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---
title: "Capillary action"
firstLetter: "C"
publishDate: 2021-03-29
categories:
- Physics
- Fluid mechanics
- Fluid statics
- Surface tension
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.
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