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 yy-axis be a flat wall, and the fluid tend to y=0y = 0 when xx \to \infty. Close to the wall, i.e. for small xx, the liquid curves up or down to touch the wall at a height y=dy = d.

Three forces are at work here: the first two are the surface tension α\alpha of the fluid surface, and the counter-pull αsinϕ\alpha \sin\phi of the wall against the tension, where ϕ\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 ϕ<π/2\phi < \pi/2 so that d>0d > 0):

0dρgydy=12ρgd2\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 zz-direction here too. This results in the following balance equation for the forces at the wall:

α=αsinϕ+12ρgd2\begin{aligned} \alpha = \alpha \sin\phi + \frac{1}{2} \rho g d^2 \end{aligned}

We isolate this relation for dd and use some trigonometric magic to rewrite it:

d=αρg2(1sinϕ)=αρg4sin2 ⁣(π4ϕ2)\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 Lc=α/(ρg)L_c = \sqrt{\alpha / (\rho g)}, yielding an expression for dd that is valid both for ϕ<π/2\phi < \pi/2 (where d>0d > 0) and ϕ>π/2\phi > \pi/2 (where d<0d < 0):

d=2Lcsin ⁣(π/2ϕ2)\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 θ\theta relative to the y=0y = 0 plane. The curve θ(s)\theta(s) is a function of the arc length ss, where ds2=dx2+dy2\dd{s}^2 = \dd{x}^2 + \dd{y}^2, and is governed by:

dxds=cosθdyds=sinθdθds=1R\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 RR of the surface along the xx-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 pp, where p0p_0 is the external air pressure:

p0ρgy=p0αR\begin{aligned} p_0 - \rho g y = p_0 - \frac{\alpha}{R} \end{aligned}

Rearranging this yields that R=Lc2/yR = L_c^2 / y. Inserting this into the curvature equation gives us:

dθds=yLc2\begin{aligned} \dv{\theta}{s} = \frac{y}{L_c^2} \end{aligned}

By differentiating this equation with respect to ss and using dy/ds=sinθ\idv{y}{s} = \sin\theta, we arrive at:

Lc2d2θds2=sinθ\begin{aligned} \boxed{ L_c^2 \dvn{2}{\theta}{s} = \sin\theta } \end{aligned}

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

Lc2d2θds2dθds=dθdssinθ\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 ss and set the integration constant to 11, such that we get zero when θ0\theta \to 0 away from the wall:

Lc22(dθds)2=1cosθ\begin{aligned} \frac{L_c^2}{2} \Big( \dv{\theta}{s} \Big)^2 = 1 - \cos\theta \end{aligned}

Isolating this for dθ/ds\idv{\theta}{s} and using a trigonometric identity then yields:

dθds=1Lc2(1cosθ)=1Lc4sin2 ⁣(θ2)=2Lcsin ⁣(θ2)\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 dx/ds\idv{x}{s} and dy/ds\idv{y}{s} to get θ\theta-derivatives:

dxdθ=dxdsdsdθ=(12sin2 ⁣(θ2))(dθds)1=Lcsin ⁣(θ2)Lc2sin(θ/2)dydθ=dydsdsdθ=(2sin ⁣(θ2)cos ⁣(θ2))(dθds)1=Lccos ⁣(θ2)\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 θ0=ϕπ/2\theta_0 = \phi - \pi/2 be the initial elevation angle θ(0)\theta(0) at the wall. Then, by integrating the above equations, we get the following solutions:

xLc=2cos ⁣(θ02)+log ⁣tan ⁣(θ04)2cos ⁣(θ2)+log ⁣tan ⁣(θ4)yLc=2sin ⁣(θ2)\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 y0y \to 0 for θ0\theta \to 0 away from the wall, and x=0x = 0 for θ=θ0\theta = \theta_0. This result is consistent with our earlier expression for dd:

d=y(θ0)=2Lcsin ⁣(θ02)=2Lcsin ⁣(π/2ϕ2)\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}


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