Right-wedge accelerated clocks and rays

Put $\theta =aT\mathrm{/}c$. The exponential definitions [F11] give $x^1+x^0=\rho e^{\theta }$ and $x^1-x^0=\rho e^{-\theta }$, both positive, so the image lies in $W$. Conversely, [F3] supplies the unique smooth Rindler inverse $\rho =\sqrt{(x^1{)}^2-(x^0{)}^2}$, $\theta =\frac{1}{2}\log ((x^1+x^0)\mathrm{/}(x^1-x^0))$; setting $T=c\theta \mathrm{/}a$ gives the stated inverse, which is smooth because $a,c>0$. The range agrees with [F3], and units and coordinates are those of [F1, F2].

Differentiating the map with [F9] gives $d{x}^0=\sinh \theta d\rho +(a\rho \mathrm{/}c)\cosh \theta dT$ and $d{x}^1=\cosh \theta d\rho +(a\rho \mathrm{/}c)\sinh \theta dT$. Substitution into the inertial metric of [F1, F2], using ${\cosh }^2\theta -{\sinh }^2\theta =1$, cancels the mixed terms and yields the displayed pullback metric. For $d\rho =dy=dz=0$, the proper-time line element and adopted clock rule [F4] give $d\tau \mathrm{/}dT=a\rho \mathrm{/}{c}^2$.

Along a stationary worldline set $\chi :=c\tau \mathrm{/}\rho =aT\mathrm{/}c$. Its inertial coordinates are $(\rho \sinh \chi ,\rho \cosh \chi )$. Differentiating twice with respect to $\tau$ by [F9] gives $U=(c\cosh \chi ,c\sinh \chi )$ and $A=(c^2\sinh \chi \mathrm{/}\rho ,c^2\cosh \chi \mathrm{/}\rho )$. Thus $\eta (U,U)=-c^2$ and $\eta (A,A)=c^4\mathrm{/}{\rho }^2$ by [F1]; $U$ is future timelike and has the proper-time normalization of [F5]. The definition [F6] makes the proper-acceleration magnitude $c^2\mathrm{/}\rho$, consistently with [F3].

For a radial null curve, set $dy=dz=0$ and $d{s}^2=0$ in the metric of step 2.1. Since $a,\rho ,c$ are positive, this gives $d\rho \mathrm{/}dT=\pm a\rho \mathrm{/}c$. By [F10], $d(\log \rho )\mathrm{/}dT=\pm a\mathrm{/}c$; integration between its endpoint radii gives $\mathrm{\Delta }T=(c\mathrm{/}a)\mathrm{\mid }\log ({\rho }_R\mathrm{/}{\rho }_E)\mathrm{\mid }$, finite for positive finite radii. [F4] identifies these null curves with the ideal vacuum rays.

Over a common nonzero Rindler-time interval $\mathrm{\Delta }T$, step 2.1 gives $\mathrm{\Delta }{\tau }_E=(a{\rho }_E\mathrm{/}{c}^2)\mathrm{\Delta }T$ and $\mathrm{\Delta }{\tau }_R=(a{\rho }_R\mathrm{/}{c}^2)\mathrm{\Delta }T$. Their ratio is ${\rho }_R\mathrm{/}{\rho }_E$. If the radii coincide the two increments agree; for distinct positive radii the ratio is positive and finite.

For an outward radial wave take dimensionless phase $\vartheta =\kappa (x^1-x^0)=\kappa \rho e^{-aT\mathrm{/}c}$ with $\kappa >0$ in inverse metres; for an inward wave take $\vartheta =-\kappa (x^1+x^0)=-\kappa \rho e^{aT\mathrm{/}c}$, using [F11]. Each is an inertial affine phase with future null covector as in [F8], and its $T$ derivative is constant along its corresponding null ray because $\rho e^{\mp aT\mathrm{/}c}$ is constant there by step 3.2. A stationary observer has $U_n=(c^2\mathrm{/}(a\rho )){\mathrm{\partial }}_T$ by step 2.1, so [F7] yields ${\omega }_n=-k(U_n)=-(c^2\mathrm{/}(a\rho )){\mathrm{\partial }}_T\vartheta$, proportional to $1\mathrm{/}\rho$ with the same positive proportionality constant at the two events. Hence ${\omega }_R\mathrm{/}{\omega }_E={\rho }_E\mathrm{/}{\rho }_R$ as a contraction identity. Under [A2] these contractions are the instantaneous received angular frequencies; their finite-apparatus interpretation retains the local ray-regime and short-observation qualifications of that postulate. This is separate from [F8] and does not extend its scope.

