= Solution
The condition $L\ll na$ implies $|\tau a|\ll1$. The <small-argument expansion of the hyperbolic tangent> gives $\tanh x=x+O(x^3)$, so
$$
\frac{L}{na}\simeq\tau a.
$$
Using $f=k_BT\tau$ yields the <entropic Hooke law for a one-dimensional chain>
$$
\boxed{f\simeq\frac{k_BT}{na^2}L}.
$$
Thus the effective spring constant is $k_BT/(na^2)$.
At fixed tension, the exact extension is
$$
L=na\tanh\!\left(\frac{fa}{k_BT}\right).
$$
Increasing the <temperature> decreases the positive argument of the <hyperbolic tangent>, so the chain contracts. In the small-extension regime this becomes
$$
\boxed{L\simeq\frac{na^2f}{k_BT}},
$$
which is inversely proportional to $T$.
Back to article page
