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\heading{E3570: Fermi gas in 2D+3D connected boxes with gravitation}
\auname{Avry Shirakov \& Rotem Kupfer}

\textbf{The problem:}

\Dn

Consider a mesoscopic box that has dimensions ${L\times L\times\ell}$,
such that ${\ell\ll L}$. In the box there are ${N}$ spin $1/2$
electrons. The mass of an electron is $m$. In items (a-d) assume
that the temperature is ${T=0}$. In items (1-4) the box is attached
to a tank that has dimensions ${L\times L\times L}$, that is placed
at height ${D}$ relative to the box, and you have to take into account
the gravitational field $g$. Express your answers using ${m,L,\ell,g,T}$.

\Dn

(1) Describe the single particle density of states. Specify the energy
range over which it is the same as for a two dimensional box.

(2) Find the fermi energy ${E_{F}}$ assuming that it is in the range
defined above. What is the maximum number ${N_{max}}$ of electrons
that can be accommodated without violating the $2D$ description?

(3) Assuming ${N<N_{max}}$ find the pressure ${P}$ on the side walls
of the box, and the force ${F}$ on the horizontal walls.

(4) Assume ${N=N_{max}}$. What is the minimum height ${D_{min}}$
to place the tank such that all of the electrons stay in the box?

(5) Assume ${N=N_{max}}$ and ${D>D_{min}}$. The temperature of the
system is raised a little bit. As a result some of the particles that
were in the $2D$ box are transferred to the tank. Estimate their
number ${N'}$. You are allowed to use any reasonable approximation.

\Dn\Dn

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\textbf{The solution:}

\Dn

(1) $3D$ density of states for spin $\frac{1}{2}$ fermions is:

\begin{equation}
g_{3D}(\varepsilon)=2L^{2}l\cdot\frac{(2m)^{\frac{3}{2}}}{(2\pi)^{2}}\varepsilon^{\frac{1}{2}}
\end{equation}


In $2D$ however, density of states is independent of energy:

\begin{equation}
g_{2D}(\varepsilon)=\frac{mL^{2}}{\pi}
\end{equation}


The condition on the energy for $2D$ approximation of $3D$ box is
that it is lower then the first excited state of the ``short'' axis:

\begin{equation}
\frac{\pi^{2}}{mL^{2}}+\frac{\pi^{2}}{2ml^{2}}\leq\varepsilon\leq\frac{4\pi^{2}}{2ml^{2}}+\frac{\pi^{2}}{mL^{2}}
\end{equation}


\Dn

(2) The maximal number of particles in $2D$ box is given by:

\begin{equation}
N=\int_{0}^{\infty}g(\varepsilon)f(\varepsilon-\mu)d\varepsilon=\int_{0}^{\infty}\frac{mL^{2}}{\pi}\cdot\frac{1}{\frac{1}{z}e^{\beta\varepsilon}+1}d\varepsilon=\frac{L^{2}m}{\pi}kT\ln(1+e^{\beta\mu})
\end{equation}


For $T\rightarrow0$:

\begin{equation}
\lim_{T\rightarrow0}kT\ln(1+e^{\beta\mu})=\mu
\end{equation}


So the number of particles at temprature $T=0$ is:

\begin{equation}
N=\frac{L^{2}m\mu}{\pi}
\end{equation}


Also at $T=0$ the chemical potential is defined as: 
\begin{equation}
\varepsilon_{F}=\mu
\end{equation}


So:

\begin{equation}
\varepsilon_{F}=\frac{\pi N}{mL^{2}}
\end{equation}


and the maximal number of particles under this approximation is:

\begin{equation}
N_{max}=\frac{L^{2}m}{\pi}(\frac{4\pi^{2}}{2ml^{2}}+\frac{\pi^{2}}{mL^{2}})\approx2\pi\frac{L^{2}}{l^{2}}
\end{equation}


\Dn

(3) First we notice that at $T=0$, $F=E$.

