Special relativity and uniform circular motion

Suppose we have two observers A and B. A is in some inertial frame. B is undergoing uniform circular motion with some speed $v$ as observed by A. They each have a clock.

My concern is with what A sees when looking at B's clock and what B sees when looking at A's clock.

I expect A will see B's clock slowed down... according to A it will take $\gamma$ seconds for B's clock to tick 1 second. This is the standard time dilation formula. Is this right?

But how does one analyze how B sees A's clock using only special relativity?

• Possible duplicate: physics.stackexchange.com/q/2554 Commented Feb 14, 2016 at 17:03
• @dmckee, Yes it's a variation on the twin paradox. so suppose B starts out at a speed of 0 matching A's clock.... then speeds up to v keeping circular motion... then slows back down to 0. Special relativity will say B's clock is behind A's clock? So this scenario really makes no difference from the standard twin paradox... Commented Feb 14, 2016 at 17:13

Recall in Euclidean geometry how the shortest path between two points was a straight line?

A similar (but opposite) thing happens in relativity. The curve between two events that has the most proper time is the straight line (or geodesic).

So if observer B has coordinates $(R\cos( \omega t),R\sin(\omega t),0)$ and observer A has coordinates $(R,0,0)$ then A ages the most. And they can compare ages when they meet. It's just that A will age more between times they meet than B will age.

Observer B will compute the rate A ages (the computation is based on when you get a light signal, and then correcting for the travel time of the signal, and yes to do the calculation B will have to take into account that they are not inertial) and see them aging faster.

You can even offset A so they never meet, e.g. let A be at $(5R,0,0)$ or $(0,0,0)$ and the results stay the same.