# Simultaneity:Confused and almost confident about my own logic

1. Sep 27, 2010

### ambarish

A train is speeding to east and in the frame of train two people simultaneously look towards the top in the frame of reference of train. One is sitting at the west end and the other at the east end. In the ground frame will the events be simultaneous.

My Logic: Consider a frame whose axes and origin coincide with the frame of reference of the train at the instant the person at the west end looks towards the top however this frame is fixed wrt ground. Further Let the origin coincide with the person at the west end. Let both the clocks at the origin (one for the train frame and the other for the ground frame) read zero at the instant the person at the west end looks to the top. Why should the two frames record a different spacetime coordinate for the person at the east end of the train. (When he looks towards the top).

The given answer : the person at the west end Would appear to have looked to the top before the person at east end.

I am completely baffled and very enthusiastic to know the answer. I will be extremely grateful for a convincing answer.

(I don't want to introduce any bulb at the bisector of two persons in the carriage which sends signals to both the end of the train)

2. Sep 27, 2010

### JesseM

Why not? The bulb can't change the coordinates each frame assigns to the event of them looking up, and it's crucial to understand that in relativity clocks at rest in a given frame are "synchronized" in such a way as to ensure that all light beams will have a measured speed of c in that frame. How would you feel about introducing clocks at the East and West end of the train that had previously been synchronized using light signals, long before the events of the people looking up? I suppose if you don't like that either you could just verify that if two events occur at the same time but different locations (like x=0,t=0 and x=10,t=0) then when you plug the coordinates of the events into the Lorentz transformation, you find that the t' coordinate of each event in the other frame is different...you probably wouldn't find this to be a very satisfying answer though.

Last edited: Sep 27, 2010
3. Sep 28, 2010

### PAllen

Try this explanation. For any effect to be measurable, you have to be talking about a train long enough for light to 'take a moment' to cross the train. The two people on the ends of the train can't look up simultaneously by looking at each other; instead, they look up at 3pm exactly by prior agreement. But first they have to make sure their clocks exactly match. So suppose they started in the center of the train, synchronized clocks, moved to the ends of the train, and look up at agreed time. All simultaneous from their definitions.

However, a ground observer sees them synchronize clocks (fine), but as they move apart, the one moving to the east end of the train is going faster than the train (relative to the ground), while the one moving west is going slower. As a result, for the ground observer, the east end person has had a slower clock than the west end person. The west end person's clock therefore reaches 3pm earlier, so they look up first.

However you try to define this operationally, you will get the same conclusion.

4. Sep 28, 2010

### ghwellsjr

PAllen--suppose the two people on the train synchronize their clocks as you say. When they get to their respective ends of the train and 3pm rolls around they will not both observe the other one looking up at exactly 3pm but rather a very short time later due to the finite speed of light, correct? Will each of them measure the exact same delay from when they looked up until they see the other one look up? If not, will the magnitudes of the delays make any difference in how fast they each moved to their respective end of the train?

5. Sep 28, 2010

### PAllen

They will each necessarily see the other look up with the same delay. Each sees the other moving away from them symmetrically. The effect of this plus finite lightspeed will be the same for both. The each will 'see' the other look up the same amount late no matter how they move appart.

6. Sep 28, 2010

### JesseM

The point of using synchronized clocks is so you won't need to worry about light delays, you can just look at the time on the synchronized clock that was right next to the event when it happened. So for example if both passengers look up simultaneously in the frame of the train, that means if there are synchronized clocks next to passenger 1 and passenger, then if passenger 1 looks up when the clock next to him reads 3pm, that means that when he receives the light from the event of passenger 2 looking up he will see that the clock next to passenger 2 also read 3pm at the moment passenger 2 looked up (even if passenger 1 does not actually see this image until his own clock reads a later time).

7. Sep 28, 2010

### ghwellsjr

Okay, good, now I want to ask about an ever so slightly different scenario. Let's consider before the train has left the station. The two passengers synchronize their clocks as before and then the train accelerates up to its final high speed. Now when they both look up at 3pm and then measure the delay to when they see the other one look up, will the delays be identical?

8. Sep 28, 2010

### JesseM

Depends on the details of how the train accelerates. If each point on the train begins to accelerate simultaneously in the ground frame, and each point's coordinate acceleration is the same at each time coordinate in the ground frame, then the clocks will remain synchronized in the ground frame (and the train will get increasingly stretched in the new rest frame of any point on the train, possibly causing the train to break apart--see the Bell spaceship paradox). If you choose some different type of acceleration, like Born rigid acceleration, then the behavior of the clocks will be different. In general there is no guarantee that accelerating clocks that are initially synchronized in their instantaneous inertial rest frame will remain synchronized in their new instantaneous inertial rest frame at a later moment, if you want to make sure two clocks are synchronized in a given inertial frame you need to manually synchronize them using either Einstein's method involving light signals or the method of bringing the clocks together and synchronizing them at a common location and then moving them apart at very low speeds in the frame where you want them to remain synchronized.

