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\exnumber{3042}
\heading{Oscillations of a piston in a cylinder filled with gas} 
\auname{Lior Bodek} 

{\bf The problem:}

\Dn
Consider a vertically aligned cylinder whose basis has an area~$\mathsf{A}$. 
A piston that has mass~$M$ is pushed from above. 
The piston is held by a spring that has an elastic constant~$K$. 
If the cylinder is empty the piston is down at zero height (${x=0}$).
The cylinder is filled with~$N$ gas particles. 
Each particle has mass $\mass$ and the temperature is~$T$.
Consequently the the piston goes up a distance~$x$, 
such that the gas occupies a volume~$\mathsf{A}x$.
Consider the following 3~cases:
\Dn

\begin{enumerate}
\item[(a)] The temperature is high, such that Boltzmann approximation can be applied. 
\item[(b)] The particles are condensed Bosons, $T$ is lower than the condensation temperature.  
\item[(c)] The particle are spinless Fermions, and the temperature is zero. 
\end{enumerate}

Answer the following questions, relating to each case separately.

\begin{enumerate}
\item What is the equilibrium position $x_{eq}$ of the piston?
\item What is the frequency $\omega$ of small oscillations?
\item Plot schematic drawing of $\omega$ versus $T$.
\end{enumerate}

Express answers using $\mathsf{A}$, $M$, $K$, $N$, $T$.
The schematic drawing is required to be be clearly displayed.
\Dn
\Dn

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{\bf The solution:}
\Dn
\Dn

\underline {\bf Part 1:  {Finding $x_{eq}$ }}

\Dn
\Dn

 \underline { The Boltzman Approximation }

\[\tag{1}PV=NT  \Rightarrow P=\frac{NT}{V}=\frac{NT}{Ax}\]
\Dn

 Where $ P $ is the pressure $ V $ is the volume $ N $ is the number of particles and $ T $ is the temperature.
\Dn
\Dn

  In order to find the equilibrium position $x_{eq}$ of the piston , we will write the force:
\Dn

\[\tag{2} F=PA\]


  We also know that a spring force is:
\Dn

\[\tag{3} F=-~$K$~$x$\]

   appling them together, we recive:
\Dn

\[\tag{4} ~$K$ x_{eq}=\frac{NTA}{V} \Rightarrow x_{eq}=\frac{NTA}{KAx_{eq}} \Rightarrow x_{eq}^2=\frac{NTA}{KA} \Rightarrow x_{eq}^2=\frac{NT}{K}\]

Which leads to the result that:
\Dn

\[\tag{5}  x_{eq}=\sqrt{\frac{NT}{K}}\]

\underline { Condensed Bosons Case}
\Dn
\Dn

We know that the standard resolts for the pressure $ P $ is:

\[\tag{6}P=\zeta \left( \frac{5}{2} \right) \left( \frac{m}{2\pi} \right)^{\frac{3}{2}} T^{\frac{5}{2}} \]

 In order to find the equilibrium position $x_{eq}$ of the piston , we will use equations 2 and 3:

\[\tag{7} F=PA=~$K$x_{eq} \Rightarrow ~$K$x_{eq}=\zeta \left( \frac{5}{2} \right) \left( \frac{m}{2\pi} \right)^{\frac{3}{2}} T^{\frac{5}{2}}A \]

Hence we get

\[\tag{8} x_{eq}=\zeta \left( \frac{5}{2} \right) \left( \frac{m}{2\pi} \right)^{\frac{3}{2}} T^{\frac{5}{2}} \frac{A}{K} \]

\Dn
\Dn

\underline {Spinless Fermions, at zero temperature }
\Dn
\Dn

In this case we know that the standard resolts for the pressure $ P $ is:

\[\tag{9}P= \frac{(6\pi^2)^{\frac{2}{3}}}{5m}  \left( \frac{N}{V} \right)^{\frac{5}{3}} \]

since

\[\tag{10}V= Ax $ $ $   and $ $ $  F=PA\]

