Exact finite-frequency transverse Maxwell wave

For $\theta =\kappa (x^0-z)$ and $a_0=A_{\ast }dx$, $k=\kappa (d{x}^0-dz)$ is nowhere zero and $g^{-1}(k,k)=-{\kappa }^2+{\kappa }^2=0$, while $g^{-1}(k,a_0)=0$. In the stated inertial coordinates $a_0$ and $k$ have constant components and Minkowski curvature is zero, so $L{a}_0=0$ and $P{a}_0=0$; also ${\mathrm{\nabla }}^a(a_0{)}_a=0$. Since $a_1=0$, both $L{a}_1=0$ and $i{k}^a(a_1{)}_a=0$. These verify all four constraints. On each compact coordinate set the coefficient bounds of [F3] hold with $K_0=\kappa$ and $D_0=C_0=B_0=B_1=0$. The inherited ${\mathrm{A}\mathrm{C}}_{\omega }$ hypothesis in [A2] matches the exact mathematical supplier framework [F5]. Finally $a_0$ is not proportional to $k$ because $A_{\ast }\mathrm{\ne }0$ and $k$ has no $dx$ component, so its transverse polarization class is nonzero. Hence [F3] supplies the affine null rays and that nonzero polarization class for this ansatz.

Write $\psi =\kappa (x^0-z)\mathrm{/}ϵ$. Differentiating the real potential in [A1] gives $F_{0x}=-\kappa A_{\ast }\sin \psi$ and $F_{zx}=\kappa A_{\ast }\sin \psi$, with all other independent components zero. To check $dF=0$ directly, its only possibly nonzero three-form component is $(dF{)}_{0zx}={\mathrm{\partial }}_0{F}_{zx}+{\mathrm{\partial }}_z{F}_{x0}$; these terms are ${\kappa }^2{A}_{\ast }{ϵ}^{-1}\cos \psi$ and $-{\kappa }^2{A}_{\ast }{ϵ}^{-1}\cos \psi$, so they cancel. Raising indices with $\eta =\mathrm{diag}(-1,1,1,1)$ gives $F^{0x}=F^{zx}=\kappa A_{\ast }\sin \psi$. Thus ${\mathrm{\partial }}_a{F}^{ax}={\mathrm{\partial }}_0{F}^{0x}+{\mathrm{\partial }}_z{F}^{zx}=0$ by the same cancellation. For the other divergence components, ${\mathrm{\partial }}_a{F}^{a0}={\mathrm{\partial }}_x{F}^{x0}=0$, ${\mathrm{\partial }}_a{F}^{ay}=0$, and ${\mathrm{\partial }}_a{F}^{az}={\mathrm{\partial }}_x{F}^{xz}=0$, since the field is independent of $x,y$. This proves $d{F}_{ϵ}=0$ and the divergence-form equation. To verify the source-free form equation directly, put $f=\kappa A_{\ast }\sin \psi$. The existing Hodge formula [F2], with the stated orientation and raised components, gives $({\star }_g{F}_{ϵ}{)}_{02}=f$ and $({\star }_g{F}_{ϵ}{)}_{23}=f$, all other independent components zero; hence ${\star }_g{F}_{ϵ}=f(d{x}^0-dz)\wedge dy$. Moreover $df=({\kappa }^2{A}_{\ast }\mathrm{/}ϵ)\cos \psi (d{x}^0-dz)$. The coordinate exterior derivative therefore gives $d({\star }_g{F}_{ϵ})=df\wedge (d{x}^0-dz)\wedge dy=0$, since the repeated covector wedges to zero. Thus both differential-form Maxwell equations hold, without appealing to a form/divergence equivalence. From [F1], $E^x=c{F}^x{}_0=c{F}_{x0}=c\kappa A_{\ast }\sin \psi$. Also $B_y=-({\star }_{g}F{)}_{y0}$ by [F1]. The Hodge formula in [F2], $({\mathrm{v}\mathrm{o}\mathrm{l}}_g{)}_{y0xz}=+1$, and $F^{xz}=F_{xz}=-\kappa A_{\ast }\sin \psi$ give $({\star }_{g}F{)}_{y0}=F^{xz}=-\kappa A_{\ast }\sin \psi$, hence $B_y=\kappa A_{\ast }\sin \psi$. All other observer components vanish because the only field components are the stated $0x$ and $zx$ components.

Substitution of the components from step 1.2 into the tensor formula of [F1] gives $T_{\mathrm{E}\mathrm{M}}^{00}=(E_x^2\mathrm{/}{c}^2+B_y^2)\mathrm{/}(2{\mu }_0)=({\epsilon }_0{E}_x^2+B_y^2\mathrm{/}{\mu }_0)\mathrm{/}2$. Because $E_x=c{B}_y$, this is $u=B_y^2\mathrm{/}{\mu }_0>0$ whenever the sine is nonzero.

Minkowski has $G_{ab}=0$, so the source-free coupled Einstein equation with cosmological constant $\mathrm{\Lambda }=0$ on the unchanged metric would require zero Einstein residual. Apply [F4] to the source-free field of step 1.2 by taking its reference field to be $F_0=F_{ϵ}$ and scale $\delta =1$; since the field is nonzero wherever the sine is nonzero, the residual is nonzero there. This excludes that exact coupled solution while retaining the exact Maxwell test-field solution of step 1.2.