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Next: Exercises Up: Surface Tension Previous: Capillary Curves


Axisymmetric Soap-Bubbles

Consider an axisymmetric soap-bubble whose surface takes the form $r=f(z)$ in cylindrical coordinates. See Section C.3. The unit normal to the surface is
\begin{displaymath}
{\bf n} \equiv \frac{\nabla(r-f)}{\vert\nabla(r-f)\vert} = \frac{{\bf e}_r- f_z\,{\bf e}_z}{(1+f_z^{\,2})^{1/2}},
\end{displaymath} (396)

where $f_z\equiv df/dz$. Hence, from (1633), the mean curvature of the surface is given by
\begin{displaymath}
\nabla\cdot {\bf n} = \frac{1}{f\,f_z}\,\frac{d}{dz}\!\left[\frac{f}{(1+f_z^{\,2})^{1/2}}\right].
\end{displaymath} (397)

The Young-Laplace equation, (353), then yields
\begin{displaymath}
\frac{f\,f_z}{a} = \frac{d}{dz}\!\left[\frac{f}{(1+f_z^{\,2})^{1/2}}\right],
\end{displaymath} (398)

where
\begin{displaymath}
a = \frac{\gamma}{p_0}.
\end{displaymath} (399)

Here, $\gamma$ is the net surface tension, including the contributions from the internal and external soap/air interfaces. Moreover, $p_0=\Delta p$ is the pressure difference between the interior and the exterior of the bubble. Equation (398) can be integrated to give
\begin{displaymath}
\frac{f}{(1+f_z^{\,2})^{1/2}} = \frac{f^{\,2}}{2\,a} + C,
\end{displaymath} (400)

where $C$ is a constant.

Suppose that the bubble occupies the region $z_1\leq z\leq z_2$, where $z_1<z_2$, and has a fixed radius at its two end-points, $z=z_1$ and $z=z_2$. This could most easily be achieved by supporting the bubble on two rigid parallel co-axial rings located at $z=z_1$ and $z=z_2$. The net free energy required to create the bubble can be written

\begin{displaymath}
{\cal E} = \gamma\,S - p_0\,V,
\end{displaymath} (401)

where $S$ is area of the bubble surface, and $V$ the enclosed volume. The first term on the right-hand side of the above expression represents the work needed to overcome surface tension, whilst the second term represents the work required to overcome the pressure difference, $-p_0$, between the exterior and the interior of the bubble. Now, from the general principles of statics, we expect a stable equilibrium state of a mechanical system to be such as to minimize the net free energy, subject to any dynamical constraints. It follows that the equilibrium shape of the bubble is such as to minimize
\begin{displaymath}
{\cal E} = \gamma\int_{z_1}^{z_2} 2\pi\,f\,(1+f_z^{\,2})^{1/2}\,dz-p_0\int_{z_1}^{z_2}\pi\,f^{\,2}\,dz,
\end{displaymath} (402)

subject to the constraint that the bubble radius, $f$, be fixed at $z=z_1$ and $z=z_2$. Hence, we need to find the function $f(z)$ that minimizes the integral
\begin{displaymath}
\int_{z_1}^{z_2} {\cal L}(f,f_z)\,dz,
\end{displaymath} (403)

where
\begin{displaymath}
{\cal L}(f,f_z) = 2\pi\,\gamma\,f\,(1+f_z^{\,2})^{1/2} - \pi\,p_0\,f^{\,2},
\end{displaymath} (404)

subject to the constraint that $f$ is fixed at the limits. This is a standard problem in the calculus of variations. (See Appendix D.) In fact, since the functional ${\cal L}(f,f_z)$ does not depend explicitly on $z$, the minimizing function is the solution of [see Equation (1687)]
\begin{displaymath}
{\cal L}-f_z\,\frac{\partial {\cal L}}{\partial f_z} = C',
\end{displaymath} (405)

where $C'$ is an arbitrary constant. Thus, we obtain
\begin{displaymath}
2\pi\,\gamma\left[\frac{f}{(1+f_z^{\,2})^{1/2}}-\frac{f^{\,2}}{2\,a}\right] = C',
\end{displaymath} (406)

which can be rearranged to give Equation (400). Hence, we conclude that application of the Young-Laplace equation does indeed lead to a bubble shape that minimizes the net free energy of the soap/air interfaces.

