Heaviside-Feynman formula derivation I want to discuss derivation of Feynman-Heaviside formula.
The topic has already been discussed here but I can not put there any question that's why I'm making new post.
Deriving Heaviside-Feynman formula for the electric field of an arbitrarily moving charge from Lienard-Wiechert potential
Could anyone help me and explain how "guillefix" user gets his corrected gradient of r?
$$\vec{\nabla} (r) = \frac{\vec{r}}{r}-\frac{\vec{r}}{r}\cdot\frac{d\vec{r}_{2}}{d t'}\bigg(\frac{-\vec{\nabla} (r)}{c}\bigg)$$
And why is it correct instead of
$$\vec{\nabla} (r) = \frac{\vec{r}}{r}~?$$
 A: See if this can help. You place yourself in the point $\boldsymbol{r}$ and you see the $i$ particle with motion $\boldsymbol{r}_i(t)$ parametrized in your clock with time $t$. You simply define the position vector $\boldsymbol{d}_i(t)\doteq\boldsymbol{r}-\boldsymbol{r}_i(t)$, the normalized position vector $\boldsymbol{n}_i(t)\doteq\boldsymbol{d}_i(t)/|\boldsymbol{d}_i(t)|$ and the retarded time $\tau$
\begin{equation*}
\tau
+\frac{|\boldsymbol{d}_i(\tau)|}{c}
=
t
\end{equation*}
so that when you have a generic function $f(\boldsymbol{r},\tau)$ (such as the electric field in a point) and you try to change the observation position you must consider that also the $\tau$ will change when you move in another position, such that
\begin{gather*}
\frac{\text{d}f}{\text{d}\boldsymbol{r}}
=
\frac{\partial f}{\partial\boldsymbol{r}}
+
\frac{\partial\tau}{\partial\boldsymbol{r}}
\frac{\partial f}{\partial\tau}
\end{gather*}
where you take back the definition of $\tau$ ($t,\boldsymbol{r}$ are unrelated parameters)
\begin{equation*}
\frac{\partial\tau}{\partial\boldsymbol{r}}
+\frac{1}{c}\frac{\partial}{\partial\boldsymbol{r}}|\boldsymbol{r}
-\boldsymbol{r}_i(\tau)|
=
\frac{\partial t}{\partial\boldsymbol{r}}
=
0
\end{equation*}
and you consider a component to obtain the general result
\begin{gather*}
\frac{\partial \tau}{\partial x}
+\frac{1}{c}\frac{\partial }{\partial x}\sqrt{\left(x-x_i(\tau)\right)^2
+\left(y-y_i(\tau)\right)^2
+\left(z-z_i(\tau)\right)^2}
=
0
\\
\frac{\partial \tau}{\partial x}
+\frac{1}{2c}\frac{1}{|\boldsymbol{r}-\boldsymbol{r}_i(\tau)|}
\left(2\left(x-x_i(\tau)\right)
\left(1-\frac{\partial x_i(\tau)}{\partial\tau}\frac{\partial \tau}{\partial x}\right)
+2\left(y-y_i(\tau)\right)
\left(-\frac{\partial y_i(\tau)}{\partial\tau}\frac{\partial \tau}{\partial x}\right)
+2\left(z-z_i(\tau)\right)\left(-\frac{\partial z_i(\tau)}{\partial\tau}\frac{\partial \tau}{\partial x}\right)\right)
=
0
\\
\frac{\partial \tau}{\partial x}
=
\frac
{\displaystyle{
\frac{1}{c}\frac{\left(x-x_i(\tau)\right)}{|\boldsymbol{r}-\boldsymbol{r}_i(\tau)|}
}}
{\displaystyle{
\frac{1}{c}\boldsymbol{n}_i(\tau)\cdot\dot{\boldsymbol{r}}_i(\tau)-1
}}
\end{gather*}
\begin{equation*}
\frac{\partial \tau}{\partial \boldsymbol{r}}
=
\frac{\boldsymbol{n}_i(\tau)}
{\boldsymbol{n}_i(\tau)\cdot\dot{\boldsymbol{r}}_i(\tau)-c}
\end{equation*}
and defining $\kappa_i(\tau)\doteq 1-\boldsymbol{n}_i(\tau)\cdot\boldsymbol{\beta}_i(\tau)$ (where $\boldsymbol{\beta}_i(\tau)$ is $\text{d}\boldsymbol{r}_i(\tau)/(c\text{d}\tau)$) there you go, you just obtained
\begin{equation*}
\nabla
=
\frac{\partial}{\partial\boldsymbol{r}}-\frac{1}{c}\frac{\boldsymbol{n}_i(\tau)}{\kappa_i(\tau)}\frac{\partial}{\partial\tau}
\end{equation*}
If you're wondering the temporal derivative is obtained more easily
\begin{equation*}
\frac{\partial}{\partial t}
=
\frac{1}{\kappa_i(\tau)}\frac{\partial}{\partial\tau}
\end{equation*}
Hope this helps
