# What is the Rindler wedge?

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?

• $c^2/\alpha$ is subtracted to make $x=0$ at $t=0$. – G. Smith Sep 2 at 21:56
• 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? – Eddward Sep 3 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 $$\tau=\frac{c}{\alpha} acosh(2)$$