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I refer to such an identity as outlined here:

See the 1st and 4th terms, as well as the 2nd and 3rd terms

If $\mathbf A(\nabla \cdot \mathbf B)$ was equal to $(\mathbf A \cdot \nabla)\mathbf B$ then all terms of the first line of this identity would cancel out, leaving zero, so surely this cannot be the case? Else it would be simpler to simply write zero.

I am taught identities like this at my physics degree. According to everything I know, terms like the 1st and 4th terms should be equal. I have exhausted every attempt to find an answer to this before coming here.

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  • $\begingroup$ @Rahul Why are you answering in a comment? $\endgroup$
    – Arthur
    May 3, 2019 at 7:00
  • $\begingroup$ Yes. Also, I understand there is a particular standard of presenting mathematical notation on here, but I'm unable to meet this standard using phone browser. If anyone could edit my question to change "nabla dot etc." into proper mathematical notation, I would be very grateful! $\endgroup$
    – Alexander Kalian
    May 3, 2019 at 7:00
  • $\begingroup$ Short answer: $\nabla\cdot f$ isn’t a “real” dot product, but a useful mnemonic for the $\operatorname{div}$ operator. Ditto for $\nabla\times f$ and $\operatorname{curl}$. $\endgroup$
    – amd
    May 3, 2019 at 7:23
  • $\begingroup$ Even if a "normal" vector $C$ replaced $\nabla$, $A(C\cdot B)$ wouldn't in general equal $(A\cdot C)B$; one's parallel to $A$, the other to $B$. $\endgroup$
    – J.G.
    May 3, 2019 at 7:49

1 Answer 1

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For one, $\nabla \cdot \mathbf u \neq \mathbf u \cdot \nabla$:

The LHS is the divergence of $\mathbf u$, which is an expression, whereas the RHS is still an operator (in fact, $\mathbf u \cdot \nabla$ is called the advection operator, seen in the Navier-Stokes equations).

The issue here is that the commutative property of the dot product doesn't hold, because the dot product is supposed to be an operation between two vectors; $\nabla$ is an operator .

You're actually looking at an abuse of notation: you can interpret $\nabla \cdot \mathbf u$ intuitively, but need to be extra careful when performing algebraic manipulations.

The above also answers why the first term is not equal to the third term in your example; as for $\mathbf A(\nabla \cdot \mathbf B)$ and $(\mathbf A \cdot \nabla)\mathbf B$: the former is simply a scalar multiple of $\mathbf A$, whereas the latter is the result of some operation on the vector $\mathbf B$, which is much more complicated. Your impression that the two might be equal also involves "moving" the dot elsewhere, which can't be done either, even in the "usual" case.

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    $\begingroup$ Just to make the $i$th components of each vector explicit, with implicit summation over repeated indices, $$[A(\nabla\cdot B)]_i=A_i\partial_j B_j,\,[(A\cdot\nabla)B]_i=A_j\partial_j B_i.$$ $\endgroup$
    – J.G.
    May 3, 2019 at 7:47

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