# What is the meniscus shape?

What is the shape of a meniscus?

I suppose that the problem is very complex, but is the solution known at least for a liquid that wets the wall in a big vessel? (exponential, maybe?)

• From the wikipedia link you gave: When a capillary tube, is dipped into a liquid and the liquid wets the tube (with zero contact angle), the liquid surface inside the tube forms a concave meniscus, which is a virtually spherical surface having the same radius as the inside of the tube
– user65081
Jan 1, 2015 at 20:21

The shape of the curve must be such that the pressure difference across the meniscus exactly counters the force of gravity on the additional column of water.

Now if we write the height of the liquid (density $\rho$) as a function of radial distance $h(r)$, and surface tension as $\sigma$, then we can write the force balance on an annular column with thickness $dr$ as:

$$\rho \cdot 2\pi r\cdot dr\cdot h\cdot g = \frac{\sigma}{\sqrt{1+\left(\frac{dh}{dr}\right)^2}}\frac{d^2h}{dr^2}dr\cdot 2 \pi r\tag1\\ \frac{d^2h}{dr^2}=\sqrt{1+\left(\frac{dh}{dr}\right)^2}\frac{\rho g}{\sigma}h$$

The following diagram helps understand how that is derived:

When you have a certain curvature in the surface, then there will be a net force on the liquid. Just like with a rope with tension, you compute the net force by looking at the difference in the gradient between two points on the curve. In the above, the vertical component of the surface tension on the left is given by

$$F'_1=-\frac{F_1 \frac{dy}{dr}}{\sqrt{1+\left(\frac{dh}{dx}\right)^2}}$$

and a similar expression can be written for $F_2$ - except that since $F_1$ points to the left, we need a minus sign, and for $F_2$ we don't. Now $F_1=F_2=F_s$, the force of the surface tension (we need a "per unit length somewhere - will get to that in a minute).

This means that the net force is given by

\begin{align}F_{net} &= F'_1 + F'_2 \\ &= \frac{F_s}{{\sqrt{1+\left(\frac{dh}{dx}\right)^2}}}\left(\left.\frac{dh}{dr}\right|_2- \left.\frac{dh}{dr}\right|_1\right)\\ &=\frac{F_s}{{\sqrt{1+\left(\frac{dh}{dx}\right)^2}}} \frac{d^2h}{dr^2}dr\end{align}

Now the force of the surface tension is the surface tension multiplied by the length of the surface over which it acts, so for a given annulus,

$$F_s = \sigma 2\pi r$$

Combining these two gives us the right hand side of equation (1) above.

The left hand side of the equation (1) is just the mass of the annulus of water - the density times the volume - multiplied by the gravitational acceleration: this is the weight of the column of water supported.

When we solve this differential equation, we find that the shape of the curve is a catenary - the sum of two exponentials with opposite signs. The solution is described in detail in many places - in essence you first substitute $z=\frac{dh}{dr}$ and solve the first order differential equation that results, then undo the substitution and solve the new first order DE. Detailed steps can be found here or do it yourself with the hints given here .

All that remains is to get the boundary conditions. We define the characteristic length

$$\ell = \sqrt{\frac{\sigma}{\rho g}}$$

and the contact angle $\theta$ at radius $R$ (for a given combination of liquid and wall, the contact angle is a given), then

$$h(r) = \frac{\cot\theta}{\sinh{\frac{R}{\ell}}} \cosh \frac{r}{\ell}$$

When $R >> \ell$, this will be mostly flat with a slope at the edges; when the two are comparable, it will be the more familiar parabolic shape that you see on a small capillary.

• Is $\sigma$ surface tension, with units of force per unit area? Do you mean to have both $\ell$ and $l$ in your final expression?
– rob
Jan 2, 2015 at 4:38
• @rob - thanks. I fixed the equation and cleaned up the definition of terms used. $\sigma$ should have units of $N/m$ - where did I end up with force per unit area? Jan 2, 2015 at 13:45
• @Floris It sound very interesting! I don't understand why did you equal the weight of the annular column to $\sigma h'' dr 2 \pi r$. Can you explain to me? Tank you. Jan 3, 2015 at 22:29
• @Floris The idea you had to study the problem (showed in the graph you sketched) is divine. But in my opinion there is something wrong in developing the proof: it works only if angles are everywhere much smaller that 45$^\circ$, but in general near the wall this is not true (look to the way you project the forces in vertical: derivatives give tangent, not sine). I fear the general solving differential equation is much harder, and your works good only, as approximation, if surface is almost flat everywhere. Jan 5, 2015 at 0:10
• @FaustoVezzaro - you are right, I did play a bit fast and loose with the equation. However, if you check mypages.iit.edu/~maslanka/LN3.pdf (google "catenary equation derivation", pick any link) you will see that the proper treatment will in fact give the same result. I'll try to update the solution - thanks for pointing it out. Jan 5, 2015 at 4:06

