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I learned electrodynamics. According to the vector potential determination, $ \mathbf B = [\nabla \times \mathbf A ], $ Coulomb gauge, $ \nabla \mathbf A = 0, $ and one of Maxwell's equations, $ [\nabla \times \mathbf B ] = \frac{1}{c}4\pi \mathbf j, $

I can assume, that

$ [\nabla \times \mathbf B ] = \nabla (\nabla \mathbf A) - \Delta \mathbf A = -\Delta \mathbf A = \frac{1}{c}4 \pi \mathbf j. $ How to prove that the one of the solutions of this equation is solution like newtonian potential, $ \mathbf A = \frac{1}{c}\int \limits_{V} \frac{\mathbf j (r) d^{3}\mathbf r}{|\mathbf r - \mathbf r_{0}|}? $

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    Yes, I know it.2012-06-14

1 Answers 1

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Skipping the part about the Green's function, you should apply Fourier transformation on your equation

$-\Delta \mathbf A = \frac{1}{c}4 \pi \mathbf j.$

to change it into

$k^2 \tilde{\mathbf{A}} = \frac{1}{c}4 \pi \tilde{\mathbf{j}}$

or

$\tilde{\mathbf{A}} = \frac{1}{c}4 \pi \tilde{\mathbf{j}}\cdot\frac{1}{k^2}$

The left hand side is a product, so the inverse Fourier transform will be a convolution. The inverse Fourier transform of $1/k^2$ is $1/|r|$ in 3D, as you can see in formula 502. Therefore one gets

$\mathbf A = \frac{1}{c}\int \limits_{V} \frac{\mathbf j (r) d^{3}\mathbf r}{|\mathbf r - \mathbf r_{0}|} \; .$

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    I understand. But also direct differentiation might be a good exercise.2012-06-18