# Thread: SR : time of clocks after simultaneous experiment

1. We know the case of the twin paradox and the explanation provided for it.

What about the following simplified example : A train moves relatively to a station (and vice versa):

-Station, with station guy in it and his clock, located at point B
-Train, with train guy in it and his clock, sitting at the exit door of the train.

>The train starts to move at point A at the beginning of the station and stops at point B at the end of the station.
(Let's just say there's no accellaration going on, it happened at a constant speed.)

>Say station guy observes a time dilatation of 1 minute as a result of this :
The station clock says 11.00h, and the train clock says 10.59h according to station guy, at point B.

At the same time train guy does the same experiment : He watches the station come closer :
The train clock says 11.00h and the station clock says 10.59h, according to train guy, at point B.

Nice,now the experiment is over, both guys sit down at the bar in the station, precisely located where the train stopped : point B.
The train is standing still in the station, and both guys can observe both clocks in the same instant.

>> What times do the clocks show ?

My personal view :

During both experiments :
Only the train moved relatively to : the station and the spacetime surrounding the earth.
The station did not move relatively to the spacetime surrounding the earth.

Therefore only the train clock has been influenced, and both guys will read the train clock at 10.59h and the station clock at 11.00h (or 10 minutes later at the bar they will read 11.09h and 11.10h)

Precisely the fact that both clocks indicate a different time, show that both experiments are not interchangeable.
I know of experiments proving the first experiment (or experiment of similar context), such as Hafele-Keating,
but i don't know of any experiments proving the second experiment, the one done by the train guy that is.

-(additions in italic to avoid misunderstandings)

-added image of the situation :

2.

3. Originally Posted by Noa Drake
We know the case of the twin paradox and the explanation provided for it.

What about the following simplified example : A train moves relatively to a station (and vice versa):

-Station, with station guy in it and his clock
-Train, with train guy in it and his clock

>The train starts to move at point A at the beginning of the station and stops at point B at the end of the station.
(Let's just say there's no accellaration going on, it happened at a constant speed.)

>Say station guy observes a time dilatation of 1 minute as a result of this :
The station clock says 11.00h, and the train clock says 10.59h according to station guy, at point B.

At the same time train guy does the same experiment : He watches the station come closer :
The train clock says 11.00h and the station clock says 10.59h, according to train guy, at point B.

Nice,now the experiment is over, both guys sit down at the bar in the station, precisely located where the train stopped.
The train is standing still in the station, and both guys can observe both clocks in the same instant.

>> What times do the clocks show ?

My personal view :

During both experiments :
Only the train moved relatively to : the station and the spacetime surrounding the earth.
The station did not move relatively to the spacetime surrounding the earth.

Therefore only the train clock has been influenced, and both guys will read the train clock at 10.59h and the station clock at 11.00h (or 10 minutes later at the bar they will read 11.09h and 11.10h)

Precisely the fact that both clocks indicate a different time, show that both experiments are not interchangeable.
I know of experiments proving the first experiment (or experiment of similar context), such as Hafele-Keating,
but i don't know of any experiments proving the second experiment, the one done by the train guy that is.

Two quick comments, and please go easy on me - I am no scientist or mathematician by any means :-)

1) Twin paradox - what IS the explanation ? Couldn't it be said that either could claim to be travelling while the other was not (relative motion) and therefore, either can claim to be slower than the other ?

2) Your .. Only the train moved relatively to : the station and the spacetime surrounding the earth.. in your above additional remarks; Question - why should the earth and it's spacetime enjoy a privileged position in the equation ?

edit - changed 1)

4. Originally Posted by Noa Drake
>The train starts to move at point A at the beginning of the station and stops at point B at the end of the station.
(Let's just say there's no accellaration going on, it happened at a constant speed.)
If it starts and stops, you can't say there is no acceleration. You could get round this, by the train moving continuously and transferring the time information to stationary observers at A and B. But then you will be switching frames of reference, which has exactly the same effect. I can't see how this differs from the standard twins experiemnt.

I would suggest you learn the (quite simple) mathematics of SR, work the answer out and then come back here to check it.

5. @Strange

It was just to simplify the case, the experiment stands just as well with the accellaration you suggest :

Experiment 1:
Station guy located at B watches the train clock at A.
Train clock travels until the train exit faces B (the bar)
Station guy reads train clock at B indicating 10.59h to him, not 11.00h.

Experiment 2:
Train guy located at A watches the station clock at B.
Station clock travels until the train exit faces B (the bar)
Train guy reads station clock at B indicating 10.59h to him, not 11.00h.

6. When they are both stationary on the station, they will both (obviously) see the clock show the same time. One of them will have experience less time to get to that point though. Please show us your workings and we can see if you have understood it or not.

7. Originally Posted by Noa Drake

Therefore only the train clock has been influenced, and both guys will read the train clock at 10.59h and the station clock at 11.00h (or 10 minutes later at the bar they will read 11.09h and 11.10h)
The train had to accelerate both up (when starting) and down (when stopping at the end of the platform). Therefore the situation is not symmetrical between train and platform. If you knew how to do the calculations, you could prove to yourself that the train clock shows less elapsed time than the platform clock.

8. Originally Posted by Noa Drake

>Say station guy observes a time dilatation of 1 minute as a result of this :
The station clock says 11.00h, and the train clock says 10.59h according to station guy, at point B.

