# Relativity of Measurements

1. Dec 2, 2015

### Ut-Napishtim

Not long ago a very thoughtful thread cleared my mind on the Relativity of Simultaneity. I am very thankful and hope to get here help also on a question related to the RELATIVITY OF MEASUREMENTS in the moving reference frames.

Suppose that innumerous bars (or rods) are loaded on a very long standing still train and aligned with their length strictly along the railroads. Several ones are given to an observer on the station. All bars are made with exactly the same length.

The train moves along a straight line and when the observer on the station measures, by an ingenious indirect method, the bars on the train he determines that the bars on the moving train are shorter than the ones he has on the station, the more so the faster the train moves, which is explained by the Special Relativity.

The train stops and the observer on the station can now measure and compare the bars from the train with the ones on the station directly. What does he find?

Are the train bars still shorter that the station ones or have they recovered their original length?

Many thanks for attention.

2. Dec 2, 2015

### HallsofIvy

Staff Emeritus
I am surprised at you asking this. You seem to know that, according to the theory of relativity the length of the moving bars, as measured by a person stationary with respect to the station, depends upon the speed of the moving bars relative to the observer. Since the train and the observer are now stationary with respect to one another, the bars on the train and at the station will be of the same length again. Do you understand that, will the train is moving away from the station, an observer on the train will measure the bars on the train as still having their original length while he will measure the bars still at the station as shorter?

3. Dec 2, 2015

### Mister T

It may help to think of length contraction, not as a shortening of the physical objects like these rods, but of the space occupied by these objects. It's not as if there are forces pushing on the ends of the rod making it physically shorter, there are no such forces.

The figure is my drawing of a pair of clocks joined by a stout metal rod through their centers. The hands of the clock rotate in sync because the clocks are synchronized. A delicate glass rod connects the tips of the clock hands. If one of the clock hands advanced more than the other, the glass rod would twist, stretch, and break. But as long as the clock hands move in sync the glass rod will not be stressed and it won't break.

When this device is placed in motion in a direction parallel to the connecting rod, relative to the observer who synchronized the clocks when they were at rest, the clocks will no longer be synchronized (relativity of simultaneity). This will cause the glass rod to break. But if another observer moves with the clocks, they would remain synchronized in his frame of reference and the glass rod wouldn't break. So, does the glass rod break or not?

Last edited: Dec 2, 2015
4. Dec 3, 2015

### Staff: Mentor

If you mean another inertial observer, this is impossible; no inertial observer can move with the clocks the whole time, because the clocks are accelerated when they are placed in motion. That is, as I'm understanding the scenario, the clocks start out at rest in one inertial frame, undergo a period of acceleration, and end up at rest in a different inertial frame, moving with respect to the original one. An observer in the first inertial frame will see the clocks start out synchronized, and then go out of sync. An observer in the second inertial frame will see the clocks start out not synchronized, and then come into sync.

It breaks. The observer in the first inertial frame explains this as you have explained it. The observer in the second inertial frame explains it by the fact that, in his frame, the glass rod was oriented to connect the two clocks, not in a sychronized state, but in a particular unsynchronized state (the one the clocks start out in, in his frame). When the clocks are forcibly changed to the synchronized state, in his frame, the rod breaks. (Note that to fully understand this, you also have to take into account that, in the second observer's frame, the rod starts out length contracted and ends up not length contracted; so even though it looks like the rod "untwists", which would shorten the glass rod, it also expands, and that more than overcomes the untwisting effect.)

(There are other complications lurking here as well, to do with the way the clocks are accelerated, but I don't know that we need to go into those here.)

5. Dec 3, 2015

### Mister T

Sorry. I worded it poorly. It was never my intention to have the apparatus undergo an acceleration. One observer at rest relative to the apparatus sees the clocks synchronized. Another observer in motion relative to the apparatus sees the clocks not synchronized. In the rest frame the glass rod is straight, in the moving frame the glass rod is bent.

6. Dec 3, 2015

### Staff: Mentor

Then I don't understand what you mean by "when this device is placed in motion". You need to more carefully specify exactly what scenario you are envisioning.

7. Dec 3, 2015

### Ut-Napishtim

My best apologies for asking stupid questions and thanks for clearly explaining them nevertheless.

May I ask one more of this kind?

Together with the bars there were loaded on the train, and given to the observer on the station, innumerous clocks that were precisely synchronized before the train started moving.

The train moved, the observer on and from the station determined by his ingenuous indirect method that the clocks on the train were behind the ones left on the station, as explained by the Special Relativity, the more so the faster the train.

The train stopped and the observer compared the clocks. What did he find? Are the clocks still behind or regained synchronization with those on the station?

Dear Hallsoflvy! I understand wasting your and others precious time. But please be kind with me and explain.

8. Dec 3, 2015

9. Dec 3, 2015

### Staff: Mentor

But then why are you asking whether the glass rod breaks? Since the state of motion of the entire apparatus does not change at all (it's in inertial motion, at rest in the first observer's frame, the whole time), there is no force acting on the rod, so it's not going to do anything; it's going to stay in whatever initial state you specified forever. As I understand it, you specified an initial state in which the glass rod was not broken, so it's not broken. But that's a matter of your initial specification of the problem; it has nothing at all to do with the physics of glass rods or relativity of simultaneity or length contraction or anything else.

