Controlled weak/slow charged test trajectory comparison

The external ODE supplier [F2] applies at each fixed $ϵ$ because $g^{(ϵ)}$ is smooth Lorentzian and $F^{(ϵ)}$ is a prescribed smooth two-form; it gives local uniqueness from future timelike proper-time data. The comparison vector field is smooth on an open set, so its derivative is bounded on a sufficiently small closed coordinate box about each initial state; the one-variable mean-value theorem [F6] along segments makes it locally Lipschitz, and [F3] gives local existence and uniqueness. The existence and common containment through $S_{\ast }$ are retained as [A1], rather than inferred from either local theorem.

Write the spatial metric block as $H=(g_{ij})$, the off-diagonal row as $b=(g_{0i})$, and $a=g_{00}$. The stated bounds give $H=dI+R$, where $d=1-2ϵ\varphi$ and $R=O({ϵ}^2)$, while $b=O({ϵ}^{3\mathrm{/}2})$ and $a=-1-2ϵ\varphi +O({ϵ}^2)$ uniformly. For sufficiently small $ϵ$, $d$ stays bounded away from zero. The determinant is polynomial in the entries, so $\det H=d^3+O({ϵ}^2)$ is bounded away from zero; its adjugate is bounded, hence $H^{-1}$ is bounded. The identity $H^{-1}-d^{-1}I=-d^{-1}{H}^{-1}R$ gives $H^{-1}=d^{-1}I+O({ϵ}^2)$, and, with $u:=2ϵ\varphi$, $(1-u{)}^{-1}=1+u+u^2\mathrm{/}(1-u)$ gives $d^{-1}=1+2ϵ\varphi +O({ϵ}^2)$. Its Schur complement $s=a-b{H}^{-1}{b}^T=-1-2ϵ\varphi +O({ϵ}^2)$ is bounded away from zero. Direct multiplication verifies the block-inverse entries $g^{00}=s^{-1}$, $g^{0i}=-s^{-1}(b{H}^{-1}{)}_i$, and $g^{ij}=(H^{-1}{)}_{ij}+(H^{-1}{b}^T{)}_i{s}^{-1}(b{H}^{-1}{)}_j$: multiplying these entries by the blocks $a,b,H$ gives the identity matrix, using $s=a-b{H}^{-1}{b}^T$. $g^{00}=-1+2ϵ\varphi +O({ϵ}^2),g^{0i}=O({ϵ}^{3\mathrm{/}2}),g^{ij}=(1+2ϵ\varphi ){\delta }_{ij}+O({ϵ}^2)\mathrm{.}$ All constants in these estimates depend only on the supplied bounds, not on $ϵ$.

Put $K^{\mu }=d{x}^{\mu }\mathrm{/}dt=(c,w^i)$ and $\lambda =d\tau \mathrm{/}dt>0$. Proper time gives ${\lambda }^2=-g(K,K)\mathrm{/}{c}^2=1+O(ϵ)$, since $\mathrm{\mid }w\mathrm{\mid }\mathrm{/}c=O(\sqrt{ϵ})$ and $g_{0i}=O({ϵ}^{3\mathrm{/}2})$. Shrinking ${ϵ}_{\ast }$ makes $1\mathrm{/}2\le \lambda \le 3\mathrm{/}2$, and $(\lambda -1)=({\lambda }^2-1)\mathrm{/}(\lambda +1)$ gives $\lambda =1+O(ϵ)$. The exact covariant equation with $U=K\mathrm{/}\lambda$ yields $\frac{d{w}^i}{dt}=-{\mathrm{\Gamma }}^i{}_{\mu \nu }{K}^{\mu }{K}^{\nu }+\frac{w^i}{c}{\mathrm{\Gamma }}^0{}_{\mu \nu }{K}^{\mu }{K}^{\nu }+\alpha \lambda (F^i{}_{\nu }{K}^{\nu }-\frac{w^i}{c}{F}^0{}_{\nu }{K}^{\nu })\mathrm{.}$ Indeed, subtracting $w^i\mathrm{/}c$ times the time component of the proper-time equation from its $i$th component cancels the derivative of $\lambda$; multiplying the result by ${\lambda }^2$ gives the display.

In the Christoffel formula of [F5], time derivatives vanish by stationarity and each spatial derivative contributes $L^{-1}{\mathrm{\partial }}_X$. Substitution of the metric and inverse estimates of step 1.2, together with the first-derivative remainder bounds, gives ${\mathrm{\Gamma }}^i{}_{00}=\frac{ϵ}{L}{\mathrm{\partial }}_{X^i}\varphi +O({ϵ}^2\mathrm{/}L),{\mathrm{\Gamma }}^i{}_{0j}=O({ϵ}^{3\mathrm{/}2}\mathrm{/}L),{\mathrm{\Gamma }}^i{}_{jk}=O(ϵ\mathrm{/}L),$ ${\mathrm{\Gamma }}^0{}_{00}=O({ϵ}^{5\mathrm{/}2}\mathrm{/}L),{\mathrm{\Gamma }}^0{}_{0j}=O(ϵ\mathrm{/}L),{\mathrm{\Gamma }}^0{}_{jk}=O({ϵ}^{3\mathrm{/}2}\mathrm{/}L)\mathrm{.}$ For example, ${\mathrm{\Gamma }}^i{}_{00}=-\frac{1}{2}{g}^{ij}{\mathrm{\partial }}_j{g}_{00}$, so its leading term is $+ϵL^{-1}{\mathrm{\partial }}_{X^i}\varphi$; every other displayed order follows by multiplying the corresponding inverse-metric order by the first derivatives of $g_{00}$, $g_{0i}$ or $g_{ij}$.

