Leading Clocks Lag: Twin Paradox Explained

[Grasshopper]

TL;DR Summary

Leading clocks lag, but what happens when they stop moving? How is it that there is no differential aging? I can't see how there should be any at all, but I can't explain how.

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Two distant observers at rest with one another synchronize their clocks utilizing the travel time of light (by first moving the clocks to their locations, then using a light signal to time when the second clock should be turned on, and setting the second clock ahead to account for the previously calculated travel time of light). Call one observer A, the other B. They are very far apart from one another.

An extremely long space train moves with respect to both of them at a very high speed (call these guys observer C1, C2, etc; Furthermore, they have also synchronized their clocks using the same method, and since they are in uniform motion, their clocks remain in sync for the duration of the trip according to their reference frame).

Observers A and B ought see the clocks at the front of the train lag behind the clocks at the back of the train, with each clock lagging behind the ones behind it by an amount of $\frac{x'v}{c^2}$. The leading clock will pass observer A, and then when it reaches observer B, the long space train stops. This will ensure that the space train travels for a non-zero interval of time with the leading clocks lagging.

For the entire trip, according to both A and B, the leading clocks will lag behind the trailing clocks. But for those on the train, the clocks will remain in sync for the duration.How is it that there is no differential aging for people in the space train?
Is it that the space train is not a rigid body, so that while the front stops (according to A and B), the rest of the train is still moving for a finite duration of time, allowing the clocks in the back of the train to re-synchronize with the clocks at the front (according to A and B)? Or is it something else?

Thanks as always!

 

[jbriggs444]

when it reaches observer B, the long space train stops

You will have to fill in more details on the stopping mechanism.

You could stop the train all at once according to the [pre-stop] train synchronization standard (e.g. with pre-programmed rocket thrusters in all of the cars).

You could stop the train all at once according to the embankment synchronization standard. (just adjust the programming on the thrusters).

You could let it stop by itself as the front crashes into a rigid, and stationary wall.

You could let it stop by itself when a chain hanging out the back suddenly snags on a rigid and stationary hook.

Yes, in that last cases, the one end keeps on going after the other end stops. This is a correct observation from the point of view of both frames. There is no such thing as a perfectly rigid body in special relativity. The deceleration caused when the one end makes an abrupt stop propagates at the speed of sound toward the other end. The speed of sound is guaranteed to be less than the speed of light.

 

[Dale]

and since they are in uniform motion, their clocks remain in sync

the long space train stops

These two parts contradict each other

 

[Grasshopper]

These two parts contradict each other

Just to be clear, you are referring only to the clocks on the moving train, correct?

I am under the impression that you are saying that once the train stops, the on-train clocks that were initially synced (because prior to this moment, those clocks on the train are justified in saying they are at rest and so can sync their clocks), are no longer in sync (and are no longer in the same uniform motion). Is this what you are saying or am I misunderstanding you?

 

[Ibix]

Just to be clear, you are referring only to the clocks on the moving train, correct?

Yes. Clocks on the train de-synchronise when it accelerates. To be precise, there isn't a clear definition on "synchronised" while the clocks are accelerating, and they will generally not be synchronised in whatever frame they end up in when they stop accelerating.

As @jbriggs444 says, the exact details of where the train ends up and what its clocks read depend on how you decelerate the train.

 

[Grasshopper]

You will have to fill in more details on the stopping mechanism.

You could stop the train all at once according to the [pre-stop] train synchronization standard (e.g. with pre-programmed rocket thrusters in all of the cars).

You could stop the train all at once according to the embankment synchronization standard. (just adjust the programming on the thrusters).

You could let it stop by itself as the front crashes into a rigid, and stationary wall.

You could let it stop by itself when a chain hanging out the back suddenly snags on a rigid and stationary hook.

Yes, in that last cases, the one end keeps on going after the other end stops. This is a correct observation from the point of view of both frames. There is no such thing as a perfectly rigid body in special relativity. The deceleration caused when the one end makes an abrupt stop propagates at the speed of sound toward the other end. The speed of sound is guaranteed to be less than the speed of light.

Regarding these crashing stops, how do observers at point B (the at rest “platform” observers outside the space train) see the train clocks behave? Initially they’d see the leading clocks lagging behind.

So I’m assuming they’d measure the leading clocks to quickly come to tick at the same rate as their own clocks, since the leading train clocks would slow and stop first. But since the clocks in the back would eventually feel the same deceleration as the leading clocks, and would still have previously been ahead of the leading clocks according to observers on the platform, should there not be differential aging between the train clocks?

Or is it that they become the leading clocks themselves, as the original leading clocks stop moving with respect to the platform observers, long enough to compensate for previously being the trailing clocks, thereby making all train clocks match each other?

 

[Ibix]

The crashing stops are hardest to analyse because they depend on the mechanical model of the train - its elasticity, essentially. Qualitatively, the clock in the nose comes to rest almost instantaneously so abruptly transitions to A/B frame's tick rate. The abrupt stop sends a mechanical wave propagating backwards through the train, which brings the second clock to rest slightly more gently and the third one slightly more gently still and so on. Each subsequent clock will start braking later even in the train frame, and will take longer about braking. The end result will be all the clocks at rest in the A/B frame and ticking at the same rate, but with a pattern of offsets that I don't think we can calculate without a lot of work with completely imaginary super-materials that can survive abrupt deceleration from near light speed.

