# 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 $$x$$.

• Consider $\frac{d(\gamma v)}{dt}=\alpha$, where $\alpha$ is the proper acceleration, integrating over time : $\gamma v =at$ or $v=\frac{at}{\sqrt{1+\frac{a^2t^2}{c^2}}}$. From this, after integrating again over $\int_0^tdt$ and space $\int_0^xdx$ we get $x=\frac{c^2}{a}\sqrt{1+\frac{a^2t^2}{c^2}}-1$. Rearranging the terms yields $(x+1)^2-c^2t^2=\frac{c^4}{\alpha ^2}$. Can't I just say: $x+1=\sinh(\tau \alpha)$ and $ct=\cosh(\tau \alpha)$. Perhaps the $1$ fromin $x+1$ can be absorbed in some initial condition? $\tau$ is the proper time measured in the inst. rest frame. Does this make sense? Commented Nov 7, 2021 at 14:06
• @AlexanderCska I think you're describing the Minkowski coordinates along a single uniformly-accelerating worldline (after switching sinh and cosh, which I assume was just a typing mistake). The question was asking about the relationship between the Minkowski coordinate system and another coordinate system (Rindler), both of which cover the whole wedge $|X|\geq T$, not just a single worldline. Commented Nov 7, 2021 at 15:07
• @AlexanderCska You can indeed define another Rindler coordinate system in which my equation (2) is replaced by $X+1=x\cosh(at)$. There are infinitely many different Rindler coordinate systems, all offset from each other by shifts in one or more of the coordinates on either side of the transformation. Commented Nov 7, 2021 at 15:07
• i think that these are the same. I need $\frac{c^2}{\alpha}$ to get $x+1= \frac{c^2}{\alpha} \cosh(\tau \alpha)$ and $ct= \frac{c^2}{\alpha} \sinh(\tau \alpha)$. Then we could identify $\frac{c^2}{\alpha}=\tilde{x}$ and the hyperbolic angle as $\tilde{t}=\alpha \tau$. I mixed up $\sinh$ and $\cosh$ and forgot $\frac{c^2}{\alpha}$ before. Having done so, this becomes exactly the same to the form you derived. I'm not sure what $t$ and $x$ mean in your case. These should not be proper distance and time? Commented Nov 7, 2021 at 17:11
• @AlexanderCska You're describing the Minkowski coordinates along just one worldline, parameterized by its proper time $\tau$. The question is about the relationship between two different coordinate systems in a whole region of spacetime: the $X,T$ coordinate system and the $x,t$ coordinate system. (Beware that your lowercase $x,t$ correspond to my uppercase $X,T$, except for the $+1$ shift in $X$.) The question isn't about relating either set of coordinates to proper distance or proper time. Those are only defined along a given worldline. Coordinates aren't limited to any individual worldline. Commented Nov 7, 2021 at 19:59