The displayed scalar and vector potentials are smooth on all of Minkowski space. With the covector convention $A_0=-\varphi \mathrm{/}c$ and $A_i={\mathbf{A}}_i$, they give $F_{01}=-E_0\mathrm{/}c$, $F_{ij}=0$, hence exactly the field in [F1]. Thus the action potential-domain hypothesis [A1] is met, and under the exact assumption [A3], [F2] gives its covariant Lorentz-force equation.

Apply [F3] under [A3] to this field: ${\dot{p}}_1=F_0$, ${\dot{\mathbf{p}}}_{\perp }=0$, and $\dot{\mathcal{E}}=F_0{v}_1$. Integration with the initial data gives $p_1(t)=p_{01}+F_{0}t$ and constant ${\mathbf{p}}_{\perp }$. By adopted [A2] and the inertial momentum relations [F4], ${\mathcal{E}}^2=m^2{c}^4+c^2\mathrm{\mid }\mathbf{p}{\mathrm{\mid }}^2$; since $\mathcal{E}=\gamma m{c}^2>0$, this is $\mathcal{E}=c\sqrt{m^2{c}^2+\mathrm{\mid }\mathbf{p}{\mathrm{\mid }}^2}$, while $\mathbf{p}=\gamma m\mathbf{v}$ gives $\mathbf{v}=c^2\mathbf{p}\mathrm{/}\mathcal{E}$. Thus the definitions of $\mathrm{\Pi },\mathcal{E},\mathbf{v}$ in the statement follow. Differentiating the shell identity with $\dot{\mathbf{p}}=F_0{e}_1$ gives $2\mathcal{E}\dot{\mathcal{E}}=2c^2{p}_1{F}_0$, hence $\dot{\mathcal{E}}=c^2{p}_1{F}_0\mathrm{/}\mathcal{E}=F_0{v}_1$, as required by [F3]. Since $m,c>0$, the square-root denominator is at least $mc>0$, and [F5] gives subluminal motion.

Suppose $F_0\mathrm{\ne }0$. Differentiating the proposed $x^1(t)$ and using ${\dot{p}}_1=F_0$ gives ${\dot{x}}^1=c{p}_1\mathrm{/}\sqrt{{\mathrm{\Pi }}^2+p_1^2}=v_1$. For the transverse formula, the derivative of $\log (p+\sqrt{{\mathrm{\Pi }}^2+p^2})$ is $1\mathrm{/}\sqrt{{\mathrm{\Pi }}^2+p^2}$: the argument is positive since $\sqrt{{\mathrm{\Pi }}^2+p^2}>\mathrm{\mid }p\mathrm{\mid }$, and the identity follows from [F6, F7] and the chain rule. Hence ${\dot{\mathbf{x}}}_{\perp }=c{\mathbf{p}}_{\perp }\mathrm{/}\sqrt{{\mathrm{\Pi }}^2+p_1^2}={\mathbf{v}}_{\perp }$. At $t=0$ both position increments vanish, so these formulas have the prescribed initial position and velocity.

If $F_0=0$, [F3] makes $\mathbf{p}$ and $\mathcal{E}$ constant, so the displayed free position has derivative $c^2{\mathbf{p}}_0\mathrm{/}\mathcal{E}(0)=\mathbf{v}$ and the initial position. Its speed is below $c$ by [F5].

If $F_0\mathrm{\ne }0$, the constructed curve satisfies the force equation and initial data by steps 2.1 and 3.1; if $F_0=0$, those facts follow from steps 2.1 and 3.2. Any other solution has the same momentum by integrating [F3]; subtracting its position from the displayed one gives a differentiable vector with zero derivative and zero initial value, so each component vanishes by [F8]. Thus it agrees throughout any common connected interval. The formulas have positive square-root arguments and positive logarithm arguments for every finite $t$, so they extend smoothly over all $\mathbb{R}$.

