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\exnumber{4450}
\heading{Beta decay n==e+p+nu} 
\auname{Guy Stein, Roy Perry}


{\bf The problem:}

In this problem we regard (anti)neutrino as massless spin 1/2 fermions.
If we place neutrons in some confined region, some of them $\beta$-decay 
into protons, electrons and antineutrinos via the reaction
${n \Leftrightarrow  p + e^{-} + \bar{\nu}}$. 
The masses are ${\mass_{n}, \mass_{p}, \mass_{e} }$. 
The objective is to estimate the final equilibrium density of~$n$ 
of the neutrons, given that that initially density was~$n_0$.   

\begin{itemize}

\item[(a)]
Find the equations of states for gas of massless spin 1/2 fermions.
Namely, express the density of particles and the density of energy as a function
of the chemical potential $\mu$, the temperature $T$, and the speed of light $c$.
Evaluate the integrals for $T=0$ and for large $T$. Define what does it mean large $T$.


\item[(b)]
Write equations for the neutron density $n$, 
given that initially there were only neutrons with density $n_0$, 
and the temperature is $T$. 
Consider appriximation under the following assumptions:

\item[(b1)]
The particles are non-relativistic (except the nutrino)

\item[(b2)]
The particles are hyper relativistic (negligible mass).

\item[(b3)]
The temperature is zero.

\item[(b4)]
The temperature is high (Boltzmann approximation).


\end{itemize}

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


\begin{itemize}
	
	\item[(a)]
	$\varepsilon = {\hbar}ck$
	
	$\varepsilon^{2} = {\hbar}^{2}c^{2}k^{2}= \frac{(2\pi)^{2}{\hbar}^{2}c^{2}}{{L}^{2}}(n_{x}^{2}+n_{y}^{2}+n_{z}^{2})$
	
	$N(\varepsilon) =2\times\frac{3\pi}{4}(\frac{L\varepsilon}{2\pi{\hbar}c})^{3} = \frac{V\varepsilon^{3}}{3\pi\hbar^{3}c^{3}}
	$
	
	$g(\varepsilon) = \frac{V\varepsilon^{2}}{\pi\hbar^{3}c^{3}}
	$	
	
	$\frac{E}{V}= \int_{0}^{\infty}\frac{\varepsilon^{3}d\varepsilon}{\pi\hbar^{3}c^{3}(1+e^{\beta(\varepsilon-\mu)})} = \frac{T^{4}}{\pi\hbar^{3}c^{3}}\int_{0}^{\infty}\frac{x^{3}dx}{1+e^{x-\mu\beta}}
	$
	
	$\frac{N}{V}= \int_{0}^{\infty}\frac{\varepsilon^{2}d\varepsilon}{\pi\hbar^{3}c^{3}(1+e^{\beta(\varepsilon-\mu)})} = \frac{T^{3}}{\pi\hbar^{3}c^{3}}\int_{0}^{\infty}\frac{x^{2}dx}{1+e^{x-\mu\beta}}
	$	
	
	When $T=0$:
	
	$\frac{E}{V}= \int_{0}^{\infty}\frac{\varepsilon^{3}d\varepsilon}{\pi\hbar^{3}c^{3}(1+e^{\beta(\varepsilon-\mu)})} \approx \int_{0}^{\mu}\frac{\varepsilon^{3}d\varepsilon}{\pi\hbar^{3}c^{3}} = \frac{\mu^{4}}{4\pi\hbar^{3}c^{3}} 
	$
	
	$\frac{N}{V}= \int_{0}^{\infty}\frac{\varepsilon^{2}d\varepsilon}{\pi\hbar^{3}c^{3}(1+e^{\beta(\varepsilon-\mu)})} \approx \int_{0}^{\mu}\frac{\varepsilon^{2}d\varepsilon}{\pi\hbar^{3}c^{3}} = \frac{\mu^{3}}{3\pi\hbar^{3}c^{3}} 
	$
	
	When $T>>\mu$ ($\beta\mu<<1$):
	
	$\frac{E}{V}=\frac{T^{4}}{\pi\hbar^{3}c^{3}}\int_{0}^{\infty}\frac{x^{3}dx}{1+e^{x-\mu\beta}} \approx \frac{T^{4}}{\pi\hbar^{3}c^{3}}\int_{0}^{\infty}\frac{x^{3}dx}{1+e^{x}} 
	$
	
	$\frac{N}{V}=\frac{T^{3}}{\pi\hbar^{3}c^{3}}\int_{0}^{\infty}\frac{x^{2}dx}{1+e^{x-\mu\beta}} \approx \frac{T^{3}}{\pi\hbar^{3}c^{3}}\int_{0}^{\infty}\frac{x^{2}dx}{1+e^{x}} 
	$
	
	$\int_{0}^{\infty}\frac{x^{n}dx}{1+e^{x}} = \int_{0}^{\infty}(\frac{-2x^{n}}{e^{2x}-1}+\frac{x^{n}}{e^{x}-1})dx=(1-\frac{1}{2^{n}})\int_{0}^{\infty}\frac{x^{n}dx}{e^{x}-1}
	$
	
