synchronized clocks with respect to rest frame

mananvpanchal

Hello,

Suppose, A and B is clocks at both end of train. A is at left and B is at right. Observer O is at middle of train at point M. Observer R is on platform.
Train is at rest and O synchronize both clock. The clock is synchronized with respect to both observer.

Now, train starts moving to right. It accelerate and after some time it runs with constant speed. Now, the clock is still synchronized with respect to O.

But what about R? Is clocks synchronized with respect to R? If no, then which clock is ahead A or B?

Thanks.

harrylin

Yes, both clocks are synchronised for both reference systems according to the standard synchronisation convention.
No, that is wrong: according to the standard synchronisation convention, the clocks are now out of synch with respect to O.
- Both moving clocks are now very slightly behind according to R.
- According to O, clock B in the front is now ahead on clock A in the rear.

This is quickly understood with a simplified analysis from the platform: neglecting the small effect from length contraction, both clocks are about equally behind. If O sends a signal to both A and B, clock A is moving towards the signal while B is running away from it. Thus the signals will reach A before B. Consequently, A will indicate less time than B at these events which O defines as simultaneous.

Harald

DaleSpam

I agree with harrylin's analysis, but want to point out that the answer depends on the details of the acceleration. This analysis assumes that in the station's frame A, B, and O all have the same acceleration profile. If they have different acceleration profiles (e.g. if the train is being pulled from the front or pushed from the back or undergoes Born-rigid acceleration) then the answer will typically be that they do not remain synchronized in the station's frame either.

AdrianMay

SR explicitly chickens out of any cases involving acceleration. As soon as the train started to accelerate, we were out of our depth. You can synchronise clocks when you're still or when you're moving, but you can't accelerate anything during the thought experiment unless you want to learn GR.

ghwellsjr

Somebody forgot to tell Einstein that. In his 1905 paper introducing Special Relativity near the end of section 4 he describes what happens to an accelerating clock compared to an inertial clock. This was the origin of the Twin Paradox which is routinely handled by SR even though at least one of the Twins accelerates.

DaleSpam

This is not correct. SR can handle acceleration just fine. All it cannot handle is gravitation.

AdrianMay

Four lines. Hardly a sufficient treatment.

Acceleration and gravitation are indistinguishable under GR, at least over short intervals where tidal affects aren't observable.

harrylin

I hoped that it was clear from my analysis that the clocks will be slightly out of synch in the station's rest frame. Note also that the usual assumption of mechanics is that no plastic deformation occurs so that after some time running at constant speed, the acceleration profile doesn't matter.

DaleSpam

Yes, it was clear, I was just adding emphasis.

mananvpanchal

Thanks guys for your replies

Ok, so moral of the story is:

Clocks synchronized in train frame at rest remain synchronized for platform frame, but not for train frame when train starts moving. (neglecting the small effect from length contraction, which will reduce more if train travels with constant speed for much more time)

Clocks synchronized in train frame at motion remain synchronized for train frame, but not for platform.

Ok, so here I am confused with two questions.

Clocks is synchronized in train frame at rest, but now train is moving with constant speed. clocks is no longer synchronized for O, but they are for R.
If O brings all clocks together, are they synchronized? If no which one is ahead?

Clocks is synchronized in train frame at motion. Clocks is not synchronized for R, but they are for O.
If O brings all clocks together, are they synchronized? If no which one is ahead?

Thanks

EDIT: please, replace "all clocks" with "both clocks"

AdrianMay

What do you mean by "bring all clocks together"? You mean the train is driving through the station? That wouldn't matter. The deal is: if you hit the gas or the brakes, the synchronisation goes wrong and you have to resync. Choose your speed, then synchronise, then stick to that speed.

harrylin

Not exactly. The effect from length contraction on clock time is what you will hardly measure after the train travels for a long time: it does not go away but it is a small correction on the time dilation effect because B has at any time very slightly less advanced than A. Ideally, according to the platform frame clock A in the rear remains extremely slightly more delayed than clock B in the front. However, I can imagine that if the train is pushed, B might be doing some additional swings in the process which brings them perfectly in tune again or even inverses the effect (is there by chance a math enthusiast in the room?).
You could try to answer all those questions by calculation (a little complex) or by using the relativity principle (easy): O can assume to be "in rest", so that the situation is symmetrical. The clocks are not synchronized at the start, next they move exactly the same, thus they must stay out of synch.

