# Wigner Rotation of static E & M fields is dizzying

I am looking for a formula to describe the set of inertial frames in which an electromagnetic field with non-parallel, non-perpendicular, and non-zero $$E$$ and $$B$$ transform to an electromagnetic field with parallel $$E'$$ and $$B'$$. It should be a curve in velocity space.

The problem seems straightforward: there's a formula for finding one such inertial frame; and any boosts from that frame in a direction which is parallel (in that frame) to the $$E'$$ and $$B'$$ fields (in that frame) will produce new frames in which the $$E''$$ and $$B''$$ fields are parallel.

In response to this Physics SE question, @MichaelSeivert pointed out that consecutive boosts can add up in a messy way, resulting in Wigner rotation. I have looked into this, and haven't waded through the math in detail, but it appears that the bottom line is that in each of the continuum of frames in which the $$E'$$ and $$B'$$ fields are parallel, although the magnitudes of the E and B fields will not change, the direction of the transformed $$E$$ and $$B$$ fields will be different in the various frames. While that is a good thing to know, it doesn't tell me (yet) how to find the formula I'm looking for.

From the perspective of an observer who experiences non-parallel, non-perpendicular, and non-zero $$E$$ and $$B$$, what are the boosts $$\vec{v}(\tau)$$ that produce new frames in which $$E'$$ and $$B'$$ are parallel?

I suspect that the easiest way to get the formula might be to do an integral in "Rhodes-Semon Rapidity Space": start with one inertial frame and keep adding infinitesimal boosts in whatever direction the $$E$$ and $$B$$ fields are pointing in the resultant frame. But it seems there should also be a purely algebraic approach, too.

• I'm confused about what you're looking for. The answer you accepted on that other question gives you a boost that results in parallel $E'$ and $B'$. Then you can just boost arbitrarily along those directions. What more are you looking for? – Mike Feb 20 at 20:49
• @DanYand's answer is very close to what I was looking for: a formula that, by varying one parameter, yields the full set of such boosts. – S. McGrew Feb 21 at 1:21

SECTION A : A 3+1-Lorentz Transformation

In above Figure-01 an inertial system $$\:\mathrm S'\:$$ is translated with respect to the inertial system $$\:\mathrm S\:$$ with constant velocity
\begin{align} \boldsymbol{\upsilon} & =\left(\upsilon_{1},\upsilon_{2},\upsilon_{3}\right)=\left(\upsilon n_{1},\upsilon n_{2},\upsilon n_{3}\right)=\upsilon \mathbf n\,, \qquad \upsilon \in \left(-c,c\right) \tag{A-01a}\label{A-01a}\\ \Vert \mathbf{n} \Vert^2 & = n^2_1 +n^2_2 + n^2_3 = 1 \tag{A-01b}\label{A-01b} \end{align} The Lorentz transformation is \begin{align} \mathbf{x}^{\boldsymbol{\prime}} & = \mathbf{x}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \mathbf{x})\mathbf{n}-\gamma \boldsymbol{\upsilon}t \tag{A-02a}\label{A-02a}\\ t^{\boldsymbol{\prime}} & = \gamma\left(t-\dfrac{\boldsymbol{\upsilon}\boldsymbol{\cdot} \mathbf{x}}{c^{2}}\right) \tag{A-02b}\label{A-02b}\\ \gamma & = \left(1-\dfrac{\upsilon^2}{c^2}\right)^{\boldsymbol{-}\frac12} \tag{A-02c}\label{A-02c} \end{align} in differential form \begin{align} \mathrm d\mathbf{x}^{\boldsymbol{\prime}} & = \mathrm d\mathbf{x}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \mathrm d\mathbf{x})\mathbf{n}-\gamma\boldsymbol{\upsilon}\mathrm dt \tag{A-03a}\label{A-03a}\\ \mathrm d t^{\boldsymbol{\prime}} & = \gamma\left(\mathrm d t-\dfrac{\boldsymbol{\upsilon}\boldsymbol{\cdot} \mathrm d\mathbf{x}}{c^{2}}\right) \tag{A-03b}\label{A-03b} \end{align} and in matrix form $$$$\mathbf{X}^{\boldsymbol{\prime}}= \begin{bmatrix} \mathbf{x}^{\boldsymbol{\prime}}\vphantom{\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c}}\\ c t^{\boldsymbol{\prime}}\vphantom{\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c}} \end{bmatrix} = \begin{bmatrix} \mathrm I+(\gamma-1)\mathbf{n}\mathbf{n}^{\boldsymbol{\top}} & -\dfrac{\gamma\boldsymbol{\upsilon}}{c} \vphantom{\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c}}\\ -\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c} & \hphantom{-}\gamma \end{bmatrix} \begin{bmatrix} \mathbf{x}\vphantom{\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c}}\\ c t\vphantom{\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c}} \end{bmatrix} =\mathrm L\mathbf{X} \tag{A-04}\label{A-04}$$$$ where $$\:\mathrm L\:$$ the real symmetric $$\:4\times 4\:$$ matrix $$$$\mathrm L \equiv \begin{bmatrix} \mathrm I+(\gamma-1)\mathbf{n}\mathbf{n}^{\boldsymbol{\top}} & -\dfrac{\gamma\boldsymbol{\upsilon}}{c} \vphantom{\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c}}\\ -\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c} & \hphantom{-}\gamma \end{bmatrix} \tag{A-05}\label{A-05}$$$$ and $$$$\mathbf{n}\mathbf{n}^{\boldsymbol{\top}} = \begin{bmatrix} \mathrm n_{1}\vphantom{\dfrac{}{}}\\ \mathrm n_{2}\vphantom{\dfrac{}{}}\\ \mathrm n_{3}\vphantom{\dfrac{}{}} \end{bmatrix} \begin{bmatrix} \mathrm n_{1} & \mathrm n_{2} & \mathrm n_{3} \end{bmatrix} = \begin{bmatrix} \mathrm n_{1}^{2} & \mathrm n_{1}\mathrm n_{2} & \mathrm n_{1}\mathrm n_{3}\vphantom{\dfrac{}{}}\\ \mathrm n_{2}\mathrm n_{1} & \mathrm n_{2}^{2} & \mathrm n_{2}\mathrm n_{3}\vphantom{\dfrac{}{}}\\ \mathrm n_{3}\mathrm n_{1} & \mathrm n_{3}\mathrm n_{2} & \mathrm n_{3}^{2}\vphantom{\dfrac{}{}} \end{bmatrix} \tag{A-06}\label{A-06}$$$$
a matrix representing the vectorial projection on the direction $$\:\mathbf{n}$$.