Acceleration, per se, has nothing to do with what the clocks read at the end. If a clock is moving at speed $v(t)$ in A/B's frame and its reading at time $t_1$ is $T_1$ then its reading at time $t_2$ is $T_1+\int_{t_1}^{t_2}\frac 1{\gamma(t)}dt$, where $\gamma(t)=1/\sqrt{1-v^2(t)/c^2}$. It's only the velocity profile that matters, in other words, and unless you engineer the velocity profiles of all the clocks extremely carefully they will not be synchronised when they come to rest.

 

[jartsa]

Maybe we can postulate that because of the very high speed the lag between front and rear clocks is thousand years. And because of the very high speed the length of the train is one nano-meter.

Now we can say that after the train has crashed on to a wall, the lag is still very close to thousand years.

 

[Dale]

I am under the impression that you are saying that once the train stops, ...

I am just saying that the description is self contradictory. On the one hand you say that they are in uniform motion so they remain synchronized and then you say that they stop which is not uniform motion. If they are in uniform motion then they cannot stop. If they stop then you cannot claim that they stay in sync (or any other claim based on them remaining in uniform motion).

Before we can analyze your scenario you need to present us with a self-consistent scenario.

 

[Grasshopper]

I am just saying that the description is self contradictory. On the one hand you say that they are in uniform motion so they remain synchronized and then you say that they stop which is not uniform motion. If they are in uniform motion then they cannot stop. If they stop then you cannot claim that they stay in sync (or any other claim based on them remaining in uniform motion).

Before we can analyze your scenario you need to present us with a self-consistent scenario.

I will try to make my set up less ambiguous.

The scenario is that the space train initially moves uniformly with speed v along the x-axis of the platform observers, and later comes to a stop.

During this INITIAL uniform motion, the space train clocks are synchronized in the space train frame. That way the train clocks will all initially read the same time as each other according to space train observers, while the observers on the platform watching the train go by will see the leading clocks lag (while the train observers will presumably see the same type of lagging for the platform clocks). Basically it’s just to establish symmetry in the initial set up. (Clocks are NOT synchronized between the space train and platform. Maybe they could be, but I don’t think it’s important for what I’m trying to find out here).

Later on, after an amount of time moving at speed v according to the platform observers, the space train will come to a stop (suppose after that amount of time the train begins a constant deceleration until the velocity between the reference frames equals 0).

I am assuming this will break the synchronization of the train clocks according to the space train, so that both the observers on the space train and platform will agree upon what the clocks on the space train read (whereas initially they couldn’t, because prior to deceleration, the space train observers were justified in claiming to be at rest with synchronized clocks, while the platform observers claimed the leading clocks were lagging).The main questions here are, once that symmetry is broken and the space train is finally at rest with respect to the platform, what do the train clocks read with respect to each other at the end, and how do they come to that reading? Presumably everyone involved will agree upon the final reading of the (now broken) space train clocks, since I assume that would be independent of coordinate choice.

So, will the final reading of the space train clocks disagree with each other? (after they are no longer in uniform motion, the space train having decelerated until v = 0) If so, why/how?

Or, will the final reading of the space train clocks be the same? If so, why/how?

EDIT: I am not concerned about if the final reading of the space train clocks match the platform clocks. I am only concerned about if the final reading of the space train clocks match each other.
.Conversely, I am assuming observers on the space train must have initially saw the platform moving toward them, seeing clocks in front on the platform also lag. However after the space train comes to a stop, they must see the clocks on the platform as synchronized (because the observers on the platform remained in uniform motion throughout. Observers on the space train did not). What is the physical justification for observers on the space train for why the clocks on the platform are synchronized when the space train comes to a stop?

 

[Dale]

Thanks, that is a much clearer description.

the space train will come to a stop (suppose after that amount of time the train begins a constant deceleration until the velocity between the reference frames equals 0).

As @jbriggs444 mentioned the answer will depend on the details of "come to a stop". It seemed that you favored the crashing approach, in which the proper distance between clocks on the train will change. You would need to specify the speed of sound in the material of the train to fully analyze that.

You could also do what is called a Born rigid deceleration, where the proper distance between clocks on the train does not change.

You could also do a simultaneous stop (which is the easiest to analyze), but you would need to specify in which frame everything stops simultaneously.

 

[Ibix]

So, will the final reading of the space train clocks disagree with each other?

Yes. I don't think it's possible to arrange things so that they show the same time.

What is the physical justification for observers on the space train for why the clocks on the platform are synchronized when the space train comes to a stop?

To calculate the time a clock shows now you take the time you see and correct for the light speed delay. After the stop the train observers no longer see the platform clocks as moving, so the correction they apply is different.

So the observers will see the platform clocks reading (almost) the same before and after the stop. But they will apply different corrections to calculate what they show right now.

