Set $\sigma_0=0$. Equating the cylindrical tether area $2\pi rL$ to $A\Delta\alpha$ gives
$$
\log\left(1+\frac{\sigma A}{\pi^2k_c}\right)
=\frac{16\pi^2k_crL}{Ak_BT},
$$
or
$$
\sigma=\frac{\pi^2k_c}{A}
\left[\exp\left(\frac{16\pi^2k_crL}{Ak_BT}\right)-1\right].
$$
Because a cylinder has $2H=1/r$, the energy becomes
$$
E(r,L)=\frac{\pi k_cL}{r}
+\frac{\pi k_BT}{8}\left[
e^{16\pi^2k_crL/(Ak_BT)}-1\right]-FL.
$$
The two stationarity equations imply
$$
\boxed{F=\frac{2\pi k_c}{r},
\qquad
e^{16\pi^2k_crL/(Ak_BT)}=\frac{A}{2\pi^2r^2}.}
$$
Eliminating $r$ gives the implicit <entropic membrane-tether force--extension relation>
$$
\boxed{L=\frac{Ak_BTF}{32\pi^3k_c^2}
\log\left(\frac{AF^2}{8\pi^4k_c^2}\right).}
$$
As tether length grows, fluctuation area is depleted, tension rises, the tether narrows, and the required force increases.

For a membrane connected to a reservoir at fixed tension, its energy per length is $\pi k_c/r+2\pi\sigma r-F$. Minimization instead gives
$$
\boxed{r=\sqrt{\frac{k_c}{2\sigma}},
\qquad F=2\pi\sqrt{2k_c\sigma}.}
$$
The reservoir supplies area without changing $\sigma$, so the force is independent of extension; the finite vesicle has an entropic, strain-stiffening response.

Solution

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

When a tether draws area from the thermal undulations of a finite vesicle, extension raises membrane tension rather than drawing area from a fixed-tension reservoir. The tether therefore narrows and its required pulling force increases with length.


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 Incoming links: Entropic membrane-tether force--extension relation