\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}

%extra packages Barakaz
\usepackage{mathtools}
\usepackage{amsmath}


%New symbols Barakaz
\newcommand\underrel[3][]{\mathrel{\mathop{#3}\limits_{%
      \ifx c#1\relax\mathclap{#2}\else#2\fi}}}

% 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}}

% 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{3711}
\heading{Fermions in gravitation field of a star} 
\auname{Barak Azoulai}


{\bf The problem:}

\Dn

Consider a neutron star as non-relativistic gas of non-interacting neutrons of mass ${m}$ in a spherical symmetric equilibrium configuration. The neutrons are held together by a gravitational potential $-mMG/r$ of a heavy object of mass ${M}$ and radius ${r_0}$ at the center of the star (${G}$ is the gravity constant and ${r}$ is the distance from the center).

\Dn

(a) Give an expression for $n(r)$ at $T > 0$ using Li functions.\\

(b) Consider the neutrons as fermions at $T = 0$ and find $n(r)$, for a given $n(r_0)$.\\

(c) Calculate it explicitly in the Boltzmann approximation.\\

(d) Repeat items (b) and (c) for a general potential $-A/r^\alpha$. \\

(e) For the case of T=0, what is the upper bound on $n(r_0)$ and on the total number $N$ of neutrons if the chemical potential is increased towards zero. Distinguish,  $a>2$ from $a<2$.

\Dn\Dn


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

This problem is about neutrons (which are fermions) in a gravitational field, one can define a small volume $dV$, take advantage of the fact that the neutrons are in an equilibrium configuration and demand that the chemical potential $\mu$ is the same for each volume $dV$.\\
\\
(a) The density of particles is defined such that the number of particles in a volume $dV$ is $N(r) = n(r)dV$, which is calculated by $N(r) = \int g(\epsilon)f(\epsilon-\mu ) d\epsilon$.
The energies of the one particle Hamiltonian at a specific radius $r$ in a box of volume $dV$ are: 
\begin{equation}
\epsilon(r) = \frac{p^2}{2m} - U(r).
\end{equation}
With $U(r) = \frac{mMG}{r}$.
The density of states in a volume $dV$ is:
%, where $g(\epsilon)d\epsilon$ is the number of energies between $[\epsilon,\epsilon+d\epsilon]$,
\begin{equation}
g(\epsilon) = 2\cdot 4\pi m^{3/2}\left[2\left(\epsilon +U(r)\right)\right]^{1/2} \cdot \frac{dV}{(2\pi)^3} = \frac{(2m)^{3/2}}{2\pi^2}\left(\epsilon +U(r)\right)^{1/2}dV.
\end{equation}
For fermions the particle distribution $f(\epsilon - \mu) = \left(e^{\beta (\epsilon - \mu)}+1\right)^{-1}$, and the general solution for the density of the particles is given by:
\begin{equation}
n(r) = \frac{(2m)^{3/2}}{2\pi^2} \int\limits_{-U(r)}^{\infty} \frac{\left(\epsilon +U(r)\right)^{1/2}}{e^{\beta(\epsilon-\mu)}+1}d \epsilon =  
\frac{(2mT)^{3/2}}{2\pi^2} \int\limits_{0}^{\infty} \frac{x^{\alpha-1}dx}{e^{x-u}+1} =  -\frac{(2mT)^{3/2}}{2\pi^2}\Gamma(3/2)\text{Li}_{3/2}(-e^{u})
\end{equation}
With $x = \beta\left(\epsilon+U(r)\right)$, $\alpha = 3/2$, $u = \beta\left(\mu +U(r)\right)$ and using $ \int\limits_{0}^{\infty} \frac{x^{\alpha-1}dx}{e^{x-u}+1} = -\Gamma(\alpha)\text{Li}_\alpha(-e^u)$\\
\\
(b) At the limit of $T=0$, one can take Eq. (3) and use the expansion $\Gamma(\alpha)\text{Li}_\alpha(-e^u) \approx - \frac{1}{\alpha}u^\alpha $, or alternatively calculate $n(r)$ explicitly and noting that $f(\epsilon-\mu)$ becomes a step function $\Theta(\mu-\epsilon)$.
%\begin{equation}
%n(r) =  \frac{(2mT)^{3/2}}{2\pi^2} \frac{2}{3}u^{3/2} = \frac{(2mT)^{3/2}}{3\pi^2}\left[\beta\left(\mu +U(r)\right)\right]^{3/2} = \frac{(2m)^{3/2}}{3\pi^2}\left(\mu +U(r)\right)^{3/2}
%\end{equation}
%Here we took only the first term, and used $u = \beta\left(\mu +U(r)\right)$, $\alpha = 3/2$\\
%This result can be obtained by explicit calculations. For $T=0$, $f(\epsilon-\mu)$ becomes a step function $\Theta(\mu-\epsilon)$,
\begin{align}
%\begin{split}
n(r) =\frac{(2m)^{3/2}}{2\pi^2} \int\limits_{-U(r)}^{\infty} \Theta(\mu-\epsilon) \left(\epsilon +U(r)\right)^{1/2}d\epsilon = \frac{(2m)^{3/2}}{3\pi^2} \left(\mu +U(r)\right)^{3/2}
%\end{split}
\end{align}
To obtain $\mu$ we set $r=r_0$ in Eq. (4) and express it with $n(r_0)$, one obtains:
\begin{equation}
\mu = \frac{\left(3\pi^2n(r_0)\right)^{2/3}}{2m} - U(r_0)
\end{equation}
For $T=0$ one gets the density of particles as:
\begin{equation}
n(r) =\frac{(2m)^{3/2}}{3\pi^2}  \left[\frac{\left(3\pi^2n(r_0)\right)^{2/3}}{2m} +U(r)- U(r_0)\right]^{3/2}
\end{equation}
(c) At the Boltzmann approximation, one can take the limit of Li function $\text{Li}(-e^u) \approx -e^u$ and get:
\begin{align}
n(r) &=  -\frac{(2mT)^{3/2}}{2\pi^2}\Gamma(3/2)\text{Li}_{3/2}(-e^{u}) \\&\approx \frac{(2mT)^{3/2}}{2\pi^2}\frac{\sqrt{\pi}}{2}\exp{\left[\beta\left(\mu+U(r)\right)\right]} = \frac{2}{\lambda_T^{3}}\exp{\left[\beta\left(\mu+U(r)\right)\right]}
\end{align}
Where we used $(\lambda_T)^2 = 2\pi/mT$, and $\Gamma(3/2) = \sqrt{\pi}/2$.
The same result can be obtained by performing the integral for $n(r)$ explicitly and noting that $f(\epsilon-\mu) = \left(e^{\beta (\epsilon - \mu)}+1\right)^{-1} \approx e^{-\beta (\epsilon - \mu)}$.
%\begin{equation}
%n(r) = \frac{(2m)^{3/2}}{2\pi^2} \int e^{-\beta (\epsilon - \mu)}\left(\epsilon +U(r)\right)^{1/2} \cdot d \epsilon = \frac{2}{\lambda_T^{3}}\exp{\left[\beta\left(\mu+U(r)\right)\right]}
%\end{equation}
Expressing $\mu$ using $n(r_0)$ will give us:
\begin{equation}
\mu = T\ln\left(\frac{1}{2}n(r_0)\lambda_T^3\right) - U(r_0)
\end{equation}
At the Boltzmann approximation one gets the density of particles as:
\begin{equation}
n(r) = n(r_0)\exp{\left[\beta \left(U(r) - U(r_0)\right)\right]}
\end{equation}
Noting that this is the expected result because in canonical equlibrium $p(r) \propto e^{-\beta(\epsilon)}$.
\\
\\
(d) Repeating items (b) and (c) will give the same results but changing $r$ to $r^\alpha$, so we get the same Eq. (6) \& (10) but with
\begin{equation}
U(r) = \frac{A}{r^\alpha}
\end{equation}
\\
(e) Increasing $\mu$ toward zero is done by increasing $n(r_0)$ as one can see from Eq. (5). At the limit of $\mu\rightarrow 0$ one obtains:
\begin{equation}
n(r)_{_{T=0}} \ \underrel[c]{\mu\to 0}{=\ } \frac{(2mA)^{3/2}}{3\pi^2} r^{-3\alpha/2}
\end{equation}
The total number of neutrons (for $\alpha \ne 2$):
\begin{align}
N_{\text{total}} &= \int\limits_{r_0}^{\infty} n(r)4\pi r^2dr = \frac{2^{7/2}(mA)^{3/2}}{3\pi} \int\limits_{r_0}^{\infty} r^{2-3\alpha/2}dr \\&= \left.\frac{2^{7/2}(mA)^{3/2}}{9\pi}\left(1-\frac{\alpha}{2}\right)^{-1}r^{3(1-\alpha/2)} \right|_{r_0}^\infty
\end{align}
The integral diverges for $\alpha \leq 2$ and converges for $\alpha > 2$ to:
\begin{equation}
N_{\text{total}} = \frac{2^{7/2}(mA)^{3/2}}{9\pi}\left(1-\frac{\alpha}{2}\right)^{-1}r_0^{3(1-\alpha/2)}
\end{equation}
For $\mu = 0$ the upper bound for the density $n(r_0)$ is given by
%$$\mu = [\frac{3\pi^2n(r_0)}{(2m)^{\frac{3}{2}}}]^{\frac{2}{3}} - \frac{mMG}{r_0}$$
\begin{equation}
n(r_0)_{_{T=0,\mu=0}} =  \left(\frac{2mA}{r_0^\alpha}\right)^{3/2}\frac{1}{3\pi^2}
\end{equation}


