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Retarded Potential of Moving Charge

  1. Apr 6, 2013 #1
    1. The problem statement, all variables and given/known data

    Calculate [itex] \phi_{ret} [/itex] 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 [itex]x_o[/itex] is the solution of [itex]f(x_o) = 0.[/itex]

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

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

    Which results in (I think),

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

    Then I suppose I would substitute [itex] t' = t - |\vec{r} - \vec{r}_o|/c.[/itex] 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 [itex] \delta(x' - v't') [/itex] instead of [itex] \delta(x' - vt')? [/itex]
    Last edited: Apr 7, 2013
  2. jcsd
  3. Apr 8, 2013 #2
    Ah, I think I figured it out. We are looking at the case where [itex] f(x_o) = 0 [/itex], so we just divide by v.
  4. Apr 9, 2013 #3
    If anyone is keeping track, it turns out I should have been taking the derivative w.r.t x. . . [itex] f'(x_o) = \frac{d}{dx}f(x_o) [/itex]. Woops!
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