For the Lorentz transformation \eqref{A-02a}-\eqref{A-02b}, the vectors $$\:\mathbf{E}\:$$ and $$\:\mathbf{B}\:$$ of the electromagnetic field are transformed as follows \begin{align} \mathbf{E}' & =\gamma \mathbf{E}\!-\!\left(\gamma\!-\!1\right)\left(\mathbf{E}\boldsymbol{\cdot}\mathbf{n}\right)\mathbf{n}+\gamma\upsilon\left(\mathbf{n}\boldsymbol{\times}\mathbf{B}\right) \tag{A-07a}\label{A-07a}\\ \mathbf{B}' & = \gamma \mathbf{B}\!-\!\left(\gamma\!-\!1\right)\left(\mathbf{B}\boldsymbol{\cdot}\mathbf{n}\right)\mathbf{n}\!-\!\dfrac{\gamma\upsilon}{c^2}\left(\mathbf{n}\boldsymbol{\times}\mathbf{E}\right) \tag{A-07b}\label{A-07b} \end{align}

SECTION B : An effort to answer the question

Based on the conditions of non-parallel, non-perpendicular and non-zero $$\:\mathbf{E},\mathbf{B}\:$$ we define a convenient coordinate system as follows (see Figure-02) \begin{align} \mathbf e_1 & \boldsymbol{\equiv}\dfrac{\mathbf E}{\Vert\mathbf E\Vert} \boldsymbol{=}\dfrac{\mathbf E}{E} \tag{B-01.1}\label{B-01.1}\\ \mathbf e_2 & \boldsymbol{\equiv}\dfrac{\mathbf e'_2}{\Vert\mathbf e'_2\Vert}\boldsymbol{=}\dfrac{E\mathbf B\boldsymbol{-}B\cos\phi\mathbf E}{EB\sin\phi} \tag{B-01.2}\label{B-01.2}\\ \mathbf e_3 & \boldsymbol{\equiv}\mathbf e_1\boldsymbol{\times}\mathbf e_2\boldsymbol{=}\dfrac{\mathbf E}{E}\boldsymbol{\times}\dfrac{E\mathbf B\boldsymbol{-}B\cos\phi\mathbf E}{EB\sin\phi}\boldsymbol{=}\dfrac{\mathbf E\boldsymbol{\times}\mathbf B}{EB\sin\phi}\boldsymbol{=}\dfrac{\mathbf E\boldsymbol{\times}\mathbf B}{\Vert\mathbf E\boldsymbol{\times}\mathbf B\Vert} \tag{B-01.3}\label{B-01.3} \end{align} where $$\:\mathbf e'_2\:$$ the projection of $$\:\mathbf{B}\:$$ on the direction normal to $$\:\mathbf{E}\:$$ $$$$\mathbf e'_2\boldsymbol{\equiv}\mathbf B\boldsymbol{-}\left(B\cos\phi\right)\dfrac{\mathbf E}{E}\boldsymbol{=}\dfrac{E\mathbf B\boldsymbol{-}B\cos\phi\mathbf E}{E}\,, \quad \Vert\mathbf e'_2\Vert\boldsymbol{=}B\sin\phi \tag{B-02}\label{B-02}$$$$ and $$$$\phi \in \left(0,\dfrac{\pi}{2}\right) \cup \left(\dfrac{\pi}{2},\pi\right) \tag{B-03}\label{B-03}$$$$ the angle between the vectors $$\:\mathbf{E},\mathbf{B}$$.

