# Rotating frames desynchronization

1. Feb 16, 2014

### johnny_bohnny

I was reading a little about the behavior of non-inertial frames in relativity, since I'm interested in knowing how can we measure time on Earth and the sequence of events here. So as we know earth is rotating and therefore the clocks on surface that are mutually at rest get desynchronized. Can somebody explain this to me better, because I don't understand it. I've always thought that frames that are mutually at rest will agree on simultaneity, but it isn't the case. How do we describe the differences and by which criteria regarding clocks that are at rest with respect to Earth's surface. What are the criteria?

It's very confusing for me to understand this so I hope somebody can simplify it. Regards, johnny

2. Feb 16, 2014

### A.T.

At the same distance from the axis they have the same rate in a rotating frame. But this gets superimposed with the gravitational time dilation of the Earth.

Only inertial frames.

3. Feb 16, 2014

### WannabeNewton

You don't need to consider rotating frames for that. Even in linearly accelerating frames there will be clock desynchronization of spatially separated ideal clocks situated along the line of acceleration. The difference between rotating frames and linearly accelerating frames is the latter allows global Einstein synchronization in principle if you're willing to use non-ideal clocks whereas for the former there is a fundamental obstruction preventing the establishment of global Einstein synchronization even for non-ideal clocks. More generally, global Einstein synchronization cannot be achieved in any rotating frame because the rotation breaks transitivity of the synchronization. The reason for this is a famous rotating frame phenomena known as the Sagnac effect: http://en.wikipedia.org/wiki/Sagnac_effect

4. Feb 16, 2014

### WannabeNewton

By the way we had a related thread some while back on clock synchronization in rotating frames that might prove helpful to you: https://www.physicsforums.com/showthread.php?t=732892

I can go into more of the mathematical details regarding the impossibility of global Einstein synchronization in rotating frames but I know not of your mathematical background so if you want the explicit calculations just say the word :)

5. Feb 16, 2014

### johnny_bohnny

I would prefer a concrete example to clarify this conceptual mess in my head. So clocks on earth that are at rest, when we consider them as the frames of reference, disagree on simultaneity. I get this, but what is the criteria for this. All clocks on the line of rotation have different perspectives on simultaneity? How does their perspective differ? There are many questions in my head and I doubt maths would help it since I'm not an excellent mathematician like most of you guys. Can you give me an example that is based with some clocks on earth, or something like that?

6. Feb 16, 2014

### johnny_bohnny

Can somebody compare the disagreement between frames with the situation regarding the agreement of the frames mutually at rest in the inertial scenario?

7. Feb 16, 2014

### DrGreg

If you had lots of clocks all around the Equator, at rest relative to the Earth's surface, and used Einstein synchronisation to sync the 2nd clock to the 1st clock, then the 3rd to the 2nd, then the 4th to the 3rd, and so on all round the Equator until you got back where you started, you would find that the last clock and the 1st clock, which are side-by-side, would be out of sync by about 200 nanoseconds.$$\frac{ \left( \frac{40 \times 10^6}{24 \times 60 \times 60} \right) \times \left( 40 \times 10^6 \right) } {\left( 3 \times 10^8 \right) ^2} \approx 2 \times 10^{-7}$$For an object rotating much faster than the Earth, the effect would be greater.

8. Feb 16, 2014

### WannabeNewton

For now forget about clock synchronization in relation to Earth bound orbits because this also involves gravitational effects which will necessarily complicate the matter. Let's consider something simpler.

Imagine we have a circular ring in free space rotating with some constant angular velocity about its symmetry axis relative to an inertial observer at the center of the ring. At each point on the ring we've placed an ideal clock, a concave mirror, and a radar set; we want to try and synchronize all these clocks with one another using Einstein synchronization. One way to go about this is using radar signals. Say we take some clock $A_1$ on the ring. From $A_1$ a light signal is emitted counter-clockwise towards an infinitesimally neighboring clock $A_2$ on the ring whereupon the light signal is reflected back to $A_1$-the placement of concave mirrors alongside each clock guarantees that this light signal will circulate along the ring. We then synchronize $A_1$ and $A_2$ using the Einstein synchronization formula i.e. we define the event at which the light signal reaches $A_2$ to be simultaneous with the event in the vicinity of $A_1$ that lies halfway in $A_1$-time between the round trip time of the light signal. Now a light signal is emitted by $A_2$ counter-clockwise towards an infinitesimally neighboring clock $A_3$ on the ring and we synchronize $A_2$ and $A_3$ using the same operational definition as above. We then repeat this process for each consecutive infinitesimally neighboring clock on the ring until we complete a full circuit around the entire ring and come back to $A_1$.

