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Potential from a charge moving at constant velocity

  1. Mar 18, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the electric potential of a point charge with constant velocity ##v##.

    2. Relevant equations
    $$V(\mathbf{r}, t) = \frac{1}{4\pi\epsilon_0} \int \frac{\rho\left(\mathbf{r}', t - \frac{| \mathbf{r}- \mathbf{r}'| }{c}\right)}{| \mathbf{r}- \mathbf{r}'|}d^3r' $$

    3. The attempt at a solution
    We look for the intersection of a lightcone into the past from ##\mathbf{r}, \,t## with a particle with constant velocity
    $$\mathbf{r}'(t') = \mathbf{v}t'$$
    Solving for the intersection
    $$t-t' = \frac{\left|\mathbf{r} - \mathbf{r}'(t')\right|}{c} $$
    $$t-t' = \frac{\sqrt{r^2 -2\mathbf{r}\cdot\mathbf{r}(t') + (r'(t'))^2}}{c} = \frac{\sqrt{r^2 -2\mathbf{r}\cdot\mathbf{v}t' + v^2(t')^2}}{c} $$
    $$(c^2 - v^2 )(t')^2 +2(\mathbf{r}\cdot\mathbf{v} -c^2t)t' + c^2t^2-r^2 = 0$$
    $$t' = \frac{-(\mathbf{r}\cdot\mathbf{v} -c^2t) \pm \sqrt{(\mathbf{r}\cdot\mathbf{v} - c^2t)^2 - (c^2 -v^2)(c^2t^2-r^2 )}}{(c^2 -v^2)} $$
    This agrees with (10.41) in Griffiths. However when I used the retarded potential directly, evaluating ## \left|\mathbf{r} - \mathbf{r}'\right|## where##\mathbf{r}'## is at the retarted time ##t'##. I'd expect
    $$V(\mathbf{r}, t) = \frac{1}{4\pi\epsilon_0} \frac{q}{\left|\mathbf{r} - \mathbf{r}'(t')\right|} = \frac{1}{4\pi\epsilon_0} \frac{q}{c\left(t -t' \right)}$$
    $$= \frac{q}{4\pi c\epsilon_0}\frac{(c^2 -v^2)}{t(c^2-v^2)-(\mathbf{r}\cdot\mathbf{v} -c^2t) + \sqrt{(\mathbf{r}\cdot\mathbf{v} - c^2t)^2 - (c^2 -v^2)(c^2t^2-r^2 )}} $$
    However according to (10.42) in Griffith's we should have
    $$V(\mathbf{r}, t) = \frac{q}{4\pi \epsilon_0}\frac{c}{\sqrt{(c^2t - \mathbf{r}\cdot\mathbf{v} )^2 + (c^2-v^2)(r^2 - c^2t^2)}} $$
    I do see a way that these expressions are equal. I suppose I ignored the change of variables for the delta function when evaluating the retarded potential intergral. Is that really what I'm missing?
    $$\rho\left(\mathbf{x}', t'(\mathbf{x}') - \frac{|\mathbf{x} - \mathbf{x}'|}{c} \right) = q\delta\left(\mathbf{x}-\mathbf{v}\left(t'(\mathbf{x}') -\frac{|\mathbf{x} - \mathbf{x}'|}{c}\right)\right) $$
     
    Last edited: Mar 18, 2015
  2. jcsd
  3. Mar 20, 2015 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Yes, you need to account for the fact that the argument of the delta function is a function of ##\mathbf{x}'##. I'm a little confused with how you are using primes in your notation. Shouldn't ##t'(\mathbf{x}')## just be the present time ##t##? And shouldn't the first ##\mathbf{x}## appearing in the delta function be the integration variable ##\mathbf{x}'##?
     
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