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

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- Solution c

Solution

The quadratic energy is

E_{2} = \sum_{n = 1}^{\infty} \frac{A π ^{2} n ^{2} - F L ^{2}}{4 L} a_{n}^{2} .

Each mode is a centered Gaussian, so

⟨ a_{n} ⟩ = 0, ⟨ a_{n}^{2} ⟩ = \frac{2 k _{B} T L}{A π ^{2} n ^{2} - F L ^{2}} .

The

n = 1

variance diverges as

F ↑ 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) ≃ \frac{1}{4} \sum_{n = 1}^{\infty} ⟨ a_{n}^{2} ⟩ = \frac{k _{B} T L}{2 A π ^{2}} \sum_{n = 1}^{\infty} \frac{1}{n ^{2}} = \frac{k _{B} T L}{12 A} .

With the common persistence length convention

L_{p} = A / (k_{B} T)

,

C (0) = \frac{L}{12 L _{p}} .

Thermal bending therefore shortens the mean projected length even without compression. Under the strictly planar tangent-correlation convention

L_{p}^{(2 D)} = 2 A / (k_{B} T)

, the same result is

L / (6 L_{p}^{(2 D)})

.

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