## Physics Friday 106

Consider two parallel rings of radius R, separated by a perpendicular distance 2d, with a soap bubble (in equilibrium) stretching between them. What is the shape of the soap bubble?

Due to symmetry, we note that the bubble will be a surface of revolution, formed by rotating a function r(z) about the z axis, which will be the perpendicular axis through the centers of the rings Further, mirror symmetry between the rings is such that if we choose z=0 to be halfway between the rings, then r(z) must be an even function. And, of course, the rings tell us that rd)=R. So, we just need to find r(z).

There are two methods of solving this. First, we can note that the bubble will be the minimum area surface fitting the above, and thus use the calculus of variations, as I did in Physics Friday 6 and
Physics Friday 19.
The surface area of the surface of revolution is given by the integral

(see equations 1-3 here).
Thus, to minimize this, we take “Lagrangian”  and apply the Euler-Lagrange equation. Actually, since , we can apply the Beltrami Identity:
,
where C is a constant of integration.
Differentiating L with respect to r‘ gives us
; plugging into the Beltrami Identity,
.
Note that this first order non-linear differential equation is separable:
,
and our bubble is a catenoid.

The second method of solving this is to note that for equilibrium, the surface tension γ (units: force/length) must be constant throughout the surface. Further, the net force due to this surface tension must be zero on any segment of the surface. Thus, we consider the forces on the edges of a thin ring segment of the bubble, from z1 to z2.

Due to symmetry, components perpendicular to the z direction of the tension acting on a portion of one edge are cancelled by other portions of the edge; the net force on one edge of this ring is simply the z component of the surface tension times the length of the edge. Considering the angle of the surface θ, we see the component of the tension is γcosθ, and the net force is thus . The requirement that this cancel on both sides of the ring is
, and thus rcosθ must be a constant. But calculus tells us that , and thus, via trigonometry, , and thus we have  is a constant, which we had in the previous method, namely, , giving the same solution .

To find our specific catenoid, we first note that by the reflection symmetry, z0=0, and so . Plugging in either of our ring constraints rd)=R gives us
,
which can only be solved for C numerically. More importantly, there is a critical value of  above which there is no (real) solution for C; the rings are too far apart for a stable bubble to bridge them. Secondly, for  below this critical value, there are two solutions. Noting that C has units of length, let us define dimensionless quantities
 and . Then our equation for C becomes . Comparing the graphs of  and , the critical value of η, η0 occurs when the two curves are tangent, so that
 and ; solving numerically, the two equations have simultaneous solution η0≈0.6627434, ζ0≈1.810171.

Now, to determine which solution of  for C is the physically valid one for , we find the area of the surface. As noted before,
; plugging in  (and ), we find
.
Using  and , we can rewrite the above as
.

Let us examine the numerical results for η=0.5.  has numerical solutions ζ≈1.178776 and ζ≈4.253600. If we plug these into our surface area results, we get for the lower solution

and for the higher solution
.

An examination of addional values will allow one to confirm that the smaller value of C solving  gives the minimum area surface; the other solution is not an extremum, but a saddle point; it corresponds to a bubble in an unstable equilibrium.