Deducing the Heaviside-Feynman formulae from Jefimenko's equations

Question

I've tried to deduce the one point charge Heaviside–Feynman formula from the Jefimenko's equations. This should be possible, by replacing the densities with Dirac deltas, somehow, but I failed. Could you provide it or sketch it?

Heaviside–Feynman formula: $$ \mathbf{E} = \frac{-q}{4\pi \varepsilon_0} \left[ \frac{\mathbf{e}_{r'}}{r'^2} + \frac{r'}{c} \frac{d}{dt} \left(\frac{\mathbf{e}_{r'}}{r'^2}\right) +\frac{1}{c^2} \frac{d^2}{dt^2} \mathbf{e}_{r'} \right] $$ $$\mathbf{B} = - \mathbf{e}_{r'} \times \frac{\mathbf{E}}{c}.$$ Here, $\mathbf{E}$ and $\mathbf{B}$ are the electric and magnetic fields respectively, $ q$ is the electric charge, $\varepsilon_0$ is the vacuum permittivity and $c$ is the speed of light. The vector $\mathbf{e}_{r'}$ is a unit vector pointing from the observer to the charge and $r'$ is the distance between observer and charge. Since the electromagnetic field propagates at the speed of light, both these quantities are evaluated at the retarded time $t - r'/c$.

Answer 1

Too long for a comment, and I don't have time to work out the kinks for a complete answer just now:

I've attempted to work through this by rewriting Jefimenko's equations as $$ \mathbf{E}(\mathbf{r},t) = \frac{1}{4 \pi \epsilon_0} \int d^4x' \left\{ \rho(\mathbf{r}', t') \hat{\mathscr{r}}\left[ \frac{\delta'(t - t' - \mathscr{r}/c)}{c\mathscr{r}} + \frac{\delta(t - t' - \mathscr{r}/c)}{\mathscr{r}^2} \right] - \frac{\mathbf{J}(\mathbf{r}',t')}{c^2 \mathscr{r}} \delta'(t - t' - \mathscr{r}/c)\right\} $$ $$ \mathbf{B}(\mathbf{r},t) = \frac{\mu_0}{4 \pi} \int d^4x' \left\{ (\mathbf{J}(\mathbf{r}', t') \times \hat{\mathscr{r}})\left[ \frac{\delta'(t - t' - \mathscr{r}/c)}{c\mathscr{r}} + \frac{\delta(t - t' - \mathscr{r}/c)}{\mathscr{r}^2} \right] \right\} $$ where $\vec{\mathscr{r}} \equiv \mathbf{r} - \mathbf{r}'$, $\mathscr{r} \equiv |\vec{\mathscr{r}}|$, and $\hat{\mathscr{r}} \equiv \vec{\mathscr{r}}/\mathscr{r}$. Note that these integrals are over $\mathbf{r}'$ and $t'$. To see that this is the case, do the integrals over $t'$.¹

It should then be possible to derive the Heaviside-Feynman formula by putting in the sources $$ \rho(\mathbf{r}', t') = q \delta^3(\mathbf{r}'-\mathbf{w}(t')) \qquad \mathbf{J}(\mathbf{r}', t') = q \mathbf{v}(t') \delta^3(\mathbf{r}'-\mathbf{w}(t')) $$ where $\mathbf{w}(t)$ is the trajectory of the particle and $\mathbf{v}(t)$ is its velocity, and then integrating over $\mathbf{r}'$ before integrating over $t'$. Doing this gives some expressions that are tantalizingly close to the Heaviside-Feynman formula, but there appear to be some stray terms as well. I suspect I've just been careless with my derivatives and/or signs; hopefully I will have some time in the coming days and be able to revisit this, straighten things out, and post a more complete answer.

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¹ True confession: I actually obtained these expressions by writing out the Lorenz-gauge potentials $\phi$ and $\mathbf{A}$ in terms of the retarded Green's function and the sources, and then explicitly differentiating those expressions with respect to $\mathbf{r}$ and $t$.