Divergence Operator on Vector Space is Dot Product of Del Operator

Theorem

Let $R$ be a region of Cartesian $3$-space $\R^3$.

Let $\map {\mathbf V} {x, y, z}$ be a vector field acting over $R$.

Let $\tuple {i, j, k}$ be the standard ordered basis on $\R^3$.

Then

$\operatorname {div} \mathbf V = \nabla \cdot \mathbf V$

where:

$\operatorname {div} \mathbf V $ denotes the divergence of $\mathbf V$

$\nabla$ denotes the del operator.

Proof

We have by definition of divergence of $\mathbf V$:

$\operatorname {div} \mathbf V = \dfrac {\partial V_x} {\partial x} + \dfrac {\partial V_y} {\partial y} + \dfrac {\partial V_z} {\partial z}$

Now:

\(\ds \nabla \cdot \mathbf V\)

\(=\)

\(\ds \paren {\mathbf i \dfrac \partial {\partial x} + \mathbf j \dfrac \partial {\partial y} + \mathbf k \dfrac \partial {\partial z} } \cdot \paren {V_x \mathbf i + V_y \mathbf j + V_z \mathbf k}\)

Definition of Del Operator

\(\ds \)

\(=\)

\(\ds \dfrac {\partial V_x} {\partial x} + \dfrac {\partial V_y} {\partial y} + \dfrac {\partial V_z} {\partial z}\)

Definition of Dot Product

$\blacksquare$

Sources

• 1951: B. Hague: An Introduction to Vector Analysis (5th ed.) ... (previous) ... (next): Chapter $\text {IV}$: The Operator $\nabla$ and its Uses: $3 a$. The Operation $\nabla \cdot \mathbf V$