The four-force equation reproduces the coordinate Lorentz force and power

(Clause (a): the coordinate force) By [F1] and [F5], $U=\gamma (c,v)$, $U^i=\gamma v_i$ and $\mathrm{d}\mathrm{/}\mathrm{d}\tau =\gamma \mathrm{d}\mathrm{/}\mathrm{d}t$ (chain rule for the inverse parametrization $t\mapsto \tau (t)$), and $p=m\gamma v$ by [F4]. For a spatial index $i$, the spatial component of the equation of motion [F2] reads $md{U}^i\mathrm{/}d\tau =q\mathrm{\Phi }(U{)}^i=q\gamma (E_i+(v\times B{)}_i)$ by [F3]. Since $U^i=\gamma v_i$ and $p_i=m\gamma v_i$, the left side is $md{U}^i\mathrm{/}d\tau =m\gamma \frac{d}{dt}(\gamma v_i)=\gamma \frac{d{p}_i}{dt}$, so $\gamma d{p}_i\mathrm{/}dt=q\gamma (E_i+(v\times B{)}_i)$ and, dividing by $\gamma >0$, $d{p}_i\mathrm{/}dt=q(E_i+(v\times B{)}_i)$ for $i=1,2,3$; componentwise this is $dp\mathrm{/}dt=q(E+v\times B)$. The derivatives are those of the chart time, and $E,B$ are the laboratory components of $F$.

(Clause (a): the power identity) For the temporal index, [F2] and [F3] give $md{U}^0\mathrm{/}d\tau =q\mathrm{\Phi }(U{)}^0=q\gamma E\cdot v\mathrm{/}c$. By [F1] and [F5], $U^0=\gamma c$ and $d{U}^0\mathrm{/}d\tau =\gamma c\mathrm{d}\gamma \mathrm{/}\mathrm{d}t$, so $md{U}^0\mathrm{/}d\tau =\frac{\gamma }{c}m{c}^2\frac{\mathrm{d}\gamma }{\mathrm{d}t}=\frac{\gamma }{c}\frac{d(\gamma m{c}^2)}{dt}$; comparing with the right-hand side, $\frac{\gamma }{c}\frac{d(\gamma m{c}^2)}{dt}=\frac{q\gamma E\cdot v}{c}$, and multiplying by $c\mathrm{/}\gamma >0$ gives $d(\gamma m{c}^2)\mathrm{/}dt=qE\cdot v$.

(Clause (b): norm preservation and the future branch) At every interior parameter $s\in J$, the proper-time derivative in [F2] is $dU\mathrm{/}d\tau =U^{\mathrm{\prime }}(s)\mathrm{/}{\tau }^{\mathrm{\prime }}(s)$, with ${\tau }^{\mathrm{\prime }}(s)>0$. Thus $U^{\mathrm{\prime }}(s)=(q\mathrm{/}m){\tau }^{\mathrm{\prime }}(s){\mathrm{\Phi }}_{x(s)}(U(s))$. The product rule and pointwise skew orthogonality [F6] give $\frac{d}{ds}\eta (U,U)=2\eta (U,U^{\mathrm{\prime }})=2(q\mathrm{/}m){\tau }^{\mathrm{\prime }}(s)\eta (U,{\mathrm{\Phi }}_{x(s)}(U))=0$; no derivative of the field is used. The scalar $\eta (U,U)$ is continuous on $J$; the mean value theorem [F8] on every compact subinterval therefore makes it constant, including any included endpoints. Its value is $-c^2$ by [F1]. Consequently $(U^0{)}^2=c^2+\mathrm{\mid }{U}_{\mathrm{v}\mathrm{e}\mathrm{c}}{\mathrm{\mid }}^2>0$, so $U^0$ never vanishes. Continuity and the intermediate value theorem [F6] show that its sign is constant on $J$; if $U^0(s_0)>0$ at any initial parameter $s_0\in J$, then $U^0>0$ throughout. The solution remains future timelike and $\mathrm{\mid }v(t)\mathrm{\mid }<c$ wherever its chart reading is differentiable, by [F1]. No existence or continuation of solutions is asserted.

