Lorentz covariant formulation — Examples

Source-free solution: the field pair $(E,B)$ is constant, hence $C^{\mathrm{\infty }}$ with every first partial derivative continuous and equal to $0$, in particular jointly $C^k$ for every $k\ge 1$ [F1]. With the divergence and curl of [F2], $\mathrm{div}E={\mathrm{\partial }}_1{E}_1+{\mathrm{\partial }}_2{E}_2+{\mathrm{\partial }}_3{E}_3=0$, $\mathrm{div}B=0$, $\mathrm{curl}E=({\mathrm{\partial }}_2{E}_3-{\mathrm{\partial }}_3{E}_2,{\mathrm{\partial }}_3{E}_1-{\mathrm{\partial }}_1{E}_3,{\mathrm{\partial }}_1{E}_2-{\mathrm{\partial }}_2{E}_1)=0$ and $\mathrm{curl}B=0$, and ${\mathrm{\partial }}_{t}E={\mathrm{\partial }}_{t}B=0$; hence the four source-free smooth-branch equations of [F2] with $\rho =0$, $J=0$ hold on ${\mathbb{R}}^3\times \mathbb{R}$.

Boosted components: substitute $E_1=0$, $E_2=E_0$, $E_3=0$ and $B_1=B_2=B_3=0$ in the formulas of [F3]: $E_1^{\mathrm{\prime }}=E_1=0$, $E_2^{\mathrm{\prime }}=\gamma (E_0-\beta c\cdot 0)=\gamma E_0$, $E_3^{\mathrm{\prime }}=\gamma (0+\beta c\cdot 0)=0$, $B_1^{\mathrm{\prime }}=B_1=0$, $B_2^{\mathrm{\prime }}=\gamma (0+\beta \cdot 0\mathrm{/}c)=0$ and $B_3^{\mathrm{\prime }}=\gamma (0-\beta E_0\mathrm{/}c)=-\gamma \beta E_0\mathrm{/}c$; therefore $E^{\mathrm{\prime }}=\gamma E_0{e}_2$ and $B^{\mathrm{\prime }}=-(\gamma \beta E_0\mathrm{/}c)e_3=-(\gamma \mathrm{\mid }v\mathrm{\mid }{E}_0\mathrm{/}{c}^2)e_3$, since $\mathrm{\mid }v\mathrm{\mid }=\beta c$ by $v=\beta c{e}_1$ and the right-handed basis [F1]. This is the asserted pair of components; the magnetic component is nonzero exactly when $\beta >0$, and the direction is $-e_3$.

Invariants: by [F3] the invariants agree with those computed from $(E,B)$, and directly from step 1.2, $\mathrm{\mid }{E}^{\mathrm{\prime }}{\mathrm{\mid }}^2={\gamma }^2{E}_0^2$, $\mathrm{\mid }{B}^{\mathrm{\prime }}{\mathrm{\mid }}^2={\gamma }^2{\beta }^2{E}_0^2\mathrm{/}{c}^2$ and $E^{\mathrm{\prime }}\cdot B^{\mathrm{\prime }}=0$ by [F5] ($e_2\perp e_3$), while $\mathrm{\mid }E{\mathrm{\mid }}^2=E_0^2$, $\mathrm{\mid }B{\mathrm{\mid }}^2=0$ and $E\cdot B=0$; hence $\mathrm{\mid }{B}^{\mathrm{\prime }}{\mathrm{\mid }}^2-\mathrm{\mid }{E}^{\mathrm{\prime }}{\mathrm{\mid }}^2\mathrm{/}{c}^2={\gamma }^2{E}_0^2({\beta }^2-1)\mathrm{/}{c}^2=-E_0^2\mathrm{/}{c}^2$ by ${\gamma }^2(1-{\beta }^2)=1$ [F4], which equals $\mathrm{\mid }B{\mathrm{\mid }}^2-\mathrm{\mid }E{\mathrm{\mid }}^2\mathrm{/}{c}^2$, and $E^{\mathrm{\prime }}\cdot B^{\mathrm{\prime }}=0=E\cdot B$. At $\beta =0$ one has $\gamma =1$ and the formulas of [F3] give $E^{\mathrm{\prime }}=E$, $B^{\mathrm{\prime }}=B=0$, so the boosted description is the laboratory description and no magnetic field appears.

Closing: step 1.1 verifies that the constant data solve the source-free smooth branch; step 1.2 gives the exact boosted components and step 2.1 the two invariant checks, with the $\beta =0$ reduction. The computation is exact algebra for the constant field and uses only the dictionary and boost formulas of [F3]; no measurement, source model, finite-energy, boundary or empirical claim is made, and no choice principle is used.

