# Relationship between $\nabla\cdot n$ and the normal $n$

If $k=\nabla\cdot n$, what is the geometric relationship between $k$ and $n$? In terms of size and direction?

Is it true that $n$ is an outward-pointing normal iff $k>0$ and $n$ is inward-pointing iff $k<0$?

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Do you mean to say $k = \nabla N$? – badatmath May 4 '12 at 6:51
Please take a moment to consider what specific question you want to ask. – Austin Mohr May 4 '12 at 6:52
I mean that "∇.N" – narges May 4 '12 at 6:52
Ok,Thank you... – narges May 4 '12 at 6:53
Are you assuming an implicitly defined surface here? I would use a local orthogonal coordinate system to see that $\nabla\cdot n$ is independent of the implicit surface equation, and equals the sum of the two principal curvatures. This implies that your $k$ equals twice the mean curvature. I think $n$ must be outward-pointing normal if you want to get the "correct" sign for the mean curvature. – Thomas Klimpel May 4 '12 at 8:31