Categories: Fluid mechanics, Fluid statics, Physics, Surface tension.

# 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 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 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 $$\dv*{y}{s} = \sin\theta$$, 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.