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Twin paradox without accelerative frame.

  1. Jan 14, 2014 #1
    I've asked this question a few times. This time i have tiny change in the question. Instead of people I'm using stopwatch & i have removed the accelerative frame.

    If you have any doubts I'll clear it but do not skip the topic.

    All i want is an answer that can be agreeable for all observers.

    Imagine one observer and two apparatus.

    Observer A who is coordinator and two boxes box 1 & box 2 which contain inactive stopwatches and automatic propulsion system which is motion dependent.
    Observer A sets the two boxes away from eachother equidistant from him. It forms an equilateral triangle. The distance between box 1 & 2 is 4 Lightyears.
    The coordinator first send some laser signals towards each box to confirm (to analyse in future) they were equal distance away.
    Then he sends another laser pulses to both of the boxes to activate their propulsion system and they starts to move towards each other. Both boxes accelerates and stop automatically when they reach a velocity .5c

    Then the coordinator sends another pulse to activate their stopwatches.

    So now we have two boxes moving towards eachother with same velocity. No acceleration included.
    Suppose if there were observers in each box, they thinks they are at rest and the other one is moving towards them at .99c. Therefore they thinks the others stopwatch has run slower.

    In this case whose watch will have run longer when they meet each other?
     
  2. jcsd
  3. Jan 14, 2014 #2

    HallsofIvy

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    By the symmetry of the situation, their stop watches will be exactly the same. Why do you ask?
     
  4. Jan 14, 2014 #3

    Demystifier

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    Good one!

    Since this is a completely symmetrical situation, both observers will agree that the two clocks show the same time when they meet. Yet, each of the observers will think that his clock is faster. But how is that possible? The catch is that:
    1) The two observers will agree that the two clocks do not show the same time BEFORE they meet.
    2) Before the meeting they will not agree whick clock shows larger time and which clock shows smaller time.
     
  5. Jan 14, 2014 #4
    The symmetry is only from the 3rd observers point of view. Observers in box do not realise they are moving so they thinks others clock has become slower
     
  6. Jan 14, 2014 #5
    How do this situation become symmetrical? because both are sharing half light speed?
    In each persons frame they think they are at rest because they are in an inertial frame so they can only perceive the others as moving.
    In this case the box 2 will have nearly the speed of light from box 1’s point of view or not?
     
  7. Jan 14, 2014 #6

    ZapperZ

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    Actually, the problem here is the definition of "meet". Are you simply defining it when they pass next to each other without stopping, or are you saying they then stop and be at rest in some reference frame? The latter requires a more explicit description of the deceleration of each one of them.

    Zz.
     
  8. Jan 14, 2014 #7
    They are not stopping. They passes by but at the point of intersection they send eachothers stopwatch data via laser for analysing
     
  9. Jan 14, 2014 #8

    ZapperZ

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    Then I don't see what the problem here is, or why you are going to such an extent with this example. Everyone in his/her own frame will see the clock in another moving frame as being slower. This is plainly illustrated in the S-S' frame example without having to resort to a 3rd observer. Someone in S will see S' clock slower, and someone in S' will see S clock slower. It is very democratic.

    Zz.
     
  10. Jan 14, 2014 #9

    PeterDonis

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    Yes, it will. But from box 1's point of view, box 2 will have started moving a long time *before* box 1's stopwatch starts, and box 2's stopwatch will have been running during all that time. So from box 1's point of view, box 2's stopwatch runs slower but also starts running sooner, and the two effects just cancel out so that box 2's stopwatch reads the same as box 1's when they meet.

    A good general rule in all these types of problems is to make sure you're taking into account relativity of simultaneity.
     
  11. Jan 14, 2014 #10

    Dale

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    It became symmetrical because of how you set up the problem. If you start with A and rotate by 180 degrees then you get B. Since the laws of physics are invariant under rotations the situation is symmetric.
     
  12. Jan 14, 2014 #11
    Let me ask you a thing
    Apply the same circumstance and imagine the box 1 is not moving and box 2 is moving with a uniform velocity towards box 1. The speed is .99c
    We have the similar situation in diferent way
    Now how do you distinguish between the situations? I mean how can you say which one is symmetrical and which one is not?
     
  13. Jan 14, 2014 #12
    I believe the actual closing rate between the two boxes, each traveling 0.5c relative to the observer, is 0.80c.
    Each box would see the clock in the other box ticking at 36 seconds/minute.
    The unboxed observer would see each of the clocks ticking at about 52 seconds per minute.

    As PeterDonis explained, only the unboxed observer sees the clocks activating simultaneously.
     
  14. Jan 14, 2014 #13
    Agreed only unboxed observer sees time ticking simultaniously. The moving observers sees others clock started later ie each observer perceives others clock has slower working rate. So both are true but when they intersect they both will dasagree on their recordings
     
  15. Jan 14, 2014 #14

    Ibix

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    You are correct that the situation is symmetrical in one frame only. In that frame, then, the analysis is straightforward and the stopwatches must agree.

    Therefore, the stopwatches must agree at the meeting as analysed in any other frame. As you pointed out, from the frame of either stopwatch, the other is ticking more slowly. Therefore, to get to the same time at the meeting the moving stopwatch must have started first.

