# From rest to relative motion

1. Feb 15, 2009

### Saw

Two MM-devices A and B (=two perpendicular arms x and y being the paths for light signals) are at rest in the same frame. Both arms are of equal length, say, 1 light-second long. B is accelerated away along the positive x axis and reaches relative velocity, say, 0.5 c. I asked: will B have to move the milestone marking 1 light-second to another location, according to SR? The answer was negative. B's milestone is still where it was before, in B frame B ship is still measured as 1 light-second long, although in A frame it starts being measured as 1/gamma = 0.86654 light-second long. Reciprocally, in B frame A ship starts being measured as 0.86654 light-second long.

I have difficulties in understanding why. Why should A change its view on the length of B ship? I only see two possible explanations:

(i) the "Lorentzian" view: B's length has physically changed or
(ii) what we could call a "conventional" approach: B has "recalibrated" its measurement instrument (the MM device) in the x arm so that the two light signals meet after their respective round trips.

In the absence of any of these "physical" changes (one automatic and given by nature, the other conventional and carried out by a human hand), I insist: why should A start measuring a change in the length of B?

The same applies to synchronisation of clocks. When they were at rest with each other, A and B synchronised the clocks located at both ends of their ships using the Einstein convention. After some time, all their A and B clocks mark 10 s. At this instant, B is accelerated away along the positive X axis. I would also expect that B's clock located at the right edge becomes desynchronised in B frame, unless "something" happens: nature's or human intervention...

2. Feb 15, 2009

### tiny-tim

Hi Saw!

But the MM apparatus relies on the two arms staying together …

you have one of the arms disappearing at c/2.

And if the other arm goes with B also, then the stationary observer sees the light along that arm follow a zigzag path, while the light along the 'B' arm doesn't bother to return to where it started.

3. Feb 15, 2009

### Al68

The fact that something is longer as measured from one reference frame than another doesn't mean that "something" happened to it, it just means that length is frame dependent. Physics has had frame dependent quantities since Newton. For example, kinetic energy is frame dependent. The fact that an object's kinetic energy will increase if I increase my relative velocity doesn't mean that "something" happened to the object. Ditto for clocks.

Al

4. Feb 15, 2009

### Saw

Thanks, tiny-tim. I wasn’t precise enough. Of course, I assume that the two arms of each apparatus stay together: A’s y and x arms stay in A frame and B’s x and y arms are accelerated, thus becoming B frame.

AI68, thanks, too. I see that is the point. It is just that I find it hard to understand. That is why my mind looked for the “conventional” explanation. I do not mean I am right in not seeing it. Just that I do not see it, so far…

Anyhow, forget about my above speculations.

So I assume that, if two MM apparatuses A and B are at rest with each other and working properly (light signals sent in both directions along arms of equal length return to the origins at the same time) and B is accelerated away, B will not have to recalibrate the length of any arm and the apparatus will keep working as originally. Right?

Does it mean that, if B had an array of clocks along, say, the x arm, which clocks had been synchronized when B was at rest with A, B does not have to resynchronize them after the acceleration? Is it so? I had thought that at least this “recalibration” would have to be done.

5. Feb 24, 2009

### Saw

I would like to retake this thread, in case someone can help, because the doubt still bothers me.

Let us forget about the rest and concentrate on this issue: you are on a ship and have synchronized distant clocks of your frame with yours; then you are accelerated; don't you have to resynchronize your distant clocks by means of light signals, following the Einstein convention?

Take for example a twin paradox scenario. In the standard solution, it is said that, for the twin who left the earth and returned, there are two frames, one for the outward trip and another for the return trip. These two frames have different simultaneity lines. This change of simultaneity lines..., doesn't it require a new synchronization?

For example:

Relative v between ships A and B is 0.5 c. For A, B moves left-to-right; for B, A moves right-to-left. A1 and B1 (observers at the left edges of the ships) synchronize their clocks to zero when they meet and synchronize clocks in their respective frames through light signals.
In A frame, A’s ship is 1.732 ls long, B’s ship is half of that, 0.866 ls long.
In B frame, B’s ship is 1 ls long, A’s ship is 1.5 ls long.
When B1 reaches Am (mid-point of A), B ship turns around.
At that event, Am’s clock reads 1.732 s, while B1’s clock reads 1.5 s. No discrepancy with regard to this.
What about clocks distant to that event, like B2 or A1?
There is discrepancy:
In A frame, at the time of the turnaround, B2, whose clock is unsynchronised and reads 1s, is by A2.
In B frame, B2 has already been passed by A2, whose clock reads 2.165 s and is already by B3/4.
That is confusing enough.
Anyhow, let us take A’s version. As I said, coinciding with the turnaround, B2 was reading B1’s time at the turnaround event (1.5 s) minus (since B was receding away) vx/c^2 = 1.5s – 0.5s = 1s. It seems that, from the turnaround onwards, since now B is approaching, the time of B2 should instantaneously “jump” from minus to plus vx/c^2 = 1.5s + 0.5 s = 2s…

How do you handle the twin paradox for the distant clocks? Not through a new synchronization?

