Twin Paradox: Will Car A Age Slower?

In summary: So the difference in age between the twins will be##d\tau^2/dt^2 = (1-R/r_0)*(1-v^2) = (1-R/r_0)*(1-v^2) = R/r_0##
  • #1
bcrelling
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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.
 
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  • #2
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
bcrelling said:
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.
 
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  • #4
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
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.
 
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Related to Twin Paradox: Will Car A Age Slower?

1. What is the Twin Paradox?

The Twin Paradox is a thought experiment in the theory of relativity that explores the concept of time dilation. It involves two identical twins, one who stays on Earth and the other who travels into space at high speeds, and examines how their aging process differs.

2. How does the Twin Paradox relate to time dilation?

The Twin Paradox demonstrates the principle of time dilation, which states that time passes differently for objects moving at different speeds. In this scenario, the twin who travels at high speeds experiences time passing slower than the twin who stays on Earth, resulting in a noticeable age difference when they are reunited.

3. Why does the traveling twin age slower in the Twin Paradox?

According to the theory of relativity, time is not absolute but is relative to the observer's frame of reference. When an object moves at high speeds, its frame of reference changes and time appears to pass slower for that object. Therefore, the traveling twin ages slower due to their changing frame of reference.

4. Is the Twin Paradox a real phenomenon?

No, the Twin Paradox is a thought experiment and does not actually occur in real life. It is used to illustrate the principles of time dilation and the theory of relativity. However, similar effects have been observed in experiments involving atomic clocks on airplanes and satellites.

5. Can the Twin Paradox be resolved?

Yes, the Twin Paradox can be resolved by considering the effects of acceleration and deceleration on the traveling twin. When the traveling twin changes their direction or speed, their frame of reference also changes, affecting their aging process. Accounting for these changes can help resolve the paradox and explain the age difference between the twins upon their reunion.

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