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Suppose two spaceships A and B have on them clocks A and B move along the same straight line at uniform speeds differing by 161 miles per second. And suppose they are at positions A and B in space when both clocks say noon. And suppose that after noon, they are set to collide with one another when clock A says 2 o'clock.

My question is what time will clock B say when the two spaceships collide? By symmetry, the answer should be 2 o'clock. However, according to special relativity, the answer is 1 o'clock. Which answer is right? They can't both be right.

This is a modification of a paradox found in Herbert Dingle's book Science at the Crossroads.

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    $\begingroup$ "...and are at positions A and B in space when both clocks say noon". According to whom? $\endgroup$ – WillO Jun 11 '16 at 0:19
  • $\begingroup$ there a long rod, that moves at the same speed relative to both whose ends are a and b, and both spaceships get there in such a way that when A sees itself in a, it sees that B is at b, and viceverza. At that moment both reset their clocks to noon $\endgroup$ – user83548 Jun 11 '16 at 0:33
  • $\begingroup$ or is that impossible and I am missing something? $\endgroup$ – user83548 Jun 11 '16 at 0:40
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    $\begingroup$ @bruce You are missing something. The ships do not agree on the assertion of simultaneity. Their time axes are not parallel and if they both arrive at their designated points at noon in ship A's frame, then they don't do so in ship B's frame. If they arrive at the designated points at noon is the rods POV then they don't do so in either ship's frame and each ship asserts a different ordering (A arrives first versus B arrives first). $\endgroup$ – dmckee Jun 11 '16 at 2:36
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    $\begingroup$ Also, I think everyone would be happier if you just said that the speed was $(\sqrt{3}/2) c$. $\endgroup$ – tparker Jun 14 '16 at 4:21
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It depends what you mean by their clocks both being at noon when they're at positions A and B, and their speeds differing by 161,000 miles per second. By the way the question was stated, I'll assume you meant their positions were measured in an outside observer's frame, i.e., at noon in the outside observer's frame ship A is at position A and ship B is at position B. We can also assume ship A sync's up its own clock to read noon when it's at point A, and similarly for ship B. (It is impossible for both ships to see both clocks at noon at once. If ship A sees both clocks at noon at one time, then ship B won't, and vice versa. Specifically, if ship A sees the event of itself being at point A (with its clock at noon) coincide with the event of ship B being at point B (with its clock at noon), then ship B will not see these events coincide.)

As for what frame their relative speeds are measured in, it sounds like you meant the ships' frames, i.e. ship A sees ship B moving towards it at 161,000 miles per second, and ship B sees ship A moving towards it at 161,000 miles per second. Then the $\gamma$ factor between the two frames is about $2$, meaning time appears to move half as fast for one ship as measured from the other. Unfortunately, having specified their relative speed in their own frames doesn't specify it in the outside observer's frame. This is what matters, since the outside observer's frame is the frame in which both their clocks were seen reading noon at once.

If one ship is moving faster than the other in this frame (the outside observer's frame), then its clock will elapse less time than the other's.

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  • $\begingroup$ Thank you, my rustiness with this subject is what caused me to make the mistake. I didn't account for the fact that the noon times are not really synchronized. $\endgroup$ – Craig Feinstein Jun 14 '16 at 6:18

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