# Retarded time Lienard Wiechert potential

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.

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.
The potential 1-form $$A ≡ A_x dx + A_y dy + A_z dz - φ dt = 𝐀·d𝐫 - φ dt$$ is given at the space-time point $$\left(𝐫,t\right)$$ by: $$A\left(𝐫,t\right) = {e\over{4πε_0c}} {{𝒗·d𝐫 - c² dt}\over{Rc - 𝐑·𝒗}},$$ 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: $$A = -{{e dt}\over{4πε_0r}},$$ or, component-wise, to $$𝐀 = 𝟎$$ and $$φ = e/{4πε_0r}$$, which is independent of $$t$$, so is also the advanced potential.