Distance in relativistic circular motion in invariant spacetime

Circular motion in special relativity is somewhat tricky: Note that for circular motion, the acceleration in the spaceship travelling in a circle is not zero, so the spaceship is not in a single frame of inertia.

Here is an interesting thought: Distances perpendicular to the direction of motion are not subject to contraction. Hence, if the observes on earth see the spaceship going on a circle with radius $R$, then in their frame of inertia the spaceship is always a distance $R$ away from the earth. Since the line from earth to spaceship is perpendicular to the direction of flight, the people on the spaceship will also believe that they are always a distance $R$ away from earth, so they will also fly on a circle.

They will, nonetheless, experience a different circumference! The best way to solve this is to consider a polygon with $N$ sides and then let $N$ go to infinity. If people on the earth measure each side of the polygon as $L_0/N$ where $L_0$ is the circumference of the polygon in the earth's frame of inertia, then the spaceship-people will measure each side to be $L_0/(N\gamma)$. Hence, for $N \rightarrow \infty$, the polygon becomes a circle. Measured from earth, it has circumference $L_0$, but for the spaceship it has circumference $L_0/\gamma$.

This suggests that the spaceship moves through non-Euclidean geometry, because it travels on a circle whose ratio between circumference and diameter is less than $2\pi$. This is a hint that accelerating frames have non-Euclidean geometry, which is excessively treated in General Relativity.

Reference: http://abacus.bates.edu/~msemon/WortelMalinSemon.pdf