# Can clocks ticking at different rates be synchronized?

1. Jan 21, 2015

### smoothoperator

I was reading an article about synchronizing clocks on the rim of a rotating disc with a central, inertial clock. I was wondering how is this possible or have I misinterpreted something. Clocks on the rim tick slower than the central clock because they move relative to the center, so I don't understand how can the rim clocks agree with the central clock on notion of simultaneity? Long story short, even in non-inertial frames, do clocks ticking at different rates always disagree on simultaneity? That at least seems normal to me. If the time on the rim clocks ticks slower relative to the proper time of the central clock, I don't understand how can the rim clocks adopt the simultaneity where the rim clocks tick slower, that seems like a contradiction. Can anybody explain what is really meant by 'synchronizing rim clocks with the central clock'?

2. Jan 21, 2015

### Staff: Mentor

Simultaneity and "tick rate" of clocks are two different things; a pair of clocks can agree on which sets of events are simultaneous (happen at the same time), but not agree on how much time elapses between two events that are not simultaneous. Obviously any coordinate chart in which this is true will be a non-inertial chart.

Here's one way of doing it: the central clock emits light pulses at fixed intervals (say one second each). Each rim clock uses these light pulses to determine simultaneity, while still using its own tick rate to determine how much time elapses between successive light pulses (this time will be less than one second). Of course that means that each rim clock, at a given event on its worldline, is using a (non-inertial) coordinate chart which is not the same as the momentarily comoving inertial chart at that event.

3. Jan 21, 2015

### smoothoperator

I still don't get it. Shouldn't the different clock rate imply different simultaneity? If according to the centre clock clocks A and B are simultaneous and tick slower than the proper time rate of the centre clock, if we adopt the centre clock convention, at which rate would the centre clock tick wrt to clocks A and B?

4. Jan 21, 2015

### Staff: Mentor

No, because simultaneity is a convention, not a physical thing. It's simply a matter of which events the different observers, with their different clocks, agree to say happened "at the same time". That's not the same thing as saying how much time elapsed between two events which are not at the same time.

Here's a concrete example for the rotating disc. Suppose the observer O, at the center of the disc, is emitting light pulses in a particular direction--say towards a particular distant object such as a star. Say the radius of the disc is 10 feet, so it takes light 10 nanoseconds, according to the center clock, to go from the center of the disc to the edge. We also have a particular observer D, at the edge of the disc, who is passing the radial line along which the light pulses travel at particular instants of interest. This observer takes, let us say, 1 second to complete a revolution, according to observer O (i.e., by observer O's clock, the disc revolves once per second).

Now consider the following events:

A: Observer O emits light pulse #0.
A1: The event on observer D's worldline that is simultaneous with A, according to the center observer.

B: Light pulse #0 reaches the edge of the disc. At the same instant, observer D is passing the radial line along which the light is traveling, so light pulse #0 passes him at this instant.
B1: The event on observer O's worldline that is simultaneous with B, according to the center observer.

C: The center observer emits light pulse #1.
C1: The event on observer D's worldline that is simultaneous with C, according to the center observer.

D: Light pulse #1 reaches the edge of the disc. At the same instant, observer D has made one revolution and is again passing the same radial line, so light pulse #1 passes him at this instant.
D1: The event on observer O's worldline that is simultaneous with B, according to the center observer.

Now, observer D adopting observer O's simultaneity convention just means that observer D agrees to say that events A and A1, B and B1, C and C1, and D and D1, each happen at the same time. It does not mean that observer D has to agree that 10 nanoseconds elapse between A/A1 and B/B1, or between C/C1 and D/D1. Observer D can still say that, by his clock, less than 10 nanoseconds elapse between those pairs of events; i.e., his clock can still tick slower than observer O's clock. It also does not mean that observer D has to agree that 1 second elapses between B/B1 and D/D1; he can still say that, by his clock, less than 1 second elapses between those pairs of events (i.e., that by his clock, the disc takes less than 1 second to complete a revolution). All observer D has to agree to is which pairs of events are simultaneous.

Obviously, the coordinate chart observer D is implicitly using, if he agrees on simultaneity with observer O but uses a different tick rate for time, is not an inertial chart. But there's nothing that requires him to use an inertial chart. Nothing changes, physically, about anything in this scenario if he adopts the non-inertial chart being described. All that changes is how he labels events with coordinates.

Faster. See above.

5. Jan 21, 2015

### WannabeNewton

It simply means you readjust the rate of ticking of the rim clocks so that they read the time $t$ of the central clock instead of their proper time. We say that such clocks are then non-ideal; an ideal clock always measures proper time. Since we have reparametrization invariance of physical quantities obtained from the worldlines of clocks, this is simply a gauge transformation.

6. Jan 21, 2015

### Staff: Mentor

We should note a terminology issue here. The word "synchronization" can be used to mean what you describe--having an observer adopt both a simultaneity convention and a "tick rate" for time that are different from the "natural" simultaneity and tick rate for the momentarily comoving inertial frame at a given event on his worldline--but it does not have to mean both of those things. It can also just mean adopting the same simultaneity convention, without adopting the same tick rate; that's how I was implicitly using the term above. (I believe we had a discussion some time ago about "proper time synchronization" vs. just "synchronization" that emphasized this difference.)

