From the Fourier-transformed Maxwell equations we have, with some algebraic manipulation, $$\mathbf{E}=\frac{1}{|k|^2}\left[\mathbf{k}\frac{\rho}{\epsilon_0}-\mathbf{k}\times k_0\mathbf{B}\right]$$ $$\mathbf{B}=\frac{\mu_0}{|k|^2}\mathbf{k}\times\left[\mathbf{J}+\epsilon_0k_0\mathbf{E}\right]$$

which can be uncoupled to give rather ugly equations by inserting the relevant field expression into the equations.

The issue is that when transforming back to position space it seems that it would give an inverse-linear relationship instead of a term with the familiar inverse-square due to the inverse square in the Fourier expression:

Explicitly with the electric field, the uncoupled expression is $$\mathbf{E}=\frac{1}{|k|^2}\left[\mathbf{k}\frac{\rho}{\epsilon_0}-\mathbf{k}\times \mathbf{k}\times k_0\frac{\mu_0}{|k|^2}\left[\mathbf{J}+\epsilon_0k_0\mathbf{E}\right]\right]$$

$$=\frac{1}{|k|^2}\big[\mathbf{k}\frac{\rho}{\epsilon_0}-\mathbf{k}\overbrace{\left(\mathbf{k}\cdot k_0\frac{\mu_0}{|k|^2}\left[\mathbf{J}+\epsilon_0k_0\mathbf{E}\right]\right)}^{=0 \text{ by continuity}}+k_0\mu_0\left(\mathbf{J}+\epsilon_0k_0\mathbf{E}\right)\big]$$


which persists in it decaying slower than $1/k^2$ and thus being unable to produce an inverse square relationship. Where is the mistake?


Turns out that it is there is no issue with the rather superficial degree of growth. The key lies in the relationship $-i\mathbf{k}\leftrightarrow \nabla$

For ease this can be equivalently done in the tensor/four-vector formalism, which gives $$\hat{F}^{\mu\nu}=i\mu_0\frac{p^\mu \hat{J}^\nu - p^\nu \hat{J}^\mu}{p^2}$$

That specific form is to enforce antisymmetry, as it is not required by the field equations themselves. For brevity, I write only the first part in the below formulas as the analysis is the same.

$$\implies F^{\mu\nu}=i\mu_0\int \frac{d^4p}{(2\pi)^4} \frac{p^\mu \hat{J}^\nu}{p^2}e^{ipx}$$

$$=-\mu_0\partial^\mu\int \frac{d^4p}{(2\pi)^4} \frac{\hat{J}^\nu}{p^2}e^{ipx}$$

$$=\frac{\mu_0}{64\pi^5}\int d^4x'\ \frac{\delta(x^0_r-x'^0)J^\nu(x')}{|x-x'|^2}-\frac{\partial^\mu\delta(x^0_r-x'^0)J^\nu(x') + \delta(x^0_r-x'^0)\partial^\mu J^\nu(x')}{|x-x'|}$$

In the last expression the $|x-x'|$ term refers to the spatial distance (i.e. not including $x^0$). It may be integrated over $x^0$ and then partially integrated over all variables to transfer the delta function's derivative to give (including the second part now)

$$\boxed{\frac{\mu_0}{64\pi^5}\int d^3x'\ \frac{x^{\mu\neq 0} J^\nu(x',x^0_r)- x^{\nu\neq 0} J^\mu(x',x^0_r)}{|x-x'|^2}-\frac{x^\mu_r\partial_{x^0_r} J^\nu(x',x^0_r) - x^\nu_r\partial_{x^0_r} J^\mu(x',x^0_r)}{|x-x'|}}$$

Where I have written $x^\mu_r=\left(\delta^\mu_0-\frac{1}{c}\delta^\mu_{1,2,3}x^{\mu\neq 0}\right)$.

The notation is rather messy, but this is indeed Jefimenko's equations in tensor form.

Two things to notice:

First, looking back at that expression, I think I have messed up the coefficients, but the form appears correct. Suggestions on how to fix the coefficients (or if they are correct) are welcome!

Second, I have ignored some delta functions as I do not know how to interpret them within this context. I plan on asking another question related to that when I get the time.


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