 

[stevendaryl]

Qualitatively, a relativistic train works like this:

1. Initially, the train is at rest relative to the tracks. The front and rear clocks are synchronized in the track frame. The length of the train is $L$.
2. After acceleration, the train is traveling at speed $v$, its length is $\dfrac{L}{\gamma}$ (due to length contraction). If nothing is done to the clock settings, then the front clock will be ahead of the rear clock, in both the track frame and the train frame.
3. The front clock must then be set back (or the rear clock must be set forward, or some combination) in order for the clcks to be synchronized in the train frame.
4. After the resynchronization, the clocks will NOT be synchronized in the track frame. In that frame, the front clock will now be behind the rear clock.
5. Later, when the train decelerates, the front clock will get even further behind the rear clock.
6. So when the train finally comes to a stop, the front clock will be behind the rear clock.

The exact difference between the front and rear clocks will depend on how the train accelerates and decelerates, but I think the qualitative conclusion is pretty firm: The front clock will be behind the back clock.

If there were no resynchronization step, then the results can be different: the two clocks could show the same time, or the front could be ahead, or the rear could be ahead.

 

[Ibix]

Now we can say that after the train has crashed on to a wall, the lag is still very close to thousand years.

The lag between front and rear clocks in the inertial train is $\gamma vl/c^2$ where $l$ is the contracted length length of the train, so this does work as a sanity check, yes.

 

[Dale]

Yes. I don't think it's possible to arrange things so that they show the same time.

You could have them all stop simultaneously according to the train frame.

 

[PeterDonis]

You could have them all stop simultaneously according to the train frame.

Yes, but then they would not be synchronized according to the rest frame (the frame in which they end up stopped/at rest). The clocks would read the same time at different times according to that frame; so they would need to be resynchronized.

 

[Dale]

Yes, but then they would not be synchronized according to the rest frame (the frame in which they end up stopped/at rest). The clocks would read the same time at different times according to that frame; so they would need to be resynchronized.

Right, I had in my mind that the clocks stopped (counting time) once they stopped (moving), but I realize that the OP did not actually write that.

Oops, sorry @Ibix

 

[Grasshopper]

The crashing stops are hardest to analyse because they depend on the mechanical model of the train - its elasticity, essentially. Qualitatively, the clock in the nose comes to rest almost instantaneously so abruptly transitions to A/B frame's tick rate. The abrupt stop sends a mechanical wave propagating backwards through the train, which brings the second clock to rest slightly more gently and the third one slightly more gently still and so on. Each subsequent clock will start braking later even in the train frame, and will take longer about braking. The end result will be all the clocks at rest in the A/B frame and ticking at the same rate, but with a pattern of offsets that I don't think we can calculate without a lot of work with completely imaginary super-materials that can survive abrupt deceleration from near light speed.

Acceleration, per se, has nothing to do with what the clocks read at the end. If a clock is moving at speed $v(t)$ in A/B's frame and its reading at time $t_1$ is $T_1$ then its reading at time $t_2$ is $T_1+\int_{t_1}^{t_2}\frac 1{\gamma(t)}dt$, where $\gamma(t)=1/\sqrt{1-v^2(t)/c^2}$. It's only the velocity profile that matters, in other words, and unless you engineer the velocity profiles of all the clocks extremely carefully they will not be synchronised when they come to rest.

So, that would mean the acceleration each clock feels as it comes to a stop is different, and that would mean that the time the clocks spend at higher speeds relative to the platform are also different, right? Specifically, the clocks in the front would be moving at a fast speed for less time than the clocks in the back. Unless I'm misunderstanding things entirely, wouldn't that decrease the difference in time readings between the clocks in the back vs clocks in the front according to the observers at rest with respect to the platform? (or perhaps cause the ones in the back to show readings that lag after the impact)

 

[Grasshopper]

Oh, let me add that I think I had a false misconception in the beginning. I was under the impression that according to the platform observers, that the clocks on the train would also be time dilated with respect to each other. But this can't be right. All the train clocks must tick at the same rate, but are just not synchronized according to the platform observers. Is that correct?

 

[PeterDonis]

All the train clocks must tick at the same rate, but are just not synchronized according to the platform observers. Is that correct?

Yes, that's correct.

 

[Ibix]

All the train clocks must tick at the same rate, but are just not synchronized according to the platform observers. Is that correct?

A clock at rest at $x'=X'$ in the prime frame ticks at times $t'_n=T'+n\Delta T'$. In the unprimed lab frame, therefore, the time of the $n$th tick event is$$\begin{eqnarray*}
t_n&=&\gamma(t'_n+\frac{v}{c^2}X')\\&=&\gamma(T'+n\Delta T'+\frac v{c^2}X')\end{eqnarray*}$$which does depend on where the clock is. This is the "leading clocks lag" rule - the larger the $X'$ (edit: the more positive the $X'$, to be precise) the later the $t$ at which a clock ticks a given $t'$. But the time between ticks is $t_{n+1}-t_n=\gamma\Delta T'$ which is independent of position, so they all tick at the same rate, yes.