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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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  • Are you saying this ship moves 6 times faster than the speed of light in a vacuum, how? – JMac Oct 30 '19 at 19:51
  • No. The speed of light is x/t. The speed you are talking about is x/t' where t' is the dilated ship's time from the earth's perspective. The twin paradox is about age difference due to a frame jump, no age difference results from constant relative velocity. – ralfcis Oct 30 '19 at 19:54
  • What speed am I talking about? I was asking what you are talking about saying the ship is going 6c; which generally means "6 times the speed of light in a vacuum" from what I understand of that notation. – JMac Oct 30 '19 at 19:58
  • Sorry I just noticed I put 6c in the question. I corrected to .6c. My mistake; my 1st response was a knee-jerk and I was the jerk. – ralfcis Oct 30 '19 at 19:59
  • Yeah, I had assumed it was something that trivial... but we've been getting some strange non-mainstream questions here, so I was concerned this question was going down that road. – JMac Oct 30 '19 at 20:00
  • Re, "...There will be no clock hand off..." Do you expect that handing somebody your clock as you speed past would convey different information than if you simply allowed them to read your clock as you speed past? – Solomon Slow Oct 30 '19 at 21:11
  • I meant the clock information hand off not a physical clock handoff. – ralfcis Oct 30 '19 at 21:16

3 Answers3

1

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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  • So no matter what numerical labels you stick on the clock readouts due to synchronization, the bottom line is the captain will age 4 yrs and the earth 5 from the 3 ly mark. Thanks. – ralfcis Oct 30 '19 at 21:35
  • I have to rescind my support for your answer. The correct answer is below even though no one will agree with it. – ralfcis Oct 31 '19 at 12:20
1

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.

Marco Ocram
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0

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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  • His biological clock agrees also with Clock B, which ran from -1year to +3years over the course of the journey, which is 4 years, and with clock C, which ran from 0 to 4 years. So it agrees with all the clocks, which is what is said in my answer and DrC's. I think you are confusing yourself. – Marco Ocram Oct 31 '19 at 13:20
  • You're right, the captain's duration is the same for all 3 clocks even though the clock readouts don't match each other compared to the earth clock. Also the 3 clocks have 3 different permanent age differences to the earth clock. These have nothing to do with the captain's biological ageing which is consistent across all 3 clocks as you say. That's where I'm confused. – ralfcis Oct 31 '19 at 13:32
  • So if the captain ages 4 yrs, it's the earth that ages 4 yrs for the clock C scenario, it ages 5 yrs for the clock A scenario and 6 yrs for clock B. If all the captain's clocks have aged 4, it's the earth that has aged 3 different ways. Yes? Sounds not right but that's what the math is telling me. – ralfcis Oct 31 '19 at 15:39
  • Why this is important is because in the twin paradox clock handoff example, with A outgoing at .6c, B is earth and C is incoming at .6c, A,B and C are all the same age, it's only the clock info handed off from A to C that has experienced a frame jump and that clock has aged 2 yrs less. If that clock could grow a beard, it would be half as long as C's once C reached earth after the handoff. – ralfcis Oct 31 '19 at 15:54
  • No, that's not right. The clocks on earth are in synch with the clocks on the colony. According to those clocks, the time was 0y when the captain left the colony and 5y when the captain reached earth. All the captain's clocks say that 4 years have passed since the captain left the colony. In addition, the earth and colony clocks say that the outgoing ship left earth at -5y and arrived at the colony at 0y, whereas the clock on the outgoing ship thought it left earth at -5y and arrived at the colony at -1y. – Marco Ocram Oct 31 '19 at 16:04
  • So the goal is to start all 3 captain's clocks to the colony time of t=0. At what value do you start the clock B time when it takes off from earth to intercept the incoming ship at the colony and hand off a t=0. It has to start at t=-5 at earth and this leads to a clock handoff of t=-1 not t=0 to match the colony time. – ralfcis Oct 31 '19 at 16:21
  • Ok, I didn't see your edit. Yes the colony sees the earth ship arrive at colony time t=0 but the time the earth ship hands off to clock B is t=-1. – ralfcis Oct 31 '19 at 16:24
  • Here's the STD https://photos.app.goo.gl/c51BHCF1jW5sZxpC9 – ralfcis Oct 31 '19 at 17:23
  • Earth ships can intercept at speeds from near 0 to near c. The clock handoff times to the ship from deep space for 0c is 0y, for .6c is -1y and for c is -3y. The ships own speed of -.6c is like a +1y handoff at the colony. If the ship ages 4 yrs regardless, this means the earth ages 5y, 6y, 8y and 0y correspondingly. That's the pattern plus another pattern for incoming speeds up to -c from space. – ralfcis Nov 01 '19 at 04:22
  • No, Ralfcis, earth ages by five years from the time the captain leaves the colony. You are still confused. I will try to find time today to write an edit to my answer that takes you step by step through how to model it. All the best. – Marco Ocram Nov 01 '19 at 07:18
  • Correct, it's the earth age that's fixed at 5. The fact the captain's age adds up to 4 in different ways is a red herring. Just consider the example of earth sending a light signal to the ship as it crosses the colony. The light signal will set the captain's clock to -3y. He travels 4y for the end result of earth ageing 5 and him only ageing 1 because he has aged 4y less than earth. – ralfcis Nov 01 '19 at 12:11
  • The source of my confusion is the clock hand off twin paradox is not a real twin paradox example. Clock info can't experience a frame jump so the captain's clock reading has not been affected by time itself as it would have if it had been physically transferred between ships. The earth ages 5 and the captain ages 4 and the age differences between clocks are due to clock syncs and are not real. – ralfcis Nov 01 '19 at 14:45
  • Change " physically transferred between ships" to "on a ship turning around" – ralfcis Nov 01 '19 at 14:55