Curl of Gradient is Zero

Definition

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

Let $\map U {x, y, z}$ be a scalar field on $\R^3$.

Then:

$\map \curl {\grad U} = \mathbf 0$

where:

$\curl$ denotes the curl operator

$\grad$ denotes the gradient operator.

Proof

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

$\ds \grad \mathbf U$

$=$

$\ds \nabla U$

$\ds \map \curl {\grad U}$

$=$

$\ds \nabla \times \paren {\nabla U}$

where $\nabla$ denotes the del operator.

Hence we are to demonstrate that:

$\nabla \times \paren {\nabla U} = \mathbf 0$

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

Then:

$\ds \nabla \times \paren {\nabla U}$

$=$

$\ds \nabla \times \paren {\dfrac {\partial U} {\partial x} \mathbf i + \dfrac {\partial U} {\partial y} \mathbf j + \dfrac {\partial U} {\partial z} \mathbf k}$

Definition of Gradient Operator

$\ds $

$=$

$\ds \paren {\dfrac \partial {\partial y} \dfrac {\partial U} {\partial z} - \dfrac \partial {\partial z} \dfrac {\partial U} {\partial y} } \mathbf i + \paren {\dfrac \partial {\partial z} \dfrac {\partial U} {\partial x} - \dfrac \partial {\partial x} \dfrac {\partial U} {\partial z} } \mathbf j + \paren {\dfrac \partial {\partial x} \dfrac {\partial U} {\partial y} - \dfrac \partial {\partial y} \dfrac {\partial U} {\partial x} } \mathbf k$

Definition of Curl Operator

$\ds $

$=$

$\ds \paren {\dfrac {\partial^2 U} {\partial y \partial z} - \dfrac {\partial^2 U} {\partial z \partial y} } \mathbf i + \paren {\dfrac {\partial^2 U} {\partial z \partial x} - \dfrac {\partial^2 U} {\partial x \partial z} } \mathbf j + \paren {\dfrac {\partial^2 U} {\partial x \partial y} - \dfrac {\partial^2 U} {\partial y \partial x} } \mathbf k$

From Clairaut's Theorem:

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

and the same mutatis mutandis for the other partial derivatives.

The result follows.

$\blacksquare$