Retarded Potential of Moving Charge

1. Apr 6, 2013

Shinobii

1. The problem statement, all variables and given/known data

Calculate $\phi_{ret}$ for a charge moving with constant v, along the x-axis.

2. The attempt at a solution

$$\phi = q \int_{-\infty}^{\infty} \frac{\delta(x' - vt')}{|\vec{r} - \vec{r}'|} dx'.$$

I then use the Dirac delta relation,

$$\int_{-\infty}^{\infty} \delta[f(x)]g(x) = \frac{g(x_o)}{|f'(x_o)|}.$$

Where $x_o$ is the solution of $f(x_o) = 0.$

I am not really sure how to do this, except that

$$\begin{eqnarray} f(x) &=& x' - vt', \\ g(x) &=& \frac{1}{|\vec{r} - \vec{r}'|}. \end{eqnarray}$$

Which results in (I think),

$$\frac{1}{|\vec{r} - \vec{r}_o|}\frac{-1}{\frac{d(vt')}{dt}}.$$

Then I suppose I would substitute $t' = t - |\vec{r} - \vec{r}_o|/c.$ Looking at the derivative in the denominator I would get,

$$v \bigg( 1 - \frac{1}{c}\frac{d}{dt}|\vec{r} - \vec{r}_o| \bigg).$$

But here is where I get stuck (assuming I have not already made any mistakes).

Ok I was able to reach the final result for the derivative;

$$-v \bigg( 1 - \frac{v}{c} \frac{(\vec{x} - \vec{x}_o)}{|\vec{r} - \vec{r}_o|} \bigg)$$

however, I have a pesky -v term multiple to the result! Should I have written the delta as $\delta(x' - v't')$ instead of $\delta(x' - vt')?$

Last edited: Apr 7, 2013
2. Apr 8, 2013

Shinobii

Ah, I think I figured it out. We are looking at the case where $f(x_o) = 0$, so we just divide by v.

3. Apr 9, 2013

Shinobii

If anyone is keeping track, it turns out I should have been taking the derivative w.r.t x. . . $f'(x_o) = \frac{d}{dx}f(x_o)$. Woops!