(Clause (c): the function and its derivatives) Put $z(t):=\mathrm{\mid }v(t){\mathrm{\mid }}^2\mathrm{/}{c}^2$, so that $0\le z\le {\delta }^2$ on an interval where $\mathrm{\mid }v\mathrm{\mid }\mathrm{/}c\le \delta$, $\delta \in (0,1)$, and $\gamma =f(z)$ with $f(z):=(1-z{)}^{-1\mathrm{/}2}$. The function $f$ is continuous on $[0,{\delta }^2]$ and differentiable on $(0,{\delta }^2)$: it is the composition of $z\mapsto 1-z$ with $y\mapsto y^{-1\mathrm{/}2}$, the latter continuous and differentiable on $(0,\mathrm{\infty })$ with derivative $-\frac{1}{2}{y}^{-3\mathrm{/}2}$ [F7], and $\gamma =f(z)$ agrees with the rational power of [F1] by [F7]. By the chain rule [F5], $f^{\mathrm{\prime }}(z)=\frac{1}{2}(1-z{)}^{-3\mathrm{/}2}$ and $f^{\mathrm{\prime }\mathrm{\prime }}(z)=\frac{3}{4}(1-z{)}^{-5\mathrm{/}2}$ on $(0,{\delta }^2)$. If $z=0$ then $\gamma =1$ and every difference and bound in clause (c) vanishes exactly; hence assume $0<z\le {\delta }^2$ below, and note $1-{\delta }^2>0$.

(Clause (c): first-order bound) Since $f$ is continuous on $[0,z]$ and differentiable on $(0,z)$, [F8] gives $\xi \in (0,z)$ with $f(z)-f(0)=f^{\mathrm{\prime }}(\xi )z$, and $f(0)=1$; hence $\mathrm{\mid }\gamma -1\mathrm{\mid }=\mathrm{\mid }{f}^{\mathrm{\prime }}(\xi )\mathrm{\mid }z=\frac{1}{2}(1-\xi {)}^{-3\mathrm{/}2}z$. Since $0<\xi \le z\le {\delta }^2<1$, the base satisfies $1-\xi \ge 1-{\delta }^2>0$, so $0<(1-\xi {)}^{-3\mathrm{/}2}\le (1-{\delta }^2{)}^{-3\mathrm{/}2}$ by monotonicity of the positive power and positivity of inverses [F7]; therefore $\mathrm{\mid }\gamma -1\mathrm{\mid }\le \frac{z}{2(1-{\delta }^2{)}^{3\mathrm{/}2}}=\frac{\mathrm{\mid }v{\mathrm{\mid }}^2\mathrm{/}{c}^2}{2(1-{\delta }^2{)}^{3\mathrm{/}2}}$.

(Clause (c): second-order bound) Define $q(s):=f(s)-1-\frac{s}{2}$ and $g(s):=s^2$ on $[0,z]$; both are continuous on $[0,z]$ and differentiable on $(0,z)$, with $q^{\mathrm{\prime }}(s)=f^{\mathrm{\prime }}(s)-\frac{1}{2}$ and $g^{\mathrm{\prime }}(s)=2s>0$ there [F7, F8]. Cauchy's mean value theorem [F8] gives $\xi \in (0,z)$ with $(q(z)-q(0))g^{\mathrm{\prime }}(\xi )=(g(z)-g(0))q^{\mathrm{\prime }}(\xi )$; since $q(0)=0=g(0)$, $g^{\mathrm{\prime }}(\xi )=2\xi \mathrm{\ne }0$ and $q^{\mathrm{\prime }}(\xi )=f^{\mathrm{\prime }}(\xi )-\frac{1}{2}$, this reads $f(z)-1-\frac{z}{2}=\frac{z^2}{2\xi }(f^{\mathrm{\prime }}(\xi )-\frac{1}{2})$. The mean value theorem [F8] applied to $f^{\mathrm{\prime }}$ on $[0,\xi ]$, where $f^{\mathrm{\prime }}$ is continuous on $[0,\xi ]$ and differentiable on $(0,\xi )$ (the derivative $f^{\mathrm{\prime }\mathrm{\prime }}$ is given on $(0,{\delta }^2)$ in step 1.4), gives $\eta \in (0,\xi )$ with $f^{\mathrm{\prime }}(\xi )-f^{\mathrm{\prime }}(0)=\xi f^{\mathrm{\prime }\mathrm{\prime }}(\eta )$; since $f^{\mathrm{\prime }}(0)=\frac{1}{2}$ this gives $f^{\mathrm{\prime }}(\xi )-\frac{1}{2}=\xi f^{\mathrm{\prime }\mathrm{\prime }}(\eta )$, and substitution yields $f(z)-1-\frac{z}{2}=\frac{z^2}{2\xi }\cdot \xi f^{\mathrm{\prime }\mathrm{\prime }}(\eta )=\frac{1}{2}{f}^{\mathrm{\prime }\mathrm{\prime }}(\eta )z^2=\frac{3}{8}(1-\eta {)}^{-5\mathrm{/}2}{z}^2$. Since $0<\eta \le \xi \le z\le {\delta }^2$, monotonicity of the positive power [F7] gives $(1-\eta {)}^{-5\mathrm{/}2}\le (1-{\delta }^2{)}^{-5\mathrm{/}2}$, hence $\mid \gamma -1-\frac{\mathrm{\mid }v{\mathrm{\mid }}^2}{2c^2}\mid =\mid f(z)-1-\frac{z}{2}\mid \le \frac{3}{8}(1-{\delta }^2{)}^{-5\mathrm{/}2}{z}^2=\frac{3}{8}(1-{\delta }^2{)}^{-5\mathrm{/}2}(\frac{\mathrm{\mid }v{\mathrm{\mid }}^2}{c^2}{)}^2$.

