# 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.

## 1 Answer

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.