Physics-Special Relatvity-Twin Paradox

Twin Paradox

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Debate Methodology

  • First, the Methodology asks the proponent of a reconciliation argument a few basic questions about space-time physics.

     

  • Second, the Methodology defines a classic Twin Paradox scenario which is broken into segments bounded by well defined events (see below). The Twin Paradox scenario is given some specific numeric parameters such as duration and the constant velocity. The proponent of a reconciliation argument is asked to give the proper time accumulation, the number of clock "ticks", for each segment.

     

  • Finally, the proponent's answers plus straightforward logic is used to show that the proposed reconciliation argument is untenable.

     

  • The two major classes of "accepted" reconciliation arguments are shown on the "Can't Be Due To Relative Velocity" and "Can't Be Due To Turnaround Acceleration" pages. The most recent claim that how the net proper time difference accumulates is unknowable is discussed in the Time Difference Accumulation Is Not Indeterminate page.

    One solution type that does not lead to contradictions uses clock retardation as a function of velocity with respect to a unique physics frame which, in general, yields asymmetric effects for the outbound and inbound segments. This construct is not part of currently accepted theory.

     

  • Scenario

 R is the "stay-at-home twin" always  at rest in the R-frame

 S is the "traveling twin"

                              (S’s outbound trip)
    E1a
E1   |                                                                                                                                             
->--V-------------------------------------------------------------------------------------------àE2
R                                                                                                                                                    S (E2a)
ß-----------------------------------------------------------------------------------------------<-E3
E4  |                                                                                                                                 
  E4a
                                                       (S’s inbound trip)

Description of events

E1: S starts the initial acceleration to relative velocity v

E1a: S ends the initial acceleration to relative velocity v

E2: S starts his turnaround acceleration

E2a: S is momentarily at rest in R.

E3: S ends his turnaround acceleration

E4a: S starts the final acceleration from relative velocity v to being at rest with R

E4: S ends the final acceleration from relative velocity v to being at rest with R

R’s clock records 200 billion ticks between E1 and E4We set v at 0.866c so that the time dilation factor is ½ .  Hence, S’s clock records 100 billion ticks between E1 and E4 and the net difference in proper time is 100 billion ticks. Since the constant velocity segments can be made arbitrarily long versus the period of acceleration, we choose, for this scenario, that S's clock accumulates 1,000 ticks of proper time between E2 and E3. The proponent of a reconciliation argument is asked to give the number of ticks (i.e., the accumulated proper time) for the traveling twin, S, for the segments defined by the pairs of events E1 - E1a, E1a - E2, E2 - E2a, E2a - E3, E3 - E4a and E4a - E4. The sum should add up to 100 billion. For the acceleration segments, the responder may give approximations. .  

© 2004, 2012


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