Pre-Quantum Electrodynamics

Now that we know that what one observer sees as an electric field can be viewed by another as a magnetic field, we can ask the general question of how fields transform upon Lorentz transformations.

Let's start with what is perhaps the simplest case: the electric field between the plates of an infinite parallel-plate capacitor. For a surface charge density \(\sigma_0\) on bottom and \(-\sigma_0\) on top plates (putting the plates perpendicular to \(\hat{\boldsymbol y}\)), this is \[ {\boldsymbol E}_0 = \frac{\sigma_0}{\varepsilon_0} \hat{\boldsymbol y}. \]

Let us now assume that we move to a reference frame moving at velocity \(v_0\) in direction \(\hat{\boldsymbol x}\). In this frame, the field between the plates will still be along \({\boldsymbol y}\). In terms of the surface charge density per plate \(\sigma\) as measured in this frame, \[ {\boldsymbol E} = \frac{\sigma}{\varepsilon_0} \hat{\boldsymbol y}. \] Lorentz contraction affects the length scale longitudinal to the motion (it does not affect the perpendicular length scales) so the surface charge density in the moving frame becomes \[ \sigma = \gamma_0 \sigma_0, \hspace{10mm} \gamma_0 = \frac{1}{\sqrt{1 - v_0^2/c^2}}. \] In the two frames, the electric fields perpendicular to the direction of motion are thus related by \[ {\boldsymbol E}^\perp = \gamma_0 {\boldsymbol E}_0^\perp. \] For the field parallel to the motion, we can simply repeat the argument but now with motion along \({\boldsymbol y}\). Since Lorentz contraction does not affect the surface charge density, we get \[ E^\parallel = E_0^\parallel. \]

Going back to our setup with plates in the \(xz\) plane which we started from, in the moving frame, there is now a magnetic field due to surface currents: \[ {\boldsymbol K}_{\mbox{top}} = \sigma v_0 \hat{\boldsymbol x} = -{\boldsymbol K}_{\mbox{bot}}. \] This magnetic field between the plates is thus \[ {\boldsymbol B} = -\mu_0 \sigma v_0 ~\hat{\boldsymbol z}. \]

If we now have a further referential frame \(\bar{S}\) moving at velocity \(v\) with respect to \({\cal S}\) (and \(\bar{v}\) with respect to the original one), we'd have \[ \bar{E}_y = \frac{\bar{\sigma}}{\varepsilon_0}, \hspace{10mm} \bar{B}_z = - \mu_0 \bar{\sigma} \bar{v} \] where \[ \bar{v} = \frac{v + v_0}{1 + v v_0/c^2}, \hspace{10mm} \bar{\sigma} = \bar{\gamma} \sigma_0, \hspace{10mm} \bar{\gamma} = \frac{1}{\sqrt{1 - \bar{v}^2/c^2}}. \]

We now want to express \(\bar{\boldsymbol E}, \bar{\boldsymbol B}\) in terms of \({\boldsymbol E}, {\boldsymbol B}\) and other data in frame \({\cal S}\). To start, we have \[ \bar{E}_y = \frac{\bar{\gamma}}{\gamma_0} \frac{\sigma}{\varepsilon_0}, \hspace{10mm} \bar{B}_z = - \frac{\bar{\gamma}}{\gamma_0} \mu_0 \sigma \bar{v}. \] The ratio of contraction factors is \[ \frac{\bar{\gamma}}{\gamma_0} = \frac{\sqrt{1 - v_0^2/c^2}}{\sqrt{1 - \bar{v}^2/c^2}} = \frac{\sqrt{c^2 - v_0^2} ~(1 + v v_0/c^2)}{\sqrt{c^2 (1 + vv_0/c^2)^2 - (v + v_0)^2}} = \frac{1 + v v_0/c^2}{\sqrt{1 - v^2/c^2}} = \gamma (1 + vv_0/c^2). \] We can thus write \[ \bar{E}_y = \gamma (1 + v v_0/c^2) \frac{\sigma}{\varepsilon_0} = \gamma \left( E_y - \frac{v}{c^2 \varepsilon_0 \mu_0} B_z \right) = \gamma \left( E_y - v B_z \right) \] and \[ \bar{B}_z = -\gamma (1 + vv_0/c^2) \mu_0 \sigma \frac{v + v_0}{1 + vv_0/c^2} = \gamma (B_z - \varepsilon_0 \mu_0 v E_y) = \gamma (B_z - \frac{v}{c^2} E_y). \]

To do \(E_z\) and \(B_y\), simply put the capacitor in the \(xy\) plane. Following the same argument, this gives \[ \bar{E}_z = \gamma (E_z + v B_y), \hspace{10mm} \bar{B}_y = \gamma \left( B_y + \frac{v}{c^2} E_z \right). \] We already know that \(\bar{E}_x = E_x\). For \(B_x\), we consider a solenoid with axis along \(x\). The windings get tighter, \(\bar{n} = \gamma n\) but the clock goes slower so \(\bar{I} = \frac{1}{\gamma} I\). These factors cancel so \(\bar{B}_x = B_x\).

We thus obtain the

Two special cases can be mentioned:

If \({\boldsymbol B} = 0\) in \({\cal S}\): Then, \(\bar{\boldsymbol B} = \gamma \frac{v}{c^2} (E_z \hat{\boldsymbol y} - E_y \hat{\boldsymbol z}) = \frac{v}{c^2} (\bar{E}_z \hat{\boldsymbol y} - \bar{E}_y \hat{\boldsymbol z})\) so \[ \bar{\boldsymbol B} = -\frac{1}{c^2} {\boldsymbol v} \times \bar{\boldsymbol E}. \]

If \({\boldsymbol E} = 0\) in \({\cal S}\): Then, \(\hat{\boldsymbol E} = -\gamma v (B_z \hat{\boldsymbol y} - B_y \hat{\boldsymbol z}) = -v (\bar{B}_z \hat{\boldsymbol y} - \bar{B}_y \hat{\boldsymbol z})\) so \[ \bar{\boldsymbol E} = {\boldsymbol v} \times \bar{\boldsymbol B}. \]