# Homework Help: Potential from a charge moving at constant velocity

Tags:
1. Mar 18, 2015

### MisterX

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. Mar 20, 2015

### TSny

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 $\mathbf{x}'$?