Do velocity and acceleration time dilation factors add?

Question

For a spinning space station such as in 2001, A Space Odyssey, what would be the time slowing in the perimeter of the spinning space station with respect to the center axis of the station?

The perimeter is moving at a speed such that the acceleration is $g=9.81\text{ m/s^2}$. Combining $g=\frac{{v}^{2}}{R}_{s}$ with $\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}$ gives dilation factor $$\sqrt{1-\frac{g\,{R}_{s}}{{c}^{2}}}$$ Assuming the radius ${R}_{s}$ of the space station is 500 meters, a perimeter clock would lose about 1e-6 seconds per year with respect to a clock in the center axis.

Now since the clock at the perimeter is subject to acceleration g, by the equivalence principle it would seem that the gravitational time dilation would apply, which is $$\sqrt{1-\frac{2R_e g}{c^2}}$$ where $R_e=6.38\times 10^6\text{ m}$ (source). This would make the perimeter clock slow by about 0.02 seconds per year.

So do I add the two dilation factors to get the total dilation factor? The gravitational dilation factor $\sqrt{1-\frac{2R_e g}{c^2}}$ is a function of both acceleration $g$ and radius $R_e$, unlike the formula for velocity dilation $\sqrt{1-\frac{v^2}{c^2}}$, which is only a function of velocity. So I suppose applying the principle of equivalence has some subtleties. My knowledge of special relativity is way ahead of my knowledge of general relativity.

Answer 1

These are not two different effects. They are the same effect as viewed in two different frames of reference. They shouldn't be added. If they were both calculated correctly, they'd be equal to each other.

They are not equal to each other, and that's because the calculation in the rotating frame is effectively assuming the existence of a gravitational potential $\Phi=-gr$, giving a time dilation factor $e^\Phi$ (in units with $c=1$). But it's only in a static spacetime, represented in nonrotating coordinates, that you can derive a diagonal metric from a single scalar potential.

If you transform from nonrotating coordinates to rotating ones, the metric for Minkowski space picks up off-diagonal terms. These terms are observed in the rotating frame as a Sagnac effect. If you calculate the line element for an object in these coordinates, I believe you get a term that can be interpreted as a gravitational time dilation, plus another term representing the Sagnac effect. The result should be the same as in the nonrotating frame.

In more nontechnical terms, a rotation isn't just equivalent to a gravitational field as you might expect from a naive application of the equivalence principle. It's equivalent to a gravitational field plus a Sagnac effect.

Answer 2

For the answer by PMay:

However, both the observer at the center axis and the observer at the perimeter would agree that the circumferance of the perimeter is $2\pi R$.

That is not true for the observer at the perimeter. He/she is moving with an acceleration, and from his/her point of view, space would be distorted, gravitational force, time dilation and other relativistic effects would appear. If he/she would line the perimeter with measuring rods which are motionless from his/her point of view, then he/she would find that it takes more than $2\pi R$ total length of the rods.

Refer to some solid textbook on GR for all those effects. Misner, Thorne, Wheeler is one of most popular.

Answer 3

Researching the Sagnac effect led me to Born coordinates for analysis if rigid rotation. I am still working through it, but it got me thinking about an intuitive way to understand the problem. Imagine the x' axis of the rotating frame of reference as being wrapped around the perimeter of the space station. The observer at the center axis of the space station would observe that a measuring rod at the perimeter positioned perpendicular to the radius of the station would be foreshortened by the factor $\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}$, just as if the measuring rod was undergoing straight line motion. However, both the observer at the center axis and the observer at the perimeter would agree that the circumferance of the perimeter is $2\,\pi \,R$. This seems paradoxical given that the center axis observer sees the measuring rods at the perimeter are foreshortened. The answer to the paradox is the lack of agreement on simultaneity. The "angular space-time" interval of the perimeter observer making one full rotation according to the center observer is ${c}^{2}\,{t}_{c}^{2}-4\,{\pi}^{2}\,{R}^{2}$. The spacetime interval according to the perimeter observer is ${c}^{2}\,{t}_{p}^{2}$. Solving for ${t}_{p}$ we get ${t}_{p}={t}_{c}\,\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}$. 'Angular space-time interval' would have to justified with calculus by taking small increments on the perimeter and adding them up.

This makes me think that foreshortening of measuring rods in a rotating ring is "not real" and merely an artifact of different definitions of simultaneous by the various observers. I recall reading in my textbooks that foreshortening has real physical effects, but this example makes me think otherwise.