Why Water Streams Break into Droplets — Plateau–Rayleigh Instability and Breakup Regimes

Open the kitchen faucet halfway. The first handspan is a smooth column. About 10 cm down it pinches into regular drops. You see it every day. Why a continuous liquid cuts itself is not obvious.

Plateau, in 1849, kept only the wavelengths that shrink the surface. Rayleigh, in 1879, picked the one that grows fastest from the dispersion relation. Engine nozzles and inkjets still score the result with the Weber number (inertia over surface tension) and the Ohnesorge number (viscosity over inertia and surface tension). Some jets stay fat drops. Some become mist.

Move the wavenumber $kR$ slider and the wavelength seeded on the column moves with the point on the growth-rate curve.

Increasing $kR$ from 0.2 to 0.7 speeds the growth, so the column pinches sooner, and passing $kR = 1.1$ pins the curve to the floor so the column stays smooth. Raising Ohnesorge from 0 to 0.4 only slows the growth, while the unstable $kR$ range does not move.

Why a Continuum Chooses to Fragment#

Surface tension is a force that seeks to minimize surface area. For a fixed volume of liquid, a spherical shape has the smallest surface area. This is why a mass of water forms a sphere in zero gravity.

Now, consider a long liquid cylinder versus a chain of droplets created by cutting that cylinder at regular intervals. Which has a smaller surface area? Counter-intuitively, the chain of droplets has a smaller area—but only when the cutting interval is longer than the cylinder's circumference. This is why a water column fragments; it is a natural result of falling into a lower energy state.

Plateau's Geometric Intuition#

Plateau calculated the change in surface area for a cylinder of radius $R$ when subjected to a weak axial perturbation of wavelength $\lambda$.

Here, $2\pi R/\lambda$ is the dimensionless wavenumber $kR$ (the product of wavenumber $k = 2\pi/\lambda$ and the radius). If this value is less than 1, $\Delta A < 0$, meaning the surface area decreases as the perturbation grows, which is energetically favorable. Conversely, if it is greater than 1, $\Delta A > 0$, and the perturbation subsides.

Rewriting the condition is simple:

Only wavelengths longer than the cylinder's circumference grow. This criterion is known as the Plateau limit.

Rayleigh's Linear Stability#

Plateau only answered "which wavelengths can grow." Thirty years later, Rayleigh identified "which wavelength grows the fastest" by inserting a small perturbation $\eta = \epsilon e^{\sigma t} \cos(kx)$ into the inviscid Navier–Stokes equations to find the dispersion relation.

Where $\sigma$ is the growth rate (unit $s^{-1}$), $\sigma_{s}$ is the surface tension (N/m), $\rho$ is the liquid density, and $I_{0}, I_{1}$ are modified Bessel functions. The term $1-(kR)^{2}$ in parentheses is exactly the Plateau limit.

The maximum value occurs at $kR \approx 0.697$, corresponding to a wavelength of $\lambda_{\max} \approx 9.02 R$. A water column prefers to be cut at a period of about 9 times its radius. This period explains the spacing of actual faucet droplets almost perfectly.

Including viscosity, as organized by Chandrasekhar, reduces $\sigma$, but the instability boundary $kR < 1$ remains unchanged. Viscosity only slows down the growth; it does not affect the necessary conditions for fragmentation.

Weber and Ohnesorge: The Map of Breakup Regimes#

While the above explanation suffices for slow flows like a faucet, high-speed jets in rocket combustors or diesel injectors involve the inertia of the surrounding gas. Two dimensionless numbers emerge here:

$We$ is the ratio of gas inertia to surface tension, and $Oh$ is the ratio of liquid viscosity to (inertia and surface tension). Summarizing experimental data, a jet passes through four regimes as its velocity increases.

If $Oh$ is small ($< 0.1$), the regime is determined by $We$ alone. As $Oh$ increases, viscosity corrections are needed, and in practice, boundaries are drawn using composite numbers like $Oh(We)^{0.5}$. Faucets and inkjets usually fall within the Rayleigh breakup region. Diesel injectors target Atomization, raising $We$ above $10^{5}$. Once the jet has already become drops, the gas tearing those drops apart is the subject of Weber number and Taylor analogy breakup. The $We$ here picks the regime. That note follows one drop as it oscillates.

