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Capillary Curves

Let adopt Cartesian coordinates on the Earth's surface such that $ z$ increases vertically upward. Suppose that the interface of a liquid of density $ \rho$ and surface tension $ \gamma$ with the atmosphere corresponds to the surface $ z=f(x)$ , where the liquid occupies the region $ z<f(x)$ . The shape of the interface is assumed to be $ y$ -independent. The unit normal to the interface (directed from liquid to air) is thus

$\displaystyle {\bf n} = \frac{\nabla(z-f)}{\vert\nabla(z-f)\vert}=\frac{{\bf e}_z - f_x\,{\bf e}_x}{(1+f_x^{\,2})^{1/2}},$ (3.25)

where $ f_x = df/dx$ . Hence, the mean curvature of the interface is (Riley 1974)

$\displaystyle \nabla\cdot {\bf n} = - \frac{f_{xx}}{(1+f_x^{\,2})^{3/2}},$ (3.26)

where $ f_{xx}= d^{\,2}f/dx^{\,2}$ . According to Equations (3.15) and (3.16), the shape of the interface is governed by the nonlinear differential equation

$\displaystyle f = \frac{l^{\,2}\,f_{xx}}{(1+f_x^{\,2})^{3/2}}.$ (3.27)

where the vertical height, $ f$ , of the interface is measured relative to its equilibrium height in the absence of surface tension. Multiplying the previous equation by $ f_x/l^{\,2}$ , and integrating with respect to $ x$ , we obtain

$\displaystyle \frac{1}{(1+f_x^{\,2})^{1/2}} = C-\frac{f^{\,2}}{2\,l^{\,2}},$ (3.28)

where $ C$ is a constant. It follows that

$\displaystyle C - \frac{f^{\,2}}{2\,l^{\,2}}\geq 1,$ (3.29)


$\displaystyle \frac{1}{f_x} = \mp \frac{C-f^{\,2}/2\,l^{\,2}}{[1-(C-f^{\,2}/2\,l^{\,2})^2]^{1/2}}.$ (3.30)

Figure: Capillary curves for $ \pi /4\leq \phi \leq 3\pi /4$ and (in order from the top to the bottom) $ k=0.6$ , $ 0.7$ , $ 0.8$ , $ 0.9$ , and $ 0.99$ .
\epsfysize =3.25in


$\displaystyle C = \frac{2}{k^{\,2}}-1,$ (3.31)

where $ 0< k< 1$ , and

$\displaystyle f = \pm\frac{2\,l}{k}\,(1-k^{\,2}\,\sin^2\phi)^{1/2}.$ (3.32)

Thus, from Equations (3.31) and (3.32),

$\displaystyle C - \frac{f^{\,2}}{2\,l^{\,2}} =-\cos(2\,\phi),$ (3.33)

and so the constraint (3.29) implies that $ \pi /4\leq \phi \leq 3\pi /4$ . Moreover, Equations (3.30) and (3.33) reduce to

$\displaystyle \frac{1}{f_x}= \frac{dx}{df} = \pm \frac{1}{\tan(2\,\phi)}.$ (3.34)

It follows from Equations (3.32) and (3.34) that

$\displaystyle \frac{dx}{d\phi} = \frac{dx}{df}\,\frac{df}{d\phi} = -\frac{l\,k\,\cos(2\,\phi)}{(1-k^{\,2}\,\sin^2\phi)^{1/2}},$ (3.35)

which can be integrated to give

$\displaystyle \frac{x}{l} = \int_\phi^{\pi/2} \frac{k\,\cos(2\,\phi)}{(1-k^{\,2}\,\sin^2\phi)^{1/2}}\,d\phi,$ (3.36)

assuming that $ x=0$ when $ \phi=\pi/2$ . Thus, we get

$\displaystyle \frac{x}{l} = \left(k-\frac{2}{k}\right)\tilde{F}(\phi,k) + \frac{2}{k}\,\tilde{E}(\phi,k),$ (3.37)


$\displaystyle \tilde{E}(\phi,k)$ $\displaystyle = E(\pi/2,k)- E(\phi,k),$ (3.38)
$\displaystyle \tilde{F}(\phi,k)$ $\displaystyle = F(\pi/2,k)- F(\phi,k),$ (3.39)


