The quadratic energy is
$$
\widetilde E_2=
\sum_{n=1}^\infty
\frac{A\pi^2n^2-FL^2}{4L}a_n^2.
$$
Each mode is a centered Gaussian, so
$$
\boxed{\langle a_n\rangle=0},
\qquad
\boxed{\langle a_n^2\rangle
=\frac{2k_BTL}{A\pi^2n^2-FL^2}}.
$$
The $n=1$ variance diverges as $F\uparrow F_c$ because the first bending mode becomes soft; Gaussian theory then fails and the quartic term controls the fluctuations.

At zero force,
$$
C(0)\simeq\frac14\sum_{n=1}^\infty\langle a_n^2\rangle
=\frac{k_BTL}{2A\pi^2}\sum_{n=1}^\infty\frac1{n^2}
=\frac{k_BTL}{12A}.
$$
With the common <persistence length> convention $L_p=A/(k_BT)$,
$$
\boxed{C(0)=\frac{L}{12L_p}}.
$$
Thermal bending therefore shortens the mean projected length even without compression. Under the strictly planar tangent-correlation convention $L_p^{(2D)}=2A/(k_BT)$, the same result is $L/(6L_p^{(2D)})$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 355 / 3 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-355/3/c/solution