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+---
+title: "Capillary action"
+firstLetter: "C"
+publishDate: 2021-03-29
+categories:
+- Physics
+- Fluid mechanics
+
+date: 2021-03-07T20:42:28+01:00
+draft: false
+markup: pandoc
+---
+
+# Capillary action
+
+**Capillary action** refers to the movement of liquid
+through narrow spaces due to surface tension, often against gravity.
+It occurs when the [Laplace pressure](/know/concept/young-laplace-law/)
+from surface tension is much larger in magnitude than the
+[hydrostatic pressure](/know/concept/hydrostatic-pressure/) from gravity.
+
+Consider a spherical droplet of liquid with radius $R$.
+The hydrostatic pressure difference
+between the top and bottom of the drop
+is much smaller than the Laplace pressure:
+
+$$\begin{aligned}
+    2 R \rho g \ll 2 \frac{\alpha}{R}
+\end{aligned}$$
+
+Where $\rho$ is the density of the liquid,
+$g$ is the acceleration due to gravity,
+and $\alpha$ is the energy cost per unit surface area.
+Rearranging the inequality yields:
+
+$$\begin{aligned}
+    R^2 \ll \frac{\alpha}{\rho g}
+\end{aligned}$$
+
+From the right-hand side we define the **capillary length** $L_c$,
+so gravity is negligible if $R \ll L_c$:
+
+$$\begin{aligned}
+    \boxed{
+        L_c
+        = \sqrt{\frac{\alpha}{\rho g}}
+    }
+\end{aligned}$$
+
+In general, for a system with characteristic length $L$,
+the relative strength of gravity compared to surface tension
+is described by the **Bond number** $\mathrm{Bo}$
+or **Eötvös number** $\mathrm{Eo}$:
+
+$$\begin{aligned}
+    \boxed{
+        \mathrm{Bo}
+        = \mathrm{Eo}
+        = \frac{L^2}{L_c^2}
+        = \frac{m g}{\alpha L}
+    }
+\end{aligned}$$
+
+The right-most side gives an alternative way of understanding $\mathrm{Bo}$:
+$m$ is the mass of a cube with side $L$, such that the numerator is the weight force,
+and the denominator is the tension force of the surface.
+In any case, capillary action can be observed when $\mathrm{Bo \ll 1}$.
+
+The most famous example of capillary action is **capillary rise**,
+where a liquid "climbs" upwards in a narrow vertical tube with radius $R$,
+apparently defying gravity.
+Assuming the liquid-air interface is a spherical cap
+with constant [curvature](/know/concept/curvature/) radius $R_c$,
+then we know that the liquid is at rest
+when the hydrostatic pressure equals the Laplace pressure:
+
+$$\begin{aligned}
+    \rho g h
+    \approx \alpha \frac{2}{R_c}
+    = 2 \alpha \frac{\cos\theta}{R}
+\end{aligned}$$
+
+Where $\theta$ is the liquid-tube contact angle,
+and we are neglecting variations of the height $h$ due to the curvature
+(i.e. the [meniscus](/know/concept/meniscus/)).
+By isolating the above equation for $h$,
+we arrive at **Jurin's law**,
+which predicts the height climbed by a liquid in a tube with radius $R$:
+
+$$\begin{aligned}
+    \boxed{
+        h
+        = 2 \frac{L_c^2}{R} \cos\theta
+    }
+\end{aligned}$$
+
+Depending on $\theta$, $h$ can be negative,
+i.e. the liquid might descend below the ambient level.
+
+
+An alternative derivation of Jurin's law balances the forces instead of the pressures.
+On the right, we have the gravitational force
+(i.e. the energy-per-distance to lift the liquid),
+and on the left, the surface tension force
+(i.e. the energy-per-distance of the liquid-tube interface):
+
+$$\begin{aligned}
+     \pi R^2 \rho g h
+     \approx 2 \pi R (\alpha_{sg} - \alpha_{sl})
+\end{aligned}$$
+
+Where $\alpha_{sg}$ and $\alpha_{sl}$ are the energy costs
+of the solid-gas and solid-liquid interfaces.
+Thanks to the [Young-Dupré relation](/know/concept/young-dupre-relation/),
+we can rewrite this as follows:
+
+$$\begin{aligned}
+     R \rho g h
+     = 2 \alpha \cos\theta
+\end{aligned}$$
+
+Isolating this for $h$ simply yields Jurin's law again, as expected.
+
+
+
+## References
+1.  B. Lautrup,
+    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+    CRC Press.