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Grad Operator
It is useful to define the vector operator
 |
(1362) |
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,
 |
(1363) |
For two scalar fields
and
,
 |
(1364) |
can be written more succinctly as
 |
(1365) |
Suppose that we rotate the coordinate axes through an angle
about
.
By analogy with Equations (A.1277)-(A.1279), the old coordinates (
,
,
) are related
to the new ones (
,
,
) via
Now,
 |
(1369) |
giving
 |
(1370) |
and
 |
(1371) |
It can be seen, from Equations (A.1280)-(A.1282), that
the differential operator
transforms in an analogous manner to
a vector.
This is another proof that
is a good vector.
Figure A.113:
Cylindrical polar coordinates.
 |
Next: Curvilinear Coordinates
Up: Vector Algebra and Vector
Previous: Gradient
Richard Fitzpatrick
2011-03-31