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JesseM
JesseM is offline
#5
Apr5-11, 10:19 PM
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Quote Quote by erics View Post
Suppose you have two ships departing from Earth at the same time, Ship #1 and Ship #2. Both ships depart Earth in the year 2000, both in the same direction from earth, and both Ships return to Earth in the year 2020, so 20 Earth Years for BOTH ships.....

Ship #2 does something somewhat different from Ship #1, you'll see


From Earth's Reference Frame:
Ship #1 Cruises at a Velocity of 86.6025% of C, which is a Lorentz Factor of 2.... for half the 20-year duration.... then the ship returns to Earth. Based on Twin Paradox, the ship's clock ticks half of 20 years.... so the ship ages 10 years.
OK, so in the Earth frame, ship #1 must travel a total distance of 0.866025 * 10 = 8.66025 light-years before turning around.
Quote Quote by erics View Post
Ship #2 Cruises at the same velocity as Ship #1, but only for the first 6 Earth Years of time.

Between Years 7 thru 10:
After 6 years time, Ship #2 gains additional velocity and passes ahead of Ship #1 at a Velocity of 98.7433% of C with respect to Earth (Lorentz Factor of exactly 7).

Then at some point in time, Ship #2 de-accelerates to 0 km/h with respect to Earth and waits patiently for Ship #1 to catch up to Ship #2. Ship #1 catches up to Ship #2 after 10 Earth Years Time.
From what I said earlier, if you want ship #1 to catch up with ship #2 at the same time that ship #1 turns around to return to Earth (after 10 years in the Earth frame), that means ship #2 must come to rest relative to Earth when it reaches a distance of 8.66025 light-years from Earth. After 6 years ship #2 will be at a distance of 0.866025 * 6 = 5.19615 light years from Earth, so it still has a distance of 8.66025 - 5.19615 = 3.4641 light years to go, so if it traverses that distance at 0.987433c then the time needed will be 3.4641/0.987433 = 3.5082 years, so ship #1 stops after 6 + 3.5082 = 9.5082 years in the Earth frame.
Quote Quote by erics View Post
But based on U - V / (1 + U*-V), the Ship #2 was moving ahead of Ship #1 at 86.6025 of C, Lorentz Factor of 2.... and then would have appeared to head back towards ship #1 also at 86.6025% of C.
No, your math is wrong here. In ship #1's frame, ship #2 is traveling at (0.987433c - 0.866025c)/(1 - 0.987433*0.866025) = 0.838115c.
Quote Quote by erics View Post
Now looking at Years 7 thru 10, however much Ship #1 was aging, Ship #2 would be aging half as much, because with respect to Ship #1's frame of reference, Ship #2 accelerated to Lorentz Factor 2 speed
First of all, as noted above ship #2 is actually initially moving at 0.838115c in ship #1's frame (before ship #2 comes to rest relative to Earth), for a Lorentz factor of 1.8332. Second, since ship #2 does come to rest relative to Earth at the turning point, ship #2 changes velocities in the frame of ship #1, so that ship #2 is traveling at 0.838115c for some time and then changes to moving at 0.866025c towards ship #1, until eventually they meet.

Perhaps you want to change the thought-experiment so that ship #2 initially accelerates to (0.866025c + 0.866025c)/(1 + 0.866025*0.866025) = 0.989743c relative to the Earth, then comes to rest at the turning point and waits for ship #1 to catch up, in which case in ship #1's frame ship #2 initially moves away at 0.866025c and then turns around and moves back towards ship #1 at 0.866025c until they meet at the turning point?