With respect to the above so-defined coordinate system we have \begin{align} \require{cancel} \mathbf{E} & \boldsymbol{=}\cancelto{E}{E_1}\mathbf e_1\boldsymbol{+}\cancelto{0}{E_2}\mathbf e_2\boldsymbol{+}\cancelto{0}{E_3}\mathbf e_3 \boldsymbol{=}E \mathbf e_1 \boldsymbol{=} \left(E,0,0\right) \tag{B-04a}\label{B-04a}\\ \mathbf{B} & \boldsymbol{=}B_1\mathbf e_1\boldsymbol{+}B_2\mathbf e_2\boldsymbol{+}\cancelto{0}{B_3}\mathbf e_3 \boldsymbol{=} \left(B\cos\phi,B\sin\phi,0\right) \tag{B-04b}\label{B-04b} \end{align} and from the Lorentz transforms of $$\:\mathbf{E}'\:$$ and $$\:\mathbf{B}'$$, equations \eqref{A-07a} and \eqref{A-07b} respectively, we have \begin{align} \mathbf{E}'\boldsymbol{=}&\gamma E \mathbf e_1\!-\!\left(\gamma\!-\!1\right)n_1 E\left(n_1\mathbf e_1\boldsymbol{+}n_2\mathbf e_2\boldsymbol{+}n_3\mathbf e_3\right)\boldsymbol{-}\gamma\upsilon n_3B\sin\phi\,\mathbf e_1 \nonumber\\ &\boldsymbol{+}\gamma\upsilon n_3B\cos\phi\,\mathbf e_2\boldsymbol{+}\gamma\upsilon \left(n_1\sin\phi-n_2\cos\phi\right)B\,\mathbf e_3 \tag{B-05a}\label{B-05a}\\ \mathbf{B}' \boldsymbol{=} &\gamma\left( \cos\phi\,\mathbf e_1\boldsymbol{+}\sin\phi\,\mathbf e_2\right)B \!-\!\left(\gamma\!-\!1\right)\left(n_1\cos\phi+n_2\sin\phi\right)B\left(n_1\mathbf e_1\boldsymbol{+}n_2\mathbf e_2\boldsymbol{+}n_3\mathbf e_3\right) \nonumber\\ &\boldsymbol{-}\dfrac{\gamma\upsilon}{c^2}n_3E\mathbf e_2\boldsymbol{+}\dfrac{\gamma\upsilon}{c^2}n_2E\mathbf e_3 \tag{B-05b}\label{B-05b} \end{align}
In matrix form \begin{align} \mathbf{E}' & \boldsymbol{=} \begin{bmatrix} \:\:E'_1\:\:\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \\ \:\:E'_2\:\:\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ \:\:E'_3\:\:\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \bigl[\gamma \!-\!\left(\gamma\!-\!1\right)n^2_1\bigr]E\boldsymbol{-}\gamma\upsilon n_3\sin\phi B\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ \!-\!\left(\gamma\!-\!1\right)n_1n_2 E+\gamma\upsilon n_3\cos\phi B \vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ \!-\!\left(\gamma\!-\!1\right)n_1n_3 E+\gamma\upsilon \left(n_1\sin\phi-n_2\cos\phi\right)B \vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \end{bmatrix} \tag{B-06a}\label{B-06a}\\ \mathbf{B}'& \boldsymbol{=} \begin{bmatrix} \:\:B'_1\:\:\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \\ \:\:B'_2\:\:\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ \:\:B'_3\:\:\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \biggl(\bigl[\gamma \!-\!\left(\gamma\!-\!1\right)n^2_1\bigr]\cos\phi\!-\!\left(\gamma\!-\!1\right)n_1n_2\sin\phi\biggr)B\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ \boldsymbol{-}\dfrac{\gamma\upsilon}{c^2}n_3E+\biggl(\bigl[\gamma \!-\!\left(\gamma\!-\!1\right)n^2_2\bigr]\sin\phi\!-\!\left(\gamma\!-\!1\right)n_1n_2\cos\phi\biggr)B\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ \dfrac{\gamma\upsilon}{c^2}n_2E\!-\!\left(\gamma\!-\!1\right)\left(n_1\cos\phi+n_2\sin\phi\right)n_3B\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \end{bmatrix} \tag{B-06b}\label{B-06b} \end{align} Note that anyone of the components $$\:E'_j,B'_j\:$$ is a linear combination of the positive constant magnitudes $$\:E,B\:$$ and so we could write \begin{align} \mathbf{E}'& \boldsymbol{=} \begin{bmatrix} \:\:E'_1\:\:\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \\ \:\:E'_2\:\:\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ \:\:E'_3\:\:\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} &\xi_{11}&&\xi_{12}&\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ &\xi_{21}&&\xi_{22}&\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ &\xi_{31}&&\xi_{32}&\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \end{bmatrix} \begin{bmatrix} \:\:E\:\:\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ \:\:B\:\:\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \end{bmatrix} \tag{B-07a}\label{B-07a}\\ \mathbf{B}'& \boldsymbol{=} \begin{bmatrix} \:\:B'_1\:\:\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \\ \:\:B'_2\:\:\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ \:\:B'_3\:\:\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} &\eta_{11}&&\eta_{12}&\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ &\eta_{21}&&\eta_{22}&\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ &\eta_{31}&&\eta_{32}&\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \end{bmatrix} \begin{bmatrix} \:\:E\:\:\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ \:\:B\:\:\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \end{bmatrix} \tag{B-07b}\label{B-07b} \end{align}
where \begin{align} \begin{bmatrix} &\xi_{11}&&\xi_{12}&\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ &\xi_{21}&&\xi_{22}&\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ &\xi_{31}&&\xi_{32}&\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \end{bmatrix} & \boldsymbol{=} \begin{bmatrix} &\bigl[\gamma \!-\!\left(\gamma\!-\!1\right)n^2_1\bigr]&&\boldsymbol{-}\gamma\upsilon n_3\sin\phi &\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ &\!-\!\left(\gamma\!-\!1\right)n_1n_2&&\gamma\upsilon n_3\cos\phi& \vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ &\!-\!\left(\gamma\!-\!1\right)n_1n_3&&\gamma\upsilon \left(n_1\sin\phi-n_2\cos\phi\right)& \vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \end{bmatrix} \tag{B-08a}\label{B-08a}\\ \begin{bmatrix} &\eta_{11}&&\eta_{12}&\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ &\eta_{21}&&\eta_{22}&\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ &\eta_{31}&&\eta_{32}&\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \end{bmatrix} & \boldsymbol{=} \begin{bmatrix} &0&&\biggl(\bigl[\gamma \!-\!\left(\gamma\!-\!1\right)n^2_1\bigr]\cos\phi\!-\!\left(\gamma\!-\!1\right)n_1n_2\sin\phi\biggr)&\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ &\boldsymbol{-}\dfrac{\gamma\upsilon}{c^2}n_3&&\biggl(\bigl[\gamma \!-\!\left(\gamma\!-\!1\right)n^2_2\bigr]\sin\phi\!-\!\left(\gamma\!-\!1\right)n_1n_2\cos\phi\biggr)&\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ &\dfrac{\gamma\upsilon}{c^2}n_2&&\!-\!\left(\gamma\!-\!1\right)\left(n_1\cos\phi+n_2\sin\phi\right)n_3&\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \end{bmatrix} \tag{B-08b}\label{B-08b} \end{align}

