Separately adopted fixed-background Maxwell and charged probes — Examples

Take the candidate expression of [F2], $C=mg(U,\xi )+qA(\xi )=m{U}_0+q{A}_0$, with $\xi ={\mathrm{\partial }}_0$ and $\chi =0$. The actual $C^2$ potential suffices for the following direct calculation. In Cartesian Minkowski coordinates the connection coefficients vanish; by [F1], the supplied $C^2$ curve obeys the Lorentz law, whose lowered time component is $md{U}_0\mathrm{/}d\tau =q{F}_{0i}{U}^i$. Since $A_0=-\varphi \mathrm{/}c$, $A_i=0$ and ${\mathrm{\partial }}_0\varphi =0$, exterior differentiation gives $F_{0i}=-{\mathrm{\partial }}_i{A}_0={\mathrm{\partial }}_i\varphi \mathrm{/}c$. The chain rule therefore yields $dC\mathrm{/}d\tau =q{F}_{0i}{U}^i+q{\mathrm{\partial }}_i{A}_0{U}^i=0$ at every interior parameter value. A real differentiable function with zero derivative is constant on each compact interior subinterval by the mean-value theorem; continuity extends the same value to the included endpoints of the finite segment. This proves constancy at the stated $C^2$ regularity without a smooth-potential application of [F2].

Since ${\mathrm{\partial }}_0$ is future-directed, $U^0>0$. The relation $U^i\mathrm{/}{U}^0=w^i\mathrm{/}c$ and $g(U,U)=-c^2$ imply $U^0=c(1-\mathrm{\mid }\mathbf{w}{\mathrm{\mid }}^2\mathrm{/}{c}^2{)}^{-1\mathrm{/}2}=\gamma c$, hence $U_0=-\gamma c$. Also $A(\xi )=A_0=-\varphi \mathrm{/}c$, so $C=-\gamma mc-q\varphi \mathrm{/}c=-E_{\mathrm{t}\mathrm{e}\mathrm{s}\mathrm{t}}\mathrm{/}c$. Since $c>0$ is constant, the displayed test energy is constant.

For an explicit work check, $F_{0i}=-E_i\mathrm{/}c$ and $A_0=-\varphi \mathrm{/}c$ give $\mathbf{E}=-\mathrm{\nabla }\varphi$ and $\mathbf{B}=0$, so the flat Lorentz law is $d(\gamma m\mathbf{w})\mathrm{/}dt=q\mathbf{E}$. Writing $\mathbf{a}=d\mathbf{w}\mathrm{/}dt$, differentiation gives $d\gamma \mathrm{/}dt={\gamma }^3\mathbf{w}\cdot \mathbf{a}\mathrm{/}{c}^2$ and $d(\gamma m\mathbf{w})\mathrm{/}dt=m\gamma \mathbf{a}+m{\gamma }^3(\mathbf{w}\cdot \mathbf{a})\mathbf{w}\mathrm{/}{c}^2$; hence $\mathbf{w}\cdot d(\gamma m\mathbf{w})\mathrm{/}dt=m{\gamma }^3\mathbf{w}\cdot \mathbf{a}=d(\gamma m{c}^2)\mathrm{/}dt$. Thus $d(\gamma m{c}^2)\mathrm{/}dt=q\mathbf{E}\cdot \mathbf{w}$, whereas $d(q\varphi )\mathrm{/}dt=q\mathrm{\nabla }\varphi \cdot \mathbf{w}=-q\mathbf{E}\cdot \mathbf{w}$. The two work terms cancel, agreeing with step 2.1.

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.

Put $B_z=\sqrt{ϵ}c{b}_0\mathrm{/}(\alpha L)$ and supply the real one-form $A=(B_z\mathrm{/}2)(x^{1}d{x}^2-x^{2}d{x}^1)$ on all of Minkowski space. Its coefficients are smooth and have SI units tesla metres, and direct differentiation gives $dA=B_{z}d{x}^1\wedge d{x}^2=F$, with $(dA{)}_{0i}=0$ and $(dA{)}_{12}=B_z$. Thus the local-potential hypothesis of [F6] is met on every finite curve segment, and the action model gives its proper-time Lorentz law throughout the Example's displayed range, independently of the general small-cutoff comparison in [F1]. In flat coordinates this law has $U^0=\gamma c$, $U^i=\gamma w^i$, and $\mathrm{\nabla }=\mathrm{\partial }$. With $E=0$ and $F_{ij}={ϵ}_{ijk}{B}_k$, its spatial components reduce to $d(\gamma m\mathbf{w})\mathrm{/}dt=q\mathbf{w}\times \mathbf{B}$. Its dot product with $\mathbf{w}$ is zero, so $\mathrm{\mid }\mathbf{w}\mathrm{\mid }$ and $\gamma =(1-\mathrm{\mid }\mathbf{w}{\mathrm{\mid }}^2\mathrm{/}{c}^2{)}^{-1\mathrm{/}2}=(1-ϵ\mathrm{\mid }{\mathbf{W}}_0{\mathrm{\mid }}^2{)}^{-1\mathrm{/}2}$ are constant. Since ${\mathbf{W}}_0$ is transverse to ${\mathbf{e}}_z$, the motion stays in that transverse plane. Define $J\mathbf{v}:=\mathbf{v}\times b_0{\mathbf{e}}_z$ there; $J$ is skew-symmetric and $J^2=-b_0^{2}I$. Identify the transverse plane with the real Euclidean normed space ${\mathbb{R}}^2$ using the ordered coordinate basis $({\mathbf{e}}_x,{\mathbf{e}}_y)$; it is Banach by [F3], and $\mathrm{\mid }Jv\mathrm{\mid }=\mathrm{\mid }{b}_0\mathrm{\mid }\mathrm{\mid }v\mathrm{\mid }$ proves that $J$ is bounded with $\mathrm{\parallel }J{\mathrm{\parallel }}_{2\to 2}=\mathrm{\mid }{b}_0\mathrm{\mid }$. Define $E(t):={\sum }_{n=0}^{\mathrm{\infty }}{t}^n{J}^n\mathrm{/}n!$ by [F3], and write $e^{tJ}:=E(t)$. Rescaling the exact force equation gives ${\mathbf{W}}^{\mathrm{\prime }}=J\mathbf{W}\mathrm{/}\gamma$ and ${\mathbf{X}}^{\mathrm{\prime }}=\mathbf{W}$. The formula $\mathbf{W}(S)=E(S\mathrm{/}\gamma ){\mathbf{W}}_0$ has the matched initial velocity and satisfies the velocity equation by the derivative law in [F3]; the Newton–Lorentz comparison is ${\mathbf{W}}_N(S)=E(S){\mathbf{W}}_0$. For either constant generator $aJ$ with $a={\gamma }^{-1}$ or $a=1$, any solution with this initial value is unique: differentiating $E(-aS)\mathbf{W}(S)$ gives zero, so [F4] makes it constant and multiplication by $E(aS)$ recovers the displayed formula.

