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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 $S$ of this interface that is enclosed by some closed curve $C$. Let ${\bf t}$ denote a unit tangent to the curve, and let ${\bf n}$ denote a unit normal to the interface directed from fluid 1 to fluid 2. (Note that $C$ circulates around ${\bf n}$ in a right-handed manner.) See Figure 10. Suppose that $p_1$ and $p_2$ are the pressures of fluids 1 and 2, respectively, on either side of $S$. Finally, let $\gamma$ be the (uniform) surface tension at the interface.

Figure 10: Interface between two immiscible fluids.
\epsfysize =1.5in

The net force acting on $S$ is

{\bf f} =\int_S (p_1-p_2)\,{\bf n}\,dS + \gamma \oint_C {\bf t}\times {\bf n}\,dr,
\end{displaymath} (342)

where $d{\bf S} = {\bf n}\,dS$ is an element of $S$, and $d{\bf r}= {\bf t}\,dr$ an element of $C$. 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 $S$ must be zero: i.e.,
\int_S (p_1-p_2)\,{\bf n}\,dS =- \gamma \oint_C {\bf t}\times {\bf n}\,dr.
\end{displaymath} (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 $C$, we find that

\oint_C {\bf F}\cdot d{\bf r} = \int_S \nabla\times {\bf F}\cdot d{\bf S},
\end{displaymath} (344)

where ${\bf F}$ is a general vector field. This theorem can also be written
\oint_C {\bf F}\cdot {\bf t}\,dr= \int_S \nabla\times {\bf F}\cdot {\bf n}\,dS.
\end{displaymath} (345)

Suppose that ${\bf F} = {\bf g}\times {\bf b}$, where ${\bf b}$ is an arbitrary constant vector. We obtain
\oint_C ({\bf g}\times {\bf b})\cdot {\bf t}\,dr= \int_S \nabla\times({\bf g}\times{\bf b})\cdot {\bf n}\,dS.
\end{displaymath} (346)

However, the vector identity (1484) yields
\nabla\times ({\bf g}\times {\bf b}) = - (\nabla\cdot{\bf g})\,{\bf b} + ({\bf b}\cdot\nabla)\,{\bf g},
\end{displaymath} (347)

since ${\bf b}$ is a constant vector. Hence, we get
{\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} (348)

where ${\bf b} \cdot(\nabla {\bf g})\cdot{\bf n} \equiv b_i\,(\partial g_j/\partial x_i)\,n_j$. Now, since ${\bf b}$ is also an arbitrary vector, the above equation gives
\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} (349)

Taking ${\bf g} = \gamma\,{\bf n}$, we find that
\gamma\oint_C {\bf t}\times {\bf n}\,dr= \gamma\int_S \left[... n})\cdot {\bf n} - (\nabla\cdot {\bf n})\,{\bf n}\right]dS.
\end{displaymath} (350)

But, $(\nabla{\bf n})\cdot {\bf n}\equiv (1/2)\,\nabla(n^2)=0$, because ${\bf n}$ is a unit vector. Thus, we obtain
\gamma\,\oint_C {\bf t}\times{\bf n}\,dr = - \gamma\int_S (\nabla \cdot{\bf n})\,{\bf n}\,dS,
\end{displaymath} (351)

which can be combined with (343) to give
\int_S\left[(p_1-p_2)-\gamma\,(\nabla\cdot{\bf n})\right]{\bf n}\,dS = 0.
\end{displaymath} (352)

Finally, given that $S$ is arbitrary, the above expression reduces to the pressure balance constraint
\Delta p = \gamma\,\nabla\cdot{\bf n},
\end{displaymath} (353)

where $\Delta p = p_1-p_2$. 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 $\Delta p$ is the jump in pressure seen when crossing the interface in the opposite direction to ${\bf n}$. Of course, a plane interface is characterized by $\nabla\cdot{\bf n}=0$. On the other hand, a curved interface generally has $\nabla\cdot{\bf n}\neq 0$. In fact, $\nabla\cdot{\bf n}$ 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 up previous
Next: Spherical Interfaces Up: Surface Tension Previous: Introduction
Richard Fitzpatrick 2012-04-27