Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 355 / 3 / b

Table of contents
- Solution b

Solution

Subtracting the two fluctuation areas gives

Δ α = \frac{k _{B} T}{8 π k _{c}} lo g [\frac{( π ^{2} k _{c} / a ^{2} + σ _{0} ) ( π ^{2} k _{c} / A + σ )}{( π ^{2} k _{c} / A + σ _{0} ) ( π ^{2} k _{c} / a ^{2} + σ )}] .

When both tensions are much smaller than

π^{2} k_{c} / a^{2}

, the molecular-scale factors cancel to leading order, leaving

Δ α = \frac{k _{B} T}{8 π k _{c}} lo g (\frac{π ^{2} k _{c} / A + σ}{π ^{2} k _{c} / A + σ _{0}}) .

Increasing tension suppresses thermal membrane undulations, releasing their hidden excess area into the tether.

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