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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 reduced to \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, but I still have difficulties with getting $\vec B\cdot(\vec A\cdot\nabla)\vec A$ term.

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

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    $\begingroup$ This is a math question, not a physics one. $\endgroup$
    – G. Smith
    Commented Oct 7, 2019 at 20:32
  • $\begingroup$ I just saw similar questions in this section. If this is really far from the topic, then I will delete the question. $\endgroup$
    – Kubrick
    Commented Oct 7, 2019 at 20:34
  • $\begingroup$ Updated. I thought that the divergence of scalar doesn't exist. $\endgroup$
    – Kubrick
    Commented Oct 7, 2019 at 20:39
  • $\begingroup$ Welcome to Physics! Stack Exchange posts are version controlled, so please do not make your post look like a revision table. Instead, just seamlessly integrate the new material into the post. There is an edit history button at the bottom of the post for those interested in seeing what changed. $\endgroup$
    – Kyle Kanos
    Commented Oct 8, 2019 at 23:54
  • $\begingroup$ I'll try next time. $\endgroup$
    – Kubrick
    Commented Oct 9, 2019 at 1:05

1 Answer 1

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Correct expression is $$ \begin{aligned} \operatorname{div}[(\vec A\cdot\vec B)\vec A] &= (\vec A \cdot \vec B) \operatorname{div} \vec A + \vec A\cdot(\vec A\cdot\nabla)\vec B + \vec B\cdot(\vec A\cdot\nabla)\vec A \\ &= (\vec A \cdot \vec B) \operatorname{div} \vec A + \vec A\cdot(\vec A\cdot\nabla)\vec B +\vec A\cdot(\vec B\cdot\nabla)\vec A +\vec A\cdot(\vec B\times(\nabla\times\vec A)) \end{aligned} $$ In order to see that these two are the same, one could use so-called BAC-minus-CAB rule.

Finally, the last formula of yours is wrong when $\operatorname{div} A \neq 0$.

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  • $\begingroup$ I have some difficulties: $(\vec A \cdot \vec B) \operatorname{div} \vec A+\vec A\cdot(\vec A\cdot\nabla)\vec B+\vec A\cdot(\vec B\cdot\nabla)\vec A+\vec A\cdot(\vec B\times(\nabla\times\vec A))=(\vec A \cdot \vec B) \operatorname{div} \vec A+\vec A\cdot(\vec A\cdot\nabla)\vec B+\vec A\cdot(\vec B\cdot\nabla)\vec A+\vec A\cdot(\nabla_{A}(\vec B\cdot \vec A)-(\vec B\cdot\nabla)\vec A )=(\vec A \cdot \vec B) \operatorname{div} \vec A+\vec A\cdot(\vec A\cdot\nabla)\vec B+\vec A\cdot\nabla_{A}(\vec B\cdot \vec A)$. What should I do to get $\vec B\cdot(\vec A\cdot\nabla)\vec A$ $\endgroup$
    – Kubrick
    Commented Oct 8, 2019 at 15:16
  • $\begingroup$ I also understand that I have a magnetic field as $\vec A$ in my problem, so it could explain why there wasn't $(\vec A\cdot\vec B)\operatorname{div} \vec A$, but I still have difficulties with another term. $\endgroup$
    – Kubrick
    Commented Oct 8, 2019 at 15:26
  • $\begingroup$ @Kubrick I’m sorry but it’s hard to understand what is the problem. Please do not put formulas in comments, put them in the main text, enumerate them and refer them by labels. $\endgroup$
    – David Saykin
    Commented Oct 8, 2019 at 16:31
  • $\begingroup$ I have updated the question. $\endgroup$
    – Kubrick
    Commented Oct 8, 2019 at 18:47
  • $\begingroup$ @Kubrick answer to your last question is that $\vec B\cdot(\vec A\cdot\nabla)\vec A$ is the same as $\vec A\cdot \nabla_{A}(\vec B\cdot \vec A)$. $\endgroup$
    – David Saykin
    Commented Oct 8, 2019 at 18:51

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