Several shapes for the same membrane

Choose a tractable family of surfaces

Choose a symmetry axis and measure polar angle $\theta$ from its north pole. The radial distance $r(\theta)$ defines the surface. Two low-order Legendre polynomials provide shape freedom: the second-order term mainly distinguishes elongated and flattened shapes, and the fourth-order term adjusts their contour. The exponential keeps the radius positive.

Scale $s$ controls area; $a$ and $b$ control shape. This family is smooth, axisymmetric and star-shaped about its center, so each radial ray crosses the membrane once. Indentations with multiple radial crossings, narrow-neck limits and asymmetric shapes lie outside this family. Its restricted elongated and flattened branches provide a controlled energy comparison.

r(\theta)=s\exp\!\left[aP_2(\cos\theta)+bP_4(\cos\theta)\right]

Here

P_2(x)=(3x^2-1)/2

and

P_4(x)=(35x^4-30x^2+3)/8

. Shape coefficients are dimensionless and

s

has length units.

Source: U. Seifert · W. Helfrich

Enforce area and volume before comparing energy

Temporarily set the overall scale to one and integrate area and volume over polar angle. Rescale the surface to the target area. Uniform scaling leaves reduced volume unchanged, so finding the requested reduced volume is now a problem in shape coefficients.

Fix $b$ and search separately for positive and negative $a$ . Bisection evaluates the interval midpoint, checks which side of the target reduced volume it lies on and retains the half that brackets the solution. Only after finding admissible shapes can we compare energies.

Minimize each branch over a finite range of fourth-order coefficients. The result is a constrained minimum within this family and search range. Finding the physical global minimum requires a larger shape space and checks for self-contact and instability.

A=2\pi\int_0^\pi r\sin\theta\sqrt{r^2+r_\theta^2}\,\mathrm d\theta,\qquad V=\frac{2\pi}{3}\int_0^\pi r^3\sin\theta\,\mathrm d\theta

Area sums the circumference times arclength of small rings; volume integrates radial sectors over solid angle.

s=\sqrt{\frac{A_{\rm target}}{A_{s=1}}},\qquad v(a,b)=v_{\rm target}

The first equation eliminates scale and the second removes one shape degree of freedom. Optimization follows the remaining constrained curve.

Worked example of a constraint residual

For $R_A=10\,\mu\mathrm m$ and $v=0.80$ , target area is $1.26\times10^3\,\mu\mathrm m^2$ and target volume is approximately $3.35\times10^3\,\mu\mathrm m^3$ . A relative volume residual of $10^{-4}$ means an error of about $0.335\,\mu\mathrm m^3$ .

First ask whether the candidate actually solves the requested problem. Then double angular quadrature resolution and compare energy, reduced volume and the Gaussian-curvature integral. Convergence supports attributing a persistent difference to the model instead of a coarse discretization.

The sphere supplies independent references: at zero spontaneous curvature, mean-curvature energy divided by $\kappa$ approaches $8\pi$ and integrated Gaussian curvature approaches $4\pi$ . These test local curvature and global closed-surface geometry.

Which scale is missing from a photograph?

Suppose only mean-curvature energy matters and area, volume and spontaneous curvature are specified. Multiplying $\kappa$ by any positive constant multiplies every candidate energy without moving the minimum. Even a perfect contour therefore cannot determine bending rigidity on its own.

Unknown tension, pressure and protein distributions can introduce further compensations. A fitted number is an answer within the chosen assumptions. Thermal fluctuations, a known applied force or an independent pressure measurement can add information absent from shape alone.

This explains why quite different material parameters may give almost indistinguishable contours. A useful inversion first asks which parameters affect the available observables, then chooses a complementary experiment. Thermal motion will provide an independent energy scale in the next part of the course.

F_b=\kappa\,\mathcal E[\text{shape}],\qquad \arg\min F_b=\arg\min\mathcal E\quad(\kappa>0)

This argument requires rigidity to appear as an overall positive factor. Additional independent energy scales require a new identifiability analysis.

Several shapes for the same membrane

Choose a tractable family of surfaces

Enforce area and volume before comparing energy

Worked example of a constraint residual

Which scale is missing from a photograph?

Try the reasoning

Compare two branches at the same constraints

Exercises and explanations

Sources and calculated scope