The potential components satisfy $F_{12}={\mathrm{\partial }}_1{A}_2-{\mathrm{\partial }}_2{A}_1=B_0$ and all electric components vanish, by the tensor convention [F1]. The potential is smooth on all of Minkowski space, so the local-potential assumption [A1] applies, and under [A3], [F2] supplies the covariant equation. The coordinate force and energy equations are therefore those of [F3].

With $\mathbf{E}=0$, the actual current carrier [F3] under [A3] gives $\dot{\mathcal{E}}=0$. The adopted massive-particle model [A2] and inertial relations [F4] give ${\mathcal{E}}^2=c^2(m^2{c}^2+\mathrm{\mid }\mathbf{p}{\mathrm{\mid }}^2)$ and $\mathbf{v}=c^2\mathbf{p}\mathrm{/}\mathcal{E}$. Thus $\mathcal{E}=c{S}_0$ and $\mathbf{v}=c\mathbf{p}\mathrm{/}{S}_0$. Using [F3], [F8], and $\mathbf{v}=c\mathbf{p}\mathrm{/}{S}_0$, one has $\mathbf{v}\times B_0{e}_3=B_0(v_2{e}_1-v_1{e}_2)$, so ${\dot{p}}_1=\mathrm{\Omega }{p}_2$, ${\dot{p}}_2=-\mathrm{\Omega }{p}_1$, ${\dot{p}}_3=0$ because $q{B}_{0}c\mathrm{/}{S}_0=q{B}_0\mathrm{/}({\gamma }_{0}m)=\mathrm{\Omega }$. Differentiation using [F6] verifies the stated sinusoidal momenta and their initial values. The sum and product rules [F9] show that the rotating components also satisfy $\frac{d}{dt}(p_1^2+p_2^2)=2\mathrm{\Omega }{p}_1{p}_2-2\mathrm{\Omega }{p}_1{p}_2=0$, so $\mathrm{\mid }\mathbf{p}(t)\mathrm{\mid }=\mathrm{\mid }{\mathbf{p}}_0\mathrm{\mid }$ and the displayed constant $\mathcal{E}=c{S}_0$ obeys the shell and $\dot{\mathcal{E}}=0$; when $\mathrm{\Omega }=0$, the force is zero and momentum is constant.

For $\mathrm{\Omega }\mathrm{\ne }0$, differentiating the proposed $x^1$ and $x^2$ with [F6] gives ${\dot{x}}^1=c{p}_1\mathrm{/}{S}_0$ and ${\dot{x}}^2=c{p}_2\mathrm{/}{S}_0$; differentiating $x^3$ gives $c{p}_3\mathrm{/}{S}_0$. These are exactly the velocity components from step 2.1, and all increments vanish at $t=0$. For $\mathrm{\Omega }=0$, the stated free position has constant derivative $c{\mathbf{p}}_0\mathrm{/}{S}_0=\mathbf{v}$.

For $\mathrm{\Omega }\mathrm{\ne }0$, set $A=c{p}_{01}\mathrm{/}(S_0\mathrm{\Omega })$ and $B=c{p}_{02}\mathrm{/}(S_0\mathrm{\Omega })$. The transverse coordinates relative to the fixed centre $(x_0^1+B,x_0^2-A)$ are $(A\sin \mathrm{\Omega }t-B\cos \mathrm{\Omega }t,B\sin \mathrm{\Omega }t+A\cos \mathrm{\Omega }t)$; their squared distance from that centre is $A^2+B^2$. Since $S_0={\gamma }_{0}mc$ and $\mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }=\mathrm{\mid }q{B}_0\mathrm{\mid }\mathrm{/}({\gamma }_{0}m)$, the radius is $c\mathrm{\mid }{\mathbf{p}}_{\perp }\mathrm{\mid }\mathrm{/}(S_0\mathrm{\mid }\mathrm{\Omega }\mathrm{\mid })=\mathrm{\mid }{\mathbf{p}}_{\perp }\mathrm{\mid }\mathrm{/}\mathrm{\mid }q{B}_0\mathrm{\mid }$. If ${\mathbf{p}}_{\perp }=0$, $x^1=x_0^1$ and $x^2=x_0^2$, so the transverse projection is a point. Then $x^3=x_0^3+c{p}_{03}t\mathrm{/}{S}_0$ makes the full spatial orbit an axial straight line when $p_{03}\mathrm{\ne }0$ and a stationary point when $p_{03}=0$. If ${\mathbf{p}}_{\perp }\mathrm{\ne }0$ and $p_{03}=0$, $x^3$ is constant and the circle is planar. The energy constancy in step 2.1 proves zero magnetic work.

