A soap film spans two coaxial rings of radius held apart

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Mechanics & Dynamical Systems · Lagrangian mechanics

Action and Symmetry, Ch The action, the first variation and the boundary term

Problem

A soap film spans two coaxial rings of radius a=5.0 cma = 5.0\,\mathrm{cm} held 2z0=6.0 cm2z_0 = 6.0\,\mathrm{cm} apart. Its stationary profiles are catenoids r=ccosh⁡(z/c)r = c\cosh(z/c), with the ring condition z0/a=z~/cosh⁡z~z_0/a = \tilde z/\cosh\tilde z, where z~=z0/c\tilde z = z_0/c. The area of a catenoid between the rings is 2πc2(z~+sinh⁡z~cosh⁡z~)2\pi c^2(\tilde z + \sinh\tilde z\cosh\tilde z). ((a)) Show that z~/cosh⁡z~\tilde z/\cosh\tilde z is largest where z~tanh⁡z~=1\tilde z\tanh\tilde z = 1, and find the largest ring separation, as a multiple of aa, at which any catenoid spans the rings. ((b)) Find the two roots z~\tilde z for these rings and the waist radius cc of each catenoid. ((c)) Compare the area of the shallow catenoid with the area of two flat discs, one across each ring. Which is the smaller area, and why does a film pulled slowly out to this separation still hold the catenoid?

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