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+---
+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.