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In Jackson's electrodynamics book, he says that from the Biot-Savart law in current density form B(x) = $\frac{\mu_0}{4\pi} \int$ J(x') $\times \frac{(x-x')}{|x-x'|^3}dv$ we can use some relation to obtain the equivalent form B(x) = $\frac{\mu_0}{4\pi} \nabla \times \int \frac{J(x')}{|x - x'|}dv$. He cites another section for what exactly this relation is, but there are multiple relations in that section and none of them seem clearly related to this derivation to me. I was hoping someone could explain how you might obtain this identity.

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  • $\begingroup$ hint: $\nabla \frac{1}{|\mathbf x - \mathbf x'|} = - \frac{\mathbf x - \mathbf x'}{|\mathbf x - \mathbf x'|^3}$ where the nabla is operating on the $\mathbf x$ variable while $\mathbf x'$ is kept constant. $\endgroup$
    – hyportnex
    Mar 11, 2023 at 21:04
  • $\begingroup$ Also the vector calculus identity $\nabla \times (f\vec{V}) = \nabla f \times \vec{V} + f \nabla \times \vec{V}$ might be useful. $\endgroup$
    – Jakob KS
    Mar 11, 2023 at 21:10

1 Answer 1

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Using the relation $$ \begin{align} \frac{\partial}{\partial x_j} \frac{1}{|\vec{x}-\vec{x}^\prime|} &= \frac{\partial}{\partial x_j}\left[ (x_ \ell -x_\ell^\prime)(x_\ell - x_\ell^\prime) \right]^{-1/2}= -\frac{1}{2}\left[(x_\ell - x_\ell^\prime) (x_\ell - x_\ell^\prime) \right]^{-3/2} \, 2 (x_j - x_j^\prime) \\ &=- \frac{x_j-x_j^\prime}{|\vec{x}-\vec{x}^\prime|^3},\end{align}$$ you find $$\varepsilon_{ijk} \frac{\partial}{\partial x_j}\int \!d^3 \! x^\prime \frac{J_k(\vec{x}^\prime)}{|\vec{x}- \vec{x}^\prime| }= -\varepsilon_{ijk} \int \! d^3 \! x^\prime \frac{J_k(\vec{x}^\prime) (x_j-x_j^\prime)}{|\vec{x}-\vec{x}^\prime|^3} =\varepsilon_{ikj}\int \! d^3 \! x^\prime \frac{J_k(\vec{x}^\prime) (x_j-x_j^\prime)}{|\vec{x}-\vec{x}^\prime|^3}, $$ being your desired relation in index notation.

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