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Grad Operator

It is useful to define the vector operator

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

$\displaystyle {\bf grad}\,f = \nabla f \equiv \left(\frac{\partial f}{\partial x},\, \frac{\partial f}{\partial y},\,\frac{\partial f}{\partial z}\right).$ (A.120)

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

$\displaystyle {\bf grad}\,(\phi \,\psi) = \phi\,\, {\bf grad}\,\psi +\psi\,\, {\bf grad}\,\phi$ (A.121)

can be written more succinctly as

$\displaystyle \nabla(\phi\, \psi) = \phi \,\nabla\psi + \psi\, \nabla \phi.$ (A.122)

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

$\displaystyle x$ $\displaystyle = x'\, \cos\theta - y'\,\sin\theta,$ (A.123)
$\displaystyle y$ $\displaystyle = x\,'\sin\theta +y'\,\cos\theta,$ (A.124)
$\displaystyle z$ $\displaystyle = z'.$ (A.125)

Now,

$\displaystyle \frac{\partial}{\partial x'} = \left(\frac{\partial x}{\partial x...
...eft(\frac{\partial z}{\partial x'} \right)_{y',z'} \frac{\partial}{\partial z},$ (A.126)

giving

$\displaystyle \frac{\partial}{\partial x'} = \cos\theta \,\frac{\partial}{\partial x} + \sin\theta \,\frac{\partial}{\partial y},$ (A.127)

and

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


next up previous
Next: Divergence Up: Vectors and Vector Fields Previous: Gradient
Richard Fitzpatrick 2016-01-22