Consider relativistically accelerating body along the coordinate $$x = \frac{c^2}{\alpha} \cosh\left(\frac{\alpha}{c} \tau\right) -\frac{c^2}{\alpha}$$ Why there is such "special" distance as $$d=\frac{c^2}{\alpha}$$ Anything happens at point of $x=d$? I see in Wikipedia it is called also the Rindler Horizon distance.

Is there any significannce of such time $$T=\frac{c}{\alpha}$$ What is the Rindler wedge and are these values related to it?

  • $\begingroup$ $c^2/\alpha$ is subtracted to make $x=0$ at $t=0$. $\endgroup$ – G. Smith Sep 2 '19 at 21:56
  • $\begingroup$ It is clear- the question is not on $x=0$, but on $d=c^2/\alpha$. I meant the spit horizon in a Rindler wedge occurs at distance $d=c^2/\alpha$. What does it mean? $\endgroup$ – Eddward Sep 3 '19 at 4:08

The significance of the distance $d$ for the uniformly accelerated observer (with initial $x=0$ at $t=0$) is that $\textit{his Rindler horizon}$ is located at distance $$d=\frac{c^2}{\alpha}$$ in the direction opposite to his acceleration (behind him). When he approaches x=d he won't be able to receive any information from the point of his origin. It will happen at the time when $\cosh\left(\frac{\alpha}{c} \tau\right)=2$ or $$\tau=\frac{c}{\alpha} acosh(2)$$

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