I'm reading about the del operator on wikipedia, there is a section called "Precautions" (https://en.wikipedia.org/wiki/Del#Precautions) that I'm trying to understand for the last 4 hours.

Here is what it says, I've highlighted the parts I don't quite understand.

Most of the above vector properties (except for those that rely explicitly on del's differential properties—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 relies on del's failure to commute: :\begin{align} (\vec u \cdot \vec v) f &\equiv (\vec v \cdot \vec u) f \\ (\nabla \cdot \vec 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 \\ (\vec 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 \vec v) f &\ne (\vec v \cdot \nabla) f \\ \end{align}

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

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.

1) What are other representations for del?

2) What is a "mnemonic identity"?

3) The final part I don't understand at all. Why I must use vector identities and del differential properties?

Is it all "unnecessary" pedantry? Is it worth being aware of this all? Before reading this wikipedia page I thought I knew how to use the del operator, now I'm confused.

Please help-me clarify all this mess in my head.

Thank you.

  • $\begingroup$ @Jean-ClaudeArbaut Sorry, this difference you just pointed out I understand very well, but that is not what I asked. I understand the counterexamples. $\endgroup$
    – R. Mannue
    Oct 12, 2016 at 21:31

1 Answer 1

  1. This refers to treating it as a single operator as opposed to using multiple operators, or explicitly writing some other representation of "divergence," "curl," "gradient," etc.
  2. That just means it's a mnemonic for the operations it represents, that by treating it as if it were a vector, you come to the actual formulae.
  3. Because it's not strictly speaking a vector (because its elements aren't numbers), treating it as if it were one might give you false results, and the product rule is given as an example of a way you might get into trouble.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.