Ehrenfest Paradox
Most recent answer: 01/18/2016
- Mirac (age 19)
Turkey
You're right that in the rotating frame one can't use Euclidean geometry. It's important to realize here that in the absence of enough mass density to make significant gravitational effects the fundamental geometry of this spacetime region is Euclidean. The description of it from the viewpoint of the non-rotating frame can use ordinary Euclidean geometry with ordinary Lorentz contraction effects, etc. Thus the conclusion that identically constructed meter sticks laid out on the rotating disk along a diameter and along the circumference and used as measuring sticks will not give C/D=π is robust.
That's why this traditional example for introducing general relativistic geometrical effects is of limited use. It just shows that one can pick reference frames that have non-Euclidean properties if a standard physical definition of spatial distance is used. It doesn't show that one must do so. To show that requires gravity.
This is such a thoughtful question, I'd like to encourage you to think about UIUC if you apply to physics graduate school.
Mike W.
posted without vetting until Lee gets back.
(published on 01/18/2016)
Follow-Up #1: relativistic merry-go-round
- Mehran (age 65)
Arlington Heights, IL
Thanks Mehran!
(published on 01/19/2016)