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Twin problem

  1. Dec 17, 2006 #1
    if you arte not fed up with the problem, please consider the following question.
    A realistic approach considers the following scenario:
    An unidimensional rectilinear (IN SPACE) path will be considered in which the moving clock starts its motion with zero initial velocity from the origin of the inertial reference frame I in which the rest clock denoted conventionally as (1) is located. After the velocity V is reached, the force F is instantaneously reversed. Then (2) is decelerated, stops and inverts its motion until the velocity -V is reached. At this point F is suddenly reversed again, so that, when (2) stops it meets (1) again and they compare their readings. [Found.Phys.Lett. 18 (2005) 1-19
    I consider that after the two outgoing steps clock (2) is at rest in the rest frame of the stay at home clock and in front of a clock of that frame which reads (displays), as a result of the clock synchronization (Einsteinian) performed in that frame, the same time as clock (1) does. In short, the question is if we should consider the folloowing incoming steps? Is the statement of the clock paradox violated that way?
    sine ira et studio
  2. jcsd
  3. Dec 17, 2006 #2

    Do you have a copy of the paper that you are willing to attach to your post such that we all refer to the same thing? You coud scan the paper and make it into a jpeg attachment.
    Last edited: Dec 17, 2006
  4. Dec 17, 2006 #3

    George Jones

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    http://www.arxiv.org/PS_cache/physics/pdf/0405/0405038.pdf" [Broken]
    Last edited by a moderator: May 2, 2017
  5. Dec 17, 2006 #4
    Thank you, Greg. Would you happen to have an arxiv pointer to this paper?
    Last edited by a moderator: May 2, 2017
  6. Dec 17, 2006 #5


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  7. Dec 17, 2006 #6
  8. Dec 17, 2006 #7
    Even as this shows the twins experiment works following the “SR rules correctly” when clock (2) is using an accelerating frame with equal acceleration amounts are used in both directions.

    I think your question is most concerned about when “clock (2) is at rest in the rest frame of the stay at home clock and in front of a clock of that frame”. That is concerned with the time on the clock (1) and its companion “distant clock (1)” that happens to be at the point where clock (2) stops in relation to the common frame of clock (1) and “distant clock (1)”.
    Yes it is true that clock (1) and “distant clock (1)” are always in synchronization. But that does not mean they simultaneous read the same time.
    That is the point of SR, that those two cannot give the same simultaneous time to any observer not remaining in their reference frame.
  9. Dec 17, 2006 #8
    clock paradox

    thank you for your help. my problems are inserted in your answer
  10. Dec 18, 2006 #9
    " I think that as a result of the synchronization the clocks of the rest frame display at each time the same running time and that is why the comparison of clocks 2 and distant clock 1 at the first stop is the same as at the second stop as the statement of the clock paradox requires?? "

    NO, I'm not sure you meant what you said here, "comparison of clocks 2 and distant clock 1 at the first stop is the same as at the second stop" or miss stated your point.
    Your comparing two events "A" when the clock is mid way though having the reverse force applied at it for an instant "stops" as it is accelerating back towards the start. And "B" when the force is stopped after having been reversed again to the original acceleration so that it stops perfectly at the beginning point. So of course event A and event B are at different times for all clocks. The keys is event A or even the start of the trip did not occur "simultaneously" at the same time for both Clock (1) and "distant clock 1" even though they are synchronized!

    " please be more specific"
    Read my last sentence as for now it should not make sense and follow me very carefully.
    - As Clock (2) travels we know it runs slow as compared to all clocks in frame 1.
    - Therefore when it reaches event A must see "distant clock 1" at a much high time as if time ran fast there and the trip took much longer by "distant clock 1" measurement.
    - But we know that cannot be true the trip had to take the same amount of time for all clocks in frame 1.
    - How can Clock 2 account for this? It must figure out exactly what time is was on "distant clock 1" when it started.
    - Easy just subtract the correct trip elapse time in frame 1 from the time Clock 2 directly sees on "distant clock 1" when it reaches event A.
    BUT be very careful at this point You are in frame 2! You must calculate this elapsed time based on the frame of Clock 2 see "distant clock 1" come moving in relation to that reference frame 2.
    - This calls for a very short interval of time
    - And that means the "Start" for when Clock 1 and 2 where together is well into the future when as measured by "distant clock 1" as compared to Clock 1.
    - That means the start event, event A, and event B are not at the same "simultaneous" time for Clock 1 and "distant clock 1" even thou they are synchronized.

    NOTE: Although I accept the thought experiment of constant acceleration switching from + to -; If you want to work though and example with hard numbers it will be much to complex. Use instead one with fixed speeds and change directions with an "instantaneous" accelerations at all three events A, B, and Start. Just use the premise that an instantaneous acceleration adds no elapsed time in any frame and you will be fine.
    Give yourself a little time with it to absorb the idea and get it.
  11. Dec 18, 2006 #10
    thanks again. my problem, stated simply is to have a yes or no answer to the following question. first I fully agree with the paper i quote and interested only in the problem considered from the reference frame of the stationary clocks. The events I consider are the first stop in the outgoing trip of the accelerated clock say A where we can compare the readings of the accelerating clock with the reading of the stationary clock in front of which the stop takes place and event say B (the second stop in the incoming trip) where again we can compare the reading of the accelerating clock with the reading of the stationary clock in front of it. Brings the second event suplementgary information for the study of the clock paradox? Is it necessary to take it into account?
  12. Dec 18, 2006 #11
    No - I don’t really understand the question.

    No part of this detail is necessary to define SR. What is necessary is the ability of SR to define the relative locations and times of all events from the view of any reference frame – and it can.

    And, Yes by successfully being able to define all those event times and locations from both frame views, SR does solve apparent paradoxes like this.
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