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\exnumber{2046}
\heading{Gas in a centrifuge} 
\auname{Isaac Weinberg}


{\bf The problem:}

\Dn

A cylinder of of radius ${R}$ rotates about its axis
with a constant angular velocity ${\Omega}$.  
It contains an ideal classical gas of $N$ particles 
at temperature ${T}$.
\\
Note that the Hamiltonian in the rotating frame 
is ${H'\left(r,p; \Omega\right) = H\left(r,p\right) - \Omega L\left(r,p\right)}$ 
where ${L\left(r,p\right)}$ is the angular momentum. 
\\
It is conceptually useful to realize that formally the Hamiltonian 
is the same as that of a charged particle in a magnetic field ("Coriolis force") 
plus centrifugal potential ${V\left(r\right)}$.  
Explain how this formal equivalence can be used in order 
to make a shortcut in the above calculation.
\Dn

(1) Find the density distribution as a function of the radial distance from the axis. 

(2) Write what is the pressure on the walls.


\Dn\Dn


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

\Dn

First, we would like to rearrange the Hamiltonian in order to simplify the partition function calculation:
\begin{equation}
H=\frac{p^2}{2\bmsf{m}} -\Omega \cdot L=\frac{p^2}{2\bmsf{m}} -\Omega \cdot r\times p=\frac{p^2}{2\bmsf{m}} - p \cdot \Omega \times r
=\frac{\left(p-\bmsf{m}\Omega \times r\right)^2}{2\bmsf{m}}-\frac{1}{2}\bmsf{m}\left(\Omega \times r\right)^2
\end{equation}
One can see that the term $\bmsf{m}\Omega \times r \rightleftharpoons eA(r)$ and that $V(r)=-\frac{1}{2}\bmsf{m}\left(\Omega \times r\right)^2$ which corresponds to a charged particle in electromagnetic field:
\[
H=\frac{1}{2\bmsf{m}}(p-eA(r))^2 + V(r)
\]


%%%%%%%%%%%%%%%
\Dn

(1)
In order to find the density distribution we will find the partition function of a very thin $\mathrm{d}r$ layer of gas distant $r$ from the center of the cylinder:


\begin{equation}
Z_1(r) = \int{\frac{\mathrm{d}r \mathrm{d}p}{(2\pi)^3}\eexp{-\beta\mathcal{H}}}
= \int{\frac{\mathrm{d}r \mathrm{d}p}{(2\pi)^3}\eexp{-\beta\left[\frac{1}{2\bmsf{m}}(p-\bmsf{m}\Omega \times r)^2-\frac{1}{2}\bmsf{m}\left(\Omega \times r\right)^2\right]}}
\end{equation}


We will change the variable $p'= p-\bmsf{m}\Omega \times r$ and since we are working in a cylindrical coordinates the integral of equation (2) will become:


\begin{equation}
Z_1(r) = \int{\frac{\mathrm{d}p'}{(2\pi)^3}\eexp{-\beta\frac{1}{2\bmsf{m}}p'^2}}
\int\limits_{z=0}^{z=L}\mathrm{d}z\int\limits_{\theta=0}^{\theta=2\pi}\mathrm{d}\theta\, r \mathrm{d}r\,\eexp{\frac{1}{2}\beta\bmsf{m}(\Omega r)^2}
=\left(\frac{1}{\lambda_T}\right)^3V\eexp{\frac{1}{2}\beta\bmsf{m}(\Omega r)^2}
\end{equation}

\begin{equation}
Z_N(r)=\frac{Z_1^N}{N!} \quad\mbox{N equal the number of atoms in the thin layer}
\end{equation}


Because each thin layer is in contact with the layer around it after equilibrium is archived all the layers have the same chemical potential which we'll calculate next:


\begin{equation}
\mu=\frac{\partial F}{\partial N}
=\frac{1}{\beta}(-\ln(Z_1(r))+\ln(N(r)))
=\frac{1}{\beta}\ln \left(\lambda_T^3\frac{N(r)}{V} \eexp{-\frac{1}{2}\beta\bmsf{m}(\Omega r)^2}  \right)
=\frac{1}{\beta}\ln(\left(\lambda_T^3n(r)\right) - \frac{1}{2}\bmsf{m}(\Omega r)^2
\end{equation}


