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A time dilation thought experiment

  1. Jun 2, 2012 #1
    Let's say we have two objects in space, far from any gravitational source. In each "object", there's an observer, and a clock. Each observer is constantly looking at both clocks(his and the other object's). Now, for observer A, observer B is moving at 0.9c, and for observer B it's the exact opposite. Looking at each other's clocks, each one sees something different. Observer A sees that his clock is moving regularly, while observer B's clock is slower. Observer B sees the opposite. When they meet, the clocks stop(while the objects are moving). Then, they compare their results. But each one is "stuck" with an image of a different time on each clock, so how could it be that looking at the same clocks(after stopping their own movements) they each see a different result?

    I have come up with a solution to this that I think might be right, but a friend of mine disagrees, so I come here for help. Maybe I'm wrong. I came up with this:
    When they see each other meeting, they aren't actually meeting. Since they are moving fast, they see each other moving at different angles, so when they "meet", they really don't. Then, my friend said that we can calculate the angular differences, and stop the clocks accordingly. To this I have said that when they actually meet and have no angular differences, their position on the T axis must be the same, so they see the same results on the clocks. However, I am merely estimating using some logic.

    What is the actual solution to this? Can somebody post a detailed analysis to this problem? I am certain this isn't a paradox.

    Thanks for reading.
  2. jcsd
  3. Jun 2, 2012 #2
    They can never really 'meet' again without undergoing a non-inertial reference frame. Then time dilation will behave differently.
  4. Jun 2, 2012 #3


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    According to your scenario, they do not "meet", they just pass each other. From each's frame of reference, he has never moved at all, the other one just whizzes past, so of course they continue to see each other's clocks as slow.
  5. Jun 2, 2012 #4
    I am afraid it is nothing to do with angular differences, which are just an unnecessary complication. We can arrange for the clocks to be practically touching when they pass each other, so that the stopped time when they pass each other is the same as when they compare clocks when they come to rest alongside each other. You can take it to a further extreme and imagine the clocks are designed to freeze the last recorded time when they collide and send them on a collision course. The situation is totally symmetrical so there are no differential elapsed times. If they start their clocks at the same time as seen in a neutral reference frame where they have identical speeds then when they pass each other they will see identical elapsed times.

    You have not really stated an initial condition which makes the final result meaningless. I recommend you redo the scenario, starting form the point of view of a neutral observer who sees A going at +v and B going at -v and then try to analyse the whole thing from a space time diagram, which is definitely worth learning about if you want to understand these sort of problems.

    Another aspect you need to get to grips with is understanding that what they literally "see" involves light travel times and classical Doppler shift and time dilation calculations normally remove these effects so that we can calculate the actual physical relativistic time dilation, without the purely optical effects that would be present even in a Newtonian analysis.
    Last edited: Jun 2, 2012
  6. Jun 2, 2012 #5


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    Sorry, you're wrong. I may post a detailed analysis later if someone else doesn't beat me to it, but in the meantime, consider that the result on a clock (the position of its hands, the numbers it displays, the amount of sand left in the top half of the hourglass, ...) depend not just on how much time has passed but also what the clock read when it was initially set running. There's nothing paradoxical about A seeing B's clock running slow, and B seeing A's clock running slow, yet both clocks reading the same when they meet - that just tells us something about how the clocks were initially set.

    The apparent paradox comes from a problem with your thought experiment - you have a hidden assumption that there was a starting time when both clocks read the same (or equivalently, that both clocks started running at the same time). But relativity of simultaneity says that if the clocks started running simultaneously for one observer, or if they both read the same at a given time for one observer.... They won't for the other observer. Indeed, the only time when the observers must agree about the readings on the two clocks at a given time is at the moment of the meeting, when they're at the same point in space-time.

    This problem is much easier to analyze if you start with a space-time diagram and label the relevant points with their (x,t) coordinates in one observers frame of reference; then use the Lorentz transforms to see how they appear to the other observer.
  7. Jun 2, 2012 #6
    Why is that?

    Then how should I define "meet"? And again, if they continue to see each other's clocks as slow and then stop them, and then only stop themselves, wouldn't they see different things on the same clock?

    I will try to reform the experiment, but I'm afraid I still don't understand this one. The way I see it, even if the situation is symmetrical, they should still see different times on the clocks, and if as you have said the clocks freeze when they pass each other, then each observer should see the clock freezing at a different time, consequently seeing different times when the observers stop.
    Also, why do I need an initial situation?

