Rindler motion and Horizon 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?   
 A: 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 $x$. 
