The electric and magnetic fields $\mathbf{E}, \mathbf{B}$ in an inertial frame $\mathcal{S}$ are related to the fields $\mathbf{E}^{\prime}, \mathbf{B}^{\prime}$ in a frame $\mathcal{S}^{\prime}$ by a Lorentz transformation. Given that $\mathcal{S}^{\prime}$ moves in the $x$-direction with speed $v$ relative to $\mathcal{S}$, and that

$$E_{y}^{\prime}=\gamma\left(E_{y}-v B_{z}\right), \quad B_{z}^{\prime}=\gamma\left(B_{z}-\left(v / c^{2}\right) E_{y}\right),$$

write down equations relating the remaining field components and define $\gamma$. Use your answers to show directly that $\mathbf{E}^{\prime} \cdot \mathbf{B}^{\prime}=\mathbf{E} \cdot \mathbf{B}$.

Give an expression for an additional, independent, Lorentz-invariant function of the fields, and check that it is invariant for the special case when $E_{y}=E$ and $B_{y}=B$ are the only non-zero components in the frame $\mathcal{S}$.

Now suppose in addition that $c B=\lambda E$ with $\lambda$ a non-zero constant. Show that the angle $\theta$ between the electric and magnetic fields in $\mathcal{S}^{\prime}$ is given by

$$\cos \theta=f(\beta)=\frac{\lambda\left(1-\beta^{2}\right)}{\left\{\left(1+\lambda^{2} \beta^{2}\right)\left(\lambda^{2}+\beta^{2}\right)\right\}^{1 / 2}}$$

where $\beta=v / c$. By considering the behaviour of $f(\beta)$ as $\beta$ approaches its limiting values, show that the relative velocity of the frames can be chosen so that the angle takes any value in one of the ranges $0 \leqslant \theta<\pi / 2$ or $\pi / 2<\theta \leqslant \pi$, depending on the sign of $\lambda$.