# Frame changing and Clock Sync

1. Mar 26, 2012

### mananvpanchal

Suppose, a long object is at rest in S frame. There is A (On the long object) and B (Near the long object, on the ground) observer in the middle of the long object. There are two clocks located on both end of long object. We suppose, both A and B is at origin of S frame and clocks at some -x and +x location. The clocks is synchronized for both observer.

Now, at t=0 the object starts moving to right with respect to B observer. Now the long object is at rest in S' frame and moving relative to S frame.

We can see from the diagram that clocks is still synchronized for A observer, but no longer synchronized for B observer. The time reading changed instantly for B. Time decrease in left clock and increase in right clock for B. After the frame change, the difference between the time readings of clocks remains constant for B. We can see that both clocks is far from B in this case.

Now, suppose a situation where observer A, B and left clock are on origin of S. Now A and B is at left end of long object. We can see that left clock reading is not changed instantly for B on frame change, but right clock's reading increases for B. We can see that left clock is near to B in this case.

Now, suppose a situation where observer A, B and right clock are on origin of S. Now A and B is at right end of long object. We can see that right clock reading is not changed instantly for B on frame change, but left clock's reading decreases for B. We can see that right clock is near to B in this case.

In all the case clock reading changed instantly when the clock is not near to the B observer. I think the distance of observer from clocks is the cause of instant change in readings, and direction of motion of clocks is the cause of in which way the clocks desyncs, up or down.

Please, feel free to point out any error if have made here.

Last edited: Mar 26, 2012
2. Mar 26, 2012

### Mentz114

I'm not sure what point you are making, but instantaneous change in velocity is not possible, requiring an infinite force to be applied for zero time. Clearly not possible.

3. Mar 28, 2012

### mananvpanchal

Yes, instantaneous change in frame is not possible if we describe this with very high speed. If we talk about very low speed then we can describe the diagrams. After all moving condition from steady condition is surely not a smooth transition. We can differentiate the two condition even the speed is very low.

4. Mar 28, 2012

### Mentz114

I suppose an infinitesimal change could be thought of as instantaneous, but for observable effects one must integrate these small changes.

I'm sorry but I still don't know what point you are making about the clock synching.

5. Mar 28, 2012

### mananvpanchal

Yes, we have to integrate small changes of frames to observe the effect. But, if we talk about small speed (like firing the object from a gun), we can change the frame. Though the effect would not be observable, but still there is theoretical desync would be occured for B.
Sorry, I cannot get where is your confusion lies.
If A changes his frame instantly the clocks still would be synchronized for A, but wouldn't be synchronized for B.

Last edited: Mar 28, 2012
6. Mar 28, 2012

### harrylin

I 'm not sure about what you were saying (sorry), but it sounds as if you misunderstand it, as if you interpret it the wrong way round. No clock reading changes instantly from for example 0:00 to 1:00.

The point is that after acceleration, if one checks the clocks in accordance with the synchronization convention, the clocks will be found to be out of synch according to the definition of this "moving" frame.
Does that help?

7. Mar 28, 2012

### Mentz114

Not for long, I suspect

Regarding the synchronisation, I'm not surprised that spatial separation plays a part in desynching.

8. Mar 28, 2012

### darkhorror

Are you just wanting to know if your thinking is correct, and that the clocks won't be in sync if frames of reference are changed?

9. Mar 28, 2012

### pervect

Staff Emeritus
I have the feeling (perhaps I'm incorrect), that you are envisioning the clock A, and the (unnamed?) clock, which I will call clock "X", that's at the other end of "the long object" from A as being two clocks on a single long and rigid object.

You didn't specify that the object was 'rigid', so perhaps I'm misinterpreting you.

Of course, there is no such thing as a rigid object in special relativity. If you push on one end of a steel rod, the other end will start to move only after a time delay, the time being based on the speed of sound in the steel. You can look it up - it's a lot slower than the speed of light.

I find it less confusing to avoid mentioning extended objects at all, and to talk about two rocketships, A, and X. Both of them are initially at rest, sharing a frame together, and both of them start accelerating really hard when they receive a light signal emitted from their midpoint (the midpoint in their shared initial frame). They accelerate so hard that they burn through their fuel in a very small amount of proper time - and they have a certain delta-v capability, so when they both stop accelerating, they have the same velocity,.

Taking this line of approach, you'll see that the proper distance between A and X changes in this scenario - it's a variant of what's commonly called the "Bell Spaceship Paradox".

Other than possible confusions arising about the notion of "objects", which may or may not be being presumed to be rigid, I don't see anything wrong that I noticed with the space-time diagrams.

If you think of A and X as separate rocketships, it's correct to note that they start accelerating at the "same time" only in the initial frame, and that in the final frame (the frame shared by both after both rocketships have burned all their fuel), they did NOT start accelerating at the same time.

