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---
title: "Young-Dupré relation"
firstLetter: "Y"
publishDate: 2021-03-07
categories:
- Physics
- Fluid mechanics
date: 2021-03-07T15:05:50+01:00
draft: false
markup: pandoc
---
# 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](/know/concept/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.
## References
1. B. Lautrup,
*Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
CRC Press.
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