Special Relativity to Magnetism -

One of the little known but ubiquitous aspects of special relativity is that it leads to the phenomena of magnetism. This page will develop this idea mathematically. ¹

Forces between two infinitely long charged threads

To begin this discussion we will calculate the forces between two infinitely long uniformly charged non conducting threads in a vacuum moving to the right in the laboratory inertial reference frame.

We will do this in two different ways the first using the Lorentz force equation from classical Electromagnetics and the second using the transformation of force between reference frames from Special Relativity.

Figure 1: Infinitely long charged wires in lab frame

Lorentz Force Method:

Electric Force:

$\\\nabla\otimes\textbf{E}=-\dfrac{\partial\textbf{B}}{\partial\textit{t}}$

$\\\nabla\odot\textbf{D}=\rho$

For our case $-\partial\textbf{B}/\partial\textit{t}=0$ therefore:

$\nabla\otimes\textbf{E}=0$

In a vacuum $\textbf{D}=\varepsilon_0\textbf{E}$ thus:

$\\\nabla\odot\textbf{E}=\rho/\varepsilon_0$

In addition to the above equations we will need Gauss’ law.

$\oint\textbf{E}\odot\textit{d}\textbf{A}=\dfrac{Q}{\varepsilon_0}$

By symmetry we can argue that the electric field due to each wire is radial and perpendicular to the charged wire.

For the gaussian surface shown in Figure 1 the end caps of the cylinder do not contribute to the dot product because the normal to those surfaces is perpendicular to the electric field.

In this case Gauss’ law reduces to:

$2\pi\textit{r}\textit{h}E_{r}=\lambda\textit{h}/\varepsilon_0$

And the static electrical force per meter on wire 2 due to the field from wire 1 is:

$f_{elec}=\dfrac{{\lambda}^{2}}{2\pi\varepsilon_0\textit{r}}newtons/meter$

Magnetic Force:

$\nabla\otimes\textbf{H}=\textbf{J}+\dfrac{\partial\textbf{D}}{\partial\textit{t}}$

$\nabla\odot\textbf{B}=0$

For wires moving at constant velocity in a vacuum: $\partial\textbf{D}/\partial\textit{t}=0$ and $\textbf{B}=\mu_{0}\textbf{H}$ Thus:

$\nabla\otimes\textbf{B}=\mu_{0}\textbf{J}$

Now integrate both sides of this equation over the surface S and use Stokes theorem to derive:

$\int(\nabla\otimes\textbf{B})\odot\textbf{n}dS={\oint\textbf{B}\odot}d\textbf{l}=\mu_{0}\int\textbf{J}\odot\textbf{n}dS=\mu_{0}\lambda\textit{v}$

It’s instructive to note here the units of current density J are $Am^{-2}$ . The surface integral give us Amps (coulombs per second). The units of $\lambda\textit{v}$ are coulombs per second.

With B radially symmetric the line integral is as follows:

${\oint\textbf{B}\odot}d\textbf{l}=2\pi\textit{r}\textit{B}$

Thus: $B_{z}=\dfrac{\mu_{0}\lambda\textit{v}}{2\pi\textit{r}}$

The magnetic force per meter is given by:

$f_{mag}={q\textbf{u}\otimes}\textbf{B}={\int_{0}^{1m}\lambda}dlv*\dfrac{\mu_{0}\lambda\textit{v}}{2\pi\textit{r}}=-\dfrac{\mu_{0}{\lambda}^{2}v^{2}}{2\pi\textit{r}}$

The total force per meter of wire 2 due to the presence of wire 1 is:

$f_{y}=f_{elec}+f_{mag}=\dfrac{{\lambda}^{2}}{2\pi\varepsilon_0\textit{r}}-\dfrac{\mu_{0}{\lambda}^{2}v^{2}}{2\pi\textit{r}}=\dfrac{{\lambda}^{2}}{2\pi\varepsilon_0\textit{r}}(1-\mu_{0}\varepsilon_0{v^{2}})=\dfrac{{\lambda}^{2}}{2\pi\varepsilon_0\textit{r}}(1-{v^{2}}/{c^{2}})$

Relativity Method:

Figure 2: Charged wires showing Lorentz contraction

In the inertial reference frame $\Sigma^{'}$ the two charged threads are at rest and the force between them is only electrostatic. In this reference frame since charge is relativistically invariant $\lambda^{'}=n^{'}q$ where $n^{'}$ is the number of charges per meter length.

The number charges in one meter of thread in the $\Sigma^{'}$ frame is thus $n^{'}l_{0}$ . In the $\Sigma$ frame $l_{0}$ is measured as $l_{0}\sqrt{1-v^{2}/c^{2}}$ therefore:

${n}l_{0}\sqrt{1-v^{2}/c^{2}}=n^{'}l_{0} \Rightarrow n=\dfrac{n^{'}}{\sqrt{1-v^{2}/c^{2}}}$

$\lambda=nq=\dfrac{n^{'}}{\sqrt{1-v^{2}/c^{2}}}=\dfrac{\lambda^{'}}{\sqrt{1-v^{2}/c^{2}}}$

The electric field between charged threads was determined in the section above. In this case all we need to do is reference $\lambda^{'}$ so that the electric in the $\Sigma^{'}$ frame is:

$E_{y}^{'}=\dfrac{\lambda^{'}}{2\pi\varepsilon_0\textit{r}}$

Since r is perpendicular to the motion of the reference frames $r^{'}=r$ and the force on one of the charges is:

$f_{y}^{'}=\dfrac{\lambda^{'}q}{2\pi\varepsilon_0\textit{r}}$

The Lorentz force transformation for our scenario is:

$f_{y}=\sqrt{1-v^{2}/c^{2}}f_{y}^{'}$

Therefore the force per charge is:

$f_{y}^{0}=\sqrt{1-v^{2}/c^{2}}\dfrac{\lambda^{'}q}{2\pi\varepsilon_0\textit{r}}$

The force per meter of thread 2 measured in the $\Sigma$ frame is the number of charges per meter measured in $\Sigma$ times the force on each charge $f_{y}$ :

$force/meter=nf_{y}^{0}=n\sqrt{1-v^{2}/c^{2}}\dfrac{\lambda^{'}q}{2\pi\varepsilon_0\textit{r}}$

Since $\lambda^{'}=\lambda\sqrt{1-v^{2}/c^{2}}$ and $nq=\lambda$ :

$f_{y}=\dfrac{{\lambda}^{2}}{2\pi\varepsilon_0\textit{r}}(1-{v^{2}}/{c^{2}})$

This is more than a remarkable coincidence, it’s built into the fabric of spacetime and Maxwells equations can be derived from Special Relativity as will be discussed in further additions to this page.

This development is entirely based upon the text: Classical Electromagnetism via Relativity by WGV Rosser, New York Plenum Press 1968. Images are photo’s from this text as well