The uniformly accelerated worldline: hyperbolic motion, constant four-acceleration magnitude, and identification as a Rindler flow line

(The tangent.) Writing $x$ in the chart, the coordinate functions are $\tau \mapsto R\sinh (\alpha \tau \mathrm{/}c)$, $\tau \mapsto R\cosh (\alpha \tau \mathrm{/}c)$, and the two constants $0$; differentiating with the chain rule and the hyperbolic derivative formulas gives $x^{\mathrm{\prime }}(\tau )=(R\frac{\alpha }{c}\cosh \theta ,R\frac{\alpha }{c}\sinh \theta ,0,0)=(c\cosh \theta ,c\sinh \theta ,0,0)$, because $R\alpha \mathrm{/}c=c$ by $R=c^2\mathrm{/}\alpha$. In particular the coordinate functions are $C^{\mathrm{\infty }}$, since each derivative has the same form with $\sinh$ and $\cosh$ exchanged.

(Elementary hyperbolic identities.) From the exponential definitions, $\cosh \theta -\sinh \theta =e^{-\theta }$ and $\cosh \theta +\sinh \theta =e^{\theta }$; combined with ${\cosh }^2\theta -{\sinh }^2\theta =1$ and $\cosh \theta >0$ this gives $\cosh \theta >\mathrm{\mid }\sinh \theta \mathrm{\mid }\ge 0$, hence $\mathrm{\mid }\tanh \theta \mathrm{\mid }<1$ for every $\theta$.

(The curve is a Rindler flow line.) Substituting $\rho =R$, $y=z=0$ into $\mathrm{\Phi }$ gives $\mathrm{\Phi }(\theta ,R,0,0)={\chi }^{-1}(R\sinh \theta ,R\cosh \theta ,0,0)=x(\tau )$ with $\theta =\alpha \tau \mathrm{/}c$; hence $x$ is exactly the flow line $\mathrm{\Phi }(\cdot ,R,0,0)$, reparametrized by the affine map $\tau \mapsto \theta =\alpha \tau \mathrm{/}c$, and at $\rho =R$ the Rindler field is $n=R^{-1}{\mathrm{\partial }}_{\theta }$ with acceleration $a_{\mathrm{c}\mathrm{g}}=R^{-1}{\mathrm{\partial }}_{\rho }$.

($x$ is a $C^{\mathrm{\infty }}$ regular future timelike worldline.) By step 1.1 the coordinate functions are $C^{\mathrm{\infty }}$, so $x$ is $C^{\mathrm{\infty }}$ by [L2]; $\eta (x^{\mathrm{\prime }},x^{\mathrm{\prime }})=c^2(-{\cosh }^2\theta +{\sinh }^2\theta )=-c^2<0$ by step 1.2 and the matrix of $\eta$ in the frame, so $x^{\mathrm{\prime }}$ is timelike; and its time component is $x^{\mathrm{\prime }0}=c\cosh \theta >c\mathrm{\mid }\sinh \theta \mathrm{\mid }=\mathrm{\mid }(x^{\mathrm{\prime }}{)}_s\mathrm{\mid }$ by step 1.2, so $x^{\mathrm{\prime }}\in C^{+}$ by the coordinate criterion [L6]. Hence $x^{\mathrm{\prime }}$ is future timelike and nonzero, and $x$ is a future timelike worldline.

(Chart speed and the wedge.) The chart time is $x^0=ct$ with $x^{\mathrm{\prime }0}=c\cosh \theta$, so $dt\mathrm{/}d\tau =\cosh \theta$ and, by step 1.1, $d{x}^1\mathrm{/}dt=(d{x}^1\mathrm{/}d\tau )\mathrm{/}(dt\mathrm{/}d\tau )=c\sinh \theta \mathrm{/}\cosh \theta =c\tanh \theta$; step 1.2 gives $\mathrm{\mid }d{x}^1\mathrm{/}dt\mathrm{\mid }=c\mathrm{\mid }\tanh \theta \mathrm{\mid }<c$. Also $x^1=R\cosh \theta >R\mathrm{\mid }\sinh \theta \mathrm{\mid }=\mathrm{\mid }{x}^0\mathrm{\mid }$ by step 1.2, so the worldline lies in $W=\{x^1>\mathrm{\mid }{x}^0\mathrm{\mid }\}$.

