# Simultaneity in circular motion

1. Mar 2, 2014

### johnny_bohnny

Hey guys

based on the Einstein train thought experiment we can say that different inertial frames disagree on simultaneity. In that specific example, the train observer is moving towards one thunder, and away from the second and in his frame the thunder in front of him occurs first followed by the thunder on the rear. Now my question is how to make sense of this while doing circular motion, like we do it on the Earth. It is vague to say that we move towards something because we might as well say we are moving away from it, from a different perspective of the circle. Can anybody explain, or compare the train scenario with events on Earth, and in which sequence do space like events occur for moving frames here on Earth?

2. Mar 2, 2014

3. Mar 2, 2014

4. Mar 2, 2014

### johnny_bohnny

Yeah, the question is similar, but this one is more concrete, or should I say it needs a more concrete answer. When thinking about relative simultaneity I always consider the example with the thunders for better understanding so I was hoping for a concrete example with something like that in the scenario.

5. Mar 2, 2014

### johnny_bohnny

The basic point of this question is to find out on which way does simultaneity calculation vary between moving observers and observers at rest while doing this sort of circular motion.

6. Mar 2, 2014

### FactChecker

I got a beautifully convincing answer from 'pervect' a couple of weeks ago. It shows that Einstein synchronization does not work in a rotating circle. The center observes a drift out of synchronization around the circle till it comes full circle and has to be out of sync with itself. I don't know how to link to that post. So here is a copy of it.

7. Mar 2, 2014

### WannabeNewton

I still have no idea what you're asking for. Could you provide a more concrete problem for us to work with?

8. Mar 2, 2014

### johnny_bohnny

Okay, no problem.

From what pervect said last time around, every point on the surface of Earth has a different meaning of simultaneity from its point of view. What does that really mean? If we take a 'trip' around the equator and try to compare the simultaneity 'surfaces' of each point, how would they differ? And what about moving observers (or points)? What I'm seeking for is basically the comparison of this situation and how it effects considerations of simultaneity and the classical one that is the basis for Einstein's train thought experiment and so on.

9. Mar 2, 2014

### pervect

Staff Emeritus
I dont' know how to describe "what it means" any more clearly. To recap quickly:

Simultaneity in special relativity is observer dependent. When you ask "what does this really mean", I suspect that you may have lingering notions that simultaneity is "real" and henceforth independent of the observer. If you do have such notions, they will be incompatible with special relativity and cause you a great deal of confusion when trying to understand the theory.

The lines of simultaneity wrap around a cylinder, like the space-time diagram in http://en.wikipedia.org/wiki/File:Langevin_Frame_Cyl_Desynchronization.png, forming a non-closed helix.

I'm not sure what you mean about "comparing simultaneity surfaces" (other than the above diagram). If you take a trip around the equator at sea level, though, the total elapsed time on the clock taking the trip (the proper time) will depend on whether you travel clockwise or anti-clockwise. This is an experimental consequence of the synchronization issues, and it's easiest to see that this behavior is expected by carrying out the analysis in an inertial (non-rotating) frame.

This was tested on the Earth by the Hafele-Keating experiment, where the clocks were flown. Because the clock were flown and didn't maintain a constant altitude above sea level, some GR effects that are beyond the scope of the SR experiment occur, basically gravitational time dilation due to height.

10. Mar 3, 2014

### A.T.

That's why the Einstein synchronization convention doesn't work here, and you have to use a different convention, for example a signal from the center.

11. Mar 3, 2014

### chingel

Simultaneity can't be just assumed, you have to figure out a way how are you going to test it. If you have two clocks at different places, how do you know that they are simultaneous, ie show the same time at the same "time"?
One thing yoiu could do is go next to one of the clocks, synchronize your watch with it, then walk slowly to the other clock and sync it with your watch. Or you could send a light pulse between the clocks, and bounce it back. The first clock sends out a signal at its time 0, second one receives it at its time 1 and the reflection is received at time 2. Then the clocks are synced. This works fine in inertial frames.
In a rotating frame if you would do either of those with a lot of clocks around the circle, then the last clock and the first clock would show different times right next to each other. So the same kind of syncing doesn't have the same meaning anymore because it will still be out of sync in the end...