    I think I'm starting to understand this... I need to think.
    Last edited: Jun 2, 2012
  8. Jun 2, 2012 #7
    I will define "meet" as being in the same place as the same time. This definition can include a passing event if the gap between the two clocks is negligible. Now consider a situation where each observer has two clocks. At the passing event they stop one clock and let the other continue to run. If they both turn around after passing and return to each other and both stop in a symmetrical fashion then the same amount of time will have passed on the running clocks since the passing event which occurred at the time the stopped clocks were stopped. If only one observer turns around to meet the other, while the other continues to move inertially, then the observer that turns around will record the least elapsed time since the passing event.

    Lets say at the passing event A's clock is stopped showing 10 seconds and B's clock is stopped showing 1 second, then the same amount of time has elapsed for A and B if A initially had 9 seconds on his clock and B had zero time indicated on her clock. If A and B both initially had zero time on their clock then different amounts of time will have elapsed on their clocks. In relativity, expressions like "start at the same time" will have different meanings according to different observers especially where events are spatially separated as in your scenario. This is known as the "relativity of simultaneity".
  9. Jun 2, 2012 #8


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    Are you thinking that A sees B moving away from him, or towards him, or on a trajectory that remains far away from him, or something entirely different?
    What do you mean, "it's the exact opposite"? For each of the possibilities I just asked you about, why wouldn't it be exactly the same for B?
    Again, I have the same question. What do you mean by "opposite"? Do you merely mean that A sees the A clock ticking regularly and the B clock ticking slower but B sees the B clock ticking regularly and the A clock ticking slower? Or do you mean that A sees the other clock ticking slower than his own while B sees the other clock ticking faster than his own? Or do you mean something entirely different?
    For the three options that I mentioned in terms of motion, I cannot come up with a consistent scenario that incorporates your statement about how each observer sees the others clock. If they are moving away from each other and seeing the others clock ticking slower than their own, then how do they meet? If they are moving towards each other, why would they see each others clock ticking slower than their own? If they are on a trajectory where they remain far removed from each other, then how do they meet?
    Why should it be surprising that the clocks might have different times on them? You never mentioned any earlier point in time when they had the same time on them.

    We can't read you mind, you have to be precise in your setup of your scenario or you will get all kinds of different answers depending on how each person fills in all the missing details.
    Do you see the confusion you are presenting? And where did the T axis come from?
    No, nobody can post a detailed analysis to this problem because haven't provided enough details in the problem. Please fill in all the details that you have in your mind and then we can give you an analysis.
  10. Jun 2, 2012 #9
    I understand that the same amount of time has passed, and that the running clocks will show the same time, but the stopped clocks still show different times to each, and then when they look at it they'll see different times on each's stopped clock. This is because in this scenario you can't know who's moving and who's stationary. But as I understand it, such an experiment is invalid, because I didn't define an initial condition. Correct?

    A sees B moving towards him, and B sees A moving towards him.

    A sees B moving towards him, B sees A moving towards him. That's what I meant by opposite.

    I mean that A sees his own clock ticking regularly and B's ticking slowly, and B sees his own ticking regularly and A's ticking slowly.

    I defined "meet" badly. When they pass by each other.

    It's not surprising that they have different times on them. I just don't understand how it's possible that after both observers stopped moving completely, they look at the same clock and see different things.

    Yes, I am sorry for being unclear here. It's a bit hard for me to express my thoughts since English is not my first language.

    I am referring to a diagram where each observer sees something different as simultaneous. Forgot to attach it, will do now.

    Again, I am sorry for being unclear. I hope I made it clearer now.

    Attached Files:

    • diag.png
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    Last edited: Jun 2, 2012
  11. Jun 2, 2012 #10
    If the initial conditions are symmetrical, then the running clocks will show the same time, the stopped clocks will show the same time and when they look at it they will see the the same times on each other's stopped clock.

    If the initial conditions are not symmetrical, then the running clocks will show different times, the stopped clocks will show different times and when they look at it they will see the the different times on each other's stopped clock.

    Do you want the symmetrical or the unsymmetrical situation?

    Also, be aware when you say something like "I understand that the same amount of time has passed" in relativity, you should say according to who, because different observers will measure different times.

    Also, when you say something like "and then when they look at it" you should specify where they are looking at "it" from, what "it" is, and when they are looking at "it" and how fast "it" is moving relative to "they".
  12. Jun 2, 2012 #11
    They do not see different things when they look at the same clock, when they are both next to the clock. They see the same time indicated on the clock. If one observer is next to the clock and stationary and the other is next to the same clock but moving, the moving observer will see the same clock ticking slower. Please be aware that the time indicated on a clock and the ticking rate of a clock are two different things.

