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Q & A: Ehrenfest Paradox

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Most recent answer: 01/19/2016
As a result of the Ehrenfest Paradox, the geometry of a rotating disc is non-Euclidean.However, while reaching this conclusion, we assumed that "the radius doesn't undergo Lorentz contraction", because "the radius is always perpendicular to the velocity vector", which is equivalent to : "in a circle, the radius is always perpendicular to the tangent at that point."But this is an Euclidean assumption (We have to use the parallel postulate to prove it.) Therefore it doesn't work in Non-Euclidean geometries, and we shouldn't be able to use it.
- Mirac (age 19)

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

See relativistic merry-go-round at:
- Mehran (age 65)
Arlington Heights, IL

Thanks Mehran!

(published on 01/19/2016)

Follow-up on this answer.