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If I am correct, then $\operatorname{div}[(\vec A\cdot \vec B)\vec C] = (\vec A \cdot \vec B) \operatorname{div} \vec C + \vec C \cdot \nabla (\vec A\cdot\vec B)= (\vec A \cdot \vec B) \operatorname{div} \vec C+ \vec C\cdot(\vec A\cdot\nabla)\vec B+\vec C\cdot(\vec B\cdot\nabla)\vec A+\vec C\cdot(\vec A\times(\nabla\times\vec B))+ \vec C\cdot(\vec B\times(\nabla\times\vec A))$

When $\vec A=\vec C$ it will be reduced to $\operatorname{div}[(\vec A\cdot \vec B)\vec A]=(\vec A \cdot \vec B) \operatorname{div} \vec A + \vec A\cdot(\vec A\cdot\nabla)\vec B+\vec A\cdot(\vec B\cdot\nabla)\vec A+\vec A\cdot(\vec B\times(\nabla\times\vec A))$

Is there any possibility that it could be $$\begin{equation}\operatorname{div}[(\vec A\cdot\vec B)\vec A]=\vec A\cdot(\vec A\cdot\nabla)\vec B + \vec B\cdot(\vec A\cdot\nabla)\vec A\qquad (*)\end{equation}$$ ? I've seen this expression, but I'm not sure whether it's correct or not.

UPD: I understood that I have a magnetic field as $\vec A$ in my task (where I've seen (∗) expression), so it could explain why there wasn't $(\vec A \cdot\vec B )\operatorname{div}\vec A$ term as $\operatorname{div}\vec A=0$, but I still have some difficulties with getting $\vec B\cdot(\vec A\cdot\nabla)\vec A$ term.

If I expand the cross product, I get $\vec A\cdot(\vec B\times(\nabla\times\vec A))=\vec A\cdot(\nabla_{A}(\vec B\cdot \vec A)-(\vec B\cdot\nabla)\vec A )$

When I checked it by coordinates, I got that $$\vec B\cdot(\vec A\cdot\nabla)\vec A=\vec A\cdot \nabla_{A}(\vec B\cdot \vec A),\qquad(**)$$ so it really reduces to $(*)$ if $\operatorname{div}\vec A=0$.

But, I guess, it is possible to notice this without checking by coordinates and immediately get equality $(*)$ or at least get equality $(**)$.

Could you give me any hints, please?

UPD: My guess is that $\vec A\cdot \nabla_{A}(\vec B\cdot \vec A)$ is similiar to $\vec A\cdot\nabla (kf)=(\vec A\cdot\nabla)kf=k(\vec A\cdot\nabla)f$, where $k$ is a constant.

Is this a valid analogy?

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Equation (*) follows from $$ {\rm div\,}((A\cdot B)C) = (A\cdot B){\rm div\,}C + B\cdot(C\cdot\nabla A)+A\cdot(C\cdot\nabla B)$$ when ${\rm div\,}A = 0$. A short proof, but not coordinate-free, does one term at a time: $${\rm div\,}(a_1b_1C) = a_1b_1{\rm div\,}C+c_1(a_{1,1}b_1+a_1b_{1,1})+c_2(a_{1,2}b_1+a_1b_{1,2})+c_3(a_{1,3}b_1+a_1b_{1,3})$$ $$ = a_1b_1{\rm div\,}C+(C\cdot\nabla a_1)b_1+(C\cdot\nabla b_1)a_1.$$ Add three such terms to complete the proof.

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  • $\begingroup$ Could you explain, please, what the designation of a vector element with two indices means? $\endgroup$
    – Kubrick
    Commented Oct 10, 2019 at 18:05
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    $\begingroup$ $a_{1,2}$ means the partial derivative of the first component of $A$ with respect to $x_2$, for example. The line where that occurs is based on the product rule for derivatives. $\endgroup$
    – Bob Terrell
    Commented Oct 10, 2019 at 20:57
  • $\begingroup$ Is there any, maybe, mnemonic rule how to get it without checking by coordinates? $\endgroup$
    – Kubrick
    Commented Oct 11, 2019 at 18:17
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    $\begingroup$ Since the product $(A\cdot B)C$ is linear in each of $A$, $B$ and $C$, you would expect any derivative of it, such as div, to consist of three such terms each involving a derivative of one of the functions, just like you have with the ordinary product rule $(fgh)' = f'gh+fg'h+fgh'$. But you don't find out the detailed form until you calculate it. $\endgroup$
    – Bob Terrell
    Commented Oct 11, 2019 at 23:02

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