Divergence of Curl is Zero
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
- 1951: B. Hague: An Introduction to Vector Analysis (5th ed.) ... (previous) ... (next): Chapter $\text {V}$: Further Applications of the Operator $\nabla$: $5$. The Operator $\operatorname {div} \curl$: $(5.6)$
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 22$: Miscellaneous Formulas involving $\nabla$: $22.44$