Editing Del (section)

==Precautions==

Most of the above vector properties (except for those that rely explicitly on del's differential properties&mdash;for example, the product rule) rely only on symbol rearrangement, and must necessarily hold if the del symbol is replaced by any other vector. This is part of the value to be gained in notationally representing this operator as a vector.

Though one can often replace del with a vector and obtain a vector identity, making those identities mnemonic, the reverse is ''not'' necessarily reliable, because del does not commute in general.

A counterexample that demonstrates the divergence (<math>\nabla \cdot \mathbf v </math>) and the [[Advection#Mathematics of advection|advection operator]] (<math>\mathbf v \cdot \nabla </math>) are not commutative:
:<math>\begin{align}
              (\mathbf u \cdot \mathbf v) f &\equiv (\mathbf v \cdot \mathbf u) f \\
              (\nabla \cdot \mathbf v) f &= \left (\frac{\partial v_x}{\partial x} + \frac{\partial v_y}{\partial y} + \frac{\partial v_z}{\partial z} \right)f
                                        = \frac{\partial v_x}{\partial x}f + \frac{\partial v_y}{\partial y}f + \frac{\partial v_z}{\partial z}f \\
              (\mathbf v \cdot \nabla) f &= \left (v_x \frac{\partial}{\partial x} + v_y \frac{\partial}{\partial y} + v_z \frac{\partial}{\partial z} \right)f
                                        = v_x \frac{\partial f}{\partial x} + v_y \frac{\partial f}{\partial y} + v_z \frac{\partial f}{\partial z} \\
  \Rightarrow (\nabla \cdot \mathbf v) f &\ne (\mathbf v \cdot \nabla) f \\
\end{align}</math>

A counterexample that relies on del's differential properties:
: <math>\begin{align}
  (\nabla x) \times (\nabla y) &= \left (\mathbf e_x \frac{\partial x}{\partial x}+\mathbf e_y \frac{\partial x}{\partial y}+\mathbf e_z \frac{\partial x}{\partial z} \right) \times \left (\mathbf e_x \frac{\partial y}{\partial x}+\mathbf e_y \frac{\partial y}{\partial y}+\mathbf e_z \frac{\partial y}{\partial z} \right) \\
                               &= (\mathbf e_x \cdot 1 +\mathbf e_y \cdot 0+\mathbf e_z \cdot 0) \times (\mathbf e_x \cdot 0+\mathbf e_y \cdot 1+\mathbf e_z \cdot 0) \\
                               &= \mathbf e_x  \times \mathbf e_y \\
                               &= \mathbf e_z \\
  (\mathbf u x)\times (\mathbf u y) &= x y (\mathbf u \times \mathbf u) \\
                               &= x y \mathbf 0 \\
                               &= \mathbf 0
\end{align}</math>

Central to these distinctions is the fact that del is not simply a vector; it is a [[vector operator]]. Whereas a vector is an object with both a magnitude and direction, del has neither a magnitude nor a direction until it operates on a function.

For that reason, identities involving del must be derived with care, using both vector identities and ''differentiation'' identities such as the product rule.