Wigner Rotation of static E & M fields is dizzying - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-07-17T11:02:19Z https://physics.stackexchange.com/feeds/question/460253 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/460253 2 Wigner Rotation of static E & M fields is dizzying S. McGrew https://physics.stackexchange.com/users/183212 2019-02-12T01:54:39Z 2019-05-10T23:58:59Z <p>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 <strong><span class="math-container">$E$</span></strong> and <strong><span class="math-container">$B$</span></strong> transform to an electromagnetic field with parallel <strong><span class="math-container">$E'$</span></strong> and <strong><span class="math-container">$B'$</span></strong>. It should be a curve in velocity space.</p> <p>The problem seems straightforward: there's a <a href="https://www.chegg.com/homework-help/certain-inertial-frame-s-electric-field-e-magnetic-field-b-n-chapter-12-problem-66p-solution-9780321972101-exc" rel="nofollow noreferrer">formula</a> for finding <em>one</em> such inertial frame; and any boosts from that frame in a direction which is parallel (in that frame) to the <strong><span class="math-container">$E'$</span></strong> and <strong><span class="math-container">$B'$</span></strong> fields (in that frame) will produce new frames in which the <strong><span class="math-container">$E''$</span></strong> and <strong><span class="math-container">$B''$</span></strong> fields are parallel.</p> <p>In response to <a href="https://physics.stackexchange.com/questions/460082/the-curious-case-of-parallel-e-and-b-fields-and-inertial-frames/460140#460140">this Physics SE question</a>, @MichaelSeivert pointed out that consecutive boosts can add up in a messy way, resulting in <a href="https://en.wikipedia.org/wiki/Wigner_rotation" rel="nofollow noreferrer">Wigner rotation</a>. 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 <strong><span class="math-container">$E'$</span></strong> and <strong><span class="math-container">$B'$</span></strong> fields are parallel, although the magnitudes of the E and B fields will not change, the <em>direction</em> of the transformed <strong><span class="math-container">$E$</span></strong> and <strong><span class="math-container">$B$</span></strong> 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.</p> <p><strong>From the perspective of an observer who experiences non-parallel, non-perpendicular, and non-zero <span class="math-container">$E$</span> and <span class="math-container">$B$</span>, what are the boosts <span class="math-container">$\vec{v}(\tau)$</span> that produce new frames in which <span class="math-container">$E'$</span> and <span class="math-container">$B'$</span> are parallel?</strong> </p> <p>I suspect that the easiest way to get the formula might be to do an integral in "<a href="http://demonstrations.wolfram.com/BoostCompositionAndWignerRotationInRhodesSemonRapiditySpace/" rel="nofollow noreferrer">Rhodes-Semon Rapidity Space</a>": start with one inertial frame and keep adding infinitesimal boosts in whatever direction the <strong><span class="math-container">$E$</span></strong> and <strong><span class="math-container">$B$</span></strong> fields are pointing in the resultant frame. But it seems there should also be a purely algebraic approach, too. </p> https://physics.stackexchange.com/questions/460253/-/461344#461344 2 Answer by Chiral Anomaly for Wigner Rotation of static E & M fields is dizzying Chiral Anomaly https://physics.stackexchange.com/users/206691 2019-02-17T16:19:43Z 2019-02-18T00:12:11Z <h1> A classification of simple Lorentz transformations </h1> <p>The <strong>Cartan-Dieudonné theorem</strong> says that for any number of dimensions and any signature, every origin-preserving isometry may be expressed as a composition of reflections. (Reference  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.</p> <p>A <strong>proper</strong> origin-preserving Lorentz transformation is one that can be expressed as an <em>even</em> number of reflections, and a <strong>simple</strong> origin-preserving Lorentz transformation is one that can be expressed as exactly <em>two</em> 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 <span class="math-container">$P$</span> defined by the two lines:</p> <ul> <li><p><strong>Ordinary rotations</strong>: <span class="math-container">$P$</span> does not contain any lightlike lines through the origin (and therefore does not contain any timelike lines through the origin, either).</p></li> <li><p><strong>Null rotations</strong>: <span class="math-container">$P$</span> contains exactly one lightlike line through the origin. This is the borderline case between an ordinary rotation and a boost.</p></li> <li><p><strong>Boosts</strong>: <span class="math-container">$P$</span> contains exactly two lightlike lines through the origin (and therefore contains infinitely many timelike lines through the origin).