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\exnumber{2340}
\heading{Tension of a chain molecule} 
\auname{Asaf Szulc}


{\bf The problem:}

\Dn
A chain molecule consists of ${N}$ units, each having a length ${a}$. The units are joined so as to permit free rotation about the joints. At a given temperature ${T}$, derive the relation between the tension ${f}$ acting between both ends of the three-dimensional chain molecule and the distance ${L}$ between the ends. \\


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{\bf The solution:}
\Dn


{\Large Problematic solution: independence of $x$, $y$, $z$ is questionable, and relation between $P(r)$ and $F$ is not established.} \\


In one-dimension, the average step length of the projection on the x-axis is $\langle a^2 cos^2\theta\rangle^{1/2} = a/\sqrt{3}$, which by using the "central limit theorem" for a long polymer ($N \gg 1$) leads to:
\begin{equation}
\label{1d_prob}
	P(x,N) \propto e^{-3x^2/(2Na^2)}
\end{equation}
with similar results for $P(y,N)$ and $P(z,N)$. 
Now, because each bond is oriented randomly, the $x$, $y$ and $z$ projections are independent of each other, and the probability density for the chain to end at $\vec{r}$  is, therefore, the product of independent factors:
\begin{equation}
\label{3d_prob}
	P(\vec{r},N) = P(x,N)P(y,N)P(z,N) \propto exp \left[ -\frac{3}{2}\frac{r^2}{Na^2}\right]
\end{equation}

In order to get the probability that the chain has a length $L$, one must multiply eqn.(\ref{3d_prob}) by the surface of a sphere with radius $L$ and get:
\begin{equation}
	P(L,N)  \propto 4\pi L^2 e^{-3L^2/(2Na^2)}
\end{equation}

Finally, by using the entropy $S$, one can write the free energy as:
\begin{equation}
	F = -TS = -Tln(P(L,N)) = \frac{3TL^2}{2Na^2} -2Tln(L) + constant
\end{equation}
and calculate the tension:
\begin{equation}
	f = -\frac{\partial F}{\partial L} = \frac{2T}{L} - \frac{3TL}{Na^2} = 
	\frac{2T}{L} \left( 1 - \frac{3L^2}{2Na^2}\right)
\end{equation}
It is also nice to see that we can find the relaxed length of the chain to be $L_0 = \sqrt{2Na^2/3}$.

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