eduKateSG · How Science Works
Choose your reading route
A cylindrical liquid thread can lower its surface energy by developing sufficiently long bulges and necks. Surface tension can then amplify those disturbances until the thread pinches into drops, with viscosity and inertia controlling the dynamics.
Contents
Understand the mechanism · Chapters 1–3
1. Surface tension can break a thread as well as round a drop
Work the calculation · Chapters 4–5
4. Compare two wavelengths with the same starting radius
5. Estimate a drop size without claiming to predict every drop
Test the explanation · Chapters 6–8
6. Viscosity and inertia decide how the shape changes
A smooth liquid jet leaving a nozzle can develop bulges, thin necks and separate droplets. Surface tension is often introduced as something that holds a liquid together, so its role in breakup may seem surprising. Its more precise tendency is to reduce interfacial area under the available constraints.
A long cylindrical thread is not always the lowest-area way to hold a fixed liquid volume. Some disturbances reduce its area, allowing surface tension to drive their growth. This is the Rayleigh–Plateau instability.
The question here is the capillary breakup of a liquid cylinder. Fast sprays can also involve aerodynamic forces, turbulence and other instabilities. Seeing droplets at the end does not identify which mechanism initiated the breakup.
Imagine a cylindrical thread of radius R over one repeating length L. Its volume is πR²L and its lateral area is 2πRL. We ignore ends by considering a periodic segment, so end caps do not contaminate the comparison.
Now add a small sinusoidal variation in radius: r(z) = a + ε cos(kz), where k = 2π/L and ε is much smaller than R. Averaging r² over one wavelength gives a² + ε²/2. To preserve volume, a² must equal R² − ε²/2.
To second order, a ≈ R − ε²/(4R). The average radius decreases slightly as bulges and necks develop. Omitting that adjustment would compare different volumes and could give the wrong surface-energy conclusion.
This is an idealised geometric construction, not a claim that a real nozzle launches a single perfect sine wave. It isolates one disturbance so we can ask whether it is initially favoured.
The surface area is A = 2π∫r√(1 + r′²) dz. For a small disturbance, expanding the square root and using the volume correction gives ΔA ≈ (πLε²/2R)[(kR)² − 1].
The sign is controlled by kR. If kR is less than one, this disturbance reduces area; if greater than one, it increases area. In wavelength language, the area-reducing condition is L greater than 2πR.
MIT’s surface-tension treatment develops the cylinder-stability argument. The assumptions matter: small perturbations, a cylindrical reference state and conserved volume. Other materials or boundary constraints need different analyses.
Area reduction identifies an energetic opportunity. To know how quickly it occurs, solve the fluid dynamics as well. An energy comparison alone does not supply a growth rate, a breakup time or a complete droplet-size distribution.
Original calculation. Let R = 0.50 mm. The threshold circumference 2πR is approximately 3.14 mm. A 2.0 mm disturbance has kR ≈ 1.57 and raises area at small amplitude. A 5.0 mm disturbance has kR ≈ 0.628 and lowers it.
For ε/R = 0.05, the fractional area change is approximately (ε²/4R²)[(kR)² − 1]. The 2.0 mm disturbance gives about +0.000917, or +0.0917%. The 5.0 mm disturbance gives about −0.000378, or −0.0378%.
These small changes are appropriate to a small-amplitude calculation. They should not be extrapolated until the radius at a neck reaches zero, because the expansion was never valid for such a large deformation.
A helpful reasoning check is to double R while leaving L fixed. The same wavelength can move from the area-reducing side to the area-increasing side. A wavelength is not unstable in isolation; its relation to the thread radius matters.
| Wavelength | kR for R = 0.50 mm | Small-disturbance area change |
|---|---|---|
| 2.0 mm | 1.57 | Increases |
| 5.0 mm | 0.628 | Decreases |
Suppose, solely for a volume account, that one 5.0 mm segment of our R = 0.50 mm thread becomes one spherical drop. Equate πR²L with 4πrd³/3. Then rd = (3R²L/4)1/3 ≈ 0.979 mm.
The resulting drop diameter is about 1.96 mm, nearly twice the original thread diameter. No liquid was created: the drop collected the volume spread along a much longer segment.
This construction does not say that the chosen wavelength grows fastest or that one clean drop forms per wavelength. Satellite drops, merging, imposed disturbances and variations in jet radius can alter the outcome.
Experiments and simulations of jets formed after droplet impact investigate how fluid properties and impact conditions affect secondary drops. Their geometry is more complicated than our static cylinder, which is exactly why the volume account should be distinguished from a full prediction.
A capillary–inertial time scale is tc = √(ρR³/γ), with liquid density ρ and surface tension γ. Using assigned values ρ = 1000 kg/m³, R = 0.00050 m and γ = 0.072 N/m gives about 1.32 milliseconds.
This is a characteristic scale, not the time from nozzle exit to breakup. The actual growth depends on wavelength, viscosity, surrounding fluid and initial disturbance amplitude. A disturbance that begins extremely small needs more amplification before a visible neck forms.
In a sufficiently viscous regime, μR/γ provides another useful time scale, where μ is dynamic viscosity. Comparing viscous and inertial effects helps determine which description fits rather than assuming every liquid behaves like water.
Research on falling viscous jets combined experiments and analysis of gravity-stretched threads. A long intact jet is compatible with a capillary instability if deformation and growth compete along its path.
Use time-resolved images to measure the radius profile, disturbance wavelength and amplitude before pinch-off. Record nozzle geometry, flow rate, temperature and fluid properties. A final droplet photograph has discarded much of the evidence needed to distinguish mechanisms.
If a small mode follows ε(t) = ε₀ exp(st), plotting ln ε against time should give a straight region of slope s while linear growth remains valid. This is a proposed analysis method, not a promise that every image sequence will show one constant rate.
A moving jet also requires a careful frame of reference. Downstream distance may approximate elapsed time for a material segment only when its transport speed is known and sufficiently uniform. A stretching jet makes that shortcut less reliable.
Work on a liquid jet in a coflowing stream distinguishes disturbances that grow while being transported from those that grow at a fixed location. Boundary conditions and the surrounding flow can change what an observer sees.
Does surface tension always stabilise a liquid shape? No. It favours area reduction, and an elongated cylinder can lower area through sufficiently long bulges and necks.
Does every wavelength longer than 2πR break up at the same speed? No. The threshold identifies the sign of the small-disturbance tendency in the ideal model; the dynamics determine different growth rates.
Can one sphere per wavelength predict satellite drops? No. It is a volume-conservation exercise with an assumed final arrangement, not a nonlinear pinch-off calculation.
Continue to Fluid Dynamics for forces and transport. The broader skill is to conserve what must be conserved before deciding whether a change of shape is favoured.