How to prove a Minimal Surface minimizes Surface Tension I know minimal surfaces often minimize surface area.  I know its mean curvature is found by setting the divergence of the second fundamental form equal to zero.  Is this the same or analogous to showing tension too is minimized by assuming the shape the surface assumes is perfectly elastic (springing back to its unperturbed state when a perturbing force is removed)?  A reference at the advanced undergraduate level would be greatly appreciated.  Thanks.
 A: It is proved by Laws of physics and in that sense no separate proof is needed. We should look to the scalar divergence whether it is constant or zero.
Surface tension is a property, an invariant physical constant that should not be changed. It is the area which is to be minimized.
In the language of calculus of variations under action of variable parameters... concepts of 1) minimal ( or maximal) and 2) constant are same... respectively  local or global their derivatives should vanish.
If $N$ denotes force per unit length (aka surface tension) and $p$ the outside pressure then with $1/R_1= \kappa1, 1/R2=\kappa2, $ we sum up two perpendicular planes forces per unit length with pressure normal to surface we have equilibrium
$$ \dfrac{N_1}{R_1} + \dfrac{N_2}{R_2} =p $$
We then have as a property of isotropic films. Taking
$$N_1= N_2 = N ,\, \dfrac{ \kappa1 +  \kappa1}{2}= H$$
$$ H= \dfrac{p}{2N} \, ; $$
When pressure differential $ p=0,H=0 $ and we have minimal surfaces as the free soap films and $p= const$ as Constant Mean Curvature $(CMC)$ soap films under action of uniform pressure on one of the two sides. Surfaces of revolution of the latter situation  $ (H= const.) $ are the Delaunay Unduloids.
And btw side interest is that it is also

*

*locus of ellipse focus rolling on a line tracing out such a  curve.


*Maximum volume of given surface area.
If you speak German.. Differential Geometrie, Wilhelm Blaschke 1921 First Edition  German.