Looking to the diagram

we see that the net force (vertical, obviously) on the annular column is (here $F=\sigma 2\pi r$) $$F \sin (\theta + d\theta) -F\sin \theta \approx F \cos \theta d \theta$$ $\theta = \tan^{-1} \left( \frac{dh}{dr} \right)$ so $$d\theta = \frac{\frac{d^2 h}{dr^2}}{1+\left( \frac{dh}{dr} \right)^2} dr \qquad \qquad \cos \theta = \frac{1}{\sqrt{1+\left( \frac{dh}{dr} \right)^2}}$$ Doing the substitution and equaling this to the weight of annular column we obtain $$h \left( 1+\left( \frac{dh}{dr} \right)^2 \right)^{\frac{3}{2}}= \ell^2 \frac{d^2 h}{dr^2}$$ where $\ell = \sqrt{\frac{\sigma}{\rho g}}$. Meniscus should be ruled by this differential equation, but it looks very hard. Boundary condition itself are problematic, surely they depend on the wall material too.

• You are right - I have updated my expressions. It does not change the final result. The wall material comes into the expression in the form of the contact angle $\theta$ which is a constant for a certain combination of liquid/wall, and is a measure of their relative attraction. That contact angle shows up in my expression (from the boundary conditions) Jan 5, 2015 at 13:18
• Our equation are different, in the exponent (3 in mine and 1 in your). Maybe in your derivation in calculating net force we cannot use the same radical for the two forces? (it contains derivative) Anyway in my opinion the final result change: nor mine neither your equation have catenary as solution. The more I think, the more I feel that this problem is very difficult, the best I can do is write the differential equation that rules the liquid surface, I can't solve it. Jan 5, 2015 at 18:24
• Introducing the adimensional function $y =\frac{h}{\ell}$ and the adimensional variable $x=\frac{r}{\ell}$ we can adimensionalize the equation finding $$y \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^{\frac{3}{2}} = \frac{d^2 y}{dx^2}$$ Now the equation looks just a bit nicer, but find the general solution still looks very difficult! Jan 12, 2015 at 7:59
• You are right, my treatment is insufficient. See for example archive.org/stream/staticsincluding00lambuoft#page/276/mode/2up - I am trying to figure out exactly what it is saying. I am beginning to think we need to take into account the curvature in the horizontal direction as well- this adds another term to the differential equation. Tricky. I will think about it more. Jan 12, 2015 at 16:40
• You are decidedly right. Reflecting upon what you wrote, I understood: the force that carry the weight of the annular column derive not only from the greater slope (this is all only for flat wall) but also from the greater length of the $r+dr$ circle (all is always so simple after having understood). On this basis I found (it looks like one equation of your wikilink) $$h=\ell^2 \left( \frac{h''}{(1+(h')^2)^{\frac{3}{2}}} + \frac{h'}{r \sqrt{1+(h')^2}} \right)$$ I'll write the derivation soon (one of next days). Jan 14, 2015 at 1:42

The shape of the meniscus is not described by an elementary function (limiting case can be an exception: we will see that for a wetting liquid in large tube, if the surface of the liquid is almost flat the profile is approximately exponential near the border), but we can write the differential equation that rules the profile (although probably any finite combination of elementary function solve it, maybe the DE can be exploited to obtain the solution numerically or with infinite sums). To do that, we will limit ourself to study the case of a wetting liquid in a cylindrical tube partially immersed in a very big vessel (we can call "sea" the outer liquid: the level far away is not influenced by what happens to the tube).

## Estimate of the forces acting on the annular column

Let's consider an infinitesimal annular column, $r$ is the distance from the center and $R$ the radius of the tube. With the exception of the layer adjacent the wall, the only forces involved are two: the weight of the annular column and the force due to the surface tension. At the equilibrium the two forces are equal in magnitude. If $h$ is the height of the annular column (respect to the liquid level in the external sea), the weight is easy: $$mg=\rho dV g=\rho \cdot (2\pi r dr h) \cdot g = 2\pi\rho g r h dr$$ A bit more laborious is the calculus of the net force due to surface tension. To estimate it we have to concentrate on the fact that this force is due to two effects: in the external circle the force is greater (because of the greater length of the circle) and the slope too is (obviously) greater. This effects are showed in figure, where I call $F$ a infinitesimal force ($\sigma r d\phi$, if you want use the polar angle) and $dF$ a second order infinitesimal force ($\sigma dr d\phi$).

