Acceleration calculation in Special Relativity

[Bob Walance]

TL;DR

Comparing inertial frame elapsed time with four accelerated frames.

Alex and Babs are on the ISS. Babs departs in her spaceship, and her engine always imparts either +1g or -1g on her body until the journey ends.

There are four phases to Babs' journey. Each phase lasts for 1 year according to Babs.

Phase 1: Babs departs and turns on her engine to its NORMAL position. At the end of this phase Babs has reached maximum velocity away from Alex.
Phase 2: Babs changes her engine to REVERSE. At the end of this phase Babs has reached zero velocity with respect to Alex.
Phase 3: Babs' engine continues in its REVERSE setting. At the end of this phase Babs has reached maximum velocity toward Alex.
Phase 4: Babs changes her engine back to NORMAL. At the end of this phase Babs has reached zero velocity with respect to Alex. She turns off her engine, and they can compare their clocks.

The Rindler equation that I use is to calculate the elapsed times for Alex is:

t = (c/A)sinh(AT/c)
where
t is Alex's proper time.
A is the proper acceleration for Babs in her rocketship
T is the proper time for Babs in her rocketship
c is the speed of light

When I use this formula for Phase 1 (A=g and T=1 year), the calculation for t yields 1.19 years elapsed on Alex's clock.

Will the calculation used for phase 1 be accurate for phase 2? I question this only because the starting velocities for these two phases are very different.

 

[DaveC426913]

Will the calculation used for phase 1 be accurate for phase 2?

Yes.

I question this only because the starting velocities for these two phases are very different.

The four phases all have symmetrical profiles, just different choices of frames of reference.

The starting velocity of phase 2 is zero - relative to a nearby co-moving object, and Babs' rocket accelerates away from it at the same rate (albeit with opposite sign) as other phases, reaching the same relative velocity (again, with opposite sign) after the same duration.


Let me throw in the caveat that I may be expressing this poorly. Let's wait for someone to weigh in who can express it better.

 

[Ibix]

TL;DR: Comparing inertial frame elapsed time with four accelerated frames.

Will the calculation used for phase 1 be accurate for phase 2? I question this only because the starting velocities for these two phases are very different.

Yes. In this case, the deceleration phase is just the time-reverse of the acceleration phase, and the return leg is the same as the outbound leg. So the total time is four times your calculation for the outbound acceleration phase.

 

[Histspec]

Will the calculation used for phase 1 be accurate for phase 2? I question this only because the starting velocities for these two phases are very different.

As the others already pointed out, the answer is yes. You find a more detailed description in
https://arxiv.org/abs/physics/0411233
where you find the formula $T=\frac{4}{g}\sinh\frac{g\tau}{4}$ (case c).

In case not everything is symmetric, you can use a more general formula that also includes non-zero initial velocities and times: $$
T=T_{0}+\frac{c}{g}\left\{ \sinh\left[\tanh^{-1}\left(\frac{u_{0}}{c}\right)+\frac{g\tau}{c}\right]-\frac{u_{0}\gamma_{0}}{c}\right\} $$
see
https://en.wikipedia.org/wiki/Hyperbolic_motion_(relativity)