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The Gradient-Your First Operator

Take a look at the gradient of a scalar function f(x,y,z):

\begin{displaymath}
\nabla f = \hat{i} \frac{\partial f}{\partial x} + \hat{j}
\...
 ...partial f}{\partial y} + \hat{k} \frac{\partial f}{\partial z}.\end{displaymath} (1)

If we ``factor out'' the function f the gradient looks like:

\begin{displaymath}
\nabla f = \left( \hat{i} \frac{\partial}{\partial x} + \hat...
 ...al}{\partial y} + \hat{k} \frac{\partial}{\partial z}\right) f.\end{displaymath} (2)

This has the appearance of a vector multiplying the function f. The term in parentheses is called ``del'' or ``nabla'' and is written as

\begin{displaymath}
\nabla = \hat{i} \frac{\partial}{\partial x} + \hat{j}
\frac{\partial}{\partial y} + \hat{k} \frac{\partial}{\partial z}.\end{displaymath} (3)

However, $\nabla$ is not really a vector. In fact, if it sits by itself it has no meaning whatsoever. It only takes on meaning when there is a scalar function for it to work with. Thus, rather than thinking of $\nabla$ as a vector multiplying a function, we should think of $\nabla$ as an object which operates on scalar functions by taking their derivatives and combining them into the gradient.

Thus, we say that $\nabla$ is a vector operator acting on scalar functions. More generally, you can almost think of operators as functions whose domain is other functions. On their own, operators have little or no meaning. Only when they are paired with some function in their domain do the have meaning.


next up previous
Next: What else can we Up: The Del Operator Previous: The Del Operator
Vector Calculus
8/19/1998