Divergence of Vector Cross Product

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Definition

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

Let $\mathbf A$ and $\mathbf B$ be vector fields over $R$.


Then:

$\map {\operatorname {div} } {\mathbf A \times \mathbf B} = \mathbf B \cdot \curl \mathbf A - \mathbf A \cdot \curl \mathbf B$

where:

$\operatorname {div}$ denotes the divergence operator
$\curl$ denotes the curl operator
$\times$ denotes vector cross product
$\cdot$ denotes dot product.


Proof

From Divergence Operator on Vector Space is Dot Product of Del Operator and Curl Operator on Vector Space is Cross Product of Del Operator:

\(\ds \operatorname {div} \mathbf V\) \(=\) \(\ds \nabla \cdot \mathbf V\)
\(\ds \curl \mathbf V\) \(=\) \(\ds \nabla \times \mathbf V\)

where $\nabla$ denotes the del operator.


Hence we are to demonstrate that:

$\nabla \cdot \paren {\mathbf A \times \mathbf B} = \mathbf B \cdot \paren {\nabla \times \mathbf A} - \mathbf A \cdot \paren {\nabla \times \mathbf B}$


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


Let $\mathbf A$ and $\mathbf B: \R^3 \to \R^3$ be expressed as vector-valued functions on $\R^3$:

$\mathbf A := \tuple {\map {A_x} {\mathbf r}, \map {A_y} {\mathbf r}, \map {A_z} {\mathbf r} }$
$\mathbf B := \tuple {\map {B_x} {\mathbf r}, \map {B_y} {\mathbf r}, \map {B_z} {\mathbf r} }$

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


Then:

\(\ds \nabla \cdot \paren {\mathbf A \times \mathbf B}\) \(=\) \(\ds \nabla \cdot \paren {\paren {A_y B_z - A_z B_y} \mathbf i + \paren {A_z B_x - A_x B_z} \mathbf j + \paren {A_x B_y - A_y B_x} \mathbf k}\) Definition 1 of Vector Cross Product
\(\ds \) \(=\) \(\ds \map {\dfrac \partial {\partial x} } {A_y B_z - A_z B_y} + \map {\dfrac \partial {\partial y} } {A_z B_x - A_x B_z} + \map {\dfrac \partial {\partial z} } {A_x B_y - A_y B_x}\) Definition of Divergence Operator
\(\ds \) \(=\) \(\ds \paren {\dfrac {\partial A_y B_z} {\partial x} - \dfrac {\partial A_z B_y} {\partial x} } + \paren {\dfrac {\partial A_z B_x} {\partial y} - \dfrac {\partial A_x B_z} {\partial y} } + \paren {\dfrac {\partial A_x B_y} {\partial z} - \dfrac {\partial A_y B_x} {\partial z} }\) rearranging
\(\ds \) \(=\) \(\ds \paren {A_y \dfrac {\partial B_z} {\partial x} + \dfrac {\partial A_y} {\partial x} B_z - A_z \dfrac {\partial B_y} {\partial x} - \dfrac {\partial A_z} {\partial x} B_y} + \paren {A_z \dfrac {\partial B_x} {\partial y} + \dfrac {\partial A_z} {\partial y} B_x - A_x \dfrac {\partial B_z} {\partial y} - \dfrac {\partial A_x} {\partial y} B_z} + \paren {A_x \dfrac {\partial B_y} {\partial z} + \dfrac {\partial A_x} {\partial z} B_y - A_y \dfrac {\partial B_x} {\partial z} - \dfrac {\partial A_y} {\partial z} B_x}\) Product Rule for Derivatives
\(\ds \) \(=\) \(\ds \map {B_x} {\dfrac {\partial A_z} {\partial y} - \dfrac {\partial A_y} {\partial z} } + \map {B_y} {\dfrac {\partial A_x} {\partial z} - \dfrac {\partial A_z} {\partial x} } + \map {B_z} {\dfrac {\partial A_y} {\partial x} - \dfrac {\partial A_x} {\partial y} }\) rearranging
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \map {A_x} {\dfrac {\partial B_y} {\partial z} - \dfrac {\partial B_z} {\partial y} } + \map {A_y} {\dfrac {\partial B_z} {\partial x} - \dfrac {\partial B_x} {\partial z} } + \map {A_z} {\dfrac {\partial B_x} {\partial y} - \dfrac {\partial B_y} {\partial x} }\)
\(\ds \) \(=\) \(\ds \map {B_x} {\dfrac {\partial A_z} {\partial y} - \dfrac {\partial A_y} {\partial z} } + \map {B_y} {\dfrac {\partial A_x} {\partial z} - \dfrac {\partial A_z} {\partial x} } + \map {B_z} {\dfrac {\partial A_y} {\partial x} - \dfrac {\partial A_x} {\partial y} }\) further rearranging
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \map {A_x} {\dfrac {\partial B_z} {\partial y} - \dfrac {\partial B_y} {\partial z} } + \map {A_y} {\dfrac {\partial B_x} {\partial z} - \dfrac {\partial B_z} {\partial x} } + \map {A_z} {\dfrac {\partial B_y} {\partial x} - \dfrac {\partial B_x} {\partial y} }\)
\(\ds \) \(=\) \(\ds \paren {B_x \mathbf i + B_y \mathbf j + B_z \mathbf k} \cdot \paren {\paren {\dfrac {\partial A_z} {\partial y} - \dfrac {\partial A_y} {\partial z} } \mathbf i + \paren {\dfrac {\partial A_x} {\partial z} - \dfrac {\partial A_z} {\partial x} } \mathbf j + \paren {\dfrac {\partial A_y} {\partial x} - \dfrac {\partial A_x} {\partial y} } \mathbf k}\) Definition of Dot Product
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \paren {A_x \mathbf i + A_y \mathbf j + A_z \mathbf k} \cdot \paren {\paren {\dfrac {\partial B_z} {\partial y} - \dfrac {\partial B_y} {\partial z} } \mathbf i + \paren {\dfrac {\partial B_x} {\partial z} - \dfrac {\partial B_z} {\partial x} } \mathbf j + \paren {\dfrac {\partial B_y} {\partial x} - \dfrac {\partial B_x} {\partial y} } \mathbf k}\)
\(\ds \) \(=\) \(\ds \mathbf B \cdot \paren {\nabla \times \mathbf A} - \mathbf A \cdot \paren {\nabla \times \mathbf B}\) Definition of Curl Operator

$\blacksquare$


Also presented as

This result can also be presented as:

$\nabla \cdot \paren {\mathbf A \times \mathbf B} = \mathbf B \cdot \paren {\nabla \times \mathbf A} - \mathbf A \cdot \paren {\nabla \times \mathbf B}$

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


Sources

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