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I have problem to calculate the proper time for Rindler coordinates: the coordinates in Minkowski space with constant acceleration are given by:$$ \begin{alignat}{7} t' &~=~ \frac{c}{g} \, && \sinh{\left(\frac{g}{c\tau}\right)} \\ x' &~=~ \frac{c^2}{g} \, && \cosh{\left(\frac{g}{c\tau}\right)} \end{alignat} $$

The proper time is $$ \tau_{\text{p}}~=~\int{\sqrt{1 - \frac{v^2 \left(t' \right)}{c^2}}}\, \mathrm{d}t' \,,$$with$$ \begin{alignat}{7} t' & ~=~ \frac{c}{g} \, && \sinh{\left(\frac{g}{c\tau}\right)} \\ \mathrm{d}t' & ~=~ && \cosh{\left(\frac{g}{c\tau}\right)} \,. \end{alignat} $$

How can I calculate $v \left(t' \right)$?

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  • $\begingroup$ $v(t')=dx'/dt'$ ? $\endgroup$ – Arturo don Juan May 19 '18 at 20:55
  • $\begingroup$ Also, I think you could use the fact that $\tau=t'/\gamma$. $\endgroup$ – Arturo don Juan May 19 '18 at 20:56
  • $\begingroup$ $ dx'/dt'=c$ so $\tau_p=0$ ???? $\endgroup$ – Eli May 19 '18 at 21:14
  • $\begingroup$ so $v(\tau)=dx'/dt'=c\tanh(g/c \,\tau)$ and $\tau_p=\int\sqrt(1-(v^2(\tau))/c^2)\,\cosh(g/c\,\tau)\,d\tau$ I think this is o.k. Tank you $\endgroup$ – Eli May 19 '18 at 21:26
  • $\begingroup$ (1) Physically, $v(t)$ should tend towards $c$ as $\tau\rightarrow\infty$, so getting $v=c\tanh(.)$ seems right. (2) I think your LaTeX messed up. $\endgroup$ – Arturo don Juan May 19 '18 at 21:59
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Velocity is defined as $dx'/dt'$, which in your case comes out to be:

$$v'=\frac{dx'}{dt'}=c\tanh\left(\frac{g\tau}{c}\right)$$

where $g$ is the proper acceleration. Notice that as $\tau\rightarrow\infty$, $v\rightarrow c$, as we would expect.

From this, you can calculate the proper time itself $d\tau=dt'/\gamma$.

$$\begin{align} d\tau&=dt'\sqrt{1-v'^2/c^2}\\ &=\left(d\tau\cosh\left(\frac{g\tau}{c}\right)\right)\sqrt{1-\tanh^2\left(\frac{g\tau}{c}\right)}\\ &=d\tau \end{align}$$

as expected. If you want to find $\tau$ as a function of $t'$, you would integrate:

$$\begin{align} \tau&=\int_{t'_0}^{t'} dt''\sqrt{1-v'(t'')^2/c^2}\\ &=\int_{t'_0}^{t'} dt''\left(1+\sinh^2(g\tau(t'')/c)\right)^{-1/2}\\ &=\int_{t'_0}^{t'} dt''\left(1+(gt''/c)^2\right)^{-1/2}\\ &=\frac{c}{g}\left[\sinh^{-1}(gt'/c)-\sinh^{-1}(gt'_0/c)\right] \end{align}$$

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  • $\begingroup$ Tank you. I have this Problem: You go on the Rindler geodesic for 10 [s] with g=10 [m/s^2] and the next 20 [s] with g=5 [m/s^2] . How many meter you travel on the geodesic $\endgroup$ – Eli May 20 '18 at 8:26
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    $\begingroup$ @Eli Split the problem in two parts: first, 10 seconds of g=10 m/s^2 acceleration starting from rest; second, 20 seconds of g=5 m/s^2 acceleration starting from the position and velocity from which you left the previous (first) motion. $\endgroup$ – Arturo don Juan May 20 '18 at 19:21

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