What does the surface gradient dot product with the normal vector mean?

Question

The surface gradient is defined as $$ \nabla_s = \nabla - \bar n \frac{\partial}{\partial n} $$

if applied on a function $\sigma$ we get $$ \nabla_s \sigma = \nabla \sigma - \bar n \frac{\partial \sigma}{\partial n} $$

What happens when this operator is dotted with the normal vector $\bar n$? $$ \nabla_s \cdot \bar{n}$$

Answer 1

In that case, and provided your surface is axisymmetric, $\nabla_s \cdot \bar{n} = \kappa$ where $\kappa$ is the local curvature. For a proof, you can refer to this post: Surface gradient and curvature