I'm aware that the twin 'paradox' is generally just a mistake of (mis)treating the traveler's frame as an inertial one, whereas it isn't, because he accelerates to make the turnaround. There just on detail I can't get. I apologize for the length, and thanks to anyone who bears with me.

I've been looking at a great website for relativity simulations, https://www.refsmmat.com/jsphys/relativity/relativity.html. It shows multiple observers, and all clocks from each one's frame of reference.

For the twin 'paradox', it shows the duration of the trip from two inertial frames: 1) the twin who we call the one 'at rest' 2) an observer that passes #1 as the traveler takes off and 'moves' in sync with him, but does not make the turnaround with him. The simulation assumes instantaneous acceleration, and the traveler makes the turnaround at t=50 on his own clock - we can say that this was agreed upon from the beginning.

From #1's perspective, the traveler accelerates to .866c and makes the turnaround at t=50, which is t=100 for #1. The traveler continues at .866c until he comes home. His clock shows t=100, and #1's shows t=200.

From #2's perspective, the 'home' is travelling away at .866c, the traveler decelerates to 0, waits t=50 on his clock, then accelerates to .99c. He arrives home at t=400 on #2's clock. From #2's perspective, the turnaround happened at t=25 on #1's clock. The math works out approximately: according to #2, the 'home' was travelling at .866c * 400 units of time, giving a distance travelled of 346.4, and the traveler raced towards #1 at .99c * 350 units of time, giving 346.5.

What bothers me is like this:

Presumably, all parties can know how much fuel was taken on board and what amounts were expended. From the home perspective, 33% of the fuel was used accelerating to .866c, 33% reversing the acceleration, and 33% accelerating home. However, from #2's frame, 33% was used decelerating from .866c to 0, and after a wait of 50, 66% was used to accelerate to .99c. For arguments sake, let's give the spacecraft a mass of 1kg. At .866c its kinematic energy is 8.986×10^16J. So, 33% of the fuel was expended to transfer that much energy on the first leg. At .99c is kinematic energy is 5.472×10^17J - about 6x the energy of .866c. How could #2 explain how just 2x the fuel (the 66%) imparts 6x the energy?

  • $\begingroup$ Why does one frame see the traveler accelerate initially and another frame see the traveler already at speed? $\endgroup$ – Asher Nov 2 '17 at 20:08
  • $\begingroup$ @Asher, because the other frame is moving at .866c in frame #1, but is inertial in his own frame, watching #1 move away at .866c. If you take a look at the simulation, it will be easier to see. $\endgroup$ – afuna Nov 2 '17 at 20:10
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    $\begingroup$ acceleration is not relative. All inertial observers will see Twin 2 accelerate, even if the acceleration happens to bring him into the same frame as a particular observer. $\endgroup$ – Asher Nov 2 '17 at 20:17
  • $\begingroup$ @Asher - sorry I misread your comment; either way, I wrote that frame #2 sees him decelerate to 0 - decelerating is a form of acceleration. $\endgroup$ – afuna Nov 2 '17 at 20:23
  • $\begingroup$ ah, yes... which means it is I who misread. I didn't keep up with the change of frames so I didn't consider that observer 1's "accelerate to 0.866c" is the same as observer 2's "decelerate to 0." It's clear now $\endgroup$ – Asher Nov 2 '17 at 21:36

When a rocket motor accelerates a reaction mass, the kinetic energy of the reaction mass increases.

When a rocket motor decelerates a reaction mass, the kinetic energy of the reaction mass decreases.

Now let us use heat pump terminology: A motor burns fuel to produce energy, which is used to pump kinetic energy from a reaction mass to a spacecraft. One joule of energy can pump many joules of energy, if there is a reservoir from where kinetic energy can be taken.

In the OP there was no mention about reaction mass, so the scenario is very unphysical. I recommend using electric trains on the surface of a planet instead of rockets.


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