# Lienard-Wiechert potential

1. Nov 3, 2011

### QuArK21343

Just to remind you, the Lienard-Wiechert potential, as far as the scalar potential is concerned, is:

$$\phi(\textbf{r},t)=\frac{q}{|\textbf r-\textbf r'(t')|\left(1-\frac{v'(t')\cdot \textbf{u}}{c}\right)}$$

, where $\textbf u$ is the unit vector in the direction of $\textbf r-\textbf r'$. This equation gives the scalar potential at position $\textbf{r}$ and at time t (the prime indicates that those quantities are to be evaluated at the retarded time). From here, and the similar expression for the vector potential, we can derive the electric and magnetic fields. To do so, we need to evaluate the gradient of $t'$ where:

$$t'=t-\frac{|\textbf r-\textbf r'|}{c}$$

This i can't do; in fact, I don't understand with respect to what I have to take the gradient. Answer should be:

$$grad (t')=\frac{-\textbf u}{c-\textbf v'\cdot \textbf u}$$

Any help is much appreciated!

2. Nov 4, 2011

### vanhees71

I guess, what you need is the gradient of the retarded time $t'$ wrt. $\vec{x}$ at $t=\text{const}$ since this is what is needed to obtain the electromagnetic field,

$$F_{\mu \nu} = \partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}.$$

This is somewhat tedious. I'd rather keep the four-dimensional form of the potential written in terms of the covariant retarded Green's function

$$A^{\mu}(x)=\int \mathrm{d}^4 x \Delta_{\text{ret}}(x-x') j^{\mu}(x')$$

with (in Heaviside-Lorentz units and $c=1$)

$$\Delta_{\text{ret}}(x-x')=\frac{1}{4 \pi} \delta \left [t'-(t-|\vec{r}-\vec{r}'|)\right ].$$

For the four-current density of a particle with charge, $q$, you have

$$(j^{\mu}(x))=\begin{pmatrix} q \delta^{(3)}[\vec{x}-\vec{y}(t)] \\ q \frac{\mathrm{d}}{\mathrm d t} \vec{y}(t) \delta^{(3)}[\vec{x}-\vec{y}(t)] \end{pmatrix},$$

where $\vec{y}(t)$ is the trajectory of the particle is function of the coordinate time, $t$.