At the same time train guy does the same experiment : He watches the station come closer :
The train clock says 11.00h and the station clock says 10.59h, according to train guy, at point B.
You've got this wrong. The two observers will not disagree as to what time is on each others clocks when they meet at point B. They will disagree when the train observer is at A. (Since you assert that there is no acceleration, we must assume that the train was already in motion when the train observer passes A.)

So when they meet, if the B observer says that his clock reads 11:00 and train observer's clock reads 10:59, then train observer will concur. When the train stops and they have their drinks, the train clock will be behind B's clock by 1 min. You can never come up with a situation where two clocks that are co-located will disagree as to their respective readings.

Neither will both observers agree as to how much of a time difference accumulated between the clocks during the trip. (Both cannot say that the other lost a minute compared to their own.) Both measure the same time dilation rate in the other clock, However they do not measure the same distance between A and B. For the train observer, the distance A-B is length contracted, so it takes less time by his clock from the time A passes him and B reaches him, than it does as measured by B's clock for the train observer to travel from A to B. So if the B observer says that the train observer's clock lost 1 minute during the trip, the train observer will say that B's clock will have lost less than one minute. Again, they will agree on what time they each read when they meet, but won't at the start of the experiment.

You have to take all relativistic effects into account (time dilation, length contraction, and relativity of simultaneity), you can't just use one and ignore the rest.

So to reiterate, the clocks agree as to what they read respective to each other when they meet, but don't when they are separated.

9. @Janus

You have assumed that the train was already in motion when the train observer passes A,
as you say at the beginning of your comment.
But that is not intended here.

Can we not say the following, i'm not sure of this, i'm asking :
The train starts from 0 km/h at A, speeds up to, say 1.5 km/h (just any example here), then slows down to 0 km/h when arrived at B.
So overall the train moves at a speed of , say 1 km/h avarage to go from A to B.
Can this avarage speed not be regarded as the relative speed v to implement in the time dilatation formula ?

(Obviouslly we're not talking about a dilatation of 1 minute, that is just symbolic to keep it readable.)

10. Originally Posted by xyzt
Originally Posted by Noa Drake

Therefore only the train clock has been influenced, and both guys will read the train clock at 10.59h and the station clock at 11.00h (or 10 minutes later at the bar they will read 11.09h and 11.10h)
The train had to accelerate both up (when starting) and down (when stopping at the end of the platform). Therefore the situation is not symmetrical between train and platform. If you knew how to do the calculations, you could prove to yourself that the train clock shows less elapsed time than the platform clock.
Why would that not be symmetric ?
Wouldn't the train guy observe the station to accellarate both up and down as well ?

11. Originally Posted by Noa Drake
[Why would that not be symmetric ?
Wouldn't the train guy observe the station to accellarate both up and down as well ?
If you look at someone else (with no other clues) you can't tell if they are moving and you are still or vice versa. However, you can always tell if you are accelerating. You don't get pushed back in your seat when another car accelerates, do you?

12. Originally Posted by Noa Drake
Originally Posted by xyzt
Originally Posted by Noa Drake

Therefore only the train clock has been influenced, and both guys will read the train clock at 10.59h and the station clock at 11.00h (or 10 minutes later at the bar they will read 11.09h and 11.10h)
The train had to accelerate both up (when starting) and down (when stopping at the end of the platform). Therefore the situation is not symmetrical between train and platform. If you knew how to do the calculations, you could prove to yourself that the train clock shows less elapsed time than the platform clock.
Why would that not be symmetric ?
The train accelerates wrt. the ground. The station does not.

Wouldn't the train guy observe the station to accellarate both up and down as well ?
That's exactly what I told you. Are you trying to read and comprehend what I tell you?

13. Originally Posted by Noa Drake
@Janus

You have assumed that the train was already in motion when the train observer passes A,
as you say at the beginning of your comment.
But that is not intended here.

Can we not say the following, i'm not sure of this, i'm asking :
The train starts from 0 km/h at A, speeds up to, say 1.5 km/h (just any example here), then slows down to 0 km/h when arrived at B.
So overall the train moves at a speed of , say 1 km/h avarage to go from A to B.
Can this avarage speed not be regarded as the relative speed v to implement in the time dilatation formula ?

(Obviouslly we're not talking about a dilatation of 1 minute, that is just symbolic to keep it readable.)
In your original post, you claimed no acceleration, just a constant speed. Now you are claiming an acceleration. Which is it? The point is that while relative motion is relative, acceleration isn't Even if you try to invoke instantaneous acceleration for the train, there will be a difference in what the train observer determines for the time at B before and after the acceleration while there will be no difference in the train's clock reading for observer B. IOW, according to B, the time on the train clock has the same reading both before and after the instantaneous acceleration, while according to the train observer, the clock at B will have different readings before and after the acceleration.

14. Originally Posted by Noa Drake

Why would that not be symmetric ?
Wouldn't the train guy observe the station to accellarate both up and down as well ?
Say persons A and B are both holding full cups of coffee during your experiment. Which one will spill coffee?

Speed is relative, but only one of A or B actually accelerates.

15. Originally Posted by pzkpfw
Say persons A and B are both holding full cups of coffee during your experiment. Which one will spill coffee?
Like still isn't working!

16. I have learned, thank you.

17. i wouldn't wanna be the one holding the cup inside the train, unless suing would make me rich.

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