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This is excercise 3.6.7 from Arfken. From the exercise just before we have: $$\bf \nabla(A\cdot B)=(A\times\nabla)\times B+(B\times\nabla)\times A+A(\nabla\cdot B)+B(\nabla\cdot A)$$ So with $\bf B=A$, we have: $$\bf \nabla(A\cdot A)=(A\times\nabla)\times A+(A\times\nabla)\times A+A(\nabla\cdot A)+A(\nabla\cdot A)$$ $$\bf \frac{1}{2} \nabla(A\cdot A)=(A\times\nabla)\times A+A(\nabla\cdot A)$$ $$\bf \frac{1}{2} \nabla(A\cdot A)-A(\nabla\cdot A)=(A\times\nabla)\times A$$

I am having difficulties with with getting from the last equality to the desired form, any help is appreciated.

Labbsserts
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  • $A\times B=-B\times A$ and hence $(A\times\nabla)\times A=-A\times (A\times\nabla)=A\times (\nabla\times A)$. Does this answer your question? – B.Hueber Jul 30 '21 at 08:12
  • @Udalricus.S. Is it correct to use $\nabla$ as a vector in this manner? – Labbsserts Jul 30 '21 at 08:18
  • You already write $A\times\nabla $ in the question, what is this? Is it the operator that you get by formally doing the cross product? – Calvin Khor Jul 30 '21 at 10:16
  • @CalvinKhor Yes. My only point with regards to Udalricus.S,'s comment was that even though $A \times B=-B \times A$, it is not necessarily true that $A \times \nabla= - \nabla \times A$. – Labbsserts Aug 02 '21 at 09:35

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One way to show this relation is via an explicit calculation: $$ A \times [\nabla \times A] = \begin{pmatrix}A_x\\A_y\\A_z\end{pmatrix} \times \left[ \begin{pmatrix}\partial_x\\ \partial_y\\ \partial_z\end{pmatrix} \times \begin{pmatrix}A_x\\A_y\\A_z\end{pmatrix} \right] = \begin{pmatrix}A_x\\A_y\\A_z\end{pmatrix} \times \begin{pmatrix}\partial_y A_z-\partial_z A_y\\ \partial_z A_x-\partial_x A_z\\ \partial_x A_y-\partial_y A_x\end{pmatrix} \\= \begin{pmatrix} A_y (\partial_x A_y-\partial_y A_x) - A_z (\partial_z A_x-\partial_x A_z) \\ A_z (\partial_y A_z-\partial_z A_y) - A_x (\partial_x A_y-\partial_y A_x) \\ A_x (\partial_z A_x-\partial_x A_z) - A_y (\partial_y A_z-\partial_z A_y) \\ \end{pmatrix} \\ = \begin{pmatrix}(A_y \partial_x A_y + A_z \partial_x A_z) - (A_y \partial_y A_x + A_z \partial_z A_x) \\ (A_z \partial_y A_z + A_x \partial_y A_x) - (A_z \partial_z A_y + A_x \partial_x A_y) \\ (A_x \partial_z A_x + A_y \partial_z A_y) - (A_x \partial_x A_z + A_y \partial_y A_z) \\ \end{pmatrix} \\= \begin{pmatrix}(A_x \partial_x A_x + A_y \partial_x A_y + A_z \partial_x A_z) - (A_x \partial_x A_x + A_y \partial_y A_x + A_z \partial_z A_x) \\ (A_x \partial_y A_x + A_y \partial_y A_y + A_z \partial_y A_z) - (A_x \partial_x A_y + A_y \partial_y A_y + A_z \partial_z A_y) \\ (A_x \partial_z A_x + A_y \partial_z A_y + A_z \partial_z A_z) - A_x \partial_x A_z + A_y \partial_y A_z + A_z \partial_z A_z) \\ \end{pmatrix} \\ = \begin{pmatrix} \frac12 \partial_x (A_x^2+A_y^2+A_z^2) - (A_x\partial_x+A_y\partial_y+A_z\partial_z) A_x \\ \frac12 \partial_y (A_x^2+A_y^2+A_z^2) - (A_x\partial_x+A_y\partial_y+A_z\partial_z) A_y \\ \frac12 \partial_z (A_x^2+A_y^2+A_z^2) - (A_x\partial_x+A_y\partial_y+A_z\partial_z) A_z \\ \end{pmatrix} \\= \frac12 \nabla (A\cdot A) - (A\cdot\nabla) A $$

md2perpe
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    Just a small side remark: for ordinary vectors there is the Grassmann identity $A\times(B\times C)=(A\cdot C)B-(A\cdot B)C$ which has no factor 1/2. It is nice to see from that direct proof where the factor 1/2 comes from when $B$ is a differential operator. – Kurt G. Jul 30 '21 at 10:27
  • Generally, when $A,B,C$ do not commute, I get $$A\times(B\times C) = A\cdot_{AC}BC - (A\cdot B)C,$$ where $\cdot_{AC}$ denotes that the inner product is taken between $A$ and $C$. If $B$ and $C$ commute then $$A\cdot_{AC}BC = A\cdot_{AC}CB=(A\cdot C)B$$ so that $$A\times(B\times C) = (A\cdot C)B - (A\cdot B)C$$ just as @KurtG. writes. – md2perpe Jul 30 '21 at 14:14
  • Interestingly, the other equation $\nabla\times(\nabla\times A)=\nabla(\nabla\cdot A)-\Delta A$ has no factor 1/2. It is Grassmann like. – Kurt G. Dec 09 '21 at 09:50