--- title: "How a soft membrane holds a shape" description: "Use a deflated vesicle to connect curvature, area and volume, calculate reduced volume and distinguish osmotic pressure from tension." canonical_url: "https://sandboxphysics.com/stories/membrane-mechanics/curvature-and-constraints/" language: "en" page_type: "story" alternate_languages: zh-CN: "https://sandboxphysics.com/zh/stories/membrane-mechanics/curvature-and-constraints/" last_modified: "2026-10-28" --- [Membrane mechanics](https://sandboxphysics.com/stories/membrane-mechanics/) L1 · Chapter 1 # How a soft membrane holds a shape Sandbox Physics · 2026-10-28 A vesicle can hold a shape through bending elasticity and pressure without a rigid skeleton. Its lipid bilayer resists stretching, and the water inside occupies a volume. The amount of water enclosed by a given membrane area determines how much room the surface has to deform. ## Before you begin The area and volume of a sphere, the radius of a circle and elementary algebra. Curvature is introduced geometrically. Later chapters explain derivatives and integrals when they are first needed. - Calculate the area radius and reduced volume, then predict how deflation changes geometric freedom. - Explain the two principal curvatures using a sphere, cylinder and saddle. ## Deflation gives the surface room to move Picture a small bag full of water. The water barely compresses and the bag material barely stretches. Seal it and both volume and surface area are nearly fixed. A lipid vesicle shares these restrictions, but its surface is a fluid bilayer whose thickness is much smaller than the vesicle radius. For a fixed area, a sphere encloses the largest volume. At that maximum, a closed surface without self-intersection must be spherical. Remove water and the same membrane has area to spare. It can elongate, flatten or develop an indentation. Comparing bending energies will tell us which allowed shape it prefers. $$ R_A=\sqrt{\frac{A}{4\pi}},\qquad v=\frac{V}{(4\pi/3)R_A^3}=\frac{6\sqrt\pi V}{A^{3/2}} $$ The area radius $R_A$ belongs to a sphere with the same area. Reduced volume $v$ divides the actual volume by that sphere’s volume. Area has units of length squared and volume of length cubed, so the ratio is dimensionless. ## A surface bends in two directions At a point on the membrane, cut through the surface in different directions. Two perpendicular cuts give its greatest and least signed curvatures, called principal curvatures. A circle has curvature equal to its inverse radius: a smaller circle turns more sharply. A surface needs two values because it has a second direction in which to bend. We choose the outward normal and a convention in which both sphere curvatures are positive. A sphere of radius $R$ has $c_1=c_2=1/R$. A cylinder is straight along its axis and bends around its circumference. A saddle bends outward in one direction and inward in the other, so its principal curvatures have opposite signs. Mean curvature averages these two bends; Gaussian curvature multiplies them. The first enters the bending energy in the next chapter. The second becomes useful when we hold topology fixed. Here topology describes whether a closed surface has holes and whether it remains one connected surface. $$ 2H=c_1+c_2,\qquad K=c_1c_2 $$ Mean and principal curvatures have units of inverse length; Gaussian curvature has units of inverse length squared. Some papers reverse the curvature sign. Convert spontaneous curvature along with that convention. A sphere bends in two directions, a cylinder in one and a saddle in opposite directions. Original geometric schematic; no measured data. ## Worked example with one fifth of the water removed Take $R_A=10\,\mu\mathrm m$. The area is $A=400\pi\,\mu\mathrm m^2$ and the largest enclosed volume is $4000\pi/3\,\mu\mathrm m^3$, approximately $4.19\times10^3\,\mu\mathrm m^3$. For an actual volume of $3.35\times10^3\,\mu\mathrm m^3$, dividing by the maximum gives $v\simeq0.800$. A nonspherical shape is geometrically possible. This number alone cannot choose an elongated or flattened shape; both may satisfy the same area and volume. Double every length. Area increases fourfold and volume eightfold, leaving reduced volume unchanged. Reduced volume separates size from shape, which will simplify the energy comparison. ## Why water crosses the membrane When a membrane passes water more readily than solute, different solute concentrations drive water flow. A dilute ideal-particle model estimates the osmotic pressure. The eventual water content of a closed vesicle depends on the external solution, the number of trapped solute particles and mechanical pressure. A particular experiment must specify temperature, concentrations and permeability. For a spherical surface with uniform tension, temporarily neglect bending forces. The outward pressure difference $\Delta p=p_{\rm in}-p_{\rm out}$ then obeys the Laplace relation. It checks dimensions and signs; a general nonspherical membrane also requires local bending forces. $$ \Pi\simeq k_BT n,\qquad \Delta p=\frac{2\sigma}{R} $$ Number density $n$ counts solute particles per volume and $k_BT$ is thermal energy. Tension has units of force per length and pressure of force per area. The first expression assumes a dilute ideal solution; the second assumes the stated spherical limit. Source: W. Helfrich · U. Seifert ## Try the reasoning ### [Remove water at fixed membrane area](https://sandboxphysics.com/labs/vesicle-shapes/) 1. Predict how the elongated and flattened branches approach each other when reduced volume approaches one. 2. Lower only reduced volume. Inspect both branches and their area and volume residuals. 3. At the same reduced volume, change spontaneous curvature. Explain why equal area and volume can still favor different shapes. ## Exercises and explanations 1. Does scaling every length by three change reduced volume? Read the explanation No. Volume increases by twenty-seven and the area raised to the three-halves power increases by the same factor. Reduced volume describes how full the enclosure is relative to its membrane area. 2. At $R_A=10\,\mu\mathrm m$, reduce the water content to half the maximum volume. Calculate reduced volume. Must the vesicle be flat? Read the explanation It has $v=0.5$. Area and volume restrict allowed shapes; energy, topology and contact conditions select among them. This value also lies outside the sliders of our finite-family shape experiment. 3. Do a sphere and cylinder with equal mean curvature have equal radii? Read the explanation No. The sphere bends in two directions and the cylinder in one. At equal mean curvature, the cylinder radius is half the sphere radius. [Next: What it costs to bend a membrane](https://sandboxphysics.com/stories/membrane-mechanics/bending-energy/) ## Sources and calculated scope 1. [W. Helfrich, Elastic Properties of Lipid Bilayers: Theory and Possible Experiments, Z. Naturforsch. C 28, 693–703 (1973)](https://doi.org/10.1515/znc-1973-11-1209) Original curvature-elasticity paper, pages 693–703. We use its local quadratic curvature energy with an explicitly stated sphere-positive curvature convention. The sphere, tube and small-slope examples are derived here; no original figure or measured material parameter is copied. 2. [U. Seifert, K. Berndl and R. Lipowsky, Shape transformations of vesicles, Physical Review A 44, 1182–1202 (1991)](https://doi.org/10.1103/PhysRevA.44.1182) Published paper: spontaneous-curvature and bilayer-coupling ensembles and axisymmetric shape branches. Our shape experiment minimizes within a finite radial variational family at fixed area and volume. It does not reproduce the complete published phase diagram, neck limits or stomatocytes.