Replies

Re:"The "twins paradox" persists, even in the absence of accelerations."

I think I've introduced some confusion here. If the problem is looked at as assuming the frames of the spaceship and the Earth are symmetric, each inertial observer concludes the other has aged less. The only way to identify the correct frame for time dilation is to note which one has accelerated, or is moving with respect to the turnaround point.

The Earth observer sees the distant turnaround point as fixed and at a greater distance than the observer on the spaceship, after the spaceship hits cruising velocity. In your examples, the spaceship was always the one moving and frame symmetry was not mentioned.

If the problem is looked at as assuming the frames of the spaceship and the Earth are symmetric, each inertial observer concludes the other has aged less.

That's not an assumption. That's a requirement of the principle of relativity. If it's not symmetric, it's not relative. You seem to want to believe that there is a "slow frame" and a "fast frame". That can't be right, unless one frame is preferred.

The only way to identify the correct frame for time dilation is to note which one has accelerated, or is moving with respect to the turnaround point.

Stop right there. Each observer notes correctly that time is slower in the other frame. All frames are equally "correct for time dilation". The final difference in the counters is caused by the fact that the traveller is not in a frame, but changes frames in transit. The difference between the frames, leveraged over the distance, causes the difference.

In your examples, the spaceship was always the one moving and frame symmetry was not mentioned.

Oh, no problem. Just turn the problem around: after one month, the Earth sends out a significantly faster ship to overtake the traveller, along with the homebound twin's time counter. From the traveller's POV, the Earth counter went outwards at some velocity from his (stationary) ship, and then came back to the same spot via another ship. When the ships meet, the counters are compared. Now it's the traveller's counter that has more counts, and by the same amount as in the first problem. The people back on Earth aren't surprised by the result, as the time dilation on the faster ship would be greater, of course.