# The curious case of parallel $E$ and $B$ fields and inertial frames

In a comment to this Physics SE question, @MichaelSeifert stated,

For the more general case, IIRC there's always a frame in which $$\vec{E}$$ and $$\vec{B}$$ are parallel when $$\vec{E}\cdot \vec{B}\neq0$$. (There's an exercise in Griffiths about this.) The frame isn't unique, though, since boosting any such frame along the common axis of the fields yields another frame in which the fields point along that same axis.

The second sentence makes sense, since a boost in the common direction of E and B would change the apparent direction of neither. However, I haven't been able to visualize it clearly yet. Here's what I've got thus far:

For an observer Alice sitting in a frame in which E and B are neither perpendicular nor parallel, the statement suggests that there will be a continuum of different inertial frames in which a second observer Bob could be, such that (to Bob) the E and B fields appear to be parallel.

Bob, on the other hand, when in any of those frames, will perceive that there is one unique direction in which he can boost, such that E and B fields remain parallel. However, any such boost will be perceived by Alice as a change of both speed and direction, so there will be a 1-parameter family of values of $$\vec{v} = \vec{v}(\tau)$$ (with $$\tau$$ the parameter) as perceived by Alice, such that if Bob boosts to any of those velocities, he will experience only parallel E and B fields. Also, we know that the family of values does not include a zero relative speed because in that case Bob would see the same fields Alice sees.

I imagine that from Bob's perspective it would be possible to boost to any arbitrary speed, positive or negative, as long as it's along the common direction as the parallel E and B he experiences. However, he would never be able to see either E or B reduce to zero, because that would mean parallel E&B fields could be transformed to perpendicular fields, which we know is not possible because $${E}\cdot {B}$$ is invariant. So Bob can keep trying different boosts in that direction which is special to him; and no matter how fast he goes, $$E^2 - B^2$$ does not change but neither E nor B can reach zero.

I'll boil this down to one question: How does $$E^2 - B^2$$ change for Bob as $$\tau$$ varies?

• It doesn't. $E^2-B^2$ is a relativistic invariant just like $E\cdot B$ is. Or what is the core question here? – Emilio Pisanty Feb 11 at 9:11

If $$\vec{E}$$ and $$\vec{B}$$ are parallel in some frame, and you boost in that direction, then the fields have the same values in the new frame. This means that the entire one-parameter family of observers see the same values of $$\vec{E}$$ and $$\vec{B}$$.

This is because of two things: first, the Lorentz transformations do not affect the components of $$E_\parallel$$ and $$B_\parallel$$ in the direction of the boost. Second, the Lorentz transformation is linear on the perpendicular components $$\vec{E}_\perp$$ and $$\vec{B}_\perp$$; so if these are zero in one frame, they're zero in any other frame boosted along that axis.

EDIT: To go from a frame $$\mathcal{S}_0$$ where $$\vec{E} \cdot \vec{B} \neq 0$$ to another frame $$\mathcal{S}$$ where they are parallel, you first must boost with velocity $$\vec{v}$$ given by $$\frac{\vec{v}}{1 + v^2} = \frac{\vec{E} \times \vec{B}}{E^2 + B^2}.$$ This will yield a frame where $$\vec{E}$$ and $$\vec{B}$$ are parallel, and (notably) perpendicular to $$\vec{v}$$, the relative velocity direction between $$\mathcal{S}_0$$ and $$\mathcal{S}$$. You can then boost $$\mathcal{S}$$ with any velocity parallel to $$\vec{E}$$ and $$\vec{B}$$ (in $$\mathcal{S}$$) to obtain a new frame $$\mathcal{S}'$$ where $$\vec{E}$$ and $$\vec{B}$$ are also parallel.

Of note, however, is that $$\mathcal{S}_0$$ and $$\mathcal{S}'$$ are not related to each other by a simple boost. This is a fun (?) fact about Lorentz transformations: the composition of two boosts in different directions is not itself a boost, but instead a boost composed with a rotation. It's probably still possible to figure out the "boost portion" of this combined transformation to get the one-parameter family $$\vec{v}(\tau)$$ of relative velocities you're looking for, but it's not a simple exercise either.

• Perfect! I'd overlooked that first fact. Here is proof that Lorentz transformations do not affect the components of E and B in the direction of the boost: [home.thep.lu.se/~malin/LectureNotesFYTA12_2016/SR6.pdf] – S. McGrew Feb 11 at 14:28
• In [chegg.com/homework-help/…, a specific-looking solution is given, to finding a frame in which E and B are parallel. I misread the solution as specifying a specific frame; in fact there is a free parameter which amounts to $\gamma^2$. – S. McGrew Feb 11 at 14:42
• I am still a bit confused. The formula in [chegg.com/homework-help/… seems to indicate that any velocity in Alice's E cross B direction provides a frame in which E and B are parallel. That can't be true because zero speed is no boost at all. Bob's direction as well as his speed, as seen by Alice, must be different for the different frames in which he perceives E and B to be parallel. Zero-speed boosts can't be an option. – S. McGrew Feb 11 at 19:58
• So what I'm hoping to find is a formula that specifies the full set of frames, from Alice's perspective and dependent on the E and B she sees, in which Bob sees the E and B fields to be parallel. It should give $v$ as a function of a parameter- which might be speed but could be arbitrary since it should describe a curve in velocity space. – S. McGrew Feb 11 at 20:06
• The edits to your answer are very helpful. Guess the right thing to do is formulate a new, more targeted question- after some more digging and thinking. – S. McGrew Feb 11 at 20:58