Now, the vectors $$\:\mathbf{E}',\mathbf{B}'\:$$ are parallel in the system $$\:\mathrm S'\:$$ if $$$$\dfrac{E'_1}{B'_1}\boldsymbol{=}\dfrac{E'_2}{B'_2}\boldsymbol{=}\dfrac{E'_3}{B'_3} \tag{B-09}\label{B-09}$$$$ equation in many respects equivalent to $$$$\mathbf E'\boldsymbol{\times}\mathbf B'\boldsymbol{=0} \tag{B-10}\label{B-10}$$$$ We express equation \eqref{B-09} in terms of the $$\:\xi$$s and $$\:\eta$$s $$$$\dfrac{\xi_{11}E\boldsymbol{+}\xi_{12}B}{\eta_{11}E\boldsymbol{+}\eta_{12}B}\boldsymbol{=}\dfrac{\xi_{21}E\boldsymbol{+}\xi_{22}B}{\eta_{21}E\boldsymbol{+}\eta_{22}B}\boldsymbol{=}\dfrac{\xi_{31}E\boldsymbol{+}\xi_{32}B}{\eta_{31}E\boldsymbol{+}\eta_{32}B} \tag{B-11}\label{B-11}$$$$ The first equality in \eqref{B-11} yields $$$$\begin{vmatrix} \xi_{11}&\eta_{11}\\ \xi_{21}&\eta_{21} \end{vmatrix} E^2 \boldsymbol{+} \biggl( \begin{vmatrix} \xi_{11}&\eta_{11}\\ \xi_{22}&\eta_{22} \end{vmatrix} \boldsymbol{+} \begin{vmatrix} \xi_{12}&\eta_{12}\\ \xi_{21}&\eta_{21} \end{vmatrix} \biggr)EB \boldsymbol{+} \begin{vmatrix} \xi_{12}&\eta_{12}\\ \xi_{22}&\eta_{22} \end{vmatrix}B^2 \boldsymbol{=}0 \tag{B-12}\label{B-12}$$$$ where for the coefficients of $$\:E^2,EB,B^2$$ \begin{align} \begin{vmatrix} \xi_{11}&\eta_{11}\\ \xi_{21}&\eta_{21} \end{vmatrix} &\boldsymbol{=}\dfrac{\gamma}{c^2}\Bigl[\left(\gamma\!-\!1\right)n^2_1\boldsymbol{-}\gamma \Bigr]\upsilon n_3 \tag{B-13a}\label{B-13a}\\ \biggl( \begin{vmatrix} \xi_{11}&\eta_{11}\\ \xi_{22}&\eta_{22} \end{vmatrix} \boldsymbol{+} \begin{vmatrix} \xi_{12}&\eta_{12}\\ \xi_{21}&\eta_{21} \end{vmatrix} \biggr) &\boldsymbol{=}\Bigl[\left(2\gamma^2\!\boldsymbol{-}\!\gamma \!\boldsymbol{-}\!1\right)n^2_3\boldsymbol{+}\gamma\Bigr]\sin\phi \tag{B-13b}\label{B-13b}\\ \begin{vmatrix} \xi_{12}&\eta_{12}\\ \xi_{22}&\eta_{22} \end{vmatrix} &\boldsymbol{=}\gamma\Bigl[\left(\gamma\!-\!1\right)\left(n_1\cos\phi\boldsymbol{+}n_2\sin\phi\right)^2\boldsymbol{-}\gamma\Bigr]\upsilon n_3 \tag{B-13c}\label{B-13c} \end{align} while the second equality in \eqref{B-11} yields $$$$\begin{vmatrix} \xi_{21}&\eta_{21}\\ \xi_{31}&\eta_{31} \end{vmatrix} E^2 \boldsymbol{+} \biggl( \begin{vmatrix} \xi_{21}&\eta_{21}\\ \xi_{32}&\eta_{32} \end{vmatrix} \boldsymbol{+} \begin{vmatrix} \xi_{22}&\eta_{22}\\ \xi_{31}&\eta_{31} \end{vmatrix} \biggr)EB \boldsymbol{+} \begin{vmatrix} \xi_{22}&\eta_{22}\\ \xi_{32}&\eta_{32} \end{vmatrix}B^2 \boldsymbol{=}0 \tag{B-14}\label{B-14}$$$$ where for the coefficients of $$\:E^2,EB,B^2$$ \begin{align} \begin{vmatrix} \xi_{21}&\eta_{21}\\ \xi_{31}&\eta_{31} \end{vmatrix} &\boldsymbol{=}\boldsymbol{-}\dfrac{\gamma\left(\gamma\!-\!1\right)\upsilon}{c^2}n_1\left(n^2_2 \boldsymbol{+}n^2_3\right)\boldsymbol{=}\boldsymbol{-}\dfrac{\gamma\left(\gamma\!-\!1\right)\upsilon}{c^2}n_1\left(1 \boldsymbol{-}n^2_1\right) \tag{B-15a}\label{B-15a}\\ \biggl( \begin{vmatrix} \xi_{21}&\eta_{21}\\ \xi_{32}&\eta_{32} \end{vmatrix} \boldsymbol{+} \begin{vmatrix} \xi_{22}&\eta_{22}\\ \xi_{31}&\eta_{31} \end{vmatrix} \biggr) &\boldsymbol{=}\boldsymbol{-}\gamma n_1 n_3\sin\phi \tag{B-15b}\label{B-15b}\\ \begin{vmatrix} \xi_{22}&\eta_{22}\\ \xi_{32}&\eta_{32} \end{vmatrix} &\boldsymbol{=}\gamma\left(\gamma\!-\!1\right)\upsilon\left(n_1\cos\phi\boldsymbol{+}n_2\sin\phi\right)^2 n_1\boldsymbol{+}\gamma\upsilon n_2 \cos\phi\sin\phi \tag{B-15c}\label{B-15c} \end{align}

For $$\:\mathbf{E}',\mathbf{B}'\:$$ to be parallel the four (4) unknown variables $$\:\upsilon,n_1,n_2,n_3\:$$ must satisfy simultaneously the system of three (3) equations \eqref{A-01b}, \eqref{B-12} and \eqref{B-14}.

First note that the variable $$\:\upsilon\:$$ is a real number in $$\:\left(-c,c\right)\:$$ and not the non-negative magnitude of the velocity $$\:\boldsymbol{\upsilon}$$, see equation \eqref{A-01a}. Second note that if a quadruple $$\:\left(\upsilon,n_1,n_2,n_3\right)\:$$ satisfies above system then the quadruple $$\:\left(\boldsymbol{-}\upsilon,\boldsymbol{-}n_1,\boldsymbol{-}n_2,\boldsymbol{-}n_3\right)\:$$ satisfies also this system. But these two quadruples represent the same velocity $$$$\upsilon\left(n_{1},n_{2},n_{3}\right)\boldsymbol{=}\boldsymbol{\upsilon}\boldsymbol{=-}\upsilon\left(\boldsymbol{-}n_{1},\boldsymbol{-}n_{2},\boldsymbol{-}n_{3}\right) \tag{B-16}\label{B-16}$$$$ We must take care about this in order to avoid double counting of the solutions.

SECTION C : A first solution

Looking carefully in equation \eqref{B-14} we note that if $$$$n_1\boldsymbol{=}0\,, n_2\boldsymbol{=}0 \tag{C-01}\label{C-01}$$$$ then its coefficients of $$\:E^2,EB,B^2\:$$ given by equations \eqref{B-15a},\eqref{B-15b} and \eqref{B-15c} respectively are all equating to zero, so equation \eqref{B-14} is satisfied. Then from equation \eqref{A-01b} we have $$$$n^2_3\boldsymbol{=}1 \quad \boldsymbol{\Longrightarrow} \quad n_3\boldsymbol{=\pm} 1 \tag{C-02}\label{C-02}$$$$ In case that
$$$$n_1\boldsymbol{=}0\,, n_2\boldsymbol{=}0\,, n_3\boldsymbol{=+} 1 \tag{C-03}\label{C-03}$$$$ equations \eqref{B-06a},\eqref{B-06b} yield

