#### Magnetism as a Relativistic Effect
red.rem.mre

Electromagnetism is by construction fully compatible with special relativity. Going further, special relativity means that the existence of electricity implies the existence of magnetism, and vice-versa.

To illustrate this, we take a simple example

See Purcell \& Morin, section 5.9.

Consider a wire at rest in the lab frame. This wire carries a current made of a (positive) line charge density \(\lambda\) moving towards the right at velocity \(v\), and a (negative) line charge density \(-\lambda\) moving towards the left at velocity \(-v\). In the lab frame, the current is thus \[ I_{\mbox{lab}} = 2 \lambda v \] Additionally, in the lab frame, there is a charge \(q\) situated at a distance \(s\) from the cable and moving with a velocity \(u < v\) parallel to the wire. In the lab frame, there is no electrical force between the wire and the charge, since the wire carries no net charge.

Let us now examine this situation in the rest frame of the moving particle. In this frame, the velocities of the line charges in the wire are given by Einstein's velocity addition rule: \[ v_\pm = \frac{v \mp u}{1 \mp uv/c^2} \] Since \(v_- > v_+\), Lorentz contraction is stronger for the negative line charges than for the positive ones. In this frame, the wire thus carries a nonzero net charge. The densities are \[ \lambda_\pm = \pm \gamma_\pm ~\lambda_0 \] where \(\lambda_0\) is the line charge density in the rest frame of the line charges, and \[ \gamma_\pm = \frac{1}{\sqrt{1 - v_\pm^2/c^2}}. \] The relationship between \(\lambda\) and \(\lambda_0\) is \[ \lambda = \gamma \lambda_0, \hspace{10mm} \gamma = \frac{1}{\sqrt{1 - v^2/c^2}}. \] The dilation factors in the test particle's frame are thus

\begin{align} \gamma_\pm &= \frac{1}{\sqrt{1 - (v \mp u)^2/(c^2 \mp uv)^2}} = \frac{c^2 \mp uv}{\sqrt{(c^2 \mp uv)^2 - c^2 (v \mp u)^2}} \nonumber \\ &= \frac{c^2 \mp uv}{\sqrt{(c^2 - v^2)(c^2 - u^2)}} = \gamma \frac{1 \mp uv/c^2}{\sqrt{1 - u^2/c^2}} \end{align}

so the resultant line charge in the test frame is \[ \lambda_{\mbox{test}} = \lambda_+ + \lambda_- = \lambda_0 ( \gamma_+ - \gamma_-) = -\frac{2\lambda u v}{c^2 \sqrt{1 - u^2/c^2}}. \] Therefore, as a result of Lorentz contraction, a current-carrying wire which is neutral in one reference frame, can appear to be charged in another.

In the frame of the test particle, there is an electric field equal to that of a uniformly charged wire: \[ E = \frac{\lambda_{\mbox{test}}}{2\pi \epsilon_0 r} \] so the force (in the test frame) is \[ F_{\mbox{test}} = q E = -\frac{\lambda v}{\pi \epsilon_0 c^2 r} \frac{q u}{\sqrt{1 - u^2/c^2}}. \] If there is a force on our test charge in this test frame, there must also be one in the lab frame. Using equation Ftr0, \[ F = \sqrt{1 - u^2/c^2} ~F_{\mbox{test}} = -\frac{\lambda v}{\pi \varepsilon_0 c^2} \frac{qu}{r} \] which upon recognizing \(c^2 = \frac{1}{\varepsilon_0 \mu_0}\) and the current \(I = 2 \lambda v\) becomes \[ F = - q u ~\frac{\mu_0 I}{2\pi r} \] which you will recognize as the magnetic part of the Lorentz force for a charge \(q\) moving at velocity \(u\) in the presence of the magnetic field of a long straight wire carrying current \(I\).

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Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42