This lead to this net force that carry the weight of the slice of the annular column under the infinitesimal element of surface: $$(F+dF) \sin (\theta + d\theta) - F \sin \theta$$ We ignore the horizontal components because for symmetry reason they cancel when we sum all infinitesimal elements: to find the pushing up force on the annular column, in the expression above we can use the total forces on smaller and bigger circle: $$|\mathbf{F}_{\mathrm{up}}| = 2 \pi (r+dr) \sigma \qquad |\mathbf{F}_{\mathrm{down}}| = 2\pi r \sigma$$ Now we are ready to estimate the force due to surface tension, exploiting infinitesimal calculus: we know that $\sin (\theta + d\theta) \cong \sin \theta + \cos \theta d \theta$, so the pushing up force can be written in this way: $$2 \pi \sigma (r \cos \theta d \theta + \sin \theta dr + \cos \theta dr d\theta)$$ Now, if $h=h(r)$, defined in $[0,R]$, is the function that gives the profile of the meniscus, we have $\theta=\tan^{-1} (h')$. So we have (in the last two we exploit $\cos (\tan^{-1} x)=\frac{1}{\sqrt{1+x^2}}$ and $\sin (\tan^{-1} x)=\frac{x}{\sqrt{1+x^2}}$) $$d \theta = \frac{h''}{1+(h')^2} dr \qquad \cos \theta = \frac{1}{\sqrt{1+(h')^2}} \qquad \sin \theta = \frac{h'}{\sqrt{1+(h')^2}}$$ So the last term is an higher order infinitesimal and can be ignored: the force became $$2 \pi \sigma \left( \frac{r h''}{(1+(h')^2)^{\frac{3}{2}}} + \frac{h'}{\sqrt{1+(h')^2}} \right) dr$$

## The solving differential equation and a limiting case

Equaling to the weight $2\pi\rho g r h dr$ we can rearrange in this differential equation, that express the equilibrium condition $$r h = \ell^2 \left( \frac{ r h''}{(1+(h')^2)^{\frac{3}{2}}} + \frac{h'}{\sqrt{1+(h')^2}} \right)$$ where $\ell = \sqrt{\frac{\sigma}{\rho g}}$ (for water $\ell \approx 3$ mm). If we suppose (but this is false, in general) that the surface is almost flat ($h'$ very small everywhere) this DE became simply $h = \ell^2 h''$ and symmetry implies that liquid surface is given by a catenary rotated (so if surface is almost flat and $R \gg \ell$, near the border the shape is approximately exponential).

## Energetic considerations

Finally, without write explicitly all calculus, let's do some energetic considerations. We will do the reasonable hypothesis that the energetic gain due to the adhesion of the liquid in the wall is directly proportional to the wet wall (let be $\xi$ the gain by unit of surface, depending on liquid and wall material). In this case the solution of the DE must satisfy this conservation equation $$\Delta E_s + \Delta E_g = 2 \pi R h(R) \cdot \xi$$ that is: the energetic gain due to the adhesion (the increase of wet surface, multiplied by $\xi$) is spent to increase liquid surface ($\Delta E_s = \sigma \Delta A$ is the work done against surface tension if surface is increased by $\Delta A$) and to lift the liquid. Let's calculate this two energy:

• it is easy to see that for a column, if $\rho$ and $g$ are constant, we can calculate gravitational potential energy by considering all mass in the middle, and this works for annular column too, because the work to raise it, is the sum of the work done to raise each slice, which is an ordinary column. On this basis we can find the total gravitational energy stored in the meniscus: $$\Delta E_g = \pi \rho g \int_0^R r h^2 dr$$

• The surface of a cone truncated is given by $\pi (r+R) a$ ($a$ is the sloped segment): in our case $a=\sqrt{(dr)^2 + (dh)^2} = dr \sqrt{1+(h')^2}$, so ignoring higher order infinitesimal we have that the surface of the strip of the top border of the annular column is $2 \pi r \sqrt{1+(h')^2} dr$. Integrating from $0$ to $R$ and subtracting $\pi R^2$ we have the increase $\Delta A$ of the surface. Multiplying by $\sigma$ we have the energy stored in the curved surface of the liquid: $$\Delta E_s = \sigma \left( 2 \pi \int_0^R r \sqrt{1+(h')^2} dr - \pi R^2 \right)$$

All this considerations allow to rewrite the conservation equation in this way: $$\int_0^R r \left( \sqrt{1+(h')^2} + A h^2 \right) dr = B h(R) + C$$ where $$A = \frac{\rho g}{2 \sigma} \qquad B = \frac{R \xi}{\sigma} \qquad C = \frac{R^2}{2}$$ The solution $h=h(r)$ of the differential equation must satisfy this condition. Another condition is, as said previously, $h'(0) = 0$. Notice that $h(0) \neq 0$, it is the difference in level between the center of the meniscus and the external "sea" (surely $h(0)$ is small if $R$ is big).