Relation between divergence of unit normal and radius of curvature I don't understand how does the divergence of a unit normal vector to a curve at a point gives the local radius of curvature. For simplicity consider a 2-D curve.
$$\nabla.n=\frac{1}{R}$$
I want to understand the mathematical proof for the expression and also some physical intuition to understand why is this true.
 A: To give you feel of what is involved consider the diagram shown below.

For small angle $d \theta$, $n=n'=1 \Rightarrow dn = n\,d\theta=d\theta$ and $R=R' \Rightarrow dR = R\,d\theta$
Thus $\frac {dn}{dR} = \frac 1R$.
A: I remember a Tensor calculus component proof in Pavel Grinfeld's book but a much more I've attempted at a simpler Geometric explanation of the formula using definition of divergence via integral in this post of MSE adapted from Tristan Needham's book. You may also see linked post to Math Overflow for more detailed discussion.
Anyhow, given the formula:

$$ \nabla \cdot X = \partial_s |X| + \kappa_p |X|$$

If we have unit normal to curve $\hat{n}$ , then the first term of RHS is zero as unit normal doesn't change length. Also, the second term is just $\kappa_p$ i.e: curvature of the curve we are considering. Hence, we can write:
$$ \nabla \cdot n= \kappa_p $$
A more detailed derivation and conditions for extending this result into higher dimensions is in linked.

Here is an example calculation using this formula: We have a circle of radius $r$: $x^2 +y^2 = r^2$, one can find it's unit normal as $\frac{<x,y>}{\sqrt{x^2 +y^2}}$. Plugging this into formula we have:
$$ \nabla \cdot \left[ \frac{<x,y>}{\sqrt{x^2 +y^2}}\right] =  \kappa_p (1)$$
The left side, it is equivalent to taking divergence of the unit radial vector $\hat{r}$. By computing divergence in polar:
$$ \nabla \cdot \left[ \frac{<x,y>}{\sqrt{x^2 +y^2}}\right]=\nabla \cdot (1 \hat{r}) = \frac{1}{r} \frac{\partial }{\partial r} ( 1 \cdot r) = \frac{1}{r}$$
So, $\kappa_p = \frac{1}{r}$ which is what we need.
A: I'm really bad at drawing but have a look at this:

The bottom part can be seen as a segment of the original surface $S_o$ all points on it flow according to the surface normal, so after some time $t$ they from the top face. If we were to integrate the surface normal over this this cube and divide it by it's volume, we have the definition of divergence:
$$ \lim_{dV \to 0} \frac{\oint n \cdot dS}{dV} = \nabla \cdot n$$
We need to work out LHS, so here $dV$ we can take to be the volume increase of the oscullating sphere times some angle propotionality constant $\kappa$. If the radius of osculating sphere is given as $\kappa d(\frac{4}{3} \pi r^3)$,.  For the actual surface integral, it is easy to see that only the top most and bottom most face contribute to it (from prespective we see it in photo). By how much? Well by surface area of patch at $S_{t+dt} $ and negative of that of $S_t$. This would be $ d(4 \pi r^2) \kappa$ where $\kappa$ is the same angle propotionality factor. We have:
$$  \lim_{ dV \to 0} \frac{\oint n \cdot dS}{dV} = \frac{d( 4 \pi r^2) \kappa}{ d( \frac{4}{3} \pi r^3) \kappa} = \frac{1}{r} = \nabla \cdot n$$