</p></li> </ul> <p>Given an orthogonal coordinate system, a plane spanned by two of the spacelike basis vectors will be called a <strong>canonical spacelike plane</strong>, and a plane containing the timelike basis vector will be called a <strong>canonical timelike plane</strong>. A <strong>canonical rotation</strong> is a rotation in a canonical spacelike plane, and a <strong>canonical boost</strong> is a boost in a canonical timelike plane. When we talk about a Lorentz boost specified by a single velocity vector <span class="math-container">$\vec v$</span>, we are implicitly referring to a canonical boost, with the timelike basis vector being the second vector that, together with <span class="math-container">$\vec v$</span>, defines the plane <span class="math-container">$P$</span>. </p> <p>The question addressed here is how to determine all <em>canonical</em> boosts that convert a given EM field to one in which the electric and magnetic fields are parallel to each other.</p> <hr> <h1> Using Clifford algebra to describe Lorentz transformations of the EM field </h1> <p>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.</p> <p>Let <span class="math-container">$\gamma^0$</span>, <span class="math-container">$\gamma^1$</span>, <span class="math-container">$\gamma^2$</span>, <span class="math-container">$\gamma^3$</span> be mutually orthogonal basis vectors, with <span class="math-container">$\gamma^0$</span> 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: <span class="math-container">$$(\gamma^0)^2=I \hskip2cm (\gamma^k)^2=-I \\ \gamma^a\gamma^b=-\gamma^b\gamma^a \tag{1}$$</span> for all <span class="math-container">$k\in\{1,2,3\}$</span> and all <span class="math-container">$a,b\in\{0,1,2,3\}$</span>. I'm using <span class="math-container">$I$</span> to denote the identity element of the algebra, and I'll use <span class="math-container">$\Gamma$</span> to denote the special element <span class="math-container">$$\Gamma\equiv \gamma^0\gamma^1\gamma^2\gamma^3. \tag{2}$$</span> The product of any vector with itself is proportional to the identity element <span class="math-container">$I$</span>, 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.</p> <p>One of the advantages of using Clifford algebra here is that reflections are easy to describe. Given any vector <span class="math-container">$w=\sum_a w_a\gamma^a$</span>, a reflection along any non-lightlike line through the origin defined by another vector <span class="math-container">$u=\sum_a u_a\gamma^a$</span> is proportional to <span class="math-container">$uwu$</span>. (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 <span class="math-container">$w$</span> can be written <span class="math-container">$w=w_\parallel+w_\perp$</span>, where <span class="math-container">$w_\parallel$</span> is parallel to <span class="math-container">$u$</span> and <span class="math-container">$w_\perp$</span> is orthogonal to <span class="math-container">$u$</span>. Since orthogonal vectors anti-commute with each other, this gives <span class="math-container">$$uwu=u(w_\parallel+w_\perp)u=u^2(w_\parallel-w_\perp)\propto -w_\parallel+w_\perp, \tag{3}$$</span> which is the desired result. </p> <p>Only constant-and-uniform EM fields will be considered here. Any such EM field <span class="math-container">$F$</span> may be represented by a <strong>bivector</strong> <span class="math-container">$$F = \sum_{a,b}\gamma^a\gamma^b F_{ab} \tag{4}$$</span> with <span class="math-container">$F_{ab}=-F_{ba}$</span>. The <strong>electric field</strong> <span class="math-container">$F_E$</span> is the part involving the timelike basis vector <span class="math-container">$\gamma^0$</span>, and the <strong>magnetic field</strong> <span class="math-container">$F_B$</span> is the part involving only the spacelike basis vectors. Explicitly, <span class="math-container">$$F_E=\frac{F-\gamma^0 F\gamma^0}{2} \hskip2cm F_B=\frac{F+\gamma^0 F\gamma^0}{2}. \tag{4b}$$</span> The quantity <span class="math-container">$F$</span> satisfies <span class="math-container">$$F^2=(F^2)_I + (F^2)_\Gamma \tag{5}$$</span> where subscripts <span class="math-container">$I$</span> and <span class="math-container">$\Gamma$</span> denote terms proportional to <span class="math-container">$I$</span> and <span class="math-container">$\Gamma$</span>, respectively. Both parts, <span class="math-container">$(F^2)_I$</span> and <span class="math-container">$(F^2)_\Gamma$</span>, are invariant under proper Lorentz transformations. Explicitly, <span class="math-container">$$(F^2)_I=F_E^2+F_B^2 \hskip2cm (F^2)_\Gamma = F_E F_B+F_BF_E. \tag{5b}$$</span> Beware that equation (5) is specific to four-dimensional spacetime, because only then is the quantity <span class="math-container">$F_E F_B+F_B F_E$</span> guaranteed to be proportional to the product of all of the basis vectors, denoted <span class="math-container">$\Gamma$</span>.