(Clause (c): the momentum and energy bounds) Multiplying the first-order bound of step 2.1 by $m\mathrm{\mid }v\mathrm{\mid }\ge 0$ and using $\mathrm{\mid }xy\mathrm{\mid }=\mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }y\mathrm{\mid }$ [F8] gives $\mathrm{\mid }p-mv\mathrm{\mid }=\mathrm{\mid }m\gamma v-mv\mathrm{\mid }=m\mathrm{\mid }v\mathrm{\mid }\mathrm{\mid }\gamma -1\mathrm{\mid }\le m\mathrm{\mid }v\mathrm{\mid }\frac{\mathrm{\mid }v{\mathrm{\mid }}^2}{2c^2(1-{\delta }^2{)}^{3\mathrm{/}2}}$. Multiplying the second-order bound of step 2.2 by $m{c}^2>0$ gives $\mid \gamma m{c}^2-m{c}^2-\frac{1}{2}m\mathrm{\mid }v{\mathrm{\mid }}^2\mid =m{c}^2\mid \gamma -1-\frac{\mathrm{\mid }v{\mathrm{\mid }}^2}{2c^2}\mid \le \frac{3}{8}(1-{\delta }^2{)}^{-5\mathrm{/}2}\frac{m\mathrm{\mid }v{\mathrm{\mid }}^4}{c^2}$. The constants depend only on $\delta$, so both bounds are uniform on any interval where $\mathrm{\mid }v\mathrm{\mid }\mathrm{/}c\le \delta$. These are pointwise algebraic bounds; no approximation of the trajectory over a time interval is asserted.

(Clauses (c) and (d): status and scope) Clauses (a) and (b) are steps 1.1-1.3, and clause (c) is steps 1.4, 2.1, 2.2 and 3.1: the two first-order and second-order bounds, the resulting momentum and energy bounds and their uniformity in $\mathrm{\mid }v\mathrm{\mid }\mathrm{/}c\le \delta$ are established there, including the exact vanishing at $z=0$. The charge $q$ is a constant parameter of Charge parameter, frame charge and current densities, volume charge and flux, and the four-current (a), $m>0$ is an externally supplied rest mass, and $F$ is a prescribed exterior field $C^1$ near the graph: the particle's own field, radiation reaction and any self-force are excluded, as are magnetic monopoles and zero or variable mass [F2]. The nonrelativistic coordinate law is not adopted as exact; clause (c) is a controlled kinematic expansion and asserts no trajectory approximation over time, which would require a separate stability argument. The argument used the metric contraction [F3], the coupling [F2], the momentum and four-force identification [F4], the proper-time chain rule [F5], the orthogonality and norm-conservation lemma [F6] (whose varying-field clause was used so that no derivative of $F$ enters), and the power and mean value tools [F7], [F8]; it is conditional on the adopted postulates, makes no empirical claim and uses no choice principle.