$\displaystyle E(\phi,k)$ $\displaystyle = \int_0^\phi (1-k^{\,2}\,\sin^2\phi)^{1/2},$ (3.40)
$\displaystyle F(\phi,k)$ $\displaystyle = \int_0^\phi (1-k^{\,2}\,\sin^2\phi)^{-1/2},$ (3.41)

are types of incomplete elliptic integral (Abramowitz and Stegun 1965). In conclusion, the interface shape is determined parametrically by

$\displaystyle \frac{x}{l}$ $\displaystyle = \left(k-\frac{2}{k}\right)\tilde{F}(\phi,k) + \frac{2}{k}\,\tilde{E}(\phi,k),$ (3.42)
$\displaystyle \frac{z}{l}$ $\displaystyle = \pm\frac{2}{k}\,(1-k^{\,2}\,\sin^2\phi)^{1/2},$ (3.43)

where $ \pi /4\leq \phi \leq 3\pi /4$ . Here, the parameter $ k$ is restricted to lie in the range $ 0< k< 1$ .

Figure: Liquid/air interface for a liquid trapped between two vertical parallel plates located at $ x=\pm l$ . The contact angle of the interface with the plates is $ \theta =30^\circ $ .
\epsfysize =3.25in

Figure 3.4 shows the capillary curves predicted by Equations (3.42) and (3.43) for various different values of $ k$ . Here, we have chosen the plus sign in Equation (3.43). However, if the minus sign is chosen then the curves are simply inverted: that is, $ x\rightarrow x$ and $ z\rightarrow -z$ . In can be seen that all of the curves shown in the figure are symmetric about $ x=0$ : that is, $ z\rightarrow z$ as $ x\rightarrow -x$ . Consequently, we can use these curves to determine the shape of the liquid/air interface which arises when a liquid is trapped between two flat vertical plates (made of the same material) that are parallel to one another. Suppose that the plates in question lie at $ x=\pm d$ . Furthermore, let the angle of contact of the interface with the plates be $ \theta $ , where $ \theta <\pi/2$ . Because the angle of contact is acute, we expect the liquid to be drawn upward between the plates, and the interface to be concave (from above). This corresponds to the positive sign in Equation (3.43). In order for the interface to meet the plates at the correct angle, we require $ f_x=-1/\tan\theta$ at $ x=-d$ and $ f_x=1/\tan\theta$ at $ x=+d$ . However, if one of these boundary conditions is satisfied then, by symmetry, the other is automatically satisfied. From Equation (3.34) (choosing the positive sign), the latter boundary condition yields $ \tan(2\,\phi)=1/\tan\theta$ at $ x=+d$ , which is equivalent to $ x=+d$ when $ \phi=3\pi/4-\theta/2$ . Substituting this value of $ \phi$ into Equation (3.42), we can numerically determine the value of $ k$ for which $ x=d$ . The interface shape is then given by Equations (3.42) and (3.43), using the aforementioned value of $ k$ , and $ \phi$ in the range $ \pi/4+\theta/2$ to $ 3\pi/4-\theta/2$ . For instance, if $ d=l$ and $ \theta =30^\circ $ then $ k=0.9406$ , and the associated interface is shown in Figure 3.5. Furthermore, if we invert this interface (i.e., $ x\rightarrow x$ and $ z\rightarrow -z$ ) then we obtain the interface which corresponds to the same plate spacing, but an obtuse contact angle of $ \theta=180^\circ-30^\circ=150^\circ$ .

Consider the limit $ k\ll 1$ , which is such that the distance between the two plates is much less than the capillary length. It is easily demonstrated that, at small $ k$ (Abramowitz and Stegun 1965)

$\displaystyle \tilde{E}(\phi,k)$ $\displaystyle \simeq \skew{5}\tilde{\phi} - \frac{k^{\,2}}{4}\,(\skew{5}\tilde{\phi}+\sin\skew{5}\tilde{\phi}\,\cos\skew{5}\tilde{\phi}),$ (3.44)
$\displaystyle \tilde{F}(\phi,k)$ $\displaystyle \simeq \skew{5}\tilde{\phi} + \frac{k^{\,2}}{4}\,(\skew{5}\tilde{\phi}+\sin\skew{5}\tilde{\phi}\,\cos\skew{5}\tilde{\phi}),$ (3.45)

where $ \skew{5}\tilde{\phi}=\pi/2-\phi$ . Thus, Equations (3.42) and (3.43) reduce to