The prescribed field is source-free: $E\equiv E_0{e}_1$ and $B\equiv 0$ are constant, so every first partial derivative of every component vanishes; by the component formulas of [F2], $\mathrm{div}E=\mathrm{div}B=0$, $\mathrm{curl}E=\mathrm{curl}B=0$ and ${\mathrm{\partial }}_{t}E={\mathrm{\partial }}_{t}B=0$, so the four source-free smooth-branch equations hold on ${\mathbb{R}}^3\times \mathbb{R}$.

The trajectory and its velocity ($h\mathrm{\ne }0$): put $g(t):=\sqrt{1+h^2{t}^2}=(1+h^2{t}^2{)}^{1\mathrm{/}2}$, a composition of the positive polynomial $t\mapsto 1+h^2{t}^2$ with the power $x\mapsto x^{1\mathrm{/}2}$; by [F4] and the chain rule [F5], $g$ is differentiable with $g^{\mathrm{\prime }}(t)=\frac{1}{2}(1+h^2{t}^2{)}^{-1\mathrm{/}2}\cdot 2h^{2}t=h^{2}t\mathrm{/}g(t)$, and since every derivative is a rational expression with nonzero denominator $g$, iterating the same rules shows $g$ and hence $r(t)=r_0+\frac{c}{h}(g(t)-1)e_1$ are $C^{\mathrm{\infty }}$ on $\mathbb{R}$. Then $r^{\mathrm{\prime }}(t)=\frac{c}{h}{g}^{\mathrm{\prime }}(t)e_1=\frac{cht}{g(t)}{e}_1=v(t)$ by [F5], so $\mathrm{\mid }v(t)\mathrm{\mid }=c\mathrm{\mid }ht\mathrm{\mid }\mathrm{/}g(t)<c$ because $h^2{t}^2<1+h^2{t}^2$ for all real $t$ [F5], and $\mathrm{\mid }v(t)\mathrm{\mid }\to c$ as $\mathrm{\mid }t\mathrm{\mid }\to \mathrm{\infty }$ because $\mathrm{\mid }ht\mathrm{\mid }\mathrm{/}g(t)\to 1$.

The branch $h=0$: here $q=0$ since $E_0>0$, $m>0$ and $h=q{E}_0\mathrm{/}(mc)$; the definition gives $r\equiv r_0$, $v\equiv 0$, $\gamma \equiv 1$ and $p\equiv 0$, so $dp\mathrm{/}dt=0=q(E+v\times B)$ and $d(\gamma m{c}^2)\mathrm{/}dt=0=qE\cdot v$ hold trivially, and the Newtonian comparison trajectory of clause (iii) is $r_N\equiv r_0$ as well, with $v_N\equiv 0$; the formulas containing $1\mathrm{/}h$ are not used in this branch.

Force and power identities ($h\mathrm{\ne }0$): by step 1.2, $\gamma (t)=g(t)$ equals $(1-\mathrm{\mid }v(t){\mathrm{\mid }}^2\mathrm{/}{c}^2{)}^{-1\mathrm{/}2}$, since $1-\mathrm{\mid }v{\mathrm{\mid }}^2\mathrm{/}{c}^2=1-\frac{h^2{t}^2}{1+h^2{t}^2}=\frac{1}{g(t{)}^2}$; hence $p(t)=m\gamma (t)v(t)=mg(t)\cdot \frac{cht}{g(t)}{e}_1=mcht{e}_1=q{E}_{0}t{e}_1$, because $mch=q{E}_0$. Therefore $dp\mathrm{/}dt=q{E}_0{e}_1=q(E+v\times B)$, since $E=E_0{e}_1$, $B=0$ and $v\times 0=0$ [F5]. For the power identity, $d(\gamma m{c}^2)\mathrm{/}dt=m{c}^2{g}^{\mathrm{\prime }}(t)=m{c}^2{h}^{2}t\mathrm{/}g(t)=q{E}_0\cdot v_1(t)$ because $q{E}_0=mch$ and $v_1(t)=cht\mathrm{/}g(t)$. These are the two coordinate identities of [F3]. The covariant equation $mdU\mathrm{/}d\tau =q\mathrm{\Phi }(U)$ of [F3] also holds: with $U^0=g(t)c$, $U^1=cht$ and $U^2=U^3=0$ one has $d{U}^0\mathrm{/}d\tau =gd(gc)\mathrm{/}dt=cg{g}^{\mathrm{\prime }}=c{h}^{2}t$ and $d{U}^1\mathrm{/}d\tau =gd(cht)\mathrm{/}dt=cgh$, so $md{U}^0\mathrm{/}d\tau =mc{h}^{2}t=q{E}_{0}ht=q\mathrm{\Phi }(U{)}^0$ and $md{U}^1\mathrm{/}d\tau =mgch=qg{E}_0=q\mathrm{\Phi }(U{)}^1$, while the components $2,3$ vanish on both sides; hence the displayed curve is an exact solution of the adopted coupling in this special model.

