Twin Paradox - Difference in Elapsed Time As A Result of Differences in Twins' Spacetime Paths

Difference in Elapsed Time As A Result of Differences in Twins' Spacetime Paths

The following paragraph shows several things:

  • how to employ a precise mathematical approach in calculating the differences in the elapsed time
  • how to prove exactly the dependency of the elapsed time on the different paths taken through spacetime by the two twins
  • how to quantify the differences in elapsed time
  • how to calculate proper time as a function (integral) of coordinate time

Let clock K be associated with the "stay at home twin". Let clock K' be associated with the rocket that makes the trip. At the departure event both clocks are set to 0.

Phase 1: Rocket (with clock K') embarks with constant proper acceleration a during a time Ta as measured by clock K until it reaches some velocity V.
Phase 2: Rocket keeps coasting at velocity V during some time Tc according to clock K.
Phase 3: Rocket fires its engines in the opposite direction of K during a time Ta according to clock K until it is at rest with respect to clock K. The constant proper acceleration has the value −a, in other words the rocket is decelerating.
Phase 4: Rocket keeps firing its engines in the opposite direction of K, during the same time Ta according to clock K, until K' regains the same speed V with respect to K, but now towards K (with velocity −V).
Phase 5: Rocket keeps coasting towards K at speed V during the same time Tc according to clock K.
Phase 6: Rocket again fires its engines in the direction of K, so it decelerates with a constant proper acceleration a during a time Ta, still according to clock K, until both clocks reunite.

Knowing that the clock K remains inertial (stationary), the total accumulated proper time Δτ of clock K' will be given by the integral function of coordinate time Δt

where v(t) is the coordinate velocity of clock K' as a function of t according to clock K, and, e.g. during phase 1, given by

This integral can be calculated for the 6 phases:

Phase 1
Phase 2
Phase 3
Phase 4
Phase 5
Phase 6

where a is the proper acceleration, felt by clock K' during the acceleration phase(s) and where the following relations hold between V, a and Ta:

So the traveling clock K' will show an elapsed time of

which can be expressed as

whereas the stationary clock K shows an elapsed time of

which is, for every possible value of a, Ta, Tc and V, larger than the reading of clock K':

Read more about this topic:  Twin Paradox

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