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In a potential which needs to be evaluated at the retarded time, is this the time which represents the actual time the "physics" occurred? So $t_{\text{ret}}=t-\frac{r}{c}$, not just because it may be that you are receiving a signal at light speed but because "causality" spreads out at the maximum speed, $c$, is this correct?

The Lienard-Wiechert 4-potential for some point charge ($q$): $A^\mu=\frac{q u^\mu}{4\pi \epsilon_0 u^\nu r^\nu}$ where $r^\nu$ represents the 4-vector for the distance from the observer. In the rest frame of the charge $A^i$ for $i=1,2,3$ is clearly zero but from what has been said about the retarded time we have that $A^0=\frac{q}{4\pi\epsilon_0c(t-r/c)}$.

Obviously I would like to get $A^0=-\frac{q}{4\pi\epsilon_0 r}$, so where is the misunderstanding of retarded time and instantaneous time? Unless we would like the time since the signal was emitted which is $r/c$? Or if $t$ itself is already $t'-r/c$ and we need to return to the instantaneous time $t$, when the signal was emitted.

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Actually since charge is at rest $u_{\nu}r^{\nu} = u_0 r^0 = ct'$ where $t'$ is retarded time, $t'=r/c$, where $r$ is the (constant) distance to the charge.

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The potential 1-form $A ≑ A_x dx + A_y dy + A_z dz - Ο† dt = 𝐀·d𝐫 - Ο† dt$ is given at the space-time point $\left(𝐫,t\right)$ by: $$A\left(𝐫,t\right) = {e\over{4πΡ_0c}} {{𝒗·d𝐫 - cΒ² dt}\over{Rc - 𝐑·𝒗}},$$ where $t ↦ 𝒓(t)$ is the trajectory of the point source, and $$𝒗 = \dot{𝒓}(t_-),$$ $$t_- = t - {R \over c},$$ $$R = |𝐑|,$$ $$𝐑 = 𝐫 - 𝒓(t_-).$$ The last three of these relations determine $t_-$, $R$ and $𝐑$ as implicit functions of $\left(𝐫,t\right)$, and a functional of the trajectory $t ↦ 𝒓(t)$. It is well-defined if $|𝒗| < c$.

If the charge is stationary, with $𝒓(t) = 𝟎$, then $𝐑 = 𝐫$, $R = r ≑ |𝐫|$, $t_- = t - r/c$ and $𝒗 = 𝟎$, with the potential 1-form thus reducing to the following: $$A = -{{e dt}\over{4πΡ_0r}},$$ or, component-wise, to $𝐀 = 𝟎$ and $Ο† = e/{4πΡ_0r}$, which is independent of $t$, so is also the advanced potential.

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