Product Rule for Curl

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Let $\mathbf A$ be a vector field over $\mathbf V$.

Let $U$ be a scalar field over $\mathbf V$.


Then:

\(\ds \map \curl {U \mathbf A}\) \(=\) \(\ds U \curl \mathbf A + \grad U \times \mathbf A\)
\(\ds \) \(=\) \(\ds U \curl \mathbf A - \mathbf A \times \grad U\)

where:

$\curl$ denotes the curl operator
$\grad$ denotes the gradient operator
$\times$ denotes vector cross product


Proof

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

\(\ds \curl \mathbf A\) \(=\) \(\ds \nabla \times \mathbf A\)
\(\ds \grad \mathbf U\) \(=\) \(\ds \nabla U\)

where $\nabla$ denotes the del operator.


Hence we are to demonstrate that:

\(\ds \nabla \times \paren {U \mathbf A}\) \(=\) \(\ds \map U {\nabla \times \mathbf A} + \paren {\nabla U} \times \mathbf A\)
\(\ds \) \(=\) \(\ds \map U {\nabla \times \mathbf A} - \mathbf A \times \paren {\nabla U}\)


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

$\mathbf A := \tuple {\map {A_x} {\mathbf r}, \map {A_y} {\mathbf r}, \map {A_z} {\mathbf r} }$

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 $\R^3$.


Then:

\(\ds \nabla \times \paren {U \mathbf A}\) \(=\) \(\ds \paren {\dfrac {\partial U A_z} {\partial y} - \dfrac {\partial U A_y} {\partial z} } \mathbf i + \paren {\dfrac {\partial U A_x} {\partial z} - \dfrac {\partial U A_z} {\partial x} } \mathbf j + \paren {\dfrac {\partial U A_y} {\partial x} - \dfrac {\partial U A_x} {\partial y} } \mathbf k\) Definition of Curl Operator
\(\ds \) \(=\) \(\ds \paren {U \dfrac {\partial A_z} {\partial y} + \dfrac {\partial U} {\partial y} A_z - U \dfrac {\partial A_y} {\partial z} - \dfrac {\partial U} {\partial z} A_y} \mathbf i\) Product Rule for Derivatives
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {U \dfrac {\partial A_x} {\partial z} + \dfrac {\partial U} {\partial z} A_x - U \dfrac {\partial A_z} {\partial x} - \dfrac {\partial U} {\partial x} A_z} \mathbf j\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {U \dfrac {\partial A_y} {\partial x} + \dfrac {\partial U} {\partial x} A_y - U \dfrac {\partial A_x} {\partial y} - \dfrac {\partial U} {\partial y} A_x } \mathbf k\)
\(\ds \) \(=\) \(\ds \map U {\paren {\dfrac {\partial A_z} {\partial y} - \dfrac {\partial A_y} {\partial z} } \mathbf i + \paren {\dfrac {\partial A_x} {\partial z} - \dfrac {\partial A_z} {\partial x} } \mathbf j + \paren {\dfrac {\partial A_y} {\partial x} - \dfrac {\partial A_x} {\partial y} } \mathbf k}\) rearrangement
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {\dfrac {\partial U} {\partial y} A_z - \dfrac {\partial U} {\partial z} A_y} \mathbf i + \paren {\dfrac {\partial U} {\partial z} A_x - \dfrac {\partial U} {\partial x} A_z} \mathbf j + \paren {\dfrac {\partial U} {\partial x} A_y - \dfrac {\partial U} {\partial y} A_x} \mathbf k\)
\(\ds \) \(=\) \(\ds \map U {\nabla \times \mathbf f} + \paren {\dfrac {\partial U} {\partial y} A_z - \dfrac {\partial U} {\partial z} A_y} \mathbf i + \paren {\dfrac {\partial U} {\partial z} A_x - \dfrac {\partial U} {\partial x} A_z} \mathbf j + \paren {\dfrac {\partial U} {\partial x} A_y - \dfrac {\partial U} {\partial y} A_x} \mathbf k\) Definition of Curl Operator
\(\ds \) \(=\) \(\ds \map U {\nabla \times \mathbf f} + \paren {\dfrac {\partial U} {\partial x} \mathbf i + \dfrac {\partial U} {\partial y} \mathbf j + \dfrac {\partial U} {\partial z} \mathbf k} \times \paren {A_x \mathbf i + A_y \mathbf j + A_z \mathbf k}\) Definition of Cross Product
\(\ds \) \(=\) \(\ds \map U {\nabla \times \mathbf f} + \paren {\nabla U} \times \mathbf f\) Definition of Gradient Operator, Definition of Vector

$\blacksquare$


Also presented as

This result can also be presented as:

$\nabla \times \paren {U \mathbf A} = \map U {\nabla \times \mathbf A} + \paren {\nabla U} \times \mathbf A$

or:

$\nabla \times \paren {U \mathbf A} = \map U {\nabla \times \mathbf A} - \mathbf A \times \paren {\nabla U}$

presupposing the implementations of $\curl$ and $\grad$ as operations using the del operator.


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Product_Rule_for_Curl&oldid=510209"