</p> <p>Here is where Clifford algebra really shines: the effect on <span class="math-container">$F$</span> of a reflection along the line through the origin defined by a vector <span class="math-container">$u$</span> is proportional to <span class="math-container">$uFu$</span>. Therefore, the result of a simple Lorentz transformation defined by reflections along <span class="math-container">$u$</span> and <span class="math-container">$w$</span> is proportional to <span class="math-container">$wuFuw$</span>. Such expressions can be evaluated easily using equations (1). We never need to write down any unweildy <span class="math-container">$4\times 4$</span> matrices, and we never need to use any awkward cross-product identities.</p> <hr> <h1> Application to the question </h1> <p>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 <span class="math-container">$$[F_E,\, F_B]=0 \hskip2cm \text{(iff E and B are "parallel")}. \tag{6}$$</span> A field that satisfies this condition will be called a <strong>diagonal</strong> field. A diagonal field can be written in the form <span class="math-container">$$F=(\alpha+\beta\Gamma)\gamma^0 f \hskip2cm \text{(iff E and B are "parallel")}, \tag{7}$$</span> where <span class="math-container">$f$</span> is a vector orthogonal to <span class="math-container">$\gamma^0$</span> and where the values of the scalars <span class="math-container">$\alpha,\beta$</span> can be determined by squaring both sides and using (5b). Use <span class="math-container">$\Gamma^2=-I$</span> to deduce the useful identity <span class="math-container">$$(\alpha-\beta\Gamma)(\alpha+\beta\Gamma) =(\alpha^2+\beta^2)I. \tag{8}$$</span> Now, consider an arbitrary EM field <span class="math-container">$F$</span>, not necessarily diagonal. The goal is to find all timelike unit vectors <span class="math-container">$u$</span> for which the boosted field has the form <span class="math-container">$$\gamma^0 uFu\gamma^0 = (\alpha+\beta\Gamma)\gamma^0 f \tag{9}$$</span> for some vector <span class="math-container">$f$</span> that is orthogonal to <span class="math-container">$\gamma^0$</span> and for some scalars <span class="math-container">$\alpha,\beta$</span>. Without loss of generality, we can take <span class="math-container">$f$</span> to be a unit vector: <span class="math-container">$$f^2=-1.$$</span> Sandwich equation (9) between <span class="math-container">$u\gamma^0\cdots\gamma^0 u$</span> and use <span class="math-container">$u^2=1$</span> to get <span class="math-container">\begin{align} F &amp;= u\gamma^0(\alpha+\beta\Gamma)\gamma^0 f \gamma^0 u \\ &amp;= -(\alpha+\beta\Gamma)u\gamma^0 f u. \tag{10} \end{align}</span> The fact that <span class="math-container">$\Gamma$</span> commutes with all bivectors was used to get the last line. The values of <span class="math-container">$\alpha,\beta$</span> can be determined by squaring both sides, which gives <span class="math-container">\begin{align} F^2 &amp;= (\alpha+\beta\Gamma)^2(u\gamma^0 f u)^2 \\ &amp;= (\alpha+\beta\Gamma)^2 \tag{11} \end{align}</span> because <span class="math-container">$(u\gamma^0 fu)^2 = 1$</span>. Equation (11) determines the values of <span class="math-container">$\alpha$</span> and <span class="math-container">$\beta$</span> in terms of <span class="math-container">$F$</span>, up to an overall sign that can be absorbed into the vector <span class="math-container">$f$</span>. Now that <span class="math-container">$\alpha$</span> and <span class="math-container">$\beta$</span> have been determined, multiply equation (10) by <span class="math-container">$\alpha-\beta\Gamma$</span> and use the identity (8) to get <span class="math-container">$$\frac{\alpha-\beta\Gamma}{\alpha^2+\beta^2}F = -u\gamma^0 f u. \tag{12}$$</span> Equation (12) is the key to addressing the question. As long as <span class="math-container">$(F^2)_\Gamma\neq 0$</span>, the field <span class="math-container">$F$</span> can be diagonalized by a canonical boost. Equation (12) says that any such field determines a unique bivector representing a single plane <span class="math-container">$P$</span> that contains <em>some</em> timelike direction, not necessarily canonical. (If it is canonical, then the field is already diagonal.) </p> <p>The question has thus been reduced to a relatively intuitive geometric problem: Find all ways of expressing this plane <span class="math-container">$P$</span> in the form <span class="math-container">$u\gamma^0 f u$</span>, where <span class="math-container">$\gamma^0 f$</span> is a canonical timelike plane <span class="math-container">$P_0$</span> and where the transformation <span class="math-container">$u\cdots u$</span> 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 <span class="math-container">$P$</span> along a timelike direction <span class="math-container">$u$</span> so that the resulting plane contains <span class="math-container">$\gamma^0$</span>. This intuitive picture strongly suggests that a continuous family of solutions exists.