10. Mar 28, 2012

### yuiop

The bold part is not correct. When A and the long rod accelerates the clocks at rest with the rod do not remain synchronised from the point of view of A (or B), if the rod maintains its proper length. The diagrams you have drawn are for the situation where A and B have had constant relative motion forever (no acceleration) and have synchronised their respective clocks in their own rest frames. Each time a system accelerates to a new relative velocity, the clocks have to be resynchronised. Clocks do not automatically jump into the future at the front and into the past at the rear. When a system accelerates while maintain proper length (Born rigid motion) the clocks at the rear naturally advance more slowly than the clocks at the front, because length contraction requires the clocks at the back to accelerate more rapidly than those at the front. No clocks go backwards in time even though a naive interpretation of spacetime diagrams might give that impression.

You need to take care when talking talking about what B (or any observer) sees about events that are not local. Far away events are subject to light travel times and the times of those events require synchronised clocks that are not present in an accelerating system, because they go out of sync during the acceleration and require finite time to resynchronise the clocks and this resynchronisation time is longer than the time intervals under consideration in an accelerating system which is effectively a moving target. It is better to talk about times in a given reference frame which is effectively a matrix of observers that are individually local to all events rather than talk about the observations of a single observer in isolation. Most formulas in relativity such as the Lorentz transformations assume this matrix of an infinite number of local synchronised clocks and so do not involve light travel times.

As above they do not desync down if by "down" you mean backwards in time. Some clocks advance at a slower rate than others during acceleration but never backwards in time. They only go backwards when somebody resynchronises the clocks manually and winds the clock hands backwards. There is nothing magical going on here.

11. Mar 29, 2012

### mananvpanchal

There may be two possibilities the diagrams is wrong or I have interpreted it wrong way. Please, tell me what might be the problem with my interpretation.
Which one? A or B? A is on long object and B is on ground near the object. Now B changes its frame. So what does A see? And what does B see?

12. Mar 29, 2012

### mananvpanchal

Yes, desync is not a surprise. But I am surprised with left clock's decreased reading and right clock's increased reading.

13. Mar 29, 2012

### mananvpanchal

Yes.

14. Mar 29, 2012

### harrylin

See also the elaboration by yuiop. I did not carefully look at your diagrams, and I told you what might be the problem but I'll elaborate.
The point is that after acceleration, if one checks the clocks in accordance with the synchronization convention, the clocks will be found to be out of synch according to the definition of A's new rest frame (due to the object's length contraction also very slightly out of synch according to B's rest frame, but that's not the issue here).
Likely each only sees a blur. :tongue: However, they use measurements to draw conclusions. A will simply see and measure that B is moving, as B is not on a long object; so I don't know what you mean with that. And B will measure that the clocks of A are out of synch, if B adapts its measurements to that of a "moving" frame. Except if the motion is perpendicular, for then A's clock times will be measured as equally shifted.

Of course you could ask an infinite number of variant questions like that, so it's not useful to ask more - you will be able to reply them all once you understand these two.

15. Mar 29, 2012

### Staff: Mentor

Hi Mentz114, just looking at this thread. Although you are correct that it isn't possible, it is a reasonable approximation. Suppose we were talking about a journey of several light years and an acceleration to .6 c in a matter of a few minutes. The approximation would be reasonable as long as we just keep in the back of our head that it is an approximation and we don't look too carefully at those few minutes of acceleration.

16. Mar 29, 2012

### Staff: Mentor

This is not true, in general. It depends on the details of the acceleration of the clocks and A. In fact, under the acceleration profile as described in the OP the opposite is true.

Btw, there is never a gap in a worldline, and clock readings never change instantaneously in any frame. Your diagrams are wrong.

Last edited: Mar 29, 2012
17. Mar 29, 2012

### mananvpanchal

Actually, here the A and B is observer's name.

Yes, we can generate the scenario of three rockets. There are three rockets left, middle and right. Left and right rocket has one one clock. A observer in middle rocket. And B observer on ground near middle rocket. Now both clocks is synchronized for both observers. Now rocket starts moving (ignore gravitation force) with constant speed (I know high speed from rest state is not possible, so we can guess it as less speed). Now all three rockets in S' frame. The diagram shows that both clock is still synchronized for A, but not for B.

Actully, here I don't want very high speed, and I don't want to accelerate for much time. I need here less speed with instant change.

18. Mar 29, 2012

### mananvpanchal

Yes, if we see diagram after t=0, it will look like B has came from far with his synchronized clocks. I have combined two diagram, in one A is at rest in S frame and in second A is moving relative to S frame.
So, can you explain me in which pattern the clocks become desync for A and B (with scenario of constant acceleration and with scenario of instant frame change)?
That is why I need your thoughts.

19. Mar 29, 2012

### Staff: Mentor

mananvpanchal, I already explicitly derived the synchronization for this exact scenario for you (assuming an instantaneous acceleration from 0 to .6c) here starting with post 42: