I say let's break out the old turntable and test this.

The clock nearest the disk center will most closely match the reference clock since it is pretty much stationary. Time dilation has nothing to do with rotational speed. The ones at the edge will have the least elapsed time because those are accelerating the most.

This experiment has already been done, using Earth as the disk.

What would you say any of this implies about how spacetime is structured and how I should know which frame is at rest? — JulianMau

There is no such thing as 'the frame that is at rest'.

More interesting: If the disk is manufactured already spinning and infinitely rigid, spatial dilation will occur at the edges, so the circumference of the disk would be more than Pi times the radius. The disk cannot slow down because that would change the circumference, and that cannot happen due to it being infinitely rigid. So it can't stop. Unlimited energy source! Of course this only illustrates the absurdity of positing an infinitely rigid thing.

The further away from the center of the rotating disk the clocks are, the most they will lag. This is in accordance with the familiar special relativity equation for the ratio of time dilation as a function of speed (dt = gamma*dt_0). That the trajectory is circular rather than linear is irrelevant to the applicability of this equation for a moving clock.

An interesting reference frame to consider, though, is the frame in which the disk is at rest and the rest of the world is rotating around its center. (But this is not an inertial frame and hence the Lorentz coordinate transformation equations don't apply to it.) It's a non-inertial frame in which a stationary observer slowly moving along the axis would measure a variable "pseudo-force" (the radial centrifugal force locally indistinguishable from a gravitational force) as well as a small lateral Coriolis force. It is due to this pseudo-gravitational force that such an observer would be able to explain why the clocks that are located closer to the center of the disk are running faster than the clocks that are located further away. They are effectively located higher up in a pseudo-gravitational field, and in accordance with general relativity, clocks down a gravitational well run slower than clocks higher up. (And due to the "principle of equivalence", a pseudo-gravitational force due to acceleration can't locally be distinguished from a "real" one due to gravitational attraction.)

The ones at the edge will have the least elapsed time because those are accelerating the most. — noAxioms

Not quite. From the point of view of special relativity, as analysed with respect to an inertial frame, only the scalar speed along the tangent of the trajectory, and not the acceleration, is relevant to the ratio of time dilation of a moving clock. Of course, it's true in this case that the clocks located further away from the center of the disk are accelerating more (centripetally), but it's not because of that that they are running slower. It's just because of their higher scalar (tangential) speed.

Isn't linear speed synonymous to tangential speed in this case? I'm not convinced that the spatial distance between the clocks located away from the centre of the disk would change considering that we don't even know whether the disk itself is Born rigid.

This experiment has already been done, using Earth as the disk. — noAxioms

Do you perceive the earth as a flat disc? Did someone put a clock at the centre of the earth?

Not quite. From the point of view of special relativity, as analysed with respect to an inertial frame, only the scalar speed along the tangent of the trajectory, and not the acceleration, is relevant to the ratio of time dilation of a moving clock. — Pierre-Normand

Yes, but look at it from the perspective of the edge. According to SR, a clock on the edge, in its own frame, is stationary and thus runs faster, not slower, than the moving clock at the center. But that frame is not maintained for even a moment, so SR doesn't really apply. The change of reference frame, and thus the change of the instant with which the center-clock is simultaneous, is what makes the center clock steadily gain time, not lose it. The effect is a moment-of-acceleration thing: acceleration component multiplied by the distance of the reference object in the direction of said acceleration component.

I had written an example once of time dilation (twins-experiment, sort of) that involved no acceleration or gravity, and hence needed to invoke only SR rules. Despite a clock in an unpowered ship being stationary in its own frame, it shows how the sum of the times for a trip log less cumulative time than a moving clock that is always running slower. I could post that if you're interested. I cannot illustrate the disk thing with just SR since the edge clocks never maintain any particular IRF.

Isn't linear speed synonymous to tangential speed in this case? I'm not convinced that the spatial distance between the clocks located away from the centre of the disk would change considering that we don't even know whether the disk itself is Born rigid. — TimeLine

Yes, linear speed (in the frame of center) is tangential speed and is constant, and the spatial distance from the center is fixed, since they're all welded to this disk and have really no choice about it. The radius should remain constant for the duration of the experiment.

Yes, but look at it from the perspective of the edge. According to SR, a clock on the edge, in its own frame, is stationary and thus runs faster, not slower, than the moving clock at the center. — noAxioms

In the proper reference frame of a point on the edge the entire disk, including the center, is stationary. But it is, as you note, non-inertial, which makes SR calculations messy.

Yes, but look at it from the perspective of the edge. According to SR, a clock on the edge, in its own frame, is stationary and thus runs faster, not slower, than the moving clock at the center. But that frame is not maintained for even a moment, so SR doesn't really apply. The change of reference frame, and thus the change of the instant with which the center-clock is simultaneous, is what makes the center clock steadily gain time, not lose it. The effect is a moment-of-acceleration thing: acceleration component multiplied by the distance of the reference object in the direction of said acceleration component. — noAxioms

Yes, that's true if you construe the rotating observer at the edge constantly shifting from one ("co-moving") inertial frame to another, and the simultaneity between this observer's clock and the moving clock at the center of the disk thus being constantly redefined. This constant redefining of simultaneity between the two clocks accounts for the fact that the accelerating observer sees the center clock running faster in spite of the fact that it is lagging as seen from the co-moving inertial frames temporarily occupied by the observer at the edge. This is also how one can account for the fact that, in the twin-paradox, the twin traveler sees her stationary twin aging more overall in spite of the fact that she sees her aging slower during the two legs of her trip (assumed to take place at constant speed) away and back. It's when the travelling twin rapidly reverses course at the far point of her trip, and she becomes stationary relative to a new inertial frame, that the other twin (and the whole Earth) skips to an older age due to simultaneity being redefined in the new inertial frame of the far away travelling twin.

The frame that I was discussing in my previous post is non-inertial, and both the center clock and the clock at the edge are stationary in this frame, since the whole frames rotates with the disk.

Isn't linear speed synonymous to tangential speed in this case? I'm not convinced that the spatial distance between the clocks located away from the centre of the disk would change considering that we don't even know whether the disk itself is Born rigid. — TimeLine

Yes, linear speed just is tangential speed in this case. I was not picturing a moving clock but rather an observer skipping from one clock to the next at various distances from the center of rotation of the disk and comparing their rates to the rate of the clock at the center. Such an observer could explain the various rates of those clocks as a result of their various positions in the apparent gravitational field. There is no issue with rigidity if the disk is assumed to be rotating at constant angular velocity.

There is no issue with rigidity if the disk is assumed to be rotating at constant angular velocity. — Pierre-Normand

Is that taking into consideration the paradigm between local and global spatial geometry?

Is that taking into consideration the paradigm between local and global spatial geometry? — TimeLine

I am unsure what issue you are trying to raise. The rotating disk is in a stationary state. It is not undergoing any sort of deformation as seen from the inertial frame in which its center is at rest, or the non-inertial rotating frame that is constituted by measuring rods and clocks affixed to it.

I am having a problem with the conditions of the rigidity and local synchronisation, but it is quite late after a really long day, so maybe if you have a read of this and we can chat more about later: http://pubs.sciepub.com/ijp/1/5/4/