Divergence of Curl is Zero

From ProofWiki
Jump to navigation Jump to search

Definition

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

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


Then:

$\map {\operatorname {div} } {\curl \mathbf V} = 0$

where:

$\curl$ denotes the curl operator
$\operatorname {div}$ denotes the divergence operator.


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:

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

where $\nabla$ denotes the del operator.


Hence we are to demonstrate that:

$\nabla \cdot \paren {\nabla \times \mathbf V} = 0$


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 $\R^3$.

Hence:

\(\ds \nabla \cdot \paren {\nabla \times \mathbf V}\) \(=\) \(\ds \nabla \cdot \paren {\paren {\dfrac {\partial V_z} {\partial y} - \dfrac {\partial V_y} {\partial z} } \mathbf i + \paren {\dfrac {\partial V_x} {\partial z} - \dfrac {\partial V_z} {\partial x} } \mathbf j + \paren {\dfrac {\partial V_y} {\partial x} - \dfrac {\partial V_x} {\partial y} } \mathbf k}\) Definition of Curl Operator
\(\ds \) \(=\) \(\ds \dfrac \partial {\partial x} \paren {\dfrac {\partial V_z} {\partial y} - \dfrac {\partial V_y} {\partial z} } + \dfrac \partial {\partial y} \paren {\dfrac {\partial V_x} {\partial z} - \dfrac {\partial V_z} {\partial x} } + \dfrac \partial {\partial z} \paren {\dfrac {\partial V_y} {\partial x} - \dfrac {\partial V_x} {\partial y} }\) Definition of Divergence Operator
\(\ds \) \(=\) \(\ds \dfrac {\partial^2 V_z} {\partial x \partial y} - \dfrac {\partial^2 V_y} {\partial x \partial z} + \dfrac {\partial^2 V_x} {\partial y \partial z} - \dfrac {\partial^2 V_z} {\partial y \partial x} + \dfrac {\partial^2 V_y} {\partial z \partial x} - \dfrac {\partial^2 V_x} {\partial z \partial y}\)


From Clairaut's Theorem:

$\dfrac {\partial^2 V_z} {\partial x \partial y} = \dfrac {\partial^2 V_z} {\partial y \partial x}$

and the same mutatis mutandis for the other partial derivatives.

The result follows.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Divergence_of_Curl_is_Zero&oldid=568570"