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----
-title: "Capillary action"
-sort_title: "Capillary action"
-date: 2021-03-29
-categories:
-- Physics
-- Fluid mechanics
-- Fluid statics
-- Surface tension
-layout: "concept"
----
-
-**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 $$R \ll L_c$$:
-
-$$\begin{aligned}
-    \boxed{
-        L_c
-        \equiv \sqrt{\frac{\alpha}{\rho g}}
-    }
-\end{aligned}$$
-
-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** $$\mathrm{Eo}$$:
-
-$$\begin{aligned}
-    \boxed{
-        \mathrm{Bo}
-        \equiv \mathrm{Eo}
-        \equiv \frac{L^2}{L_c^2}
-        = \frac{m g}{\alpha L}
-    }
-\end{aligned}$$
-
-The rightmost 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.