# Umsynchronization of clocks.

1. Oct 24, 2008

### matheinste

Hello all.

If two clocks are at rest relative to each other and are synchronized using the usual Einstein synch procedure, under what conditions will they appear to be unnsynchronized to another observer who has always been at rest relative to them? Also, if an observer previously moving relative to the synchronized clocks ( and so seeing them out of synch with each other ) accelerates ( decelerates ) to become at rest relative to them, will these same two clocks then appear to be in synch with each other to the previously "moving" observer?

My answers would be that i am unsure of the first part because accelerations would be involved to make any changes, and yes for the second part.

Matheinste.

2. Oct 24, 2008

### Fredrik

Staff Emeritus
Were they synchronized by someone who's at rest relative to them? In that case, "any". Otherwise "none".

Yes, assuming that he starts the experiment that attempts to verify that they are in synch after he stops, and finishes it before he starts moving again.

3. Oct 24, 2008

### matheinste

Hello Fredrik.

Yes i should have said that the clocks were synchronized by someone at rest relative to them. I suppose this person could also act as the observer at rest relative to them.

Am i misreading something because you say that in this case the clocks will appear to be UNsynchronized to the observer under any conditions.

The second part is clear.

If the second situation is reversed and the clocks accelerate ( decelerate ) to become at rest relative to an observer who was previously moving relative to them ( and so were out of synch with each other according to him ), will they now appear to him to be in synch with each other. Or does the fact that the sychronized clocks have undergone acceleration affect them.

Matheinste.

4. Oct 24, 2008

### Fredrik

Staff Emeritus
Oops, I got the alternatives confused somehow. They will always be synchronized in an inertial frame that has the same velocity as the clocks.

If instead they were synchronized by someone who wasn't at rest with respect to the clocks, they will never be synchronized in an inertial frame that has the same velocity as the clocks, unless the velocity (in the clocks' rest frame) of the guy who synchronized them was perpendicular to the line connecting them.

What a clock measures is the integral of $\sqrt{-dt^2+dx^2}$ along the curve in spacetime that represents its motion, and if I understand your description correctly, the world lines of the two clocks are identical in the observer's rest frame (except for starting position), so the difference between the times they are showing at simultaneous events will not have changed.

5. Oct 24, 2008

### matheinste

Thanks Fredrik.

Thats just what i thought.

Matheinste.

6. Oct 28, 2008

### matheinste

Hello all.

Just to clear up a specific point which may well have ben covered by previous answers in this thread. I have searched several books but cannot find an answer to this specific point.

Take two clocks in the same inertial reference frame synchronized to each other by an observer in the same frame using the Einstein synch method. If the clocks, along with the obserever are accelerated/decelerated and return to the previous, or another state of inertial motion, will the observer consider them to still be synchronized.

The answers so far in this trhread suggest that they will still be in synch, but i have a feeling that somewhere else i have seen a different answer.

Matheinste.

7. Oct 28, 2008

### Fredrik

Staff Emeritus
Assuming that the world lines of the two clocks and the world line of this observer guy all look the same in the original rest frame (except for their starting position), the answer is definitely yes. This is an immediate consequence of the definition of proper time and the postulate that clocks measure proper time.

You may have read about the scenario where the clocks are attached to opposite ends of a rigid rod. (It's been discussed in this forum a few times). The rod will be getting shorter (Lorentz contracted) when its speed increases (in the original rest frame). So the rear must be accelerating faster than the front, and later it must be decelerating faster than the front. In this case, the world lines will not be identical.

They might still have the same proper time though, e.g. if the deceleration profile is the "opposite" of the acceleration profile, so that the first half of the world line of the rear is a mirror image of the second half of the world line of the front and vice versa. But they may not have the same proper time in general. (I'm too lazy to check).

8. Oct 28, 2008

### matheinste

Hello Frerik.

The attachment to a rigid rod scenario rings a bell. I think thats where i have seen the clocks as being out of synch before, but the case of clocks at either end of an accelerating space ship. I think that answers my question prety well. Thanks again.

Matheinste

9. Nov 11, 2008

### Austin0

Maybe I am unclear on what you mean here. Wouldn't the acceleration and deceleration
automatically be reciprocal no matter what the route taken to return to the original frame? How could you get back to the original velocity without exactly equal deceleration.?

My understanding of the asynchronicity effect has been that it is not due to action on the clocks themselves but is associated with the location on the moving frame. Is this incorrect???
Ie: if you move a clock while in motion from the back of the frame to the front wouldn't it then be in synch with a clock that had been there from the beginning???? And vice versa.

10. Nov 11, 2008

### Fredrik

Staff Emeritus
In order to get back to the original velocity, the integral of the acceleration as a function of time over the time interval involved must be zero. If a solid rod that's initially at rest is accelerated for a while and ends up at rest again after a while, then the accelerations of both endpoints will satisfy that condition. I'm just saying that I haven't tried to prove or disprove that this implies that the integral of $\sqrt{dt^2-dx^2}$ along the world line of the front is the same as the integral of $\sqrt{dt^2-dx^2}$ along the world line of the rear.

As I've been saying, what a clock displays is the integral of $\sqrt{dt^2-dx^2}$ along the curve in Minkowski space that represents its motion (i.e. its "world line"). In the scenario you're describing, the two clocks have very different world lines. dx is =0 everywhere on one of them, but not the other, so the accelerating clock measures a shorter time.