Let's say we have a frame of reference at rest $R'$ and a frame of reference uniformly accelerated $R'$ with a constant acceleration $\alpha$.

I understand that we can show that the coordinates $(x',ct')$ in the Minkowski spacetime diagram are :

$$ \begin{equation} \begin{array} xx'(\tau) = \frac{c^2}{\alpha}\left(\cosh\left(\frac{\alpha \tau}{c}\right)\right) \quad ;& ct'(\tau) = \frac{c^2}{\alpha}\left(\sinh\left(\frac{\alpha \tau}{c}\right)\right) \end{array} \end{equation} $$

From that point, we see that the path followed by an observer placed in the accelerated frame of reference seen from the frame of reference at rest is an hyperbolic motion.

As far as I understand, Rindler coordinates are used to describe this hyperbolic motion. However, I don't understand how I can derive them from the two relations ($x'(\tau), ct'(\tau)$) I wrote above.


1 Answer 1


You consider a family of observers parametrised by $\xi$ with trajectories coordinates $$ x(\xi,\tau) =\xi \cosh\sigma, \\ t(\xi, \tau) = \xi \sinh \sigma, $$ then Rindler coordinates are the pair $\xi$,$\tau$. (I set $c=1$). The metric is $$ d\tau^2= dt^2-dx^2 = \xi^2 d\sigma^2 -d\xi^2 $$ We see that along the curve $\xi= constant$ the elapsed proper time is $\tau= \xi \sigma$ so on the trajectory $\xi=1$ the coordinate $\sigma$ coincides with the proper time of the observer moving along that trajectory. Further, the observer on the trajectory with parameter $\xi=1$ has constant (in hs frame) acceleration $a=1$. One can see as the trajactory becomes $$ x= 1-\frac 12 \sigma^2\\ t=\sigma $$ for small $\sigma$. That the acceleration is constant follows from the fact that the hyperbola is the Minkowski geometry equivalent of the circle, and the Rindler metric $$ d\tau^2= \xi^2 d\sigma^2 -d\xi^2 $$ is the hyperbolic version of polar coordinates' $$ ds^2= r^2 d\theta^2+dr^2. $$
Just as every bit of the circle is equivalent to every other bit, to the observer on the hyperbola every bit of the $\xi=1$ trajectory is the same as every other bit, so the $\xi=1$ observer feels an eternal acceleration of $a=1$.

What is great about Rindler coordinates is that you can see that our accelerating obserever can outrun any photon that starts more that "1" to her left --- She has an event horizon! Restoring $c$ into usual, units this means that at a steady acceleration of one $g$ the event horizon turns out to be one light year away. This weird numerical coincidence is T.J. Bass's famous Godwhale equation "$gy=c$" (One year accelerating at one $g$ takes you to the speed of light in Newton/Galileo kinematics). It is comparable to the weirdness that if we draw a map in which the earth/sun distance (1au) is reprented one inch, then one light year is represented by one mile.

  • $\begingroup$ Google says (1 AU)*(1 mile)/((1 inch)*c) is 1.00190187 years. Close enough for government work. ;) (It looks like the Google Calculator uses the tropical year, 365.242198781 days). $\endgroup$
    – PM 2Ring
    Commented May 10, 2020 at 7:05
  • $\begingroup$ So basically we consider that $\xi = \frac{c^2}{\alpha}$ and $\sigma = \frac{\alpha \tau}{c}$ ? $\endgroup$ Commented May 10, 2020 at 7:13

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