10. Dec 3, 2015

### Mister T

It's a puzzle for beginners.

11. Dec 3, 2015

### Staff: Mentor

It doesn't look like a puzzle to me; it looks like a deliberately misleading question, inviting the reader to assume that some change in the state of motion of the glass rod has occurred, when it hasn't. After all, I'm not a beginner, and I misinterpreted it that way. Also, it invites the reader to assume that there must be some physical principle involved to determine whether the glass rod breaks or not, when it's actually just an initial assumption of the problem.

12. Dec 3, 2015

### Mister T

As I said, it was written poorly. Let me try again.

13. Dec 3, 2015

### Mister T

Shown in the figure is a pair of clocks joined by a stout metal rod through their centers. The clocks are synchronized in their rest frame so that the hands rotate in sync. A delicate glass rod connects the tips of the clock hands. If one of the clock hands advances more than the other, the glass rod would twist, stretch, and break. But as long as the clock hands move in sync the glass rod will not be stressed and it won't break.

An observer in motion along the line joining the clocks will conclude that the clocks are not synchronized and so expects the glass rod to be broken and disconnected. But an observer at rest with respect to the clocks will see them in sync and so expect the glass rod to be intact and unbroken. The glass rod can't be both broken and unbroken. Can you explain which it is and why?

14. Dec 3, 2015

### Staff: Mentor

No, he doesn't. He just expects the glass rod to be tilted, with respect to the connecting rod in the center. In other words, instead of connecting the two clock hands when they are both at, say, twelve o'clock, the moving observer sees the glass rod connecting the hands when one is at twelve o'clock and the other is at, say, ten o'clock (or whatever the amount of non-sychronization between the clocks is in his frame). But the glass rod still has a constant length, and still rotates around perfectly fine without breaking.

15. Dec 3, 2015

### Mister T

But the premise is that if, in the rest frame, the clock hands were at 10 and 12, that would cause enough stress to break the glass rod. The rod ends would have to pivot where they're connected to the hands, and the rod would have to stretch. Either of those would be enough to cause the glass to shatter.

It's like the relativistic cookie cutter puzzle. In both cases it's the space occupied by the objects that is altered, there is no stress on the objects themselves.

16. Dec 3, 2015

### Staff: Mentor

But they aren't there in the rest frame. They are at 12 and 12. They are at 10 and 12 in the moving frame, but that's not the only thing that changes in the moving frame. See below.

No, they just are at an angle, because the rod is moving in this frame, and the rotation of the two clock hands looks different in the moving frame because of that. The rod is under zero stress in the moving frame, because stress is an invariant; if the rod is under zero stress in its rest frame, it's under zero stress in every frame.

No, it wouldn't, because "stretching" would mean stress was being applied to the rod, and as above, stress is an invariant.

It is true that the rod's coordinate length in the moving frame will be different from its coordinate length in the rest frame of the assembly. (Note that the glass rod is not at rest in the latter frame either, since it is attached to the rotating clock hands.) The whole assembly is length contracted in the moving frame; plus, the rod, being connected to the rotating clock hands, has an additional motion that must be taken into account. But all of this is just the usual relativistic effect of changing frames; it has nothing whatever to do with any actual physical stress applied to the rod.

You're contradicting yourself. If there is no stress on the glass rod, then obviously it won't break, and that's true in every frame. But you said the observer in the moving frame will expect the rod to break. So once again, you're using misleading language, describing incorrectly what the observer at rest in the moving frame will expect.

17. Dec 5, 2015

### Ut-Napishtim

With respect and deference – Could someone answer my second question (repeated from above)?

Together with the bars there were loaded on the train, and given to the observer on the station, innumerous clocks that were precisely synchronized before the train started moving.

The train moved, the observer on and from the station determined by his ingenuous indirect method that the clocks on the train were behind the ones left on the station, as explained by the Special Relativity, the more so the faster the train.

The train stopped and the observer compared the clocks. What did he find? Are the clocks on the train still behind or regained synchronization with those on the station?

18. Dec 5, 2015

### Mister T

If the observer stayed put, and the train returned to him after having moved around for awhile, then the train clock will be behind the observer's clock. This is the well-known twin paradox.

19. Dec 5, 2015

### HallsofIvy

Staff Emeritus
Once the train has stopped, the clocks aboard it be running at the same rat as the clocks that were always "stationary" but will be behind because they were running slower while the train was moving. Note that an observer on the train would have seen the stationary clocks as slower, because they were moving relative to that observer. But the crucial point here is that the clocks on the train accelerated and decelerated while the "stationary" clocks did not.

20. Dec 5, 2015

### Ut-Napishtim

Thanks for attention, but it is NOT what I did ask.

The innumerous synchronized clocks were spread along an immensely long train before it moved. When the train stopped there happened to be right near the observer, that stood on the station while the train moved, one clock out of these innumerous and previously exactly synchronized (between themselves and with those given to the station observer) clocks that were on the train.

My question is about a comparison (after the train stopped) of the observer's clock with the one of the train clocks (as described above) and NOT as in the scenario of your response.

The train moved in one direction along a straight line only. It did NOT change direction as in the Twin Paradox various scenari.