From 6f560e87fa5cd10852e073adbedfc3d46e5da8bb Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Sat, 13 Mar 2021 21:50:05 +0100
Subject: Expand knowledge base

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+---
+title: "Young-Laplace law"
+firstLetter: "Y"
+publishDate: 2021-03-11
+categories:
+- Physics
+- Fluid mechanics
+
+date: 2021-03-07T14:54:41+01:00
+draft: false
+markup: pandoc
+---
+
+# Young-Laplace law
+
+In liquids, the **Young-Laplace law** governs surface tension:
+it describes the tension forces on a surface
+as a pressure difference between the two sides of the liquid.
+
+Consider a small rectangle on the surface with sides $\dd{\ell_1}$ and $\dd{\ell_2}$,
+orientated such that the sides are parallel to the (orthogonal)
+principal directions of the surface' [curvature](/know/concept/curvature/).
+
+Surface tension then pulls at the sides with a force
+of magnitude $\alpha \dd{\ell_2}$ and $\alpha \dd{\ell_2}$,
+where $\alpha$ is the energy cost per unit of area,
+which is the same as the force per unit of distance.
+However, due to the surface' curvature,
+those forces are not quite in the same plane as the rectangle.
+
+Along both principal directions,
+if we treat this portion of the surface as a small arc of a circle
+with a radius equal to the principal radius of curvature $R_1$ or $R_2$,
+then the tension forces are at angles $\theta_1$ and $\theta_2$
+calculated from the arc length:
+
+$$\begin{aligned}
+    \theta_1 R_1
+    = \frac{1}{2} \dd{\ell_2}
+    \qquad \qquad
+    \theta_2 R_2
+    = \frac{1}{2} \dd{\ell_1}
+\end{aligned}$$
+
+Pay attention to the indices $1$ and $2$:
+to get the angle of the force pulling at $\dd{\ell_1}$,
+we need to treat $\dd{\ell_2} / 2$ as an arc,
+and vice versa.
+
+Since the forces are not quite in the plane,
+they have a small component acting *perpendicular* to the surface,
+with the following magnitudes $\dd{F_1}$ and $\dd{F_2}$
+along the principal axes:
+
+$$\begin{aligned}
+    \dd{F_1}
+    &= 2 \alpha \dd{\ell_1} \sin\theta_1
+    \approx 2 \alpha \dd{\ell_1} \theta_1
+    = \alpha \dd{\ell_1} \frac{\dd{\ell_2}}{R_1}
+    = \frac{\alpha}{R_1} \dd{A}
+    \\
+    \dd{F_2}
+    &= 2 \alpha \dd{\ell_2} \sin\theta_2
+    \approx 2 \alpha \dd{\ell_2} \theta_2
+    = \alpha \dd{\ell_2} \frac{\dd{\ell_1}}{R_2}
+    = \frac{\alpha}{R_2} \dd{A}
+\end{aligned}$$
+
+The initial factor of $2$ is there since
+the same force is pulling at opposide sides of the rectangle.
+We end up with $\alpha / R_{1,2}$ multiplied by
+the surface area $\dd{A} = \dd{\ell_1} \dd{\ell_2}$ of the rectangle.
+
+Adding together $\dd{F_1}$ and $\dd{F_2}$ and
+dividing out $\dd{A}$ gives us the force-per-area (i.e. the pressure)
+added by surface tension,
+which is given by the **Young-Laplace law**:
+
+$$\begin{aligned}
+    \boxed{
+        \Delta p
+        = \alpha \Big( \frac{1}{R_1} + \frac{1}{R_2} \Big)
+    }
+\end{aligned}$$
+
+The total excess pressure $\Delta p$ is called the **Laplace pressure**,
+and fully determines the effects of surface tension:
+a certain interface shape leads to a certain $\Delta p$,
+and the liquid will flow (i.e. the surface will move)
+to try to reach an equilibrium.
+
+
+## References
+1.  B. Lautrup,
+    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+    CRC Press.
+2.  T. Bohr,
+    *Surface tension and Laplace pressure*, 2021,
+    unpublished.
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