7. Jan 21, 2015

### smoothoperator

I understand you argumentation in the case of the rotating disc, but this still sounds very very weird to me. This whole process of coordinate adaption. So for instance, in our, Earth frame that is slightly non-inertial we may adopt coordinates of a frame that is moving with the speed of 0.9c relative to us and say that's a valid process. I don't see objects get contracted and times dilated in everyday life so this seems pretty absurd. We're travelling at very low speeds and I don't understand how possibly can we adopt the coordinate system of something that travels at very high speeds.

8. Jan 21, 2015

### phinds

This is completely incorrect. You, right now as you read this, are traveling at .9999999 c, NOT some "low speed", in one FoR. You are also traveling at a low speed. You are also not moving at all. It just depends on what FoR you choose and to say that one FoR is preferred over another is incorrect.

9. Jan 21, 2015

### Staff: Mentor

Sure, we could do this; our description of events happening in our vicinity in terms of those coordinates would look really weird, as you say, but it would still be perfectly consistent.

In fact, we routinely do something else here on Earth that amounts to adopting a simultaneity convention other than the "natural" one for our state of motion. The Earth is rotating; that means that, for example, if I'm sitting in New York and a friend of mine is sitting Australia, we are moving relative to each other at roughly a kilometer per second. This is pretty slow compared to the speed of light but atomic clocks can easily detect the difference in clock rates that each of us would see in the other's clock, and the difference in "natural" simultaneity conventions between our two momentarily comoving inertial frames. However, all of us ignore this in our daily lives and we all use a common time standard that includes a common simultaneity convention, which is basically the "natural" simultaneity convention of a non-rotating inertial frame centered on the Earth.

(We also all adopt the same "tick rate" for our clocks, which basically amounts to the tick rate of a clock at rest at sea level on the rotating Earth--not that this is not the same as the "natural" tick rate of an inertial clock at the center of the Earth, so our commonly used time standard already has a simultaneity convention and a tick rate that don't "match up" as they would for an inertial frame.)

If we were routinely moving at speeds close to the speed of light relative to one another, our common scheme might not work as well; or we might just get used to the fact that our common standard of simultaneity means our description of some events looks weird from the standpoint of our momentarily comoving inertial frame. But either way, it is all perfectly consistent.

10. Jan 21, 2015

### phinds

Peter, you lost me on that one. What am I missing? Seems to me that you maintain exactly the same spatial relationship to each other and are not moving relative to each other.

11. Jan 21, 2015

### Staff: Mentor

Remember that "moving" is frame-dependent. The motion I'm referring to is relative to an inertial frame. In the non-inertial rotating frame of the Earth, yes, the two observers are both at rest; but smoothoperator appears to be viewing inertial frames and their rules for simultaneity as being somehow "privileged", so I'm illustrating how we don't in fact use those rules in everyday life.

12. Jan 21, 2015

### phinds

Got it. Thanks.

13. Jan 21, 2015

### smoothoperator

Could you explain these weird consequences to me in short terms? Would the Earth get length contracted or would the object that is travelling at 0.9c get contracted, according to our frame and adopted coordinates?

14. Jan 21, 2015

### phinds

SmoothOperator, one thing I should have added to my previous post is that of course we don't "experience" time dilation. No one ever does. It is not a physical thing, it is a result of remote observations. Were it otherwise, then we would all be experiencing an infinite number of different time dilations simultaneously, which obviously doesn't make sense.

15. Jan 21, 2015

### smoothoperator

I know that, that's one of the basic laws of relativity, that our clock and proper time tick at a standard rate and others differently relative to us depending on their state of motion. Thank you anyway for you reply.

16. Jan 21, 2015

### phinds

OK, good. Just wanted to make sure since my interpretation of your earlier post led me to believe that you were a bit shaky on frames of reference. Sorry if I was wrong.

17. Jan 21, 2015

### smoothoperator

No prob, in fact I am shaky, but not because of your mentioned reasons but because the fact that we can adopt anything in a non-inertial frame seems absurd to me. In fact, it seems that I may be 90 percent length contracted from the Earth frame just because of different coordinate adoptions. Confusing.

18. Jan 21, 2015

### phinds

Well, it's not something you experience, it's just how you look to someone else and how you look to them depends on the FoR, so you can look an infinite number of different ways to different observers, but you remain unchanged in your FoR.

19. Jan 22, 2015

### Ibix

A frame of reference is just a choice of coordinates. Imagine you draw a graph of $y=x^2$ on a piece of graph paper, and so does the person next to you. Imagine extending your graph paper out to cover your neighbour's graph. In your coordinate system, hers can be written with a rotation and a translation:
$$\begin{eqnarray} x'&=&(x-x_0)\cos\theta-(y-y_0)\sin\theta\\ y'&=&(y-y_0)\cos\theta+(x-x_0)\sin\theta \end{eqnarray}$$
That means that she's drawn a graph of$$(y-y_0)\cos\theta+(x-x_0)\sin\theta=((x-x_0)\cos\theta-(y-y_0)\sin\theta)^2$$
Wow.

There's nothing wrong with that formula. It changes nothing about the physical reality of the graph. But your neighbour would be daft if she adopted your coordinates, your frame of reference, because the maths is so much more complex. But she could do.

Similarly, we could adopt a frame of reference moving at .9c relative to Earth. We'd be idiots, much like your nelghbour using your coordinates, but we wouldn't change anything physical.

20. Jan 22, 2015

### smoothoperator

But we on Earth perceive objects and their spatial dimensions at something close like their proper dimensions, right? By perception I mean 'seeing'. Why don't we see them contracted if we adopt a convention where they are very contracted?