{title2=$r_0=\sqrt{k_c/(2\sigma)}$}

Minimizing the <Helfrich energy> per length of an unconstrained cylindrical membrane gives $r_0=\sqrt{k_c/(2\sigma)}$, balancing the area cost against bending curvature.

Cylindrical membrane equilibrium radius

For a nearly planar membrane, the quadratic <Helfrich energy> of a Fourier mode is
$$
E_{\mathbf q}=\frac A2(k_cq^4+\sigma_0q^2)|h_{\mathbf q}|^2.
$$
The <equipartition theorem> and $A-A_{\rm proj}=\int|\nabla h|^2d^2x/2$ give
$$
\alpha(\sigma_0)=\frac{k_BT}{2}\int_{q_{\min}}^{q_{\max}}
\frac{d^2q}{(2\pi)^2}\frac1{k_cq^2+\sigma_0}.
$$
Using the stated molecular and system-size cutoffs, $q_{\max}=\pi/a$ and $q_{\min}=\pi/\sqrt A$, gives
$$
\boxed{\alpha(\sigma_0)=\frac{k_BT}{8\pi k_c}
\log\left(\frac{\pi^2k_c/a^2+\sigma_0}
{\pi^2k_c/A+\sigma_0}\right).}
$$

Solution

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

{c}
{wiki}

The Helfrich energy of a fluid membrane combines surface tension with curvature elasticity. Without spontaneous curvature or a topological term,
$$
E=\int dS\left[\sigma+\frac{k_c}{2}(2H)^2\right].
$$


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