- #1

Snaab

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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:

[itex]j^\mu (\mathbf{r},t) = e(1,\mathbf{v_0}) \delta(\mathbf{r'} - \mathbf{r_0}(t))[/itex].

Now comes my question.

In the stationary frame, the 4-current density should be given by:

[itex]j^\mu(\mathbf{r}) = (c e \delta(\mathbf{r}),0)[/itex]

So, when this gets boosted, shouldn't one have the following current density:

[itex]j^\mu = (\gamma c e \delta(\mathbf{r'} - \mathbf{r_0}(t)), \gamma e \mathbf{v_0}\delta(\mathbf{r'} - \mathbf{r_0}(t)))[/itex]

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

Cheers,

Snaab

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:

[itex]j^\mu (\mathbf{r},t) = e(1,\mathbf{v_0}) \delta(\mathbf{r'} - \mathbf{r_0}(t))[/itex].

Now comes my question.

In the stationary frame, the 4-current density should be given by:

[itex]j^\mu(\mathbf{r}) = (c e \delta(\mathbf{r}),0)[/itex]

So, when this gets boosted, shouldn't one have the following current density:

[itex]j^\mu = (\gamma c e \delta(\mathbf{r'} - \mathbf{r_0}(t)), \gamma e \mathbf{v_0}\delta(\mathbf{r'} - \mathbf{r_0}(t)))[/itex]

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

Cheers,

Snaab

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