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There is a colony 3 ly away from earth and relatively stationary. The incoming ship will sync his 1st clock to the colony's time which has perviously sync'd to earth time. This is a twin paradox clock hand off example which will result in the 1st clock ageing 4 years and the earth ageing 5 from the clock hand off at the colony.

The 2nd clock will be sync'd to an earth ship intercepting the incoming ship at the colony where the incoming ship takes a clock hand off from the earth ship also travelling at .6c. This is a twin paradox clock hand off example which will result in the 2nd clock ageing 3 years and the earth ageing 5 from the clock hand off at the colony.

The 3rd clock will be sync'd to earth time when it reaches earth. There will be no clock hand off so there will be no frame jump or permanent age difference, it will just be an example of constant relative velocity. The start of the incoming ship's ageing process can be worked backward to agree with the result that there will be no age difference between the earth and ship at unification. Working backward, the ship's and earth's clocks would have been zeroed 3.75 ly away from earth in order for the two clocks to agree on unification.

So, my question is why does the simple act of when you sync the clocks affect how the captain ages differently relative to earth time?

Qmechanic
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ralfcis
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3 Answers3

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If I understand the question correctly: Both end points are in the same inertial frame and always agree on the time. The journey takes 5y in that frame and 4y on the ship. The sync clock leaves Earth at time zero taking 5 Earth years, 4 Journey years to complete the outward journey. Its time is copied to clock B for the return journey which takes a further 5 Earth years and 4 journey years to complete. 10 Earth years elapse during the round trip, but only 8 journey years.

Clock A starts its one-way journey showing 5y and ends up showing 9y.

Clock B starts its return journey showing 4y and ends showing 8y.

Clock C starts its one way journey at 6y to end up showing 10y.

The Captains beard is 4y longer when he arrives than when he left on his one-way journey. All the on-board clocks agree with that.

Or did I misunderstand your question?

DrC
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Your second example is the one that has misled you.

Regardless of how you synch your clocks, you will believe it has taken you 4 years to get from the colony to Earth, while the clocks on Earth will say that five years have passed since you left the colony.

In your second example, while it is true that the end-to-end round trip recorded by the outgoing space-ship and your incoming ship will be 8 years, only four will have passed according to the clock on the outgoing ship, so you will still have four years to travel.

Professor Sushing
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This will be a highly unpopular answer but it is in agreement with the following answer from this thread:

What is the proper way to explain the twin paradox?

This is the answer protected by Qmechanic yet my question has been downgraded and so, undoubtedly, will my answer.

In that answer, the Rindler metric determines age difference after a frame jump has been made. Before a frame jump is made, clocks are engaged in constant relative velocity and so reciprocal time dilation applies and no age difference occurs.

If the captain syncs his clock with the earth clock (t=5) on earth and you work backwards to what his time would have been on the colony, his sync'd clock would have read 1. So he aged 4 of his years to reach earth.

Earth's perspective of this event is t=0 so from earth's perspective, earth has aged 5 yrs. But from the capatin's perspective, earth's clock was not at t=0 when his clock was t'=1. According to his line of simultaneity and reciprocal time dilation, earth's clock was t=1.8. So from the captain's perspective, the earth only aged 3.2, not 5 yrs, while the captain aged 4yrs. Also, from a half speed intermediary perspective of 1/3c, both earth and captain aged 4 years. What's the true time for earth's ageing?

Perspective makes it impossible to determine how much the earth aged relative to the captain until the captain reaches earth and perspective can no longer be an issue (because of co-location). The captain syncing his clock to earth time t=5 means the captain had aged 4 of his years from the colony and since both aged at the same rate (no frame jump) the earth had to have also aged 4 yrs like the captain. So the 1/3c perspective is correct while the others distort the answer.

As for the other 2 clocks on the captain's ship that were subject to the Rindler metric, the age difference from the earth clock is permanent, not reciprocal. Clock B will indeed age 2 years less than the earth clock due to the Rindler metric after the frame jump on the inbound journey and clock A will age 1 year less.

For clock B, which will be sync'd to earth time t=-1 at the colony (because the earth ship's clock started at t= -5 in order to end at t=5), the clocks at unification will be t=5 on the earth clock and t= 3 on clock B.

For clock A, which will be sync'd to earth time t=0 at the colony, the clocks at unification will be t=5 on the earth clock and t= 4 on clock A. Remember Clock C ends with t=5 matching the earth clock t=5.

So 3 different clocks on the same ship with 3 different times at unification but the captain's biological clock agrees with clock C (minus 1 yr from the start of clock C until the captain reached the colony) for how much he aged from the colony. On the face of it, clock A had the right answer but for the wrong reason.

ralfcis
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