simple time dilation question

Hi,

I am confused about this simple time dilation question:

  1. For a journey from Earth, accelerating to 0.999C over a 0.5 light year distance, then decelerating over a 0.5 light year distance, and heading back towards earth: accelerating to 0.999C over a 0.5 light year distance, then decelerating over a 0.5 light year distance. (total trip distance 2 light years)

  1. For a journey from Earth, accelerating to 0.999C over a 0.5 light year distance, then coasting for a 1 light year distance, then decelerating over a 0.5 light year distance, and heading back towards earth: accelerating to 0.999C over a 0.5 light year distance, then coasting for a 1 light year distance, then decelerating over a 0.5 light year distance. (total trip distance 4 light years)

  2. For a journey from Earth, accelerating to 0.999C over a 0.5 light year distance, then decelerating over a 0.5 light year distance, and continuing away from Earth: accelerating to 0.999C over a 0.5 light year distance, then decelerating over a 0.5 light year distance. (total trip distance 2 light years)

  1. For a journey from Earth, accelerating to 0.999C over a 0.5 light year distance, then coasting for a 1 light year distance, then decelerating over a 0.5 light year distance, and continuing away from Earth: accelerating to 0.999C over a 0.5 light year distance, then coasting for a 1 light year distance, then decelerating over a 0.5 light year distance. (total trip distance 4 light years)

Those four examples all have the same rates of acceleration for the same overall time dilation if time dilation only occurs during their acceleration time. The final distance to Earth is known (2ly or 4ly), so to check the clocks for 3 and 4, the communication time can be added to the measured clock time on the distant clock to see what the time dilation was.

What is the time differential between the Earth clock and each of

1,2,3,4? Also at the end of the journey are 1,2,3,4 clocks equal when compared, taking into account the communication time (ie light speed).

cheers, Jamie

Reply to
Jamie M
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You're confused about a lot of things. One of them is about posting on this newsgroup: you should not.

Bye Jack

Reply to
jack4747

You are very confused.

Time dilation occurs all the time that they are in relative motion.

There is only a unique answer to the question for cases 1 and 3 where the clocks being compared have been returned to their original position.

In the case of 3 the returnee will have spent an extra period of time travelling at 0.999c whilst the Earthlings all aged an extra 2 years. It is left as an exercise to the reader to work out by how much the spaceman moving at a constant 0.999c has aged.

In cases 2 & 4 where the clocks are widely separated by 2ly and 4ly respectively the answer depends on the frame of reference of the observer who is asked to do the observations. Simultaneity is only well defined for objects at the same coordinates in spacetime.

Acceleration just makes it a bit harder to calculate the proper time interval for the moving observer. See for example:

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"paradox"

ISTR from part II relativity that you can get to the centre of our galaxy with a sustained 1g acceleration in about 20 years travel time in the frame of the spaceship. The people left back on Earth age considerably more whilst the intrepid spacemen fly off there (about

30000 years if memory serves). Here is a version of the same thread:

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Also considers some of the practical difficulties.

--
Regards, 
Martin Brown
Reply to
Martin Brown

Hi,

Only case 1 and case 2 return to earth. Case 3 is 2 light years from earth at the end of the journey, and case 4 is 4 light years from earth at the end of the journey.

Final distance between clocks:

case1: 0 light years case2: 0 light years case3: 2 light years case4: 4 light years

It is a simple question asking the absolute clock differentials taking into account the light speed communication time to communicate the clock times.

cheers, Jamie

Reply to
Jamie M

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