\begin{align} \mathbf{E}' & \boldsymbol{=} \begin{bmatrix} \:\:E'_1\:\:\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \\ \:\:E'_2\:\:\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ \:\:E'_3\:\:\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \gamma E\boldsymbol{-}\gamma\upsilon \sin\phi B\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ \gamma\upsilon\cos\phi B \vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ 0 \vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \end{bmatrix} \boldsymbol{=} \gamma \begin{bmatrix} E\boldsymbol{-}\upsilon \sin\phi B\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ \upsilon\cos\phi B \vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ 0 \vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \end{bmatrix} \tag{C-04a}\label{C-04a}\\ \mathbf{B}'& \boldsymbol{=} \begin{bmatrix} \:\:B'_1\:\:\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \\ \:\:B'_2\:\:\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ \:\:B'_3\:\:\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \gamma\cos\phi B \vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ \boldsymbol{-}\dfrac{\gamma\upsilon}{c^2}E+\gamma\sin\phi B\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ 0\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \end{bmatrix} \boldsymbol{=} \gamma \begin{bmatrix} \cos\phi B \vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ \boldsymbol{-}\dfrac{\upsilon}{c^2}E+\sin\phi B\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ 0\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \end{bmatrix} \tag{C-04b}\label{C-04b} \end{align} and equation \eqref{B-12}, with its coefficients determined from equations \eqref{B-13a}-\eqref{B-13b}-\eqref{B-13c}, yields $$$$\left(\boldsymbol{-}\dfrac{\gamma^2\upsilon}{c^2}\right)E^2\boldsymbol{+}\left[\left(2\gamma^2-1\right)\sin\phi\right]EB\boldsymbol{+}\left(\boldsymbol{-}\gamma^2\upsilon\right)B^2\boldsymbol{=}0 \tag{C-05}\label{C-05}$$$$ or $$$$\boldsymbol{-}\dfrac{\upsilon}{c^2}E^2\boldsymbol{+}\left(1\boldsymbol{+}\dfrac{\upsilon^2}{c^2}\right)EB\sin\phi\boldsymbol{-}\upsilon B^2 \boldsymbol{=}0 \tag{C-06}\label{C-06}$$$$ This last equation is of course the condition of parallelism of the vectors $$\:\mathbf{E}',\mathbf{B}'\:$$ given by equations \eqref{C-04a},\eqref{C-04b} \begin{align} \dfrac{\gamma E\boldsymbol{-}\gamma\upsilon \sin\phi B}{\gamma\cos\phi B}&\boldsymbol{=}\dfrac{\gamma\cos\phi B}{\boldsymbol{-}\dfrac{\gamma\upsilon}{c^2}E+\gamma\sin\phi B} \quad \boldsymbol{\Longrightarrow}\quad \dfrac{\upsilon}{1\boldsymbol{+}\dfrac{\upsilon^2}{c^2}} \boldsymbol{=}\dfrac{EB\sin\phi }{B^2\boldsymbol{+}\dfrac{E^2}{c^2}}\quad \boldsymbol{\Longrightarrow} \nonumber\\ \dfrac{\upsilon}{1\boldsymbol{+}\dfrac{\upsilon^2}{c^2}}& \boldsymbol{=}\dfrac{\Vert\mathbf E\boldsymbol{\times}\mathbf B\Vert}{B^2\boldsymbol{+}\dfrac{E^2}{c^2}} \tag{C-07}\label{C-07} \end{align} so that $$$$\dfrac{\boldsymbol{\upsilon}}{1\boldsymbol{+}\dfrac{\upsilon^2}{c^2}} \boldsymbol{=}\dfrac{\mathbf E\boldsymbol{\times}\mathbf B}{B^2\boldsymbol{+}\dfrac{E^2}{c^2}} \tag{C-08}\label{C-08}$$$$ But now we must check if $$\:\upsilon\:$$ given implicitly by equation \eqref{C-07} is in the range $$\:\left(\boldsymbol{-}c,\boldsymbol{+}c\right)$$.

Note that \eqref{C-07} is a quadratic equation with respect to $$\:\upsilon/c$$ $$$$\left(\dfrac{\upsilon}{c}\right)^2 \boldsymbol{-}\left(\dfrac{E^2+c^2B^2}{cEB\sin\phi}\right)\left(\dfrac{\upsilon}{c}\right)\boldsymbol{+}1\boldsymbol{=}0 \tag{C-09}\label{C-09}$$$$ with real roots $$$$\left(\dfrac{\upsilon}{c}\right)_{\boldsymbol{\pm}} \boldsymbol{=}\dfrac{\left(E^2+c^2B^2\right)\boldsymbol{\pm}\sqrt{\left(E^2+c^2B^2\right)^2\boldsymbol{-}\left(2cEB\sin\phi\right)^2}}{2cEB\sin\phi} \tag{C-10}\label{C-10}$$$$ Since their sum is positive and their product is 1 $$$$\left(\dfrac{\upsilon}{c}\right)_{\boldsymbol{+}}\boldsymbol{+}\left(\dfrac{\upsilon}{c}\right)_{\boldsymbol{-}}\boldsymbol{=}\left(\dfrac{E^2+c^2B^2}{cEB\sin\phi}\right)>0\,, \quad \left(\dfrac{\upsilon}{c}\right)_{\boldsymbol{+}}\cdot\left(\dfrac{\upsilon}{c}\right)_{\boldsymbol{-}}\boldsymbol{=}1 \tag{C-11}\label{C-11}$$$$ acceptable is the root $$$$0<\left(\dfrac{\upsilon}{c}\right)_{\boldsymbol{-}} \boldsymbol{=}\dfrac{\left(E^2+c^2B^2\right)\boldsymbol{-}\sqrt{\left(E^2+c^2B^2\right)^2\boldsymbol{-}\left(2cEB\sin\phi\right)^2}}{2cEB\sin\phi} <1 \tag{C-12}\label{C-12}$$$$ Now it could be proved that the case $$$$n_1\boldsymbol{=}0\,, n_2\boldsymbol{=}0\,, n_3\boldsymbol{=-} 1 \tag{C-13}\label{C-13} %\tag{B-29}\label{B-29}$$$$ gives as result the negative of \eqref{C-12} so the solution is the same, see equation \eqref{B-16} to avoid double counting.

SECTION D : A continuous 1-parametric set of solutions

From the fact that the number of the variables (=4) is greater by 1 than the number of equations (=3) we must check if there exists a 1-parametric set of solutions.

