Grad Operator

It is useful to define the vector operator

$\displaystyle \nabla \equiv \left( \frac{\partial}{\partial x},\, \frac{\partial}{\partial y},\, \frac{\partial }{\partial z}\right),$

(1.119)

which is usually called the grad or del operator. This operator acts on everything to its right in a expression, until the end of the expression or a closing bracket is reached. For instance,

$\displaystyle {\bf grad}\,f = \nabla f \equiv \left(\frac{\partial f}{\partial x},\, \frac{\partial f}{\partial y},\,\frac{\partial f}{\partial z}\right).$

(1.120)

For two scalar fields $\phi$ and $\psi$ ,

$\displaystyle {\bf grad}\,(\phi \,\psi) = \phi\,\, {\bf grad}\,\psi +\psi\,\, {\bf grad}\,\phi$

(1.121)

can be written more succinctly as

$\displaystyle \nabla(\phi\, \psi) = \phi \,\nabla\psi + \psi\, \nabla \phi.$

(1.122)

Suppose that we rotate the coordinate axes through an angle $\theta$ about . By analogy with Equations (A.17)–(A.19), the old coordinates (, , ) are related to the new ones (, , ) via

$\displaystyle x$	$\displaystyle = x'\, \cos\theta - y'\,\sin\theta,$	(1.123)
$\displaystyle y$	$\displaystyle = x\,'\sin\theta +y'\,\cos\theta,$	(1.124)
$\displaystyle z$	$\displaystyle = z'.$	(1.125)

Now,

$\displaystyle \frac{\partial}{\partial x'} = \left(\frac{\partial x}{\partial x... ...eft(\frac{\partial z}{\partial x'} \right)_{y',z'} \frac{\partial}{\partial z},$

(1.126)

giving

$\displaystyle \frac{\partial}{\partial x'} = \cos\theta \,\frac{\partial}{\partial x} + \sin\theta \,\frac{\partial}{\partial y},$

(1.127)

and

$\displaystyle \nabla_{x'} = \cos\theta\, \nabla_x + \sin\theta \,\nabla_y.$

(1.128)

It can be seen, from Equations (A.20)–(A.22), that the differential operator $\nabla$ transforms in an analogous manner to a vector. This is another proof that $\nabla f$ is a good vector.