	$=(1-\frac{1}{2^{n}})\Gamma(n+1)\zeta(n+1)=(1-\frac{1}{2^{n}})n!\zeta(n+1)
	$
	
	$
	\zeta(4)=\frac{\pi^{4}}{90}
	$
	
	$\frac{E}{V}=\frac{7\pi^{3}T^{4}}{120\hbar^{3}c^{3}} 
	$
	
	$\frac{N}{V}=\frac{\zeta(3)T^{3}}{\pi\hbar^{3}c^{3}}
	$
	
	
	\item[(b)]
	
	$\varepsilon^{2} = \hbar^{2}c^{2}k^{2}+m_{0}^2c^{4}
	$
	
	$N(\varepsilon) = 2\times\frac{4\pi}{3}(\frac{L}{2\pi{\hbar}c})^{3}(\varepsilon^{2}-m_{0}^{2}c^{4})^{\frac{3}{2}}= \frac{V}{3\pi\hbar^{3}c^{3}}(\varepsilon^{2}-m_{0}^{2}c^{4})^{\frac{3}{2}}$
	
	$
	g(\varepsilon) = \frac{V\varepsilon}{\pi\hbar^{3}c^{3}}(\varepsilon^{2}-m_{0}^{2}c^{4})^{\frac{1}{2}}
	$
	
	$
	n = \frac{1}{\pi\hbar^{3}c^{3}}\int_{m_{n}c^{2}}^{\infty}\frac{\varepsilon(\varepsilon^{2}-m_{n}^{2}c^{4})^{\frac{1}{2}}d\varepsilon}{1+e^{\beta(\varepsilon-\mu_{n})}}
	$
	
	$
	n_{0}-n = \frac{1}{\pi\hbar^{3}c^{3}}\int_{m_{e}c^{2}}^{\infty}\frac{\varepsilon(\varepsilon^{2}-m_{e}^{2}c^{4})^{\frac{1}{2}}d\varepsilon}{1+e^{\beta(\varepsilon-\mu_{e^-})}}
	$
	
	$
	n_{0}-n = \frac{1}{\pi\hbar^{3}c^{3}}\int_{m_{p}c^{2}}^{\infty}\frac{\varepsilon(\varepsilon^{2}-m_{p}^{2}c^{4})^{\frac{1}{2}}d\varepsilon}{1+e^{\beta(\varepsilon-\mu_{p})}}
	$
	
	
	$
	n_{0}-n = \frac{1}{\pi\hbar^{3}c^{3}}\int_{0}^{\infty}\frac{\varepsilon^{2}d\varepsilon}{1+e^{\beta(\varepsilon-\mu_{\nu})}}
	$
	
	$ \mu_{n}(n_0 - n) =  \mu_{p} (n) +  \mu_{e^-} (n) +  \mu_{\bar{\nu}} (n) $
\end{itemize}

The beta decay reaction is: 
$$ n \Leftrightarrow e^{-} + p + \bar{\nu} $$

This means that at equilibrium, the following relation between the chemical\\
potentials of the different gases must hold : 

$$ \mu_{n}(n_0 - n) =  \mu_{p} (n) +  \mu_{e^-} (n) +  \mu_{\bar{\nu}} (n) $$

where n is the number of beta decays.

Since all of the involved particles are spin $1/2$ fermions, therefore obey the same 
particle statistics and dispersion relations (besides the anti-neutrinos in most cases), we only need to evaluate one integral at a time of the form : 

$$ N = \int _{0}^{\infty} {f(\epsilon-\mu) g(\epsilon) d\epsilon} $$  

\Dn

$$ g(\vec{p})=\frac{V}{\pi^2 \hbar^3} p^2 $$
$$ f(\epsilon-\mu) = \left(e^{\beta(\epsilon-\mu)}+1\right)^{-1} $$ \\

{\bf (b1) Non - relativistic :} $ E = p^2/2m $ 
$$ g(\epsilon)d\epsilon =\frac{\sqrt{2m^3}\;V}{\pi^2 \hbar^3} \sqrt{\epsilon} $$

{\bf (b2) Hyper - relativistic :} $ E = pc $ 

$$ g(\epsilon)d\epsilon = \frac{V}{\pi^2 (\hbar c)^3} \epsilon^2 $$

{\bf (b3) Zero temperature : }$\beta \to \infty $ 
\newline
\newline
$ f(\epsilon-\mu) = \theta(\mu-\epsilon) $
\newline
\newline
For the non-relativistic fermions :
\newline
\newline
$ A = \displaystyle { \frac{\sqrt{2}}{\pi^2 \hbar^3} } $

$ \displaystyle{\frac{N}{V}} =  n = A m^{\frac{3}{2}} \int_{0}^{\mu}{\sqrt{\epsilon} d\epsilon} = \frac{2}{3}A \;(m \mu)^{\frac{3}{2}} $

$ \displaystyle { \mu = \frac{1}{m} \left ( \frac{3n}{2A} \right) ^{2/3} } $
\newline
\newline
For the hyper-relativistic fermions :  \\

