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A thought experiment. A person on the Earth has a clock. Another person who also has a clock is in a rocket ship passing the Earth travelling at 50% the speed of light. They synchronise watches as rocket man goes past. Rocket man (1) travels for a year and at that point he encounters another man (2) in a rocket travelling towards Earth at 50% the speed of light. As they pass Rocket Man (2) who also has a clock, sets the date and time to that of Rocket man (1) (information is passed). A year later, as Rocket Man (2) passes the Earth (he doesn't slow down) he compares his clock to that of Earth Man.

According to special relativity Rocket man's clock should be (a lot) slower than Earth man due to the speed travelling. However also under special relativity one should be able to say that it was Earth man (along with the solar system etc) and Rocket Man (1) that were moving, while Rocket man (2) was fixed. Therefore Earth man's clock should be slower.

I had another earlier version of this question which had some naive elements re clocks changing direction so I have tried to avoid any acceleration elements in this formulation.

So the question is how is the apparent paradox that both clocks should be slow resolved ?

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    $\begingroup$ The situation is not symmetric. In frame (1) rocket man 2 is moving at 0.5c and rocket man 3 is moving at -0.5. In frame (2) the Earth is moving at -0.5c and rocket man 3 is moving at -0.8c. In frame (3) the Earth is moving at 0.5c and rocket man 2 is moving at 0.8c. Since there is no symmetry there is no paradox. $\endgroup$ – John Rennie Jan 24 '15 at 7:34
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    $\begingroup$ Avoiding the acceleration of a material object does not avoid the key point that the out-n-back path is not straight in four-space (in any inertial frame of reference). The core feature of this problem is exactly the same as it was before. Draw the space-time diagram. Draw the space-time diagram. Draw the space-time diagram, already! That is the whole resolution of the mis-named paradox. Every time. $\endgroup$ – dmckee Jan 24 '15 at 16:35
  • $\begingroup$ I have to admit that the first time through relativity, I did not get the importance of this approach because I had my head all wrapped up in the complexities of the algebra arising from the Lorentz transformation, so let me be explicit in recommending it to beginning students. Accordingly there is much to be said for Tatsu Takeuchi's book. And the drawings are cute, too. $\endgroup$ – dmckee Jan 24 '15 at 16:39
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The basic answer is simple. In the rest frame of Rocket man (2), The Earth clock and the clock of Rocket man (1) are both running slower than his own, but Rocket man (1)'s velocity is even higher than that of the Earth and so the clock of Rocket man (1) is running even slower than that of Earth. Thus, at the moment that Rocket man (2) passes Rocket man (1) and they synchronize clocks, the Earth's clock is already significantly ahead of that of the Rocket clocks, so even though Rocket man (2)'s clock ticks faster for the remainder of the trip, it doesn't manage to catch up with the time on the Earth clock and thus the Earth clock is still ahead when Rocket man (2) reaches Earth.

If that isn't clear, though, I can show how it works in the numerical example you suggested. It's not quite clear from the way you phrased things whether Rocket man (1) meets Rocket man (2) after a year has passed on his own clock, or after a year has passed in the Earth frame (the frame you used to specify his velocity). But let's assume you mean the latter. Then when Rocket man (1) meets Rocket man (2) they will both be $0.5$ light years away from Earth in the Earth frame--let's assume there is some object at rest relative to the Earth there, like an asteroid. And assume Earth clock and the clock of Rocket man (1) were both set to $T=0$ years when they passed, which means when Rocket man (1) reaches this asteroid 1 year later in the Earth frame, his clock reads $T=1 \sqrt{1 - 0.5^2} = 0.8660254$ years. Then rocket man (2) travels back towards Earth for another year in the Earth frame, so the Earth clock reads $T=2$ years and Rocket man (2)'s clock reads $T = 2*0.8660254 = 1.7320508$ years when they meet.

Now you can analyze things in the frame of Rocket man (2). In this frame he is at rest while the Earth is moving towards him at $0.5c$, while by the relativistic velocity addition formula, Rocket man (1) is moving towards him at $(0.5c + 0.5c)/(1 + 0.5^2) = 0.8c$. Meanwhile the asteroid is also moving towards him at $0.5c$, and due to length contraction it's at a distance of $0.5*0.8660254 = 0.4330127$ light years ahead of the Earth. So if we say that the Earth and Rocket (1) were at position $x=0$ in this frame when they were beside each other, and this happened at $t=0$, then the asteroid's position as a function of time is given by $x(t) = t*0.5c + 0.4330127$ and Rocket man (1)'s position as a function of time is given by $x(t) = t*0.8c$, so in this frame they will meet when $t*0.5c + 0.4330127 = t*0.8c$, solving for $t$ gives a time of $t=1.4433757$ years in this frame. But in this frame Rocket man (1)'s clock is running slow by a factor of $\sqrt{1 - 0.8^2} = 0.6$, so his clock reads $T=1.4433757*0.6=0.8660254$ years when he reaches the asteroid and passes Rocket man (2) (the same value that was found for his clock reading in the Earth frame, naturally). And in this frame Earth's clock has been running slow by a factor of $\sqrt{1 - 0.5^2}=0.8660254$, so the Earth clock reads $T=1.4433757*0.8660254=1.25$ years "at the same moment that the two Rocket men pass in this frame (keep in mind that simultaneity is relative, so the event on Earth that's simultaneous with the Rocket men passing in Rocket man (2)'s frame is different from the event on Earth that's simultaneous with the Rocket men passing in Earth's own frame--in that frame, the event of Earth's clock reading $T=1$ year was the one that was simultaneous with the Rocket men passing).

So, you can see that in Rocket man (2)'s frame, the Earth clock is ahead of Rocket man (1)'s clock at the moment the two Rocket men pass (and Rocket man (2) sets his clock to read the same as (1)), with Earth reading $T=1.25$ years and the Rocket clocks reading $T=0.8660254$ years. And the Earth is $0.4330127$ light-years away from the asteroid where the Rocket men meet, so since Earth is traveling towards Rocket man (2) at $0.5c$, it takes an additional $0.4330127/0.5 = 0.8660254$ years for the Earth to reach Rocket man (2) in this frame. Since Rocket man (2) is at rest in this frame his clock ticks an additional $0.8660254$ years during this period, and since Earth's clock is running slow by a factor of $0.8660254$ it ticks forward by $0.8660254*0.8660254 = 0.75$ years, so that when they meet the Earth clock will show a time of $T=1.25 + 0.75 = 2$ years, while Rocket man (2)'s clock will show a time of $T = 0.8660254 + 0.8660254 = 1.7320508$ years. So you can see that we got exactly the same answer for what the two clocks read when they meet using Rocket man (2)'s frame that we did when we used the Earth frame, even though they disagree about which clocks were running slower than others.

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So the question is how is the apparent paradox that both clocks should be slow resolved ?

Two clocks are in relative motion? And the "paradox" is that both clocks are slower relative to the other clock?

OP seems to be talking about the "paradox" of time dilation being symmetric.

So here's a solution:

Instead of saying "both clocks are slower than the other clock", we say: "In first clock's frame second clock is slower, and in second clocks frame first clock is slower"

We "solved" the "paradox" by saying that slowness a is a frame dependent thing.

In other words slowness is not as real as OP thinks.

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  • $\begingroup$ Sorry - what is OP? $\endgroup$ – Mark Kelleher Jan 29 '15 at 21:40
  • $\begingroup$ "OP" means "original poster" also "original post" $\endgroup$ – stuffu Feb 3 '15 at 6:29

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