Special Relativity and Clocks on a Rotating Disk
I was reading about something called the Ehrenfest Paradox and it got me thinking about something (that I think is) similar:
Suppose we take a large, flat, and rigid disk, and we attach to various parts of it a number of clocks (some very close to the center of the disk, some along the edge, others in between). We synchronize all these clocks using some master clock that is at reference with respect to them, and then set the disk spinning very, very fast. Suppose we then stop the disk after the master clock reads, say, 12 hours, and compare the times of the clocks on the disk with the master clock.
What should we expect to find, and what are the implications about the structure of spacetime? How do we even know it wasn't the reference frame of the master clock that was spinning the whole time?
Do you agree that I am correct in suspecting that less time will have ticked off the clocks on the disk because they are moving relative to the master clock? And that the clocks nearer the center of the disk will show less time having past than those at the outer edge, due to their higher rotational velocity? What would you say any of this implies about how spacetime is structured and how I should know which frame is at rest?
Suppose we take a large, flat, and rigid disk, and we attach to various parts of it a number of clocks (some very close to the center of the disk, some along the edge, others in between). We synchronize all these clocks using some master clock that is at reference with respect to them, and then set the disk spinning very, very fast. Suppose we then stop the disk after the master clock reads, say, 12 hours, and compare the times of the clocks on the disk with the master clock.
What should we expect to find, and what are the implications about the structure of spacetime? How do we even know it wasn't the reference frame of the master clock that was spinning the whole time?
Do you agree that I am correct in suspecting that less time will have ticked off the clocks on the disk because they are moving relative to the master clock? And that the clocks nearer the center of the disk will show less time having past than those at the outer edge, due to their higher rotational velocity? What would you say any of this implies about how spacetime is structured and how I should know which frame is at rest?
Comments (13)
I say let's break out the old turntable and test this.
This experiment has already been done, using Earth as the disk.
Quoting JulianMauThere 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.
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.)
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.
Do you perceive the earth as a flat disc? Did someone put a clock at the centre of the earth?
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.
Quoting TimeLineYes, 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.
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, 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.
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.
Is that taking into consideration the paradigm between local and global spatial geometry?
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.