    P.S> Thanks for posting a diagram. That will help everyone a lot. Unfortunately I do not have time to go into it right now.
  13. Jun 2, 2012 #12
    I'm talking about an unsymmetrical situation, yes. When both observers look at the stopped clocks(as you have said, where both stopped clocks show different times to each observer) after reaching a stationary state. Is such a situation even possible?

    Added another diagram to try and describe my experiment.

    Attached Files:

    Last edited: Jun 2, 2012
  14. Jun 2, 2012 #13
    Not sure I follow. The time on a clock is a result of its ticking rate, so shouldn't they be proportionate(so I could treat them the same)? Or do you mean something else?
  15. Jun 2, 2012 #14


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    One is an instant, the other is a rate.
    Of two clocks on a table, if one clock went on a relativistic trip to Pluto and back to rejoin its twin, they would end up ticking at the same rate, yet the time on each would be different.

    More aptly, if two spaceships pass each other, they might each see the same time on their clocks (say, 11:01:01), yet the clocks may not be ticking at the same rate (one is ticking at half the rate of the other). One must then ask if and when they were calibrated together.
  16. Jun 2, 2012 #15


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    That would be true if (a) both clocks had been synchronised so they both read t=0 at the same time, and (b) from that point onwards the relative rate of each clock relative to the other was constant.

    You haven't specified exactly how and when the clocks were synced. If the two clocks were not right next to each other at that time, we have a problem, because under those conditions observers will disagree over what "at the same time" actually means. If the two clocks were right next to each other when they were synced, then there's no problem with (a), but there's a problem with (b): how exactly do the clocks move between being started and being stopped? They can't be moving at a constant velocity relative to each other.
  17. Jun 3, 2012 #16
    Let's imagine you have two clocks on your shelf. The one one the left is working correctly and the other is broken and always shows 3PM. At 2PM as shown on the left clock, the right clock is showing 3PM, but that does not mean the clock on the right is ticking faster, (in fact it is not ticking at all). At 3 PM, both clock show the same time, but that does not mean they are now ticking at the same rate, because again the one on the right is still not ticking at all. The instantaneous indication on a clock does not tell you anything about its ticking rate. You need an initial indication and a final indication (and another clock to compare it to) to obtain any information about ticking rates. Hence the requirement for initial conditions.

    Another example. Let's say a race track is 10 miles long and the winning car finishes at 2.30 PM. What was the average speed of the winning car? You can only answer this question if you know the initial conditions, i.e. what time the race started.
  18. Jun 3, 2012 #17


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    You say that it's unsymmetrical but your diagram looks symmetrical to me. Everything in frame 1 for Observer A is the same as in frame 2 for Observer B. Where's the asymmetry?
  19. Jun 3, 2012 #18
    As an old thought experiment contructor I would suggest that first two clocks are next to each other, and they are put to the same time. Then one or both clocks are SLOWLY moved where ever we want them to be.

    Then at some specific time both clocks are put to time zero, and one or both clocks are shot at high speed into some direction.

    Both clocks will agree that both clocks were shot at time zero.
  20. Jun 3, 2012 #19
    This is an analysis based on your sketch in #9, which shows the paths of clocks A and B as seen in the reference frame of a neutral observer that I will call C. Lets assume in C's reference frame both A and B start their stop watches t=0 and A is at x=-1 light seconds and B is at x = +1 light seconds. A has a velocity of +0.5c and B has a velocity of -0.5c as measured in frame C and they are heading towards each other. In frame C they pass each other at t = 2 seconds. According to observer C, the stop watches of A and B will both show 1.732 seconds at the passing event when they both stop their watches. It is irrelevant to the results whether A and B actually stop instantaneously when they meet or continue with constant velocity. Using the relativistic velocity addition equation we can work out that A sees B coming towards him at -0.8c and B sees A coming towards him at +0.8c.

    Using the Lorentz transformations we can work out what observers A and B see in their rest frames. See http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/ltrans.html

    I am a bit short of time right now, so I will continue this a bit later, but this should give some solid initial conditions to start with.
    Last edited: Jun 3, 2012
  21. Jun 3, 2012 #20
    I figured the asymmetry is in the third situation(2 final frames).

    Thank you, I'll try continuing this analysis as well using the link you provided.

    Aha, now it's clear.
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