Consider the case $p_0=0$, in which there is no pressure difference across the surface of the bubble. In this situation, writing $C=b>0$, Equation (400) reduces to

\begin{displaymath}
f = b\,(1+f_z^{\,2})^{1/2}.
\end{displaymath} (407)

Moreover, according to the previous discussion, the bubble shape specified by (407) is such as to minimize the surface area of the bubble (since the only contribution to the free energy of the soap/air interfaces is directly proportional to the bubble area). The above equation can be rearranged to give
\begin{displaymath}
f_z=\pm \left(\frac{f^{\,2}}{b^{\,2}}-1\right)^{1/2},
\end{displaymath} (408)

which leads to
\begin{displaymath}
z-z_0 = \int_b^r\,\frac{df}{f_z} = \pm\int_b^r\frac{df}{(f^{\,2}/b^{\,2}-1)^{1/2}}=\pm b\,\cosh^{-1}(r/b),
\end{displaymath} (409)

or
\begin{displaymath}
r = b\,\cosh(\vert z-z_0\vert/b),
\end{displaymath} (410)

where $z_0$ is a constant. This expression describes an axisymmetric surface known as a catenoid.

Figure 16: Radius versus axial distance for a catenoid soap bubble supported by two parallel co-axial rings of radius $c$ located at $z=\pm 0.65\,c$.
\begin{figure}
\epsfysize =3.5in
\centerline{\epsffile{Chapter04/fig6.07.eps}}
\end{figure}

Suppose, for instance, that the soap bubble is supported by identical rings of radius $c$ that are located a perpendicular distance $2\,d$ apart. Without loss of generality, we can specify that the rings lie at $z=\pm d$. It thus follows, from (410), that $z_0=0$, and

\begin{displaymath}
r = b\,\cosh(z/b).
\end{displaymath} (411)

Here, the parameter $b$ must be chosen so as to satisfy
\begin{displaymath}
c = b\,\cosh(d/b).
\end{displaymath} (412)

For example, if $d=0.65\,c$ then $b=0.6416\,c$, and the resulting bubble shape is illustrated in Figure 16.

Let $d/c=\zeta$ and $d/b=u$, in which case the above equation becomes

\begin{displaymath}
G(u) = u - \zeta\,\cosh u = 0.
\end{displaymath} (413)

Now, the function $G(u)$ attains a maximum value
\begin{displaymath}
G(u_0) = u_0-\frac{1}{\tanh u_0},
\end{displaymath} (414)

when $ u_0= \sinh^{-1}(1/\zeta)$. Moreover, if $G(u_0)>0$ then Equation (413) possesses two roots. It turns out that the root associated with the smaller value of $u$ minimizes the interface system energy, whereas the other root maximizes the free energy. Hence, the former root corresponds to a stable equilibrium state, whereas the latter corresponds to an unstable equilibrium state. On the other hand, if $G(u_0)<0$ then Equation (413) possesses no roots, implying the absence of any equilibrium state. The critical case $G(u_0)=0$ corresponds to $u=u_c$ and $\zeta=\zeta_c$, where $u_c\,\tanh u_c=1$ and $\zeta_c=1/\sinh u_c$. It is easily demonstrated that $u_c=1.1997$ and $\zeta_c=0.6627$. We conclude that a stable equilibrium state of a catenoid bubble only exists when $\zeta\leq \zeta_c$, which corresponds to $d\leq 0.6627\,c$. If the relative ring spacing $d$ exceeds the critical value $0.6627\,c$ then the bubble presumably bursts.

Figure 17: Radius versus axial distance for an unduloid soap bubble calculated with $k=0.95$.
\begin{figure}
\epsfysize =3.5in
\centerline{\epsffile{Chapter04/fig6.08.eps}}
\end{figure}

Consider the case $p_0\neq 0$, in which there is a pressure difference across the surface of the bubble. In this situation, writing

$\displaystyle 2\,a$ $\textstyle =$ $\displaystyle \alpha+\beta,$ (415)
$\displaystyle 2\,a\,C$ $\textstyle =$ $\displaystyle \alpha\,\beta,$ (416)

Equation (400) becomes
\begin{displaymath}
\frac{(\alpha+\beta)\,f}{(1+f_z^{\,2})^{1/2}}= f^{\,2} + \alpha\,\beta,
\end{displaymath} (417)

which can be rearranged to give
\begin{displaymath}
f_z = \mp \frac{(\alpha^2-f^{\,2})^{1/2}\,(f^{\,2}-\beta^{\,2})^{1/2}}{f^{\,2}+\alpha\,\beta}.
\end{displaymath} (418)

