Does an observer moving in a circle with constant angular velocity in space experience GR gravitational time dilation?

Question

Assuming that there are no other planets or other gravitational sources around the observer in empty space, would the observer's very fast circular motion create GR gravitational or else called acceleration time dilation?

I am not interested on the SR or else called kinetic constant velocity time dilation, only for the acceleration time dilation? Even if the angular velocity is kept constant, since this is a circular motion the velocity vector direction is changing all the time and therefore must be considered as an accelerated motion. Right? So therefore there must be also a GR gravitational time dilation component together with the SR kinetic time dilation due to the constant angular velocity of this assumed circular motion.

The only reason I could think of for no gravitational time dilation to exist is because the observer's stable path circular motion and angular velocity, the centripetal cancels out with the centrifugal and therefore there is no gravitational time dilation component.

The reason why I am asking this question is because I was puzzled lately by the fact that by the recent g-2 Fermilab muons experiment lecture, in the experiment's muon storage ring (i.e. cyclotron) the only time dilation present they have announced was about (64-2.2)=61.8μs and was purely a SR kinetic time dilation. They never said anything about any GR acceleration time dilation effect due the circular motion, present on the experiment?

Answer 1

So therefore there must be also a GR gravitational time dilation component together with the SR kinetic time dilation due to the constant angular velocity of this assumed circular motion ... the only time dilation present they have announced was about (64-2.2)=61.8μs and was purely a SR kinetic time dilation. They never said anything about any GR acceleration time dilation effect due the circular motion, present on the experiment?

There is no accelerational time dilation. This known as the clock hypothesis and has been experimentally verified up to about $10^{18}$ g (see: Bailey et al., “Measurements of relativistic time dilation for positive and negative muons in a circular orbit,” Nature 268, July 28, 1977, pg 301).

Gravitational time dilation does exist, but it is not determined by acceleration. Gravitational time dilation is determined by gravitational potential. In an inertial frame there is no gravitational potential, so regardless of the acceleration there is only the kinematic time dilation. The analysis you mention is correct.

Now, per the equivalence principle, it is possible to treat a uniformly accelerating reference frame like a uniform gravitational field. This does not help with uniform circular motion because the acceleration is not uniform, and in particular the Coriolis force complicates things. You can use the mathematical framework of tensors etc. to calculate time dilation in a rotating reference frame. You will get a term that has the form of a gravitational time dilation due to the centrifugal force, but the remainder of the terms do not follow the usual kinematic time dilation formula. So overall it is not particularly helpful to think of a rotating reference frame in terms of gravitational time dilation.

Answer 2

Different worldlines have different lengths, and the elapsed time on a worldline is the length of the worldline. Names like "special-relativistic time dilation" and "gravitational time dilation" are given to ratios between lengths of worldlines in specific geometric arrangements. They're special cases, not different phenomena.

A circling particle has a helical worldline. In Euclidean 3-space, if you've got a straight line and a helix heading in the same direction, the helix is longer. If the slope of the helix relative to the straight line is $m$, then it's longer by a factor of $\sqrt{1+m^2}$. In spacetime, by convention the slope is called $v/c$, and because of the flipped sign in the metric, the helix is longer by a factor of $\sqrt{1-(v/c)^2}$ (which is less than $1$, so it's actually shorter).

General relativity adds nothing to special relativity except that spacetime curvature is related to energy-momentum density. In this problem, we're taking curvature to be negligible, so the geometry is the same as it was in the special-relativistic case, and the ratio of lengths is the same.

See this answer for more about the geometry of gravitational time dilation and this answer for special-relativistic time dilation.