# Why does a rotating circle near light speed increase in circumference, not decrease (contract)?

Say you have a circle that's rotating with a linear velocity of $v$ and radius $r$. It's circumference at rest is $2\pi r$. Speeding the disc up should (in my mind) cause this circumference to contract according to $$2\pi r \sqrt{1-\beta^2}$$However in the book I am reading (The Elegant Universe by Brian Greene) it states that the circumference doesn't contract, but increase in length, as in trying to measure it the 'rods' used to measure contract instead. Why is it that the circumference doesn't contract also?

(I don't know if there's much maths involved, but I'm pretty competent in maths, so understanding any that may come up shouldn't be too much of a problem)

EDIT: Also there's a picture of space-time warping in the shape of a saddle (apologies if that's not correct terminology). Does that mean that there are varying values of $r$ in relation to the stationary $r$?

## 1 Answer

This is an interesting problem, it happens that according to special relativity a rotating reference frame "precesses" when looked from an outside inertial frame, the phenomenon it's known as Thomas preccesion and it causes, among other things, the arc length increase you are referring to.

• Thanks. If i understand correctly (hopefully?) the effect comes around because the velocity vector is changing over an infinitesimally small time period? – Jordan Abbott Aug 5 '17 at 20:26
• Exactly, this paper does an easy derivation on section 3.4 using the velocity addition formula (but in the complex plane) to add up the the infinitesimal changes on the velocity, resulting in a 2$\pi\gamma$R arc length. – David Leonardo Ramos Aug 5 '17 at 20:34
• Unfortunately I don't have access to that paper but I'll dig around with some of the maths :) Much appreciated. – Jordan Abbott Aug 5 '17 at 21:28