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\exnumber{2353}
\heading{Tension of a stretched chain} 
\auname{Daniel Dahan}


{\bf The problem:}

\Dn
A rubber band is modeled as a single chain of $N\gg 1$ massless non-interacting links,
each of fixed length $a$. Consider a one-dimensional model where the links are restricted 
to point parallel or anti-parallel to a given axis, while the endpoints are constraint 
to have a distance $X=(2n-N)a$, where $n$ is an integer. Later you are requested to use 
approximations that allow to regard $X$ as a continuous variable. Note that the body 
of the chain may extend beyond the length $X$, only its endpoints are fixed. 
In items (c,d) a spring is pushed between the two endpoints, such that the additional 
potential energy $-KX^2$ favors large $X$, and the system is released (i.e. $X$ is free to fluctuate).

\Dn

\begin{itemize}

\item[(a)] Calculate the partition function $Z(X)$.
Write the exact combinatorial expression. Explain 
how and why it is related trivially to the entropy $S(X)$. 

\item[(b)] Calculate the force $f(X)$ that the chain applies 
on the endpoints. Use the Stirling approximation 
for the derivatives of the factorials. 
  
\item[(c)] Determine the temperature $T_c$ below which 
the ${X=0}$ equilibrium state becomes unstable.

\item[(d)] For ${T<T_c}$ write an equation for the 
stable equilibrium distance $X(T)$. Find an explicit 
solution by expanding $f(X)$ in leading order.  

\end{itemize}

\Dn\Dn


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{\bf The solution:}

\Dn
\begin{itemize}
\item[(a)] The partition function is defined: 
\[
Z=\sum e^{-\beta E_r}
\] 
If all $E_r=0$:
\[
Z=\#\text{states}
\]
\[
Z(X) =\frac{N!}{n!(N-n!)}=\frac{N!}{((Na+X)/2a)!((Na-X)/2a)!} 
\]
The entropy:
\begin{equation*}
\begin{aligned}
S(X)&=\ln (\#\text{states})=\ln (Z(X))\\&= \ln (N!)-\ln \left(\left(\frac{Na+X}{2a}\right) !\right)-\ln \left( \left( \frac{Na-X}{2a}\right)!\right)
\end{aligned}
\end{equation*}
%%%%%%%%%%%%%%%

\item[(b)]
Realizing that the force is conjugate to $X$: 
\[
f(X)  =T\frac{\partial \ln Z}{\partial X}=-\frac{T}{2a}\ln \left[ \frac{Na+X}{Na-X}\right]
=-\frac{T}{a}\tanh ^{-1} \left( \frac{X}{Na}\right)
\]

\pagebreak

%%%%%%%%%%%%%%


\item[(c)]
The free energy of the system:
\[
F=E-TS=-KX^2-T\ln(Z(X))
\]
\[
\left.\frac{d^2F(T=T_c)}{dX^2}  \right\vert _{X=0}=0
\]
\[
0=-2K+\frac{T_c}{2a}\left( \frac{1}{Na}+\frac{1}{Na}\right)
\]
\[
T_c=2NKa^2
\]

%%%%%%%%%%%%%


\item[(d)] While $T<T_c$ there is a stable solution were  $X\neq 0$: 

\[ 
\frac{dF}{dX}=0 
\]
\[
0=-2KX+\frac{T}{a} \tanh ^{-1} \left(\frac{X}{Na} \right) 
\]
\[
\frac{X}{Na}=\frac{T}{T_c}\tanh ^{-1} \left(\frac{X}{Na} \right)
 \]
Setting a new variable $\xi=\frac{X}{Na}$ one can graphically analyse the solution. \\ 

\begin{tikzpicture}
begin{tikzpicture}
\pgfplotsset{every axis legend/.append style={
		at={(0.5,1.03)},
		anchor=south}}
\begin{axis}[
	ylabel=$\ $ ,
	xlabel=$\xi$,]
\legend{$\xi$,$\frac{T}{T_c} \tanh^{-1}(\xi) $}	
\addplot
	[mark=none,color=black,ultra thick,samples=250,
	 domain=0:1] 
	{x};	
\addplot
	[mark=none,color=blue,samples=250,
	 domain=0:1] 
	{0.25*(ln(1+x)-ln(1-x))	};
\addplot
	[mark=none,color=blue,samples=250,
	 domain=0:0.88] 
	{0.5*(ln(1+x)-ln(1-x))};
\addplot
	[mark=none,color=blue,samples=250,
	 domain=0:0.65] 
	{0.75*(ln(1+x)-ln(1-x))};	
\node[coordinate,pin=180:{${{_{T<T_c}}}$}] 
		at (axis cs:0.995,1.50)	{};
\node[coordinate,pin=180:{${{_{T=T_c}}}$}] 
		at (axis cs:0.860,1.30)	{};
\node[coordinate,pin=180:{${{_{T>T_c}}}$}] 
		at (axis cs:0.623,1.1)	{};
\end{axis}
\end{tikzpicture}\\ 
Expanding while using $\tanh ^{-1}(\varepsilon)	\approx \varepsilon+\frac{\varepsilon^3}{3}$ (were $\frac{X}{Na}\ll 1$):

 \[
f(X)\approx \frac{T}{a}\left(\frac{X}{Na}+\frac{X^3}{3(Na)^3} \right)
\]
Finding $X$ that is non zero:

\[
\frac{X}{Na}= \frac{T}{T_c}\left(\frac{X}{Na}+\frac{X^3}{3(Na)^3} \right)
\]
\[
X=\sqrt{3}Na\sqrt{ \frac{T_c}{T}-1}
\]
\end{itemize}
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\end{document} 