But what we find upon coming back to $A_1$ is that there is a gap $\Delta t_{\text{desynch}}$ between the original time $t_0$ that $A_1$ read and the time $t_f$ that it reads after performing the above synchronization full circuit around the entire ring starting from $A_1$ and ending at $A_1$. What this means is the above clock synchronization procedure fails to be transitive i.e. if $A_1$ is Einstein synchronized with $A_2$ and $A_2$ is Einstein synchronized with $A_3$ then $A_1$ will not be Einstein synchronized with $A_3$.

Compare this with what happens to ideal clocks at rest in an inertial frame. In such a case if $A_1$ is Einstein synchronized with $A_2$ and $A_2$ is Einstein synchronized with $A_3$ then $A_1$ will be Einstein synchronized with $A_3$ i.e. the procedure is transitive. Because it is transitive, all the clocks at rest in the inertial frame will agree on simultaneity and we can build a global time coordinate $t$ such that the surfaces $t = \text{const}$ correspond to global simultaneity surfaces shared by all of the clocks at rest in this inertial frame.

However in the case of the clocks at rest on the rotating ring, the synchronization procedure outline above is as already mentioned not transitive. Therefore distant clocks on the ring will not agree on simultaneity and we cannot build a global synchronous time $t$ such that the $t = \text{const}$ surfaces represent global simultaneity surfaces shared by all of the clocks on the ring.

In the case of the rotating ring in free space, the reason for the discontinuity $\Delta t_{\text{desynch}}$ that arises after attempting Einstein synchronization in full circuit around the entire ring is the already mentioned Sagnac effect. In our case this is very easy to understand. Imagine you're sitting at some point on the rotating ring and you place two clocks, one separated from you in the clockwise direction and one separated from you in the counterclockwise direction, such that both are equidistant from you. We still have concave mirrors setup at each point on the ring. You set the hands of the two clocks to the same position and temporarily lock the hands in place. Then you emit a light signal in the prograde direction and a light signal in the retrograde direction and have the clocks start ticking when the respective light signals reach them. Clearly the clock separated from you in the clockwise direction starts ticking before the clock separated from you in the counterclockwise direction starts ticking because for the former the light signal in the retrograde direction catches up to an approaching target whereas for the latter the light signal in the prograde direction catches up to a receding target. This means however that clock synchronization on the ring is path dependent and this is exactly why we get the discontinuity $\Delta t_{\text{desynch}}$ after attempting the above synchronization procedure starting and ending at the same clock on the ring. Notice that this wouldn't happen if the ring was non-rotating i.e. if the clocks at rest on the ring were also at rest in the inertial frame of the observer at the center of the ring.

Aha, DrGreg beat me to it!

9. Feb 16, 2014

### johnny_bohnny

10. Feb 16, 2014

### bcrowell

Staff Emeritus
11. Feb 17, 2014

### A.T.

Note that there are different issues here:

- Clocks at rest in a rotating frame which have different distances to the axis, cannot be synchronized at all, because they run at different rates.

- Clocks at rest in a rotating frame which have the same distance to the axis, cannot be synchronized using a certain convention (Einstein synchronization). But they run at the same rate, so they can be synchronized by other means like sending a signal from the center.

12. Feb 17, 2014

### bcrowell

Staff Emeritus
They can be synchronized with the signal, but then they're not synchronized with each other according to Einstein synchronization. The basic issue is that Einstein synchronization isn't transitive in a rotating frame.

