Spacetime diagram in Rindler coordinates I am currently studying the Rindler coordinates 
$$T = x \sinh(a t) , \, X = x \cosh(at).$$
I am trying to understand the connection between the Rindler coordinates and their Minkowski diagram. 


*

*Why is $t= \pm \infty$ in the case of $x=0$ and why does that correspond to a line with a $45^\circ$ angle? 

*What are the straight lines for constant $t$ and how can I obtain them from the Rindler coordinates? 

*Why are the lines for constant $x$ curved?


I feel like I am failing to understand this Minkowski diagram at a very basic level and clearing that up will help a lot. I already read the wikipedia page but I couldn't wrap my heard around the connection between the coordinates and the diagram yet.
 A: Start with the coordinate system $(T,X)$, so that every point in two-dimensional spacetime is labeled by a pair of numbers $(T,X)$. Define $t$ and $x$ implicitly by
$$
  T =x\,\sinh(at)
\tag{1}
$$
$$
  X =x\,\cosh(at).
\tag{2}
$$
To depict the relationship graphically, consider a graph whose vertical and horizontal axes are labeled by $T$ and $X$, respectively. 
To the questions, consider the identities
$$
  \frac{T}{X} = \frac{\sinh(at)}{\cosh(at)}
\tag{3}
$$
$$
  X^2-T^2=x^2.
\tag{4}
$$
Now, here are the answers to the specific questions, in a different order.

What are the straight lines for constant $t$ and how can I obtain them from the Rindler coordinates? 

Equation (3) says that the ratio $T/X$ is completely determined by $t$, regardless of $x$. Therefore, the points with a given value of $t$ are the same as the points with a given ratio $T/X$. These are straight lines through the origin in the $T,X$ plane. The maximum and minimum possible slopes are $+1$ and $-1$, respectively, because these are the maximum and minimum possible values of the right-hand side of equation (3). 

Why are the lines for constant $x$ curved?

Equation (4) says that the combination $X^2-T^2$ is completely determined by $x$, regardless of $t$. Therefore, the points with a given value of $x$ are the same as the points with a given value of $X^2-T^2$. To visualize this, choose a particular value for $x$ and write equation (4) like this:
$$
  X = \pm\sqrt{T^2 + x^2}.
\tag{5}
$$
Since $x$ is a constant (we chose its value), this equation gives us $X$ as a function of $T$. We get two curves, one for each sign of the square root. For $x\neq 0$, each of these curves is a hyperbola. The one on the negative-$X$ side passes through the point $(T,X)=(0,-|x|)$, and the one on the positive-$X$ side passes through the point $(T,X)=(0,|x|)$. To see that the asymptotic slopes of these hyperbolas are $\pm 1$, consider equation (5) in the limit $T\rightarrow \pm\infty$. In this limit, the term $x^2$ is negligible, which means that the asymptotes are $X=\pm\sqrt{T^2}=\pm|T|$.  The smaller the value of $x^2$, the more closely the hyperbolas "hug" these asymptotes. In the limiting case $x=0$, the hyperbolas become the asymptotes themselves. And this answers the next question...

Why is $t=\pm \infty$ in the case of $x=0$ and why does that correspond to a line with a $45^\circ$ angle? 

Equation (4) says that if $x=0$, then $X=\pm T$. These are the two $45^\circ$ lines through the origin. And if $X=\pm T$, then equation (3) says $t=\pm \infty$. As described above, this is just a limiting case of the pair of hyperbolas we get for any given non-zero value of $x$.