We get

\[\tag{11} F=PA=~$K$x_{eq} \Rightarrow ~$K$x_{eq}= \frac{(6\pi^2)^{\frac{2}{3}}}{5m}  \left( \frac{N}{Ax} \right)^{\frac{5}{3}}A  \]

And eventually we have

\[\tag{12} (x_{eq})^\frac{8}{3}= \left(\frac{6\pi^2}{A}\right)^\frac{2}{3}\frac{N^\frac{5}{3}}{5Km}  \]

Or

\[\tag{13} x_{eq}= \left(\frac{6\pi^2}{A}\right)^\frac{1}{4} \left(\frac{1}{5Km}\right)^\frac{3}{8} N^{\frac{5}{8}}\]

\Dn
\Dn


\underline {\bf Part 2 :Finding $ \omega $ for small oscillations}

\Dn
\Dn

 \underline { The Boltzman Approximation }

\Dn

The force in equalibrium is:

\[\tag{14} F(x_{eq})=PA-~$K$ x_{eq}   \Rightarrow  F(x_{eq})=\frac{NT}{x_{eq}}-~$K$ x_{eq}\]

For small oscilations

\[\tag{15} F(x_{eq}+\Delta x)=\frac{NT}{x_{eq}+\Delta x}-~$K$ (x_{eq}+\Delta x )\]

Which leads to the fact that

\[\tag{16} F(\Delta x)= F(x_{eq}+\Delta x)-F(x_{eq})=\frac{NT}{x_{eq}+\Delta x}-~$K$ (x_{eq}+\Delta x )-\frac{NT}{x_{eq}}+~$K$ x_{eq}\]

Using some math and Eq (4) we get

\[ F(\Delta x) = \frac{NT}{x_{eq}+\Delta x}-~$K$ x_{eq}-~$K$\Delta x -\frac{NT}{x_{eq}}+~$K$ x_{eq}=\frac{-NT\Delta x}{x_{eq}(x_{eq}+\Delta x)}-K\Delta x\] 
\[\tag{17} \simeq \frac{-NT\Delta x}{x_{eq}^2}-K\Delta x = -K\Delta x-K\Delta x = -2K\Delta x\]

We also know that for small harmonic osillations:

\[\tag{18}  F(\Delta x)=-m\omega^2\Delta x\]

Taking Eq (17) and (18) together -

\[\tag{19} - m\omega^2\Delta x  = -2K\Delta x  \Rightarrow \omega^2 = \frac{2K}{m}  \]

Eventually we get:

\[\tag{20} \omega=\sqrt{\frac{2K}{m}}\]


\Dn
\Dn

\underline { Condensed Bosons Case}
\Dn
\Dn

We will do the same as we did for boltzman approximation using the force:

\[\tag{21} F(x_{eq})=PA-~$K$ x_{eq}   \Rightarrow  F(x_{eq})=\zeta \left( \frac{5}{2} \right) \left( \frac{m}{2\pi} \right)^{\frac{3}{2}} T^{\frac{5}{2}}A-~$K$ x_{eq}\]

For small oscilations

\[\tag{22} F(x_{eq}+\Delta x)=\zeta \left( \frac{5}{2} \right) \left( \frac{m}{2\pi} \right)^{\frac{3}{2}} T^{\frac{5}{2}}A -~$K$ (x_{eq}+\Delta x )\]

Which leads to 

\[\tag{23} F(\Delta x)= F(x_{eq}+\Delta x)-F(x_{eq})=-$K$\Delta x\]

Using  Eq (18) we get

\[\tag{24} - m\omega^2\Delta x  = -K\Delta x  \Rightarrow \omega^2 = \frac{K}{m}  \]

Eventually we get:

\[\tag{25} \omega=\sqrt{\frac{K}{m}}\]