12. Mar 3, 2014

### WannabeNewton

The OP isn't asking whether or not the clocks on the equator can be Einstein synchronized with one another in order to build global Einstein simultaneity surfaces for the entire family of clocks. This is impossible and the OP already knows this. Rather the OP is asking what the simultaneity surfaces look like for each point on the equator when using Markze-Wheeler simultaneity which is defined by the averaging of round-trip light signals between local events and causally connected events relative to each point. These will look rather complicated. Note that if we do as A.T. says and instead synchronize to the central clock then the simultaneity surfaces of all points on the equator will be the exact same and will just be the simultaneity hyperplanes of the central clock. Both of these simultaneity conventions yield non-orthogonal simultaneity surfaces for all points on the equator.

To see what the Markze-Wheeler simultaneity surfaces look like for each point on the equator, see p.8 of the following: http://arxiv.org/pdf/gr-qc/0006095v2.pdf

13. Mar 5, 2014

### johnny_bohnny

Thank you very much, WannabeNewton. You got my point, but I'm also wondering what will the simultaneity surfaces look like for points that are in motion on a rotating body. In the classical setup of inertial frames to show relative simultaneity we have a stationary observer and a moving observer, I wonder what will it look like if the similar setup is applied on Earth, for example. Let's say one point that is at rest with Earth and another one that is moving in some direction. What determines their perspectives of simultaneity and how do they look like? Thanks in advance.

14. Mar 5, 2014

### WannabeNewton

It's in the link I provided. But please keep in mind that this is just one choice of simultaneity convention. When you talk about non-inertial observers the statement "what will the simultaneity surfaces look like" is completely ambiguous. There is no canonical choice of simultaneity convention for non-inertial observers. The link calculates the surfaces for the Marzke-Wheeler simultaneity relative to an observer in uniform circular motion, which is the generalization of Einstein simultaneity relative to an inertial observer, but this is not the only choice as explained in the same post the link was provided in.

15. Mar 5, 2014

### johnny_bohnny

I understand that, thanks for the help and the reference. I've quickly looked and I thought it was only the description of stationary observers relative to a rotating body, but it's great to know that it describes even the situation which I needed help with it. Now it's time to examine it.

16. Mar 5, 2014

### WannabeNewton

The simultaneity surfaces of the stationary observers are just the usual orthogonal hyperplanes. These observers are inertial. The calculations and images in the link refer specifically to the Marzke-Wheeler simultaneity surfaces of the observers in uniform circular orbit.

17. Apr 9, 2014

### analyst5

Sorry, for bringing this up after a few weeks, but how can the stationary observers be inertial since the Earth is not only undergoing rotation, but also non-inertial circular motion around the Sun? Just curios about it.

18. Apr 9, 2014

### WannabeNewton

We are considering the Earth as an isolated system fixed relative to the distant stars, uniformly rotating about an axis. The stationary observers in this case are the inertial observers fixed with respect to the distant stars.

19. Apr 9, 2014

### analyst5

So this is allowed in relativity? We may only 'focus' on the earth as a rotating, isolated system without considering its revolution around the sun. So it's true to say that those stationary observers are inertial in one sense, but non-inertial in another? And the relativistic effects can be determined wrt to those points in the context of them being inertial. I hope you can explain this to me because it's a new concept. Regards.

20. Apr 9, 2014

### Staff: Mentor

It's a matter of neglecting additional effects which are much smaller than the one under discussion. We do this all the time in physics. The earth's angular speed in its orbit around the sun is about 1/365 of the angular speed of a point on the earth's surface as it rotates around the earth's axis.