Our starting point is the electric and magnetic fields of an infinite straight wire, which are derived in most introductory textbooks on electrodynamics, such as Griffiths [ 6 ], and which we state here without proof.

The electric field of an infinite straight wire (in vacuum) with charge density $\lambda$ points away from the wire with magnitude \begin{equation} E=\frac{\lambda}{2\pi\epsilon_0 r} \label{Eline} \end{equation} where $r$ is the perpendicular distance from the wire and $\epsilon_0$ is the permittivity constant. The magnetic field of such a wire with current density $I$ has magnitude \begin{equation} B=\frac{\mu_0 I}{2\pi r} \label{Bline} \end{equation} with $r$ as above, and where $\mu_0$ is the permeability constant, which is related to $\epsilon_0$ by \begin{equation} \epsilon_0 \mu_0 = \cmsq \label{MKS} \end{equation} (The direction of the magnetic field is obtained as the cross product of the direction of the current and the position vector from the wire to the point in question.)

We will also need the Lorentz force law, which says that the force $\FF$ on a test particle of charge $q$ and velocity $\vv$ is given by \begin{equation} \FF = q (\EE + \vv\times\BB) \end{equation} where $\EE$ and $\BB$ denote the electric and magnetic fields (with magnitudes $E$ and $B$, respectively).

Consider an infinite line charge, consisting of identical particles of charge $\rho$, separated by a distance $\ell$. This gives an infinite wire with (average) charge density \begin{equation} \lambda_0 = \frac{\rho}{\ell} \end{equation} Suppose now that the charges are moving to the right with speed \begin{equation} u = \cc\,\tanh\alpha \end{equation} Due to length contraction, the charge density seen by an observer at rest increases to \begin{equation} \lambda = {{\rho}\above1pt{\hbox{$\frac{\ell}{\cosh\alpha}$}}} = \lambda_0 \cosh\alpha \label{line} \end{equation}

Suppose now that there are positively charged particles moving to the right, and equally but negatively charged particles moving to the left, each with speed $u$. Consider further a test particle of charge $q$ situated a distance $r$ from the wire and moving with speed \begin{equation} v = \cc\,\tanh\beta \end{equation} to the right. Then the net charge density in the laboratory frame is 0, so that there is no electrical force on the test particle in this frame. There is of course a net current density, however, namely \begin{equation} I = \lambda u + (-\lambda)(-u) = 2\lambda u \end{equation}

What does the test particle see? Switch to the rest frame of the test particle; this makes the negative charges appear to move faster, with speed $u_→u$, and the positive charges move slower, with speed $u_+<u$. The relative speeds satisfy \begin{eqnarray} \overc{ u_+} &=& \tanh(\alpha-\beta) \\ \overc{ u_-} &=& \tanh(\alpha+\beta) \end{eqnarray} resulting in current densities \begin{equation} \lambda_\pm = \lambda\cosh(\alpha\mp\beta) = \lambda(\cosh\alpha\cosh\beta\mp\sinh\alpha\sinh\beta) \end{equation} resulting in a total charge density of \begin{eqnarray} \lambda' &=& \lambda_+ - \lambda_- \nonumber\\ &=& -2\lambda_0 \sinh\alpha \sinh\beta \\ &=& -2\lambda \tanh\alpha \sinh\beta \nonumber \end{eqnarray} According to ($\ref{Eline}$), this results in an electric field of magnitude \begin{equation} E' = \frac{\lambda'}{2\pi\epsilon_0 r} \end{equation} which in turn leads to an electric force of magnitude \begin{eqnarray} F' &=& qE' = \frac{\lambda}{\pi\epsilon_0 r} \, q\tanh\alpha\sinh\beta \nonumber\\ &=& \frac{\lambda u}{\pi\epsilon_0\csq r} \, qv \cosh\beta \\ &=& \frac{\mu_0 I}{2\pi r} \, qv \cosh\beta \nonumber \end{eqnarray}

To relate this to the force observed in the laboratory frame, we must consider how force transforms under a Lorentz transformation. We have 1) \begin{equation} \FF' = \frac{d\PP'}{dt'} \end{equation} and of course also \begin{equation} \FF = \frac{d\PP}{dt} \end{equation} But since in this case the force is perpendicular to the direction of motion, we have \begin{equation} d\PP=d\PP' \end{equation} and since $dx'=0$ in the comoving frame we also have \begin{equation} dt = dt' \cosh\beta \end{equation} Thus, in this case, the magnitudes are related by \begin{equation} F = \frac{F'}{\cosh\beta} = \frac{\mu_0 I}{2\pi r} \, qv \end{equation} But this is just the Lorentz force law \begin{equation} \FF = q \,\vv \times \BB \end{equation} with $B = |\BB|$ given by ($\ref{Bline}$)!

We conclude that in the laboratory frame there is a magnetic force on the test particle, which is just the electric force observed in the comoving frame!