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Physics
Special and General Relativity
Understanding twin paradox without math
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[QUOTE="Lluis Olle, post: 6824236, member: 728827"] An (enough) good explanation for me is: - The Alice (Earth) observer has a lattice of synchronized clocks that fill the whole space. Specifically there's an "Earth's clock" at the star position, that reports back its readings to Earth (at each time, what happened there). - In Bob's (traveler) perspective, during the outbound path and just when he observes the star reach his position, his clock reads 2.0 and as he knows that clocks run slower in Earth, he predicts Earth's clock to be ##2 \cdot \frac{1}{\gamma}=1.6##. - Leading clocks lag, which means that for two synchronized clocks separated by a distance of D in the same frame, the leading clock (respect the velocity) will lag ##\frac{D \cdot v}{c^2}## respect the rear clock, as observed by a rest frame. So the "Earth's clock" at the star position is ahead its correspondent sibling at Earth by 0.9, and it records 2.5 (1.6+0.9) at the event of the star reaching Bob at his time of 2. When the Earth's observer collect the reported data, all fits nicely. - The velocity is then suddenly inverted and the inbound path started. As soon as this happens, and Bob is put abruptly into another frame, that "Earth's clock" at the star is now the leading clock of the play, so it lags by 0.9 respect its correspondent sibling clock at Earth. Obviously, the reading of that clock in that "after" instant is the same as "before", but now lags by 0.9 respect the correspondent sibling at Earth. So that Earth's clock must be 3.4. In no time for Alice, Earth clock has gone from 1.6 to 3.4 - Then, it takes another 1.6 Earth-years for Earth to reach Bob's position. And that accounts for the 5 years elapsed in Earth, but only 4 for the traveler twin. [/QUOTE]
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Understanding twin paradox without math
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