Consider a thin disk with radius R at z = 0 in a Lorentz reference frame. The disk rotates rigidly with angular velocity . In

Consider a thin disk with radius R at z = 0 in a Lorentz reference frame. The disk rotates rigidly with angular velocity Ω. In the early years of special relativity there was much confusion over the geometry of the disk: In the inertial frame it has physical radius (proper distance from center to edge) R and physical circumference C = 2πR. But Lorentz contraction dictates that, as measured on the disk, the circumference should be

and the physical radius, R, should be unchanged. This seemed weird. How could an obviously flat disk in flat spacetime have a curved, non-Euclidean geometry, with physical circumference divided by physical radius smaller than 2π? In this exercise you will explore this issue.

(a) Consider a family of observers who ride on the edge of the disk. Construct a circular curve, orthogonal to their world lines, that travels around the disk (at

This curve can be thought of as lying in a 3-surface of constant time x^0̂ of the observers’ proper reference frames. Show that it spirals upward in a Lorentz-frame spacetime diagram, so it cannot close on itself after traveling around the disk. Thus the 3-planes, orthogonal to the observers’ world lines at the edge of the disk, cannot mesh globally to form global 3-planes.

(b) Next, consider a 2-dimensional family of observers who ride on the surface of the rotating disk. Show that at each radius

the constant-radius curve that is orthogonal to their world lines spirals upward in spacetime with a different slope. Show this means that even locally, the 3-planes orthogonal to each of their world lines cannot mesh to form larger 3-planes—thus there does not reside in spacetime any 3-surface orthogonal to these observers’ world lines. There is no 3-surface that has the claimed non-Euclidean geometry.

1-v C (with v = S2R),