%let's take the first order term using $(1+x)^\alpha \approx 1+\alpha x $
%\begin{equation}
%\begin{split}
%n(r)_{_{T=0}} = \frac{(2m)^{\frac{3}{2}}}{3\pi^2} [(\frac{mMG}{r^\alpha})^{\frac{3}{2}}(\frac{r^\alpha\mu}{mMG} +1)^{\frac{3}{2}} - (\frac{mMG}{r^\alpha})^{\frac{3}{2}}] \approx  \\ \approx \frac{(2m)^{\frac{3}{2}}}{3\pi^2}[(\frac{mMG}{r^\alpha})^{\frac{3}{2}}\frac{3}{2}\frac{r^\alpha\mu}{mMG}]  =\frac{\sqrt{2MG}m^2}{\pi^2}\mu r^{-\frac{\alpha}{2}}
%\end{split}
%\end{equation}
%For the total number of neutrons we have
%\begin{equation}
%N_{Total} = \int_{r_0}^{\infty} n(r)4\pi r^2dr = \frac{4\sqrt{2MG}m^2}{\pi}\mu \int_{r_0}^{\infty} r^{2-\frac{\alpha}{2}}dr
%\end{equation}
%Where $\mu$ as before is a function of $n(r_0)$:
%\begin{equation}
%(\mu + \frac{mMG}{r_0})^{\frac{3}{2}} = \frac{3\pi^2n(r_0)}{(2m)^{\frac{3}{2}}}+(\frac{mMG}{r_0})^{\frac{3}{2}}
%\end{equation}



\Dn

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