# How to prove the existence of the following equation?

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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The usual procedure is to find the solution of $\Delta A(r) = \delta(r)$ (the green function of the equation) and to make a convolution with the inhomogeneous part. In this case it seems that the green function in $1/|r|$. – guaraqe Jun 13 '12 at 22:55
I don't know about the Green function, because I don't study math on this level. Can you explain if it won't be hard? – John Taylor Jun 13 '12 at 23:03
It is based on Fourier transform, do you know it? – guaraqe Jun 13 '12 at 23:14
Yes, I know it. – John Taylor Jun 14 '12 at 9:36

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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From a mathematician's viewpoint, a cleaner approach would be to insert $\mathbf{A}$ into the equations and check. Fourier analysis requires assumptions that might be heavier than those for being a pointwise solution. – Siminore Jun 18 '12 at 11:06
Sure, but I used Fourier analysis precisely because the OP knows it. – Raskolnikov Jun 18 '12 at 11:07
I understand. But also direct differentiation might be a good exercise. – Siminore Jun 18 '12 at 11:09