Curl of Curl is Gradient of Divergence minus Laplacian

Theorem
Let $\map {\R^3} {x, y, z}$ denote the real Cartesian space of $3$ dimensions.

Let $\mathbf V$ be a vector field on $\R^3$.

Then:
 * $\curl \curl \mathbf V = \grad \operatorname {div} \mathbf V - \nabla^2 \mathbf V$

where:
 * $\curl$ denotes the curl operator
 * $\operatorname {div}$ denotes the divergence operator
 * $\grad$ denotes the gradient operator
 * $\nabla^2 \mathbf V$ denotes the Laplacian.

Proof
From Curl Operator on Vector Space is Cross Product of Del Operator, and Divergence Operator on Vector Space is Dot Product of Del Operator and the definition of the gradient operator:

where $\nabla$ denotes the del operator.

Hence we are to demonstrate that:
 * $\nabla \times \paren {\nabla \times \mathbf V} = \map \nabla {\nabla \cdot \mathbf V} - \nabla^2 \mathbf V$

Let $\mathbf V$ be expressed as a vector-valued function on $\mathbf V$:


 * $\mathbf V := \tuple {\map {V_x} {\mathbf r}, \map {V_y} {\mathbf r}, \map {V_z} {\mathbf r} }$

where $\mathbf r = \tuple {x, y, z}$ is the position vector of an arbitrary point in $R$.

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

Also presented as
This result can also be presented as:


 * $\nabla \times \paren {\nabla \times \mathbf V} = \map \nabla {\nabla \cdot \mathbf V} - \nabla^2 \mathbf V$

presupposing the implementations of $\operatorname {div}$, $\curl$ and $\grad$ as operations using the del operator.