# 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.

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 $$x$$.