Zero Mean Curvature

I heard somewhere or another the claim that soap film forms a "surface of zero mean curvature".  I wasn't sure exactly what that meant until I read this book which gave me the tools to understand what that means, and prove it.  It turns out to be a simple consequence of Gauss's Divergence theorem.  In tensor notation Gauss's Divergence theorem states

$$
\int_{\Omega}\nabla_iT^idV = \int_{\partial{\Omega}}T_iN^idA
$$
where

• $\Omega$ is some volume of space
• $\nabla_i$ is the covariant derivative along the $i$th coordinate $z^i$
• $\delta{\Omega}$ is the surface of the volume
• $T^i$ is any single index tensor defined over the whole volume
• $N^i$ is the unit vector normal to the surface in contravariant form

 This works in any number of dimensions, so if you take surface embedded in 3 dimensional euclidean space $\vec{z} = \vec{z}(s^1,s^2)$ and cut it, then the theorem tells us

$$
\int_{S}\nabla_{\alpha}T^{\alpha}dA = \int_{\partial{S}}T_{\alpha}n^{\alpha}dl
$$
where

• $S$ is a region of the embedded surface
• $\partial{S}$ is its edge
• $\alpha$ indicates indices that range over surface not ambient coodinates
• $T^{\alpha}$ is any single index tensor defined over the whole region.
• $n^{\alpha}$ is the unit vector lying in the surface normal to the edge.

Using the notation $\vec{s_\alpha}$ for the covariant basis $\frac{\partial{\vec{R}}}{\partial{s^\alpha}}$ of the surface, the curvature $B^\alpha_\beta$ is defined by
$$
 B^\alpha_\beta \vec{N} = \nabla_\beta \vec{s^\alpha}
$$
which works because you can show that the RHS is always normal to the surface.  The mean curvature is just the contraction
$$
 B^\alpha_\alpha
$$
We can now prove the theorem by noting that $\vec{s^\alpha}$ is a single index tensor defined over the whole surface.  (It doesn't invalidate Gauss' theorem that its values are vectors rather than numbers.)  Substituting $\vec{s^\alpha}$ for $T^\alpha$ we get
$$
\int_{S}\nabla_{\alpha}\vec{s^{\alpha}}dA = \int_{\partial{S}}\vec{s_\alpha}n^{\alpha}dl
$$
which we can rewrite as
$$
\int_{S}B^\alpha_\alpha \vec{N}dA = \int_{\partial{S}}\vec{n}dl
$$
But since the surface tension is everywhere constant on the surface of a soap film the RHS is proportional to the overall force on the surface region due to tension.  If the region is stationary then the overall force must be zero, meaning that the RHS is zero (assuming that the air pressure is the same either side of the surface).  Thus for non-closed soap film surface regions
$$
\int_{S}B^\alpha_\alpha \vec{N}dA =0
$$
And since this is true for any sub region S we must have
$$
B^\alpha_\alpha =0
$$
QED.

POSTSCRIPT

Since the energy required to stretch a soap film over a fixed boundary is proportional to its area, it is clear that the solution must also minimize the surface area.  It turns out that the requirement that the surface area is minimized is equivalent to the requirement that the mean curvature is everywhere zero.  It is slightly more involved to prove this but a proof of it can be found in Chapter 15 of Introduction to Tensor Analysis and the Calculus of Moving Surfaces.