Why does divergence represent expansion or contraction?

Question

Why does $\mathrm{div}\ V$ represent how much $V$ is expanding or contracting? By its definition I know that diverging means deviating from its original path, but what about $\mathrm{div}\ V$ makes it so $V$ expands or contracts, is there a $\mathrm{div}$ formula that explains it?

Answer 1

A common definition of the divergence is $$ \operatorname{div}{F(x)} = \lim_{\text{Vol}(S) \to 0} \frac{1}{\text{Vol}(S)} \int_S V \cdot dS, $$ where $S$ is some set of sensible (i.e. smoothish, bounded, and if you want an easy life, convex) surfaces that enclose the point $x$. An obvious set of $S$ to take is spheres, which gives the slightly more concrete $$ \operatorname{div}{F(x)} = \lim_{r \to 0} \frac{3}{4\pi r^3} \int_{\lvert x'- x \rvert=r} V(x') \cdot dS', $$ or one can show straightforwardly that this gives the usual expression $ \sum_i \partial_i V_i$ by taking cubes or cuboids and expanding $F$ in a power series about $x$.

The whole point of this definition is that the divergence is then obviously defined as a limit of the flux of $V$ through surfaces surrounding the point, (and is coordinate-independent, which is also a good idea in general). Hence the divergence clearly relates to expansion/contraction caused by the vector field.

Answer 2

Let $D$ be a small spherical region centered at point $P$. By the divergence theorem, $$(\nabla\cdot{\bf V})_P \approx \frac{1}{\mathrm{vol}(D)} \iint_S {\bf V}\cdot{\bf n}\, dS.$$ But $\iint_S {\bf V}\cdot{\bf n}\, dS$ is the net flux of ${\bf V}$ through the surface $S$ of $D$. Thus, $\nabla\cdot{\bf V}$ is a measure of source of the field ${\bf V}$.