Divergence, curl of normal vector

[techmologist]

How do you interpret the divergence or curl of the unit normal defined on a surface? This sometimes comes up when applying Stokes' theorem. A simple example would be

Surface area =
[tex]\int_{S} \hat{n} \cdot \hat{n} dA = \int_{V} \nabla \cdot \hat{n} dV[/tex]

where S is the closed surface that bounds a volume V. Since the normal n is defined on S, how do you interpret div n in the interior region? Do you just extend the field n on S to a field N on V in such a way that it is continuously differentiable and satisfies N = n on S?

I am assuming that there's nothing wrong with applying Stokes'/Divergence theorem when the vector field being integrated depends on the region of integration.

 

[LCKurtz]

Stokes' theorem does not apply if the vector field is not defined on an open region containing the volume V. And the idea of extending the definition of the normal to the region wouldn't work because it wouldn't be unique anyway.

For what it's worth, that surface integral on the left just evaluates to the area of the surface.

 

[techmologist]

Thanks for the reply. Does that mean that the derivation here

https://www.physicsforums.com/showthread.php?t=525080

is invalid? At first I didn't think about it, but then I realized that what I'm calling div n in that derivation is only the divergence due to the surface coordinates. If the normal contribution to the divergence,

[tex]\mathbf{n} \cdot \frac{\partial}{\partial n} \mathbf{n}[/tex]

is not zero, then div n is not the mean curvature. The expression above is zero if you assume that the field n remains unit length as it extends from the surface into the surrounding volume.

EDIT: Well, at least the integrand in

∫(∇n)⋅n−(∇⋅n)n da

depends only on how the normal vector field changes on the surface S. That integrand is really what I calculated as twice the mean curvature. And of course the mean curvature can only depend on the values of n on the surface, not on how n (its extension, actually) varies away from the surface.