6. Feb 24, 2009

### Fredrik

Staff Emeritus
Why would you even bother? The distant clock isn't going to stay synchronized with yours anyway. If you want to know what the distant clock displays in your "now" (in the new co-moving inertial frame), just use a Lorentz transformation to calculate it.

7. Feb 24, 2009

### Saw

Fredrik, I don´t follow you. Why should I bother? Because they are my clocks, they are in my frame, comoving with me and I may need them for the usual purposes: to measure the speed of passing by objects, to ring my wife and remind her it's time to go and fetch the children..., whatever. I should not need a LT for that: I would use a LT if I transformed from the time reading of a passing frame, but that would be quite cumbersome. I prefer my clocks and for this purpose I need to know if, due to my recent acceleration, they have gone out-of-sync in this new context or, on the contrary, they have automatically adjusted to the new situation...

8. Feb 24, 2009

### Staff: Mentor

Learn about Rindler coordinates. I am not 100% sure, but I think that your clocks will wind up automatically synchronized in the new frame. But you can derive it yourself from Rindler coordinates.

9. Feb 24, 2009

### matheinste

Hello all.

Sorry to butt in but this may be relevant. I asked a question in the forum some time ago but unfortunately I cannot now find it and so cannot give the exact wording but the idea was: Given two clocks, synchronized by the standard procedure, mounted on a ship, do they need to be resynchronized if, after a period of acceleration they are returned to an inertial frame. The answer was they do not need resynching.

Later I read about the equivalence principle which suggested that clocks in a gravitational field run at different rates and so presumably become out of synch, depending on the gravitational potential, and that as gravitation is indistinguishable from acceleration the same is true for accelerated clocks. So accelerated clocks go out of synch.

Assuming both parts are correct and putting them together, would it be correct or not to say that clocks go out of synch during acceleration but return to synch when returned to an inertial frame?

I am also assuming that the same applies to the global acceleration of a frame to which saw seems to be referring. Would that be correct?

Matheinste

10. Feb 24, 2009

### Fredrik

Staff Emeritus
They are? You didn't say that. If you meant that they are both in the rocket, e.g. one in the back and one in the front, then I think DaleSpam is right about the Rindler coordinates.

11. Feb 24, 2009

### Fredrik

Staff Emeritus
Fortunately you suck at typing , so it was easy to find. It's the only thread with "umsynchronization" in the title .

I think I interpreted your question that time as "if they're at rest in some inertial frame, and then accelerated for a while, and finally brought back to rest in that same frame..." And even in that case, I wasn't sure if it's always true that they will be back in synch. (It would take some work to check it for an arbitrary acceleration, and I haven't done it).

It's much easier to do the math the other way round. Do it in SR and use the EP to carry the result over to GR.

I haven't done the math on this one.

I don't think I understand this question.

12. Feb 25, 2009

### Ich

If you accelerate clocks (e.g. at the tip and at the tail of a rocket), they get out of synch and stay so.

13. Feb 25, 2009

### matheinste

Hello Ich,

Quote:-
---If you accelerate clocks (e.g. at the tip and at the tail of a rocket), they get out of synch and stay so. -----

Does this mean that they appear out of synch to a comoving observer to who they appeared to be in synch before the acceleration.

Matheinste

14. Feb 25, 2009

### Ich

Not sure I understand correctly, so let me re-state:
If two clocks are in synch, as measured in their common rest frame, and then accelerate to an different common v, they are no longer in synch as measured in their new common rest frame.
If they had both exactly the same acceleration profile, they still would be in synch in the original frame, but not in the new one.

15. Feb 25, 2009

### matheinste

Thanks Ich.

So i can say that the clocks as you described them will be out of synch as measured by an observer in their new common rest frame, but in synch if measured by an observer in their original common rest frame. Assuming a common acceleration profile.

Matheinste

16. Feb 25, 2009

### Ich

Exactly.

17. Feb 25, 2009

### matheinste

Hello Ich.

Thats clear then.

Now let me be an observer in an inertial frame. When i see clocks in another inertial frame moving relative to me, i would normally measure them out of synch with respect to each other and running slow with respect to my clocks. So does this not apply if i view clocks moving relative to me that were originally synched in the frame from which i am measuring.

Matheinste.

18. Feb 25, 2009

### Ich

They would still run slow, but in synch.
I you made them artificially run faster, you can introduce a common coordinate time that clocks in both states of motion show. A similar procedure is used to define http://en.wikipedia.org/wiki/Terrestrial_Time" [Broken].

Last edited by a moderator: May 4, 2017
19. Feb 25, 2009

### matheinste

Hello Ich.

So in the case of clocks originally synched in a common frame with my clocks, they would, when moving inertially relative to my frame, of course run slow but would, as measured by me, be in synch with each other in their common frame, but not in synch with my clocks because they are running at a different rate.

Matheinste.

20. Feb 25, 2009

### Ich

That's correct in a sense, but might cause some cofusion.
"In synch" would usually mean that if I measure two events to be simultaneous, clocks in another frame would measure them to be simultaneous, too. And that is the case in your setup.
But this is only semantics, where different people have different views, and it might well be possible that my understanding differs from some mainstream definition that I'm unaware of.