m_e^2} \right)^{3/2}}$. The other particles would occupy as bosons the ground state as a BEC as noted in the right scenario in Figure 1 and their density would be $n_\pi=\bar n-n_e=\bar n-\frac{1}{{3{\pi ^2}}}{\left( {m_\pi ^2 - m_e^2} \right)^{3/2}}$.
		\begin{figure}
			\centering
			\includegraphics[width=0.7\linewidth, height=0.25\textheight]{"energylevels"}
			\caption[]{Energy levels and occupation for different mass relation and overall number of particles. The red particles indicate bosons as a BEC, the blue ones indicate fermions.}
			\label{fig:this ias the label}
		\end{figure}
	\end{itemize}
	
	
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