What time does an observer see when watching a clock on a rotating circle?

[chubert]

Hi,
I'm new here. This question may already have an answer but I didn't find it. Sorry if there's already one. It's a static version of the twin paradox, without travel and without twins.

We have 2 circles C and C' that are superposed. C is fixed, with time t and co-ordinate x along the circle, and has circumference S. C' is rotating at constant speed v, with time t' and co-ordinate x'. Following relativity, and ignoring the centrifugal acceleration that can be as small as wanted by increasing the circle size, we have t'=$\gamma$ (t-vx/c²).

And now the question: an observer is located at x=S/2. He can be also considered located at x=-S/2. The observation is made at t=0. What time t' does this observer see when he watches the clock of C' located at the same point? Is it -$\gamma$ vS/(2c²) or $\gamma$ vS/(2c²)?

Or does he see infinitely many clocks from C' showing $\gamma$ (n-1/2)vS/c² for n integer?

 

[PAllen]

Hi,
I'm new here. This question may already have an answer but I didn't find it. Sorry if there's already one. It's a static version of the twin paradox, without travel and without twins.

We have 2 circles C and C' that are superposed. C is fixed, with time t and co-ordinate x along the circle, and has circumference S. C' is rotating at constant speed v, with time t' and co-ordinate x'. Following relativity, and ignoring the centrifugal acceleration that can be as small as wanted by increasing the circle size, we have t'=$\gamma$ (t-vx/c²).

And now the question: an observer is located at x=S/2. He can be also considered located at x=-S/2. The observation is made at t=0. What time t' does this observer see when he watches the clock of C' located at the same point? Is it -$\gamma$ vS/(2c²) or $\gamma$ vS/(2c²)?

Or does he see infinitely many clocks from C' showing $\gamma$ (n-1/2)vS/c² for n integer?

Welcome to physics forums!

The Lorentz transform only applies to orthonormal coordinates in inertial frames. Your x coordinate is not 'straight', and your primed system is not inertial, so you cannot use the Lorentz transform.

The easiest way to analyze such a problem is simply to pick one inertial frame for all the analysis. You can figure out what any observer sees from this one frame.

In addition, your problem has no real answer in principle. You are not asking about a duration, but what time is seen on a clock at an arbitrary moment. The answer depends on when and how initial setting and synchronization is done - which you don't specify. Further, for a non-inertial frame there is no uniquely preferred method of synchronization (as there is for inertial frames). Which one you choose will produce different answers.

 

[bcrowell]

Hi, chubert,

Welcome to PF!

It sounds to me like you're groping your way toward an idea that is explained very clearly in this paper:

Gron, Relativistic description of a rotating disk, Am. J. Phys. 43 (1975) 869

The idea is that clocks can't be globally synchronized in a rotating frame of reference. Unfortunately Gron's paper isn't freely available online. I've given a shorter explanation of some of the same ideas here: http://www.lightandmatter.com/html_books/genrel/ch03/ch03.html#Section3.4 (subsection 3.4.4).

What Gron does is to take Lorentz transformations between neighboring points on the circle, which, if done carefully, avoids the problems PAllen pointed out with your approach.

-Ben

 

[chubert]

Welcome to physics forums!

The Lorentz transform only applies to orthonormal coordinates in inertial frames. Your x coordinate is not 'straight', and your primed system is not inertial, so you cannot use the Lorentz transform.

The easiest way to analyze such a problem is simply to pick one inertial frame for all the analysis. You can figure out what any observer sees from this one frame.

In addition, your problem has no real answer in principle. You are not asking about a duration, but what time is seen on a clock at an arbitrary moment. The answer depends on when and how initial setting and synchronization is done - which you don't specify. Further, for a non-inertial frame there is no uniquely preferred method of synchronization (as there is for inertial frames). Which one you choose will produce different answers.

Thank you Pallen for your welcome and your prompt reply!

I agree that the coordinate is not straight, but it can be made 'almost' straight by increasing the circle size, and for S large enough, the transformation will necessarily lead for t=0 to something like t'=-ax along the circle for some positive a, which is the main point and produces the paradox. The paradox can disappear only if a=0.

Regarding synchronization: each circle synchronizes his own clocks in the usual way (one can consider polygones to avoid infinitesimal calculus), and at x=0 and t=0 we let x'=0 and t'=0.

 

[bcrowell]

Regarding synchronization: each circle synchronizes his own clocks in the usual way (one can consider polygones to avoid infinitesimal calculus), and at x=0 and t=0 we let x'=0 and t'=0.

This is what the Gron paper proves is impossible.

-Ben

 

[chubert]

Hi, chubert,

Welcome to PF!

It sounds to me like you're groping your way toward an idea that is explained very clearly in this paper:

Gron, Relativistic description of a rotating disk, Am. J. Phys. 43 (1975) 869

The idea is that clocks can't be globally synchronized in a rotating frame of reference. Unfortunately Gron's paper isn't freely available online. I've given a shorter explanation of some of the same ideas here: http://www.lightandmatter.com/html_books/genrel/ch03/ch03.html#Section3.4 (subsection 3.4.4).

What Gron does is to take Lorentz transformations between neighboring points on the circle, which, if done carefully, avoids the problems PAllen pointed out with your approach.

-Ben

Thank you Ben, and nice feedback here!

OK, the problem lies then in the C' synchronization. I'm reading your section 3.4.4, thanks for the link.

 

[chubert]

This is what the Gron paper proves is impossible.

-Ben

You were too fast and posted while I was replying to your first post. Your lecture is very interesting and seems to respond to my question, although it will take me some time to assimilate :) Thanks again Ben.

 

[PAllen]

I agree that the coordinate is not straight, but it can be made 'almost' straight by increasing the circle size, and for S large enough, the transformation will necessarily lead for t=0 to something like t'=-ax along the circle for some positive a, which is the main point and produces the paradox. The paradox can disappear only if a=0.

Regarding synchronization: each circle synchronizes his own clocks in the usual way (one can consider polygones to avoid infinitesimal calculus), and at x=0 and t=0 we let x'=0 and t'=0.

The problem is that you can make nearby points on the circle effectively straight, but for a point half way around the circle, you cannot pretend you have a straight line.

As for synchronization, the issue is in your C' frame - as a whole, you cannot make it nearly inertial. In an inertial frame, any 'reasonable' way of actually synchronizing clocks produces the same result. In a non-inertial frame, different operational ways of doing it produce different results, and there is no clear way to prefer one over the other.

 

[chubert]

The problem is that you can make nearby points on the circle effectively straight, but for a point half way around the circle, you cannot pretend you have a straight line.

As for synchronization, the issue is in your C' frame - as a whole, you cannot make it nearly inertial. In an inertial frame, any 'reasonable' way of actually synchronizing clocks produces the same result. In a non-inertial frame, different operational ways of doing it produce different results, and there is no clear way to prefer one over the other.

Yes, it is clear for me now, I was naively believing that C' (which I now know from Ben as being the carousel) could be synchronized. I could have guessed that the paradox was a proof of this impossibility (which it is I believe) :). Thanks to both of you, I can now sleep well.