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Young-Laplace Equation
Consider an interface separating two immiscible fluids that are in equilibrium with one another. Let these two
fluids be denoted 1 and 2. Consider an arbitrary segment
of this interface that is enclosed by some closed curve
.
Let
denote a unit tangent to the curve, and let
denote a unit normal to the
interface directed from fluid 1 to fluid 2. (Note that
circulates around
in a
right-handed manner.) See Figure 10. Suppose that
and
are the
pressures of fluids 1 and 2, respectively, on either side of
. Finally, let
be the (uniform)
surface tension at the interface.
Figure 10:
Interface between two immiscible fluids.
 |
The net force acting on
is
 |
(342) |
where
is an element of
, and
an element of
. Here, the first term on the
right-hand side is the net normal force due to the pressure difference across the interface, whereas the
second term is the net surface tension force.
Note that body forces play no role in (342), because the interface has zero volume. Furthermore,
viscous forces can be neglected, since both fluids are static.
Now, in equilibrium,
the net force acting on
must be zero: i.e.,
 |
(343) |
(In fact, the net force would be zero even in the absence of equilibrium, because the interface has zero mass.)
Applying Stokes' theorem (see Section A.22) to the curve
, we find that
 |
(344) |
where
is a general vector field. This theorem can also be written
 |
(345) |
Suppose that
, where
is an arbitrary constant vector. We obtain
 |
(346) |
However, the vector identity (1484) yields
 |
(347) |
since
is a constant vector.
Hence, we get
![\begin{displaymath}
{\bf b}\cdot\oint_C {\bf t}\times {\bf g}\,dr = {\bf b}\cdot...
...bla {\bf g})\cdot{\bf n} - (\nabla\cdot{\bf g})\,{\bf n}]\,dS,
\end{displaymath}](img1038.png) |
(348) |
where
.
Now, since
is also an arbitrary vector, the above equation gives
![\begin{displaymath}
\oint_C {\bf t}\times {\bf g}\,dr = \int_S\left[(\nabla {\bf g})\cdot{\bf n} - (\nabla\cdot{\bf g})\,{\bf n}\right]dS.
\end{displaymath}](img1040.png) |
(349) |
Taking
, we find that
![\begin{displaymath}
\gamma\oint_C {\bf t}\times {\bf n}\,dr= \gamma\int_S \left[...
...bf n})\cdot {\bf n} - (\nabla\cdot {\bf n})\,{\bf n}\right]dS.
\end{displaymath}](img1042.png) |
(350) |
But,
, because
is a unit vector. Thus, we obtain
 |
(351) |
which can be combined with (343) to give
![\begin{displaymath}
\int_S\left[(p_1-p_2)-\gamma\,(\nabla\cdot{\bf n})\right]{\bf n}\,dS = 0.
\end{displaymath}](img1045.png) |
(352) |
Finally, given that
is arbitrary, the above expression reduces to the pressure balance constraint
 |
(353) |
where
.
The above relation is generally known as the Young-Laplace equation, and
can also be derived by minimizing the free energy of the interface. (See Section 4.8.) Note that
is
the jump in pressure seen when crossing the interface in the opposite direction to
.
Of course, a plane interface is characterized by
. On the other hand, a curved
interface generally has
. In fact,
measures the local
mean curvature of the interface. Thus, according to the Young-Laplace equation, there is a pressure jump across
a curved interface between two immiscible fluids, the
magnitude of the jump being proportional to the surface tension.
Next: Spherical Interfaces
Up: Surface Tension
Previous: Introduction
Richard Fitzpatrick
2012-04-27