# Liénard-Wiechert potential derivation

1. Jun 21, 2011

### Snaab

Hi there,

I'm studying some classical field theory, and I have a question about the derivation of the Liénard-Wiechert potential for a moving point charge, see http://en.wikipedia.org/wiki/Liénard–Wiechert_potential#Derivation" for the derivation and the meaning of the symbols I use.

The current density in the observing frame is said to be given by:
$j^\mu (\mathbf{r},t) = e(1,\mathbf{v_0}) \delta(\mathbf{r'} - \mathbf{r_0}(t))$.

Now comes my question.
In the stationary frame, the 4-current density should be given by:
$j^\mu(\mathbf{r}) = (c e \delta(\mathbf{r}),0)$

So, when this gets boosted, shouldn't one have the following current density:
$j^\mu = (\gamma c e \delta(\mathbf{r'} - \mathbf{r_0}(t)), \gamma e \mathbf{v_0}\delta(\mathbf{r'} - \mathbf{r_0}(t)))$

In other words, why isn't there a factor $\gamma$ in the density like posed on wikipedia? Also in my own literature there is no such factor.

Cheers,

Snaab

Last edited by a moderator: Apr 26, 2017
2. Jun 21, 2011

### Bill_K

The missing γ has been swallowed by the delta function. There's a γ in the transformation from r to r', and one must use the identity δ(x) = c δ(cx). In other words, δ(r) = (1/γ)δ(r').