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A lipid vesicle is a closed lipid bilayer enclosing fluid. Its shape and thermal membrane undulations depend on membrane tension, membrane bending modulus, and constraints such as fixed enclosed volume.

For small-slope height fluctuations of a nearly planar membrane, $A$ is projected area, $\sigma$ is membrane tension, $\kappa$ is membrane bending modulus, and $q$ is a wavenumber. With a Fourier convention in which the mode amplitude $h_{\mathbf{q}}$ has units of length, the quadratic Helfrich membrane energy and the equipartition theorem give the displayed thermal variance. Bending dominates for $q\gg \sqrt{\sigma /\kappa }$, giving a $q^{-4}$ spectrum. Tension dominates at smaller $q$, giving $q^{-2}$. The corresponding membrane elastic length is $\sqrt{\kappa /\sigma }$ when length is measured as $1/q$; a wavelength convention multiplies it by $2\pi$.

Take a long periodic section of length $L$, or neglect end effects and choose an integer number of wavelengths. The amplitude $u_q$ here is a single real sine amplitude, fixing the normalization of the thermal membrane undulation spectrum. For a lipid vesicle with an axisymmetric membrane deformation, conservation of the enclosed volume givesThe mean radius therefore decreases at second order. This adjustment is essential: fixing the mean radius instead of the volume would miss the unstable term in the fixed-volume capillary spectrum of a cylindrical membrane.

The surface area of an axisymmetric graph is $\mathcal{S}=2\pi \int R\sqrt{1+R_z^2}dz$. Expanding for $|u_q|\ll R_0$ and $q|u_q|\ll 1$ givesMultiplication by the membrane tension yields a quadratic potential energy $\Delta E=K_q{u}_q^2/2$, whereFor $q{R}_0>1$, the equipartition theorem gives the stable-mode fixed-volume capillary spectrum of a cylindrical membraneHere $k_B$ is the Boltzmann constant and $T$ is the temperature. The cosine amplitude has the same variance and is an independent real coordinate at quadratic order. If instead $R-{\rho }_0={\sum }_{q\ne 0}{a}_q{e}^{iqz}$ with $a_{-q}=a_q^{\ast }$, the corresponding positive-$q$ complex coefficient has $⟨|a_q{|}^2⟩=k_{B}T{R}_0/[2\pi \sigma L((q{R}_0{)}^2-1)]$. This is the same thermal membrane undulation spectrum in a different Fourier mode normalization.

For $q{R}_0<1$, the cylinder is unstable rather than having a negative fluctuation variance. The quadratic surface area change is negative: a sufficiently long-wavelength modulation reduces area while retaining volume. This is the Rayleigh–Plateau instability, expressed here as pearling of a tension-dominated lipid vesicle. A cylinder supporting such modes has no unconstrained harmonic thermal equilibrium about the uniform state. The $q=0$ uniform radius change is forbidden by fixed volume, and $q{R}_0=1$ is marginal in the tension-only approximation. The finite length and endpoint constraints determine which nonzero Fourier modes are allowed.

The large membrane tension assumption controls modes with $q{R}_0$ of order one. For large $q$ the omitted membrane bending modulus contribution grows approximately as $\kappa q^4$ and overtakes the tension term when $\kappa q^2\gtrsim \sigma$. Thus the displayed tension-only spectrum requires $q\ll \sqrt{\sigma /\kappa }$ as well as small amplitudes and a positive stiffness. Close enough to the marginal wavelength, bending corrections must also be retained; their relative scale at $q{R}_0$ of order one is $\kappa /(\sigma R_0^2)$.

For the circular phase boundary in a lipid bilayer, the energy is its perimeter times the line tension $\gamma$. Let $\phi$ be the polar angle and write $r(\phi )={\rho }_0+u_m\sin (m\phi )$, where the periodicity requires an integer $m=q{R}_0\ge 1$. Fixed enclosed area impliesThe perimeter, including the mean-radius change, isThus the fixed-area capillary spectrum of a circular boundary, again for a single real sine or cosine amplitude, isThe complex Fourier mode coefficient convention again divides this result by two. The mode $m=0$ is excluded by the fixed-area constraint, so the negative-stiffness interval $0<m<1$ is not an allowed mode of a closed circular boundary.

At $q{R}_0=1$, the zero stiffness is the translation mode of a circular boundary, not a shape instability. Displacing the centre by a small distance $h$ changes the radius to first order by $h\cos \phi$ or $h\sin \phi$ while leaving area and perimeter unchanged. For example, an exactly translated circle hasThe higher harmonics complete a true translation mode of a circular boundary. The $m=1$ denominator therefore represents free centre motion: an unconfined domain's centre has no restoring force or finite equilibrium variance in an infinite membrane. Radius fluctuations about a recentered domain omit this mode; external confinement would give it a separate restoring stiffness.

Write the membrane height above its mean plane as a sum of Fourier modes, $h(\mathbf{r})={\sum }_{\mathbf{q}}{h}_{\mathbf{q}}{e}^{i\mathbf{q}\cdot \mathbf{r}}$. The supplied average is a thermal mean-squared modal height; for complex Fourier coefficients it is understood as $⟨|h_{\mathbf{q}}{|}^2⟩$, with $h_{-\mathbf{q}}=h_{\mathbf{q}}^{\ast }$ for a real height field. The terms are:

In the small-slope approximation, Helfrich membrane energy containsThe equipartition theorem, with the independent real modes counted consistently, gives the stated membrane fluctuation spectrum under tension. Its units are length squared: $\sigma q^2$ and $\kappa q^4$ both have units $\mathrm{J}{\mathrm{m}}^{-4}$, so $k_{B}T/[A(\sigma q^2+\kappa q^4)]$ has units ${\mathrm{m}}^2$.

At small length scales, $q$ is large and the bending term dominates:The fluctuation amplitude is controlled mainly by the membrane bending modulus and thermal energy, rather than tension. The crossover occurs when $\sigma q_c^2=\kappa q_c^4$, henceThe bending-dominated regime extends up to the membrane elastic length when scale means $1/q$, or up to ${\lambda }_c$ when it means wavelength. At still larger scales tension gives a $q^{-2}$ spectrum. At the opposite end, the continuum formula requires wavelengths larger than the microscopic membrane thickness or molecular cutoff. Without numerical $\kappa$ and $\sigma$, no numerical crossover scale is determined.