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A Hamiltonian Look at a Soap Film

A soap film stretched between two rings that share a common axis—as in Figure 1—has a surface of minimal area that, when divided by $$2 \pi$$, is

$\int_{-a}^{a} \underbrace{y \sqrt{1+(y ^\prime) ^2 }}_{L(y,y ^\prime )} dx = \int_{x=-a}^{x=a} y\,ds\tag1$

and subject to boundary conditions

$y( \pm a ) = r.\tag2$

Alternatively, $$(1)$$ is the potential energy of the film whose surface tension1 is $$1/ 2\pi$$.

Figure 1. Soap film — a minimal surface that spans two circular hoops.

The Euler-Lagrange equation for the minimizer of $$(1)$$ does not look appealing, and I relegate it to a footnote later on. But interestingly, the equivalent Hamiltonian system turns out to be very symmetric and simple. Indeed, the standard definition2 of momentum $$p$$ yields

$p \overset{def}{=} L_{y ^\prime }(y,y ^\prime ) = \frac{y y ^\prime }{\sqrt{1+(y ^\prime) ^2 }} = y \sin \theta, \tag3$

where $$\theta$$ is marked in Figure 1. And by a short computation, the Hamiltonian $$H\overset{def}{=} p y ^\prime - L$$ comes out to be

Figure 2. Trajectories of the Hamiltonian system $$(5)$$. Since $$H= {\rm const.}$$ along each trajectory, every solution behaves like a solution to a linear system (although $$(5)$$ is nonlinear).
$H = - \sqrt{ y ^2 - p ^2 } =- y \cos \theta.\tag4$

Trajectories of the Hamiltonian system, i.e., the level curves of $$H$$, are hyperbolas $$y^2 - p^2 = {\rm const.}$$ (see Figure 2). And Hamilton’s equations $$y ^\prime = H_p$$, $$p ^\prime =- H_y$$ are pleasantly simple:3

$y ^\prime = -p/H, \ \ p ^\prime = -y/H.\tag5$

Since $$H = {\rm const.}$$ along solutions, we conclude that $$y ^{\prime\prime} = H^{-2}y$$. By the symmetry of the boundary conditions $$(2)$$, $$y = y_0 \cosh (x/H)$$. Setting $$x=0$$ in the expression for $$H$$ makes $$y=y_0$$ and $$p \overset{(3)} = 0$$. Consequently, $$|H| = y_0$$ and thus

$y=y_0 \cosh(x/y_0).\tag6$

Here, $$y_0$$ must be chosen so that $$(2)$$ holds.

Some Observations

1. All minimal surfaces spanning two rings are dilations of one another. This is evident even without $$(6)$$ from the fact that dilations preserve the property of area minimization.
2. If the rings are spread too far apart for fixed $$r$$, the soap film will snap. How far is too far? The graphs of $$(6)$$ fill the sector $$y \geq s |x|$$, where $$s=1.996 \ldots$$ is the slope of the line through the origin tangent to the graph of $$y= \cosh x$$ (see Figure 3b). A soap film can therefore span two rings iff $$r/a \geq s \approx 2$$. The maximal possible distance between the rings is almost equal to their diameter!

Figure 3. 3a. Graphs of $$(6)$$ fill the sector. 3b. The definition of slope $$s=\min_{x>0}x ^{-1} \cosh x=1.996...$$ 3c. Points $$A$$ and $$B$$ satisfy $$r/a>s$$ and two possible shapes exist for the soap film, but only $$AmB$$ is stable and physically observable. Both curves $$AnB$$ and $$AmB$$ are geodesics in the metric $$y \,ds$$, but only $$AnB$$ has a conjugate pair of points $$a$$ and $$b$$. As such, it cannot be minimal; in fact, its Morse index is $$1$$ [1]. In contrast, $$AmB$$ has no conjugate points and is thus a minimizer of the “length” $$\int y\,ds$$. The case of $$r/a=s$$ is critical and unstable, as the two solutions coalesce into a saddle node of potential energy.

3. Two critical functions of $$(1)$$ exist if $$r/a > s$$ in Figure 3. However, only one is physically realizable (the other is unstable). If we had somehow managed to give the film an initial shape that was close to $$AnB$$, it would either snap to $$AmB$$ or pinch off.
4. Equilibrium shapes are geodesics in the metric $$y\;ds$$, and we can think of them as rays of light propagating through the medium with the speed of light $$c=1/y$$. Indeed, the travel time for the light is $$\int dt= \int ds/c = \int y\,ds$$ — the same expression as the area in $$(1)$$. So the area can be interpreted as either the potential energy of the film or as the travel time in the artificial optical medium.
5. Figure 4. 4a. $$H$$ is the horizontal force that acts on any section of the soap “tube,” up to a constant factor that depends on the surface tension. $$H= {\rm const.}$$ as a function of $$x$$ is the consequence of Newton’s first law. 4b. Here, $$y \sin \theta = p$$ is the cumulative force on a portion of the tube that acts in the plane of the section.
More generally, if $$(1)$$ is replaced by $$\int_{a}^{b} F(y) ds$$, then the corresponding Hamiltonian becomes $$H = F(y) \cos \theta$$ and the momentum is $$p = F(y) \sin \theta$$.

Physical Meaning of $$H$$ and $$p$$

In dynamics, $$H$$ and $$p$$ are the energy and momentum. But in our static problem where $$x$$ is the static counterpart of time, the meanings of $$H$$ and $$p$$ are different; instead, they are the forces that are described in the caption of Figure 4.

A GeometricalInterpretation of $$H$$ and $$p$$

Figure 5a depicts an alternative interpretation of $$H$$ and $$p$$ (here, $$y(x)$$ is any function). In Figure 5b, $$y$$ is of the form $$(6)$$; as $$F$$ slides along the $$x$$-axis, the length $$|FR| = {\rm const.}$$ and the velocity of $$R$$ points at $$F$$ (proving the latter is left as a challenge).

Figure 5. 5a. Geometrical meaning of the momentum and the Hamiltonian. 5b. The moving segment $$RF$$ of constant length $$|H|$$ is a “bike” whose “front wheel” $$F$$ follows the $$x$$-axis; the velocity of the “rear wheel” $$R$$ is aligned with the segment $$RF$$ and follows a tractrix, the involute of the catenary.

1 Surface tension is the force that is applied to each side of an incision to hold the slit together per the incision’s unit length.

2 In almost all textbooks, the definitions of $$p$$ and $$H$$ are pulled out of thin air, although they do have a very natural and eye-opening motivation [1].

3 In contrast to the messier Euler-Lagrange equation: $$\frac{d}{dx} \frac{y y ^\prime }{\sqrt{1+(y ^\prime) ^2 }} - \sqrt{ 1+(y ^\prime) ^2 } = 0$$.