1. Nov 14, 2014

### JanEnClaesen

In other words, when we take for potential function instead of F the square root of (6F/6x)²+(6F/6y)² (in the particular case of two-dimensions). Does this lead to anything interesting?

2. Nov 16, 2014

### Simon Bridge

Define "interesting".
Why don't you follow it and see?

Note: I think you mean $$U=\sqrt{\frac{\partial F}{\partial x^2}+\frac{\partial F}{\partial y^2}}$$ ... the Σ button on the toolbar has symbols includeing ∂