Subtracting the two fluctuation areas gives
$$
\Delta\alpha=\frac{k_BT}{8\pi k_c}\log\left[
\frac{(\pi^2k_c/a^2+\sigma_0)(\pi^2k_c/A+\sigma)}
{(\pi^2k_c/A+\sigma_0)(\pi^2k_c/a^2+\sigma)}\right].
$$
When both tensions are much smaller than $\pi^2k_c/a^2$, the molecular-scale factors cancel to leading order, leaving
$$
\boxed{\Delta\alpha=\frac{k_BT}{8\pi k_c}
\log\left(\frac{\pi^2k_c/A+\sigma}
{\pi^2k_c/A+\sigma_0}\right).}
$$
Increasing tension suppresses <thermal membrane undulations>, releasing their hidden excess area into the tether.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-355/3/b/solution