As shown in SECTION C in a system $$\:\mathrm S'\:$$ moving uniformly along the $$\:x_3$$-axis of system $$\:\mathrm S\:$$ with velocity $$\:\boldsymbol{\upsilon}$$, see equations \eqref{C-08} and \eqref{C-12}, the new vectors $$\:\mathbf E'\:$$ and $$\:\mathbf B'\:$$ are parallel. Let $$\:\mathbf m\boldsymbol{=}\left(m_1,m_2,\cancelto{0}{m_3}\right)\:$$ the unit vector along the common axis of these two vectors. As pointed out in related answers and comments, if in system $$\:\mathrm S'\:$$ we have a frame $$\: \Sigma\:$$ moving along the unit vector $$\:\mathbf m$$, that is along the common axis of $$\:\mathbf E'\:$$ and $$\:\mathbf B'$$, with ANY SUBLUMINAL constant velocity
$$$$\mathbf u \boldsymbol{=} u\left(u_{1},u_{2},u_{3}\right)=\left(u m_{1},u m_{2},u m_{3}\right)=u \mathbf m\,, \qquad u\in \left(-c,c\right) \tag{D-01}\label{D-01}$$$$ see Figure-03, then the new vectors $$\:\mathbf E^{^{\boldsymbol \Sigma}}\:$$ and $$\:\mathbf B^{^{\boldsymbol \Sigma}}$$ remain parallel.

So, if we compose the two Lorentz transformations $$$$\text{system}\:\:\:\mathrm S\:\:\: \stackrel{\boldsymbol{\upsilon}}{\boldsymbol{=\!=\!=\!=\!=\!\Longrightarrow}} \:\:\:\text{system}\:\:\:\mathrm S' \:\:\: \stackrel{\mathbf u }{\boldsymbol{=\!=\!=\!=\!=\!\Longrightarrow}}\:\:\:\text{system}\:\:\:\Sigma \tag{D-02}\label{D-02}$$$$ then we'll have a parametric set of solutions with parameter the continuous variable $$\:u\in \left(-c,c\right)$$.

But the so derived composition of two Lorentz transformations will introduce in our solutions the undesired rotations we want to exclude since they trivially keep the parallelism invariant.

To avoid rotations we use a system $$\:\mathrm S''\:$$ moving with constant velocity $$\:\mathbf w\:$$ with respect to $$\:\mathrm S$$, see Figure-04. The velocity $$\:\mathbf w\:$$ is the velocity $$\:\mathbf u\:$$ as seen from $$\:\mathrm S$$, that is the relativistic addition of the perpendicular velocities $$\:\boldsymbol \upsilon\:$$ and $$\:\mathbf u$$ $$$$\mathbf w \boldsymbol{=} \dfrac{\mathbf u}{\gamma_{\boldsymbol \upsilon}}\boldsymbol{+ \upsilon}\boldsymbol{=}\dfrac{u}{\gamma_{\boldsymbol \upsilon}}\left(m_{1},m_{2},0\right)\boldsymbol{+} \upsilon\left(0,0,1\right) \tag{D-03}\label{D-03}$$$$ where the subscript $$\boldsymbol{\upsilon}$$ in $$\gamma_{\boldsymbol \upsilon}$$ is inserted to remind us that we talk about that of equation \eqref{A-02c}.

So, instead of the composition of Lorentz transformations \eqref{D-02} we make use of one Lorentz transformation $$$$\text{system}\:\:\:\mathrm S\:\:\: \stackrel{\mathbf{w}\boldsymbol{=}\tfrac{\mathbf{u} }{\gamma_{\boldsymbol{\upsilon}}}\boldsymbol{+\upsilon}}{\boldsymbol{=\!=\!=\!=\!=\!\Longrightarrow}} \:\:\:\text{system}\:\:\:\mathrm S'' \tag{D-04}\label{D-04}$$$$

So, the tips of the velocity vectors $$\:\mathbf w\:$$ is a straight segment $$\rm AA'$$, a 1-parameter set of solutions. The parameter is $$\:u\:$$ and runs along this segment, see Figure-05 below.

SECTION E : Edit

From \eqref{C-08} and \eqref{C-12} for the velocity vector we have $$$$\boldsymbol{\upsilon}\boldsymbol{=}\dfrac{\left(E^2+c^2B^2\right)\boldsymbol{-}\sqrt{\left(E^2+c^2B^2\right)^2\boldsymbol{-}4c^2\Vert\mathbf E\boldsymbol{\times}\mathbf B\Vert^2}}{2\Vert\mathbf E\boldsymbol{\times}\mathbf B\Vert^2}\left(\mathbf E\boldsymbol{\times}\mathbf B\right) \tag{E-01}\label{E-01}$$$$ For its magnitude $$$$\Vert\boldsymbol{\upsilon}\Vert=\upsilon\boldsymbol{=} \dfrac{\left(E^2+c^2B^2\right)\boldsymbol{-}\sqrt{\left(E^2+c^2B^2\right)^2\boldsymbol{-}4c^2\Vert\mathbf E\boldsymbol{\times}\mathbf B\Vert^2}}{2\Vert\mathbf E\boldsymbol{\times}\mathbf B\Vert} \tag{E-02}\label{E-02}$$$$ Note that since \begin{align} \mathcal{E} & \boldsymbol{=}\tfrac12 \epsilon_{0}\left(E^2+c^2B^2\right)\boldsymbol{=}\text{energy density of EM field in empty space} \tag{E-03a}\label{E-03a}\\ \boldsymbol{\mathcal{P}} & \boldsymbol{=}\epsilon_{0}c^2\left(\mathbf E\boldsymbol{\times}\mathbf B\right) \boldsymbol{=}\text{Poynting vector of EM field in empty space} \tag{E-03b}\label{E-03b} \end{align} equation \eqref{E-02} for the magnitude of the velocity vector yields $$$$\boxed{\:\Vert\boldsymbol{\upsilon}\Vert=\upsilon\boldsymbol{=}\left( \dfrac{c\mathcal{E}\boldsymbol{-}\sqrt{c^2\mathcal{E}^2\boldsymbol{-}\left\Vert \boldsymbol{\mathcal{P}}\right\Vert^2}}{\Vert\boldsymbol{\mathcal{P}}\Vert}\right)c\:} \tag{E-04}\label{E-04}$$$$

