Suppose that there is a point particle with charge $q$ moving at the speed of light, $c$, in the positive $x$ direction. The particle crosses the origin at time $t=0$. Using the general definition of the Lorenz gauge potential: $$\phi(\textbf{r}, t)=\frac{1}{4\pi\epsilon_0}\int\delta\left(t'-t+\frac{|\textbf{r}-\textbf{r}'|}{c}\right)\frac{\rho(\textbf{r}', t')}{|\textbf{r}-\textbf{r}'|}dt'd^3\textbf{r}'$$ Show that: $$\phi(\textbf{r}, t)= \begin{cases} \frac{q}{4\pi\epsilon_0}\frac{1}{ct-x} & \text{if } ct>x, \\ 0 & \text{if } ct < x. \end{cases}$$

An attempt at a solution:

The particle is moving in the positive x direction at a speed c, meaning it will have a trajectory $\textbf{r}_q(t)=ct\hat{\textbf{x}}$. Because it is a point particle, its charge density is given by $\rho(\textbf{r}', t)=q\delta^3({\textbf{r}-\textbf{r}_q(t)})=q\delta(x-ct)\delta(y)\delta(z)$. Additionally, I can write $|\textbf{r}-\textbf{r}'|=\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}$ I can rewrite the integral above with these substitutions:

$$\phi(\textbf{r}, t)=\frac{1}{4\pi\epsilon_0}\int\delta\left(t'-t+\frac{\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}}{c}\right)\frac{q\delta(x'-ct')\delta(y')\delta(z')}{\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}}dt'dx'dy'dz'$$

$$\phi(\textbf{r}, t)=\frac{1}{4\pi\epsilon_0}\int\delta\left(t'-t+\frac{\sqrt{(x-ct')^2+(y)^2+(z)^2}}{c}\right)\frac{1}{\sqrt{(x-ct')^2+(y)^2+(z)^2}}dt'$$

Now I use the general formula: $$\delta(f(x))=\frac{\delta(x-x_0)}{|f'(x_0)|}$$

In this case, we have $f(t')=t'-t+\frac{\sqrt{(x-ct')^2+(y)^2+(z)^2}}{c}$. I will denote the roots of $f(t')$ as $t_{ret}=t-\frac{\sqrt{(x-ct_{ret})^2+(y)^2+(z)^2}}{c}$. The derivative of $f(t')$ is $f'(t')=1-\frac{x-ct'}{\sqrt{(x-ct_{ret})^2+(y)^2+(z)^2}}=\frac{\sqrt{(x-ct_{ret})^2+(y)^2+(z)^2}-(x-ct_{ret})}{\sqrt{(x-ct_{ret})^2+(y)^2+(z)^2}}$. Substituting in the simplified delta function yields:

$$\phi(\textbf{r}, t)=\frac{1}{4\pi\epsilon_0}\int\frac{\delta(t'-t_{ret})}{\frac{\sqrt{(x-ct_{ret})^2+(y)^2+(z)^2}-(x-ct_{ret})}{\sqrt{(x-ct_{ret})^2+(y)^2+(z)^2}}}\frac{1}{\sqrt{(x-ct')^2+(y)^2+(z)^2}}dt'$$

We can now integrate and simplify to get:

$$\phi(\textbf{r}, t)=\frac{1}{4\pi\epsilon_0}\left[\frac{1}{\sqrt{(x-ct_{ret})^2+(y)^2+(z)^2}-(x-ct_{ret})}\right]$$

I do not know how to proceed from here. How do I finish it?

  • 2
    $\begingroup$ The solution you wrote for $\phi(\vec r ,t)$ seems to only depend on $x$ and $t$ on the RHS, but not $y$ and $z$. Is that a typo? Are you evaluating $\phi$ at $y=z=0$? $\endgroup$
    – hft
    Sep 23 at 20:57
  • 2
    $\begingroup$ You haven’t calculated the retarded time. $\endgroup$
    – Ghoster
    Sep 23 at 21:11
  • 3
    $\begingroup$ I assume you mean a massless particle traveling at the speed of light? $\endgroup$
    – Bob D
    Sep 23 at 21:14
  • $\begingroup$ The question asked us to prove that $\phi(\textbf{r}, t)=\frac{q}{4\pi\epsilon_0}\frac{1}{ct-x}$ for ct>x, I also find it weird that it has only an x and t dependence, the question does not specify whether I should evaluate it at y=z=0. It is an old exam question, so I assumed that it is not necessary. Yes, it would have to be a massless particle. $\endgroup$
    – NX37B
    Sep 23 at 22:09
  • $\begingroup$ We don’t actually know of any massless, charged particles, so it’s a peculiar problem. $\endgroup$
    – Ghoster
    Sep 23 at 22:13