For R, the easy way to solve it is to do just the same: look at the problem from O's perspective, and as the clocks are together the result must be the same from R's perspective.
See above; I'm sure that you can now answer that yourself.

mananvpanchal

I cannot understand this. I don't need any calculation. I need just answer of my confusion.

I restate this if it is hard to understand.

Clocks is synchronized in train frame at rest, but now train is moving with constant speed. clocks is no longer synchronized for O, but they are for R.
If O brings both clocks together in their frame, are they synchronized? If no which one is ahead?

Clocks is synchronized in train frame at motion. Clocks is not synchronized for R, but they are for O.
If O brings both clocks together in their frame, are they synchronized? If no which one is ahead?

Please, tell me the answer.... synchronized or not???

harrylin

I gave you the answer to your first question ("they must stay out of synch"); your question was clear!

The answer to your second question is similar, which you could have figured out for yourself without any calculation, as I explained. What is the use of hearing answers if it doesn't make you understand?

Note also that for O the two clocks remain in/out of synch with each other, but will be slightly behind on clocks that didn't move.

AdrianMay

Aha! I think there's a terminology mismatch going on here. Synchronised might mean that zero on one corresponds with zero on the other, or it might mean they're going at the same speed. That's two different things.

We can all just define zero of time and space (origin) as the place and time where/when the middle of the train passes through the middle of the station. If we'd been writing down numbers before the train arrived on a different basis, we can all just note when in our old sync the origin happened and correct our notes afterwards. This is not the problem.

The problem is how fast the clocks run and how long the sticks are. It's not sensible to neglect length contraction compared to time dilation or vice versa because they both happen to the same degree, namely, gamma.

So let's just assume we can define the origin clearly. We can because there are no distances to worry about. We put a big red spot on the middle of the train and the middle of the platform, and when they coincide, that's everybody's origin of time and space. Sorted.

Now let's get started with the real problem. We don't know how to synchronise clocks that are not at the origin. We can't carry them to the origin, sync them there and carry them elsewhere because WE DONT HAVE AN AXIOM to tell us what happens when we move clocks around. In fact, the only axiom we have that might come in useful here is the one about the constancy of c. We can use that as follows: at time t1, launch a light ray from the origin to the remote clock and back. See it arrive back at t2. Set the remote clock to (t1+t2)/2. Because of the axiom, we can even do that on the train. That's the only method we have that's guaranteed by the axioms.

Now you can draw the rhombus diagram to show that the guy on the platform will think that the guy on the train screwed up. You draw the train as a world line at about 1 o'clock, and the light lines at 45 degrees. You see that the light beam reaches the front clock late because the clock is running away from the beam. So the guy on the platform thinks that readings on the front clock underestimate the actual time.

Now imagine that the guy on the train wants to measure the platform. Assume that both train and platform are 100m long at rest. He'll put cameras at the front and the back of the train and tell them to go off at t=0. What happens? Options:
* both cameras see the exact end of the platform
* both cameras see green fields
* both cameras see platform with no end in sight
* one camera sees fields while the other sees platform.

Figure it out for yourself. The correct answer is that both cameras see green fields and conclude that the platform is shorter than the train. Now we reverse the whole argument and see that the platform also thinks the train is too short.

We now have three out of four pieces of the puzzle in place:
1) Obviously, the time axis of the train looks wonky from the point of view of the platform. That's just because the train defines his own centre of x as the centre of the train, which the platform can see is moving.
2) We've also established that the space axis of the train is wonky, that's the bad sync of the front and back clocks.
3) We see lorentz contraction resulting from the bad sync of the clocks.

To complete the square, we'd like to see time dilation resulting from the disagreement about the centre of x. I'll leave that for homework ;-)

mananvpanchal

Ok, so my first question's answer is : clocks is out of synch.

Can you tell me what is the reason for it? (length contraction or any other)
Because, slowly bringing together is one of Einstein's conventions.

harrylin

I wonder if you're right that slow clock transport is one of Einstein's conventions. Anyway, I already gave you an "Einsteinian" reason: it obeys the relativity principle. However, you seem to want a "Lorentzian" reason.
From the point of view of O, I also gave that reason to you: both clocks are equally affected by their motion according to O, and thus they will equally slow down according to O. As a result, clocks that were out of synch stay out of synch.

You can surely fill in what the result is for clocks that were in synch, especially as I gave the answer in post #16 (if you cannot, then I'm very sorry, but I won't reply anymore! ).

Is that good enough for you or do you want to hear the more complex reason according to R?