By [F3], $Q(t):=E(t{)}^{T}E(t)$ is smooth and $Q^{\mathrm{\prime }}(t)=E(t{)}^T(J^T+J)E(t)=0$, with $Q(0)=I$. Clause 2 of [F4], applied to its four entries as a real vector on any compact interval, gives $Q(t)=I$, so every $E(t)$ preserves the Euclidean norm. For fixed $S>0$, put $r={\gamma }^{-1}$ and $P(s):=E(rs)E(S-s)$ on $[0,S]$. The semigroup and derivative laws in [F3] imply that these factors commute with $J$ and $P^{\mathrm{\prime }}(s)=(r-1)JE(rs)E(S-s)$. Its derivative is continuous, so the componentwise integral identity in [F4] gives $E(rS)-E(S)={\int }_0^S(r-1)JE(rs)E(S-s)ds$. Apply this identity to ${\mathbf{W}}_0$ and then [F5]. Both exponential factors preserve norm, so $\mathrm{\mid }\mathbf{W}-{\mathbf{W}}_N\mathrm{\mid }\le S\mathrm{\mid }{b}_0\mathrm{\mid }\mathrm{\mid }1-{\gamma }^{-1}\mathrm{\mid }\mathrm{\mid }{\mathbf{W}}_0\mathrm{\mid }$. At $S=0$ the difference and integral vanish directly. The exact identity $1-{\gamma }^{-1}=ϵ\mathrm{\mid }{\mathbf{W}}_0{\mathrm{\mid }}^2\mathrm{/}(1+\sqrt{1-ϵ\mathrm{\mid }{\mathbf{W}}_0{\mathrm{\mid }}^2})\le ϵ\mathrm{\mid }{\mathbf{W}}_0{\mathrm{\mid }}^2$ proves the velocity bound throughout the displayed parameter range, including ${\mathbf{W}}_0=0$.

Define $J^{-1}=-J\mathrm{/}{b}_0^2$ and the positions explicitly by $\mathbf{X}(S)=\mathbf{X}(0)+\gamma J^{-1}(E(S\mathrm{/}\gamma )-I){\mathbf{W}}_0$ and ${\mathbf{X}}_N(S)=\mathbf{X}(0)+J^{-1}(E(S)-I){\mathbf{W}}_0$. Their derivatives are the respective velocities by [F3], and their initial positions match. Clause 1 of [F4] and [F5] give $\mathrm{\mid }\mathbf{X}(S)-{\mathbf{X}}_N(S)\mathrm{\mid }\le {\int }_0^S\mathrm{\mid }\mathbf{W}(s)-{\mathbf{W}}_N(s)\mathrm{\mid }ds\le \mathrm{\mid }{b}_0\mathrm{\mid }ϵ\mathrm{\mid }{\mathbf{W}}_0{\mathrm{\mid }}^3{\int }_0^{S}sds=\frac{1}{2}\mathrm{\mid }{b}_0\mathrm{\mid }ϵ\mathrm{\mid }{\mathbf{W}}_0{\mathrm{\mid }}^3{S}^2$, with $S=0$ immediate. The smooth orthogonal flows exist for all real $S$ and give $\mathrm{\mid }\mathbf{W}\mathrm{\mid }=\mathrm{\mid }{\mathbf{W}}_N\mathrm{\mid }=\mathrm{\mid }{\mathbf{W}}_0\mathrm{\mid }$ and $\mathrm{\mid }\mathbf{X}-\mathbf{X}(0)\mathrm{\mid },\mathrm{\mid }{\mathbf{X}}_N-{\mathbf{X}}_N(0)\mathrm{\mid }\le S_{\ast }\mathrm{\mid }{\mathbf{W}}_0\mathrm{\mid }$ on the stated finite interval, hence common compact containment. Their constant speed is strictly below $c$ because $ϵ\mathrm{\mid }{\mathbf{W}}_0{\mathrm{\mid }}^2\le 1\mathrm{/}2$, and setting proper time $\tau =t\mathrm{/}\gamma$ reconstructs the future-directed proper-time test worldline satisfying the force law used in step 1.1. These direct calculations establish the displayed example throughout its stated range; for sufficiently small $ϵ$ they also lie within the general theorem's hypotheses [F1]. The metric and field remain prescribed inputs by [F2], with no coupled-background equation.