The four-potential of a moving charge
Although it is implicit in what we have already said, let us write down the transformation laws which give $\phi$ and $\FLPA$ in a moving system in terms of $\phi$ and $\FLPA$ in a stationary system. Since $A_\mu=(\phi,\FLPA)$ is a four-vector, the equations must look just like Eqs. (25.1), except that $t$ is replaced by $\phi$, and $\FLPx$ is replaced by $\FLPA$. Thus, \begin{equation} \begin{alignedat}{2} \phi'&=\frac{\phi-vA_x}{\sqrt{1-v^2}},&\quad A_y'&=A_y,\\[1ex] A_x'&=\frac{A_x-v\phi}{\sqrt{1-v^2}},&\quad A_z'&=A_z. \end{alignedat} \label{Eq:II:25:24} \end{equation} This assumes that the primed coordinate system is moving with speed $v$ in the positive $x$-direction, as measured in the unprimed coordinate system.  We will consider one example of the usefulness of the idea of the four-potential. What are the vector and scalar potentials of a charge $q$ moving with speed $v$ along the $x$-axis? The problem is easy in a coordinate system moving with the charge, since in this system the charge is standing still. Let’s say that the charge is at the origin of the $S'$-frame, as shown in Fig. 25–2. The scalar potential in the moving system is then given by \begin{equation} \label{Eq:II:25:25} \phi'=\frac{q}{4\pi\epsO r'}, \end{equation} $r'$ being the distance from $q$ to the field point, as measured in the moving system. The vector potential $\FLPA'$ is, of course, zero.  Now it is straightforward to find $\phi$ and $\FLPA$, the potentials as measured in the stationary coordinates. The inverse relations to Eqs. (25.24) are \begin{equation} \label{Eq:II:25:26} \begin{alignedat}{2} \phi&=\frac{\phi'+vA_x'}{\sqrt{1-v^2}},&\quad A_y&=A_y',\\[1.5ex] A_x&=\frac{A_x'+v\phi'}{\sqrt{1-v^2}},&\quad A_z&=A_z'. \end{alignedat} \end{equation} Using the $\phi'$ given by Eq. (25.25), and $\FLPA'=\FLPzero$, we get \begin{align*} \phi&=\frac{q}{4\pi\epsO}\,\frac{1}{r'\sqrt{1-v^2}}\\[.5ex] &=\frac{q}{4\pi\epsO}\, \frac{1}{\sqrt{1-v^2}\sqrt{x'^2+y'^2+z'^2}}. \end{align*} This gives us the scalar potential $\phi$ we would see in $S$, but, unfortunately, expressed in terms of the $S'$ coordinates. We can get things in terms of $t$, $x$, $y$, $z$ by substituting for $t'$, $x'$, $y'$, and $z'$, using (25.1). We get  \begin{equation} \label{Eq:II:25:27} \phi=\frac{q}{4\pi\epsO}\, \frac{1}{\sqrt{1-v^2}}\, \frac{1}{\sqrt{[(x-vt)/\sqrt{1-v^2}]^2+y^2+z^2}}. \end{equation} 
 \begin{align} \label{Eq:II:25:27} \phi=\frac{q}{4\pi\epsO}\, &\frac{1}{\sqrt{1-v^2}}\;\times\\[-1ex] &\frac{1}{\sqrt{[(x-vt)/\sqrt{1-v^2}]^2+y^2+z^2}}.\notag \end{align}  Following the same procedure for the components of $\FLPA$, you can show that \begin{equation} \label{Eq:II:25:28} \FLPA=\FLPv\phi. \end{equation} These are the same formulas we derived by a different method in Chapter 21. 