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The Rindler coordinates are: $$ \begin{align} x'(\tau)&={\frac {\cosh \left( g\tau \right) }{g}} \\ t'(\tau)&={\frac {\sinh \left( g\tau \right) }{g}} \end{align} \tag{1} $$

$$ \begin{align} x'\left(\tau,\,\xi\right)&=\left( \xi+{g}^{-1} \right) \cosh \left( g\tau \right) \\ t' \left(\tau,\,\xi\right)&=\left( \xi+{g}^{-1} \right) \sinh \left( g\tau \right) \end{align} \tag{2} $$

Questions:

  • Why do equations (1) have the horizon problem and equations (2) don't?

  • How can I calculate the horizon line?

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I simplified definition of the horizon (enough for this case, but by no means general enough) is regions of space-time where the metric becomes singular. In this setting you begin with the Minkowski metric: $$ds^2 = -dT^2 + dX^2 + dY^2 + dZ^2$$ First you should realize the coordinates in (1) cannot be correct because there is no dependence on $\xi$. Have a look at (https://en.wikipedia.org/wiki/Rindler_coordinates) for the standard Rindler coordinates.

As stated in the Wikipedia article, once you employ a transformation of the form of (1), $$\begin{array}{r} T = x\sinh(gt) \\ X = x\cosh(gt) \end{array}\tag{1}\label{eq:rind1},$$ the metric becomes: $$ds^2 = -(gx)^2dt^2 + dx^2 + dy^2 + dz^2$$ which becomes singular if either $x=0$ or $g=0$. If you use a transformation like in (2) $$\begin{array}{r} T = \left(x+\frac{1}{g}\right)\sinh(gt) \\ X = \left(x+\frac{1}{g}\right)\cosh(gt) \end{array}\tag{2}\label{eq:rind2},$$ you obtain the following: $$ds^2 = -(1+gx)^2dt^2+dx^2+dy^2+dz^2$$ whose components don't vanish for any $g$ or $x$.

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  • $\begingroup$ Tank you for your answer. So if the metric is singular we have problem. $\endgroup$ – Eli May 27 '18 at 18:36

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