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Sandbox Physics
Dynamic interfaces

L1 · Chapter 1

Why surface tension pulls water into beads

Sandbox Physics ·

A drop becomes round because a sphere encloses its volume with the least area. A long water thread can save area by separating into drops. Surface tension lowers interfacial energy; whether that smooths a shape depends on its geometry.

Making more surface costs work

Imagine pulling a sliding wire across a liquid film. The wire creates new area, and your hand must do work. At fixed temperature and composition, the reversible work per added area is the surface tension. The same quantity is a force per unit length acting along the surface, normal to its boundary.

We write it as γ\gamma, measured in N m−1\mathrm{N\,m^{-1}} or equivalently J m−2\mathrm{J\,m^{-2}}. Count each interface: a soap film has two liquid–air faces, so increasing its projected area creates surface on both sides.

dE=γ dA\mathrm dE=\gamma\,\mathrm dA
The tension is uniform, with temperature and composition held fixed.

Sources: John W. M. Bush

Curvature turns tension into a pressure jump

Cut a spherical drop mentally into two hemispheres. The pressure difference pushes the halves apart across a disk. Surface tension pulls them together along the rim. Disk area grows as radius squared; rim length grows as radius. A smaller drop therefore needs a larger pressure difference.

Define Δp=pliquid−pgas\Delta p=p_{\mathrm{liquid}}-p_{\mathrm{gas}}, with a sphere convex towards the gas having positive curvature. A sphere bends in two perpendicular directions. A cylinder bends around its circumference but remains straight along its axis.

Δp πR2=2πRγ⟹Δp=2γR\Delta p\,\pi R^2=2\pi R\gamma\quad\Longrightarrow\quad\Delta p=\frac{2\gamma}{R}
A force balance on a hemisphere gives the spherical Laplace pressure.
Δp=γ(1R1+1R2)\Delta p=\gamma\left(\frac1{R_1}+\frac1{R_2}\right)
For a general surface, add the two signed principal curvatures.
Δpsphere=2γ/R\Delta p_{\rm sphere}=2\gamma/RΔpcylinder=γ/R\Delta p_{\rm cylinder}=\gamma/R
Equal radii, different curvature sums. Both surfaces are schematic; the cylinder is considered far from its ends.

Sources: John W. M. Bush

Worked example: the pressure inside a millimetre drop

Use approximate room-temperature teaching values for water: γ=0.072 N m−1\gamma=0.072\,\mathrm{N\,m^{-1}}, ρ=1000 kg m−3\rho=1000\,\mathrm{kg\,m^{-3}}, and radius R=1.00 mmR=1.00\,\mathrm{mm}. Convert the radius to metres before calculating. The excess pressure is 144 Pa144\,\mathrm{Pa}. This is a local pressure difference, not a net external force on the drop.

Gravity changes the pressure over this height by roughly ρgR=9.81 Pa\rho gR=9.81\,\mathrm{Pa}. Compare gravitational and capillary pressure scales through the Bond number. Here it is 0.1360.136, consistent with a fairly rounded drop. Increasing the radius tenfold increases this ratio a hundredfold.

Bo=ρgR2γ=(Rℓc)2,ℓc=γρg≃2.71 mm\mathrm{Bo}=\frac{\rho gR^2}{\gamma}=\left(\frac{R}{\ell_c}\right)^2,\qquad \ell_c=\sqrt{\frac{\gamma}{\rho g}}\simeq2.71\,\mathrm{mm}
Gravity changes the shape gradually; the capillary length marks comparable scales of gravity and surface tension.

A solid surface also helps choose the contact angle

A drop on a solid has three interfaces: solid–liquid, solid–gas and liquid–gas. Moving the contact line replaces some solid–gas area with solid–liquid area and changes the free surface. Minimizing their combined energy gives the equilibrium angle, measured inside the liquid.

Young’s relation assumes a smooth, chemically uniform solid in equilibrium. Roughness and chemical variation can pin the line, allowing a range between advancing and receding angles. Calculating a moving contact line requires a model for the region near the solid.

γSG−γSL=γcos⁡θe\gamma_{\mathrm{SG}}-\gamma_{\mathrm{SL}}=\gamma\cos\theta_e
The subscripts denote solid–gas and solid–liquid interfacial energies.

Sources: John W. M. Bush · Oron

Predict, change and explain

Compare curvature and wetting in capillary tubes

  1. Predict the height change when the tube radius halves, then compare two tubes.
  2. Keep the liquid fixed and switch advancing and receding angles. Identify the parameter combination fixed by height, then state the additional measurement needed to determine tension separately.

Exercises and explanations

  1. A spherical drop and a long cylindrical thread have the same radius. Which has the larger pressure jump?

    Read the explanation

    The sphere has twice the jump: it curves in two principal directions, whereas the cylinder curves in one. Observe the cylinder far from its ends.

  2. Reduce the example radius to 0.10 mm0.10\,\mathrm{mm}. What happens to the pressure jump and Bond number?

    Read the explanation

    The jump becomes 1440 Pa1440\,\mathrm{Pa} and the Bond number becomes 0.001360.00136. Smaller drops have larger excess pressure and weaker relative gravitational deformation.

Sources and calculation scope

Examples and diagrams are newly authored. The paper comparison recalculates analytic relations; the research reading distinguishes those curves from experimental photographs and molecular simulations.

  1. John W. M. Bush, Surface Tension Module, MIT lecture notes

    Lecture 1, sections 1.1–1.3: interfacial energy, normal and tangential stress, and capillary scaling. Teaching source; the examples here are newly calculated.

  2. Oron, Davis & Bankoff, Long-scale evolution of thin liquid films, Rev. Mod. Phys. 69, 931 (1997)

    Published review and framework for long-wave reductions. Specific implemented equations are cross-checked against the accessible author papers below.

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