Variation on twins thought experiment

  • Thread starter bcrelling
  • Start date
  • #1
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2

Main Question or Discussion Point

If we have two twins driving cars.
They remain at the same height above sea level so gravitational time dilation is equal.
Both cars travel at the same speed so time dilation due to speed is equal for both.

However...
Car A drives around a small circular race track and experiences centripetal acceleration. Car B drives is a straight line and does a circuit of the globe.

Will car A age slower due increased angular acceleration it experiences?

Thanks.
 

Answers and Replies

  • #2
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Are you considering the planet to be rotating or non rotating?

Are you comparing their age using standard Schwarzschild simultaneity or some other method?
 
  • #3
A.T.
Science Advisor
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Will car A age slower due increased angular acceleration it experiences?
Angular acceleration is not something that you "experience" in a frame invariant sense. It's computed with respect to some point in some coordinates. And it has nothing to do with aging.

Did you maybe mean proper (centripetal) acceleration? Proper acceleration itself also doesn't affect the instantaneous aging rate:
http://en.wikipedia.org/wiki/Clock_hypothesis
So assuming the planet is not rotating, and they have the same speed relative to the planet, they will age by the same amount between meetings.
 
  • #4
69
2
To be exact:
The planet is not rotating.
Both riders start at the same time and same place.
Time is compared by the twins carrying clocks which are compared after driver B completes a circuit of the globe and both drivers meet at the starting point.

But I think you've already answered my question- there would be no difference between them.

Thanks!
 
  • #5
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5,319
Yes, there would be no difference. Here is how you would calculate it:

The Schwarzschild metric is ##d\tau^2 = (1-R/r) dt^2 - (1-R/r)^{-1}dr^2 - d\Omega^2## where ##d\Omega^2=r^2 (d\theta^2+\sin^2\theta~d\phi^2)## is the metric of a 2-sphere. On the surface of the planet let ##r=r_0## and ##v=d\Omega/dt##. So

##d\tau^2/dt^2 = (1-R/r_0) - v^2##

This does not depend on the shape of the path as long as it stays on the surface and goes at a constant speed.
 
Last edited:

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