---
title: "Young-Laplace law"
firstLetter: "Y"
publishDate: 2021-03-11
categories:
- Physics
- Fluid mechanics
- Surface tension

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.