From 922a0bbeb81f9a0297c6a728d243cbec75cf9c3b Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Mon, 29 Mar 2021 09:15:42 +0200
Subject: Expand knowledge base, move WKB approximation

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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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