If we equate $\mu(r)=\mu(0)$ we can find the distribution of the density:


\begin{eqnarray}
\mu(r)&=&\mu(0)\nonumber\\
\ln \left(\lambda_T^3n(r) \eexp{-\frac{1}{2}\beta\bmsf{m}(\Omega r)^2}  \right)&=&\ln \left(\lambda_T^3n(0) \eexp{-\frac{1}{2}\beta\bmsf{m}(\Omega\cdot 0)^2}\right)\nonumber\\
n(r) & = & n(0)\eexp{\frac{1}{2}\beta\bmsf{m}(\Omega r)^2}
\end{eqnarray}

%%%%%%%%%%%%%%
\Dn

(2)
To find the pressure on the walls of the cylinder we will calculate the force on the walls from the partition function of the entire cylinder, not just a thin layer as we did in the previous section.


\begin{equation}
Z_1 = \int{\frac{\mathrm{d}p'}{(2\pi)^3}\eexp{-\beta\frac{1}{2\bmsf{m}}p'^2}}
\int\limits_{z=0}^{z=L}\mathrm{d}z\int\limits_{\theta=0}^{\theta=2\pi}\mathrm{d}\theta\int\limits_{r=0}^{r=R}\mathrm{d}r\, r\eexp{\frac{1}{2}\beta\bmsf{m}(\Omega r)^2}
=\left(\frac{1}{\lambda_T}\right)^3\frac{2\pi L}{\bmsf{m}\Omega^2\beta}\left(\eexp{\frac{1}{2}\beta\bmsf{m}(\Omega R)^2}-1\right)
\end{equation}

\begin{equation}
Z_N(r)=\frac{Z_1^N}{N!} \quad\mbox{now N equal the number of atoms in the cylinder}
\end{equation}

\begin{equation}
\langle F\rangle_R=\frac{1}{\beta}\frac{\partial\ln(Z_N)}{\partial R}
=N\bmsf{m}\Omega^2R\frac{1}{1-\eexp{-\frac{1}{2}\beta\bmsf{m}(\Omega R)^2}}
\end{equation}

\begin{equation}
P = \frac{F}{A}=\frac{N\bmsf{m}\Omega^2R}{2\pi RL}\frac{1}{1-\eexp{-\frac{1}{2}\beta\bmsf{m}(\Omega R)^2}}
\end{equation}


We will replace $N$ with $nV=n\pi R^2L$ and get:


\begin{equation}
P =\frac{n\pi R^2L\bmsf{m}\Omega^2R}{2\pi RL}\frac{1}{1-\eexp{-\frac{1}{2}\beta\bmsf{m}(\Omega R)^2}}
=n\frac{\bmsf{m}R^2}{2}\Omega^2\frac{1}{1-\eexp{-\frac{1}{2}\beta\bmsf{m}(\Omega R)^2}}
\end{equation}


If $\Omega \rightarrow 0$ we can expand the exponent $\mathrm{e}^x=1+x...$ and get:


\begin{equation}
P \stackrel{\Omega \rightarrow 0}{\approx}	 n\frac{\bmsf{m}R^2}{2}\Omega^2\frac{1}{1-\left(1-\frac{1}{2}\beta\bmsf{m}(\Omega R)^2\right)}
=\frac{n}{\beta}
\end{equation}


Which corresponds to the equation of state of an ideal gas.\\
On the Other hand if $\Omega \rightarrow \infty$ the force on the walls equals the Centripetal force exerted by the wall on all the atoms:


\begin{equation}
\langle F\rangle_R\stackrel{\Omega \rightarrow \infty}{\approx}N\bmsf{m}\Omega^2R
\end{equation}




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