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I'm following the derivation for the Field of a moving point charge, from Griffiths' Introduction to Electrodynamics. Particularly, on page 458 the author states that

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I find this familiar except for the $\frac{1}{2}$ that vanished from the first line. I know the 2nd line comes from a triple product but I can't see why there's a $2$ factor and where does it come from.

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  • $\begingroup$ Aren't these identities mentioned on the inside covers and the first chapter of that book? $\endgroup$
    – Triatticus
    Nov 20, 2020 at 3:59

2 Answers 2

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The full product rule for the gradient of a dot product is $$ \nabla(A\cdot B) = (A\cdot\nabla)B + (B\cdot\nabla)A + A \times(\nabla\times B) + B \times(\nabla\times A) $$ where I've elided the vector symbols. This means that in $\nabla(r'\cdot r')$, there will be two terms each of $(r'\cdot\nabla)r'$ and $r'\times(\nabla\times r')$, hence the factor of $2$.

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This is a common identity in vector calculus for the gradient of a dot product. For two vectors $\mathbf{A}$ and $\mathbf{B}$, we have: $$\nabla(\mathbf{A}\cdot\mathbf{B})=(\mathbf{A}\cdot\nabla)\mathbf{B}+(\mathbf{B}\cdot\nabla)\mathbf{A}+\mathbf{A}\times(\nabla\times\mathbf{B})+\mathbf{B}\times(\nabla\times\mathbf{A})$$

Note what happens in the case where $\mathbf{A}=\mathbf{B}$: \begin{align} \nabla(\mathbf{A}\cdot\mathbf{A})&=(\mathbf{A}\cdot\nabla)\mathbf{A}+(\mathbf{A}\cdot\nabla)\mathbf{A}+\mathbf{A}\times(\nabla\times\mathbf{A})+\mathbf{A}\times(\nabla\times\mathbf{A})\\ &= 2\left[(\mathbf{A}\cdot\nabla)\mathbf{A}+\mathbf{A}\times(\nabla\times\mathbf{A})\right] \end{align}

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