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Grad Operator
It is useful to define the vector operator
![$\displaystyle \nabla \equiv \left( \frac{\partial}{\partial x},\, \frac{\partial}{\partial y},\, \frac{\partial }{\partial z}\right),$](img6559.png) |
(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,
![$\displaystyle {\bf grad}\,f = \nabla f \equiv \left(\frac{\partial f}{\partial x},\, \frac{\partial f}{\partial y},\,\frac{\partial f}{\partial z}\right).$](img6560.png) |
(A.120) |
For two scalar fields
and
,
![$\displaystyle {\bf grad}\,(\phi \,\psi) = \phi\,\, {\bf grad}\,\psi +\psi\,\, {\bf grad}\,\phi$](img6561.png) |
(A.121) |
can be written more succinctly as
![$\displaystyle \nabla(\phi\, \psi) = \phi \,\nabla\psi + \psi\, \nabla \phi.$](img6562.png) |
(A.122) |
Suppose that we rotate the coordinate axes through an angle
about
.
By analogy with Equations (A.17)-(A.19), the old coordinates (
,
,
) are related
to the new ones (
,
,
) via
Now,
![$\displaystyle \frac{\partial}{\partial x'} = \left(\frac{\partial x}{\partial x...
...eft(\frac{\partial z}{\partial x'} \right)_{y',z'} \frac{\partial}{\partial z},$](img6567.png) |
(A.126) |
giving
![$\displaystyle \frac{\partial}{\partial x'} = \cos\theta \,\frac{\partial}{\partial x} + \sin\theta \,\frac{\partial}{\partial y},$](img6568.png) |
(A.127) |
and
![$\displaystyle \nabla_{x'} = \cos\theta\, \nabla_x + \sin\theta \,\nabla_y.$](img6569.png) |
(A.128) |
It can be seen, from Equations (A.20)-(A.22), that
the differential operator
transforms in an analogous manner to
a vector.
This is another proof that
is a good vector.
Next: Divergence
Up: Vectors and Vector Fields
Previous: Gradient
Richard Fitzpatrick
2016-01-22