Why a thin stream grows into beads

Hold the volume fixed before comparing areas

A long cylinder can enclose its volume with less area by becoming a series of rounded bulges. To decide whether an initially tiny ripple starts that process, compare an entire wavelength of surface. Looking only at the narrowest point misses the axial curvature.

Let the original radius be $R_0$, with coordinate $z$ along the axis and dimensional wavenumber $q=2\pi/\lambda$. Adding a cosine alone increases the average cross-sectional area. A second-order shift in the mean radius removes that extra volume.

Sources: John W. M. Bush · Eggers & Dupont

Two curvatures decide which way liquid moves

Write the dimensionless wavenumber as $k=qR_0$. The area falls when $k<1$, or when the wavelength exceeds the circumference of the cylinder. This is the Rayleigh–Plateau threshold. Short ripples add too much axial curvature to lower the area.

The same argument appears locally as a pressure difference. A narrow neck curves more tightly around the axis, increasing one contribution to capillary pressure. The side profile supplies the other signed curvature. For a short ripple, that axial contribution can reverse the pressure difference.

Take a cylinder convex towards the gas as having positive curvature. In the dimensional expression below, subscripts denote derivatives, $\kappa$ has units of $\mathrm{m^{-1}}$, and the capillary pressure contribution is $\gamma\kappa$.

Sources: Eggers & Dupont

Evolve the radius and velocity together

Liquid leaving a neck must enter its neighbours. In the leading slender, axisymmetric approximation, axial velocity is nearly uniform across a section. Multiplying that velocity by cross-sectional area gives the volume flux. Radius squared is therefore a useful conserved variable.

Measure length in $R_0$ and time in the inertial capillary scale $t_c=\sqrt{\rho R_0^3/\gamma}$. Coordinates, radius and velocity in the following equations are dimensionless. Positive $u$ points along increasing axial coordinate. The Ohnesorge number $\mathrm{Oh}=\mu/\sqrt{\rho\gamma R_0}$ compares viscosity with inertial capillary effects.

Mass conservation relates radius change to incoming and outgoing flux. Axial momentum balances acceleration, capillary pressure gradients and viscous extensional stress. The factor of three comes from axisymmetric extension. We use the Eggers–Dupont equations with full curvature, zero gravity and periodic boundaries.

Sources: Eggers & Dupont

Extract the growth rate from the coupled equations

Set $r=1+\delta$ near a uniform cylinder and retain first-order terms. Conservation gives $\delta_t=-u_z/2$, and the curvature disturbance is $\delta\kappa=-\delta-\delta_{zz}$. Differentiate the first relation in time and insert the momentum equation to eliminate velocity.

For a disturbance proportional to $e^{st}\cos(kz)$, two spatial derivatives produce $-k^2$ and a time derivative produces $s$. The growth rate follows from a quadratic. One root is positive for $k<1$. More viscosity reduces that root without shifting the threshold.

This dispersion relation belongs to the one-dimensional slender model. The full inviscid three-dimensional Rayleigh theory contains modified Bessel functions and gives different rates at finite wavenumber. They agree in the long-wave limit. A numerical check must use the reference associated with the equations being solved.

Sources: Eggers & Dupont · John W. M. Bush

Worked example: long and short initial ripples

Take $\mathrm{Oh}=0.25$ and $k=0.65$, corresponding to $\lambda/R_0\simeq9.67$. The growing root is approximately $s=0.225$; the other is $-0.542$. The growing branch gains a factor of the natural exponential base in about $4.44t_c$.

An initially resting thread needs both roots: its initial radius derivative must vanish. Starting with cosine amplitude $0.0100R_0$, the combined linear solution gives $0.0145R_0$ at $t=3t_c$. Raising viscosity to $\mathrm{Oh}=1$ reduces the growing root to $0.0899$.

Now choose $k=1.3$. The right-hand side is negative. At $\mathrm{Oh}=0.25$, the roots are complex with real part about $-0.634$. The same initially resting ripple has amplitude about $0.00256R_0$ after $3t_c$. It has largely flattened.

Sources: Eggers & Dupont

Follow the neck beyond the linear shape

A growing disturbance eventually stops resembling a scaled cosine. Bulges round out, the neck narrows, and higher spatial harmonics appear. The first cosine amplitude extracted from radius records then leaves the linear reference. Total volume remains a useful check. Without external work, viscosity should reduce the sum of kinetic and surface energy.

The browser solver stores radius squared in cells and velocity between cells. Each cell gains its incoming flux and loses its outgoing flux. Liquid removed from a neck is therefore transferred to adjacent regions. Refining the grid checks whether the trajectory converges before the neck becomes unresolved.

Breakup creates new free surfaces and needs a continuation rule. This calculation stops when the neck approaches its resolution cutoff or the axial slope exceeds the specified range. Its last frame shows the final neck this calculation resolves. Compare different grids over their common resolved interval.

Sources: Eggers & Dupont