11 May 2025
Surface Tension, Contact Angle and Virtual Work
A molecule well inside a liquid is attracted in all directions. A molecule near the surface has fewer neighbours on the vapour side, so work must be done to create new surface.
If reversible work $dW$ creates an area $dA$ at fixed temperature,
\[\boxed{\gamma=\frac{dW}{dA}},\]where $\gamma$ is the surface energy per unit area. Since
\[[\gamma]=\mathrm{J\,m^{-2}}=\mathrm{N\,m^{-1}},\]the same quantity is also the tangential force per unit length across a line drawn in the surface. This force description is surface tension.
Principle of virtual work
A virtual displacement is an infinitesimal change compatible with the constraints. In quasistatic equilibrium, the first-order work supplied by the applied forces equals the corresponding increase of potential or surface energy:
\[\boxed{\delta W_{\rm applied}-\delta U=0}.\]Consider a rectangular frame carrying a soap film, with a movable side of length $L$. If the side is displaced through $\delta x$, the two faces of the film gain area
\[\delta A=2L\,\delta x.\]The surface-energy increase is $\delta U=2\gamma L\,\delta x$. Virtual work gives
\[F\,\delta x=2\gamma L\,\delta x,\]so
\[\boxed{F=2\gamma L}.\]
Excess pressure across a curved surface
For one interface whose principal sections bulge toward the lower-pressure phase, take $R_1$ and $R_2$ as their positive radii of curvature. The Young-Laplace relation gives the pressure excess on the opposite, higher-pressure side:
\[\boxed{\Delta p=p_{\rm high}-p_{\rm low} =\gamma\left(\frac{1}{R_1}+\frac{1}{R_2}\right)}.\]The spherical result follows directly from virtual work. Let a liquid drop of radius $R$ expand virtually by $dR$. Then
\[dV=4\pi R^2\,dR, \qquad dA=8\pi R\,dR.\]The pressure work creates surface energy:
\[\Delta p\,dV=\gamma\,dA.\]Therefore
\[\Delta p(4\pi R^2\,dR) =\gamma(8\pi R\,dR),\]and for a liquid drop or an air bubble in a liquid, each of which has one interface,
\[\boxed{\Delta p=\frac{2\gamma}{R}}.\]A soap bubble has an inner and an outer interface. Its surface-energy change is $2\gamma\,dA$, so
\[\boxed{\Delta p=\frac{4\gamma}{R}}.\]For a cylindrical interface of radius $R$ and fixed length $L$,
\[dV=2\pi RL\,dR, \qquad dA=2\pi L\,dR.\]Virtual work then gives
\[\boxed{\Delta p=\frac{\gamma}{R}},\]in agreement with the general relation because the two principal radii are $R$ and $\infty$.
Angle of contact
At the line where solid, liquid and vapour meet, the angle of contact $\theta$ is measured through the liquid between the tangent to the liquid-vapour surface and the solid surface.
Let $\gamma_{SV}$, $\gamma_{SL}$ and $\gamma_{LV}$ denote the solid-vapour, solid-liquid and liquid-vapour interfacial tensions. A virtual advance $dx$ of the contact line replaces solid-vapour area by solid-liquid area. Per unit length of the contact line, the surface-energy change is
\[dU=(\gamma_{SL}-\gamma_{SV})\,dx +\gamma_{LV}\cos\theta\,dx.\]At equilibrium $dU=0$. Hence Young’s equation is
\[\boxed{\gamma_{SV}-\gamma_{SL} =\gamma_{LV}\cos\theta}.\]Thus $\theta<90^\circ$ corresponds to preferential wetting of the solid, while $\theta>90^\circ$ corresponds to non-wetting.
The editable three-part diagram is available as a TikZ file. The film, curved-surface and contact-angle balances are checked in the Maxima worksheet; every displayed residual is zero.
Discussion