Why can the divergence of a vector field be expressed as the dot product between the gradient and the vector?

Question

I understand the intuition of divergence, but want to understand this part of its mechanics. Thanks

Answer 1

The simple answer is just because it fits with how divergence is conceptually defined, really. Divergence is a quantity that is related to flux. In 2D, divergence is flux per unit area, and in 3D, divergence is flux per unit volume. More succinctly, divergence is flux density. That means

• 2D: flux = ∫∫ div(F) dA
• 3D: flux = ∫∫∫ div(F) dV

 

However we already have notions of how to calculate flux as a line integral (2D) or surface integral (3D), so it would seem like using the idea of flux density is unnecessary.

 

It's through either Green's Theorem (2D) or the Divergence theorem (3D) that we can equate flux integrals into forms that naturally bring about ∇•F. I'll show you the 2D case; the 3D case follows exactly the same logic.

 

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We measure flux as a line integral around a closed curve: flux = ∫F•ndl. Green's Theorem is a way to change a closed-loop line integral into a corresponding area integral over the loop's interior (How it gets to this point is beyond the scope of your question, and it is a bit of a lengthy proof. The result of Green's theorem does not ever presuppose some notion of "divergence" though--we just get something based on the derivatives of the components of F). Green's theorem states

 

∫ F•n dl = ∫∫ (dF_x/dx+dF_y/dy) dA

 

It is from our definitions of the dot product and ∇ themselves that allow us to condense the results of Green's theorem into the traditional notation:

 

∫ F•n dl = ∫∫ (dF_x/dx+dF_y/dy) dA = ∫∫ (∇•F) dA

 

And remember, each one of those integrals is equal to flux. Remember we already established the notion that divergence is the area-density of flux. Because of this, the quantity in the area integral above must be divergence by definition.