fields, they are mutually orthogonal; besides, the magnetic field is transverse, ( h ( 1 $^{)}$k ) = 0, while the electric field, generally, is not: ( e ( 1 $^{)}$k ) = 0; it is transverse for the special case of propagation along the external fields k$_{⊥}$ = 0. It is seen again that a massless excitation is possible in mode 1 as long as it propagates along the external field, otherwise the fields (32) vanish if k 2 = 0, unless k$_{⊥}$ = 0, when the square root turns to unity. The electric and magnetic fields carried by the vectors ( ˜ Fk )$_{µ}$ and ( Fk )$_{µ}$, respectively, are The electric fields carried by the vectors c + µ and c − $_{µ}$, respectively, are while their magnetic fields are The orthogonality of the electric fields ( e $^{+}$e $^{−}$) = ( e $^{(1)}$e $^{−}$) = 0 , seen in eqs. (32), (35), (36), originates from the fact that, in the special frame, the time-component of one of their mutually orthogonal ( c $^{+}$c − = 0), and 4-transversal ( c $^{±}$k = 0) vector-potentials c ± µ disappears: c − 0 = 0. The 3-vector e − is directed along the axis (call it axis 1 ), orthogonal to the plane, where the external fields and the propagation vector k lie (the plane (3,2) ). The vector e + lies in that plane. It makes the universal angle α = arctan( k$_{3}$k$_{⊥}$/ ( k 2 0 − k 2 $_{3}$)) with the direction of the external fields 3 . Also the magnetic field h − lies in the plane spanned by the external fields and the propagation direction, while h + is orthogonal to this plane. The electric fields in the eigenmodes 2 and 3, e (2 , $^{3)}$, lie both in the common plane spanned by the vectors e $^{±}$: 19 e (2 , 3) µ = − 2Λ$_{3}$ e − µ + [ Λ$_{2}$ − Λ$_{4}$ ± √ (Λ$_{2}$ − Λ$_{4}$) 2 + 4Λ 2 3 ] e + $_{µ}$. (38) h + = [k$_{⊥}$ × ϵ ] k$_{0}$ √ k 2 3 − k 2 0 , h − = − k$_{⊥}$ k$_{3}$ k$_{⊥}$ + ϵ k$_{⊥}$, ( h $^{±}$k ) = ( h $^{+}$h $^{−}$) = ( h $^{±}$e$^{±}$) = 0 . (37) e − = [k$_{⊥}$ × ϵ ] k$_{0}$ √ ( B 2 + E $^{2}$) ( k $^{2}$E2 − kF $^{2}$k ) = [k$_{⊥}$ × ϵ ] k$_{0}$ | k$_{⊥}$ | , ( e $^{−}$k ) = 0 , (36) e + = { k$_{⊥}$ k$_{3}$ + ϵ ( k 2 3 − k 2 $_{0}$) } √ ( B 2 + E $^{2}$) ( k $^{2}$B2 + kF $^{2}$k ) = k$_{⊥}$ k$_{3}$ √ k 2 3 − k 2 0 + ϵ √ k 2 3 − k 2 $_{0}$, ( e $^{+}$e $^{−}$) = 0 , ( e $^{+}$k ) = 0 , (35) e ( ˜ F ) = − k$_{⊥}$ Bk$_{3}$ + [k$_{⊥}$ × ϵ ] i k$_{4}$E − B ( k 2 4 + k 2 $_{3}$) , h ( ˜ F ) = − k$_{⊥}$ Ek$_{3}$ − [k$_{⊥}$ × ϵ ] Hk$_{0}$ + ϵ Ek 2 $_{⊥}$, (33) e ( F ) = k$_{⊥}$ Ek$_{3}$ + [k$_{⊥}$ × ϵ ] i k$_{4}$B + E ( k 2 4 + k 2 $_{3}$) , h ( F ) = − k$_{⊥}$ Bk$_{3}$ + [k$_{⊥}$ × ϵ ] Ek$_{0}$ + ϵ Bk 2 $_{⊥}$. (34)