Liénard-Wiechert Fields for a static particle

Question

Note: There was already a similar question to mine, but it did not actually answer my question: Retarded time Lienard Wiechert potential

When considering the Liénard-Wiechert fields, which are the electromagnetic fields of a moving charge carrying particle and taking the special case, that the particle is static, i.e. v=0, then we end up with the following expression for the electric field: $$ \vec{E}(\vec{r},t) = \frac{e}{4\pi\varepsilon_0}\frac{\vec{r}}{r^2}\vert_{ret}, $$ which is almost the expression we would get from the Coulomb potential, except for the fact that it is evaulated at $t=t_{ret}$.

My question is: What is the physical explanation of this? Is it just the fact, that when placing an electron somewhere, even the static Coulomb Field does not instantaneously "fill space", but takes finite time for that? Evaluating $t_{ret} =t-\frac{\vert\vec{r}-\vec{r'}\vert}{c}$ would mean that $\vec{r'}$ is just the position of the particle itself, correct?

Answer 1

Everything you've said is correct. Consider if we somehow made the charge suddenly appear, then for us at some $\vec{r}$, we still need to wait a finite amount of time to feel its influence, or we would have superluminal communication. Essentially, in the static Coulomb case, we're within the light-cone of the charge having become stationary, and then we demand it stays stationary in the future.

As for the meaning of evaluating it at a retarded time, well you can simply say $\vec{r}$ is not a function of time, and so it will be the same no matter when you evaluate it, which will recover the mathematical form of Coulomb's law.