In a certain inertial frame S, the electric field E and the magnetic field B are neither parallel nor perpendicular, at a particular space-time point. Show that in a different inertial system \overline{ S }, moving relative to S with velocity v given by
\frac{ v }{1+v^{2} / c^{2}}=\frac{ E \times B }{B^{2}+E^{2} / c^{2}},
the fields \overline{ E } \text { and } \overline{ B } are parallel at that point. Is there a frame in which the two are perpendicular?
Choose axes so that E points in the z direction and B in the yz plane: E = (0, 0, E); B =(0, B \cos \phi, B \sin \phi).
Go to a frame moving at speed v in the x direction:
\overline{ E }=(0,-\gamma v B \sin \phi, \gamma(E+v B \cos \phi)) ; \overline{ B }\left(0, \gamma\left(B \cos \phi+\frac{v}{c^{2}} E\right), \gamma B \sin \phi\right).
(I used Eq. 12.109.) Parallel provided \frac{-\gamma v B \sin \phi}{\gamma\left(B \cos \phi+\frac{v}{c^{2}} E\right)}=\frac{\gamma(E+v B \cos \phi)}{\gamma B \sin \phi} , or
-v B^{2} \sin ^{2} \phi=\left(B \cos \phi+\frac{v}{c^{2}} E\right)(E+v B \cos \phi)=E B \cos \phi+v B^{2} \cos ^{2} \phi+\frac{v}{c^{2}} E^{2}+\frac{v^{2}}{c^{2}} E B \cos \phi0=v B^{2}+\frac{v}{c^{2}} E^{2}+E B \cos \phi\left(1+\frac{v^{2}}{c^{2}}\right) ; \frac{v}{1+v^{2} / c^{2}}=-\frac{E B \cos \phi}{B^{2}+E^{2} / c^{2}}
NowE \times B =\left|\begin{array}{ccc}\hat{ x } & \hat{ y } & \hat{ z } \\0 & 0 & E \\0 & B \cos \phi & B \sin \phi\end{array}\right|=-E B \cos \phi . \text { So } \frac{ v }{1+v^{2} / c^{2}}=\frac{ E \times B }{B^{2}+E^{2} / c^{2}} . qed
\begin{array}{ll}\bar{E}_{x}=E_{x}, \quad \bar{E}_{y}=\gamma\left(E_{y}-v B_{z}\right), & \bar{E}_{z}=\gamma\left(E_{z}+v B_{y}\right) \\\bar{B}_{x}=B_{x}, \quad \bar{B}_{y}=\gamma\left(B_{y}+\frac{v}{c^{2}} E_{z}\right), & \bar{B}_{z}=\gamma\left(B_{z}-\frac{v}{c^{2}} E_{y}\right)\end{array} (12.109)
No, there can be no frame in which E \perp B, for (E·B) is invariant, and since it is not zero in S it can’t be zero in \overline{ S }.