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---
title: "Meniscus"
firstLetter: "M"
publishDate: 2021-03-11
categories:
- Physics
- Fluid mechanics
date: 2021-03-11T14:39:56+01:00
draft: false
markup: pandoc
---
# Meniscus
When a fluid interface, e.g. the surface of a liquid,
touches a flat solid wall, it will curve to meet it.
This small rise or fall is called a **meniscus**,
and is caused by surface tension and gravity.
In 2D, let the vertical $y$-axis be a flat wall,
and the fluid tend to $y = 0$ when $x \to \infty$.
Close to the wall, i.e. for small $x$, the liquid curves up or down
to touch the wall at a height $y = d$.
Three forces are at work here:
the first two are the surface tension $\alpha$ of the fluid surface,
and the counter-pull $\alpha \sin\phi$ of the wall against the tension,
where $\phi$ is the contact angle.
The third is the [hydrostatic pressure](/know/concept/hydrostatic-pressure/) gradient
inside the small portion of the fluid above/below the ambient level,
which exerts a total force on the wall given by
(for $\phi < \pi/2$ so that $d > 0$):
$$\begin{aligned}
\int_0^d \rho g y \dd{y}
= \frac{1}{2} \rho g d^2
\end{aligned}$$
If you were wondering about the units,
keep in mind that there is an implicit $z$-direction here too.
This results in the following balance equation for the forces at the wall:
$$\begin{aligned}
\alpha
= \alpha \sin\phi + \frac{1}{2} \rho g d^2
\end{aligned}$$
We isolate this relation for $d$
and use some trigonometric magic to rewrite it:
$$\begin{aligned}
d
= \sqrt{\frac{\alpha}{\rho g}} \sqrt{2 (1 - \sin\phi)}
= \sqrt{\frac{\alpha}{\rho g}} \sqrt{4 \sin^2\!\Big(\frac{\pi}{4} - \frac{\phi}{2}\Big)}
\end{aligned}$$
Here, we recognize the definition of the capillary length $L_c = \sqrt{\alpha / (\rho g)}$,
yielding an expression for $d$
that is valid both for $\phi < \pi/2$ (where $d > 0$)
and $\phi > \pi/2$ (where $d < 0$):
$$\begin{aligned}
\boxed{
d
= 2 L_c \sin\!\Big(\frac{\pi/2 - \phi}{2}\Big)
}
\end{aligned}$$
Next, we would like to know the exact shape of the meniscus.
To do this, we need to describe the liquid surface differently,
using the elevation angle $\theta$ relative to the $y = 0$ plane.
The curve $\theta(s)$ is a function of the arc length $s$,
where $\dd{s}^2 = \dd{x}^2 + \dd{y}^2$,
and is governed by:
$$\begin{aligned}
\dv{x}{s}
= \cos\theta
\qquad
\dv{y}{s}
= \sin\theta
\qquad
\dv{\theta}{s}
= \frac{1}{R}
\end{aligned}$$
The last equation describes the curvature radius $R$
of the surface along the $x$-axis.
Since we are considering a flat wall,
there is no curvature in the orthogonal principal direction.
Just below the liquid surface in the meniscus,
we expect the hydrostatic pressure
and the Young-Laplace law agree about the pressure $p$,
where $p_0$ is the external air pressure:
$$\begin{aligned}
p_0 - \rho g y
= p_0 - \frac{\alpha}{R}
\end{aligned}$$
Rearranging this yields that $R = L_c^2 / y$.
Inserting this into the curvature equation gives us:
$$\begin{aligned}
\dv{\theta}{s}
= \frac{y}{L_c^2}
\end{aligned}$$
By differentiating this equation with respect to $s$
and using $\dv*{y}{s} = \sin\theta$, we arrive at:
$$\begin{aligned}
\boxed{
L_c^2 \dv[2]{\theta}{s} = \sin\theta
}
\end{aligned}$$
To solve this equation, we multiply it by $\dv*{\theta}{s}$,
which is nonzero close to the wall:
$$\begin{aligned}
L_c^2 \dv[2]{\theta}{s} \dv{\theta}{s}
= \dv{\theta}{s} \sin\theta
\end{aligned}$$
We integrate both sides with respect to $s$
and set the integration constant to $1$,
such that we get zero when $\theta \to 0$ away from the wall:
$$\begin{aligned}
\frac{L_c^2}{2} \Big( \dv{\theta}{s} \Big)^2
= 1 - \cos\theta
\end{aligned}$$
Isolating this for $\dv*{\theta}{s}$ and using a trigonometric identity then yields:
$$\begin{aligned}
\dv{\theta}{s}
= \frac{1}{L_c} \sqrt{2 (1 - \cos\theta)}
= \frac{1}{L_c} \sqrt{4 \sin^2\!\Big( \frac{\theta}{2} \Big)}
= - \frac{2}{L_c} \sin\!\Big( \frac{\theta}{2} \Big)
\end{aligned}$$
We use trigonometric relations on the equations
for $\dv*{x}{s}$ and $\dv*{y}{s}$ to get $\theta$-derivatives:
$$\begin{aligned}
\dv{x}{\theta}
&= \dv{x}{s} \dv{s}{\theta}
= \bigg( 1 - 2 \sin^2\!\Big( \frac{\theta}{2} \Big) \bigg) \bigg( \dv{\theta}{s} \bigg)^{-1}
= L_c \sin\!\Big( \frac{\theta}{2} \Big) - \frac{L_c}{2 \sin(\theta/2)}
\\
\dv{y}{\theta}
&= \dv{y}{s} \dv{s}{\theta}
= \bigg( 2 \sin\!\Big(\frac{\theta}{2}\Big) \cos\!\Big(\frac{\theta}{2}\Big) \bigg) \bigg( \dv{\theta}{s} \bigg)^{-1}
= - L_c \cos\!\Big( \frac{\theta}{2} \Big)
\end{aligned}$$
Let $\theta_0 = \phi - \pi/2$ be the initial elevation angle $\theta(0)$ at the wall.
Then, by integrating the above equations, we get the following solutions:
$$\begin{gathered}
\boxed{
\frac{x}{L_c}
= 2 \cos\!\Big(\frac{\theta_0}{2}\Big) + \log\!\bigg| \tan\!\Big(\frac{\theta_0}{4}\Big) \bigg|
- 2 \cos\!\Big(\frac{\theta}{2}\Big) + \log\!\bigg| \tan\!\Big(\frac{\theta}{4}\Big) \bigg|
}
\\
\boxed{
\frac{y}{L_c}
= - 2 \sin\!\Big(\frac{\theta}{2}\Big)
}
\end{gathered}$$
Where the integration constant has been chosen such that $y \to 0$ for $\theta \to 0$ away from the wall,
and $x = 0$ for $\theta = \theta_0$.
This result is consistent with our earlier expression for $d$:
$$\begin{aligned}
d
= y(\theta_0)
= - 2 L_c \sin\!\Big(\frac{\theta_0}{2}\Big)
= 2 L_c \sin\!\Big( \frac{\pi/2 - \phi}{2} \Big)
\end{aligned}$$
## References
1. B. Lautrup,
*Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
CRC Press.
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