2
$\begingroup$

The following is a variation on the twin paradox, with no acceleration and no 3rd reference point (i.e. described purely in terms of relative motion).

Imagine two observers initially located right next to each other in empty space. Then, imagine they move away from each other at constant velocity for a period of time, instantly going from being at rest relative to each other to moving away from each other at constant velocity (i.e. without accelerating). After a period of time, the two observers change direction and start moving toward each other with constant velocity, again without any deceleration or acceleration. When the observers reach each other, they instantly come to a stop, again without any deceleration.

This thought experiment can be divided into 4 stages:

  1. Observers at rest next to each other
  2. Observers moving away from each other
  3. Observers moving toward each other
  4. Observers at rest next to each other

The experiences of the observers are exactly symmetrical. From the perspective of each observer, he is always at rest and the other observer moves back and forth.

From the perspective of any one of the observers, there are 3 relevant inertial frames: 1) his own, which includes the other observer at the 1st and 4th stage; 2) the inertial frame of the other observer as he is moving away; and 3) the inertial frame of the other observer as he is coming back.

According to my understanding of special relativity, in the first stage, each observer's passage through time is thought to proceed at the same rate and they exist within the same "now." In the second stage (moving away from each other), each observer perceives the other observer's passage through time as slower in comparison to his own. Also during the second stage, the "now" in which each observer exists includes what is the past to the other observer, according to the relativity of simultaneity. During the third stage (moving toward each other), each observer once again perceives the other observer's passage of time as slower in comparison to his own. And since they are moving toward each other, the "now" in which each observer exists includes what is the future to the other observer. In the fourth stage (when they meet), each observer's passage through time is thought to proceed at the same rate and they exist within the same "now."

Based on the above analysis:

  • The experience of each observer is exactly symmetrical.
  • At no point did one observer perceive another observer's passage through time as faster in comparison to his own. In fact, each observer only ever perceived another observer's passage through time as the same or slower than his own.
  • The changes that occurred to each observer's "now" were symmetrical between stages two and three. In other words, each observer's "now" became shifted to the other observer's past by X amount in stage two and then became shifted to the other observer's future by X amount in stage three. If shifts such as these could possibly influence the state of either observer in the final stage, they should cancel each other out in the case of this thought experiment.
  • When the observers are rejoined, they are no longer moving relative to each other and once again share the same "now" (i.e. one's "now" is not shifted to include the other's future, etc.).

At the end of the experiment, has time passed differently for one of the observers relative to the other? If so, which one (i.e. the one or the other)?

$\endgroup$
6
  • 3
    $\begingroup$ you can't change direction to meet again without acceeration $\endgroup$ Commented Mar 19, 2016 at 18:44
  • $\begingroup$ @MartinBeckett I think that is beside the point. I could have involved two additional actors in this thought experiment who would synchronize their clocks with the original observers upon passing them on their way to meet each other. The outcome would be the same. $\endgroup$ Commented Mar 19, 2016 at 18:53
  • $\begingroup$ If they both have the same path, their clocks agree, no paradox. If you argue that either could consider that they were stationary and only the other moved then the fact that they felt acceleration prohibits this $\endgroup$ Commented Mar 19, 2016 at 19:15
  • 4
    $\begingroup$ If the velocity changes, there is still acceleration. If it changes instantaneously, then the acceleration is infinite. $\endgroup$
    – Javier
    Commented Mar 19, 2016 at 19:44
  • 6
    $\begingroup$ "i.e. without accelerating" Err ... I think you're having trouble with the notion of acceleration. A change in velocity is acceleration. Making the change in less time is a higher acceleration. Doing it instantaneously is an infinite acceleration. $\endgroup$ Commented Mar 19, 2016 at 19:49

2 Answers 2

3
$\begingroup$

As long as the observers are moving with respect to each other, they must see each others' clocks moving slower.

By symmetry, their clocks must agree when they meet again.

So here's the sort of thing that happens:

They start with their clocks synchronized at (say) noon.

Now they begin their outward journeys.

As A travels outward, he says "B's clock is traveling at half-speed".

At 4:00 by A's clock, A says "I've been flying for four hours. During that time, B's clock moved two hours. It started at noon and now says 2PM".

At that instant, A turns around (so does B) and A says "B's clock now reads 6PM".

A flies for another 4 hours, at which point his clock reads 8PM. During that time, he says that B's clock moves only two hours, from 6PM to 8PM. When they meet, both clocks say 8PM.

(B says all the same things about A that A says about B.)