Suppose that on a point in an inertial system $$\:\mathrm S\:$$ the vectors $$\:\mathbf{E},\mathbf{B}\:$$ are non-parallel, non-perpendicular and non-zero, see Figure-02. If by a Lorentz transformation in the new inertial system $$\:\mathrm S'\:$$ the vectors $$\:\mathbf{E'},\mathbf{B'}\:$$ are parallel then their magnitudes $$\:E',B'\:$$ are fixed and could be determined through the two invariants of the electromagnetic field \begin{align} E\,'^{\,2}\boldsymbol{-}c^2 B\,'^{\,2} & \boldsymbol{=} E^2\boldsymbol{-}c^2 B^2 \tag{E-05a}\label{E-05a}\\ E\,'B\,' &\boldsymbol{=}\mathbf E\cdot \mathbf B \boldsymbol{=}EB\cos\phi \tag{E-05b}\label{E-05b} \end{align} Solving with respect to $$\:E\,',B\,'\:$$ we have \begin{align} E\,'^{\,2}&\boldsymbol{=}\tfrac12\left[\sqrt{\left(E^2\boldsymbol{-}c^2 B^2\right)^2\boldsymbol{+}4c^2\left(\mathbf E\cdot \mathbf B \right)^2}\boldsymbol{+}\left(E^2\boldsymbol{-}c^2 B^2\right)\right] \tag{E-06a}\label{E-06a}\\ c^2 B\,'^{\,2} & \boldsymbol{=}\tfrac12\left[\sqrt{\left(E^2\boldsymbol{-}c^2 B^2\right)^2\boldsymbol{+}4c^2\left(\mathbf E\cdot \mathbf B \right)^2}\boldsymbol{-}\left(E^2\boldsymbol{-}c^2 B^2\right)\right] \tag{E-06b}\label{E-06b} \end{align} Adding \eqref{E-06a}, \eqref{E-06b} we find the new energy density $$\:\mathcal{E'}\:$$ of the electromagnetic field $$$$\mathcal{E'} \boldsymbol{=}\tfrac12 \epsilon_{0}\left(E\,'^{\,2}+c^2 B\,'^{\,2}\right)\boldsymbol{=}\tfrac12 \epsilon_{0}\sqrt{\left(E^2\boldsymbol{-}c^2 B^2\right)^2\boldsymbol{+}4c^2\left(\mathbf E\cdot \mathbf B \right)^2} \tag{E-07}\label{E-07}$$$$ expressed also as $$$$\mathcal{E'} \boldsymbol{=} \sqrt{\mathcal{E}^2\boldsymbol{-}\dfrac{\left\Vert \boldsymbol{\mathcal{P}}\right\Vert^2}{c^2}} \tag{E-08}\label{E-08}$$$$

The energy density $$\:\mathcal{E'}\:$$ and the magnitudes $$\:E',B'\:$$ are the same in any system $$\:\mathrm S'\:$$ with parallel $$\:\mathbf{E'},\mathbf{B'}$$, so invariants in a strict sense.

• Excellent, very clear answer. – S. McGrew Feb 24 at 14:44

# A classification of simple Lorentz transformations

The Cartan-Dieudonné theorem says that for any number of dimensions and any signature, every origin-preserving isometry may be expressed as a composition of reflections. (Reference [1] gives a relatively friendly proof of this theorem.) When the signature is Lorentzian, an origin-preserving isometry is called a Lorentz transformation. The "origin" here is a point in spacetime, not just in space; and "reflection" means a reflection along a single line through the origin.

A proper origin-preserving Lorentz transformation is one that can be expressed as an even number of reflections, and a simple origin-preserving Lorentz transformation is one that can be expressed as exactly two reflections. Since each reflection is defined by a line through the origin, simple Lorentz transformations can be classified into three types, based on the plane $$P$$ defined by the two lines:

• Ordinary rotations: $$P$$ does not contain any lightlike lines through the origin (and therefore does not contain any timelike lines through the origin, either).

• Null rotations: $$P$$ contains exactly one lightlike line through the origin. This is the borderline case between an ordinary rotation and a boost.

• Boosts: $$P$$ contains exactly two lightlike lines through the origin (and therefore contains infinitely many timelike lines through the origin).

Given an orthogonal coordinate system, a plane spanned by two of the spacelike basis vectors will be called a canonical spacelike plane, and a plane containing the timelike basis vector will be called a canonical timelike plane. A canonical rotation is a rotation in a canonical spacelike plane, and a canonical boost is a boost in a canonical timelike plane. When we talk about a Lorentz boost specified by a single velocity vector $$\vec v$$, we are implicitly referring to a canonical boost, with the timelike basis vector being the second vector that, together with $$\vec v$$, defines the plane $$P$$.

The question addressed here is how to determine all canonical boosts that convert a given EM field to one in which the electric and magnetic fields are parallel to each other.

# Using Clifford algebra to describe Lorentz transformations of the EM field

An algebraic approach is requested. I'll use Clifford algebra, which is a wonderful computational tool for handling Lorentz transformations of the EM field. I'll review the basics here to establish my notation and conventions.

Let $$\gamma^0$$, $$\gamma^1$$, $$\gamma^2$$, $$\gamma^3$$ be mutually orthogonal basis vectors, with $$\gamma^0$$ being timelike and the others being spacelike. In Clifford algebra, vectors can be multiplied, and the product is associative. The basis vectors generate the whole algebra, through these relations: $$(\gamma^0)^2=I \hskip2cm (\gamma^k)^2=-I \\ \gamma^a\gamma^b=-\gamma^b\gamma^a \tag{1}$$ for all $$k\in\{1,2,3\}$$ and all $$a,b\in\{0,1,2,3\}$$. I'm using $$I$$ to denote the identity element of the algebra, and I'll use $$\Gamma$$ to denote the special element $$\Gamma\equiv \gamma^0\gamma^1\gamma^2\gamma^3. \tag{2}$$ The product of any vector with itself is proportional to the identity element $$I$$, and two vectors anti-commute with each other if and only if they are orthogonal to each other. Equation (1) says that the basis vectors satisfy both of these general rules.

One of the advantages of using Clifford algebra here is that reflections are easy to describe. Given any vector $$w=\sum_a w_a\gamma^a$$, a reflection along any non-lightlike line through the origin defined by another vector $$u=\sum_a u_a\gamma^a$$ is proportional to $$uwu$$. (We won't need to worry about the proportionality factor here, because the present question is concerned only with directions, not with magnitudes.) To see this, simply note that $$w$$ can be written $$w=w_\parallel+w_\perp$$, where $$w_\parallel$$ is parallel to $$u$$ and $$w_\perp$$ is orthogonal to $$u$$. Since orthogonal vectors anti-commute with each other, this gives $$uwu=u(w_\parallel+w_\perp)u=u^2(w_\parallel-w_\perp)\propto -w_\parallel+w_\perp, \tag{3}$$ which is the desired result.

