# Duration of an accelerated movement in special relativity

Consider a point $$P$$ which does the following (one dimensional) movement:

1. Uniform acceleration $$a_0$$ (proper aceleration) from $$t=0$$ to $$t=T$$ (time measured in lab's frame).

2. Uniform deceleration $$a_0$$ from $$t=T$$ to $$t=3T$$.

3. Uniform acceleration $$a_0$$ from $$t=3T$$ to $$t=4T$$.

I want to know how much time has passed in $$P$$'s frame. That is, what is $$\Delta\tau$$?

Since the movement starts with no velocity, we can easily calculate its four-velocity: $$u^\mu=c(\cosh(a_0\tau/c),\sinh(a_0\tau/c),0,0).$$

As $$u^0=\gamma c$$, we conclude that $$\gamma=\cosh(a_0\tau/c)$$ and thus the duration of the first part can be calculated as $$T=\int_0^{\Delta\tau_i} \cosh\left(\frac{a_0}{c}\tau\right)\mathrm{d}\tau\quad\implies\quad \Delta\tau_i=\frac{c}{a_0}\mathrm{arcsinh }\left(\frac{a_0}{c}T\right).$$

However, for the other parts the initial velocity is not zero and then the function I get when calculating $$\Delta t=\int\gamma\:\mathrm{d}\tau$$ is not easily invertible.

Is there a better way to do this?

• Can you see that there are four congruent legs of your trip? Split the trip from T to 3T. – robphy Feb 1 '19 at 21:22
• The proper time differences between $t=0$-$t=T$, $t=T$-$t=2T$, $t=2T$-$t=3T$ and $t=3T$-$t=4T$ are all the same? That is not clear to me. – Gabriel Feb 1 '19 at 21:24
• Can you sketch it on a spacetime diagram? – robphy Feb 1 '19 at 21:26
• @robphy To be honest, I never found spacetime diagrams to be that useful. Probably I don't understand them well yet. – Gabriel Feb 1 '19 at 21:27

From your setup, there is symmetry to exploit. If you break up the T-to-3T trip in half, you can see [by drawing a spacetime diagram] that you have four congruent legs to your trip. You just calculated the duration of the first leg. So, by symmetry, multiply that by 4 to get the total duration.

update2:

In this second update, I have shown three uniformly accelerated observers as with dashed worldlines, which are hyperbolas on a spacetime diagram. Since the magnitudes of their proper-accelerations are equal, these hyperbolas are congruent--related by reflections in space, reflections in time, and translations on this spacetime diagram.
[If we were doing Galilean relativity, these would be congruent parabolas.]

This round-trip starting and ending at rest in this frame is essentially the splicing of four congruent portions.

[As a simpler case, suppose the traveler departed with a nonzero outgoing speed and returned at the same speed.... essentially the magenta observer. Is it clear that if that trip were split into halves, those halves have equal elapsed times.] Note that the 0-to-T leg is the mirror image in space of the 2T-to-3T leg.
Note that the T-to-2T leg is the mirror image in space of the 3T-to-4T leg.

The 0-to-T leg is the "time-reverse" of the 3T-to-4T leg.
The T-to-2T leg is the "time-reverse" of the 2T-to-3T leg.

update3:

Consider the first leg of the trip... uniform-acceleration $$a_0$$ to the right.

Suppose instead the traveler went in the opposite direction in space.
After the same time $$\Delta t_1$$, is it clear that the traveler would have the opposite velocity (same speed, opposite direction) as the original?
This space-reflected segment is a translation of the leg that starts [back] at the turnaround event.

Similarly, if you ran time in reverse [retrodiction, if you will], then at earlier time $$-\Delta t_1$$, is it clear that the traveler would have had the opposite velocity (same speed, opposite direction) as the original?
This time-reflected segment is a translation of the leg that returns at the reunion event.

• I would apreciate any further explanation about what the durations are the same. – Gabriel Feb 1 '19 at 21:31
• I added a spacetime diagram to my answer. – robphy Feb 1 '19 at 22:01
• Thank you for you answer but I still don't think it is clear that the durations are the same. I feel like I have just drawed the diagram in such a way. – Gabriel Feb 2 '19 at 9:14
• I updated the spacetime diagram to show three uniformly-accelerated observers (with the same magnitude of proper-acceleration) that your traveler essentially hitched rides with to complete the round-trip. – robphy Feb 2 '19 at 13:03

You correctly wrote the expression for the four-velocity and $$\gamma$$ but it'd be much easier to use straight forward the time coordinate (Rindler) transform for a uniformly accelerated frame (without integral calculus and 4-velocities) as $$t=\frac{c}{a_0}\mathrm{asinh }\left(\frac{a_0}{c}\tau\right)$$ where $$\tau$$ is proper time in (in P's frame) and $$t$$ is time in "lab frame" or at rest frame. So, using this general transform for the first part $$t_1=\frac{c}{a_0}\mathrm{asinh }\left(\frac{a_0}{c}T\right)$$ which is the same as you obtained. The next part can be derived by the use of the same time transform but with the negative $$a_0$$ and for the duration is $$2T=3T-T$$, so $$t_2=\frac{c}{-a_0}\mathrm{asinh }\left(\frac{-a_0}{c}2T\right)=\frac{c}{a_0}\mathrm{asinh }\left(\frac{a_0}{c}2T\right)$$ Then obviously $$t_2 \neq 2t_1$$. The third part is given by delta $$T=4T-3T$$, so $$t_3=\frac{c}{a_0}\mathrm{asinh }\left(\frac{a_0}{c}T\right)$$