    Indeed, this is what the Lorentz transforms show. The watches only start simultaneously in the frame where they tick at the same rate. In all other frames, the slower ticking one starts earlier, so the times are always equal when they meet.

    You need to let go of the notion of a universal "now" if you want an intutive feel for SR. The distinction between "simultaneous for me" and "simultaneous for you" keeps tripping you up.p
     
  16. Jan 14, 2014 #15

    PeterDonis

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    I'm not sure "disagree" is the right term. Certainly their recordings will be different, but that's because they are in relative motion. They will agree on the direct observable, which is that their clock readings are the same when they meet. It's true that they will "disagree" on which stopwatch started first, but that's not a direct observable.
     
  17. Jan 14, 2014 #16
    If it is like that then ill have to some additional setup to disqualify your prediction
    At the beginning the unboxed observer sends light pulses to start their stopwatches. Here the sent light pulse is reflected towards eachother. At the exact middle point of their distance, the light pulses meet together. There is an instrument kept at the midde (by the observer) it receieves the signal and analyses wheather it was meeted simultaneously. If not the lagging signal sender will be destroyed. But in the unboxed observers frame, there will be no problem at all and the experiment goes well
     
  18. Jan 14, 2014 #17

    ghwellsjr

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    Let me modify your scenario slightly, Instead of it forming an equilateral triangle, let's make them all be along a straight line. Otherwise, we have a confusing situation when later on you say they move towards each other. This change won't make any difference to the point of your scenario.

    Let me make another slight modification to your scenario. Instead of three separate sets of pulses, we only need one and we will assume that the boxes can accelerate instantaneously, OK? It won't make any difference.

    They won't think the other one is moving towards them at .99c. The correct number is 0.8c but even then it is a mixed story. There are two ways to look at the situation. If you want to know what they will think based on what they actually see and can directly measure, they will see the other observer moving towards them at 0.5c for most of the trip and then near the end they will see the other observer start to accelerate toward them at 0.8c. It is at this point that they see the other ones stopwatch start ticking.

    The other way to look at the situation is from the standpoint of their individual rest frames. In that case, the other observer is only moving at one speed 0.8c but the other observer started moving considerably earlier than they did and their stopwatch started ticking considerably earlier than their own did.

    As has been mentioned by others, their stopwatches will have the same reading when they pass each other.

    In order to make these analyses more clear, I have created some spacetime diagrams. First is the rest frame of the coordinator, shown as the thick blue line. I take up the scenario after the two boxes in red and green have been moved to a distance of 4 light-years apart. The dots represent 1-year increments of time for each observer:

    attachment.php?attachmentid=65597&stc=1&d=1389715283.png

    At the Coordinate Time of 1 year, the coordinator sends a blue signal to both remote boxes to start their instant accelerations towards each other and to start their stopwatches.

    As you can see, they each see the other ones stopwatch sitting at zero for over two years. Then they see the other one start accelerating towards them and then they see the other ones stopwatch start running at three times the rate of their own. When they finally meet, their own clocks have advanced by about three and a half years and they have seen the other ones clock advance by the same amount of time. I think this is very clear. What do you think?

    Now I will transform the scenario into the rest frame of the red box after it has accelerated to 0.5c:

    attachment.php?attachmentid=65598&stc=1&d=1389715283.png

    As you can see, the previous analysis of what they each can actually see is maintained in this space time diagram, but in addition, we can see the other analysis where the green box started accelerating way before the red box did and that's why both of them can accumulate the same amount of time (3.5 years) during their trips, but only the green box is Time Dilated in red's rest frame.

    And we can transform to green's rest frame and see that the prior analysis of what each observer sees still holds true but in addition, it is only the red observer that is Time Dilated in green's rest frame:

    attachment.php?attachmentid=65599&stc=1&d=1389715283.png

    Does this make perfect sense to you? Any questions?
     

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  19. Jan 14, 2014 #18

    PeterDonis

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    Simultaneity is relative. As you have specified the scenario, the instrument in the middle will indeed receive the return signals simultaneously, indicating that the stopwatches started simultaneously *in the observer's frame*. But by relativity of simultaneity, that means the stopwatches do *not* start simultaneously in either of the "moving" frames (box 1's rest frame while the boxes are approaching each other, or box 2's rest frame while the boxes are approaching each other).

    The spacetime diagrams posted by ghwellsjr make all this visually obvious.
     
  20. Jan 14, 2014 #19
    If the clocks do not start simultaneously in the moving frame, then that means the simultaneity analyser analyses the reflected signal has lagged and thereby destroys the moving observer, is that right?
     
  21. Jan 14, 2014 #20

    ghwellsjr

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    Yes, as PeterDonis has pointed out, the blue coordinator will see the first signals sent out by the red and green boxes (at the point of their accelerations and when their stopwatches are started) at the same time in all three frames. In fact, the blue coordinator will see each pair of the signals sent by the red and green boxes as they pass by his location. In other words, the 1-, 2-, and 3-year signals pass by him simultaneously as well as the two boxes pass by him simultaneously.
     
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