Raising the first index of $F$ with step 1.2 and using $F_{0i}=-E_i\mathrm{/}c$, $F_{ij}={ϵ}_{ijk}{B}_k$ gives $\alpha F^i{}_{\nu }{K}^{\nu }=\alpha (E+w\times B{)}^i+O({ϵ}^2{c}^2\mathrm{/}L),\alpha F^0{}_{\nu }{K}^{\nu }=O({ϵ}^{3\mathrm{/}2}{c}^2\mathrm{/}L)\mathrm{.}$ For the first equality, the diagonal inverse-metric corrections are $O(ϵ)$ times the leading electric and magnetic accelerations, each $O(ϵc^2\mathrm{/}L)$; the off-diagonal term $g^{i0}$ is $O({ϵ}^{3\mathrm{/}2})$ and multiplies $\alpha F_{0j}{w}^j=O({ϵ}^{3\mathrm{/}2}{c}^2\mathrm{/}L)$. In the second equality the leading time contraction is $\alpha E\cdot w\mathrm{/}c=O({ϵ}^{3\mathrm{/}2}{c}^2\mathrm{/}L)$, and the inverse-metric off-diagonal terms are no larger. Thus $(w^i\mathrm{/}c)\alpha F^0{}_{\nu }{K}^{\nu }=O({ϵ}^2{c}^2\mathrm{/}L)$. Multiplication by $\lambda =1+O(ϵ)$ preserves the leading force and changes it by at most $O({ϵ}^2{c}^2\mathrm{/}L)$.

With $K^i=O(\sqrt{ϵ}c)$, the first three Christoffel orders in step 2.1 give $-{\mathrm{\Gamma }}^i{}_{\mu \nu }{K}^{\mu }{K}^{\nu }+\frac{w^i}{c}{\mathrm{\Gamma }}^0{}_{\mu \nu }{K}^{\mu }{K}^{\nu }=-\frac{ϵc^2}{L}{\mathrm{\partial }}_{X^i}\varphi +O({ϵ}^2{c}^2\mathrm{/}L)\mathrm{.}$ The leading term is $-{\mathrm{\Gamma }}^i{}_{00}{c}^2$. The ${\mathrm{\Gamma }}^i{}_{0j}$ and ${\mathrm{\Gamma }}^i{}_{jk}$ terms are respectively $O({ϵ}^2{c}^2\mathrm{/}L)$ and $O({ϵ}^2{c}^2\mathrm{/}L)$; the displayed ${\mathrm{\Gamma }}^0$ terms, after multiplication by $w^i\mathrm{/}c$, are also at most $O({ϵ}^2{c}^2\mathrm{/}L)$. Since ${\mathrm{\Phi }}_{ϵ}:=ϵc^2\varphi$, the leading term is $-{\mathrm{\nabla }}_x{\mathrm{\Phi }}_{ϵ}$.

Combining steps 1.3, 3.1, and 2.2 in the exact coordinate equation gives $\frac{d\mathbf{w}}{dt}=-{\mathrm{\nabla }}_x{\mathrm{\Phi }}_{ϵ}+\alpha (\mathbf{E}+\mathbf{w}\times \mathbf{B})+{\mathbf{R}}_{ϵ},\mathrm{\parallel }{\mathbf{R}}_{ϵ}{\mathrm{\parallel }}_1\le C_r{ϵ}^2{c}^2\mathrm{/}L$ for a finite $C_r$ depending only on the stated uniform bounds. The definitions of $X,W,S$ give $dX\mathrm{/}dS=W$ and $dW\mathrm{/}dS=L(ϵc^2{)}^{-1}dw\mathrm{/}dt$. Since ${\mathrm{\nabla }}_x=L^{-1}{\mathrm{\nabla }}_X$, $\alpha E=ϵc^2{L}^{-1}e$, $\alpha B=\sqrt{ϵ}c{L}^{-1}b$, and $w=\sqrt{ϵ}cW$, the equation becomes the claimed comparison system with residual $r_{ϵ}$ bounded by $C_rϵ$.

Let $Z_{ϵ}=(X_{ϵ},W_{ϵ})$, $Z_N=(X_N,W_N)$, and $D(S)=\mathrm{\parallel }{Z}_{ϵ}(S)-Z_N(S){\mathrm{\parallel }}_1$. Their integral equations, matched initial data, the residual bound from step 4.1, and the derivative bound for $f$ on the convex set $\mathcal{Q}$ give $D(S)\le C_rϵS+L_f{\int }_0^{S}D(s)ds\le C_rϵS_{\ast }+L_f{\int }_0^{S}D(s)ds\mathrm{.}$ The Lipschitz estimate follows by applying the chain rule to $u\mapsto f(Z_N+u(Z_{ϵ}-Z_N))$ and integrating its derivative on $[0,1]$; the componentwise norm inequality follows from the scalar absolute-value triangle inequality [F7].

Define $H(S)=C_rϵS_{\ast }+L_f{\int }_0^{S}D(s)ds$. Then $D\le H$, $H(0)=C_rϵS_{\ast }$ and $H^{\mathrm{\prime }}=L_{f}D\le L_{f}H$. By [F6], $(e^{-L_{f}S}H(S){)}^{\mathrm{\prime }}=e^{-L_{f}S}(H^{\mathrm{\prime }}-L_{f}H)\le 0$; the mean value theorem in [F6] therefore makes this function nonincreasing. Hence $D(S)\le H(S)\le C_rϵS_{\ast }{e}^{L_{f}S}$ for $0\le S\le S_{\ast }$, proving the stated uniform estimate.