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 $$\vec{\nabla} (r) = \frac{\vec{r}}{r}~?$$


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

  • $\begingroup$ Yes, You made it very clear. Thank You so much. $\endgroup$ Aug 12 '20 at 8:43
  • $\begingroup$ Notice that the whole argument is made with $\tau$ not with $t$, that is not useful in describing the particle motion. $\tau$ is a sort of local or proper time, meaning that when you place yourself in a position near the particle such that $|\boldsymbol{d}_i|=0$ you will have $\tau=t$ $\endgroup$
    – Rob Tan
    Aug 12 '20 at 8:49
  • $\begingroup$ @JarogniewBorkowski If the problem was solved consider to upvote and accept the answer, thanks. $\endgroup$
    – Rob Tan
    Aug 12 '20 at 9:09
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    $\begingroup$ It is very helpful. I couldn't understand this gradient since years untill today. I check already everything and yes, I can calculate correct gradient. Now I have one independent physical case.... I need to derive that $$\vec{E} = - \vec{\nabla} \phi - \frac{\partial \vec{A}}{\partial t}$$ but I will make another post for that question. Best regards. $\endgroup$ Aug 12 '20 at 11:29
  • $\begingroup$ I'm happy I could help! I stayed a lot of time to understand it too. Post a question about the electric field and I will answer as soon as I can $\endgroup$
    – Rob Tan
    Aug 12 '20 at 11:50

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