We can assume, without loss of generality, that $\vert\alpha\vert>\vert\beta\vert$. It follows, from the above expression, that $\vert\alpha\vert\leq f\leq \vert\beta\vert$. Hence, we can write
$\displaystyle f^{\,2}$ $\textstyle =$ $\displaystyle \alpha^2\,\cos^2\phi + \beta^{\,2}\,\sin^2\phi,$ (419)
$\displaystyle k^2$ $\textstyle =$ $\displaystyle \frac{\alpha^2-\beta^2}{\alpha^2},$ (420)

where $0\leq\phi\leq \pi/2$ and $0< k\leq 1$. It follows that
$\displaystyle f$ $\textstyle =$ $\displaystyle \vert\alpha\vert\,(1-k^2\,\sin^2\phi)^{1/2},$ (421)
$\displaystyle \beta$ $\textstyle =$ $\displaystyle {\rm sgn}(\beta)\,\vert\alpha\vert\,(1-k^2)^{1/2},$ (422)

and
\begin{displaymath}
\frac{dz}{d\phi} = \frac{1}{f_z}\,\frac{df}{d\phi} = \pm \left(f + \frac{\alpha\,\beta}{f}\right),
\end{displaymath} (423)

which can be integrated to give
\begin{displaymath}
\vert z\vert= \vert\alpha\vert\left[E(\phi,k) + {\rm sgn}(\alpha\,\beta)\,(1-k^2)^{1/2}\,F(\phi,k)\right],
\end{displaymath} (424)

where $E(\phi,k)$ and $F(\phi,k)$ are incomplete elliptic integrals [see Equations (381) and (382)]. Here, we have assumed that $\phi=0$ when $z=0$. There are three cases of interest.

Figure 18: Radius versus axial distance for a positive pressure nodoid soap bubble calculated with $k=0.95$.
\begin{figure}
\epsfysize =3.5in
\centerline{\epsffile{Chapter04/fig6.09.eps}}
\end{figure}

In the first case, $\alpha>0$ and $\beta>0$. It follows that $(1-k^2)^{1/2}\leq r/\alpha\leq 1$ for $\pi/2\geq\phi\geq 0$, and $0.5\leq \gamma/(p_0\,\alpha) < 1$ for $1\geq k> 0$, where

$\displaystyle r$ $\textstyle =$ $\displaystyle \alpha\,(1-k^2\,\sin^2\phi),$ (425)
$\displaystyle \vert z\vert$ $\textstyle =$ $\displaystyle \alpha\left[E(\phi,k) +(1-k^2)^{1/2}\,F(\phi,k)\right].$ (426)

The axisymmetric curve parameterized by the above pair of equations is known as an unduloid. Note that an unduloid bubble always has positive internal pressure (relative to the external pressure): i.e., $p_0>0$. An example unduloid soap bubble is illustrated in Figure 17

Figure 19: Radius versus axial distance for a negative pressure nodoid soap bubble calculated with $k=0.95$.
\begin{figure}
\epsfysize =3.5in
\centerline{\epsffile{Chapter04/fig6.10.eps}}
\end{figure}

In the second case, $\alpha>0$ and $\beta<0$. It follows that $(1-k^2)^{1/4}\leq r/\alpha \leq 1$ for $\phi_0\geq \phi\geq 0$, and $0<\gamma/(p_0\,\alpha)\leq 0.5$ for $0< k\leq 1$, where $\phi_0= \sin^{-1}([1-(1-k^2)^{1/2}]^{1/2}/k)$, and

$\displaystyle r$ $\textstyle =$ $\displaystyle \alpha\,(1-k^2\,\sin^2\phi),$ (427)
$\displaystyle \vert z\vert$ $\textstyle =$ $\displaystyle \alpha\left[E(\phi,k) -(1-k^2)^{1/2}\,F(\phi,k)\right].$ (428)

The axisymmetric curve parameterized by the above pair of equations is known as an nodoid. This particular type of nodoid bubble has positive internal pressure: i.e., $p_0>0$. An example positive pressure nodoid soap bubble is illustrated in Figure 18.

In the third case, $\alpha<0$ and $\beta>0$. It follows that $(1-k^2)^{1/2}\vert\leq r/\vert\alpha\vert\leq \alpha\vert\,(1-k^2)^{1/4}$ for $\pi/2\geq \phi\geq \phi_0$ (or $\pi/2\leq \phi\leq \pi-\phi_0$), and $0>\gamma/(p_0\,\vert\alpha\vert)\geq -0.5$ for $0< k\leq 1$, where

$\displaystyle r$ $\textstyle =$ $\displaystyle \vert\alpha\vert\,(1-k^2\,\sin^2\phi),$ (429)
$\displaystyle \vert z\vert$ $\textstyle =$ $\displaystyle \vert\alpha\vert\left[E(\phi,k) -(1-k^2)^{1/2}\,F(\phi,k)\right].$ (430)

The axisymmetric curve parameterized by the above pair of equations is again a nodoid. However, this particular type of nodoid bubble has negative internal pressure: i.e., $p_0<0$. An example negative pressure nodoid soap bubble is illustrated in Figure 19.


next up previous
Next: Exercises Up: Surface Tension Previous: Capillary Curves
Richard Fitzpatrick 2012-04-27