One wavelength of volume sets the drop diameter#

The fastest wavelength is $\lambda_{\max} \approx 9.02 R$. One pinched segment has the cylinder volume $\pi R^{2}\lambda$, so $9.02\pi R^{3}$. The sphere of the same volume satisfies $\frac{4}{3}\pi a^{3} = 9.02\pi R^{3}$, hence $a^{3}/R^{3} = 9.02 \times 3/4 = 6.765$. The cube root is about 1.89. Drop diameter sits near 1.9 jet diameters. That is why the Rayleigh row is the only one that says the drop is thicker than the jet.

The wavelength where instability opens is the circumference, and only $\lambda > 2\pi R$ grows, about $6.28 R$. Then $9.02/6.28 \approx 1.44$. The fastest period is only 44% longer than the shortest unstable wave. Feed the nozzle a shorter wrinkle and the column stays smooth.

The leading factor in $\sigma^{2}$ is $\sigma_{s}/(\rho R^{3})$. The factor that multiplies $kR$ depends on wavenumber alone, so peak $\sigma$ falls as $R^{-3/2}$. The listing uses a 1 mm radius, $\sigma \approx 380\,\mathrm{s}^{-1}$, and about 18 ms. At 1 m/s the pinch is 1.8 cm down the jet. That is not the 10–15 cm under the sink. A thicker column stretches the same formula in time.

Finding the Fastest Wavelength with Python#

We can numerically search for the optimal wavenumber by translating Rayleigh's dispersion relation using Bessel functions. We use i0 and i1 from scipy.

import numpy as np
from scipy.special import i0, i1
from scipy.optimize import minimize_scalar
 
def rayleigh_sigma(kR, sigma_s=0.072, rho=1000.0, R=1e-3):
    """Inviscid Plateau-Rayleigh growth rate (s^-1)."""
    if kR <= 0 or kR >= 1:
        return 0.0
    coeff = sigma_s / (rho * R**3)
    return np.sqrt(coeff * kR * (1 - kR**2) * i1(kR) / i0(kR))
 
def fastest_mode(R, sigma_s=0.072, rho=1000.0):
    """Wavenumber and corresponding wavelength that grow the fastest."""
    res = minimize_scalar(
        lambda x: -rayleigh_sigma(x, sigma_s, rho, R),
        bounds=(0.01, 0.99), method='bounded'
    )
    kR_star = res.x
    wavelength = 2 * np.pi * R / kR_star
    sigma = rayleigh_sigma(kR_star, sigma_s, rho, R)
    return kR_star, wavelength, sigma
 
if __name__ == "__main__":
    R = 1e-3          # 1 mm radius water column
    kR_star, lam, sig = fastest_mode(R)
    t_break = np.log(R / 1e-6) / sig   # Assume ε_0 = 1 μm
    print(f"kR* = {kR_star:.3f}")
    print(f"lambda_max = {lam*1e3:.2f} mm  (= {lam/R:.2f} R)")
    print(f"sigma_max = {sig:.1f} /s")
    print(f"breakup time ~ {t_break*1e3:.1f} ms")

The output shows kR* ≈ 0.697, λ_max ≈ 9.02 R, σ_max ≈ 380 /s, and a breakup time of approximately 18 ms. Multiplying by $U = 1\,\text{m/s}$ gives 1.8 cm, not a check against the 10–15 cm faucet. The 18 ms assumes a 1 μm seed, a thousandfold growth on a 1 mm radius. Divide $\ln 1000 \approx 6.9$ by 380 and the same 18 ms comes back.

Using the 1 mm, 18 ms case as a faucet height misses the pinch#

1.8 cm and 10 cm are not the same column. Because $\sigma$ scales as $R^{-3/2}$, freezing 18 ms before measuring the radius under-predicts the breakup length. $\lambda_{\max} \approx 9 R$ chooses the wavelength. It is not a universal ruler for fall distance.

Leave the jet in Rayleigh and hold the wavelength near that 9 radii if the drops should stay fat. For mist, push $We$ above $10^{5}$ into atomization. How far viscosity tilts that boundary is hard to read from $Oh$ alone. The scale that splits inertia from viscosity is how the Reynolds number changes the flow.