</p> <p>To solve this algebraically, sandwich both sides of (12) between factors of <span class="math-container">$u$</span> and use the fact that <span class="math-container">$u$</span> anticommutes with <span class="math-container">$\Gamma$</span> to get <span class="math-container">$$\frac{\alpha+\beta\Gamma}{\alpha^2+\beta^2}uF u = -\gamma^0 f. \tag{13}$$</span> Now, given any candidate boost defined by a timelike vector <span class="math-container">$u$</span>, all we need to do is evaluate the left-hand side of (13) and observe whether or not the <span class="math-container">$\gamma^0$</span>-independent terms cancel. For this, we don't even need to require that <span class="math-container">$u$</span> be a unit vector; instead, we can write <span class="math-container">$u=\gamma^0+v$</span> for some vector <span class="math-container">$v$</span> that is orthogonal to <span class="math-container">$\gamma^0$</span>, and then the condition to be checked is <span class="math-container">$$(\alpha+\beta\Gamma) (\gamma^0+v)F (\gamma^0+v) \hskip1cm \text{ has no \gamma^0-independent terms}. \tag{14}$$</span> 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.</p> <hr> <p><strong>References:</strong></p> <p> Lam (2005), <em>Introduction to Quadratic Forms over Fields</em>, American Mathematical Society</p> https://physics.stackexchange.com/questions/460253/-/462148#462148 3 Answer by Frobenius for Wigner Rotation of static E & M fields is dizzying Frobenius https://physics.stackexchange.com/users/110781 2019-02-21T20:31:56Z 2019-05-10T23:58:59Z <p><strong>SECTION A : A 3+1-Lorentz Transformation</strong></p> <p><a href="https://i.stack.imgur.com/SnT3b.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SnT3b.png" alt="enter image description here"></a></p> <p>In above Figure-01 an inertial system <span class="math-container">$\:\mathrm S'\:$</span> is translated with respect to the inertial system <span class="math-container">$\:\mathrm S\:$</span> with constant velocity<br> <span class="math-container">\begin{align} \boldsymbol{\upsilon} &amp; =\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 &amp; = n^2_1 +n^2_2 + n^2_3 = 1 \tag{A-01b}\label{A-01b} \end{align}</span> The Lorentz transformation is <span class="math-container">\begin{align} \mathbf{x}^{\boldsymbol{\prime}} &amp; = \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}} &amp; = \gamma\left(t-\dfrac{\boldsymbol{\upsilon}\boldsymbol{\cdot} \mathbf{x}}{c^{2}}\right) \tag{A-02b}\label{A-02b}\\ \gamma &amp; = \left(1-\dfrac{\upsilon^2}{c^2}\right)^{\boldsymbol{-}\frac12} \tag{A-02c}\label{A-02c} \end{align}</span> in differential form <span class="math-container">\begin{align} \mathrm d\mathbf{x}^{\boldsymbol{\prime}} &amp; = \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}} &amp; = \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}</span> and in matrix form <span class="math-container">\begin{equation} \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}} &amp; -\dfrac{\gamma\boldsymbol{\upsilon}}{c} \vphantom{\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c}}\\ -\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c} &amp; \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} \end{equation}</span> where <span class="math-container">$\:\mathrm L\:$</span> the real symmetric <span class="math-container">$\:4\times 4\:$</span> matrix <span class="math-container">\begin{equation} \mathrm L \equiv \begin{bmatrix} \mathrm I+(\gamma-1)\mathbf{n}\mathbf{n}^{\boldsymbol{\top}} &amp; -\dfrac{\gamma\boldsymbol{\upsilon}}{c} \vphantom{\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c}}\\ -\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c} &amp; \hphantom{-}\gamma \end{bmatrix} \tag{A-05}\label{A-05} \end{equation}</span> and <span class="math-container">\begin{equation} \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} &amp; \mathrm n_{2} &amp; \mathrm n_{3} \end{bmatrix} = \begin{bmatrix} \mathrm n_{1}^{2} &amp; \mathrm n_{1}\mathrm n_{2} &amp; \mathrm n_{1}\mathrm n_{3}\vphantom{\dfrac{}{}}\\ \mathrm n_{2}\mathrm n_{1} &amp; \mathrm n_{2}^{2} &amp; \mathrm n_{2}\mathrm n_{3}\vphantom{\dfrac{}{}}\\ \mathrm n_{3}\mathrm n_{1} &amp; \mathrm n_{3}\mathrm n_{2} &amp; \mathrm n_{3}^{2}\vphantom{\dfrac{}{}} \end{bmatrix} \tag{A-06}\label{A-06} \end{equation}</span><br> a matrix representing the vectorial projection on the direction <span class="math-container">$\:\mathbf{n}$</span>.