(Proper-time parametrization and four-velocity.) By step 2.1 the curve is a $C^1$ future timelike worldline, so its proper-time function satisfies $T(b)-T(a)=c^{-1}{\int }_a^b\sqrt{-\eta (x^{\mathrm{\prime }},x^{\mathrm{\prime }})}du=c^{-1}{\int }_a^{b}cdu=b-a$ by [L8] and step 2.1; thus the given parameter is a proper-time parameter of $x$ up to an additive constant, in agreement with [L9]. Consequently the four-velocity is $U=x^{\mathrm{\prime }}\mathrm{/}{T}^{\mathrm{\prime }}=c(\cosh \theta ,\sinh \theta ,0,0)$ by [L10] and step 1.1, and $\eta (U,U)=-c^2$.

(Asymptotics.) From the addition formulas in exponential form of step 1.2, $x^1-x^0=R(\cosh \theta -\sinh \theta )=R{e}^{-\theta }$ and $x^1+x^0=R(\cosh \theta +\sinh \theta )=R{e}^{\theta }$, both strictly positive for every $\tau$. As $\tau \to +\mathrm{\infty }$ we have $\theta \to +\mathrm{\infty }$, so $e^{-\theta }\to 0$ by [L14] and $x^0=R\sinh \theta \to +\mathrm{\infty }$; as $\tau \to -\mathrm{\infty }$ we have $\theta \to -\mathrm{\infty }$, so $e^{\theta }\to 0$ and $x^0\to -\mathrm{\infty }$. Hence the worldline approaches both null planes asymptotically and meets neither. For $\theta \to +\mathrm{\infty }$, $\tanh \theta =(1-e^{-2\theta })\mathrm{/}(1+e^{-2\theta })\to 1$ by [L14]; oddness gives the negative-end limit $-1$. Thus the signed chart velocity tends to $\pm c$ and its magnitude to $c$.

(The four-acceleration and its magnitude.) Differentiating each component of $U$ in the proper-time parameter $\tau$ with the chain rule and the hyperbolic derivative formulas gives $A=U^{\mathrm{\prime }}=c(\sinh \theta ,\cosh \theta ,0,0)\frac{\alpha }{c}=\alpha (\sinh \theta ,\cosh \theta ,0,0)$. Then $\eta (A,A)={\alpha }^2(-{\sinh }^2\theta +{\cosh }^2\theta )={\alpha }^2$ and $\eta (U,A)=c\alpha (-\cosh \theta \sinh \theta +\sinh \theta \cosh \theta )=0$ by step 1.2; the second identity is consistent with [L10].

(Identification of the four-acceleration with $c^2{a}_{\mathrm{c}\mathrm{g}}$.) Along $x$ one has $dx\mathrm{/}d\tau =cn$ by steps 1.3 and 3.1 with $n=R^{-1}{\mathrm{\partial }}_{\theta }$; the total derivative of $n\circ x$ along this smooth curve is ${\mathrm{\nabla }}_{dx\mathrm{/}d\tau }n$, and by [L12] and [L11] it equals ${\mathrm{\nabla }}_{cn}n=c{\mathrm{\nabla }}_{n}n=c{a}_{\mathrm{c}\mathrm{g}}$. Therefore $A=dU\mathrm{/}d\tau =cd(n\circ x)\mathrm{/}d\tau =c{\mathrm{\nabla }}_{dx\mathrm{/}d\tau }n=c^2{a}_{\mathrm{c}\mathrm{g}}$. Substituting $\rho =R$ into $\eta (a_{\mathrm{c}\mathrm{g}},a_{\mathrm{c}\mathrm{g}})={\rho }^{-2}$ from [L13] gives $\eta (a_{\mathrm{c}\mathrm{g}},a_{\mathrm{c}\mathrm{g}})=1\mathrm{/}{R}^2={\alpha }^2\mathrm{/}{c}^4$.

(Scope.) Every step above is a finite-dimensional computation in the affine model from the cited definitional suppliers: the curve, its causal character, its proper-time parametrization, its four-acceleration, the identification $A=c^2{a}_{\mathrm{c}\mathrm{g}}$ and the asymptotics. No force law, mass, geodesic property, clock hypothesis or empirical input was used, and no choice principle was used.