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

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 \(d\):

\[\begin{aligned} d = y(\theta_0) = - 2 L_c \sin\!\Big(\frac{\theta_0}{2}\Big) = 2 L_c \sin\!\Big( \frac{\pi/2 - \phi}{2} \Big) \end{aligned}\]

- B. Lautrup,
*Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition, CRC Press.

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