</p> <p>For the Lorentz transformation \eqref{A-02a}-\eqref{A-02b}, the vectors <span class="math-container">$\:\mathbf{E}\:$</span> and <span class="math-container">$\:\mathbf{B}\:$</span> of the electromagnetic field are transformed as follows <span class="math-container">\begin{align} \mathbf{E}' &amp; =\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}' &amp; = \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}</span></p> <hr> <p><strong>SECTION B : An effort to answer the question</strong></p> <p><a href="https://i.stack.imgur.com/HeKxM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HeKxM.png" alt="enter image description here"></a></p> <p>Based on the conditions of non-parallel, non-perpendicular and non-zero <span class="math-container">$\:\mathbf{E},\mathbf{B}\:$</span> we define a convenient coordinate system as follows (see Figure-02) <span class="math-container">\begin{align} \mathbf e_1 &amp; \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 &amp; \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 &amp; \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}</span> where <span class="math-container">$\:\mathbf e'_2\:$</span> the projection of <span class="math-container">$\:\mathbf{B}\:$</span> on the direction normal to <span class="math-container">$\:\mathbf{E}\:$</span> <span class="math-container">\begin{equation} \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} \end{equation}</span> and <span class="math-container">\begin{equation} \phi \in \left(0,\dfrac{\pi}{2}\right) \cup \left(\dfrac{\pi}{2},\pi\right) \tag{B-03}\label{B-03} \end{equation}</span> the angle between the vectors <span class="math-container">$\:\mathbf{E},\mathbf{B}$</span>.</p> <p>With respect to the above so-defined coordinate system we have <span class="math-container">\begin{align} \require{cancel} \mathbf{E} &amp; \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} &amp; \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}</span> and from the Lorentz transforms of <span class="math-container">$\:\mathbf{E}'\:$</span> and <span class="math-container">$\:\mathbf{B}'$</span>, equations \eqref{A-07a} and \eqref{A-07b} respectively, we have <span class="math-container">\begin{align} \mathbf{E}'\boldsymbol{=}&amp;\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\\ &amp;\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{=} &amp;\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\\ &amp;\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}</span><br> In matrix form <span class="math-container">\begin{align} \mathbf{E}' &amp; \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}'&amp; \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}</span> Note that anyone of the components <span class="math-container">$\:E'_j,B'_j\:$</span> is a linear combination of the positive constant magnitudes <span class="math-container">$\:E,B\:$</span> and so we could write <span class="math-container">\begin{align} \mathbf{E}'&amp; \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} &amp;\xi_{11}&amp;&amp;\xi_{12}&amp;\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ &amp;\xi_{21}&amp;&amp;\xi_{22}&amp;\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ &amp;\xi_{31}&amp;&amp;\xi_{32}&amp;\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}'&amp; \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} &amp;\eta_{11}&amp;&amp;\eta_{12}&amp;\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ &amp;\eta_{21}&amp;&amp;\eta_{22}&amp;\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ &amp;\eta_{31}&amp;&amp;\eta_{32}&amp;\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}</span><br> where <span class="math-container">\begin{align} \begin{bmatrix} &amp;\xi_{11}&amp;&amp;\xi_{12}&amp;\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ &amp;\xi_{21}&amp;&amp;\xi_{22}&amp;\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ &amp;\xi_{31}&amp;&amp;\xi_{32}&amp;\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \end{bmatrix} &amp; \boldsymbol{=} \begin{bmatrix} &amp;\bigl[\gamma \!-\!\left(\gamma\!-\!1\right)n^2_1\bigr]&amp;&amp;\boldsymbol{-}\gamma\upsilon n_3\sin\phi &amp;\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ &amp;\!-\!\left(\gamma\!-\!1\right)n_1n_2&amp;&amp;\gamma\upsilon n_3\cos\phi&amp; \vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ &amp;\!-\!\left(\gamma\!-\!1\right)n_1n_3&amp;&amp;\gamma\upsilon \left(n_1\sin\phi-n_2\cos\phi\right)&amp; \vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \end{bmatrix} \tag{B-08a}\label{B-08a}\\ \begin{bmatrix} &amp;\eta_{11}&amp;&amp;\eta_{12}&amp;\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ &amp;\eta_{21}&amp;&amp;\eta_{22}&amp;\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ &amp;\eta_{31}&amp;&amp;\eta_{32}&amp;\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \end{bmatrix} &amp; \boldsymbol{=} \begin{bmatrix} &amp;0&amp;&amp;\biggl(\bigl[\gamma \!-\!\left(\gamma\!-\!1\right)n^2_1\bigr]\cos\phi\!-\!\left(\gamma\!-\!1\right)n_1n_2\sin\phi\biggr)&amp;\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ &amp;\boldsymbol{-}\dfrac{\gamma\upsilon}{c^2}n_3&amp;&amp;\biggl(\bigl[\gamma \!-\!\left(\gamma\!-\!1\right)n^2_2\bigr]\sin\phi\!-\!