13. Feb 17, 2014

### WannabeNewton

That won't achieve what the OP wants, as Ben already pointed out. Say we have a rotating disk, ideal clocks laid out along the rim of the disk, concave mirrors laid out along the rim of the disk, and a master clock at the center of the disk. Certainly if we have the master clock send out a spherical pulse of light at some instant that travels outwards, gets reflected back when it reaches the rim of the disk, and arrives back at the master clock individually in the form of light rays then we can set each clock on the rim so that it reads the time simultaneous with the time read by the master clock halfway between the round trip time of the radar signal to and fro each clock. Because of the axial symmetry, all the clocks on the rim will be set to read the same time and hence will be synchronized according to the master clock and the global inertial frame that the master clock is at rest in.

But the key point is the clocks on the rim will only be synchronized according to the inertial frame of the master clock. In other words on each $t = \text{const}$ simultaneity surface of the master clock, all the clocks on the rim of the rotating disk will read the same time-this won't be $t$ obviously since the clocks are not at rest relative to the master clock but the clocks will all read the same time as one another on each $t$ slice regardless. But this is not the global rest frame of the clocks on the rim. If we instead go to the global rest frame of these clocks, and we can appropriately talk about the global rest frame of the clocks because they form a Born rigid time-like congruence, then there is no way to synchronize all the clocks with one another using Einstein synchronization as pointed out by Ben. This is of course because Einstein synchronization relies on the isotropy of light which doesn't hold globally in the rest frame of the clocks-it of course holds locally even in the rest frame hence why we can always Einstein synchronize neighboring clocks on the rim with one another. Anyways, so what this means is we cannot use Einstein synchronization to build a global synchronous time $t$ whose simultaneity surfaces are adapted to the rest frame of the clocks on the rim, but this is in fact what the OP was asking for. If we modify the definition of Einstein synchronization in the global rest frame of the clocks by giving up the isotropy of light inherent in this synchronization procedure, which we know doesn't exist globally in the rest frame of the clocks, then we can synchronize them.

For a more mathematical treatment see here: http://arxiv.org/pdf/gr-qc/0204063v2.pdf

14. Feb 17, 2014

### A.T.

The clock at the center is not at rest in the rest frame of the clocks at the rim?

15. Feb 17, 2014

### WannabeNewton

Sure, it certainly is, but in the inertial rest frame of the master clock, which is the clock at the center, the clocks on the rim are circling around it, which is what I was referring to or at least meant to in case it came off as ambiguous. Regardless, the issue elucidated above remains.

Last edited: Feb 17, 2014
16. Feb 18, 2014

### A.T.

Obviously not, because the OP asks about the entire surface of the Earth, which cannot be synchronized. But if we restrict ourselves to say the equator, the clocks can be synchronized by a signal from the center, and all the observers on the equator will agree about the sequence of events on the equator.

17. Feb 18, 2014

### WannabeNewton

This still doesn't achieve what the OP wants even if restricted to the equator and Ben and I already explained why so no need to repeat.

18. Feb 18, 2014

### A.T.

The OP is asking about clock de-synchronization and agreeing on the order of events. If we restrict ourselves to the equator and synchronize the clocks by a signal from the center, then they will not de-synchronize and all the observers on the equator can agree about the sequence of events on the equator.

19. Feb 18, 2014

### WannabeNewton

What you're describing is equivalent to the Universal Time system of synchronization. However the OP explicitly stated "I've always thought that frames that are mutually at rest will agree on simultaneity, but it isn't the case." Synchronizing the clocks on the equator according to a master clock at the center won't change this because said synchronization procedure just makes the clocks agree on simultaneity of events relative to the master clock, not on simultaneity of events relative to their own rest frames ergo why we can't define global simultaneity in the extended rest frame of the clocks mutually at rest.

20. Feb 18, 2014

### A.T.

What disagreement about simultaneity of events on the rim would the observers on the rim have, if their clocks were synchronized by a signal from the center? Can you give an example?

If there are clocks placed everywhere around the rim, and every event on the rim gets a t-coordinate according the local clock, then I don't see how there could be a disagreement about the order of events.