Disc rotating close to the speed of light Imagine a large disc rotating, such that its perimeter moves close to the speed of light. Due to the tangential velocity, its perimeter should undergo length contraction. On the other hand, there is no radial velocity, therefore the radius should remain constant. But, since the perimeter is $2\pi r$, it should also remain constant.  What's wrong?
By the way, what would a distant observer on the rotation axis at infinite distance observe?
 A: The simple answer is that the circumference shrinks, because it is no longer a circle but a helix (spiral) in spacetime. As you extend the time axis around the disc, it does not join up at the same time it started, but slightly later or slightly earlier. In Euclidean space, one circuit of the helix would naturally be slightly longer than the circle's circumference. In spacetime, with its mixed sign metric, one circuit of the helix turns out to be shorter than the corresponding circle would be.

In the diagram, time is vertical and space is horizontal (in the reference frame of a static observer). At each point on the circumference of the disc, local space and time axes for the rotating observer are drawn, so they join up. Because it is moving, the space and time axes are both tilted, by a Lorentz transformation. The time axis being tilted means it advances in time as you follow the circumference around, not meeting up with itself. The length of this loop is different, and as it happens, shorter. As you get faster and approach light speed, both coordinates approach $45^\circ$ and get shorter, due to length contraction. But the volume of spacetime they trace out remains exactly inside the same cylinder of a stationary disc, and besides motion blur, will look exactly the same shape to an observer.
$\pi$ doesn't change. It doesn't require any weird hyperbolic geometry (as some sources have suggested) to understand.
