Formula for all the Lorentz boosts that result in $E$ parallel to $B$? This is a follow-up question to a previous question regarding a minimum-energy invariant of the electromagnetic field.  @ChiralAnomaly showed that there is indeed an invariant minimum energy density whenever $E\cdot B \not= 0$; and that there is a one-parameter family of inertial frames in which $E$ and $B$ are parallel.  In those frames the energy density is the minimum energy density, and is the same in all of them.
The minimum energy density is easy enough to calculate.  Is there a straightforward way to derive a formula that, given $E$ and $B$ in one frame, yields all the Lorentz boosts that result in $E$ parallel to $B$?
 A: We need a Lorentz transformation which will make $\vec{E}$ and $\vec{B}$ from inertial frame $S$ parallel in an inertial frame $S'$, i.e. $\vec{E'} \times \vec{B'} = 0$. Consider $c=1$ for calculations.
Let us try $\vec{v'} = \alpha (\vec{E} \times \vec{B})$ for some $\alpha \in \mathbb{R}/\{0\}$. Here, $v'$ denotes the velocity of the frame $S'$ with respect to $S$.
Then, $$\vec{E'} = \gamma(\vec{E}+\vec{v'}\times\vec{B}) = (1 - \alpha B^2) \vec{E}+\alpha (\vec{E} \cdot \vec{B}) \vec{B} \\ 
\vec{B'} = \gamma(\vec{B}-\vec{v'}\times\vec{E}) = (1 - \alpha E^2) \vec{B}+\alpha (\vec{E} \cdot \vec{B}) \vec{E}$$
Case 1 (a): Of course, as you mentioned, if $\vec{E} \cdot \vec{B} = 0$ and $|\vec{E}|=|\vec{B}|=0$, $\vec{E'}$ and $\vec{B'}$ cannot be made parallel.
Case 1 (b): Even if $\vec{E} \cdot \vec{B} = 0$, but $E^2 \ne B^2$, we can make either $\vec{E'}$ or $\vec{B'}$ zero, by choosing $\alpha = \displaystyle\frac{1}{\text{max}(E^2, B^2)}$
Case 2: Suppose, $\vec{E} \cdot \vec{B} \ne 0$, then $\vec{E'}$ and $\vec{B'}$ can be made parallel by choosing an $\alpha$ such that $$\alpha^2 [(\vec{E} \cdot \vec{B})^2 - E^2B^2] = 1 - \alpha(E^2+B^2).$$ The relative velocity $\vec{v'}$ of the frame $S'$ with respect to the inertial frame $S$ is given by 
$$\frac{\vec{v'}}{1+ v'^2} = \frac{\alpha(\vec{E}\times\vec{B})}{1+\alpha^2({\vec{E}\times \vec{B}})^2} = \frac{\alpha (\vec{E}\times \vec{B})}{1+\alpha^2[E^2B^2-(\vec{E} \cdot \vec{B})^2]}= \frac{\vec{E}\times\vec{B}}{E^2+B^2}.$$
