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\heading{E3745: Fermions in a gravitational field} 

\auname{Rotem Tzuk}

\textbf{The problem:}

\Dn

Consider fermions of mass $m$ and spin 1/2 in a gravitational field
with constant acceleration $g$ and at uniform temperature $T$. 

\Dn

(1) Assume first that the fermions behave as classical particles and
find their density $n(h)$ as function of the height $h\geq0$ and
the density $n(0)=n_{0}$.$\Dn$

(2) Near the bottom $h=0$ the fermions are degenerate, i.e. their
Fermi energy $\epsilon_{F}^{0}\equiv\epsilon_{F}(h=0)$ is $\epsilon_{F}^{0}\geqq T$.
Assume $T=0$, Find the local Fermi momentum $p_{F}(h)$ and $n(h)$
given $n_{0}$.$\Dn$

(3) Consider now $T\neq0$ (but still $\epsilon_{F}^{0}\geqq T$)
and estimate the height $h_{c}$ such that for $h\gg h_{c}$ the fermions
are non-degenerate. Find $n(h)$ for $h\gg h_{c}$ in terms of $n_{0}$.$\Dn$

\Dn\Dn

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\textbf{The solution:}

\Dn

(1) The Partition function for a layer of a uniform density positioned
at a height $[h,h+\delta h$ is just the regular classical gas partition
function multiplied by an exponential factor containing the dependence
in the height of the element
\begin{equation}
Z_{1}(\beta,h,\delta h)=\intop_{0}^{L}\intop_{0}^{L}\:\intop_{h}^{h+\delta h}\intop_{-\infty}^{+\infty}2(e^{-(\frac{\beta p^{2}}{2m}+\beta mgz)})\frac{dxdydzd^{3}p}{(\beta mg)(2\pi)^{3}}=\frac{2L^{2}}{(2\pi\lambda_{T})^{3}}(e^{-\beta mgh}-e^{-\beta mg(h+\delta h)})
\end{equation}


\[
=\frac{2L^{2}}{(\beta mg)(2\pi\lambda_{T})^{3}}e^{-\beta mgh}(1-e^{-\beta mg\delta h})
\]


and since $\delta h$ is infinitesimally small

\begin{equation}
Z_{1}(\beta,h,\delta h)=\frac{2L^{2}}{(2\pi\lambda_{T})^{3}}e^{-\beta mgh}(\delta h)=\frac{2L^{2}\delta h}{(2\pi\lambda_{T})^{3}}e^{-\beta mgh}
\end{equation}
We could factorize each particle in order to get 
\begin{equation}
Z_{N}(\beta,h,\delta h)=\frac{1}{N(h)!}(Z_{1})^{N}
\end{equation}


Deriving the Free energy 
\begin{equation}
F=TN(h)(-1+ln\frac{(2\pi\lambda_{T})^{3}N(h)}{2L^{2}\delta h}+\beta mgh)
\end{equation}


and the chemical potential
\begin{equation}
\mu(\beta,h)=T(ln\frac{(2\pi\lambda_{T})^{3}n(h)}{2}+\beta mgh)
\end{equation}


By demand for chemical equilibrium we get $\mu(h)=\mu(0)$ hence 
\begin{equation}
T(ln\frac{(2\pi\lambda_{T})^{3}n(h)}{2})+mgh=T(ln\frac{(2\pi\lambda_{T})^{3}n(0)}{2})
\end{equation}


\begin{equation}
n(h)=n(0)e^{-\beta mgh}
\end{equation}


\Dn

(2) While $T=0$ since $\lambda_{T}\sim\frac{1}{\sqrt{mT}}\rightarrow\infty$
we get overlap between the different fermionic wave functions a.k.a
degeneration.

Counting the Number of states in the Fermi sphere 
\begin{equation}
N=2\times\frac{1}{8}\times\frac{4}{3}\pi n_{f}^{3}
\end{equation}


from $n_{f}=(\frac{3N}{\pi})^{1/3}$ we get the Fermi energy 
\begin{equation}
\epsilon_{F}=\frac{\pi^{2}}{2mL^{2}}n_{f}^{2}=\frac{1}{2m}(\frac{3\pi^{2}N}{V})^{2/3}
\end{equation}


The chemical potential just equals $\epsilon_{F}^{0}$ so we have
\begin{equation}
\epsilon_{F}^{0}-mgh=\frac{p_{F}(h)^{2}}{2m}=\frac{1}{2m}(3\pi^{2}n(h))^{2/3}
\end{equation}


from substituting $h=0$ we get $\epsilon_{F}^{0}=\frac{1}{2m}(3\pi^{2}n(0))^{2/3}$
and
\begin{equation}
\frac{1}{2m}(3\pi^{2}n(h))^{2/3}=\frac{1}{2m}(3\pi^{2}n(0))^{2/3}-mgh
\end{equation}


\begin{equation}
3\pi^{2}n(h)=(3\pi^{2}n(0))^{2/3}-2m^{2}gh)^{3/2}=(2m)^{3/2}(\epsilon_{F}^{0}-mgh)^{3/2}
\end{equation}


\Dn

(3) The degeneration would be maintained as long as $\epsilon_{F}\gg T$,
remembering 
\begin{equation}
\epsilon_{F}^{0}\equiv\epsilon_{F}(h=0)
\end{equation}


the critical height would provide 
\begin{equation}
\epsilon_{F}=\epsilon_{F}^{0}-mgh=T
\end{equation}


but since $\epsilon_{F}^{0}\gg T$ the critical height would be 
\begin{equation}
\epsilon_{F}^{0}\approx mgh_{c}
\end{equation}
\begin{equation}
h_{c}=\frac{\epsilon_{F}^{0}}{mg}
\end{equation}


we can get $\mu$ from the condition on $h=0$ (since we are still
at a very low temperature) 
\begin{equation}
\mu=\epsilon_{F}^{0}=\frac{1}{2m}(3\pi^{2}n(0))^{2/3}
\end{equation}


substituting into the expression for the chemical potential at the
classical region $h>h_{c}$
\begin{equation}
\frac{1}{2m}(3\pi^{2}n(0))^{2/3}=T(ln\frac{(2\pi\lambda_{T})^{3}n(h)}{2}+\beta mgh)
\end{equation}
\begin{equation}
n(h)=\frac{2}{(2\pi\lambda_{T})^{3}}e^{\frac{\beta}{2m}(3\pi^{2}n(0))^{2/3}-\beta mgh}
\end{equation}

\end{document}