\left(\gamma\!-\!1\right)n_1n_2\cos\phi\biggr)&amp;\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\\ &amp;\dfrac{\gamma\upsilon}{c^2}n_2&amp;&amp;\!-\!\left(\gamma\!-\!1\right)\left(n_1\cos\phi+n_2\sin\phi\right)n_3&amp;\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}} \end{bmatrix} \tag{B-08b}\label{B-08b} \end{align}</span></p> <p>Now, the vectors <span class="math-container">$\:\mathbf{E}',\mathbf{B}'\:$</span> are parallel in the system <span class="math-container">$\:\mathrm S'\:$</span> if <span class="math-container">\begin{equation} \dfrac{E'_1}{B'_1}\boldsymbol{=}\dfrac{E'_2}{B'_2}\boldsymbol{=}\dfrac{E'_3}{B'_3} \tag{B-09}\label{B-09} \end{equation}</span> equation in many respects equivalent to <span class="math-container">\begin{equation} \mathbf E'\boldsymbol{\times}\mathbf B'\boldsymbol{=0} \tag{B-10}\label{B-10} \end{equation}</span> We express equation \eqref{B-09} in terms of the <span class="math-container">$\:\xi$</span>s and <span class="math-container">$\:\eta$</span>s <span class="math-container">\begin{equation} \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} \end{equation}</span> The first equality in \eqref{B-11} yields <span class="math-container">\begin{equation} \begin{vmatrix} \xi_{11}&amp;\eta_{11}\\ \xi_{21}&amp;\eta_{21} \end{vmatrix} E^2 \boldsymbol{+} \biggl( \begin{vmatrix} \xi_{11}&amp;\eta_{11}\\ \xi_{22}&amp;\eta_{22} \end{vmatrix} \boldsymbol{+} \begin{vmatrix} \xi_{12}&amp;\eta_{12}\\ \xi_{21}&amp;\eta_{21} \end{vmatrix} \biggr)EB \boldsymbol{+} \begin{vmatrix} \xi_{12}&amp;\eta_{12}\\ \xi_{22}&amp;\eta_{22} \end{vmatrix}B^2 \boldsymbol{=}0 \tag{B-12}\label{B-12} \end{equation}</span> where for the coefficients of <span class="math-container">$\:E^2,EB,B^2$</span> <span class="math-container">\begin{align} \begin{vmatrix} \xi_{11}&amp;\eta_{11}\\ \xi_{21}&amp;\eta_{21} \end{vmatrix} &amp;\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}&amp;\eta_{11}\\ \xi_{22}&amp;\eta_{22} \end{vmatrix} \boldsymbol{+} \begin{vmatrix} \xi_{12}&amp;\eta_{12}\\ \xi_{21}&amp;\eta_{21} \end{vmatrix} \biggr) &amp;\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}&amp;\eta_{12}\\ \xi_{22}&amp;\eta_{22} \end{vmatrix} &amp;\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}</span> while the second equality in \eqref{B-11} yields <span class="math-container">\begin{equation} \begin{vmatrix} \xi_{21}&amp;\eta_{21}\\ \xi_{31}&amp;\eta_{31} \end{vmatrix} E^2 \boldsymbol{+} \biggl( \begin{vmatrix} \xi_{21}&amp;\eta_{21}\\ \xi_{32}&amp;\eta_{32} \end{vmatrix} \boldsymbol{+} \begin{vmatrix} \xi_{22}&amp;\eta_{22}\\ \xi_{31}&amp;\eta_{31} \end{vmatrix} \biggr)EB \boldsymbol{+} \begin{vmatrix} \xi_{22}&amp;\eta_{22}\\ \xi_{32}&amp;\eta_{32} \end{vmatrix}B^2 \boldsymbol{=}0 \tag{B-14}\label{B-14} \end{equation}</span> where for the coefficients of <span class="math-container">$\:E^2,EB,B^2$</span> <span class="math-container">\begin{align} \begin{vmatrix} \xi_{21}&amp;\eta_{21}\\ \xi_{31}&amp;\eta_{31} \end{vmatrix} &amp;\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}&amp;\eta_{21}\\ \xi_{32}&amp;\eta_{32} \end{vmatrix} \boldsymbol{+} \begin{vmatrix} \xi_{22}&amp;\eta_{22}\\ \xi_{31}&amp;\eta_{31} \end{vmatrix} \biggr) &amp;\boldsymbol{=}\boldsymbol{-}\gamma n_1 n_3\sin\phi \tag{B-15b}\label{B-15b}\\ \begin{vmatrix} \xi_{22}&amp;\eta_{22}\\ \xi_{32}&amp;\eta_{32} \end{vmatrix} &amp;\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}</span></p> <blockquote> <p>For <span class="math-container">$\:\mathbf{E}',\mathbf{B}'\:$</span> to be parallel the four (4) unknown variables <span class="math-container">$\:\upsilon,n_1,n_2,n_3\:$</span> must satisfy simultaneously the system of three (3) equations \eqref{A-01b}, \eqref{B-12} and \eqref{B-14}. </p> </blockquote> <p>First note that the variable <span class="math-container">$\:\upsilon\:$</span> is a real number in <span class="math-container">$\:\left(-c,c\right)\:$</span> and not the non-negative magnitude of the velocity <span class="math-container">$\:\boldsymbol{\upsilon}$</span>, see equation \eqref{A-01a}. Second note that if a quadruple <span class="math-container">$\:\left(\upsilon,n_1,n_2,n_3\right)\:$</span> satisfies above system then the quadruple <span class="math-container">$\:\left(\boldsymbol{-}\upsilon,\boldsymbol{-}n_1,\boldsymbol{-}n_2,\boldsymbol{-}n_3\right)\:$</span> satisfies also this system. But these two quadruples represent the same velocity <span class="math-container">\begin{equation} \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} \end{equation}</span> We must take care about this in order to avoid double counting of the solutions.