Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 2 / 18D (Electromagnetism)

Use SI units and the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$. Write $x^{\mu }=(ct,x,y,z)$, ${\partial }_{\mu }=(c^{-1}{\partial }_t,\nabla )$ and the four-current $J^{\mu }=(c\rho ,\mathbf{J})$, with Greek indices from $0$ to $3$. The electromagnetic field tensor is the antisymmetric tensor whose components areIndices are raised and lowered with $\eta$. Define the dual electromagnetic field tensor by ${\tilde{F}}^{\mu \nu }=\frac{1}{2}{ϵ}^{\mu \nu \rho \sigma }{F}_{\rho \sigma }$, with ${ϵ}^{0123}=+1$. Then the covariant Maxwell equations areHere $\rho$ and $\mathbf{J}$ are charge and current density, ${\mu }_0$ is the vacuum magnetic permeability, and ${ϵ}_0{\mu }_0{c}^2=1$. For example the $\nu =0$ equation gives $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$, and the spatial equations give $\nabla \times \mathbf{B}-c^{-2}{\partial }_t\mathbf{E}={\mu }_0\mathbf{J}$. With the stated dual convention, ${\tilde{F}}^{0i}=-B_i$ and ${\tilde{F}}^{ij}={ϵ}_{ijk}{E}_k/c$; the second tensor equation gives the two homogeneous Maxwell equations.

Contractions of Lorentz tensors have no free indices and are Lorentz scalars. The antisymmetry and the components in part (a) giveThus $\mathbf{E}\cdot \mathbf{B}$ and $|\mathbf{B}{|}^2-|\mathbf{E}{|}^2/c^2$ are the two electromagnetic field invariants under proper Lorentz transformations. The second contraction uses the orientation-dependent epsilon tensor: strictly, $\mathbf{E}\cdot \mathbf{B}$ is a pseudoscalar, changing sign under an improper transformation such as spatial parity. Under the boosts in this question it is invariant.

Transform the electromagnetic field tensor by $F^{\prime \mu \nu }={\Lambda }^{\mu }{}_{\alpha }{\Lambda }^{\nu }{}_{\beta }{F}^{\alpha \beta }$. For example,Reading off the other components in the same way yields the transverse field transformationThe longitudinal components remain $E_x^{\prime }=B_x^{\prime }=0$. These signs correspond to the stated boost in which $S^{\prime }$ moves with velocity $v$ relative to $S$.

For $E_0\ne 0$, put $\beta =v/c$. Part (c) givesThe vanishing of their cross product is equivalent toFor $0<\theta <\pi /2$ the two roots are $\tan (\theta /2)$ and $\cot (\theta /2)$. Only the first has $|\beta |<1$. The continuous endpoint $\theta =0$ gives $\beta =0$. Thus the physical boost isSubstitution gives ${\mathbf{E}}^{\prime }=c{\mathbf{B}}^{\prime }$, so the fields point in the same direction. At $\theta =\pi /2$ the formal root is $v=c$, which is not an inertial observer's velocity. The nonzero perpendicular equal-strength fields then have both electromagnetic field invariants zero. If they were parallel in a finite frame, those invariants would force both fields to vanish, impossible under an invertible Lorentz transformation. Hence no finite inertial boost makes them parallel in this case. If $E_0=0$, both fields are zero in every frame and no velocity is singled out.