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\exnumber{2050}
\heading{Pressure by a particle in a spring-box system} 
\auname{Eli Gudinetsky}

\Dn

A spring that has an elastic constant ${K}$ and 
natural length ${L}$ is connected between a wall 
at $x=0$ and a piston at~${x=X}$. 
Consequently the force that acts of the piston is ${F_0=-K(X-L)}$. 
A classical particle of mass ${\mathsf{m}}$ is attached 
to the middle point of the spring. 
The system is at equilibrium, the temperature is~${T}$.

\begin {itemize}

\item[(1)]
Write the Hamiltonian (be careful).

\item[(2)]
Write an expression for the partition function ${Z\left(\beta,X\right)}$.
The answer is an expression that may contain a definite integral.

\item[(3)]
Write an expression for the force ${F}$ on the piston. 
The answer is an expression that may contain a definite integral.

\item[(4)]
Find a leading order (non-zero) expression for ${F-F_0}$
in the limit of high temperature. 

\item[(5)]
Find a leading order (non-zero) expression for ${F-F_0}$
in the limit of low temperature. 

\end {itemize}

Your answers should not involve exotic functions, 
and should be expressed using $(X,L,K,\mass,T)$.

\Dn\Dn

{\bf The Solution:}

\Dn

\begin{enumerate}
\item 
When cutting a spring in half its spring constant is doubled. This is understood by viewing the whole spring as 2 identical halves connected in series, so
\begin{align*} K^{-1}_{whole}=K^{-1}_{half}+K^{-1}_{half}\Rightarrow K_{half}=2K_{whole}\, .
\end{align*}
The Hamiltonian is therefore,
\begin{align} 
\mathcal{H}=\frac{p^2}{2m}+\frac{2K\Paren{x-L/2}^2}{2}+\frac{2K\Brack{\Paren{X-x}-L/2}^2}{2}\, .
\end{align}
Here, $x$ is the position of the particle, $p$ is its momentum, the second term is due to the left spring and the third term is due to the right spring.
\item
Writing the partition function we have,
\begin{align} 
Z=\frac{1}{2\pi}\int_{-\infty}^{\infty}{\mathrm{d}p\int_{0}^{X}{\mathrm{d}x\, e^{-\beta \Brack{p^2/2m+{2K\Paren{x-L/2}^2/}{2}+{2K\Paren{{X-x}-L/2}^2}/{2}}}}}  \, .
\end{align}
Let us rewrite the argument of the spatial exponent, 
\begin{align*} 
&\Paren{x-L/2}^2-\Paren{X-x-L/2}^2=2x^2+L^2/2+X^2-2Xx-LX=\\
&=2\Paren{x^2-xX+X^2/4}+\Paren{L^2-2LX+X^2}/2=2\Paren{x-X/2}^2+\Paren{L-X}^2/2  \, .
\end{align*}
Substituting the above and solving the kinetic part we get,
\begin{align} 
Z=\frac{1}{\lambda_T}e^{-\beta K\Paren{L-X}^2/2}{\int_{0}^{X}{\mathrm{d}x\, e^{-2\beta K \Paren{x-X/2}^2}}}  \, .
\end{align}
\item
The generalized force associated with the parameter $X$ is the force $F$ on the piston. By definition it is,
\begin{align} 
\Seq{F}&=\frac{1}{\beta}\frac{\partial \ln{Z}}{\partial X}=\frac{1}{\beta}\Brack{\beta K \Paren{L-X}+\frac{\partial}{\partial X}\ln\Paren{\int_{0}^{X}{\mathrm{d}x\, e^{-2\beta K \Paren{x-X/2}^2}}}}\\
&=F_0+\frac{1}{\beta}\frac{\partial}{\partial X}\ln\Paren{\int_{0}^{X}{\mathrm{d}x\, e^{-2\beta K \Paren{x-X/2}^2}}}  \, .
\end{align}
\item
In the limit of high temperatures, or $\beta\to0$, we can expand the exponent in the intergal and have:
\begin{align} 
\Seq{F}-F_0&\approx\frac{1}{\beta}\frac{\partial}{\partial X}\ln\Paren{\int_{0}^{X}{\mathrm{d}x\, 1}}=\frac{1}{\beta}\frac{\partial}{\partial X}\ln{X}=\frac{1}{\beta X} \, .
\end{align}
\item
Let us denote 
\begin{align}
I\Paren{X,\beta}=\int_{0}^{X}{\mathrm{d}x\, e^{-2\beta K \Paren{x-X/2}^2}}=\frac{2}{\sqrt{2\beta K}}\int_{0}^{\sqrt{\frac{\beta K}{2}}X}{\mathrm{d}y\, e^{-y^2}}\, ,
\end{align}
(in the last equality we just changed variables) giving us:
\begin{align}
\Seq{F}-F_0=\frac{1}{\beta I\Paren{X,\beta}}\frac{\partial}{\partial X}I\Paren{X,\beta}\, .
\end{align}
Notice that in the limit $\beta\to\infty$ we have $I\Paren{X,\beta}\to\sqrt{\pi}/2$ becase of the upper bound of the integral. This gives as the approximated value:
\begin{align}
\Seq{F}-F_0\approx\sqrt{\frac{2K}{\pi\beta}}e^{-\beta K \frac{X^2}{2}}\, ,
\end{align}
where the last exponent came from the fundamental theorem of calculus.
\end{enumerate}



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