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There are two spaceships, A and B, moving towards each other, such that they will eventually pass each other. At a point equidistant from both ships, their velocities are both 0.5c towards that point.

Using relativistic velocity addition, the velocity of B in the reference frame of A is (1/(1.25))c, and vice versa. In the reference frame of A, a clock on B will be ticking slower than a clock on A, due to time dilation. Likewise, in the reference frame of B, a clock on A will be ticking slower than a clock on B, due to time dilation.

At the point at which they pass each other, the pilots of the ships show their clocks to each other via windows on their respective ships. My question is, will each pilot read both clocks differently from the other pilot? Pilot A should observe that his clock has been moving faster than Pilot B's, while Pilot B should observe that his clock has been moving faster than Pilot A's. Is there a paradox here, or have I been careless about something? (Note that I have been careful to avoid any acceleration/deceleration in this problem.)

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  • $\begingroup$ Hint: The clock on my wall currently reads 3:17. From that information alone, can you infer anything about how fast it's been going? $\endgroup$ – WillO Sep 26 '15 at 13:54
  • $\begingroup$ As to the first question: At the moment they pass each other, the pilots (obviously) must agree about what their clocks show. $\endgroup$ – WillO Sep 26 '15 at 14:52
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I would not ordinarily answer a question like this, because these questions usually come from people who have made no effort. It's clear from your comment on m4r35n357's answer that you are an exception to that rule, so I'm happy to provide the spacetime diagram. You will find that it pays to get good at drawing these; as m4r35n357 says, they are always the best way to clear up confusion about relativity.

enter image description here

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  • $\begingroup$ Thank you for your answer WillO, I got essentially the same answer not too long after my last comment, but I wouldn't have had the time to flesh is out like you have. How did you make that diagram? $\endgroup$ – Meer Ashwinkumar Sep 27 '15 at 18:54
  • $\begingroup$ @MeerAshwinkumar: I'm not sure exactly what you mean by "How did you make that diagram?". Are you asking what software I used? If so, I used SmartDraw. $\endgroup$ – WillO Sep 28 '15 at 16:13
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Draw a spacetime diagram with the two spaceships approaching the origin (your equidistant point) at 0.5c. Read off from that diagram the (equal) elapsed proper times ($\tau^2 = t^2 - x^2$) for each ship when they meet (and can compare readings), and also the elapsed coordinate time (which will be longer).

Only now should you consider time dilation; it is an effect, not a cause! You might begin to doubt the need for even considering it, as you have all the numbers you need from the spacetime diagram.

TL;DR, always start with a spacetime diagram. Don't try to use Lorentz transforms, the velocity addition formula or time dilation/length contraction as your primary analysis. This is the easy way to do SR!

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  • $\begingroup$ I've made the spacetime diagram, but I can't seem to find a solution. I think I should add that when we begin to consider the motion of both spaceShips, their clocks show the same time, e.g. 5 p.m. With respect to an observer with a clock equidistant from both ships, the clocks on both spaceships will be ticking slower; but they will do so at an identical rate to each other. But if we only consider two reference frames, the frame where spaceship A is not moving, and the frame at which spaceship B is moving at 1/1.25c towards A, then definitely A and B won't agree on the time when they meet. $\endgroup$ – Meer Ashwinkumar Sep 27 '15 at 4:14
  • $\begingroup$ I think I have the solution now, after reading some other posts on similar problems. I have assumed that when we begin to consider the motion of the spaceships, both pilots have synchronized their clocks to 5 pm. However this synchronization is only simultaneous in one reference frame. If both pilots have synchronized their clocks to 5pm simultaneously in the reference frame of the equidistant observer, this synchronization will not be simultaneous in other reference frames, e.g. that of pilot A. $\endgroup$ – Meer Ashwinkumar Sep 27 '15 at 6:48
  • $\begingroup$ @MeerAshwinkumar: You've got it! $\endgroup$ – WillO Sep 27 '15 at 17:21

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