9. Sep 28, 2010

### PAllen

Just amplifying JesseM's (great) answer, if you've managed, as described, to keep the clocks synchronized with each other in the ground frame they will not be synchronized with each other in the train's final inertial frame. This follows from the expansion of the train - the clocks have been moving relative to each other in each one's frame.

10. Sep 28, 2010

### Rasalhague

EDIT: As with my post #12, I misread the original post as defining simultaneity in the ground frame, whereas it actually defines it in the train frame.

Careful! Ghwellsjr's original post, if I understood the question, concerned two passengers sitting in a train which is travelling east at some constant velocity, call that velocity v, one passenger at the east end, one at the west end. This is a different situation from the one you describe. In Ghwellsjr original scenario, neither passenger is moving relative to the other. They look up simultaneously as judged in a spacetime coordinate system (reference frame) in which the ground is at rest and both passengers are moving east with the same velocity, v, that of the train. By contrast, in a spacetime coordinate system according to which the train and its passengers have zero velocity, and the ground has velocity –v, the passenger at the back of the train (west) looks up first. Both passengers' observations will agree on the length of the time between one looking up and the other looking up, and on the order of events, namely that the rear passenger looks up first. So this is not a perfectly symmetrical scenario.

Although the relativity of simultaneity is ultimately related to the fact that there exists this universal finite speed limit, it's assumed in this scenario that any delay due to the time taken for signals to propagate from one end of the train to the other has been taken into account by the passengers and corrected for when they make their assessment of who looked up first and how long it took after that for the other to look up.

Last edited: Sep 29, 2010
11. Sep 28, 2010

### PAllen

Last edited: Sep 28, 2010
12. Sep 28, 2010

### Rasalhague

EDIT: I'll leave this post as I wrote it, because it's refered to in later posts, but I suggest Ambarish ignores it, I made some mistakes.

(1) I forgot that it was the train frame where simultaneity had been defined.

(2) I misunderstood the purpose of PAllen's synchronisation example.

(3) Worst of all, in my final paragraph, which does describe the original scenario, I said front when I should have said back.

I still say be careful. If you've created a symmetric scenario in which each passenger considers themself to look up first because there's a significant difference in velocity between them, this isn't the scenario ambarish originally asked about.

It isn't clear to me from #4 whether ghwellsjr was thinking of the passengers as having sat down at their respective ends after synchronising their clocks in this way, but the original question contrasted measurements in just two different spacetime coordinate systems: a coordinate system in which the train and passengers at rest (train frame), and one in which the train and passengers are speeding east at the same velocity as each other (ground frame). Yes, we could introduce two more coordinate systems so that we have a more complicated problem involving a total of four coordinate systems each with a different velocity (ground, train, eastern passenger, western passenger), but maybe better to tackle the simpler question first? If we want to use your synchronisation method to illustrate that original scenario, we could take the passengers' walking pace as negligible compared to the speed of the train, or we could just have them sat down, as before, after having synchronised their clocks. Then, either way, the "given answer" is correct: if the two passengers look up at the same time from the perspective of the ground frame, then, from the perspective of anyone moving with the train's velocity (their clocks synchronised as described), the person at the back (west) looks up first. Having taken account of the time for signals from the other end of the train to reach them, both passengers agree that the person at the back looked up first, and on how long it was between that event and when the passenger at the front looked up.

On the other hand, if we change just one condition in the original and say that instead of looking up at the same time as measured in the ground frame, the passengers synchronise their clocks together as you describe while the train is moving, return to their seats and sit down (with negligible differences in walking pace), and look up at the same time according to their own now synchronised clocks, then, in the ground frame, the front [EDIT: sorry, definitely the back!] passenger looks up first.

Last edited: Sep 29, 2010
13. Sep 28, 2010

### PAllen

----pallen says:
No, I believe the the ground observer will see the West, trailing passenger look up first, as quoted from references by the original poster. This follows directly from transforming from the train system to the ground system (as well as from my argument). Remember, in the train system, where we are assuming simultaneity, the ground is moving west (if the train is moving east).

I will put my further discussion of my clock synchronization method as observed on the train vs. ground in another post.
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14. Sep 28, 2010

### Rasalhague

D'oh! You're right about that. Sorry, Ambarish. In #12, I lost track of which frame the two events (each passenger looking up) were said to be simultaneous in and confused myself. Thanks for point that out, PAllen, and apologies if I misinterpreted your posts.

