The transformation of currents and charges
You may have worried about the simplification we made above when we took the same velocity $v$ for the particle and for the conduction electrons in the wire. We could go back and carry through the analysis again for two different velocities, but it is easier to simply notice that charge and current density are the components of a four-vector (see Chapter 17, Vol. I). We have seen that if $\rho_0$ is the density of the charges in their rest frame, then in a frame in which they have the velocity $\FLPv$, the density is \begin{equation*} \rho=\frac{\rho_0}{\sqrt{1-v^2/c^2}}. \end{equation*}  In that frame their current density is \begin{equation} \label{Eq:II:13:34} \FLPj=\rho\FLPv=\frac{\rho_0\FLPv}{\sqrt{1-v^2/c^2}}. \end{equation}  Now we know that the energy $U$ and momentum $\FLPp$ of a particle moving with velocity $\FLPv$ are given by \begin{equation*} U=\frac{m_0c^2}{\sqrt{1-v^2/c^2}},\quad \FLPp=\frac{m_0\FLPv}{\sqrt{1-v^2/c^2}}, \end{equation*}  where $m_0$ is its rest mass. We also know that $U/c = m_0c/\sqrt{1-v^2/c^2}$ and $\FLPp$ form a relativistic four-vector. Since $c\rho = c\rho_0/\sqrt{1-v^2/c^2}$ and $\FLPj$ depend on the velocity $\FLPv$ exactly as do $U/c$ and $\FLPp$, we can conclude that $c\rho$ and $\FLPj$ are also the components of a relativistic four-vector. This property is the key to a general analysis of the field of a wire moving with any velocity, which we would need if we want to do the problem again with the velocity $\FLPv_0$ of the particle different from the velocity of the conduction electrons. If we wish to transform $\rho$ and $\FLPj$ to a coordinate system moving with a velocity $u$ in the $x$-direction, we know that they transform just like $t$ and $(x,y,z)$, so that we have (see Chapter 15, Vol. I) \begin{alignat}{2} x'&=\frac{x-ut}{\sqrt{1-u^2/c^2}}, \quad&j_x'&=\frac{j_x-u\rho}{\sqrt{1-u^2/c^2}}\notag\\ y'&=y, &j_y'&=j_y,\notag\\[1ex] z'&=z, &j_z'&=j_z,\notag\\ \label{Eq:II:13:35} t'&=\frac{t-ux/c^2}{\sqrt{1-u^2/c^2}}, \quad&\rho'&=\frac{\rho-uj_x/c^2}{\sqrt{1-u^2/c^2}}. \end{alignat}  With these equations we can relate charges and currents in one frame to those in another. Taking the charges and currents in either frame, we can solve the electromagnetic problem in that frame by using our Maxwell equations. The result we obtain for the motions of particles will be the same no matter which frame we choose. We will return at a later time to the relativistic transformations of the electromagnetic fields.