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\exnumber{2351}
\heading{Tension of a rubber band} 
\auname{Itai Zbeda}


{\bf The problem:}

\Dn

The Elasticity of a rubber band can be described by a one dimensional model of a polymer. The polymer consists of ${N}$ monomers that are arranged along a straight line, hence forming a chain. Each unit can be either in a state of length a with energy ${E_a}$, or in a state of length b with energy ${E_b}$. We define ${f}$ as the tension, i.e. the force that is applied while holding the polymer in equilibrium.

\Dn

(1) Write expressions for the partition function ${Z_{G}(\beta,f)}$. \Dn

(2) For very high temperatures ${F_{G}(T,f)\approx F_{G}^{(\infty)}(T,f)}$, where ${F_{G}^{(\infty)}(T,f)}$ is a linear function of T. Write the expression for ${F_{G}^{(\infty)}(T,f)}$. \Dn

(3) Write the expression for ${F_G(T,f)-F_{G}^{(\infty)}(T,f)}$. Hint: this expression is quite simple - within this expression ${f}$ should appear only once in a linear combination with other parameters. \Dn

(4) Derive  an  expression  for  the  length ${L}$ of the polymer at thermal equilibrium, given the tension ${f}$. Write two separate expressions: one for the infinite temperature result ${L(\infty,f)}$ and one for the difference ${L(T,f)-L(\infty,f)}$. \Dn

(5) Assuming ${E_a=E_b}$, write a linear approximation for the function ${L(T,f)}$ in the limit of weak tension. \Dn

(6) Treating ${L}$ as a continuous variable, find the probability distribution ${P(L)}$. Write also the result that you get from this expression. \Dn

(7) Write an expression that relates the function ${f(L)}$ to the probability distribution ${P(L)}$. Write also the result that you get from this expression. \Dn

(8) Find what would be the results for ${Z_G(\beta,f)}$ if the monomer could have any length ${\in [a,b]}$. Assume that the energy of the monomer is independent of its length. \Dn

(9) Find what would be the results for ${L(T,f)}$ in the latter case. \Dn

\textit{Note:} Above a "linear function" means ${y=Ax+B}$.\\
Please express all results using ${(N,a,b,E_a,E_b,f,t,L)}$


\Dn\Dn


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

\Dn

(1)
The Hamiltonian for a single monomer in a polymer chain is:
\begin{align}
\mathcal{H}(x_i)=E_{x_i}-fx_i
\end{align}

Therefore, the partition function for a single monomer with 2 possible configurations (a and b) will be:
\begin{align}
Z^1=e^{-\beta \mathcal{H}(a)}+e^{-\beta \mathcal{H}(b)}=e^{-\beta (E_a-fa)}+e^{-\beta (E_b-fb)}
\end{align}

and for N monomers:
\begin{align}
Z_G=(Z^1)^N
\end{align}

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

(2)
Using the identity:
\begin{align}
e^x+e^y=\exp{\left(\frac{x+y}{2}\right)}\cosh\left(\frac{x-y}{2}\right)
\end{align}
We can factorize out of ${Z^1}$ the term
\begin{align}
\exp{\left[\beta\left(f\frac{a+b}{2}-\frac{E_a+E_b}{2}\right)\right]}
\end{align}
and get:
\begin{align}
Z^1=\exp{\left[\beta\frac{fa+fb}{2}-\beta\frac{E_a+E_b}{2}\right]}\cdot2\cosh\left[\beta\left(\frac{E_b-E_a}{2}-\frac{b-a}{2}f\right)\right]
\end{align}
And the Gibbs Free energy is:
\begin{align}
\frac{F}{N}=-T\ln Z^1=\left(\frac{E_a+E_b}{2}-\frac{a+b}{2}f\right)-T\ln\left(2\cosh\left[\beta\left(\frac{E_b-E_a}{2}-\frac{b-a}{2}f\right)\right]\right)
\end{align}

For high temperature, we'll mark:
\begin{align}
\cosh\left[\beta\left(\frac{E_b-E_a}{2}-\frac{b-a}{2}f\right)\right]=\cosh\left(\beta \tilde{E}\right)
\end{align}
Using ${\cosh(x)\approx1+\frac{1}{2}x^2}$, we get ${\ln\left(1+\frac{1}{2}x^2\right)\approx \frac{1}{2}x^2}$, resulting in:

\begin{align}
\frac{F^{\infty}}{N}=\left(\frac{E_a+E_b}{2}-\frac{a+b}{2}f\right)-T\ln 2-\frac{1}{2}\beta\tilde{E}^2
\end{align}


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

(3)
Using previous results:
\begin{align}
\frac{1}{N}\left(F-F^{\infty}\right)=\frac{1}{2}\beta\tilde{E}^2-T\ln\left(\cosh(\beta\tilde{E})\right)
\end{align}