</p> <hr> <p><strong>SECTION C : A first solution</strong></p> <p>Looking carefully in equation \eqref{B-14} we note that if <span class="math-container">\begin{equation} n_1\boldsymbol{=}0\,, n_2\boldsymbol{=}0 \tag{C-01}\label{C-01} \end{equation}</span> then its coefficients of <span class="math-container">$\:E^2,EB,B^2\:$</span> 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 <span class="math-container">\begin{equation} n^2_3\boldsymbol{=}1 \quad \boldsymbol{\Longrightarrow} \quad n_3\boldsymbol{=\pm} 1 \tag{C-02}\label{C-02} \end{equation}</span> In case that<br> <span class="math-container">\begin{equation} n_1\boldsymbol{=}0\,, n_2\boldsymbol{=}0\,, n_3\boldsymbol{=+} 1 \tag{C-03}\label{C-03} \end{equation}</span> equations \eqref{B-06a},\eqref{B-06b} yield </p> <p><span class="math-container">\begin{align} \mathbf{E}' &amp; \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}'&amp; \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}</span> and equation \eqref{B-12}, with its coefficients determined from equations \eqref{B-13a}-\eqref{B-13b}-\eqref{B-13c}, yields <span class="math-container">\begin{equation} \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} \end{equation}</span> or <span class="math-container">\begin{equation} \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} \end{equation}</span> This last equation is of course the condition of parallelism of the vectors <span class="math-container">$\:\mathbf{E}',\mathbf{B}'\:$</span> given by equations \eqref{C-04a},\eqref{C-04b} <span class="math-container">\begin{align} \dfrac{\gamma E\boldsymbol{-}\gamma\upsilon \sin\phi B}{\gamma\cos\phi B}&amp;\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}}&amp; \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}</span> so that <span class="math-container">\begin{equation} \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} \end{equation}</span> But now we must check if <span class="math-container">$\:\upsilon\:$</span> given implicitly by equation \eqref{C-07} is in the range <span class="math-container">$\:\left(\boldsymbol{-}c,\boldsymbol{+}c\right)$</span>. </p> <p>Note that \eqref{C-07} is a quadratic equation with respect to <span class="math-container">$\:\upsilon/c$</span> <span class="math-container">\begin{equation} \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} \end{equation}</span> with real roots <span class="math-container">\begin{equation} \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} \end{equation}</span> Since their sum is positive and their product is 1 <span class="math-container">\begin{equation} \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)&gt;0\,, \quad \left(\dfrac{\upsilon}{c}\right)_{\boldsymbol{+}}\cdot\left(\dfrac{\upsilon}{c}\right)_{\boldsymbol{-}}\boldsymbol{=}1 \tag{C-11}\label{C-11} \end{equation}</span> acceptable is the root <span class="math-container">\begin{equation} 0&lt;\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} &lt;1 \tag{C-12}\label{C-12} \end{equation}</span> Now it could be proved that the case <span class="math-container">\begin{equation} n_1\boldsymbol{=}0\,, n_2\boldsymbol{=}0\,, n_3\boldsymbol{=-} 1 \tag{C-13}\label{C-13} %\tag{B-29}\label{B-29} \end{equation}</span> gives as result the negative of \eqref{C-12} so the solution is the same, see equation \eqref{B-16} to avoid double counting. </p> <p><strong>SECTION D : A continuous 1-parametric set of solutions</strong></p> <p>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. </p> <p>As shown in <strong>SECTION C</strong> in a system <span class="math-container">$\:\mathrm S'\:$</span> moving uniformly along the <span class="math-container">$\:x_3$</span>-axis of system <span class="math-container">$\:\mathrm S\:$</span> with velocity <span class="math-container">$\:\boldsymbol{\upsilon}$</span>, see equations \eqref{C-08} and \eqref{C-12}, the new vectors <span class="math-container">$\:\mathbf E'\:$</span> and <span class="math-container">$\:\mathbf B'\:$</span> are parallel. Let <span class="math-container">$\:\mathbf m\boldsymbol{=}\left(m_1,m_2,\cancelto{0}{m_3}\right)\:$</span> the unit vector along the common axis of these two vectors. As pointed out in related answers and comments, if in system <span class="math-container">$\:\mathrm S'\:$</span> we have a frame <span class="math-container">$\: \Sigma\:$</span> moving along the unit vector <span class="math-container">$\:\mathbf m$</span>, that is along the common axis of <span class="math-container">$\:\mathbf E'\:$</span> and <span class="math-container">$\:\mathbf B'$</span>, with ANY SUBLUMINAL constant velocity<br> <span class="math-container">\begin{equation} \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} \end{equation}</span> see Figure-03, then the new vectors <span class="math-container">$\:\mathbf E^{^{\boldsymbol \Sigma}}\:$</span> and <span class="math-container">$\:\mathbf B^{^{\boldsymbol \Sigma}}$</span> remain parallel.