The single particle energy function: 
\begin{equation}
\varepsilon_{k_{x},k_{x},n_{z}}=k_{x}^{2}+k_{y}^{2}+\frac{1}{2m}\left(\frac{\pi}{l}n_{z}\right)^{2}
\end{equation}


We notice that the system is at the ground state for it's ``short''
axis (it is only dependent on $l$) with $n_{z}=1$. The force on
the horizontal walls is: 
\begin{equation}
F=-\frac{\partial E}{\partial l}=-N\frac{\partial(\frac{\pi^{2}}{2ml^{2}})}{\partial l}=\frac{N\pi^{2}}{ml^{3}}
\end{equation}


The energy is given by:

\begin{equation}
E=\int_{0}^{\infty}g_{2D}(\varepsilon)\cdot\varepsilon\cdot f(\varepsilon-\mu)d\varepsilon=\int_{0}^{\infty}\frac{mL^{2}}{\pi}\cdot\frac{\varepsilon}{\frac{1}{z}e^{\beta\varepsilon}+1}d\varepsilon
\end{equation}


By taking $T=0$ and $\mu=\varepsilon_{F}$ we get:

\begin{equation}
E=\frac{\pi}{2m}\frac{N^{2}}{L^{2}}
\end{equation}


and the force on the side walls is:

\begin{equation}
F=-\frac{\partial E}{\partial L}=-\frac{\partial(\frac{\pi}{2m}\frac{N^{2}}{L^{2}})}{\partial L}=\frac{\pi N^{2}}{mL^{3}}
\end{equation}


The pressure is given by:

\begin{equation}
P=\frac{F}{A}=\frac{\pi N^{2}}{AmL^{3}}=\frac{\pi N^{2}}{mL^{4}l}
\end{equation}


\Dn

(4) The condition for no transition between the box and the tank is
that the chemical potential of the unoccupied tank is higher then
the chemical potential of the box:

\begin{equation}
\mu_{box}=\varepsilon_{F}=\frac{\pi N_{max}}{mL^{2}}
\end{equation}


\begin{equation}
\mu_{tank}(N'=0)=mgd
\end{equation}


\begin{equation}
\mu_{box}<\mu_{tank}(N'=0)
\end{equation}


\begin{equation}
d_{min}>\frac{\pi N_{max}}{m^{2}L^{2}g}
\end{equation}


\Dn

(5) Now we assume $T>0$, $d>d_{min}$. The number of particles in
the box is given by:

\begin{equation}
N=\frac{L^{2}m}{\pi}kT\ln(1+e^{\beta\mu})
\end{equation}


\begin{equation}
\mu=kT\ln(e^{\frac{\pi N}{L^{2}mkT}}-1)=kT\ln(e^{\frac{\varepsilon_{f}}{kT}}-1)
\end{equation}


\begin{equation}
\mu=kT\ln\left(e^{\frac{\varepsilon_{f}}{kT}}\left(1-e^{-\frac{\varepsilon_{f}}{kT}}\right)\right)=kT\left(\frac{\varepsilon_{f}}{kT}\right)+kT\ln\left(1-e^{-\frac{\varepsilon_{f}}{kT}}\right)
\end{equation}


By using Taylor expansion near $T=0$, (Note:$e^{-\frac{\varepsilon_{f}}{kT}}\rightarrow0$
; taylor expansion for $\ln\left(1-x\right)\sim\ln\left(\frac{1}{1+x}\right)\sim-\ln\left(1+x\right)\sim-x$):

\begin{equation}
\mu_{box}=\varepsilon_{f}-kTe^{-\frac{\varepsilon_{f}}{kT}}
\end{equation}


We assume low occupation in the the tank so Boltzmann approximation
is valid (and also $N-N'\approx N$)

\begin{equation}
\mu_{tank}=mgd+kT\ln(\frac{N'}{V}\lambda_{T}^{3})
\end{equation}


We equate the two potential:

\begin{equation}
mgd+kT\ln(\frac{N'}{V}\lambda_{T}^{3})=\varepsilon_{F}-kTe^{-\frac{\varepsilon_{f}}{kT}}
\end{equation}


In order to get an analytical solution for $N'$ we must neglect the
exponential term in the chemical potential of the box: 
\begin{equation}
N'=\frac{V}{\lambda_{T}^{3}}e^{\frac{\varepsilon_{F}-mgd}{kT}}
\end{equation}


One can verify that for $T\rightarrow0$, $d=d_{min}$ we get $N'=0$.

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\end{document}
