The magical world of classical electrodynamics

It is well known that given classical mechanics, where Galilean relativity applies, the laws of electrodynamics are somewhat strange. There's nothing fundamentally wrong with this, as this is the same type of behaviour as waves in a medium. Nothing prevents an object from moving faster than the speed of sound in a medium, and assuming that this is true, nothing should prevent the same to apply to light.

So let's consider what happens to electrodynamics for supraluminal objects.

The Maxwell equations and Galilean transforms

First let's recap some simple things about the Maxwell equations. As we're using non-relativistic physics here, it may be useful to keep time and space split, so we'll keep to the vector form.

\begin{eqnarray} \vec{\nabla} \cdot \vec{E}(t, \vec{x}) &=& \frac{\rho(t, \vec{x})}{\varepsilon_0}\\ \vec{\nabla} \cdot \vec{B}(t, \vec{x}) &=& 0\\ \vec{\nabla} \times \vec{E}(t, \vec{x}) &=& \frac{\partial \vec{B}(t, \vec{x})}{\partial_t}\\ \vec{\nabla} \times \vec{B}(t, \vec{x}) &=& \mu_0 (\vec{J}(t, \vec{x}) + \varepsilon_0 \frac{\partial \vec{B}(t, \vec{x})}{\partial_t}) \end{eqnarray}

As a reminder of future events, we have that $c = (\varepsilon_0 \mu_0)^{-1/2}$. In the non-relativistic limit, we have that $\mu_0 = 0$, which should indicate that it is the last line that will have poor behaviour under Galilean transforms. Indeed, let's consider the Galilean transform

$$\vec{x}' = \vec{x} + \vec{v} t, \vec{x} = \vec{x}' - \vec{v} t$$

Not much changes with respect to spatial derivatives :

\begin{eqnarray} \frac{\partial}{\partial x'} f(t, x') &=& \frac{\partial x}{\partial x'} \frac{\partial}{\partial x} f(t, x + vt)\\ &=& \frac{\partial}{\partial x} f(t, x + vt)\\ \end{eqnarray}
Last updated : 2019-05-21 14:23:50
Tags : physics , electrodynamics , classical-mechanics