</p> <p><a href="https://i.stack.imgur.com/Wp6wb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Wp6wb.png" alt="enter image description here"></a></p> <p>So, if we compose the two Lorentz transformations <span class="math-container">\begin{equation} \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} \end{equation}</span> then we'll have a parametric set of solutions with parameter the continuous variable <span class="math-container">$\:u\in \left(-c,c\right)$</span>.</p> <p>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.</p> <p>To avoid rotations we use a system <span class="math-container">$\:\mathrm S''\:$</span> moving with constant velocity <span class="math-container">$\:\mathbf w\:$</span> with respect to <span class="math-container">$\:\mathrm S$</span>, see Figure-04. The velocity <span class="math-container">$\:\mathbf w\:$</span> is the velocity <span class="math-container">$\:\mathbf u\:$</span> as seen from <span class="math-container">$\:\mathrm S$</span>, that is the relativistic addition of the perpendicular velocities <span class="math-container">$\:\boldsymbol \upsilon\:$</span> and <span class="math-container">$\:\mathbf u$</span> <span class="math-container">\begin{equation} \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} \end{equation}</span> where the subscript <span class="math-container">$\boldsymbol{\upsilon}$</span> in <span class="math-container">$\gamma_{\boldsymbol \upsilon}$</span> is inserted to remind us that we talk about that of equation \eqref{A-02c}.</p> <p><a href="https://i.stack.imgur.com/hQLOW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hQLOW.png" alt="enter image description here"></a></p> <p>So, instead of the composition of Lorentz transformations \eqref{D-02} we make use of one Lorentz transformation <span class="math-container">\begin{equation} \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} \end{equation}</span></p> <blockquote> <p>So, the tips of the velocity vectors <span class="math-container">$\:\mathbf w\:$</span> is a straight segment <span class="math-container">$\rm AA'$</span>, a 1-parameter set of solutions. The parameter is <span class="math-container">$\:u\:$</span> and runs along this segment, see Figure-05 below.</p> </blockquote> <p><a href="https://i.stack.imgur.com/hQ1xC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hQ1xC.png" alt="enter image description here"></a></p> <hr> <p><strong>SECTION E : Edit</strong></p> <p>From \eqref{C-08} and \eqref{C-12} for the velocity vector we have <span class="math-container">\begin{equation} \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} \end{equation}</span> For its magnitude <span class="math-container">\begin{equation} \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} \end{equation}</span> Note that since <span class="math-container">\begin{align} \mathcal{E} &amp; \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}} &amp; \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}</span> equation \eqref{E-02} for the magnitude of the velocity vector yields <span class="math-container">\begin{equation} \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} \end{equation}</span></p> <hr> <p>Suppose that on a point in an inertial system <span class="math-container">$\:\mathrm S\:$</span> the vectors <span class="math-container">$\:\mathbf{E},\mathbf{B}\:$</span> are non-parallel, non-perpendicular and non-zero, see Figure-02. If by a Lorentz transformation in the new inertial system <span class="math-container">$\:\mathrm S'\:$</span> the vectors <span class="math-container">$\:\mathbf{E'},\mathbf{B'}\:$</span> are parallel then their magnitudes <span class="math-container">$\:E',B'\:$</span> are fixed and could be determined through the two invariants of the electromagnetic field <span class="math-container">\begin{align} E\,'^{\,2}\boldsymbol{-}c^2 B\,'^{\,2} &amp; \boldsymbol{=} E^2\boldsymbol{-}c^2 B^2 \tag{E-05a}\label{E-05a}\\ E\,'B\,' &amp;\boldsymbol{=}\mathbf E\cdot \mathbf B \boldsymbol{=}EB\cos\phi \tag{E-05b}\label{E-05b} \end{align}</span> Solving with respect to <span class="math-container">$\:E\,',B\,'\:$</span> we have <span class="math-container">\begin{align} E\,'^{\,2}&amp;\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} &amp; \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}</span> Adding \eqref{E-06a}, \eqref{E-06b} we find the new energy density <span class="math-container">$\:\mathcal{E'}\:$</span> of the electromagnetic field <span class="math-container">\begin{equation} \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} \end{equation}</span> expressed also as <span class="math-container">\begin{equation} \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} \end{equation}</span></p> <blockquote> <p>The energy density <span class="math-container">$\:\mathcal{E'}\:$</span> and the magnitudes <span class="math-container">$\:E',B'\:$</span> are the same in any system <span class="math-container">$\:\mathrm S'\:$</span> with parallel <span class="math-container">$\:\mathbf{E'},\mathbf{B'}$</span>, so invariants in a strict sense.</p> </blockquote>