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

(4)
We can derive ${L}$, using the conjugated general force ${f}$:
\begin{align}
\frac{L}{N}=T\frac{\partial}{\partial f}\ln Z^1=\frac{a+b}{2}-\frac{b-a}{2}\cdot \tanh \left(\frac{(E_b-E_a)-(b-a)f}{2T}\right)
\end{align}

Using $\tanh(x)\approx x$ we get:
\begin{align}
\frac{L^\infty}{N}=\frac{a+b}{2}+\frac{(b-a)^2}{4T}f-\frac{(b-a)(E_b-E_a)}{4T}=Af+B
\end{align}
And then:
\begin{align}
\frac{1}{N}\left(L-L^\infty\right)=\frac{(b-a)(E_b-E_a)}{4T}-\frac{(b-a)^2}{4T}f-\frac{b-a}{2}\tanh \left(\frac{(b-a)f-(E_b-E_a)}{2T}\right)
\end{align}

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

(5)
For ${E_a=E_b}$ and a weak tension (${f\rightarrow0}$):
\begin{align}
\frac{L}{N}=\frac{a+b}{2}+\frac{(b-a)^2}{4T}f
\end{align}

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

(6)
Now, since ${E_a=E_b}$, each possible configuration of a single monomer has the same probability ${P_a=P_b=\frac{1}{2}}$, so the mean value ${\left<L\right>}$ and variance ${\sigma^2}$ are (${x_i}$ representing a single monomer):
\begin{align}
&\left<L\right>=\sum_{i=1}^{N}\left<x_i\right>=N\left<x_i\right>=N\left(\frac{a+b}{2}\right)\\
&\sigma^2=\left<L^2\right>-\left<L\right>^2=N\left(\left<x_i^2\right>-\left<x_i\right>^2\right)=N\left(\frac{a-b}{2}\right)^2
\end{align}

Assuming a long monomer chain ${L}$ where ${N\gg1}$, we can now use the central limit theorem, and the probability distribution ${P(L)}$ takes the form: 
\begin{align}
P(L)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{1}{2}\left(\frac{L-\left<L\right>}{\sigma}\right)^2}
\end{align}

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

(7)
The probability ${P(L)}$ for the chain to assume a specific finite length ${L}$ is:
\begin{align}
P(L)=\frac{1}{Z}Z(L)
\end{align}

\Dn
With ${Z}$ being the summation on all possible length states of the polymer (a normalization constant). So we can derive the tension ${f(L)}$ same as before:
\begin{align}
f(L)=\frac{1}{\beta}\frac{\partial \ln P(L)}{\partial L}=-T\frac{L-\left<L\right>}{\sigma^2} 
\end{align}

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

(8)
The size of a single monomer is now a continuous variable ${x_i\in[a,b]}$ with E independent of ${x_i}$, so that the Hamiltonian of a monomer takes the form of
\begin{align}
\mathcal H(x_i)=E-fx_i
\end{align}

And the partition function becomes:
\begin{align}
Z^1=e^{-\beta E}\int_{b}^{a}e^{\beta fx_i}dx_i=\left.\frac{e^{-\beta E}}{f\beta }e^{\beta fx_i} \right|^a_b=\frac{e^{-\beta E}}{f\beta }\left(e^{\beta fa}-e^{\beta fb}\right)
\end{align}

Setting ${E=0}$ we get:
\begin{align}
Z^1=\frac{1}{f\beta }\left(e^{\beta fa}-e^{\beta fb}\right)
\end{align}

And for N monomers:
\begin{align}
Z_G=\left(Z^1\right)^N=\left[\frac{1}{f\beta }\left(e^{\beta fa}-e^{\beta fb}\right)\right]^N
\end{align}

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

(9)
Repeating the algebra in (2) and (5), we get for the new partition function:
\begin{align}
&Z_G=\left[\frac{1}{f\beta}\exp{\left(\beta f\frac{a+b}{2}\right)}\cdot 2\sinh\left(\beta f\frac{a-b}{2}\right)\right]^N\\
&\frac{F}{N}=-f\frac{a+b}{2}-T\ln \left[2\sinh\left(\beta f\frac{a-b}{2}\right)\right]+T\ln\left(f\beta\right)\\
&\frac{L}{N}=T\frac{\partial}{\partial f}\left(\ln Z^1\right)=\frac{a+b}{2}-\frac{T}{f}+\frac{a-b}{2}\coth \left(\beta f\frac{a-b}{2}\right)\\
\end{align}
And for high temperature approximation (${\coth{x}\approx x^{-1}+\frac{1}{3}x}$):
\begin{align}
\frac{L^\infty}{N}=\frac{a+b}{2}+\frac{(a-b)^2}{12}\beta f
\end{align}
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\end{document} 
