Finite WKB Maxwell residual, null rays and polarization quotient — Examples

The phase gradient is $k=\nu (d{x}^0-dz)$ and $g^{-1}(k,k)={\nu }^2(-1+1)=0$. The constant $a_0$ obeys $g^{-1}(k,a_0)=0$ and $L{a}_0=0$; $a_1=0$ makes $iL{a}_1+P{a}_0=0$ and $i{k}^a(a_1{)}_a+{\mathrm{\nabla }}^a(a_0{)}_a=0$. Here $\mathrm{\nabla }$ is the flat connection and the Ricci tensor vanishes. Thus all four hypotheses of [F2] hold.

Differentiating $A_{ϵ}^{\mathbb{R}}$ in the displayed coordinates gives $(F_{ϵ}{)}_{01}={\mathrm{\partial }}_0(A_{ϵ}^{\mathbb{R}}{)}_1=-\nu A_{\ast }\sin (\theta \mathrm{/}ϵ)$ and $(F_{ϵ}{)}_{31}={\mathrm{\partial }}_z(A_{ϵ}^{\mathbb{R}}{)}_1=+\nu A_{\ast }\sin (\theta \mathrm{/}ϵ)$; no other independent component is produced. The SI component convention in [F1] therefore gives the displayed $E_1$ and $B_2$.

In [F2], this example has $a_1=0$, hence $d_1=0$ and $P{a}_1-\mathrm{\nabla }{d}_1=0$. Its exact residual formula therefore gives ${\mathrm{\nabla }}^a(F_{ϵ}{)}_{ab}=0$, and its exact closure identity gives $d{F}_{ϵ}=0$. Raising the free index in the first equality gives ${\mathrm{\nabla }}_a{F}_{ϵ}^{ab}=0$. To check the other source-free Maxwell equation directly, put $f:=\nu A_{\ast }\sin (\theta \mathrm{/}ϵ)$. The displayed components give $F_{ϵ}=-fd{x}^0\wedge d{x}^1+fdz\wedge d{x}^1$. From the metric and orientation in the Example, [F4] yields $\star F_{ϵ}=f(d{x}^0-dz)\wedge d{x}^2$. Since $df$ is a scalar multiple of $d\theta =\nu (d{x}^0-dz)$, $d(\star F_{ϵ})=df\wedge (d{x}^0-dz)\wedge d{x}^2=0$. Therefore $d{F}_{ϵ}=0$ and $d(\star F_{ϵ})=0$ hold exactly for every fixed $ϵ$, as required by [F3].

Since $\nu A_{\ast }\mathrm{\ne }0$, there are points with $\sin (\theta \mathrm{/}ϵ)\mathrm{\ne }0$. At any such point the field is nonzero and [F3] gives positive electromagnetic observer energy. Minkowski has $G_{ab}=0$, so the source-free Einstein equation with unchanged metric and $\mathrm{\Lambda }=0$ fails there. This caveat concerns coupling only; the Maxwell calculation in step 2.1 remains exact.

In Minkowski coordinates $U=(\gamma c,\gamma \mathbf{w})$, where $\gamma =(1-\mathrm{\mid }\mathbf{w}{\mathrm{\mid }}_2^2\mathrm{/}{c}^2{)}^{-1\mathrm{/}2}>0$. Since $U^0=d{x}^0\mathrm{/}d\tau =cdt\mathrm{/}d\tau =\gamma c$, we have $d\mathrm{/}d\tau =\gamma d\mathrm{/}dt$. With $\mathbf{E}=0$, the time component of ${\mathrm{\nabla }}_{U}U=\alpha F^{\mathrm{♯}}(U)$ gives $\gamma cd\gamma \mathrm{/}dt=0$, hence $d\gamma \mathrm{/}dt=0$. The spatial component gives $\gamma d(\gamma \mathbf{w})\mathrm{/}dt=\alpha \gamma \mathbf{w}\times \mathbf{B}$; canceling $\gamma >0$ and then using constancy of $\gamma$ yields $d\mathbf{w}\mathrm{/}dt=(\alpha \mathrm{/}\gamma )\mathbf{w}\times \mathbf{B}$. The dot product with $\mathbf{w}$ is zero, so $\mathrm{\mid }\mathbf{w}{\mathrm{\mid }}_2$ and $\gamma$ stay constant. Initially, $\gamma ={\gamma }_0=(1-ϵ\mathrm{\mid }{\mathbf{W}}_0{\mathrm{\mid }}_2^2{)}^{-1\mathrm{/}2},{\gamma }_0^{-1}=\sqrt{1-ϵ\mathrm{\mid }{\mathbf{W}}_0{\mathrm{\mid }}_2^2}\mathrm{.}$ The radicand is at least $1\mathrm{/}2$, so the positive root is defined.

