---
title: "Meniscus"
sort_title: "Meniscus"
date: 2021-03-11
categories:
- Physics
- Fluid mechanics
- Fluid statics
- Surface tension
layout: "concept"
---

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](/know/concept/young-laplace-law/)
to 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 $$\idv{y}{s} = \sin\theta$$, we arrive at:

$$\begin{aligned}
    \boxed{
        L_c^2 \dvn{2}{\theta}{s} = \sin\theta
    }
\end{aligned}$$

To solve this equation, we multiply it by $$\idv{\theta}{s}$$,
which is nonzero close to the wall:

$$\begin{aligned}
    L_c^2 \dvn{2}{\theta}{s} \dv{\theta}{s}
    = \dv{\theta}{s} \sin\theta
\end{aligned}$$

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 $$\idv{\theta}{s}$$ 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 $$\idv{x}{s}$$ and $$\idv{y}{s}$$ 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.