This is all very easy to see for yourself if you draw the spacetime diagram. If you don't know how to do that, it's a skill worth acquiring.

Note: For simplicity, I'm assuming that the turnarounds are instantaneous, which is why the revisions about the timings of distant events are instantaneous. With more realistic assumptions about acceleration, these adjustments would be more gradual. But every time someone changes velocity (which includes a change of direction) he must change his mind about the timing of all distant events.

$\endgroup$
13
  • $\begingroup$ Thank you for your response. So, A starts at noon (12 PM) and B starts at 11 AM? Can we have them both start at 12 PM? $\endgroup$ Commented Mar 19, 2016 at 19:38
  • $\begingroup$ Shouldn't there be 3 sudden shifts in each observer's perception of the time at the other observer? 1) when the observers go from being at rest relative to each to moving away from each other; 2) when the observers go from moving away from each other to moving toward each other; and 3) when the observers go from moving toward each other to being at rest relative to each other? $\endgroup$ Commented Mar 19, 2016 at 19:44
  • $\begingroup$ When they go from moving toward each other to being at rest relative to each other, they are in the same place. When you change velocity, you do not change your mind about the time of the event you are currently at. You do change your mind about the times of distant events. The relevant events (i.e. the changes on B's clock) are distant during the first two shifts, but not during the third. $\endgroup$
    – WillO
    Commented Mar 19, 2016 at 19:49
  • $\begingroup$ Re your first comment: You are right! I had A and B starting from different places, so A immediately changed his mind about B's clock. But you wanted them to start from the same place. I will edit accordingly. $\endgroup$
    – WillO
    Commented Mar 19, 2016 at 19:51
  • $\begingroup$ Fixed now. At the beginning and the end, B's clock is not distant so there is no change in A's opinion of what time it says. Only at the turnaround is there a change. (As I originally misread your question, A and B took off from different places,so there was also a change in perspective at the beginning --- but that's edited out now.) $\endgroup$
    – WillO
    Commented Mar 19, 2016 at 19:53
1
$\begingroup$

Read or at least skim through the three questions:

The key point explained in this series of questions is that the elapsed time on a clock carried by some observer is equal to the length of their world line, where this length is calculated using the appropriate metric. The simplified equation:

$$ t' = \frac{t}{\gamma} $$

is a special case that applies only when the two observers are in relative motion in a straight line at constant velocity, and trying to apply it to situations where the velocity changes will not give the correct results.

As described in the questions I've linked the length of the world line is an invariant (called the proper length), where the description invariant means all observers in all frames can calculate the proper length and will get the same result. In this case the calculation is obvious easiest in the frame of the observer who remains on Earth watching the twins fly off in different directions. In that frame the symmetry makes it obvious that the twins have the same proper length between their departure and return so their elapsed times will be the same.

But the motivation for your question is obviously to try and understand the calculation from the perspective of each twin. How this calculation is done is outlined in my answer to What is the proper way to explain the twin paradox?, though in general the calculation is hard and we end up doing it numerically with a big computer rather than on paper. The key point about the calculation is that the metric the twins use to calculate the proper length depends on their motion. In particular you cannot neglect the acceleration required to reverse the velocity at the halfway points because during the turnround each twin observes the other's time to be flowing faster not slower. If I am one twin and you are the other, and you are observing me then you would observe:

  • during the outward leg my clock records less time than your

  • during the turnround my clock records more time than yours

  • during the return leg my clock records less time than yours

But when you add up the times recorded by my clock during the three stages you will find that the slow and fast stages balance out and my clock has recorded exactly the same time as your clock.

I appreciate that this probably isn't a very satisfying answer because it all feels a bit like trickery - how can we know the different time rates during the three stages balance out? The answer is that the invariance of the proper time is the fundamental principle on which relativity is built, and that applies to general relativity as well as special relativity. If the different rates of time balance out in any frame, which they do in the Earth frame, then they must balance out in all frames.

$\endgroup$
1
  • $\begingroup$ Thank you very much. I'm noticing that people seem to be imagining the two observers both moving away from each other at the same speed and then both moving back toward each other. It's actually fundamental to my question that such symmetrical motion not be assumed. I would like to understand the solution to this problem purely from the perspective of relative motion, only. Also, we can reimagine the story using as many additional actors as needed in order to avoid considering the effects of acceleration. Therefore, I disagree with your statement that we cannot neglect acceleration. Thanks. $\endgroup$ Commented Mar 21, 2016 at 17:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.