Rescaling time and velocity in the exact equation gives ${\mathbf{W}}^{\mathrm{\prime }}=(b_0\mathrm{/}{\gamma }_0)\mathbf{W}\times {\mathbf{e}}_z$. Differentiating the planar rotation matrix with [F3] verifies the solution $\mathbf{W}(S)=R_z(-b_{0}S\mathrm{/}{\gamma }_0){\mathbf{W}}_0\mathrm{.}$ The comparison equation in the Statement similarly has the solution ${\mathbf{W}}_N(S)=R_z(-b_{0}S){\mathbf{W}}_0$, with ${\mathbf{X}}^{\mathrm{\prime }}=\mathbf{W}$ and ${\mathbf{X}}_N^{\mathrm{\prime }}={\mathbf{W}}_N$. These rotations preserve speed, and $ϵ\mathrm{\mid }{\mathbf{W}}_0{\mathrm{\mid }}_2^2\le 1\mathrm{/}2$ keeps the exact speed strictly below $c$. Integrate $\mathbf{w}(t)=\sqrt{ϵ}c\mathbf{W}(S(t))$ to define $\mathbf{x}(t)$ and set $\tau =t\mathrm{/}{\gamma }_0$. The resulting curve is timelike for every finite $t$. Substitution gives ${\mathrm{\nabla }}_U{U}^0=0=\alpha F^0{}_{\nu }{U}^{\nu }$ and ${\mathrm{\nabla }}_U{U}^i={\gamma }_0^{2}d{w}^i\mathrm{/}dt=\alpha {\gamma }_0(\mathbf{w}\times \mathbf{B}{)}^i=\alpha F^i{}_{\nu }{U}^{\nu }$, so it is an exact worldline. The matrix $R_z(\theta )$ is orthogonal on the $xy$-plane by ${\sin }^2\theta +{\cos }^2\theta =1$, and the addition formulas in [F3] give $R_z({\theta }_1)R_z({\theta }_2)=R_z({\theta }_1+{\theta }_2)$.

Put $\mathrm{\Delta }\theta =b_{0}S(1-{\gamma }_0^{-1})$. Orthogonality and the rotation addition law reduce the velocity difference to $\mathrm{\mid }{R}_z(\mathrm{\Delta }\theta ){\mathbf{W}}_0-{\mathbf{W}}_0{\mathrm{\mid }}_2$. Expanding its two planar components and cancelling the cross terms gives $\mathrm{\mid }{R}_z(\mathrm{\Delta }\theta ){\mathbf{W}}_0-{\mathbf{W}}_0{\mathrm{\mid }}_2^2=2\mathrm{\mid }{\mathbf{W}}_0{\mathrm{\mid }}_2^2(1-\cos \mathrm{\Delta }\theta )=4\mathrm{\mid }{\mathbf{W}}_0{\mathrm{\mid }}_2^2{\sin }^2(\mathrm{\Delta }\theta \mathrm{/}2)\mathrm{.}$ By [F5], its square root is at most $\mathrm{\mid }{\mathbf{W}}_0{\mathrm{\mid }}_2\mathrm{\mid }\mathrm{\Delta }\theta \mathrm{\mid }$. Also $1-{\gamma }_0^{-1}=\frac{ϵ\mathrm{\mid }{\mathbf{W}}_0{\mathrm{\mid }}_2^2}{1+\sqrt{1-ϵ\mathrm{\mid }{\mathbf{W}}_0{\mathrm{\mid }}_2^2}}\le ϵ\mathrm{\mid }{\mathbf{W}}_0{\mathrm{\mid }}_2^2\mathrm{.}$ This proves the stated velocity bound.

At $S=0$, the matched initial positions give $\mathrm{\mid }\mathbf{X}(0)-{\mathbf{X}}_N(0){\mathrm{\mid }}_2=0$, which is the claimed bound. Now let $S>0$. The position difference is the integral of $\mathbf{W}-{\mathbf{W}}_N$. For any continuous vector function $v$ on $[0,S]$, Cauchy–Schwarz gives $\mathrm{\mid }u\cdot v(s)\mathrm{\mid }\le \mathrm{\mid }u{\mathrm{\mid }}_2\mathrm{\mid }v(s){\mathrm{\mid }}_2$; taking $u$ to be the unit vector in the direction of ${\int }_0^{S}v$ shows $\mathrm{\mid }{\int }_0^{S}v(s)ds{\mathrm{\mid }}_2\le {\int }_0^S\mathrm{\mid }v(s){\mathrm{\mid }}_{2}ds$ (the zero-integral case is immediate). Thus the velocity bound from step 3.1 gives $\mathrm{\mid }\mathbf{X}(S)-{\mathbf{X}}_N(S){\mathrm{\mid }}_2\le {\int }_0^S\mathrm{\mid }{b}_0\mathrm{\mid }ϵ\mathrm{\mid }{\mathbf{W}}_0{\mathrm{\mid }}_2^{3}sds$. Since $S>0$, the function $s\mapsto s$ is continuous and integrable on the nondegenerate interval $[0,S]$, and $s^2\mathrm{/}2$ is a primitive; [F6] therefore gives ${\int }_0^{S}sds=S^2\mathrm{/}2$. This proves the claimed position bound. The estimates hold on every specified finite interval; no uniform-in-time bound is asserted.

Put $B_z:=\sqrt{ϵ}c{b}_0\mathrm{/}(\alpha L)$. By the SI convention of [F1], the prescribed two-form is $F=B_{z}d{x}^1\wedge d{x}^2$; [F7] gives $\star F=B_{z}d{x}^0\wedge dz$. Both forms have constant coefficients in Minkowski coordinates, so $dF=0$, ${\mathrm{\nabla }}_a{F}^{ab}={\mathrm{\partial }}_a{F}^{ab}=0$, and $d(\star F)=0$. Also $B_z\mathrm{\ne }0$ because $b_0\mathrm{\ne }0$ and $\alpha \mathrm{\ne }0$. Thus the fixed Minkowski metric, $\mathrm{\Lambda }=0$, and $F_0=F$ satisfy the source-free hypotheses of [F2]; taking $\delta =1$ there gives positive observer energy and a nonzero residual. Hence the prescribed orbit background is not a fully coupled flat Einstein–Maxwell solution with unchanged Minkowski metric. The example establishes only the conditional trajectory calculation and its finite-time comparison.