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Q & A: twins again

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Most recent answer: 07/17/2014
Q:
This is a question relating to time dilation and relativity In this scenario twins George and Mike move away from each other at constant velocity near the speed of light. From George’s perspective, he is stationary and it is Mike who is in motion near light-speed. So George assumes Mike must age slower than the rate at which George is aging. Mike’s clock must be ticking slower than George’s clock. From Mike’s perspective, he is stationary and it is George who is in motion near light-speed. So Mike assumes George must age slower than the rate at which Mike is aging. George’s clock must be ticking slower than Mike’s clock. George aged 10 years since Mike left, and calculates by that Mike must have aged only 1 year. Mike aged 10 years since Mike left, and calculates by that George must have aged only 1 year. Now George and Mike move towards each other at constant velocity near the speed of light. George believes when he sees Mike again, Mike will have aged a total of only 2 years for that round trip. And George will have aged a total of 20 years during that round trip. Mike believes when he sees George again, George will have aged a total of only 2 years for that round trip. And Mike will have aged a total of 20 years during that round trip. So when they meet each other, are the twins still the same age? Why? Each believed the other would look 18 years younger after the round trip. Would they see the other aging rapidly as he returned from the trip?
- Gulraj Sekhon
Pacifca
A:

If they were moving apart, then they had to accelerate in order to come back together, unless the universe wraps around on itself. Those special relativistic rules only apply to inertial frames, not accelerating ones. Whoever accelerates toward the other guy sees the other guy age quickly. If they both equally accelerate back then they see each other as the same age when they meet up again.

Mike W.


(published on 07/17/2014)

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