Divergence of Product of Scalar Field with Curl of Vector Field

Theorem

Let $R$ be a region of space.

Let $U$ be a scalar field over $R$.

Let $\mathbf A = \curl \mathbf B$ be a vector field over $R$ whose vector potential is $\mathbf B$.

Then:

$\map {\operatorname {div} } {U \curl \mathbf B} = \paren {\curl \mathbf B} \cdot \paren {\grad U}$

where:

$\operatorname {div}$ denotes the divergence operator

$\grad$ denotes the gradient operator

$\curl$ denotes the curl operator.

Proof

\(\ds \map {\operatorname {div} } {U \mathbf A}\)

\(=\)

\(\ds \map U {\operatorname {div} \mathbf A} + \mathbf A \cdot \grad U\)

Product Rule for Divergence

\(\ds \leadsto \ \ \)

\(\ds \map {\operatorname {div} } {U \curl \mathbf B}\)

\(=\)

\(\ds \map U {\operatorname {div} \curl \mathbf B} + \paren {\curl \mathbf B} \cdot \paren {\grad U}\)

substituting $\curl \mathbf B$ for $\mathbf A$

\(\ds \)

\(=\)

\(\ds \paren {\curl \mathbf B} \cdot \paren {\grad U}\)

Divergence of Curl is Zero

$\blacksquare$

Sources

• 1951: B. Hague: An Introduction to Vector Analysis (5th ed.) ... (previous) ... (next): Chapter $\text {V}$: Further Applications of the Operator $\nabla$: $8$. Two Useful Divergence Formulae: $(5.10)$