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We have spaceship "A" traveling at a velocity of 99% of the speed of light for 2 seconds in the positive $x$-direction and then traveling at a velocity of half the lightspeed for 1 second in the negative x-direction.

If a spaceship "B" is to travel at a constant velocity and both spaceships leave the origin at the same exact time, at which velocity must spaceship "B" travel in order to meet spaceship "A" exactly at the same time and place in the future?

If I draw the spacetime diagram for this, then looking at the axis, it seems like spaceship B should travel at a quarter the speed of light for 4 seconds. However, the time on the clocks inside the spaceship "A" and spaceship "B" would read different lapsed time, correct? Two more questions come up:

  • a) Is there a way for spaceship "B" to travel at a certain velocity so the elapsed time would be the same in both spaceships and still meet at the same time and place?

  • b) If not possible, how would I calculate the elapsed time difference?

Qmechanic
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Blackbird
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  • Possibly useful: the spacetime diagram in my answer to https://physics.stackexchange.com/q/508931/ – robphy Jan 18 '22 at 21:44

1 Answers1

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Your question is very vague. There are four inertial reference frames involved, so the answer depends upon which of the frames the specified times and distances are related to- you really should spell that out to avoid confusion.

It is also not clear what you mean when you say that ship A travels at 0.5c for 1 second- what happens at the end of the 1 second? Does it come to rest, or do you mean that it is intercepted by ship B at that point in time?

However, I think you mean that the times and distances are given in the original frame in which both spaceships where at rest before the thought experiment started, and that you want ship B to intercept ship A at the specified time. In that case the two ships must intercept after 3s in the original frame, at which point ship A is 1.5 light years from the origin. So you need ship B to have been travelling along the x axis for 3s at 0.5c. That all assumes the accelerations are instantaneous and that ship A travels at very close to c (c being unattainable) on its outbound leg.

The elapsed times as measured onboard the ships will be less than the elapsed times in the original rest frame. You can calculate them using the time dilation formula applied to each of the three separate legs of travel involved.

The final part of your question is absurd, given the first part. There is only one speed at which B can travel if it is to intercept A at the time you specified, so why do you ask whether it can travel at another speed and still intercept A at the same point?

Ignoring that, more generally if A makes a journey with two legs and B re-meets A after a direct journey, you have a version of the twin paradox- the time measured by A will always be less than the time measured by B.

Marco Ocram
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  • Thank you, Marco. If Spaceship B is to accelerate differently, and have an average velocity of a certain value, then is it theoretically possible for Spaceship B to meet Spaceship A exactly at the same time and space when Spaceship A arrives at its final destination and for both spaceships internal clocks to read the same elapsed time? If so, which set of questions can I use to calculate and understand that more? – Blackbird Jan 21 '22 at 03:53