Suppose the passengers are now seated each at their own end of the train after having synchronised their clocks, as PAllen described (or by light signals or radio or some other means of signalling). If they look up at the same time according to their own clocks, then in the ground frame, the one in the back (west) looks up first.

15. Sep 29, 2010

### PAllen

The final point left open from my discussion with Rasalhague is the difference in how my proposed synchronization procedure is perceived by the ground observer versus the train passengers.

In the train, if the two passengers start in the center, synchronizing clocks, and move to the ends of the train, and sit down, any effects of their moving to the end of the train can be made aribitrarily small. This is explicitly because sqrt(1 - v**2/c**2) for v << c differs from 1 by O(v**2/c**2).

However, from the ground observer's point of view, the train is moving at some v close to c. Adding a very small additional velocity to this increases (say e) increases the time dilation by O(e) (straight algebra, work it out), and the time to get to the end of the train is O(length/e), thus approximately a fixed significant slowing of the east moving passenger occurs relative to the one 'walking' to the west. Thus, as I said, from the ground, the clocks are not synchronized, and the east clock ends up slow compared to the west, and thus the ground observer sees the west passenger look up first. (Consistent with direct application of the Lorentz transform).

16. Sep 30, 2010

### ghwellsjr

I'm afraid I may have caused some confusion on this thread. After I got answers to my questions in my first post (#4), I should have stated my intention before continuing on with more questions.

The original poster, ambarish, did not "want to introduce any bulb at the bisector of two persons in the carriage which sends signals to both the end of the train" in the analysis of this problem. I was trying to come up with a scheme to do just that and also not require the two passangers to move from their positions at the two ends of the train.

This is a scheme to synchronize two identical, stable, and precise clocks separated by some distance and at rest with each other. These clocks would allow the two passengers to do two things: 1) determine when it was time for them to look up and 2) measure the time they saw the other one look up. They then calculate the difference between these two times and one of them communicates that difference to the other one who then adjusts his clock in the right direction to make those differences be identical the next time around.

I believe this scheme has already been indirectly confirmed but I would like direct confirmation from some experts.

17. Sep 30, 2010

### JesseM

I don't understand, what is the "scheme" exactly? You seem to say that one of them only "adjusts his clock" after they have both looked up and then figured out the differences in times on their own clocks when they look up, but then how did they "determine when it was time to look up" in the first place? How were their clocks set before they looked up?

18. Sep 30, 2010

### PAllen

I'm not an expert, but I am sure this scheme can work. Of course it involves sending light signals, and will not synchronize the clocks from the ground frame. I was trying to satisfy the the OP in a different way, using no light signals at all, but requiring that the passengers start together to syncrhonize, then move to the ends of the train (slowly).

19. Sep 30, 2010

### PAllen

I believe the intended scheme is as follows:

Each passenger starts out with a clock whose 'rate' is sufficiently accurate. Then, by prior agreement, they both look up (send a signal) at e.g. 2pm. Then A sends to B the difference between when A sent the signal and when A got B's signal. When B receives this information, if it is the same as the difference between when B sent his signal and when B got A's signal, they were already synchronized. Otherwise, the descrepancy tells B how to adjust his clock to match A. Then they can do the intended simultaneity test at 3 pm.

Of course, if they start out with clocks synchronized on the ground before boarding the train, they will not be synchronized, in the train frame, when the train reaches cruising speed close to c. Thus the whole issue. To re-synchronize for the train, either some signalling scheme needs to be used, or something like my scheme can be used (board at center of train with synchronized clocks, move slowly to the ends of the train only when cruising speed is reached).

20. Sep 30, 2010

### ghwellsjr

PAllen, thanks, that's exactly right. I'm sorry for not filling in the details, I just assumed everyone was remembering them from previous posts.

Now I want to revisit my scenario when the train starts out stationary in the station with synchronized clocks and then accelerates to its final speed. I previously asked if the two clocks would still be synchronized. Here's JesseM's prior answer:
Now I want to explore the first method of acceleration that results in the train breaking apart. Obviously, we don't want that to happen so let's consider a different situation:

The two "passengers" are really "engineers" and they have their own identical locomotives loaded with the same amount of fuel separated by some distance on the same track. There are no cars between or anywhere on the track. They start out stationary with synchronized clocks (using my scheme) and at a previously agreed on time they each step on the gas. They both accelerate identically until they run out of fuel at which point they coast with no friction to cause them to slow down.

Now according to Jessie's answer, to an observer on the ground, their clocks will remain synchronized and they will remain the same distance apart but to the "engineers" the locomotives will have moved farther apart which I see and understand. But, after a sufficient length of time for the "dust" to settle, what can we say about what has happened to the synchronization of their clocks (using my scheme)? I think that both "delays" will have increased, but will they be different?