The displayed curves satisfy the force equation and initial data by steps 2.1 and 3.1. Any other solution with the same data has constant energy $\mathcal{E}=c{S}_0$ by [F3], so step 2.1 puts its momentum in the same constant-coefficient system ${\dot{p}}_1=\mathrm{\Omega }{p}_2$, ${\dot{p}}_2=-\mathrm{\Omega }{p}_1$, ${\dot{p}}_3=0$. For the difference $d$ of two such momenta, the product rule [F9] gives $\frac{d}{dt}(d_1^2+d_2^2)=2\mathrm{\Omega }{d}_1{d}_2-2\mathrm{\Omega }{d}_2{d}_1=0$, and ${\dot{d}}_3=0$. With zero initial difference, [F7] makes $d_1^2+d_2^2=0$ and $d_3=0$ on every common connected interval, hence both momenta agree, including $\mathrm{\Omega }=0$. Their velocities then agree by [F4], so each position difference has zero derivative and zero initial value; [F7] gives equality. The formulas are smooth for all real $t$; $S_0\ge mc>0$, and in the rotating case the denominator $\mathrm{\Omega }$ is nonzero. The resulting worldline is timelike by [F5].

The scalar and vector potentials are smooth. By [F1], their tensor components are $F_{01}=-E_0\mathrm{/}c$, $F_{12}=B_0$, and the remaining independent components vanish, so they give precisely the displayed fields. Thus [A1] applies and, under the explicit ${\mathrm{A}\mathrm{C}}_{\omega }$ premise [A3], [F2] yields the covariant force equation. Under the flat massive-particle adoption [A2] and the same countable-choice condition, [F3] gives the coordinate Lorentz law in the named inertial chart.

The component definition [F10] gives ${\mathbf{v}}_d=-E_0{e}_2\mathrm{/}{B}_0$, so $\mathrm{\mid }{\mathbf{v}}_d\mathrm{\mid }\mathrm{/}c=\mathrm{\mid }{E}_0\mathrm{\mid }\mathrm{/}(c\mathrm{\mid }{B}_0\mathrm{\mid })<1$. Orthogonality from [F12] gives ${\mathbf{v}}_d\cdot \mathbf{E}={\mathbf{v}}_d\cdot \mathbf{B}=0$ and, by the triple-product identity [F7], ${\mathbf{v}}_d\times \mathbf{B}=-\mathbf{E}$. Substitution in [F4] makes the electric parallel correction terms zero and gives ${\mathbf{E}}^{\mathrm{\prime }}={\gamma }_d(\mathbf{E}+{\mathbf{v}}_d\times \mathbf{B})=0$. The same component definition gives ${\mathbf{v}}_d\times \mathbf{E}=(E_0^2\mathrm{/}{B}_0)e_3$, so the magnetic formula gives ${\mathbf{B}}^{\mathrm{\prime }}={\gamma }_d(B_0-E_0^2\mathrm{/}(B_0{c}^2))e_3=\mathbf{B}\mathrm{/}{\gamma }_d$, using $1-\mathrm{\mid }{\mathbf{v}}_d{\mathrm{\mid }}^2\mathrm{/}{c}^2={\gamma }_d^{-2}$.

