Say I get on a Sci-Fi speed Rocket Ship and leave Earth at .999999% the Speed of Light to me I travel for 21 minutes reach Mars then U-Turn back to Earth for another 21 minutes at 0.999999% the Speed of Light again. Back on Earth if I compared my watch to someone else’s would my watch be slightly ahead or slightly behind?
Like if I’m the one traveling I’d expect their watch to be slightly ahead of mine because slightly less time has passed. But at the same time from my frame of view. Earth and my bussy with the watch just shot away from me for 21 minutes and then returned and came back 21 minutes later so my watch should be ahead of theirs since they were the one traveling.
In: Physics
This is the [twin paradox](https://en.wikipedia.org/wiki/Twin_paradox).
Nobody sees their own clock running slower, but they see all moving clocks running slower, at a rate dependent on the relative speed. The trick is, there are *three* inertial frames of reference in this problem, not two: the one in which Earth is at rest, the one in which you’re at rest on the first stage, and the one in which you’re at rest on the way back. You can work the problem in each of these and get the same answer: that your clock is a little behind the stay-at-home clock.
[edit] It didn’t make it into the article, but I think this diagram is helpful:
https://en.wikipedia.org/wiki/Talk:Twin_paradox/Archive_11#/media/File:Twin_5.png
The three frames of reference are the rest frames of the stay-at-home twin, the outbound twin, and the returning twin. The third is translated so that the traveling twin has the same coordinates after turnover as before. The arrows are the twins’ worldlines, the thin lines are their lines of simultaneity at turnover. The dashed diagonal lines show the light cone from the start.
When the traveller says his clock should be ahead of Earth’s, he’s ignoring the time that passed on Earth between points A and C.
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