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

# Young-Dupré relation

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 right-angled 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.

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