$ B = \displaystyle{ \frac{1}{\pi^2 (\hbar c)^3} } $

$ \displaystyle{\frac{N}{V}} =  n = B \int_{0}^{\mu}{\epsilon ^ 2 d\epsilon} = \frac{B}{3} \; \mu^3 $

$\mu = \displaystyle{\left(\frac{3n}{B} \right)^{\frac{1}{3}}} $ \\

In the general case : 
\newline
\newline


$ \displaystyle{E = c\sqrt{p^2+m^2 c^2} } $ 

$\displaystyle{ p = \frac{1}{c} \sqrt{E^2 - m^2c^4}} $ 

$ \displaystyle{\frac{dp}{dE} = \frac{1}{c} \frac{E}{\sqrt{E^2 - m^2 c^4}} = \frac{E}{pc^2}} $ \\

$ g(\epsilon) d\epsilon = \displaystyle {BV \epsilon \sqrt{\epsilon^2 - m^2 c^4}} \; d\epsilon $ 
\newline
\newline
$ n = \displaystyle {\int_{mc^2} ^{\mu + mc^2} { B\epsilon \sqrt{\epsilon^2 - m^2 c^4} \; d\epsilon} } = \frac{B}{3} \left(\epsilon^2 - m^2 c^4 \right) ^ {3/2} |^{\mu+mc^2}_{mc^2} = \frac{B}{3} \left(\mu( \mu+ 2m c^2 )\right) ^ {3/2} $ \\

Solving for $\mu$ : 

$ \mu(n) = \sqrt{m^2c^4 + \left(\frac{3n}{B} \right) ^ {2/3}} - mc^2 $
\\

If all the particles are non-relativistic (except the anti-neutrino), then: \\
$$ \left(\frac{1}{m_p} + \frac{1}{m_e} \right) \left ( \frac{3n}{2A} \right) ^{2/3} + 
\displaystyle{\left(\frac{3n}{B} \right)^{\frac{1}{3}}} = \frac{1}{m_n} \left ( \frac{3(n_0-n)}{2A} \right) ^{2/3} $$ \\



If all the particles are hyper-relativistic, then: \\

$ 3\displaystyle{\left(\frac{3n}{B} \right)^{\frac{1}{3}}} = \displaystyle{\left(\frac{3(n_0-n)}{B} \right)^{\frac{1}{3}}}$ \\ 

Solving for n, we get : 
$$ n = \frac{n_0}{28} $$ 
\newline
{\bf (b4) High temperature : }
\newline


$ \displaystyle{f(\epsilon-\mu) \approx e^{-\beta(\epsilon-\mu)}} $
\newline 
$ n = \displaystyle{ B\int_{mc^2} ^ {\infty} {e^{-\beta(\epsilon-\mu)} \epsilon \sqrt{\epsilon^2 - m^2 c^4} \; d\epsilon }} $ \\

change of variables : $ y = \sqrt{\epsilon^2 - m^2c^4} \quad ; \quad d\epsilon = \displaystyle{ \frac{y}{\epsilon}} dy $ \\

$ n = \displaystyle{\frac{B}{3} e^{\beta \mu} \int_{0}^{\infty} { e^{-\beta \sqrt{y^2 + m^2c^4} }y^2 dy} } $ \\

non-relativistic limit : $ \sqrt{m^2c^4 + y^2} \approx mc^2 (1 + \frac{y^2}{2mc^2}) $ \\

$ n = \displaystyle{\frac{B}{3} e^{\beta (\mu - mc^2)} \int_{0}^{\infty} { e^{-\displaystyle{\frac{\beta}{2mc^2} }y^2} y^2 dy} } $ \\

$ n = \displaystyle{\frac{B}{3} \left(\frac{2mc^2}{\beta} \right)^{3/2} e^{\beta (\mu - mc^2)} \int_{0}^{\infty} { e^{-z^2} z^2} dz} =  \displaystyle{\frac{B \sqrt{\pi}}{12} \left(\frac{2mc^2}{\beta} \right)^{3/2} e^{\beta (\mu - mc^2)}} $   \\

$ \mu(n) = mc^2 + k_{B}T \; \left[ \frac{3}{2} ln\left( \frac{\pi \hbar^2 }{m k_B T}\right)  + ln \left(3n \right) \right ]$ \\

Hyper-relativistic limit : $ \sqrt{m^2c^4 + y^2} \approx |y| $ \\

$ n = \displaystyle {\frac{B}{3} e^{\beta \mu} \int_{0}^{\infty} {e^{-\beta y} y^2 dy} } = 
\displaystyle {\frac{2B}{3\beta^3} e^{\beta \mu}} $ \\

$ \mu(n) = \displaystyle{ k_BT ln\left(\frac{3n}{2B(k_BT)^3}\right) }$ \\

The chemical potential equation is therefore: \\

$\displaystyle{3 k_BT ln\left(\frac{3n}{2B(k_BT)^3}\right) } = \displaystyle{ k_BT ln\left(\frac{3(n_0-n)}{2B(k_BT)^3}\right) } $ 

$$ n^3 = \left( \frac{2B}{3} \right ) ^2 \frac{n_0 - n}{\beta^6} $$

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\Dn



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