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Imagine a space ship orbiting the sun at near relativistic speeds, when compared to Earth (say from the reference frame of the sun) there would be x amount of time dilation.

This would mean people on the ship could live for 100's of Earth years.

My question is what would happen if a second ship flew directly away from the Sun and back (ignoring acceleration) at the same speed the orbiting ship is travelling, would the time dilation for the ship that travels in a straight line be the same as that moving in a circle?

The reason I ask is that I was imagining a scenario where intergalactic travel is possible as time dilation means even a journey of light years only ages the crew a fraction of that time, with space stations orbiting certain star systems at similar speeds so that people that don't work on ships can live a similar length of time. I understand that time dilation is all linked to a certain reference frame but working with a reference frame of the sun what would the time dilation be for the different objects taking different paths but at the same speed relative to the sun? If the ship docked back with the space station would they both say the same amount of time had passed?

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    $\begingroup$ The lifetime of a relativistic ship crew is always the same, it's their natural lifetime (if they get old, that is, relativistic flight is a dangerous proposition). The only thing that happens in a relativistic rocket is that the distance to the future universe seems to shrink in the direction the ship is traveling. The energy requirements for such a scenario are so large, though, that baryonic relativistic travel is a technical near-impossibility. In any case, baryonic travel is always worse than light-travel (during which one does not age, at all), so it's not even worth the expense. $\endgroup$ – CuriousOne Sep 8 '15 at 12:47
  • $\begingroup$ Interesting point that you wouldn't age at all during light-travel it didn't even occur to me! I understand that from the perspective of the crew their life would still be 80 years but I was trying to reconcile those that didn't move living 1000's of years (i.e. their planet might have been destroyed by war etc) :) $\endgroup$ – JackFrost Sep 8 '15 at 13:04
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    $\begingroup$ Time depends on the relative state of the moving clock and it seizes entirely for the case of light. The only quantity that has physical relevance in that case is a distance. In other words: to light the entire future happens at the same time, just not in the same place. $\endgroup$ – CuriousOne Sep 8 '15 at 13:09
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"would the time dilation for the ship that travels in a straight line be the same as that moving in a circle?"

The short answer, with a few more assumptions, is yes, or it certainly can be.

Firstly, we need to agree to leave aside the problem that one cannot relativistically orbit the Earth, at least not in a freefall orbit. The orbital speed for a stable, freefalling orbit around a planet varies as $1/\sqrt{r}$, where $r$ is the orbit radius, so once you are travelling at more than the escape velocity $v_e$ at the Earth's surface, you need to be imparting a constant inwards directed thrust to maintain an orbit at this speed.

So now let us look at the problem from an inertial observer, hanging about in deep space around the Earth and also let's think of the orbit as being far enough from the Earth so that we can neglect gravitational relativistic effects.

From this observer's standpoint, if the two spaceships always maintained the same constant speed, then their paths through spacetime would have the same arclength (the same total proper time) when they finally met again - suppose for the sake of argument that they meet again at the position of our stay-at-home, inertial observer, so that all three come together and compare clocks. Therefore, both the travelling spacefarers would have aged the same amount, and both would have aged less than the stay at home inertial observer. We would need to neglect the acceleration of the out-and-back again observer as the he/she decelerated, turned around and accelerated back to cruising speed, but as long as the turnaround period is small compared to the whole journey, this error can be made as small as we like.

The out and back again spacefarer travels a dogleg path through spacetime to join the inertial observer who travels straight along the $t$ direction, as in the conventional twin paradox. The third, orbitting spacefarer travels a helical path through spacetime, making the same, constant angle to the time axis (in the inertial observer's spacetime) as the doglegging spacefarer.

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  • $\begingroup$ I thought that it would work like this, as from the perspective of earth both would have travelled the same "distance", and I was working under the assumption most of the nitty gritty details would need to be over looked :) $\endgroup$ – JackFrost Sep 8 '15 at 13:07

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