What it costs to bend a membrane

Add the energy of small patches

Treat the membrane as a smooth surface of negligible thickness. Its preferred total curvature $c_0$ has units of inverse length. A departure costs energy proportional to $(2H-c_0)^2$. The bending rigidity $\kappa$ has units of energy; a larger value makes bending more expensive.

The integral means divide the surface into small patches $\mathrm dA$, calculate their contributions and add them. The Gaussian term has its own coefficient $\bar\kappa$. Both coefficients describe curvature elasticity. Stretching the membrane area is a separate deformation.

Source: W. Helfrich

Why a larger sphere need not cost more

Every point on a sphere has $2H=2/R$, which can be taken outside the integral. Area $4\pi R^2$ grows as radius squared, and curvature squared falls by the same factor. For $c_0=0$, the mean-curvature contribution is $8\pi\kappa$.

With spontaneous curvature it becomes $2\pi\kappa(2-c_0R)^2$. Radius now matters because the membrane has a preferred bending scale. Reversing the curvature convention requires reversing $c_0$ as well.

For a closed surface without a boundary, the Gauss–Bonnet theorem gives $\int K\,\mathrm dA=4\pi(1-g)$, where $g$ counts handles. At fixed spherical topology the Gaussian energy is constant and cannot rank different shapes. Fusion or pore formation changes that premise.

Source: U. Seifert

Tension and stretching stiffness play different roles

Tension is the energy change per small added area, or the force per length along a cut. At strictly fixed total area it can enter the calculation as a Lagrange multiplier: an auxiliary quantity that enforces the area constraint. Its value then follows from that constraint.

The area compression modulus $K_A$ measures how rapidly tension changes with relative area extension. Let $A_0$ be the relaxed area and $\alpha=(A-A_0)/A_0$ a small strain. A quadratic stretching energy gives the expressions below. Both tension and modulus have units of energy per area; tension is the current stress and the modulus is its slope with strain.

A vesicle can first increase its apparent area by flattening thermal wrinkles, before appreciably separating its lipids. The definition of area therefore matters when interpreting an experiment. Interpreting molecular stretching requires separating wrinkle flattening from true lipid-area extension.

Worked example in thermal-energy units

Choose the teaching value $\kappa=20k_BT$ and no spontaneous curvature. The spherical mean-curvature energy is $160\pi k_BT\simeq503k_BT$. At $T=298\,\mathrm K$, thermal energy is approximately $4.11\,\mathrm{pN\,nm}$, giving $2.07\times10^3\,\mathrm{pN\,nm}$.

Increasing the radius from five to ten micrometers leaves this value unchanged. Closing an initially flat membrane into a sphere still requires bending energy. Whether that sphere minimizes energy among already closed surfaces at fixed area and volume is a different comparison.

For $K_A=0.20\,\mathrm{N\,m^{-1}}$ and $\alpha=0.01$, the small-strain model gives $\sigma=2.0\,\mathrm{mN\,m^{-1}}$. These are teaching values chosen to demonstrate units.