The curve is a future timelike worldline: by step 1.2 the curve is $C^{\mathrm{\infty }}$ with coordinate velocity $v(t)$ and $\mathrm{\mid }v(t)\mathrm{\mid }<c$; its tangent $(c,v(t))$ has $\eta$-square $-c^2+\mathrm{\mid }v(t){\mathrm{\mid }}^2<0$ and positive time component $c>0$, so it is future timelike at every $t$, and the chart reading is that of a massive worldline in the sense of [F1]. At $h=0$ the constant curve has tangent $(c,0)$, also future timelike.

The unscoped Newtonian extrapolation ($h\mathrm{\ne }0$): define $v_N(t):=(q{E}_0\mathrm{/}m)t{e}_1=hct{e}_1$ and $r_N(t):=r_0+\frac{1}{2}hc{t}^2{e}_1$. Since $q{E}_0\mathrm{/}m=hc$, step 1.2 gives $v(t)=\frac{cht}{g(t)}{e}_1=\frac{v_N(t)}{g(t)}=\frac{v_N(t)}{\sqrt{1+h^2{t}^2}}$, and therefore $\mathrm{\mid }{v}_N(t)\mathrm{\mid }=g(t)\mathrm{\mid }v(t)\mathrm{\mid }=c\mathrm{\mid }ht\mathrm{\mid }$, which exceeds $c$ exactly when $\mathrm{\mid }ht\mathrm{\mid }>1$, while $\mathrm{\mid }v(t)\mathrm{\mid }<c$ for every finite $t$ by step 1.2. The two velocities agree only at $t=0$; $v_N$ is the Newtonian constant-acceleration prediction of the unscoped law.

Error bounds ($h\mathrm{\ne }0$): put $s:=ht$ and $g=\sqrt{1+s^2}\ge 1$. Rationalisation gives $g-1=\frac{s^2}{g+1}\le \frac{s^2}{2}$ and $1-\frac{1}{g}=\frac{g-1}{g}\le \frac{s^2}{2}$; hence $\mathrm{\mid }v-v_N\mathrm{\mid }=\mathrm{\mid }{v}_N\mathrm{\mid }\mid 1-\frac{1}{g}\mid =c\mathrm{\mid }s\mathrm{\mid }(1-\frac{1}{g})\le \frac{c}{2}\mathrm{\mid }s{\mathrm{\mid }}^3$, which is the velocity bound. For the displacement, $r(t)-r_N(t)=\frac{c}{h}(g-1-\frac{s^2}{2})e_1$, and the elementary identity $1+\frac{s^2}{2}-g=\frac{(g-1{)}^2}{2}$ gives $\mid g-1-\frac{s^2}{2}\mid =\frac{(g-1{)}^2}{2}\le \frac{s^4}{8}$ by the first inequality; multiplying by $c\mathrm{/}\mathrm{\mid }h\mathrm{\mid }$ gives $\mathrm{\mid }r-r_N\mathrm{\mid }\le \frac{c}{8\mathrm{\mid }h\mathrm{\mid }}\mathrm{\mid }s{\mathrm{\mid }}^4$. Consequently on $\mathrm{\mid }ht\mathrm{\mid }\le \delta$ the velocity error is at most $c{\delta }^3\mathrm{/}2$ and the displacement error at most $c{\delta }^4\mathrm{/}(8\mathrm{\mid }h\mathrm{\mid })$; these are special-model comparisons for this explicit solution and do not import a general trajectory-stability consequence of The four-force equation reproduces the coordinate Lorentz force and power (c).

Closing and scope: the prescribed field is a source-free constant smooth-branch solution (step 1.1); for $h\mathrm{\ne }0$ the displayed curve is $C^{\mathrm{\infty }}$, future timelike and subluminal at every finite time, and it satisfies the force and power identities with $p=q{E}_{0}t{e}_1$ (steps 1.2, 2.1, 2.2); the Newtonian comparison and its exact superluminal condition are step 2.3, and the special-model error bounds are step 3.1; the branch $h=0$ with $q=0$ is step 1.3. The field is prescribed and external, the particle's own field, radiation reaction and any self-force are excluded, no production of the field is described, and no measurement or empirical claim is made. The argument used the field and worldline objects [F1], the smooth-branch equations [F2], the force and power identities [F3], the power and root calculus [F4] and the chain and product rules [F5]; no choice principle is used.