Boost any laboratory initial event and future-timelike four-velocity into the primed chart; metric and future-cone preservation keep the transformed data future timelike by [F8]. Let the boosted initial event have coordinate time $t{\mathrm{\prime }}_0$ and spatial position $\mathbf{x}{\mathrm{\prime }}_0$, with spatial momentum $\mathbf{p}{\mathrm{\prime }}_0$. The primed field is uniform magnetic by step 2.1. By [A2] and [F3], its coordinate equations are $d\mathbf{x}\mathrm{\prime }\mathrm{/}dt\mathrm{\prime }=c\mathbf{p}\mathrm{\prime }\mathrm{/}\sqrt{m^2{c}^2+\mathrm{\mid }\mathbf{p}\mathrm{\prime }{\mathrm{\mid }}^2}$ and $d\mathbf{p}\mathrm{\prime }\mathrm{/}dt\mathrm{\prime }=q(d\mathbf{x}\mathrm{\prime }\mathrm{/}dt\mathrm{\prime })\times \mathbf{B}\mathrm{\prime }$; their right sides depend on momentum and the constant field, not on time. Put $s=t\mathrm{\prime }-t{\mathrm{\prime }}_0$. Apply [F5] at $s=0$ to position $\mathbf{x}{\mathrm{\prime }}_0$, momentum $\mathbf{p}{\mathrm{\prime }}_0$ and field $\mathbf{B}\mathrm{\prime }$, obtaining its global orbit $(\mathit{\xi }(s),\mathit{\pi }(s))$. Set $\mathbf{x}\mathrm{\prime }(t\mathrm{\prime })=\mathit{\xi }(t\mathrm{\prime }-t{\mathrm{\prime }}_0)$ and $\mathbf{p}\mathrm{\prime }(t\mathrm{\prime })=\mathit{\pi }(t\mathrm{\prime }-t{\mathrm{\prime }}_0)$. The chain rule leaves those autonomous equations unchanged and gives precisely the required initial data at $t\mathrm{\prime }=t{\mathrm{\prime }}_0$; future timelikeness is unchanged. Conversely, translating any primed solution by the same constant time shift gives a solution with those data at $s=0$, so [F5] gives uniqueness on every common connected interval. The action in [A1] is unchanged by the boost: [F8] preserves the metric term, and pulling the one-form potential back by the coordinate change preserves its line integral. Thus stationary curves map to stationary curves, and under [A3], [F2] gives the covariant force equation in either chart. Applying the inverse Lorentz map therefore gives a laboratory worldline with the original initial data. Conversely, any laboratory solution transforms to a primed solution and coincides with the unique [F5] orbit; hence all laboratory solutions are exactly the inverse boosts of those magnetic orbits.

With $\mathit{\beta }={\mathbf{v}}_d\mathrm{/}c$, the inverse $B(-\mathit{\beta })$ from [F11] applied to the time-coordinate formula [F9] gives $t={\gamma }_d(t^{\mathrm{\prime }}+{\mathbf{v}}_d\cdot {\mathbf{x}}^{\mathrm{\prime }}\mathrm{/}{c}^2)$. Differentiating along the magnetic-frame orbit gives $dt\mathrm{/}d{t}^{\mathrm{\prime }}={\gamma }_d(1+{\mathbf{v}}_d\cdot {\mathbf{v}}^{\mathrm{\prime }}\mathrm{/}{c}^2)$. Since its speed satisfies $\mathrm{\mid }{\mathbf{v}}^{\mathrm{\prime }}\mathrm{\mid }<c$ by [F5, F6], this derivative is at least ${\gamma }_d(1-\mathrm{\mid }{\mathbf{v}}_d\mathrm{\mid }\mathrm{/}c)>0$. The positive lower bound makes $t$ strictly increasing and unbounded in both directions as $t^{\mathrm{\prime }}$ ranges over $\mathbb{R}$; thus the inverse-boosted curve is defined as a laboratory-time orbit for all real $t$.

Finally, ${\mathbf{v}}_d\times \mathbf{B}=-\mathbf{E}$ from step 2.1, so the laboratory force in [F3] under [A3] vanishes at constant $\mathbf{v}={\mathbf{v}}_d$. The corresponding momentum is constant and its speed is subluminal by step 2.1. This exhibits one exact drift solution; step 3.1 shows that general initial data instead produce inverse-boosted magnetic orbits. No assertion is made about null or supercritical fields.