In a frame $\mathcal{F}$ the electromagnetic fields $(\mathbf{E}, \mathbf{B})$ are encoded into the Maxwell field 4-tensor $F^{a b}$ and its dual ${ }^{*} F^{a b}$, where

$$F^{a b}=\left(\begin{array}{cccc} 0 & -E_{x} & -E_{y} & -E_{z} \\ E_{x} & 0 & -B_{z} & B_{y} \\ E_{y} & B_{z} & 0 & -B_{x} \\ E_{z} & -B_{y} & B_{x} & 0 \end{array}\right)$$

and

$${ }^{*} F^{a b}=\left(\begin{array}{cccc} 0 & -B_{x} & -B_{y} & -B_{z} \\ B_{x} & 0 & E_{z} & -E_{y} \\ B_{y} & -E_{z} & 0 & E_{x} \\ B_{z} & E_{y} & -E_{x} & 0 \end{array}\right)$$

[Here the signature is $(+---)$ and units are chosen so that $c=1$.] Obtain two independent Lorentz scalars of the electromagnetic field in terms of $\mathbf{E}$ and $\mathbf{B}$.

Suppose that $\mathbf{E} \cdot \mathbf{B}>0$ in the frame $\mathcal{F}$. Given that there exists a frame $\mathcal{F}^{\prime}$ in which $\mathbf{E}^{\prime} \times \mathbf{B}^{\prime}=\mathbf{0}$, show that

$$E^{\prime}=\left[(\mathbf{E} \cdot \mathbf{B})\left(K+\sqrt{1+K^{2}}\right)\right]^{1 / 2}, \quad B^{\prime}=\left[\frac{\mathbf{E} \cdot \mathbf{B}}{K+\sqrt{1+K^{2}}}\right]^{1 / 2},$$

where $\left(E^{\prime}, B^{\prime}\right)$ are the magnitudes of $\left(\mathbf{E}^{\prime}, \mathbf{B}^{\prime}\right)$, and

$$K=\frac{1}{2}\left(|\mathbf{E}|^{2}-|\mathbf{B}|^{2}\right) /(\mathbf{E} \cdot \mathbf{B})$$

[Hint: there is no need to consider the Lorentz transformations for $\mathbf{E}^{\prime}$ and $\mathbf{B}^{\prime}$.]