We now investigate more general transformations of electric and magnetic fields between different inertial frames. Our starting point is the electromagnetic field of an infinite flat metal sheet, which is derived in most introductory textbooks on electrodynamics, such as Griffiths [ 6 ], and which we state here without proof.

The electric field of an infinite metal sheet with charge density $\sigma$ points away from the sheet and has the constant magnitude \begin{equation} E = \frac{\sigma}{2\epsilon_0} \end{equation} The magnetic field of such a sheet with current density $\vkap$ has constant magnitude \begin{equation} B = \frac{\mu_0}{2} |\vkap| \end{equation} and direction determined by the right-hand-rule.

Consider a capacitor consisting of 2 horizontal ($y=\hbox{constant}$) parallel plates, with equal and opposite charge densities. For definiteness, take the charge density on the bottom plate to be $\sigma_0$, and suppose that the charges are at rest, that is, that the current density of each plate is zero. Then the electric field is given by \begin{equation} \EE_0 = E_0 \,\yhat = \frac{\sigma_0}{\epsilon_0} \,\yhat \end{equation} between the plates and vanishes elsewhere. Now let the capacitor move to the left with velocity \begin{equation} \uu = -u \,\xhat = -\cc\tanh\alpha \>\,\xhat \label{vdef} \end{equation} Then the width of the plate is unchanged, but, just as for the line charge (5) of §11.1. the length is Lorentz contracted, which decreases the area, and hence increases the charge density. The charge density (on the bottom plate) is therefore \begin{equation} \sigma = \sigma_0\cosh\alpha \label{sigma} \end{equation} But there is now also a current density, which is given by \begin{equation} \vkap = \sigma \,\uu \end{equation} on the lower plate. The top plate has charge density $-\sigma$, so its current density is $-\vkap$. Then both the electric and magnetic fields vanish outside the plates, whereas inside the plates one has \begin{eqnarray} \EE &=& E^y \,\yhat = \frac{\sigma}{\epsilon_0} \,\yhat \\ \BB &=& B^z \,\zhat = -\mu_0 \,\sigma u \,\zhat \end{eqnarray} which can be rewritten using ($\ref{vdef}$) and ($\ref{sigma}$) in the form \begin{eqnarray} E^y &=& E_0 \cosh\alpha \label{EMmaster}\\ \end{eqnarray}\begin{eqnarray} B^z &=& B_0 \sinh\alpha \label{EMmaster2} \end{eqnarray} For later convenience, we have introduced in the last equation the quantity \begin{equation} B_0 = -c \mu_0 \sigma_0 = -c \mu_0 \epsilon_0 E_0 = - \cm\,{E_0} \label{Bdef} \end{equation} which does not correspond to the magnetic field when the plate is at rest — which of course vanishes since $\uu=0$.

The above discussion gives the electric and magnetic fields seen by an observer at rest. What is seen by an observer moving to the right with speed $v=\cc\tanh\beta$? To compute this, first use the velocity addition law to compute the correct rapidity to insert in ($\ref{EMmaster}$) and ($\ref{EMmaster2}$), which is simply the sum of the rapidities $\alpha$ and $\beta$!

The moving observer therefore sees an electric field $\EE{}'$ and a magnetic field $\BB{}'$. From ($\ref{EMmaster}$)–($\ref{Bdef}$), and the hyperbolic trig formulas (5) and (6) of §4.1. we have \begin{eqnarray} E'^y &=& E_0 \cosh(\alpha+\beta) \nonumber\\ &=& E_0 \cosh\alpha\cosh\beta + E_0 \sinh\alpha\sinh\beta \nonumber\\ &=& E_0 \cosh\alpha\cosh\beta - \cc B_0 \sinh\alpha\sinh\beta \nonumber\\ &=& E^y \cosh\beta - \cc B^z \sinh\beta \label{EL} \end{eqnarray} and similarly \begin{eqnarray} B'^z &=& B_0 \sinh(\alpha+\beta) \nonumber\\ &=& B_0 \sinh\alpha\cosh\beta + B_0 \cosh\alpha\sinh\beta \nonumber\\ &=& B^z \cosh\beta - \cm\,{E^y} \sinh\beta \end{eqnarray} The last step in each case is crucial; we eliminate $\alpha$, $E_0$, and $B_0$, since the “lab” frame is an artifact of our construction. The resulting transformations depend, as desired, only on the two “moving” frames, and their relative rapidity, $\beta$.

Repeating the argument with the $y$ and $z$ axes interchanged (and being careful about the orientation), we obtain the analogous formulas \begin{eqnarray} E'^z &=& E^z \cosh\beta + \cc B^y \sinh\beta \\ B'^y &=& B^y \cosh\beta + \cm\,{E^z} \sinh\beta \end{eqnarray} Finally, by considering motion perpendicular to the plates one can show [ 6 ] \begin{equation} E'^x = E^x \end{equation} and by considering a solenoid one obtains [ 6 ] \begin{equation} B'^x = B^x \label{BL} \end{equation}

Equations ($\ref{EL}$)–($\ref{BL}$) describe the behavior of the electric and magnetic fields under Lorentz transformations. These equations can be nicely rewritten in vector language by introducing the projections parallel and perpendicular to the direction of motion of the observer, namely \begin{eqnarray} \EE_\parallel &=& \frac{\vv\cdot\EE}{\vv\cdot\vv} \,\vv \\ \BB_\parallel &=& \frac{\vv\cdot\BB}{\vv\cdot\vv} \,\vv \end{eqnarray} and \begin{eqnarray} \EE_\perp &=& \EE-\EE_\parallel \\ \BB_\perp &=& \BB-\BB_\parallel \end{eqnarray} We then have \begin{eqnarray} \EE{}'_\parallel &=& \EE_\parallel \\ \BB{}'_\parallel &=& \BB_\parallel \end{eqnarray} and \begin{eqnarray} \BB{}'_\perp &=& \left( \BB_\perp - \cmsq\,\vv \times \EE_\perp \right) \cosh\beta \label{Bperp}\\ \end{eqnarray}\begin{eqnarray} \EE{}'_\perp &=& \left( \EE_\perp + \vv \times \BB_\perp \right) \cosh\beta \label{Eperp} \end{eqnarray}