Only constant-and-uniform EM fields will be considered here. Any such EM field $$F$$ may be represented by a bivector $$F = \sum_{a,b}\gamma^a\gamma^b F_{ab} \tag{4}$$ with $$F_{ab}=-F_{ba}$$. The electric field $$F_E$$ is the part involving the timelike basis vector $$\gamma^0$$, and the magnetic field $$F_B$$ is the part involving only the spacelike basis vectors. Explicitly, $$F_E=\frac{F-\gamma^0 F\gamma^0}{2} \hskip2cm F_B=\frac{F+\gamma^0 F\gamma^0}{2}. \tag{4b}$$ The quantity $$F$$ satisfies $$F^2=(F^2)_I + (F^2)_\Gamma \tag{5}$$ where subscripts $$I$$ and $$\Gamma$$ denote terms proportional to $$I$$ and $$\Gamma$$, respectively. Both parts, $$(F^2)_I$$ and $$(F^2)_\Gamma$$, are invariant under proper Lorentz transformations. Explicitly, $$(F^2)_I=F_E^2+F_B^2 \hskip2cm (F^2)_\Gamma = F_E F_B+F_BF_E. \tag{5b}$$ Beware that equation (5) is specific to four-dimensional spacetime, because only then is the quantity $$F_E F_B+F_B F_E$$ guaranteed to be proportional to the product of all of the basis vectors, denoted $$\Gamma$$.

Here is where Clifford algebra really shines: the effect on $$F$$ of a reflection along the line through the origin defined by a vector $$u$$ is proportional to $$uFu$$. Therefore, the result of a simple Lorentz transformation defined by reflections along $$u$$ and $$w$$ is proportional to $$wuFuw$$. Such expressions can be evaluated easily using equations (1). We never need to write down any unweildy $$4\times 4$$ matrices, and we never need to use any awkward cross-product identities.

# Application to the question

When we say that the electric and magnetic fields are "parallel" to each other, we are using a language that is specific to four-dimensional spacetime. Algebraically, this condition may be expressed simply as $$[F_E,\, F_B]=0 \hskip2cm \text{(iff E and B are "parallel")}. \tag{6}$$ A field that satisfies this condition will be called a diagonal field. A diagonal field can be written in the form $$F=(\alpha+\beta\Gamma)\gamma^0 f \hskip2cm \text{(iff E and B are "parallel")}, \tag{7}$$ where $$f$$ is a vector orthogonal to $$\gamma^0$$ and where the values of the scalars $$\alpha,\beta$$ can be determined by squaring both sides and using (5b). Use $$\Gamma^2=-I$$ to deduce the useful identity $$(\alpha-\beta\Gamma)(\alpha+\beta\Gamma) =(\alpha^2+\beta^2)I. \tag{8}$$ Now, consider an arbitrary EM field $$F$$, not necessarily diagonal. The goal is to find all timelike unit vectors $$u$$ for which the boosted field has the form $$\gamma^0 uFu\gamma^0 = (\alpha+\beta\Gamma)\gamma^0 f \tag{9}$$ for some vector $$f$$ that is orthogonal to $$\gamma^0$$ and for some scalars $$\alpha,\beta$$. Without loss of generality, we can take $$f$$ to be a unit vector: $$f^2=-1.$$ Sandwich equation (9) between $$u\gamma^0\cdots\gamma^0 u$$ and use $$u^2=1$$ to get \begin{align} F &= u\gamma^0(\alpha+\beta\Gamma)\gamma^0 f \gamma^0 u \\ &= -(\alpha+\beta\Gamma)u\gamma^0 f u. \tag{10} \end{align} The fact that $$\Gamma$$ commutes with all bivectors was used to get the last line. The values of $$\alpha,\beta$$ can be determined by squaring both sides, which gives \begin{align} F^2 &= (\alpha+\beta\Gamma)^2(u\gamma^0 f u)^2 \\ &= (\alpha+\beta\Gamma)^2 \tag{11} \end{align} because $$(u\gamma^0 fu)^2 = 1$$. Equation (11) determines the values of $$\alpha$$ and $$\beta$$ in terms of $$F$$, up to an overall sign that can be absorbed into the vector $$f$$. Now that $$\alpha$$ and $$\beta$$ have been determined, multiply equation (10) by $$\alpha-\beta\Gamma$$ and use the identity (8) to get $$\frac{\alpha-\beta\Gamma}{\alpha^2+\beta^2}F = -u\gamma^0 f u. \tag{12}$$ Equation (12) is the key to addressing the question. As long as $$(F^2)_\Gamma\neq 0$$, the field $$F$$ can be diagonalized by a canonical boost. Equation (12) says that any such field determines a unique bivector representing a single plane $$P$$ that contains some timelike direction, not necessarily canonical. (If it is canonical, then the field is already diagonal.)

The question has thus been reduced to a relatively intuitive geometric problem: Find all ways of expressing this plane $$P$$ in the form $$u\gamma^0 f u$$, where $$\gamma^0 f$$ is a canonical timelike plane $$P_0$$ and where the transformation $$u\cdots u$$ is a reflection along some timelike direction. The same geometric problem can also be posed like this: Find all ways of reflecting the given plane $$P$$ along a timelike direction $$u$$ so that the resulting plane contains $$\gamma^0$$. This intuitive picture strongly suggests that a continuous family of solutions exists.

To solve this algebraically, sandwich both sides of (12) between factors of $$u$$ and use the fact that $$u$$ anticommutes with $$\Gamma$$ to get $$\frac{\alpha+\beta\Gamma}{\alpha^2+\beta^2}uF u = -\gamma^0 f. \tag{13}$$ Now, given any candidate boost defined by a timelike vector $$u$$, all we need to do is evaluate the left-hand side of (13) and observe whether or not the $$\gamma^0$$-independent terms cancel. For this, we don't even need to require that $$u$$ be a unit vector; instead, we can write $$u=\gamma^0+v$$ for some vector $$v$$ that is orthogonal to $$\gamma^0$$, and then the condition to be checked is $$(\alpha+\beta\Gamma) (\gamma^0+v)F (\gamma^0+v) \hskip1cm \text{ has no \gamma^0-independent terms}. \tag{14}$$ Thanks to equation (14), the problem has been reduced to straightfoward associative algebra, along with a relatively simple geometric picture described in the preceding paragraph. Since this post is already long, I'm going to call that good enough.

References:

[1] Lam (2005), Introduction to Quadratic Forms over Fields, American Mathematical Society

• This looks very promising! If I understand correctly, a boost in the restrictive sense simply amounts to a new inertial frame that is moving with a velocity $\vec v$ relative to the "home" frame; while a boost in the less restrictive sense additionally includes what amounts to a 3-space rotation. In that case, what I'm looking for is the set of boosts in the more restrictive sense that make E parallel to B in the new frame. – S. McGrew Feb 17 at 16:56
• @S.McGrew I updated the answer based on your feedback, and I deleted all of my previous comments to clean up the page. – Chiral Anomaly Feb 18 at 0:13