1
$\begingroup$

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.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

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 $x$.

Improve this answer
$\endgroup$
6
  • $\begingroup$ 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? $\endgroup$
    – Alexander Cska
    Nov 7, 2021 at 14:06
  • $\begingroup$ @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. $\endgroup$
    – Chiral Anomaly
    Nov 7, 2021 at 15:07
  • $\begingroup$ @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. $\endgroup$
    – Chiral Anomaly
    Nov 7, 2021 at 15:07
  • $\begingroup$ 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? $\endgroup$
    – Alexander Cska
    Nov 7, 2021 at 17:11
  • $\begingroup$ @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. $\endgroup$
    – Chiral Anomaly
    Nov 7, 2021 at 19:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.