\Dn
\Dn

\underline {Spinless Fermions, at zero temperature }
\Dn
\Dn

Writing $  F(x_{eq}) $:

\[\tag{26}  F(x_{eq})=PA-~$K$x_{eq} \Rightarrow F(x_{eq})= \frac{(6\pi^2)^{\frac{2}{3}}}{5m}  \left( \frac{N}{Ax_{eq}} \right)^{\frac{5}{3}}A-~$K$x_{eq}=\left(\frac{6\pi^2}{A}\right)^\frac{2}{3}\frac{1}{5m}  \left( \frac{N}{x_{eq}} \right)^{\frac{5}{3}}-~$K$x_{eq}   \]

\Dn
And $  F(x_{eq}+\Delta x) $:

\[\tag{27}  F(x_{eq}+\Delta x)=\left(\frac{6\pi^2}{A}\right)^\frac{2}{3}\frac{1}{5m}  \left( \frac{N}{x_{eq}+\Delta x} \right)^{\frac{5}{3}}-~$K$(x_{eq}+\Delta x)   \]

Using substruction:

\[\tag{28}  F(x_{eq}+\Delta x)- F(x_{eq})=\left(\frac{6\pi^2}{A}\right)^\frac{2}{3}\frac{1}{5m}  \left( \frac{N}{x_{eq}+\Delta x} \right)^{\frac{5}{3}}-~$K$(x_{eq}+\Delta x)-\left(\frac{6\pi^2}{A}\right)^\frac{2}{3}\frac{1}{5m}  \left( \frac{N}{x_{eq}} \right)^{\frac{5}{3}}+~$K$x_{eq}   \]

And with some math ,we get:

\[\tag{29}  F(\Delta x)=\left(\frac{6\pi^2}{A}\right)^\frac{2}{3}\frac{N^\frac{5}{3}}{5m}  \left[(x_{eq}+\Delta x)^\frac{-5}{3} - (x_{eq})^\frac{-5}{3} \right]-~$K$\Delta x \]

Expending the expression to a series:

\[\tag{30} (a+x)^\frac{-5}{3}|_{x\rightarrow 0}=a^\frac{-5}{3}-\frac{5}{3}(a+x|_{x\rightarrow 0})^\frac{-8}{3}x+...\]

Hence

\[\tag{31} (x_{eq}+\Delta x)^\frac{-5}{3}=x^\frac{-5}{3}- \frac{5}{3}(x)^\frac{-8}{3}\Delta x\]

And so we get

\[\tag{32} \left[(x_{eq}+\Delta x)^\frac{-5}{3} - (x_{eq})^\frac{-5}{3} \right]=x^\frac{-5}{3}- \frac{5}{3}(x)^\frac{-8}{3}\Delta x-x^\frac{-5}{3}=- \frac{5}{3}(x)^\frac{-8}{3}\Delta x\]

Placing  Eq (12) inside Eq (29) and the solution from our expension ,we get

\[\tag{33}  F(\Delta x)=\left(\frac{6\pi^2}{A}\right)^\frac{2}{3}\frac{N^\frac{5}{3}}{5m}  \left[\frac{5}{3}(x)^\frac{-8}{3}\Delta x \right]-~$K$\Delta x=\left(\frac{6\pi^2}{A}\right)^\frac{2}{3}\frac{N^\frac{5}{3}}{5m}  \left[-\frac{5}{3}\frac{1}{ \left(\frac{6\pi^2}{A}\right)^\frac{2}{3}\frac{N^\frac{2}{3}}{5Km}} \Delta x \right]-~$K$\Delta x \]

\[\tag{34}  F(\Delta x)=-\frac{5}{3} K \Delta x - K \Delta x = -\frac{8}{3} K \Delta x \]

Now we will use the connection between the force and the frequency 

\[\tag{35} - m\omega^2\Delta x  =-\frac{8}{3} K \Delta x  \Rightarrow \omega^2 = \frac{8K}{3m}  \]

So we have

\[\tag{36} \omega= \sqrt{\frac{8K}{3m}}\]

\Dn
\Dn


\underline {\bf Part 3 : Schematic drawing of $ \omega $ versus T}

\Dn
\Dn

\includegraphics[width=6in]{3042pic}


\end{document} ‎‎‎ 
