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 -axis be a flat wall,
and the fluid tend to when .
Close to the wall, i.e. for small , the liquid curves up or down
to touch the wall at a height .
Three forces are at work here:
the first two are the surface tension of the fluid surface,
and the counter-pull of the wall against the tension,
where 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 so that ):
If you were wondering about the units,
keep in mind that there is an implicit -direction here too.
This results in the following balance equation for the forces at the wall:
We isolate this relation for
and use some trigonometric magic to rewrite it:
Here, we recognize the definition of the capillary length ,
yielding an expression for
that is valid both for (where )
and (where ):
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 relative to the plane.
The curve is a function of the arc length ,
and is governed by:
The last equation describes the curvature radius
of the surface along the -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 ,
where is the external air pressure:
Rearranging this yields that .
Inserting this into the curvature equation gives us:
By differentiating this equation with respect to
and using , we arrive at:
To solve this equation, we multiply it by ,
which is nonzero close to the wall:
We integrate both sides with respect to
and set the integration constant to ,
such that we get zero when away from the wall:
Isolating this for and using a trigonometric identity then yields:
We use trigonometric relations on the equations
for and to get -derivatives:
Let be the initial elevation angle at the wall.
Then, by integrating the above equations, we get the following solutions:
Where the integration constant has been chosen such that for away from the wall,
and for .
This result is consistent with our earlier expression for :
- B. Lautrup,
Physics of continuous matter: exotic and everyday phenomena in the macroscopic world, 2nd edition,