$\displaystyle \frac{x}{l}$ $\displaystyle \simeq -\frac{k}{2}\,\sin(2\,\skew{5}\tilde{\phi}),$ (3.46)
$\displaystyle \frac{z}{l}$ $\displaystyle \simeq \frac{2}{k}-\frac{k}{2}\,[1-\cos(2\,\skew{5}\tilde{\phi})].$ (3.47)

It follows that the interface is a segment of the curved surface of a cylinder whose axis runs parallel to the $ y$ -axis. If the distance between the plates is $ 2\,d$ , and the contact angle is $ \theta $ , then we require $ x=d$ when $ \phi=3\pi/4-\theta/2$ (which corresponds to $ \skew{5}\tilde{\phi}=-\pi/4+\theta/2$ ). From Equation (3.46), this constraint yields

$\displaystyle \frac{d}{l}\simeq \frac{k}{2}\,\cos\theta.$ (3.48)

Thus, the height that the liquid rises between the two plates--that is, $ h = z(x=0)=z(\skew{5}\tilde{\phi}=\pi/2)\simeq 2\,l/k$ --is given by

$\displaystyle h \simeq \frac{\gamma\,\cos\theta}{\rho\,g\,d}.$ (3.49)

This result is the form taken by Jurin's law, (3.23), for a liquid drawn up between two parallel plates of spacing $ 2\,d$ .

Figure: Liquid/air interface for a liquid in contact with a vertical plate located at $ x=0$ . The contact angle of the interface with the plate is $ \theta =25^\circ $ .
\epsfysize =3.25in

Consider the case $ k=C =1$ , which is such that the distance between the two plates is infinite. Let the leftmost plate lie at $ x=0$ , and let us completely neglect the rightmost plate, because it lies at infinity. Suppose that $ h=z(x=0)$ is the height of the interface above the free surface of the liquid at the point where the interface meets the leftmost plate. If $ \theta $ is the angle of contact of the interface with the plate then we require $ f_x=-1/\tan\theta$ at $ x=0$ . Because $ C=1$ , it follows from Equation (3.28) that

$\displaystyle \frac{h^{\,2}}{2\,l^{\,2}} = 1-\sin\theta,$ (3.50)


$\displaystyle h=2\,l\,\sin(\pi/4-\theta/2).$ (3.51)

Furthermore, again recalling that $ C=1$ , Equation (3.30) can be integrated to give

$\displaystyle x = \int_z^h \frac{df}{f_x}= l \int_{z/2l}^{h/2l}\frac{1-2\,y^{\,2}}{y\,(1-y^{\,2})^{1/2}}\,dy,$ (3.52)

where we have chosen the minus sign, and $ y=f/(2\,l)$ . Making the substitution $ y=\sin u$ , this becomes

$\displaystyle \frac{x}{l} = \int_{\sin^{-1}(z/2\,l)}^{\sin^{-1}(h/2\,l)}\left(\...
...cos u}{\sin u}\right)+2\,\cos u\right]_{\sin^{-1}(z/2\,l)}^{\sin^{-1}(h/2\,l)},$ (3.53)

which reduces to

$\displaystyle \frac{x}{l} = \cosh^{-1}\left(\frac{2\,l}{z}\right)-\cosh^{-1}\le...
...c{h^{\,2}}{l^{\,2}}\right)^{1/2}- \left(4-\frac{z^{\,2}}{l^{\,2}}\right)^{1/2},$ (3.54)

because $ \cosh^{-1}(z)\equiv \ln[z+(z^{\,2}-1)^{1/2}]$ . Thus, Equations (3.51) and (3.54) specify the shape of a liquid/air interface that meets an isolated vertical plate at $ x=0$ . In particular, Equation (3.51) gives the height that the interface climbs up the plate (relative to the free surface) due to the action of surface tension. Note that this height is restricted to lie in the range $ -2\,l\leq h\leq 2\,l$ , irrespective of the angle of contact. Figure 3.6 shows an example interface calculated for $ \theta =25^\circ $ .

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Next: Axisymmetric Soap-Bubbles Up: Surface Tension Previous: Jurin's Law
Richard Fitzpatrick 2016-03-31