Categories: Fluid mechanics, Fluid statics, Physics, Surface tension.

In fluid mechanics, the **Young-Dupré relation** relates the contact angle of a droplet at rest on a surface to the surface tensions of the interfaces. Let \(\alpha_{gl}\), \(\alpha_{sl}\) and \(\alpha_{sg}\) respectively be the energy costs of the liquid-gas, solid-liquid and solid-gas interfaces:

\[\begin{aligned} \boxed{ \alpha_{sg} - \alpha_{sl} = \alpha_{gl} \cos\theta } \end{aligned}\]

The derivation is simple: this is the only expression that maintains the droplet’s boundaries when you account for the surface tension force pulling along each interface.

A more general derivation is possible by using the calculus of variations. In 2D, the upper surface of the droplet is denoted by \(y(x)\). Consider the following Lagrangian \(\mathcal{L}\), with the two first terms respectively being the energy costs of the top and bottom surfaces:

\[\begin{aligned} \mathcal{L} = \alpha_{gl} \sqrt{1 + (y')^2} + (\alpha_{sl} - \alpha_{sg}) + \lambda y \end{aligned}\]

And the last term comes from the constraint that the volume \(V\) of the droplet must be constant:

\[\begin{aligned} V = \int_0^L y \dd{x} \end{aligned}\]

The total energy to be minimized is thus given by the following functional, where the endpoints of the droplet are \(x = 0\) and \(x = L\):

\[\begin{aligned} E[y(x)] = \int_0^L \Big( \alpha_{gl} \sqrt{1 + (y')^2} + (\alpha_{sl} - \alpha_{sg}) + \lambda y \Big) \dd{x} \end{aligned}\]

In this optimization problem, the endpoint \(L\) is a free parameter, i.e. the \(L\)-value of the optimum is unknown and must be found. In such cases, the optimum \(y(x)\) needs to satisfy the so-called *transversality condition* at the variable endpoint, in this case \(x = L\):

\[\begin{aligned} 0 &= \Big( \mathcal{L} - y' \pdv{\mathcal{L}}{y'} \Big)_{x = L} \\ &= \bigg( \alpha_{gl} \sqrt{1 + (y')^2} + (\alpha_{sl} - \alpha_{sg}) + \lambda y - \frac{(y')^2}{\sqrt{1 + (y')^2}} \bigg)_{x = L} \\ &= \bigg( \alpha_{gl} \frac{1}{\sqrt{1 + (y')^2}} + (\alpha_{sl} - \alpha_{sg}) + \lambda y \bigg)_{x = L} \end{aligned}\]

Due to the droplet’s shape, we have the boundary condition \(y(L) = 0\), so the last term vanishes. We are thus left with the following equation:

\[\begin{aligned} \alpha_{gl} \frac{1}{\sqrt{1 + (y'(L))^2}} = \alpha_{sg} - \alpha_{sl} \end{aligned}\]

At the edge of the droplet, imagine a small rectangular triangle with one side \(\dd{x}\) on the \(x\)-axis, the hypotenuse on \(y(x)\) having length \(\dd{x} \sqrt{1 + (y')^2}\), and the corner between them being the contact point with angle \(\theta\). Then, from the definition of the cosine:

\[\begin{aligned} \cos\theta = \frac{\dd{x}}{\dd{x} \sqrt{1 + (y'(L))^2}} = \frac{1}{\sqrt{1 + (y'(L))^2}} \end{aligned}\]

When inserted into the above transversality condition, this yields the Young-Dupré relation.

- B. Lautrup,
*Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition, CRC Press.

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